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ELECTROMAGNETISM
Principles and Applications
Paul Lorrain
University of Montreal
Dale R. Corson
Cornell University
W . H . F r e e m a n a n d C o m p a n y
S a n F r a n c i s c o
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Cover: Jet-ink printer. See page 108.
Sponsoring Editor: Peter Renz; Project Editor: Pearl C. Vapnek; Manuscript
Editor: Rober t Mann; Designer: Gary A. Head; Illustration Coordinator:
Batyah J anows i ; Illustrator: George V. Kelvin Science Graphics ; Compositor:
Syntax Internat ional ; Printer and Binder: The Maple -Va i l Book Manufac tur
ing Group .
Library of Congress Cataloging in Publication Data
Lorra in , Pau l .
Electromagnet ism: principles and appl icat ions .
Bib l iography: p .
Includes index.
1. Elect ro mag net ism . I . Cor son, Dale R. , joint author . I I . Ti t le .
Q C 7 6 0 . L 6 5 3 7 7 8 - 1 9 1 1
ISBN 0-7167-0064-6
The present volume is based on Electromagnetic Fields and Waves, Second
Edi t ion, by Paul Lorrain and Dale R. Corso n. Copy right © 1962, 1970 by
W. H . F reeman and Company.
Electromag netism: Principles and Applications
Copy right © 1978 , 1979 by Paul Lorrain and Da le R. Corson .
No part of this book may be reproduced by any mechanical , photographic , or e lect ronic
process , or in the form of a phonographic recording, nor may i t be s tored in a re t r ieval
sys tem, t ransmit ted, or otherwise copied for publ ic or pr ivate use , wi thout wri t ten per
miss ion from the publ isher .
Printed in the United States of America
1 2 3 4 5 6 7 8 9
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CONTENTS
PREFACE ixLIST OF SYMB OLS xiii
CHAPTER 1 VECTORS 1
CHAPTER 2 FIELDS OF STATIONARY
ELECTRIC CHARGES: I 40
Coulomb 's Law, Electric Field Intensity E , Electric
Potential V
CHAPTER 3 FIELDS OF STATIONARY
ELECTRIC CHARGES: II 65
Gauss's Law, Poisson's and Laplace's Equations,
Uniqueness Theorem
CHAPTER 4 FIELDS OF STATIONARY
ELECTRIC CHARGES: III 89
Capacitance, Energy, and Forces
CHAPTER 5 DIRECT CURRENTS
IN ELECTRIC CIRCUITS 110
CHAPTER 6 DIELECTRICS: I 156
Electric Polarization P , Bound Charges, Gauss's Law,
Electric Displacement D
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Contents
CHAPTER 7 DIELECTRICS: II 179
Continuity C onditions at an Interface, Energy Density
and Forces, Displacement Current
CHAPTER 8 MAGNETIC FIELDS: I 201
The Magnetic Induction B and the Vector Potential A
CHAPTER 9 MAGNETIC FIELDS: II 220
Ampere's Circuital Law
CHAPTER 10 MAGNETIC FIELDS: III 234
Transformation of Electric and Magnetic Fields
CHAPTER 11 MAGNETIC FIELDS: IV 260
The Faraday Induction Law
CHAPTER 12 MAGNETIC FIELDS: V 279
Mutual Inductance M and Self-Inductance L
CHAPTER 13 MAGNETIC FIELDS: VI 311
Magnetic Forces
CHAPTER 14 MAGNETIC FIELDS: VII 333
Magnetic Materials
CHAPTER 15 MAGNETIC FIELDS: VIII 357
Magnetic Circuits
CHAPTER 16 ALTERNATING CURRENTS: I 371
Comp lex Num bers and Phasors
CHAPTER 17 ALTERNATING CURRENTS: II 398
Impedance, Kirchoff' s Laws, Transformations
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Contents
CHAPTER 18
CHAPTER 19
CHAPTER 20
APPENDIX A
APPENDIX B
APPENDIX C
APPENDIX D
APPENDIX E
APPENDIX F
INDEX 499
vi i
VECTOR DEFINITIONS, IDENTITIES,
AND THEOREMS 492
SI UNITS AND THEIR
SYMBOLS 494
SI PREFIXES AND THEIR
SYMBOLS 495
CONVERSION TABLE 496
PHYSICAL CONSTANTS 497
GREEK ALPHABET 498
ALTERNATING CURRENTS: III 428
Power Transfer and Transformers
MAXWELL 'S EQUATIONS 453
ELECTROMAGNETIC WAVES 468
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PREFACE
Like Electromagnetic Fields and Waves by the same authors, th is book aims to
give the reader a working knowledge of e lect romagnet ism. Those who wil l use
it should refer to a collection of essays by Alfred North Whitehead entit led Th e
Aims of Education,^ particularly the f irst essay, which carries the same ti t le. At
the outse t , White head e xplains his point: "T he w hole book is a protest against
dead kn ow led ge , that is to say, against inert ideas . . . ideas that are merely
received into the mind without being uti l ized, or tested, or thrown into fresh
combinat ions." I t i s in th is spi r i t that we have given more than 90 examples
and set 332 problems, 27 of which are solved in detail .
The book is intended for courses at the freshman or sophomore level,
either as an introduction to the subject for physicists and engineers, or as a last
course in elect romagnet ism for s tudents in other discipl ines.
I t is assum ed that the reader has had a one-term cou rse on differential and
integral calculus. No previous knowledge of vectors , mul t iple in tegrals , dif-
ferent ia l equat ions, or complex numbers i s assumed.
The features of our previous book that have been so appreciated will be
found here as wel l : the examples ment ioned above, problems drawn f rom the
current l i terature, a summary at the end of each chapter, and three-dimensional
figures, this t ime drawn in true perspective.
The book is divided into twenty short chapters suitable for self-paced in
struction. Following an introductory chapter on vectors, the discussion of elec
t romagnet ism ranges f rom Coulomb's law through plane waves in dielect r ics .
Electric circuits are discussed at some length in three chapters: 5, 17, and 18.
Chapter 16 is ent i re ly devoted to complex n umb ers and phasors. The re is no
way of dealing properly with alternating currents without using complex num
bers, and, contrary to what i s of ten assumed, complex numbers are not much of
tF ree P res s , New York , 1976 .
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X Preface
an obstacle at this level. Our discussion of relativity occupies only part of one
chapter (Chapter 10)—just enough to explain briefly the Lorentz transformation
and the transformation of electr ic and magnetic f ields. This naturally leaves a
host of questions unanswered, but should be sufficient in the present context.
THE PROBLEMS
The longer problems concern devices and methods of measurement descr ibed in
the physics and electr ical engineering l i terature of the past few years. They are
"programmed" in the sense that they progress by smal l s teps and that the inter
mediate results are usually given.
These longer problems are meant to give the reader the opportunity tolearn how to make approximat ions and to bui ld models amenable to quant i ta
t ive analysis. I t is of course hoped that they can teach the heuristic process in
volved in solving problems in the f ield of electromagnetism. In fact, after
working through many such problems, s tudents end up surpr ised to see that
they can deal with real si tuations.
These problems contain a large amount of peripheral information and
should provide some interesting reading. They should also incite the reader to
apply his newly acquired knowledge to other f ields and should stimulate crea
t iv i ty . Many of them can serve for open-ended exper iments.
The easier problems are marked E, from C hapte r 2 on. They can be solvedin just a few lines, but they nonetheless often require quite a bit of thought.
Some problems are marked D. Those are quite diff icult .
A number of problems require curve plot t ing. This i s because curves are
so much more meaningful than formulas. I t is assumed that the student has
access to a programmable calculator , or possibly to a computer . Otherwise the
calculat ions would be tedious.
On the average, about two problems are solved in detail at the end of each
chapter .
UNITS AND SYMBOLS
The units and symbols used are those of the Systeme International d'UniteSj
designated SI in all languages.f The system originated with the proposal made
t S e e ASTMIIEEE Standard Metric Practice, publ ished by the Ins t i tute of Electr ical and
Elec tron ics Engine e rs , Inc . , 345 East 47 th S t ree t, New Yo rk , N . Y . 10017 .
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Preface xi
by the I talian engineer Giovanni Giorgi in 1901 that electr ical units be based on
the meter -ki logram-second system. The Giorgi system grew with the years and
came to be called, f irst the MKS system, and later the MKSA system (A forampere). I ts development was fostered mostly by the International Bureau of
Weights and Measures, but also by several other international bodies such as
the International Council of Scientif ic Unions, the International Electrotechnical
Commission, and the Internat ional Union of Pure and Appl ied Physics.
Appendix D provides a conversion table for passing f rom cgs to SI uni ts ,
and inverse ly. "F urth er use of the cgs units of electr icity and ma gne tism is
depreca ted , " t
SUPPLEMENTARY READING
The fol lowing books are recommended for supplementary reading:
Electromagnetic Fields and Waves: Second Edition, by the same authors and the same
publ isher , publ ished in 1970. There are several references to this book in the foot
notes .
Standard Handbook for Electrical Engineers, McGraw-Hi l l , New York , 1968 .
Electronics Engineers' Handbook, McGraw-Hi l l , New York , 1975 .
Reference Data for Radio E ngineers, Howard Sams and Company, Inc . , Ind ianapol i s ,
1970.
Electrostatics and Its Applications, A. D . Moore , Edi to r , John Wi ley , New York , 1973 .
These books can be found in most physics and engineer ing l ibrar ies . The
reader will f ind the second and third references to be inexhaustible sources of
informat ion on pract ical appl icat ions.
ACKNOWLEDGMENTS
I am indebted first of all to my students, who have taught me so much in the
course of innumerable discussions. I am also indebted to the persons who
assisted me in writ ing this book: to Francois Lorrain, who took part in the
preparation of the manuscript and who prepared sketches of the three-dimen
sional objects that appear in the f igures in true perspective; to Ronald Liboiron,
who also worked on the preparation of the f igures; to Jean-Guy Desmarais and
flbid., p . 11 .
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xi i Preface
Guy Belanger, both of whom revised the complete text; and particularly to
Lucie Lecomte, Nancy Renz, and Al ice Chenard, who typed or re typed over a
thousand pages of manuscr ipt .
I am also grateful to Robert Mann, who took such great care in editing
both this book and the previous one; and to Pearl C. Vapnek, of W. H.
Freeman and Company, for her met iculous work on the proofs.
Finally, I wish to thank collectively all those persons who wrote to me in
connection with Electromagnetic Fields and Waves, even though they all re
ceived a promp t answ er and were duly thanked individual ly . Their feedback was
invaluable .
Dale Corson's duties as President, and later as Chancellor, of Cornell Uni
versity have unfortunately prevented him from working on this offshoot of
Electromagnetic Fields and Waves, of which he is co-author.
P A U L LO RRAI N
Montreal, 1978
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LIST OF SYMBOLS
SPACE, TIME, MECHANICS
Element of length dl, ds, dr
Total length / , L, s, r
Element of area da
Total area S
Element of volume dr
Total volume T
Solid angle O
Normal to a surface n
Waveleng th A
Radian length K=\/2TT
Time /
Period T=l/f
F r eq u e n cy / = l/T
Angular f requency,
angu lar velocity w = 277/
Velocity v
Mass m
Mass densi ty p
M o m e n t u m p
Mo me nt of inertia /
Force
Torque
Pressure
Energy
P o w e r
FTP
I I '
P
ELECTRICITY AND MAGNETISM
Quantity of electr ic
i ty Q
Volume charge den
sity p
Surface charge den
sity cr
Linear charge den
sity A
Electric potential V
Induced elect romo-
tance V
Electric field inten
sity E
Electr ic displace
m e n t I)
Permitt ivity of vac
uum e 0
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xiv List of Symbo ls
Relative permitt ivity
Permittivity
Elect r ic dipole moment
Electric polariza
tion, electr ic di
pole moment per
uni t volume
Electric suscepti
bility
Mobi l i ty
Electric current
Volume cur r en t
densi ty
Surface current den
sity
Vector potent ia l
Magnet ic induct ion
Magnetic f ield in
tensity
Magnetic f luxMagnetic f lux l ink
age
Permeabili ty of vac
u um
Relat ive permeabi l
ity
Permeabi l i ty
Magnet ic dipole
moment per uni t
vo lume
Magnet ic suscept i
bil i ty
Magnet ic dipole
m o m e n t
Resistance
Capaci tance
e = e r € n =DIE
P
Xe
At
I
a
A
B
H
<T>
A
M O
Mr
M ;
M
Self - inductance L
Mutual inductance M
Impedance Z
Admi t t ance Y
Resistivity P
Conduct ivi ty a
Poynt ing vector
MATHEMATICAL SYMBOLS
Approx imate lyequa l to ~
Propor t ional to
Exponent ial of x
Average
Real part of z
Modulus o f z
Decadic log of x
Natural log of x
Arc tangent x
Complex conjugate
of z
Vector
Gradient
= BlH D i v e r
ge n c e
Curl
m
R
C
Laplacian
Unit vectors in Car
tesian coordinates
Unit vector along r
Field point
Source point
e 7
, e x p x
B
Re z
I"
lo g x
In x
arc t anx
Z *
E
V
V -
V x
V -
U i, K
r T
x, y, z
x', y', z'
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ELECTROMAGNETISM
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CHAPTER 1
VECTORS
1.1 VECTORS
1.2 SCALAR PRODUC T
1.2.1 Example: W ork Done by a Force
1.3 VECTOR PRODUC T1.3.1 Examples: Torque, Area of a Parallelogram
1.4 THE TIME DERIVA TIVE
1.4.1 Examples: Position, Velocity, and Acceleration
1.5 THE GRADIENT
1.5.1 Examples: Topographic Map, Electric Field Intensity
1.6 THE SURFACE INTEGRAL
1.6.1 Example: Area of a Circle
1.6.2 Example: Electrically Charged Disk
1.7 THE VOLUME INTEGRAL
1.7.1 Example: Volume of a Sphere
1.8 FLUX
1.8.1 Example: Fluid Flow
1.9 DIVERGENCE
1.10 THE DIVERGENCE THEOREM
1.10.1 Examples: Incompressible Fluid, Explosion
1.11 THE LINE INTEGRA L
1.11.1 Example: Work Done by a Force
1.12 THE CURL
1.12.1 Example: Water Velocity in a Stream
1.13 STOKES'S THEOREM
1.13.1 Example: Conservative Vector Fields
1.14 THE LAP LAC I AN
1.14.1 Example: The Laplacian of the Electric Potential
1.15 SUMMARY
PROBLEMS
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Electric and magnetic phenomena are described in terms of the fields of
electric charges and currents. For example, one expresses the force between
two electric charge s as the prod uct of the mag nit ud e of one of the charges
and the field of the other.
Th e object of this first chap ter is to describe the math ema tic al me tho ds
used to deal with fields. All the mat eri al in this cha pt er is essen tial for a
prop er unders tand ing of what follows.
Figure 1-1 Pres sure and wind-veloci ty f ie lds over th e N o r t h A t l a n t i c on N o v e m b e r
1, 1967, at 6 h o u r s , G r e e n w i c h M e a n T i m e . The curved l ines are isobars, or l ines
of equa l p res sure . P res sures are given in ki lopasca l s . H igh-pres sure a reas are
d e n o t e d by H, and l ow-pres sure a reas by L. In this case an a r row ind ica te s th e
di rec t ion and veloci ty of t he w ind at its t i p ; a r ro w l engths are p r o p o r t i o n a l to
wind veloci t ies (the l onges t a r row in this f igure represents a wind veloci ty of
25 me te rs pe r second) . Wind vec tors are given only at a few poin t s , where ac tua l
m e a s u r e m e n t s w e r e m a d e .
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Figure 1-2 A vec tor A and the th ree vec tors A xi, A yj, A.k, which , when p lacedend- to-end , a re equ iva len t t o A .
Ma them at i ca l ly , a f ie ld is a func t ion tha t desc r ibes a phys ica l q uan t i ty
at a l l points in space. In scalar fields, th is qu an t i ty i s speci f ied b y a s ingle
nu mb er for each po in t . P ressure , t em per a tu re , and e l ec t r i c po te n t i a l a r e
exam ples o f sca la r quan t i t i e s tha t can vary f rom one po in t to ano the r in
s p a c e . F o r vector fields, bo th a nu m be r and a d i r ec t ion a r e requ i r ed . W ind
veloci ty , gravi ta t ional force, and elect r ic f ie ld in tensi ty are examples of
such vector quant i t ies . Both types of f ie ld are i l lust rated in Fig . 1-1.
Vec to r quan t i t i e s wi l l be ind ica ted by boldface t y p e ; italic type wi l l
ind ica te e i the r a sca la r quan t i ty o r the magn i tude o f a vec to r quan t i ty .*
We shal l fol low the usual custom of using right-hand Cartesian
coordinate systems, as in Fig . 1-2; the posi t ive z di rect ion is the di rect ion
of adv anc e o f a r igh t -ha nd sc rew ro ta t ed in the sense tha t t u rns the pos i t ive
x-ax i s in to the pos i t ive v -ax i s th rough the 90° ang le .
1.1 VECTORS
A vector can be specified by its components a long any th ree mu tua l ly per
pend icu la r axes . I n the Car t es i an coord ina te sys t em of F ig . 1 -2 , f o r example ,
t h e c o m p o n e n t s o f t h e v e c t o r A a r e A x, A y, A z.
f In a ha nd wr i t te n text , it is con ven ient t o ident ify a vector by mean s of an arro w over
the symbol , fo r example , A .
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4 Vectors
SCALAR PRODUCT
T h e scalar, o r dot product, is t he sca la r qu an t i ty ob ta i ned on mu l t ip ly ing
the mag n i tu de o f the fi rs t vec to r by the ma gn i tu de o f the second a nd by the
cosin e of the ang le betw een th e tw o vector s . In Fig . 1-3, for exa mp le,
A • B = AB c o s (q> — 6). (1-5)
I t f o l lows f rom th i s de f in i t ion tha t t he usua l commuta t ive and d i s
t r ibu t ive ru les o f o rd ina ry a r i thm et i c mul t ip l i c a t ion a pp ly to the sca la r
p r o d u c t :
A • B = B - A, (1-6)
and, for any three vectors ,
A - ( B + C ) = A - B + A - C . (1-7 )
The vec to r A can be un ique ly expressed in t e rms o f i t s componen t s
th rough the use o f unit vectors i , j , k , which are def ined as vec tors of un i t
magn i tude in the pos i t ive x , y, z d i r ec t ion s , r espe c t ive ly :
A = A xi + A,} + A :k. (1-1)
Th e vec to r A is thus the sum of th r ee vec to r s o f m ag n i t ud e A x, A y, A z,
paral le l to the x-, y~, z-axes, respect ively .
T h e magnitude of A is
A = (A2
X + A) + A2
)1
'2
. (1-2)
T h e sum of two vectors is o b t a i n e d b y a d d i n g t h e i r c o m p o n e n t s :
A + B = (A x + B x)\ + (A y + B y)\ + (A z + B J k . ( 1 - 3 )
Subtraction i s s imply ad d i t io n wi th one o f the vec to r s chan ged in s ign :
A - B = A + ( - B ) = (A x - B x)i + (A, - B y)\ + [A z - B :)k . (1-4)
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The lat ter proper ty wi l l be ver i f ied graphical ly in Prob. 1-3. I t a lso fol lows
t h a t
i - i = 1, j j = l , k k = 1. (1-8)
j • k = 0, k • i = 0, i • j = 0. (1-9)
T h e n ,
A • B = (AJ + A y\ + Azk) • (B x\ + B y\ + B.k), (1-10)
= AXB X + AyB y + A-B-. (1-11)
I t is easy to chec k th at th is resul t i s cor re ct for two v ecto rs in a plane, as
in Fig. 1-3:
A • B = AB c os (<p - 0) = AB co s cp c o s 9 + AB sin q> sin Q, (1-12)
= A XB X + A yB y. (1-13)
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6 Vectors
1.2.1 EXAMPLE: WORK DONE BY A FORCE
A s imple phys ical example of the scalar product i s the work clone by a force F act ing
through a d i sp lacement s : W = F • s, as in Fig . 1-4.
Figure 1-4 T he wo rk d on e by a force F whose po int of app l icat i on mo ves by a
dis tance s i s (F co s <p)s, o r F • s.
1.3 VECTOR PRODU CT
T h e vector, o r cross product, of two vectors i s a vector whose di rect ion is
perpend icu la r to the p lane con ta in ing the two in i t i a l vec to r s and whose
ma gni tud e is t he p rod uc t o f the mag n i tu des o f those vec to r s and the s ine
of the ang le be twe en them . W e ind ica te the vec to r p rodu c t thu s :
A x B = C. (1-14)
Th e ma gn i tud e o f C is
C = \AB sin (cp - 0) | , (1-15)
w i t h <p a n d 0 defined as in Fig. 1-3. T he d ire cti on of C is given by th e r i gh t-
hand sc rew ru le : i t i s t he d i r ec t ion o f advance o f a r igh t -hand sc rew whose
axis , he ld perp end icu la r to the p lane o f A an d B, i s r o t a t ed in the sense tha t
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1.3 Vector Product 7
ro ta t es the f i r s t -named vec to r (A) in to the second -nam ed (B) th ro ugh the
smal l e r ang le .
T h e c o m m u t a t i v e r u l e i s no t fol lowed for the vector product , s inceinver t ing the o rder o f A and B inver t s the d i r ec t ion o f C:
A x B = - ( B x A) .
The d i s t r ibu t ive ru le , however , i s f o l lowed fo r any th r ee vec to r s :
A x (B + C) = (A x B) + (A x C).
This wi l l be shown in Prob. 1-7.
F r om the def in i tion o f the vec to r p ro du c t i t f o llows th a t
(1-16)
(1-17)
(1-18)x i = 0 , j x j = 0 , k x k = 0 ,
and , fo r the usua l r igh t -handed coord ina te sys t ems, such as tha t o f F ig . 1 -2 ,
i x j = k, j x k = i, k x i = j , j x i = — k , a nd so o n . (1-19)
W r i t ing ou t the vec to r p rod uc t o f A an d B in t e rm s o f the co mp on en t s ,
A x B = (A xi + A y] + A±) x (B x\ + B y\ + B zk) , (1-20)
= (A yB z - A zB y)\ + (A._B X - AXBZ)\ + (A xB y - A yB x)k , (1-21)
i j k
AxA y A :
B x B y B :
(1-22)
We can check this resul t for the two vectors of Fig . 1-3 by expanding
sin ((p — 9) an d no t ing tha t t he vec to r p ro duc t is i n the pos i t ive z d i r ec t ion .
1.3.1 EXAMPLES: TORQUE, AREA OF A PARALLELOGRAM
A good phys ica l exa mple of the vec tor p r odu c t i s t he torque T produced by a fo rce
F ac t ing wi th a moment a rm r about a po in t O, as in Fig. 1-5, where T = r x F .
A second example i s t he area of a parallelogram, as in Fig. 1-6, where the area
S = A x B. Th e area is thu s rep resen ted by a vector perp end icul ar to the surface.
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Figure 1-6 Th e area of the para l le lo gra m is A x B = S. Th e vector S is no rm al to
the pa ra l l e logram.
THE TIME DERIVATIVE
We sha l l o f t en be concerned wi th the r a t es o f change o f sca la r and vec to r
quan t i t i e s wi th bo th t ime and space coord ina tes , and thus wi th the t ime
and space der iva t ives .
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9
Figure 1-7 A vec tor A , i ts i nc rem ent AA, and the i r co mp one nt s .
Th e t ime der iva t ive o f a vec to r quan t i ty i s s t r a igh t fo rward . In a
t ime Af , a vector A, as in Fig . 1-7, may change by AA, which in general
r epresen t s a change bo th in magn i tude and in d i r ec t ion . On d iv id ing AA by
At and taking the l imi t in the usual way, we ar r ive at the def ini t ion of the
t ime der ivat ive dA/dt:
dA = H m Ajt + AO - A(r)
dt A,^O At
= Hm ^ i + A ^ j + A ^ ( 1 2 4 )
A f ^ O Af
dt dt tit
Th e t ime der iv at ive of a vec tor i s thu s equa l to the vec tor sum of the t im ed e r i v a t i v e s o f i t s c o m p o n e n t s .
1.4.1 EXAMPLES: POSITION. VELOCITY. AND ACCELERATION
The t im e de r iva t ive of the position r of a point is i ts velocity v, and the t ime derivat ive
of v is the acceleration a. See Prob. 1-10.
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10 Vectors
THE GRADIENT
Let us cons ider a sca la r quan t i ty tha t is a co n t in uo us an d d i f f er en t iab lefunct ion of the coor din ate s and ha s the value / ' a t a cer ta in poin t in spac e,
as in Fig . 1-8. Fo r exa mp le, / cou ld be the elec t ros tat ic pot en t ia l V. W e
wish to kn ow ho w / chan ges over the d i s t anc e dl measured f rom tha t po in t .
N o w
df = %- dx + 4- dy = A • dl . (1-26)cx oy
w h e r e
A = %- i + % j , a n d dl = dx i + dy j . (1-27)< .\- cy
Th e vec to r A, wh ose com po ne n t s a r e the r a t es o f cha nge o f / wi th
d i s t ance a long the coord ina te axes , i s ca l l ed the gradient of the scalar
qua n t i ty / . Th e op era t ion o n the sca la r / de f ined by the t e rm grad ie n t
is indic ated by the sym bo l V, cal led "del" . T hu s
A = \f. (1-28)
Figure 1-8 The quant i ty / , a funct ion of pos i t ion, changes from / to / ' + df over
the d i s t ance dl.
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1.5 The Gradient 11
I n t h r e e d i m e n s i o n s ,
. . . d .3, C
V = + j — + k — .C.Y dy dz
d - 2 9 )
The par t i a l d i f f e r en t i a t ions ind ica ted a r e ca r r i ed ou t on wha tever sca la r
qua n t i ty s t and s to the r igh t o f the V sym bol . T hu s
ox dy dz(1-30)
a n d
I V I + +df 1/2
d - 3 1 )
T h u s ,
df = V / • dl = |V/'| |dl| cos 9, (1-32)
w h e r e 9 i s t he ang le be tween the vec to r s V/ and dl .
We now ask wha t d i r ec t ion one shou ld choose fo r dl i n o rder tha t
df be m ax im um . Th e answe r i s : t he d i r ec t ion in which 9 = 0, that i s , the
d i r ec t ion o f Vf. Th e g rad ien t o f / i s t hus a vec to r whose ma gn i tud e a nd
d i r ec t ion a r e those o f the max imum space r a t e o f change o f / .
The gradient of a scalar funct ion at a given point i s a vector having
the fo l lowing p rop er t i e s :
1) I ts com po ne n t s a r e the r a t es o f chan ge o f the func tion a long the
d i r ec t ions o f the coord ina te axes .
2 ) I t s m ag n i t ud e i s t he m ax im um ra te o f cha nge o f the func t ion wi thd i s t a n c e .
3) I t s d i rect ion is that of the maximum rate of change of the funct ion.
4 ) I t po in t s toward larger values of the funct ion.
Th e g rad ien t i s a vec to r po in t - func t ion der ived f rom a sca la r po in t -
func t ion .
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12 Vectors
1.5.1 EXAMP LES: TOPOGR APHIC MAP. ELECTRIC FIELD INTENSITY
Fig . 1-9 s h o ws th e grad ien t of the elevat ion on a topographic map.
In th e nex t chap ter we shall see tha t , in an electrostat ic f ield , th e electric field
intensity E is e q u a l to m i n u s th e grad ien t of the e lec t r ic po ten t ia l V: E = — W.
Figure 1-9 T o p o g r a p h i c ma p of a hill. The numbers shown g ive th e e leva t ion E in
mete rs . The grad ien t of E is the slope of the hill at the po in t cons idered , and it
p o i n t s t o wa rd an mcrease in e leva t ion : V£ = (dE/dx)i + (dE/dy)\. The a r ro w s h o ws
\E at one po in t where th e e leva t ion is 400 m e t e r s .
1.6 THE SURFACE INTEGRAL
Consider the x-y plane. To locate a point in this plane, one requires two
coordinates, x and y. N ow c ons ide r a given sph eric al surface, as in Fig. 1-10.
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13
z
y
x
Figure 1-10 Spherical surface of radius r. T h e t w o c o o r d i n a t e s 8 a n d q> suffice to
de te rmine the pos i t ion of a po in t P on the sur face . How do these coord ina te s
com par e wi th the l a t i t ude and the longi tud e a t t he sur face of the ea r th ?
To f ix a po in t we aga in need two coord ina tes , 0 a n d q> . Of cour se we cou ld
u s e t h e t h r e e c o o r d i n a t e s x, y, z. B u t t h o s e t h r e e c o o r d i n a t e s a r e r e d u n d a n t ,
on the given sphere, s ince x 2 + y 2 + z 2 = r 2. So we on ly need two co
o r d i n a t e s , 0 a n d (p .Th is is a genera l ru l e : one a lw ays needs to spec ify tw o coord ina te s to
ident i fy a point on a given surface.
I f we require the area of the given surface, we must in tegrate the
element of area, say dx dy, o v e r both var i ab les . We then have a surface
integral.
More genera l ly , an in t egra l eva lua ted over two var i ab les ( tha t a r e
no t necessa r i ly spac e coord ina te s ) i s ca l l ed a double integral.
Suppose the su r f ace ca r r i es a charge . Then , to f ind the to t a l charge ,
we must in t eg ra te the su r f ace charge dens i ty , mul t ip l i ed by the e l ement o f
a r ea , over both v a r i a b l e s .
So , a su rf ace in t egra l i s a do ub le in t egra l eva lua ted o ver the tw o
coord ina tes tha t a r e r equ i r ed to f ix a po in t on a g iven su r f ace .
I f the sur face is s i tuated in the x-y plane, the sur face integral i s of
the form
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14 Vectors
w h e r e f{x,y) i s a funct ion of x and y , and where the l imi ts a a n d b o n x
m ay be funct ion s of y . Th e l imi ts c a n d d are cons tan t s . We f i r s t eva lua te
the integral between the braces. This gives ei ther a constant or a funct ionof y . The n th i s cons ta n t , o r func t ion , is i n t eg ra ted over y . The d oub le in t egra l
is therefore a n integral wi thin an integ ral .
As we shal l see in the next example, one ar r ives at the same resul t
by inver t ing the o rder o f in t eg ra t ion and wr i t ing the doub le in t egra l i n
the form
where the l imi ts of in tegrat ion are di f ferent : the l imi ts c ' and a" c a n n o wbe funct ions of x , whi le a' a n d b' a r e c o n s t a n t s .
We have shown braces in the above express ions so as to ind ica te
the m an ne r in which a dou b le in t egra l i s eva lu a ted , bu t these b races a r e
never used in pract ice .
EXAMPLE: AREA OF A CIRCLE
I t is a lways advi s ab le to t e s t one ' s sk i ll a t und e rs t an ding and us ing new c oncep t s
by applying them to extremely s imple cases . So, le t us calcula te the area of a circle
by mea ns of a surface integr al .
So as to s impl i fy the calculat ion, we shal l f ind the area of the top r ight-hand
sector of the c i rc le of Fig. 1-11, and then mul t iply by 4.
The e l ement o f a rea dA is dx dy. We proceed as follows. We first slide dA from
left to right, to generate the shaded strip in the figure. Then we slide the strip from
y = 0 to y = R, while adjus t ing i ts length correct ly. This generates the 90-degree
sec tor .
Let us first find an integral for the strip. The strip has a height dy . It starts at
x = 0 and ends a t the edge of the c i rc le , where
x2 + v
2
= R 1. (1-33)
T h u s , a t the r igh t-h and e nd of the s t r ip.
X = (R* - v 2 ) (1-34)
and the s t r ip has an area
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15
Figure 1-11 (a ) Th e infini tesimal e lemen t of area tlx dy sweeps from the v-axis to
the periphery of the c i rc le to generate the shaded s t r ip, (b ) The s t r ip now sweeps
from the .v-axis to the top of the c i rc le to generate the top-r ight quadrant .
The to t a l a rea A of the circle is given by
A4 crfr-""*}*
Se t t ing y/R = sin 8, t h e n
co s 0. dy = R cos 0 dO.
Also , when y = 0, 0 = 0, and when y = R, 8 = n/2. T h e n
— = R R4 J "
/I = TtR 2,
c o s 2 8d8 = - R 2,4
(1-35)
(1-36)
(1-37)
(1-38)
(1-39)
as expec ted .
If we used a vert ical s t r ip, it wo uld extend from
y = 0 to y = (R 2 - x2
)112
. (1-40)
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16 Vectors
(1-41)
T h e n we would have tha t
A = , , . P ( / V ( / A ,
4 Jv = o J, = o '
= \ l (R2
- x 2 ) 1 ' 2
dx . (1-42)
This integral is s imi l a r to t h a t of E q . 1-36 and the a rea of the circle is aga in nR2
.
EXAMPLE: ELECTRICALLY CHARGED DISK
If we have an electrically ch arged disk with a surface charge dens i ty a given by, say,
a = Ky2
. (1-43)
then , us ing a vert ical s t r ip and i n t egra t ing ov e r the comp le te c i rcl e , th e t o t a l cha rge is
J+ R i " v
3
^+ (R 2-* 2)" 2")K —
1 _ 3_
Using a t ab le of i n t e g r a l s / we find that
(1-44)
(1-45)
(1-46)
Q = - KR4
.v
4(1-47)
1.7 THE VOLUME INTEGRAL
In a volume integral, there are three variables, say x, y, z in Carte sian co
ordinates, or r, 0, (p in polar coordinates (Fig. 1-10). In the more general
case where the three variables can be of any nature, say velocities, or voltages,
etc . , one has a triple integral.
Here again, one integrates with respect to each variable in succession,
starting with the innermost integral.
* See, for example , Dwight , Table of Integrals and Other Mathematical Data, No . 350 .03 .
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1.7 The Volume Integral 17
1.7.1 EXAMPLE: VOLUME OF A SPHERE
To ca lcu la t e the volume of a sphere, we fi rs t f ind the e lement of volume. Referr ingto Fig. 1-12, this is
(rdO)(r si n 9 dq>){dr).
T h e n
Let us integrate , f i rs t wi th respect to cp , then with respect to 0, and finally with
respect to r. This m eans th a t we f ir st ro t a t e the e lement o f vo lu me ab out the ve r t i ca l
axis to generate a r ing. Then we s l ide the r ing from 9 = 0 to 8 = n, whi le keep ing
the rad ius r a n d dr cons tant . This generates a spherical shel l of radius r and th i ck
ness dr . Then we va ry r from 0 to R t o gene ra t e the sphe re . So
V = r p = " r= "2nr 2 sin 0 dO dr , (1-49)
J r = 0 J e = 0 ' '
(1-50)
(1-51)
Figure 1-12 Elemen t o f vo lu me in sphe r i ca l co ord in a te s .
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18 Vectors
This is a part icular ly s imple case because the l imits of integrat ion are a l l con
s tants . Indeed, the t r iple integral of Eq. 1-48 is real ly the product of three integrals .
In P ro b . 1 -19 we sha l l r epea t t he ca l cu la t ion in Car t e s i an coor d ina te s . The
l imits of integrat ion wil l then be funct ions , ins tead of being cons tants .
1.8 FLUX
I t i s of ten necessary to calculate the f lux of a vector through a sur face. For
example , one migh t wi sh to ca lcu la t e the magne t i c f lux th rough the i ron
co re of an elec t rom agn et . The flux of a vecto r A th ro ug h an inf ini tesimal
surface da is
c/O = A • da . (1-52)
where the vec to r da i s normal to the sur face. The f lux d<t> is t h e c o m p o n e n t
o f the vec to r A norm al to the su r f ace , mul t ip l i ed b y da . For a f inite surface S
we f ind the to tal f lux by int egr at in g A • da over the en t i r e su r f ace :
J >da . (1-53)
For a c losed surface the vector da i s t ake n to po in t outward.
1.8.1 EXAMPLE: FLUID FLOW
Let us cons ider fluid flow. We define a vector pv, p being the f luid dens i ty and v
the fluid velocity at a point. The flux of pv thr ou gh any closed surface is the net
ra t e a t which mass l eaves the vo lume bounded by the sur face . In an incompres s ib l e
fluid this flux is zero.
1.9 DIVERGENCE
Th e ou tw ard f lux O of a vec to r A th r ou gh a c losed su r f ace can be ca lcu
l a t ed e i the r f rom the above equa t ion o r as fo l lows . Le t us cons ider an
in f in i t es imal vo lume dx, dy, dz and the vector A, as in Fig . 1-13, whose
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19
Figure 1-13 E lem en t of vo lu m e dx, dy, dz a r o u n d a p o i n t P, where the vec tor A has
the va lue i l l us t ra t ed by the a r row.
c o m p o n e n t s A x, A A, are func t ions o f the coord i na te s x, y, z. W e c o n s i d e r
an in f in i tes imal vo lum e and f i r s t -o rder va r i a t io ns o f A.
The va lue o f A x at the cen ter of the r igh t-h an d face can be tak en to
be the average va lue over tha t f ace . Through the r igh t -hand f ace o f the
vo lume e lement , t he ou tgo ing f lux i s
dHfx
= U c + ^ ^ \ d y d z , (1-54)
s ince the no rm al co mp on en t o f A a t t he r igh t -ha nd f ace i s t he x -c om po ne n t
of A at that face.
At the lef t -hand face,
c^ L = - ^ - ^ ^ j d y d z . (1-55)
Th e minu s s ign befo re the paren th es i s i s necessa ry here because , A xi b e i n g
inward at th is face and da be ing ou tw ard , t he cos ine of the ang le b e tween
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20 Vectors
t he two vec to r s i s — 1. Th e net ou tw ar d f lux thr ou gh the tw o faces is the n
PA PAd$> R + m, = —- dx dy dz = —^ dx, (1-56)OX ox
w h e r e dx is t he vo lum e o f the in f in it es imal e l em ent .
I f we calcu late the net f lux thro ug h th e oth er pa i rs of faces in the sam e
manner , we f ind the to t a l ou tward f lux fo r the e l ement o f vo lume dx to be
\ ox Py dz J
Suppose now tha t we have two ad jo in ing in f in i t es imal vo lume e le
m en t s and tha t we add the flux th rou gh th e boun d in g su r f ace o f the fi rs t
vo lu me to the flux th rou gh th e bou nd ing su r f ace of the second . At the
co m m on f ace the fluxes a r e equ a l in m ag n i t ud e bu t op pos i t e in s ign , and
they cancel as in Fig. 1-14. The flux from the f irst volume plus that from the
secon d is the f lux th rou gh the bou nd ing su r f ace o f the comb ine d v o lum es .
To ex tend th i s ca l cu la t ion to a f in i t e vo lume, we sum the ind iv idua l
f luxes for each of the inf ini tesimal vo lum e elem en ts in the f inite volu me , an d
the to t a l ou tward f lux i s
At any g iven po in t in the vo lume, the quan t i ty
PA X PA V PA,— - H -Iox dy Pz
i s thus the outgoing f lux per uni t volume. We cal l th is the divergence of the
vec to r A a t t he po in t .
Th e d ivergence o f a vec to r po in t - fun c t ion i s a sca la r po in t - fun c t ion .
Accord ing to the ru les fo r the sca la r p roduc t ,
v . A = 8A , + 8A l + dA lt ( 1 . 5 9 )
o x dy dz
wh ere the o pe ra to r V is def ined as in Eq. 1-29. The o pe ra to r V is no t a
vec to r , of cour se , bu t i t i s conv en ien t to use the no ta t ion o f the sca la r p ro du c t
to ind ica te the oper a t io n tha t i s ca r r i ed ou t .
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21
Figure 1-14 At the common face, the flux of A out oidz l i s equa l i n magn i tude , bu t
opposi te in s ign, to the f lux of A ou t of dz 2 •
1.10 THE DIVERGENCE THEOREM
Th e total o ut w ar d f lux of Eq . 1-58 is a lso equ al to the sur face integra l of
t h e n o r m a l o u t w a r d c o m p o n e n t o f A . T h u s
This i s the divergence theorem. Note that the integral on the lef t involves
only the values of A on the sur face 5 , whereas the two integrals on the r ight
invo lve the va lues o f A th roughou t the vo lume T enc losed b y S . Th e d i
vergence the ore m i s a gene ra l i za t io n o f the fundam enta l t he ore m of the
ca lcu lus ( see P rob . 1-20).
We can now redef ine the divergence of the vector A as fol lows. I f
t h e v o l u m e T i s a l low ed t o sh r ink suf fic ient ly , so th at V • A doe s not var y
apprec iab ly over i t , t hen
Js
A • da = (V • A) , ( t - > 0 ) , ( 1 -61)
a n d
V • A = lim - f A • da . (1-62)
r ^ O T JS
The d iverg ence i s t hu s the ou tw ar d f lux per un i t vo lum e, as the vo lum e t
a p p r o a c h e s z e r o .
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22 Vectors
1.10.1 EXAMPLES: INCOMPRESSIBLE FLUID, EXPLOSION
In an incompressible fluid, V • (pv) is every wh ere eq ual to zero, s ince the ou tw ard
mass f lux per uni t volume is zero.
Wi th in an explosion, V • (pv) is positive.
1.11 THE LINE INTEGRAL
The in teg ra l
/ > • „ , .
eva lua ted f rom the po in t a to the po in t b on some specified curve, is a line
integral. Each e lemen t o f l eng th dl on the curve is mult ip l ied by the local
va lue o f A acco rd ing to the ru le fo r the sca la r p roduc t . These p roduc ts
a re then summed to ob ta in the va lue o f the in teg ra l .
A vector field A is said to be conservative if the line integral of A • dl
a round any c lo sed cu rve i s ze ro :
A • dl = 0. (1-63)
The c irc le on the in tegra l s ign indica tes tha t the pa th of in tegra t ion is
c losed . We shal l f ind in Sec . 2 .5 tha t an e lec tros ta t ic f ie ld is conservat ive .
/ ./ /. / EXAMPLE: WORK DONE BY A FORCE
T h e work done by a force F act ing from a t o b along some specif ied path is
W - £ F • dl, (1-64)
w h e r e b o t h F a n d dl mu st be kn ow n fun ct ions of the coo rdi na tes i f the integral i s
to be eva lua ted ana ly t i ca l ly .
Let us calcula te the work done by a force F that is in the y di rec t ion and has a
m a g n i t u d e p r o p o r t i o n a l t o y, as i t moves around the c i rcular path from a to b in
Fig. 1-15. Since
F = ky j a n d dl = dx i + dy j , (1-65)
W = S! F > d l = 5?ydy = kr 2/2 . (1-66)
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1.12 THE CURL
Let us calculate the value of the l ine integral of A • dl i n the mor e genera l case
where i t i s not zero.
Fo r an in fin i t esimal e l e men t o f pa t h dl i n the xy-p lane , and f rom the
def in it ion o f the sca la r p rod uc t ,
A • dl = A x dx + A y dy . (1-67)
T h u s , for any closed path in the xy-plane and for any A,
A • dl = j>A x dx + &A y dy . (1-68)
Now consider the inf ini tesimal path in Fig . 1-16. There are two con
t r ibu t ion s to the f ir st i n t eg ra l on the r igh t -h and s ide o f the abov e e qua t ion ,
one a t y — {dy/2) a n d o n e a t y + (dy/2):
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24
y
Figure 1-16 Closed , rec t angula r pa th in the xr -p lane . cen te red on the po in t
P(x, y, 0) , where the vec tor A has the com po nen t s Ax, A
y. The in t egra t ion
a ro un d the pa th is pe r formed in the d i rec tion of the a r row s .
Th ere i s a min us s ign befo re the second t e rm b ecaus e the pa th e l em ent a t
y + (cly/2) i s i n the nega t ive x d i r ec t ion . There fo re
A xdx = -~dydx. (1-70)dy
Simi la r ly ,
a n d
j>Aydy = -^-dxdy, (1-71)
A d l = [^-?£*)dxdyox dy
for the infinitesimal path of Fig. 1-16.
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1.12 The Curl 25
If we set
dA y dAx
g3 = ^ - d-7 3)
ox dy
then
A'A\ = g3da, (1-74)
where da = dx dy is the area enclosed on the xy-plane by the infinitesimal
pat h. No te t hat the above equa ti on is correct only if the line integral is
evaluated in the positive direction in the xy-plane, that is, in the direction
in which one would have to turn a right-hand screw to make it advance in
the positive direction of the z-axis. This is known as the right-hand screw
rule.
Let us now consider g3 and the other two symmetrical quantities as
the components of a vector
/dA, dA.
\dy dzI +
dA x cA\ . ^ /dAy dAx ^
dz dx I dx dv
which may be written as
V x A (1-75)
We shall call this vector the curl of A. The quantity g3 is then its z com pon ent .
If we consider the element of area as a vector da pointing in the direc
tion of advance of a right-hand screw turned in the direction chosen for
the line integral, then da = da k and
A d l = (V x A)-da. (1-76)
This means that the line integral of A • dl around the edge of an
element of area da is equal to the scalar product of the curl of A by this
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26 Vectors
element of area, as long as we observe the above sign convention. This is a
general result that applies to any element of area da, whatever i ts orientation.
Thus
(V x A)„ = l i m ^ c f A - d l : (1-77)
s-o S J
the co mp on en t of the curl of a vector n or ma l to a surface S at a given point
is equa l to the line integral of the vector ar ou nd the bo un da ry of the surface,
divided by the area of the surface when this area ap pr oa ch es zero a ro un d
the point.
EXAMPLE: WATER VELOCITY IN A STREAM
Let us cons ider a stream in which th e veloci ty v is p ro p o r t i o n a l to the d is tance
from the b o t t o m . We set the z-axis parallel to the d i rec t ion of flow, and the x-axis
p e rp e n d i c u l a r to the s t r e a m b o t t o m , as in Fig . 1-17. T h e n
i\- = 0, r v = 0, (1-78)
We shal l calculate th e curl from Eq. 1-77. For (V x v) v we c h o o s e a path paral lel
to th e yz-p lane . In evalua t ing
v d l
around such a p a t h we no te tha t the c o n t r i b u t i o n s are equal and oppos i te on the
par t s para l le l to the z-axis, hence (V x v) v = 0. Likewise, (V x v)2 = 0.
F o r the j ' - c o m p o n e n t we c h o o s e a path paral lel to the x z -p l a n e and evalua te
the in teg ra l a round it in the sense tha t wou ld advance a r igh t -hand screw in the
posi t ive y d i rec t ion . On the p a r t s of the path paral lel to the x-ax is , v • dl = 0 since
v and dl are p e rp e n d i c u l a r . On the b o t t o m p a r t of the path , at a d is tance x from
the yz-p lane ,
v • dl = cx Az,
wh e re a s at (x + Ax)
F o r th e whole pa th ,
£ v • dl = — c(x + Ax) Az.
dl -c Ax Az
(1-79)
(1-80)
(1-81)
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2~
Figure 1-17 Th e veloci ty v in a viscous fluid is assu me d to be in the di rect ion of the
r -ax i s and pro por t ion a l t o the d i s t ance .x f rom the bo t to m. Th en V x v = — c'y
and the y-com pone nt o f t he cur l is
(V x v)j, = lim§ v • dl : A x A ;
s~o S Ax Az
Ca lcul at ing V x v di rect ly from E q. 1-75,
V x v
i j k
d f 0
dx r.v fz
0 0 cx
(1-82)
(1-83)
which is the sam e resul t as abo ve.
1.13 STOKE S'S THEO REM
Eq ua t io n 1-76 is t rue only for a pa th so smal l tha t V x A can be co nsid ered
cons tan t over the su r f ace da bo un de d by the pa t h . Wh at if t he pa th C is
so l a rge tha t t h i s cond i t ion i s no t met? The equa t ion can be ex tended r ead i ly
to a rb i t r a ry pa ths . We d iv ide the su r f ace— a n y su r f ace bounded by the
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28
Figure 1-18 An a rb i t ra ry sur face bou nd ed by the curve C. Th e sum of the l i ne
integrals arou nd th e curvi l in ear squ ares show n is equ al to the l ine integral
a r o u n d C.
pa th o f in t eg ra t ion in qu es t ion — into e l em ent s o f a r ea d a ^ d a 2 , a n d s o
for th , as in Fig . 1-18. For any one of these smal l areas,
A d I , = (V x A ) - d a , , (1-84)
We add the lef t -hand sides of these equat ions for a l l the da ' s , and
the n we ad d al l the r ig ht- ha nd sides . Th e sum of the lef t -hand s ides is the
l ine in t egra l a round the ex te rna l boundary , s ince the re a r e a lways two equa l
a n d o p p o s i t e c o n t r i b u t i o n s t o t h e s u m a l o n g e v e r y c o m m o n s i d e b e t w e e n
ad jacen t da ' s . The sum of the r igh t -hand s ide i s mere ly the in t egra l o f
(V x A) • da over the f ini te sur face. Thus, for an y surface S b o u n d e d b y a
c losed curve C,
A • dl = J s (V x A) • da. (1-85)
This i s Stokes's theorem.
Figure 1 -18 i l l us t r a t es the s ign conven t ion .
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1.14 The Laplacian 29
EXAMPLE: CONSERVATIVE VECTOR FIELDS
Stokes ' s theo rem can h e l p us to u n d e r s t a n d conservative vector fields. U n d e r w h a t
c o n d i t i o n is a vector f ield conservat ive? In o ther words , under what cond i t ion is
the l ine in tegral of A • dl a r o u n d an arb i t ra ry pa th equal to zero? From Stokes ' s
t h e o re m , th e l ine in tegral of A • dl a r o u n d an arb i t ra ry c losed pa th is zero if an d
only if V x A = 0 everywhere . Th is cond i t ion is me t if A = V/ . T h e n
£1dx '
df 81dz
(1-86)
a n d
8A ZdA,
(V x A). = — i - —1
dy dz
d 2f d 2f
dy cz cz dy
= 0, (1-87)
a n d so on for the o t h e r c o m p o n e n t s of the curl .
The field of A is therefo re conserva t ive , if A can be expressed as the g ra d i e n t
of some scalar funct ion / .
1.14 THE LAPLACIAN
The d ivergence of the gradient is of great imp orta nce in electromag netism.
Since
then
\f = f i + f i + fk, (1-88)
dx dy dz
dx cy dz* (1-89)
Th e divergence of the grad ient is the sum of the secon d derivatives with
respect to the rectangul ar coord inate s. The quantit y V • V/ is abbre viated to
\2
f, and is called the Laplacian of/ . The operato r V2
is called the Laplace
operator.
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30 Vectors
1.14.1 EXAMPLE: THE LAPLACIAN OF THE ELECTRIC POTENTIAL
We shall see in Sec. 3.4 that the Luplacian of the electric potential i s p r o p o r t i o n a l
to the space charge dens i ty.
1.15 SUMMARY
I n C a r t e s i a n c o o r d i n a t e s a vector qu an t i ty i s wr i t ten in the form
A = A x i + A, j + A z k. ( / - / )
where i , j . k are unit vectors d i r ec ted a long the x, y, z axes respect ively .T h e magnitude of the ve ctor A is the scala r
A = (A 2 + A]. + A2
)112
. (1-2)
V e c t o r s c a n b e a d d e d a n d s u b t r a c t e d :
A + B = (A x + B x)i + (A y + B y)j + (A z + B :)k. (1-3)
A - B = (A x - B x)\ + (A, - B y)\ + (A z - B ._)k. (1-4)
The scalar product (Fig . 1-3) i s
A • B = AB cos (<p — 01 (1-5)
= AXBX + A yB, + A ZB._. (1-11)
I t i s c o m m u t a t i v e ,
A • B = B • A, (1-6)
and d i s t r ibu t ive ,
A • (B + C) = A • B + A • C. (1-7)
T h e vector product (Fig. 1-3) is
A x B = C (1-14)
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1.15 Summ ary 31
with
C = |/4 B si n (<p - 0)\.
An oth er equ iva len t de f in i tion o f the vec to r p rodu c t i s
(1-15)
A x B
i j k
B v B t. B.
The vec to r p roduc t fo l lows the d i s t r ibu t ive ru le
A x (B + C) = (A x B) + (A x C ),
b u t no t t h e c o m m u t a t i v e r u l e :
A x B = - ( B x A).
T h e time derivative of A is
dA dA r dA, dA. ,i + — j + — k .
dt dt dt dt
T h e del operator i s def ined as fol low s:
V0 . 0 . C
ox dy cz
a n d t h e gradient of a sc ala r func tion / is
c x o z
(/-22)
( / - / 7 )
(1-25)
(1-29)
(1-30)
Th e g rad ien t g ives the max im um ra te o f chan ge o f / wi th d i s t ance a t t he
po in t cons idered , and i t po in t s toward l a rger va lues off.
T h e surface integral i s a co m m on type o f double integral. It is used for
in t egra t ing over two coord ina tes , say x and y. I t i s co m po se d of an integra l
wi th in an integra l . Th e inner in tegral i s eva lua ted f i rs t .
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32 Vectors
T h e volume integral is one form of triple integral. It is used for in
t eg ra t in g func t ions o f th r ee space coo rd ina tes , say x, y, z. In this case, we
ha ve an integral tha t is wi th in an integral tha t is wi th in an integral . T he
inn erm ost in t eg ra l i s eva lua ted f i rs t, leav ing a doub le in t egra l , and so on .
T h e flux O of a vec to r A th rough a su r f ace S is
j :A • da . (1-53)
For a c losed surface the vector da p o i n t s o u t w a r d .
T h e divergence of A,
V • A
dA, dA,. dA ,
dx + dy + dz (1-59)
i s t he ou t wa rd f lux o f A per un i t vo lum e a t t he po in t cons id ered .
T h e divergence theorem s t a t es tha t
j * V • A dx = J*s A da, (1-60)
w h e r e x i s t he vo lume bounded by the su r f ace S.
T h e line integral
J>-ove r a speci f ied c urv e is the su m of the te rm s A • dl fo r each e l ement dl
of the cu rve be tween the po in t s a a n d b.
Fo r a c losed curve C t ha t bounds a su r f ace S, w e h a v e Stokes's theorem:
A • dl = J s (V x A) • da.
w h e r e
(1-85)
V x A
• J k
d d d
dx dy cz
/ L A r A,
(1-75)
is the curl of the vec tor funct ion A.
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Problems 33
T h e Laplacian i s t he d ivergence o f the g ra d ie n t :
d 2f ' d 2f c 2fV • V/ ' = \2f = - L + ^ 4 + U . ( 7 - 5 9 )
o x o y dz
PROBLEMS
1-1 Sho w that the two vecto rs A = 9i + j — 6k and B = 4i — 6j + 5k are perp end icul ar
to each o th e r .
1-2 Sho w that the angle betwe en the two vectors A = 2i + 3j + k an d B = i — 6j + k
is 130.5°.
1-3 The vec tors A , B, C a re coplana r . Show graphica l ly tha t
A • (B + C) = A • B + A • C.
1-4 If A and B are adjac ent s ides of a par al le log ram , C = A + B a n d D = A — B are the
d i a g o n a l s , a n d 0 i s the angle between A and B, show tha t (C 2 + D 2) = 2(A 2 + B 2)
a n d t h a t ( C 2 - D 2) = 4AB co s 9.
1-5 Cons ide r two un i t vec tors a and b, as shown in Fig. 1-19.
Find the t r igonometr ic re la t ions for the s ine and cos ine of the sum and difference
of two angles from the values of a • b and a x b.
Solution: Fi rs t ,
a = co s a i + sin a j , (1)
b = cos fix + si n [S j . (2)
Now we can wr i t e the produc t a • b in two different ways . Fro m Eq. 1-11.
a • b = cos a cos [1 + sin a sin /?, (3)
and, from Eq. 1-5,
a • b = co s (a - /?). (4)
T h u s
cos (a — ji) = cos a cos j) + sin a sin /?. (5)
If we set \V = -/?,
co s (a + /?') = co s a co s ft' — sin a sin fi'. (6)
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34
Figure 1-19 Two unit vectors, a an d b,
situated in the xv-plane. See Prob. 1-5.
Similarly, from Eq. 1-22,
a x b cos a sin a 0
cos p sin p 0
= (cos a sin p — cos p sin a)k
and, from Eqs. 1-14 and 1-15,
a x b = —sin (a — p)k.
Thus
sin (ot — P) = sin a cos p — cos a sin p.
Setting again p' = — p,
sin (x + P') = sin a cos p' + cos a sin P'.
(7)
(S)
(9)
(10)
( I D
1-6 Show that t he mag ni tude of (A x B) • C is the volume of parallelep iped whose edges
are A, B, C, and show tha t (A x B) • C = A • (B x C) .
1-7 Sh ow t ha t A x ( B + C) = A x B + A x C .
1-8 Show that a x (b x c) = b( a • c) - c( a • b).
1-9 If r • (dr/<-/>) = 0, show that r = constant.
1-10 A gun fires a bullet at a velocity of 500 meters per second and at an angle of 30" with
the horizontal.
Find the position vector r, the velocity vector v, and the acceleration vector a
of the bullet, t seconds after the gun is fired. Sketch the trajectory and show the three
vectors for some time t.
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35
V
Figure 1-20 Tw o p o i n t s P' and P in
space, wi th the vec tor r and the uni t
vec tor rt. We shall often use such
p a i r s of poin t s , w i th P' at the source ,
a n d P at the poin t where the field is
ca lcu la t ed . For e x a m p l e , one m i g h t
ca lcu la t e th e electric field intensity at
P with th e e l e m e n t of c h a r g e at P'.
S e e P r o b . 1-13.
/ - / / Let r be the rad ius vec tor f rom th e origin of c o o r d i n a t e s to any p o i n t , and let A be a
cons tan t vec tor . Show tha t V(A • r) = A.
1-12 S h o w t h a t (A • V)r = A.
1-13 The vec tor r is directed from P'(x',y',z') to P(.v.y.r), as in Fig. 1-20.
a) If t he po in t P is fixed and the p o i n t P' is a l lowe d to move , show tha t th e grad ien t
of (\/r) under these condi t ions is
w h e r e rj is the uni t vec tor a long r. Show tha t t h i s is the m a x i m u m r a t e of c h a n g e of
b) Show s imilar ly that , if P' is fixed and P is a l lowed to m o v e ,
1-14 The c o m p o n e n t s of a vec tor A are
-I . - V
r y— y
< x
w h e r e /'is a funct ion of x, y, :. S h o w t h a t
A = r x Y/ ; A • r = 0. a n d A • Y / ' = O.
1-15 a) S h o w t h a t V • r = 3.
b ) W h a t is the flux of r t h r o u g h a spherical surface of r a d i u s al
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36
II
Figure 1-21 Trunca ted cy l inde r . See
P r o b . 1-18.
1-16 S h o w t h a t
V • ( / A ) = / V A + A • v./;
wh ere / i s a scalar funct ion an d A is a vector funct ion.
1-17 Th e vec tor A = 3.vi + y j + 2zk , a n d / = x 2 + y 2 + :2.
a) Show that V • ( / A ) a t the po int (2,2,2) i s 120, by f irs t calcula t ing / A an d then
ca lcu la t ing i t s d ive rgence .
b) Ca lcu la t e V • (/A ) by first finding V/ 'an d V • A. an d then u sing the iden tity
of Prob. 1-16 above.
c) If x, y, : a re mea sured in me te rs , wha t a re the un i t s o f V • ( /A) ?
1-18 Calcu la t e the vo lume of the por t ion of cy l inde r shown in F ig . 1 -21 .
Set the xy-plane on the base of the cyl inder and the origin on the axis . Wri te
the e lement of volume of height / ; as h dx dy, wi th
1-19 Calcu la t e the vo lume of a sphe re of rad ius R i n C a r t e s i a n c o o r d i n a t e s .
1-20 Cons ide r a func t ion f(x). Th e fund ame nta l t heore m of the ca l cu lus s t a te s tha t
Le t A = / (x ) i , and cons ide r a cy l indr i ca l vo lume i pa ra l l e l t o the x-ax i s , ex tending
from x = a to x = b, as in Fig. 1 -22.
Show tha t t he d ive rgence theorem, appl i ed to the f i e ld A and the vo lume T,
yields the fundamental theorem of the calculus in the form s ta ted above.
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37
Figure 1-22 Cylindrical volume parallel to the x-axis, extending from x = a to
x = b. See Prob. 1-20.
1-21 The gravitational forces exerted by the sun on the planets are always directed toward
the sun and depend only on the distance r. This type of field is called a central force field.
Find the potential energy at a distance r from a center of attraction when the
force varies as 1 /r2
. Set the pot enti al energy equal to zero at infinity.
1-22 Show that the gravita tion al field is conser vative. Disrega rd the curvat ure of the earth
and assume that g is a constant.
Wh at if one takes int o accoun t the curv atu re of the earth and the fact that gdecreases with altitude? Is the gravitational field still conservative?
1-23 Let P an d Q be any two paths in space that have the same end points a and b.
Sho w that , if A is a conser vat ive field,
J > - d l = J > - d l .
over P over Q
1-24 Let A be a conservative field and P0 a fixed point in space. Let
f(x,y,z) = ^ A • dl,
where the line integral is calculated along any path joining P 0 to (x,y,z).
Show that A = V/.
1-25 The azimuthal force exerted on an electron in a certain betatron (Prob. 11-9) is
proportional to J- 0' 4.
Show that the force is non-conservative.
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38 Vectors
1-26 A vector field is defined by A = f(r)r.
S h o w t h a t V x A is e q u a l to zero .
1-27
1-28
1-29
1-30
1-31
1-32
Show tha t V x (/A) =
S h o w t h a t V • (A x D;
vecto rs .
S h o w t h a t V • V x A =
(Yf) x A + / (V x A), wh e re / is sca la r and A is a vecto r .
= D • (V x A) — A • (V x D), wh e re A and D are any two
0.
O n e of the fou r Maxwel l equat ions s ta tes tha t V x E = —dB/dt, wh e re E and B
are respect ively th e electric field intensity in volts pe r m e t e r and the m a g n e t i c in
d u c t i o n in tes las at a p o i n t .
Show tha t th e l ine in tegral of E • dl is 20.0 microvo l t s over a s q u a re 100 mill i
meters on the side when B is 2.00 x 10 "3
f tesla in the d i r e c t i o n n o rm a l to the s q u a re ,
wh e re t is the t ime in seconds .
In Sec. 1-14 we defined the L a p l a c i a n of a sca la r po in t - func t ion / . It is also useful
to define th e L a p l a c i a n of a vecto r po in t - func t ion A:
V2
A = V2
Axi + \2
Ay) + \2
Azk.
Show tha t V2
( V / ) = V (V2
/ ) .
Show tha t
V x V x A = V(V • A) — V2
A.
We shal l need this resul t in C h a p t e r 20.
(1)
Solution: Let us find th e .v -componen ts of these th ree quan t i t i es . Then we can
find the y- and z -c o m p o n e n t s by s y m m e t ry .
Since
V x V x A
i
8
dx
J
8
dy
K
c
dz
8A Z dAy dAx dAz
dy dz dz dx
8A,
dx
dA,
8y
12]
then its .v -componen t is
d_ fdA,
dy \ dx
8A :
~8y
dA x
The .v -componen t of V(V • A) is
8 (cAx
dx V dx
dA y dA z
~dz~(3 )
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Problems 39
while that of V2
A is
Then we shou ld have tha t
d 2Ax
82
AX
82
AX
t x cy cz*
d 2A t
cy fx 5y 2
f2
/ lv
3*4,— -I ~
f z2 fz fx
d2
A, f2
/ L 82
A,
f x dy fx f z f x2
82
AX
82
AX
dy2 f z
2(5 )
which is correct .
The y-c om pon ent of the given equat ion ca n be found from E q. 5 by replacing x
by y , y by z, an d z by x . This gives anot he r ide nti ty , and s imilarly for the z-co mp one nt .
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CHAPTER 2
FIELDS OF ST A TIONARY
ELECTRIC CHARGES: I
Coulomb''s Law, Electric Field Intensity E ,
Electric Potential V
2.1 COULO MB'S LAW
2.2 THE ELECTRIC FIELD INTENSITY E
2.3 THE PRINCIPLE OF SUPERP OSITION
2.4 THE FIELD OF AN EXTENDED
CHARGE DISTRIBUTION
2.5 THE ELECTRIC POTENTIAL V
2.5.1 Examp le: The Electric Dipole
2.6 SUMMARY
PROBLEMS
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W e beg in our s tudy of e l ec t r i c an d mag ne t i c ph en om en a by inves t iga t ing
the f ie lds of s ta t io na ry elect r ic cha rge s.We sha l l s t a r t wi th Cou lomb 's l aw and deduce f rom i t , i n th i s chap te r ,
the elect r ic f ie ld in tensi ty an d the elect r ic pote nt ia l for any dist r i but ion of
s t a t ionary e l ec t r i c charges .
COULOMB'S LAW
I t i s f ound ex per im en ta l ly t ha t t he force be tween two s t a t io nary e l ec t r i c
point c h a r g e s Q a a n d Q b (a) acts a lon g the l ine join ing th e tw o cha rges, (b) i s
p r o p o r t i o n a l t o t h e p r o d u c t QaQ
b, an d ( c) i s i nver se ly p r op or t ion a l to the
square o f the d i s t ance r s e p a r a t i n g t h e c h a r g e s .
If t he charges a r e ex tende d , the s i tua t ion is m ore co mp l ica te d in tha t
the "d i s t ance be tween the charges" has no def in i t e mean ing . Moreover , t he
presence o f Q b can modi fy the charge d i s t r ibu t ion wi th in Q a, and vice versa,
l ead ing to a compl ica ted var i a t ion o f fo rce wi th d i s t ance .
W e t h u s h a v e Coulomb's law fo r s t a t ionary po in t charges :
F
"b
" 4 ^ ^ -r i
'( 2 _ 1 )
w h e r e F ah i s the force exer ted by Q a on Q b, a n d r t i s a uni t vector point ing in
the d i r ec t ion from Q a to Q h, as in Fig . 2-1. Th e force is repu lsive if Q a a n d
Q b are of the sam e sign ; it is a t t rac t iv e i f they are of di f ferent s igns. As u sual ,
F i s m e a s u r e d in n e w t o n s , Q in coulombs, a n d r i n mete r s . Th e con s tan t e 0
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42
f a *
Figure 2-1 C h a r g e s Q a a n d Q b sepa ra t ed by a d i s t ance r. The force exerted on Q h by
Q a is F ah and is in the di rect ion of ^ a lon g the l ine joi nin g the two ch arge s .
i s the permittivity of free space:
e 0 = 8.854 187 82 x 1 < T1 2
f a r a d / m e t e r .+
(2-2)
Coulomb 's l aw app l i es to a pa i r o f po in t charges in a vacuum. I t
a l so app l i es in d i e l ec t r i cs and conduc to r s i f F a b i s the di rect force between
Q a a n d Q b, i r respe ct ive of the forces ar is ing f rom o the r cha rges wi thin the
m e d i u m .
Subs t i tu t ing the va lue o f e 0 i n t o C o u l o m b ' s l a w ,
F ^ 9 x l 0 # r i , (2-3)r
wh ere the factor of 9 i s ac cu rat e to ab ou t o ne pa r t in 1000; i t sho uld real ly
be 8.987 551 79.
T h e C o u l o m b f o rc e s in n a t u r e a r e e n o r m o u s w h e n c o m p a r e d w i t h
the g rav i t a t iona l fo rces . The g rav i t a t iona l fo rce be tween two masses m a
a n d m h separa ted by a d i s t ance r i s
F = 6.672 0 x \Q - llm am blr2. (2-4)
* F ro m Eq. 2 -1 , e 0 i s expres sed in cou lom bs squa re d pe r new ton- squa re me te r . Ho w
ever , a cou lom b squared pe r newton-m ete r , o r a cou lo mb s quare d pe r jou le , is a fa rad ,
as we sha ll see in Sec. 4.3.
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2.3 The Principle of Superposition 43
The g rav i t a t iona l fo rce on a p ro ton a t t he su r f ace o f the sun (mass =
2.0 x 1 0 3 0 k i lograms, r ad ius = 7 .0 x 10 8 meters) i s equal to the elect r ic
fo rce be tween a p ro ton and one microgram of e l ec t ron s , separa ted by ad i s t ance equa l to the sun ' s r ad ius .
I t i s r em ark ab le indee d tha t we sho u ld no t be cons c ious o f these
en or m ou s fo rces in everyday life . Th e pos i t ive an d nega t ive charg es ca r r i ed
respect iv ely by the pr ot on an d the ele ct ro n are very near ly , if no t exact ly ,
the sam e. Exp er i m en ts ha ve sho wn t ha t they dif fer, if a t a ll , by at mo st on e
par t i n 10 2 2 . O r d i n a r y m a t t e r i s t h u s n e u t r a l , a n d t h e e n o r m o u s C o u l o m b
forces p reven t the acc um ula t ion o f any appre c iab le qu an t i t y o f charg e of
ei ther s ign.
2.2 THE ELECTRIC FIELD INTENSITY E
W e th ink o f the fo rce be twe en the po in t charg es Q a a n d Q b i n C o u l o m b ' s
l aw as the p roduc t o f Q a a n d t h e field of Q b, or vice versa. We define the
electric field intensity E to be the force per uni t charge exer ted on a test
charge in the f ie ld . Thus the elect r ic f ie ld in tensi ty due to the point charge
Q a is
E
"=
7 r = 7 %r
> - <2
"5
>Qb 4 7 r e 0 r
Th e elect r ic field in ten si ty i s m ea su red in vol ts per m eter .
The elect r ic f ie ld in tensi ty due to the point charge Q a i s t he same,
whether the t es t charge Q h is in the field or not, even if Q b i s l a rg e c o m p a r e d
to Q„.
2.3 THE PRINCIPLE O F SUPERPOSITION
If t he e l ec t ri c f ie ld i s p ro du ce d by m or e tha n on e charge , each one p rod uce s
i ts ow n f ie ld , an d the res ul ta nt E is s imp ly the vector su m of the indiv idua l
E 's . This i s the principle of superposition.
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44 Fields of Stationary Electric Char ges: I
THE FIELD OF AN EXTENDED
CHARGE DISTRIBUTION
F o r a n e x t e n d e d c h a r g e d i s t r i b u t i o n ,
E - J L f . / * * ' . (2-6)
Th e mea n in g o f the var io us t e rm s und er the in t egra l s ign is i l l us t r a t ed in
Fig. 2-2: p i s t he e l ec t r i c charge dens i ty a t t he source po in t (x',y',z')\ rj is
a uni t vector point ing f rom the source point to the f ie ld point (x ,y ,z) where
E i s ca l cu la t ed ; r i s t he d i s t ance be tween these two po in t s ; dr ' i s the e lem ent
o f v o l u m e dx' dy' dz'. I f the re exist sur face dis t r ib ut io ns of cha rge, the n we
must add a su r f ace in t egra l , wi th the vo lume charge dens i ty p r ep laced by
the su r f ace charge dens i ty a a n d t h e v o l u m e x' replaced by the sur face S' .
Th e e l ec t r i c cha rge dens i ty ins ide ma cro sco p ic bod ies i s o f cour se
no t a sm oo th funct ion of x ' , y', z'. However , nuc le i and e l ec t rons a r e so
smal l and so closely packed, that one can safely use an average elect r ic
charge dens i ty p, as above , to ca l cu la t e macroscop ic f i e lds .
When the e l ec t r i c f i e ld i s p roduced by a charge d i s t r ibu t ion tha t i s
dis tur be d by the int ro du ct i on of a f in ite test ch arg e Q', E is the force per unit
charge as the magn i tude o f the t es t charge Q' t e n d s t o z e r o :
E = lim (2-7)
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2.5 The Electric Potential V 45
THE ELECTRIC POTENTIAL V
Consider a t e s t po in t charge Q' t ha t can- be mov ed a bo u t in an e l ec t r ic
f ie ld . The work W r equ i r ed to move i t a t a cons tan t speed f rom a po in t P t
t o a po in t P 2 along a given path is
The nega t ive s ign i s r equ i r ed to ob ta in the work done against the field.
H e r e a g a i n , w e a s s u m e t h a t Q' i s so smal l t ha t t he charg e d i s t r ibu t ions a r e
no t apprec iab ly d i s tu rbed by i t s p r esence .
I f the p ath is c losed , the tota l wo rk d on e is
Let us evaluate th is in tegral . To simpl i fy mat ters , we f i r s t consider the
elect r ic f ie ld produced by a s ingle point charge Q. T h e n
Figure 2 -3 shows tha t t he t e rm under the in t egra l on the r igh t i s s imply
dr/r2 o r — d(l/r). Th e sum of the inc rem en ts of (1/ r ) over a c losed pat h is
zero, s ince r has the same va lue a t t he beg inn ing and a t t he end o f the pa th .
Th en the l ine in t egra l i s ze ro , an d the ne t wo rk d on e in mo ving a p o in t
c h a r g e Q' ar ou nd a ny closed p ath in the field of a point cha rge Q, which
is fixed, is zero.
I f the elect r ic f ie ld is produced, not by a s ingle point charge Q, b u t
by some f ixed charge d i s t r ibu t ion , t hen the l ine in t egra l s co r r espond ing
to each ind iv idua l charge o f the d i s t r ibu t ion a r e a l l ze ro . Thus , f o r any
d i s t r ib u t ion o f f ixed charg es ,
(2-8)
(2-9)
(2-10)
(2-11)
An electrostatic field is therefore conservative (Sec. 1.11). I t can be
shown tha t t h i s impor tan t p roper ty fo l lows f rom the so le f ac t t ha t t he
Coulomb fo rce i s a cen t r a l f o r ce ( see P rob . 1-21).
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46
Then , f rom S tokes ' s t heorem (Sec . 1.13), at a l l points in space,
V x E = 0. (2-12)
and we can wr i t e tha t
E = -W , (2-13)
w h e r e V i s a scalar poi nt funct ion, s ince V x \V = 0.
We can thus desc r ibe an e l ec t ros t a t i c f i e ld comple te ly by means o f
the funct ion V(x,y,z), which is cal led the electric potential. T h e n e g a t i v e
sign is req uire d in ord er tha t the elect r ic field in tens i ty E poi nt to w ar d a
decrease i n po ten t i a l , accord ing to the usua l conven t ion .
I t i s impor tan t to no te tha t V i s no t un ique ly def ined ; we can ad d to
i t any qua n t i ty th a t is i nde pen den t o f the coor d in a tes w i tho u t a f fec ting E
in any way .
As sh ow n in Pro b. 1-23, the work d on e in mov ing a test ch arg e at acons tan t speed f rom a po in t P { t o a po in t P 2 i s i nde pen den t o f the pa th .
W e must r em em ber th a t we a r e dea l ing here wi th e l ec t rosf a f /cs . I f
the re were moving charges p resen t , V x E would no t necessa r i ly be ze ro ,
a n d W wo uld th en des cr ibe only pa r t of the elect r ic f ie ld in ten si ty E. W e
sha l l i nves t iga te these more compl ica ted phenomena l a t e r on , i n Chap te r 11 .
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47
According to Eq. 2-13,
E d l = - V K - d l = —cIV. (2-14)
Then, for any two points Px
and P2
as in Fig. 2-4,
V2
- V, = - J ^ E d l = £ E d l . (2 -15)
Note that the electric field intensity E(x,y,z) determines only the
difference betw een the pote nti als at tw o different poin ts. When we wish to
speak of the electric pote nti al at a given point, we mus t therefore arbi trar ily
define the potential in a given region of space to be zero. It is usually con
venient to choose the potential at infinity to be zero. Then the potential V
at the point 2 is
V = f" E d l . (2-16)
The work W required to bring a charge Q' from a point at which the
pote ntial is defined to be zero to the point co nside red is VQ'. Thus V is
WjQ' and can be defined to be the work per unit charge. The potential V
is expressed in jou les per co ul omb , or in volts.
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48
Figure 2-5 Equipotent ia l surface and l ines of force near a point charge.
When the f i e ld i s p roduced by a s ing le po in t charge Q, t h e p o t e n t i a l
a t a d i s t ance r is
I t wi l l be observed that the sign of the potent ia l V i s the same as that of Q.
The p r inc ip le o f superpos i t ion app l i es to the e l ec t r i c po ten t i a l V as
wel l as to the elect r ic f ie ld in tensi ty E. The potent ia l V a t a po in t P due to
a charge d i s t r ibu t ion o f dens i ty p i s therefore
w h e r e r i s t he d i s t ance be tween the po in t P and the e l ement o f charge p dx'.
The points in space that are a t a given potent ia l def ine an equipotential
surface. For example , an equ ipo ten t i a l su r f ace abou t a po in t charge i s a
con ce ntr i c sph ere as in Fig . 2-5. W e can see from Eq. 2-13 tha t E is every
where normal to the equipotent ia l sur faces (Sec. 1 .5) .
(2-17)
(2-18)
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2.5 The Electric Potential V 49
If we join end -to- end infinitesimal vectors rep resen ting E, we get a
curve in space—called a line of force—that is every whe re no rm al to the
equ ipot enti al surfaces, as in Fig. 2-5. The vecto r E is everyw here ta ngen t to
a line of force.
2.5.1 EXAMPLE: THE ELECTRIC DIPOLE
T h e electric ciipole s h o w n in Fig. 2-6 is one t ype of cha rge d i s t r ibu t ion tha t is en
counte red f requent ly . We sha l l r e tu rn to it in C h a p t e r 6.
The e l ec t r i c d ipo le cons i s t s of two cha rges , one pos i t ive and the othe r nega t ive ,
of th e s a m e m a g n i t u d e , s e p a r a t e d by a di s t ance s. We shal l calcula te V and E at a
di s t ance r t h a t is l a r g e c o m p a r e d to s.
A t P,
+ Q<
s
-Q
(2-19)
Figure 2-6 Th e two c h a r g e s + Q and — Q form a dipole . The e lec t r ic po ten t i a l at P
is the sum of t he po ten t i a l s due to the i nd iv idua l cha rges . The vec tor s points from
— Q to +Q, an d r, is a uni t vec tor po in t ing f rom the origin to the p o i n t of
o b s e r v a t i o n P.
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50 Fields of Stationary Electric Charges: I
w h e r e
Similar ly,
a n d
rz = r + - + rs cos 0
, s y s1 + ( — + - cos 0~lr r
r I 1 + - cos 0r
1 + - cos 0r
1 - — cos 0.*2r
r
rb
V =
1 + — cos 9,2r
4n e 0r2
cos 0.
(2-20)
(2-21)
(2-22)
(2-23)
(2-24)
(2-25)
(2-26)
This express ion is val id for r 3 » s 3 .
I t i s interes t ing to note that the potent ia l due to a dipole fa l ls off as 1/ r 2, where
as the potent ia l f rom a s ingle point charge varies only as 1/ r . This comes from the
* R e m e m b e r t h a t
n(n - 1) , n(n - l)(n - 2) ,(1 + df = 1 + na + —r.—- a 1 + - £ a 3 +
2 ' 3!
If n i s a pos i t ive integ er , the ser ies s tops wh en t he coeff ic ient i s zero . Try, for ex am ple ,
n = 1 and n = 2. If n i s not a pos i t ive integer , the ser ies converges for a 2 < 1.
This series is very often used when a « 1. Then
(1 + a)" * 1 + na .
F o r e x a m p l e .
(1 + a) 2 * 1 + 2a . (1 + aC " 2 « 1 - - , e tc .
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51
Figure 2-7 Lines of force (arrows) and equipotent ia l l ines for the dipole of Fig. 2-6.
The dipole is vert ical , a t the center of the f igure , wi th the pos i t ive charge c lose to
and above the nega t ive cha rge . In the cen t ra l r eg ion the l i nes come too c lose
toge the r to be shown.
fact that the charges of a dipole appear c lose together for an observer some dis
tance away, and that thei r f ie lds cancel more and more as the dis tance r increases .
We define the dipole moment p = Qs as a vec tor whose magni tude i s Qs a n dthat i s di rected from t he nega t ive to t he pos i t ive cha rge . Then
V = - -2. (2-27)
Figure 2-7 shows equipotent ia l l ines for an e lect r ic dipole . Equipotent ia l sur
faces a re gene ra t ed by ro t a t ing equipo ten t i a l l i nes a round the ve r t i ca l ax i s .
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52 Fields of Stationary Electric Charges: I
Let us now calculate E a t the point P(x.O.r) in the plane y = 0 as in Fig. 2-6.
We have tha t
p cos 0 p :
K
= 4 ?, 2
"2 8 )
and. a t P( . \ ,0 , r) ,
cV 3p xz 3r> sin 0 co s 0
£ = = — ? - — = —! , ( 2- 29 )
cx 4ne 0 r 5 4n e 0 r3
- 9 V 3 P y : = 0. ,2-30)dy 4ne 0 r
since y = 0, and
8V _ p (1 3 -2
<3z 4 7 i e 0 \
p 3 c o s 2 f) — 1
E r = - — = -y— I - 3 j - , ( 2- 31 )
dz 4ne0
\r r 5
(2-32)4n e 0 r 3
Figure 2-7 shows l ines offeree for the e lect r ic dipole .
SUMMARY
I t i s found empir ical ly that the force exer ted by a po in t charge Q a on a
po in t charge Q b is
4 n e 0 r 2
whe re /• i s t he d i s t an ce b e tween the charg es an d r, i s a un i t vec to r po in t ing
from Q a t o Q b. This i s Coulomb's law. We cons ider the fo rce ¥ a b as be ing thep r o d u c t o f Q b by the electric field intensity due to Q a,
E ^ T ^ S r , , (2-5)
or vice versa.
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Problems 53
According to the principle of superposition, two or more electric
field intensities acting at a given point add vectorially. For an extended
charge distribution,
E =1
cm )4ne 0 J*' r
2
The electrostatic field is conservative,
V x E = 0, (2-72)
hence
E = -\V, (2-13)
where
1 cp dx'
4 7 T € A Jz
rne0
is the electric potential; p dx' is the element of cha rge cont ain ed within the
element of volume dx', and r is the distan ce betwee n this elemen t and t he
point where V is calculated.
PROBLEMS
2-JE 1 COULOMB'S LAW
a ) Ca lcu la t e th e elect r ic f ie ld intens i ty that would be just sufficient to b a l a n c e
the gravi ta t ional force of t he ea r th on an e lec t ron .
b) If this e lect r ic f ie ld were produced by a second e l ec t ron loca ted be low th e
f i rs t one, what would be the di s t ance be tween the two e lec t rons?
T h e c h a r g e on an e lec t ron is —1.6 x 1 0 "
1 9
c o u l o m b and its m a s s is 9.1 x1 0 " 3 1 k i l o g r a m .
2-2E SEPARATION OF PHOSPHATE FROM QUARTZ
C r u s h e d F l o r i d a p h o s p h a t e ore cons i s t s of part ic les of q u a r t z m i x e d w i t h
part ic les of p h o s p h a t e r o c k . If the m i x t u r e is vibra t ed , the q u a r t z b e c o m e s c h a r g e d
+
The le t ter E i nd ica t e s tha t th e p r o b l e m is re la t ively easy.
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54
Figure 2-8 Elec t ros t a t i c s epa ra t ion of pho spha te f rom qu ar t z in c rushed pho sph a te
ore . The vibrat ing feeder VF cha rges the phospha te pa r t i c l e s pos i t ive ly , and the
quartz part ic les negat ively. See Prob. 2-2.
nega t ively and the pho sph a te pos i t ively . The ph osp ha te can then be s epa ra t ed ou t
as in Fig. 2-8.
Ov er what m ini mu m d is tanc e must the part ic les fa ll if they must be sep arat edby at leas t 100 mil l imeters?
Set £ = 5 x 1 0 5 vol ts per meter and the specif ic charge equal to 10" 5 c o u l o m b
per k i logram.
2-3E ELECTRIC FIELD INTENSITY
A charge + Q is situ ate d at x = a, y = 0, an d a cha rge — Q at x = — a, y = 0.
Ca lcul ate the e lect r ic field intens i ty a t the po int x = 0, y = a.
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Problems 55
2-4 ELECTRIC FIELD INTENSITY
A c i rcu la r d i sk of cha rge has a rad ius R and carr ies a surface charge dens i ty a.
Find the e lect r ic f ie ld intens i ty £ a t a point P on the axis a t a dis tance a.
What i s t he va lue of E w h e n a « Rl
You can solve this problem by calculat ing the f ie ld a t P due to a r ing of charge
of radiu s /• an d width dr , and th en integ rat in g from r = 0 to r = R.
CATHODE-RAY TUBE
Th e elect ron be am in the cat ho de- ray t ub e of an osci l loscope is deflected vert ical ly
and horizontal ly by two pairs of paral le l deflect ing pla tes that are mainta ined a t ap
propr i a t e vo l t ages . As the e l ec t rons pas s be tween one pa i r , t hey a re acce le ra t ed by the
elect r ic f ie ld, and their kinet ic energy increases . Thus , under s teady-s ta te condi t ions ,
we achieve an increase in the kinet ic energ y of the e lect ron s wi thou t any exp end i ture
of power in the deflect ing pla tes , as long as the beam does not touch the pla tes .
Co uld th i s ph eno me no n be used a s the bas i s o f a pe rpe tua l mo t ion ma chin e?
2-6E CATHODE-RAY TUBE
In a certa in cathode-ray tube, the e lect rons are accelerated under a di fference
of pote nt ia l of 5 ki lov ol ts . After being accelera ted, they t ravel horizo ntal ly ov er a
dis tance of 200 mil l imeters .
Ca lcu la t e the downward de f l ec t ion ove r th i s d i s t ance caused by the grav i t a
t ional force .
An e l ec t ron ca r r i e s a cha rge of —1.6 x 1 0 "1 9
coulomb and has a mass o f
9 .1 x 1 0 " 3 1 k i l o g r a m .
2-7E MACROSCOPIC PARTICLE GUN
I t i s poss ible to obtain a beam of f ine part ic les in the fol lowing manner. Figure 2-9shows a paral le l -pla te capaci tor wi th a hole in the upper pla te . I f a part ic le of dus t ,
say, i s int roduced in the space between the pla tes , i t sooner or la ter comes into contact
with one of the pla tes and acquires a charge of the same polari ty. I t then f l ies over to
the oppos i t e p l a t e , and the proces s repea t s itself. The part ic le osci l la tes back and forth,
unt i l i t i s los t e i ther a t the edges or thro ug h th e centra l hole . O ne can th us obta in a
beam emerg ing f rom the ho le by admi t t ing pa r t i c l e s s t ead i ly in to the capac i to r . In
order to achieve high veloci t ies , the gun must , of course , operate in a vacuum.
Beams of macroscopic pa r t i c l e s a re used for s tudying the impac t o f mic ro-
m e t e o r i t e s .
Now i t has been found that a spherical part ic le of radius R lying on a charged
pla t e acqui re s a cha rge
Q = 1.65 x 4n ere
0R 2E
0 c o u l o m b ,
where £ 0 i s the e lect r ic f ie ld intens i ty in the absence of the part ic le , and e r i s the re la t ive
perm it t ivi ty of the p art ic le (Sec. 6.7) . W e assu me that t he pa rt ic le is a t leas t s l ight ly
c o n d u c t i n g .
Ass um ing that the pla tes are 10 mil l im eters apar t and tha t a vol tag e di fference
of 15 ki lovo l ts i s app l ied betw een t hem , f ind the veloci ty of a spherical part ic le 1
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56
Figure 2-9 Device for p roduc ing abeam of f ine part ic les . See Prob. 2-7
mic romete r in d i amete r a s i t emerges f rom the ho le in the upper p l a t e . Assume tha t
the part ic le has the dens i ty of water and that e r = 2 .
2-8E ELECTROSTATIC SPRAYING
Wh en pa in t ing i s don e wi th an ord in a ry spra y gun , pa r t o f t he pa in t e s capes
depos i t ion. The fract ion los t depends on the shape of the surface, on drafts , e tc . , and
can be as high as 80%. Th e use of the ordi nar y spr ay gun in large-scale indu s tr ia l pr o
cesses would therefore resul t in intolerable was te and pol lut ion.
Th e efficiency of spray p aint in g can be increas ed to nearly 100%, and th e po l lu
t ion reduced by a large factor , by charging the droplets of paint , e lect r ical ly, and
applying a vol tage di fference between the gun and the object to be coated.
I t i s found that , in such devices , the droplets carry a specif ic charge of roughly
o n e c o u l o m b p e r k i lo g r a m .
Assuming that the e lect r ic f ie ld intens i ty in the region between the gun and the
par t i s a t leas t 10 ki lov ol ts per mete r , wha t i s the m in im um rat i o of the e lect r ic force
to the grav i t a t ion a l fo rce?
2-9 THE RUTHERFORD EXPERIMENT
In 1906, i n the course of a h i s to r i c expe r im ent th a t de mo ns t ra t ed the sma l l
s ize of the a to mi c nucleus , Rut herfo rd obse rved that an a lph a part ic le (Q = 2 x
1.6 x 1 0 " 1 9 coulomb) wi th a kinet ic energy of 7.68 x 10 6 e lec t ron-vol t s making a
head-on col l is ion with a gold nucleus (Q = 79 x 1.6 x 1 0 " 1 9 coulomb) i s repe l l edb a c k w a r d s .
T h e electron-volt i s t he k ine t i c ene rgy acqui red by a pa r t i c l e ca r ry ing one
elect r onic charg e when i t i s accele rated th rou gh a di fference of pot ent i a l of on e vo l t :
1 e l ec t ron-vol t = 1.6 x 1 0 " 1 9 j o u l e .
a ) W ha t i s t he d i s t ance of c losest appr oac h a t which the e l ec t ros t a t i c po ten t i a l
energy is equa l to the ini t ia l kinet ic energy ? Expres s yo ur resul t in fem tom eters
( 1 0 "1 5 mete r ) .
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5 7
Figure 2-10 Cyl indr i ca l e l ec t ros t a t i c ana lyse r . A pa r t i c l e P can reach the col lector
C only i f i t i s pos i t ive an d if i t sa t is f ies the equ at io n given in Pro b. 2-11 . Oth erw ise ,
i t s tops on one of the cyl indrical e lect rodes . For a given charge-to-mass ra t io Q/m,
one can select di fferent veloci t ies by changing V.
A circle with the letter I represents an amm eter, while a circle with a letter V
represents a voltmeter.
b) W ha t i s t he ma xim um force of repu l s io n?
c) W hat is the max im um accele rat ion in </"s?
Th e mass of the a lpha part ic le is ab ou t 4 t imes that of a prot on , or 4 x 1.7 x
1 0 "2 7 k i l o g r a m .
ELECTROSTATIC SEED SORTING
No rm al s eeds can be s epa ra t ed f rom d i s co lored on es and f rom fore ign ob jec t s
by means of an e l ec t ros t a t i c s eed-sor t ing machine tha t ope ra t e s a s fo l lows . The s eeds
are observed by a pair of photocel ls as they fa l l one by one ins ide a tube. I f the color
is not r ight , vol ta ge is app l ied to a needle tha t dep os i ts a charge on th e seed. Th e seeds
then fa ll between a pair of e lect r ical ly cha rge d pla te s tha t deflect the undes ired one s
in to a s epa ra t e b in .
On e such ma chin e can sor t d ry peas a t t he ra t e o f 100 pe r s econd , o r a bou t
2 tons pe r 24-hour day .
a) If the seeds are dr op pe d at the ra te of 100 per second, over wha t d is tanc e
mu st they fal l if they m ust be sepa rate d by 20 mil l imeters whe n they pass between the
photoce l l s ? Neglec t a i r r e s i s t ance .
b) Assuming that the seeds acquire a charge of 1.5 x 10" 9 c o u l o m b , t h a t t h e
deflect ing pla tes are paral le l an d 50 mil l im eters apar t , and tha t the pote nt ia l di fference
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58 Fields of Stationary Electric Charges: I
be tween them i s 25 k i lovol t s , how fa r should the p l a t e s ex tend be low the cha rg ing
needle i f the charg ed seeds mu st be deflected by 40 mil l im eters on leaving th e p la tes ?
Assume that the charging needle and the top of the deflect ing pla tes are c loseto the photoce l l .
2-1 IE CYLINDRICAL ELECTROSTATIC ANALYSER
Figure 2-10 shows a cyl indrical e lect ros ta t ic analyser or veloci ty se lector . I t con
s is ts of a pair of cyl indrical conductors separated by a radia l dis tance of a few mil l imeters .
Sho w that , if the radia l dis tance betw een the cyl indrica l surfaces of avera ge
r a d i u s R is a, an d if the vol tag e betwe en the m is V, then the part ic les col lected a t /
have a veloci ty
(Q VRV12
\m a J
PARALLEL-PLATE ANALYSER
Figure 2-11 shows a pa ra l l e l -p l a t e ana lyse r . The ins t rument s e rves to measure the
energy d is t r ib ut io n of the cha rged part ic les emit te d by a sou rce. I t i s of ten u sed
for sources of e lect rons . I t has been used, for example , to inves t igate the energy
dis t r ibu t ion of e lec t ron s in the pla sm a formed at the focus of a lens i l lum inated by
a powerful laser .
If the source emits part ic les of various energies , then, for a given value of V
and for a given geometry, the col lector receives only those part ic les that have the
Figure 2-11 Para l l e l -p l a t e ana lyse r fo r measur ing the ene rgy d i s t r ibu t ion of
pa r t i c l e s emerg ing f rom a source S. The part ic les fol low a bal l is t ic t ra jectory in the
uniform elect r ic f ie ld between the two pla tes . Part ic les of the correct energy and, of
cou rse , of the correc t pola r i ty are col lected a t C. Th e othe r part ic les hi t e i ther o ne
of the two pa ra l l e l p l a t e s . The po la r i t i e s shown apply to e l ec t rons or to nega t ive
ions . See Prob. 2-12.
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Problems 59
corresponding energy. O ne therefore observes th e energy spectrum of the particles
by measuring th e collector current as a function of V. As we shall see below, we
assume that th e particles all carry th e same charge.In th e figure, particles of charge Q have a kinetic energy QVQ.
This type of analyser is relatively simple to build. It can also be, at th e same time,
compact and accurate. For example, in one case, th e electrodes measured 40 mm x
115 mm, with b = 15 mm and a = 50 mm. With slit widths of 0.25 mm, the de
tector current as a function of V for mono-energetic electrons gave a sharp peak as
in Fig. 2-12 having a width at half maximum of only 1%.
a) Find a for a particle of mass m and charge Q.
Solution: We can treat the particle as a projectile of mass m, charge Q, and
initial velocity vQ. with
1\ , (1)
V2
, = QV0,
r0J
2
^) . (2,in
The particle has a vertical acceleration, downwards, equal to QE/m, or
Q(V/b)
Then th e time of flight, from slit to slit, is
Do sin 8 2v r.mb7 = 2 — = — — s i n f l , (3)
Q(V/b)/m QV
an d th e distance a is given by
2vhnba = ( r n co s 8)T = sin 8 co s 0, (4)
0
QV
AbV 0 2bV 0
= sin 6 co s 8 = sin 28. (5)V V
Note that, for a given value of V0, a is independent of both m and Q. It is propor
tional to V0. If all the particles carry th e same charge Q, a is proportional to their
energy QV 0.
b) The optimum value of 0 may be defined as that for which th e horizontal
distance a traveled by the particle is maximum. One can then tolerate a slightly
divergent beam at the entrance slit, with an average 8 of 0 o p I .
Find ()„„,.
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60
Figure 2-12 The c u r r e n t / at the
col l ec tor C of Fig. 2 - 1 1 , as a funct ion
of V.
-AV
2±
Solution: Since th e di s t ance a m u s t be m a x i m u m , da/dO m u s t be ze ro . N ow
si n 20 has a m a x i m u m at 0 = 45 , so 6op, = 45°.
Note that , s ince a is m a x i m u m at 0 = 45 this is a l so the c o n d i t i o n for m a x i m u m
reso lu t ion .
c) Now find V0 as a funct ion of V, for given values of a and b, at 0opl.
Solution: F r o m Eq. 5, wi th 0 = 45°,
V0
2bV. (6)
d) F ind da/dV0 for 0 = 0opl.
Solution: At 0 = fl„ n,.
da
'IK,
2b
~V'
(7)
e) Find th e m i n i m u m v a l u e of V as a funct ion of V0 at fl opl.
Solution: In all the a b o v e e q u a t i o n s . V is a lways d iv ided by b. To find V we
use th e fact that V m u s t be sufficient to reduce th e vert ical veloci ty to z e r o :
QV > - m(v0 sin 4 5c
)2
, (8)
V > m 2QV0
1 V0
2Q m 2(9)
C o m p a r i n g now wi th Eq. 6, we see t h a t alb m u s t be smal l e r t han 4.
This ana lyse r is prac t i ca l for ene rg ies QK 0less than, say, 10 ki loe lec t ron-vol t s .
I t would be i nconvenien t to use vol t ages V l a rge r than a few ki lovol t s . For highe r
ene rg ies one w o u l d use the ana lyse r of P r o b . 2-11.
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Problems 61
2-13 CYLINDRICAL AND PARALLEL-PLATE ANALYSERS COMPARED
Show tha t bo th the cy l indr i ca l ana lyse r o f P rob . 2 -11 and the pa ra l l e l -p l a t e
ana lyse r o f P rob . 2 -12 measure the ra t io
w h e r e m i s the mass of the part ic le , v i t s veloci ty, and Q i t s charge.
2-14 ION THRUSTER
Synchronous satellites desc r ibe c i rcu la r o rb i t s above the equa tor a t t he a l t i t ude
where thei r angular veloci ty is equal to that of the earth. Since these sate l l i tes are
ex t remely cos t ly to bu i ld and to l aunch , t hey mus t rema in ope ra t iona l fo r s eve ra l
years . However, i t i s imposs ible to adjus t thei r ini t ia l pos i t ion and veloci ty wi th suff i
c ient accuracy to mainta in them fixed with respect to the earth for long periods oft ime . Moreover , t hey a re pe r tu rbed by the moon, and the i r an tennas mus t rema in
cons tan t ly d i rec t ed towards the ea r th .
Synchronous s a t e l l i t e s a re the re fore equipped wi th thrusters whose func t ion
is to exert the smal l forces necessary to keep them properly oriented and on their
presc r ibed orb i t s . Even then , t hey keep wander ing about the i r r e fe rence pos i t ion . For
example , t he i r a l t i t ude can va ry by a s much as 15 k i lomete rs .
The th rus t i s m'v, w h e r e m i s the mass of propel lant e jected per uni t t ime and v
i s the exhaus t veloci ty wi th respect to the sa te l l i te . Clearly, m should be sma l l . Then
v should be large. However, as we shal l see , too large a r could lead to an excess ive power
consumpt ion . So the choice of ve loc i ty depends on the maximum a l lowable va lue for m'
and on the ava i l ab le power .
One way of achieving large values of v i s to e ject a beam of charged part ic les ,
as in Fig. 2-13. Th e pro pel l ant i s ionized in an ion sou rce and e jected as a p os i t ive
ion beam.
By itself, the beam cannot exert a thrus t , for the fol lowing reason. Ini t ia l ly, the
net charge on the sate l l i te i s zero. When i t e jects pos i t ive part ic les , i t acquires a negat ive
charge. The part ic les are thus held back by the e lect ros ta t ic force , and, once the
satelli te is sufficiently ch arg ed , they fall ba ck o n the satelli te.
To achieve a thrus t , the ions must be neutra l ized on leaving the sate l l i te . This
is achiev ed by m ea ns of a heated f ilament, as in Fig. 2-13. Th e f i lament emits e lect ro ns
tha t a re a t t rac t ed by the pos i t ive ly cha rged beam.
a) Th e curre nt / of pos i t ive ions in the be am is carr ie d by part ic les of ma ss m
a n d c h a r g e ne , w h e r e e i s t he e l ec t ron ic cha rge .
(1 2)mv 2
Q
Show tha t t he th rus t i s
b) W ha t i s t he va lue o f f fo r a 0 .1 ampe re beam of p ro ton s with V = 5 0 k i l o v o l t s ?
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62
S
Figure 2-13 Sch em at ic diagr am of an ion thru s ter . The prop el la nt i s adm it ted a t P
and ionized in S: A is a beam-s hapin g e l ec t rode and B i s t he acce le ra t ing e l ec t rode ,
ma in ta ined a t a vo l t age V with respect to S; t he hea ted f i l ament F injects e lect rons
in to the ion beam to make i t neu t ra l . E lec t rode B i s part of the outer skin of the
sate l l i te . See Prob. 2-14.
c) If we call P t h e p o w e r IV spent in acce le ra t ing the beam, show tha t
Note tha t , s ince F = (2m'P)1 2
, for given values of m and P, the thrus t i s in
dependent o f t he cha rge - to-mass ra t io o f t he ions .
Also, s ince F is 2P/v, the thrus t i s inversely p r o p o r t i o n a l t o v, for a given pow er
e x p e n d i t u r e P.
Final ly, the las t express ion shows that , again for a given P, i t is preferable to
use heavy io ns carryin g a s ingle char ge (» = 1) an d to use a low accelera t ing vo l tage V.
d) If the f i lament is lef t unh eate d a nd if the be am curr ent i s 0.1 amp ere , ho w
long does i t take the body of the sa te l l i te to a t ta in a vol tage of 50 ki lovol ts?
Assu me tha t the sa te l l i te is spherica l an d tha t i t has a rad ius of 1 m e t e r .
COLLOID THRUSTER
Before reading this solved problem you should f i rs t read Prob. 2-14.
F igure 2-14 shows a so-ca l l ed co l lo id th rus t e r . A conduc t ing f lu id i s pumped
into a hol low needle mainta ined a t a high pos i t ive vol tage wi th respect to the
sa te l li t e 's ou te r sk in . Mic ro scopic d ro p le t s fo rm a ro un d the edge of the op ening ,
where the e lect r ic f ie ld intens i ty and the surface charge dens i ty are extremely high.
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63
Figure 2-14 Col lo id th rus t e r . The
hol low needle TV ejects a beam of
pos i t ively charged, high-veloci ty
drople t s . E lec t rode B and f i l ament F
are as in Fig. 2-13. See Prob. 2-15.
These d rop le t s , ca r ry ing spec i fi c cha rges Q/m as large as 4 x 10 4 coulombs pe r k i lo
gra m, are acceler ated in the e lect r ic field and form a high-en ergy je t . As a rule , several
need les a re g rouped toge the r so a s to ob ta in a l a rge r th rus t .
a) Calculate the thrus t exerted by a s ingle needle for the specif ic charge given
above , when V = 10 k i lovol t s and / = 5 mic r oam peres .
Use the resul ts of Prob. 2-14.
Solution:
'2m
QV.
1/2
P. I D
1/2
1 0 4 x 5 x 1 0 " 6 , (2)V4 x 10* x 1 0 4
= 3 .5 mic rone wto ns . (3)
b) C alcu late the ma ss of f luid e jected per seco nd.
Solution: T h e m a s s in ejected per second is m/Q t imes the charge e jected per
second , which is 5 x 1 0 " 6 c o u l o m b :
tri = 5 x 10 6 ' 4 x 1 0 4 = 1.3 x 1 0 " 1 0 ki logram/second . (4 )
c) The thrus t i s so smal l that i t i s not pract ical to measure i t di rect ly. The
me tho d used is i l lus t ra ted in Fig. 2-15. Th e thru s ter i s f irst put i nto op era t ion .
Then swi t ch S i s open ed an d the vo l t age IR ac ros s R i s obse rved on an osci l losc ope,
as a funct ion of the t ime. Th e curve has the shap e show n in Fig. 2-16a.
Calculate the thrus t as a funct ion of V, / , D, T.
Solution: The th rus t i s
) = 2IV , (5)v
T V= 21V- = 2 — IT .
D D(6)
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Figure 2-15 Method used for measur ing the th rus t o f t he co l lo id th rus t e r o f F ig .2-14 . The cur ren t / p rod uce d by the cha rged d rop le t s co l l ec ted on the hem ispher i ca l
e lect rode gives a vol tage IR that can be observed as a funct ion of the t ime on an
osc i l loscope .
d) If there are several types of droplets , the curve of J as a funct ion of t h a s
several pla teaus as in Fig. 2-16b.
How can one ca l cu la t e the th rus t i n th i s case?
Solution: In the las t express ion for F in Eq. 6, we see that F is 2V/D t imes the
area under the curve of Fig. 2-16a. Thus , in the case of Fig. 2-16b,
F = 2-[m
Idt. (7)D J °
T t t
(a) (b)
Figure 2-16 (a ) Current / as a funct ion of the t ime in the se t -up of Fig. 2-15, s tar t ing
at the ins tant when switch S i s opened, when al l the droplets are of the same s ize
and carry equal charges , (b) If there are several types of droplets of di fferent s izes
or charges , the curve of / as a funct ion of t has one p la t eau for each type .
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CHAPTER 3
FIELDS OF STATIONARY
ELECTRIC CHARGES: II
Gauss's Law, Poissons and Laplace's Equations,
Uniqueness Theorem
3.1 SOLID ANGLES
3.1.1 The Angle Subtended by a Curve at a Point
3.1.2 The Solid Angle Subtended by a Closed Surface at a Point
3.2 GAUSS'S LAW
3.2.1 Examp le: Infinite Sheet of Charge
3.2.2 Examp le: Spherical Charge
3.3 CONDUCTORS
3.3.1 Examp le: Hollow Conductor
3.3.2 Examp le: Isolated Charged Conducting Plate
3.3.3 Examp le: Pair of Parallel Conducting Plates Carrying
Charges of Equal Magnitudes and Opposite Signs
3.4 POISSON'S EQUATION
3.4.1 Examp le: Flat Ion Beam
3.5 LAPLACE'S EQUATION
3.6 THE UNIQUEN ESS THEORE M
3.7 IMAGES
3.7.1 Examp le: Point Charge Near an Infinite Grou nded
Conducting Plate
3.8 SUMMARY
PROBLEMS
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No w th a t we have d i scussed the two fun dam enta l conc ep t s o f e l ec t r i c fie ld
in tens i ty E and e l ec t r i c po ten t i a l V, we sha l l s tudy Gauss ' s l aw, Po i sson ' sequa t ion , and the un iqueness theorem. Th i s wi l l g ive us th r ee power fu l
methods fo r ca l cu la t ing e l ec t ros t a t i c f i e lds .
But f i r s t we must s tudy sol id angles.
3.1 SOLID ANG LES
3.1.1 THE ANGLE SUBTEND ED BY A CURVE AT A POINT
Co nsid er the cu rve C an d the po in t P o f F ig . 3 -1 . The y a r e s i tua ted in a
plane. We wish to f ind the angle oc sub tended by C a t P . Th i s i s
The in t egra l s a r e eva lua ted over the cu rve C: fo r the second one , we mul t ip ly
each element of length dl by sinW and div ide b y r; t hen we sum the r esu l t s .
Th e a r c c sub tend s the same ang le a a t P .
I f no w th e curv e is c lose d, as in Fig . 3-2a, wi th P insid e C, the anglesubtended at P by the smal l c i rcle of radius b is 27rb/b , or 2n r a d i a n s . T h e
ang le sub tended by C a t an y inside poin t P is there fore also 2n r a d i a n s .
I f P is outside C, as in Fig . 3-2b, then the segments C x a n d C 2 s u b t e n d
ang les o f the sam e ma gn i tud e bu t o f opp os i t e s igns , s ince s in 0 i s posi t ive
o n C t and nega t ive on C 2 . Thus the ang le sub tended by C a t an y o u t s i d e
po in t i s ze ro .
(3-1)
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67
Figure 3-1 A c u r v e C and a p o i n t P s i tua t ed in a plane . The a n g l e s u b t e n d e d by C
at P is a. Th e e lement dl' is the c o m p o n e n t of dl in the di rec t ion pe rpendicu la r to
t he rad ius vec tor .
Figure 3-2 [a ) Closed curve C and a p o i n t P s i tuated ins ide . The a n g l e s u b t e n d e d by
C at P is the s a m e as t h a t s u b t e n d e d by the smal l c i rc le , namely 2n r a d i a n s .
(/)) When th e p o i n t P is o u t s i d e C. th e angle sub tended by C at P is ze ro because
segment s Cj on the r ight and C2 on the left of t he po in t s where th e t a n g e n t s to
t he curve go t h r o u g h P subtend equa l and oppos i t e ang les .
3.1.2 THE SOLID ANGLE SUBTENDED BY A CLOSED
SURFACE AT A POINT
Fig ure 3-3 sh ow s a closed surface S an d a poi nt P situate d inside. Th e small
cone intersects an element of area da on S. This small cone defines the solid
angle subtended by da at P.
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68
Figure 3-3 A closed surface S and a po in t P s i tuated ins ide . Th e cone defines the
so l id angle sub tended by the e l ement o f a rea d a at P. The so l id angle sub tended a t
P by the complete surface S i s the same as that subtended by the smal l sphere of
r a d i u s b, n a m e l y 4JI s t e r a d i a n s .
By def ini t ion, th is sol id ang le is the a rea da , pro jec ted on a p l ane
perpend icu la r to the r ad ius vec to r , and d iv ided by r 2:
dQ = da cos 9 da (3-2)
w h e r e 0 i s t he ang le be tween the r ad ius vec to r r a n d t h e v e c t o r da , n o r m a l
to the su r f ace and po in t ing ou tward .
Sol id angles are expressed in steradians.
The to t a l so l id ang le sub tended by S a t P is
Qda cos 0
(3-3)
Th i s so l id ang le is a l so the so l id ang le sub te nde d a t P by the smal l sphere
s of radius a. Thus the so l id ang le sub tended by S a t an y inside point i s
Q = Ana 2/a 2 = An s t e r a d i a n s . (3-4)
N ote th a t a so l id ang le i s d imen s ion less , l i ke an o rd ina ry ang le .
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I f the point P i s s i tuated outside the sur face as in Fig . 3-4, then r j • da
i s posi t ive over S Y a n d n e g a t i v e o v e r S 2, and the to t a l so l id ang le sub tended
a t an y ou t s ide po in t i s ze ro .
The s i tua t ion r em ain s unch an ge d if t he su rf ace is con vo lu ted in such
a way that a l ine drawn f rom P cu t s the su r f ace a t more than one po in t . The
total sol id angle subtended by a c losed surface is s t i l l 4n at an inside point ,
and ze ro a t an ou t s ide po in t .
GAUSS'S LAW
G au ss 's law relates the flux of E th ro ug h a closed surface to the total c har ge
enc losed wi th in the su r f ace .
By using this law one can f ind the elect r ic f ie ld of s imple charge
d i s t r i bu t io ns wi th a m in i m um of e f fo r t. Le t us suppos e tha t a po in t charge Q
i s located at P in Fig . 3-3. W e can ca lcula te the f lux of E thr ou gh the closed
surface as fol lows. By def ini t ion, the f lux of E through the element of area
da is
(3-5)
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70 Fields of Stationary Electric Char ges: II
w h e r e T1 • da i s the project ion of da on a p lane normal to r t. T h e n
E • da = dO , (3-6)Ane 0
w h e r e da i s t he e l ement o f so l id ang le sub tended by da a t the po in t P.
To find the total f lux of E, we in tegra te over the whole su r f ace S. Since
the po in t P i s inside the sur face, by hypothesis , the in tegral of da is An a n d
J*s E • da = Q/e 0. (3-7)
I f m ore th an o ne po in t ch arge r es ides wi th in S, the f luxes add alge
braical ly , and the total f lux of E l eav ing the vo lu me i s equ a l to the to t a l
enc losed charge d iv ided by e 0 .
I f the charge enclosed by the sur face S' is di str ib ut ed ov er a f inite
vo lume, the to t a l enc losed charge i s
Q = J*, P dz\ (3-8)
w h e r e p i s t he charge dens i ty and x i s the volume enclosed by the sur face S' .
T h e n
f, E • da' = - f, p dr'. (3-9)
This is Gauss's law stated in integral form.
Applying the divergence theorem (Sec. 1 .10) to the lef t -hand side,
J \ V • E dr' = — £ p dr'. (3-10)
Since this eq ua t io n is val id for any closed surface, the in t eg ran ds m ust b eequal and, a t any point in space,
V - E = p / e 0 . (3-11)
This is Gauss's law stated in differential form. I t concerns the derivatives of
E wi th r espec t to the coord ina tes , and no t E itself.
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3.2 Gauss's Law 71
3.2.1 EXAMPLE: INFINITE SHEET OF CHARGE
Figure 3-5 shows an infinite sheet of charge, of surface charge density a. On each
side, E = a/2e0.
Figure 3-5 Portion of an infinite sheet
of charge of density a coulombs per
square meter. The imaginary box of
cross-section A encloses a charge a A.
The flux of E leaving the box is 2EA.
Therefore, E = c/2e 0 .
EXAMPLE: SPHERICAL CHARGE
Let us calculate E, both inside and outside a spherical charge of radius R and uni
form density p, as in Fig. 3-6. The electric field intensity E is a function of the distance
r from the center 0 of the sphere to the point considered.
We shall use the subscript o to indicate that we are dealing with the field out
side the charge distrib ution, and the subscrip t / inside. The total charge Q is ( 4 /3 ) 7 iK3
p .
Consider an imaginary sphere of radius r > R, concentric with the charged
sphere. We know that E 0 must be radial. Then, according to the integral form of
Gauss's law.
4nr2
E0
= Q/e0, (3-12)
Figure 3-6 Uniform spherical charge. We can calculate the electric field at a point
P outside the sphere, as well as at a point P' inside the sphere, by using Gauss's
law.
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0 R
Figure 3-7 In the case of a uniform spherical charge dis t r ibut ion, the e lect r ic f ie ld
intens i ty E rises linearly from the center to the surface of the sphere and falls off as
the inve rse squa re of the d i s t ance ou t s ide the sphe re . On the o the r hand , t he
elect r ic potent ia l fa l ls f rom a maximum at the center in parabol ic fashion ins ide
the sphere , and as the inverse f i rs t power of the dis tance outs ide .
a n d
E „ = — % (3-13)
as i f the tota l charge Q were s i tuated a t the center of the sphere .
I f t he cha rge were no t d i s t r ibu te d wi th sphe r i ca l sym met ry , E 0 would no t be
uni form over the imagina ry sphe re . Gauss ' s l aw would then on ly g ive the ave rage
va lue of the normal component o f E„ ove r the sphe re .
To ca lcu la t e E , a t an in t e rna l po in t , we draw an imagina ry sphe re of rad ius r
t h r o u g h t h e p o i n t P. Symmet ry requi re s aga in tha t E , be rad ia l : t hus
47rc2
£, . = ( 4 / 3 ) j t r 3 p / e 0 , (3-14)
a n d
E, = p r / 3 e 0 . (3-15)
F igure 3-7 shows E a n d V as funct ions of the radia l dis tance r . Note that E t =
E„ at ;• = R. This i s r equ i red by Gauss ' s l aw . s ince the cha rge conta ined wi th in a
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3.3 Conductors 73
spherical shel l of zero thickness a t r = R i s zero. Note a lso that Vt
= V0 at r = R.
A di scont inu i ty in V would require an inf ini te e lect r ic f ie ld intens i ty, s ince E is
-dV/dr.
CONDUCTORS
A conduc to r can be denned as a mate r i a l i ns ide which charges can f low
freely.
S ince we a r e dea l ing wi th e l ec t ros t a t i cs , we assume tha t t he charges
hav e r eache d the i r equ i l ib r ium pos i t ions an d a r e fixed in space . Th en , ins ide
a con du cto r , there is zero elect r ic f ie ld , an d al l poi nts are a t the sam e p ote nt ia l .
I f a con du ct or i s p lac ed in an elect r ic f ie ld , cha rge s f low tem po rar i ly
within i t so as to produce a second f ie ld that cancels the f i r s t one at every
p o i n t i n s id e t h e c o n d u c t o r .
C o u l o m b ' s l a w a p p l i e s w i t h i n c o n d u c t o r s , e v e n t h o u g h t h e ne t field
is zero. Ga us s 's law, V • E = p/e 0, i s a lso val id . Then, s ince E = 0 , the
c h a r g e d e n s i t y p must be ze ro .
As a co ro l l a ry , any ne t s t a t i c cha rge on a co nd uc to r m us t r es ide on
its surface.
At the sur face of a conductor , E must be normal , for i f there were a
t ange n t i a l co m po ne n t o f E , charg es wou ld flow a lon g the su rf ace , which
w o u l d b e c o n t r a r y t o o u r h y p o t h e s i s . T h e n , a c c o r d i n g t o G a u s s ' s l a w , j u s t
ou t s ide the su r f ace , E = c / e 0 , as in Fig . 3-8, a be ing the su r f ace charge
dens i ty .
I t i s paradoxical that one should be able to express the elect r ic f ie ld
intens i ty at the sur face of a co nd uc to r in ter m s of the local sur face cha rge
dens i ty a alo ne , des pi te th e fact tha t the f ie ld is of cou rse d ue to al l t he charges ,
whe ther they a r e on the conduc to r o r e l sewhere .
E A
Figure 3-8 Por t ion of a cha rged
conduc tor ca r ry ing a sur face cha rge
dens i ty a. The cha rge enc losed by the
imagina ry box i s a da. There is zero
fie ld ins ide the conductor . Then, from
Gauss ' s l aw , E = ff/e 0.
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74 Fields of Stationary Electric Charg es: II
EXAMPLE: HOLLOW CONDUCTOR
A hollow conductor, as in Fig. 3-9, has a charge on its inner surface that is equal in
magni tude and oppos i t e in s ign to any cha rge tha t may be enc losed wi th in the
hol low. Thi s is r ead i ly dem ons t ra t ed by cons ide r ing a Gauss i an sur face tha t l i es
wi th in the cond uc to r and tha t enc loses the ho l low. S ince E is eve rywh ere ze ro o n
this surface, the tota l enclosed charge must be zero.
Figure 3-9 Cross -s ec t ion of a ho l low conduc tor . A c h a r g e Q i n the ho l low induces
a tota l charge - Q on the inner surface. I f the conductor has a zero net charge, a
cha rge + Q i s induced on the outer surface.
3.3.2 EXAMPLE: ISOLATED CHARGED CONDU CTING PLATE
A n isolated charged conducting plate of inf ini te extent mu st carry equa l cha rge
dens i t i e s a on i ts two faces , as in Fig. 3-10a. Outs ide , £ = 2(a/2e0) = a/e0, while ,
ins ide the pla te , the f ie lds of the two surface dis t r ibut ions cancel and £ = 0.
EXAMPLE: PAIR OF PARALLEL CO NDUCTING PLATES CARRYING
CHARGES OF EQUAL MAGNITUDES AND OPPOSITE SIGNS
If on e has a pair of parallel conducting plates, as in Fig. 3-10b, carrying charges of egual
magnitude and opposite signs, the charges pos i t ion themselves as in the f igure and
£ = 0 every whe re excep t in the region betwee n the pla tes .
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75
3.4.1
a
• :
I TO
+ <J —o
*
£ = 0
• • "T"
•
E = 0 +a
£ = — " /•: = ()
-i l l ©
£ = 0
(a ) (b)
Figure 3-10 (a) Charged conduc t ing p la t e ca r ry ing equa l sur face cha rge dens i t i e s a
on each s ide . The field inside the pla t e is ze ro , {b) P a i r of paral le l pla tes carrying
surface charge dens i t ies +a and —<j. T he electric field inside th e pla t e s and
o u t s i d e is ze ro .
3 A POISSON 'S EQUAT ION
Ac cord ing to G au ss' s law, Eq . 3-11, V • E = p/e0. Now, since E = -W,
from Eq. 2-13,
„,„ 82
V d2
V d2
V pS
2
V = — + - 3 + — y = - A (3-16)dx dy dz e 0
This is Poisson's equation.
Poisson's equation often serves as the starting point for calculating
electrostatic fields. It states that, within a constant factor, the charge density
p is equ al to the sum of the secon d de rivati ves of V with respect to x, y, z.
EXAMPLE: FLAT ION BEAM
F i g u r e 3-11 s h o w s a flat ion beam s i tua t ed be tween tw o g r o u n d e d p a r a l l e l c o n
duc t ing p la t e s . A c o m p l e t e s t u d y of the elect r ic and magnet ic f ie lds of the b e a m
w o u l d be q u i t e e l a b o r a t e , but we sha l l d i s rega rd the m o t i o n of the i ons and l imi t
ourse lves to the electric field, in the s imple case where p is uniform ins ide th e b e a m
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76
p = 0 -
P
p=0-
Figure 3-11 Flat ion beam of uniform space charge densi ty p, b e t we e n t wo g ro u n d e d
conduct ing p la tes .
and where edge effects are negligible. We shal l use the subscript i on E and V inside
the beam, and the subscr ip t o ou ts ide .
Ins ide the beam,
d 2V,
~d xT
tf
e0'(3-17)
Et = - dV , = px
dx 6A.4. (3-18)
wh e re A is a cons tant of in tegrat ion. By sym me try, E,- = 0 at x = 0. ThereforeA = 0; inside the beam,
E, = px/e 0. (3-19)
F = —~x2
+ B, (3-20)
wh e re B i s ano the r cons tan t o f in teg ra t ion .
Now consider a cyl indrical volume whose upper and lower faces are outs ide
the beam, at equal d is tances from the plane x = 0, as in Fig . 3-11. Th e m ag nitu de
of E„ is the same at both faces. Let S be the area of one face. According to Gauss 'slaw.
2£„S = 2aSp/e0, (3-21)
£„ = ap/e0 x > a, (3-22)
£
„ = - a p / e 0 x < -a. (3-23)
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3.4 Poisson's Equation 77
Therefo re , ou t s ide the beam.
apx
V„ = — — + C x > a. (3-24)
apxV. = — + D x < -a, (3-25)
where C and D are aga in cons tan t s o f in teg ra t ion . S ince V = 0 both at x = b
an d at x = — b,
C = D = apb/e0(3-26)
a n d
apV„ = —(b- x) x > a. (3-27)
x < —a. (3-28)
Now there can be no d i scon t inu i ty o f V at the surface of the beam, for o therwise
E = — \V would be infini te. Then Eqs. 3-20 and 3-27 must agree at x = a:
-^a> + B
2 e 0
ap(b - a). (3-29)
ap I, aB = —\b
2(3-30)
E q u a t i n g t h e V's from Eqs. 3-20 and 3-28 at x = -a gives the same resul t .
F ina l ly , in s ide the beam,
p r x2 + a h - ^e 0
2V: = —
£, = px/e 0.
(3-31)
(3-32)
Above, for x > a.
K = °Ab-x),
E 0 = ap/e0,
(3-33)
(3-34)
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78
Figure 3-12 The electric field intensity E and po tent ia l F a s funct ions of .v, betw eenthe tw o pla tes in Fig. 3-11.
and below, for x < — a,
V 0 = ^(b + x), (3-35)
E 0 = -ap/e 0. (3-36)
The curves for V a n d E as funct ions of x are shown in Fig. 3-12.
3.5 LA PL A CE 'S EQ UA TION
W h e n p = 0 , P o i s s o n ' s e q u a t i o n b e c o m e s
\2V = 0. (3-37)
This i s Laplace s equation. This equa t ion app l i es no t on ly to e l ec t ros t a t i cs
bu t a l so to hea t conduc t ion , hydro- and ae rodynamics , e l as t i c i ty , and so on .
3.6 THE UNIQUENESS THEOREM
Accord ing to the uniqueness theorem, P o i s s o n ' s e q u a t i o n h a s o n l y o n e
s o l u t i o n V(x,y,z), for a given charge densi ty p(x,y,z) and fo r g iven boundary
c o n d i t i o n s . 1
f For a p roof o f t h i s t heorem, s ee our Electromagnetic Fields and Waves, p. 142.
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3.7 Images 79
This is an i mp or tan t the ore m. If, so meh ow, by intui tion or by analogy,
o n e finds a function V that satisfies both Poisson's equation and the given
boundary cond ition s, then it is the correc t V.
3.7 IMAGES
T h e me tho d of images involves the conver sionof an electric field into another
equivalent electric field that is simpler to calculate.
With this method, one modifies the field in such a way that, for the
region of interest, bot h the charge distrib ution and the boun dar y condit ions
are conserved. The method is a remarkable application of the uniqueness
t h e o r e m . It is best explaine d by mea ns of an examp le.
EXAMPLE: POINT CHARGE NEAR AN INFINITE GROUNDED
CONDUCTING PLATE
Consider a point charge Q at a distance D from an infinite conducting plate connected
to ground, as in Fig. 3-13a. This plate may be taken to be at zero potential. It is
clear that if we remove the ground ed co ndu cto r and replace it by a charge — Q at a
distance D beh ind the pla ne, the n every point of the plane will be equidis tant from
Q and from — Q, and will thus be at zero potential. The field to the left of the con
duct ing sheet is theref ore unaffected.
The charge — Q is said to be the image of the charge Q in the plane.
The potential V at a point P(x,y) as in Fig. 3-13b, is given by
Q(3-38)
where
r = (x2
+ y2
)112
an d f = [(2D - x)2
+ y2
]1
'2
.
The components of the electric field intensity at P are those ofW:
dV Qx Q(2D - x)47ie nA ,. = — 47re n —— = —w - H -, ,
(3-39)
(3-40)
BV Qy Qy
- * t e 0 — = - 3 73-
dv r r
(3-41)
The lines of force and the equipotentials are shown in Fig. 3-14.
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80
(a) (b)
Figure 3-13 (a ) Poin t cha rge Q nea r a g rounded conduc t ing p la t e , (b ) T h e
conduc t ing p la t e has been rep laced by the image cha rge — Q to calculate the field
at P .
At the surface of the conduct ing pla te , x = D, r = /•',
E x = 2QD/4ne0r\ E, = 0, (3-421
and the induced cha rge dens i ty i s - e0E
x. In this part icular case , the surface charge
dens i ty is negat ive , and the e lect r ic f ie ld intens i ty points to the r ight a t the surface
of the conduc t ing p la t e .
Figure 3-14 Lines of force ( ident i f ied by arrows) and equipotent ia ls for a point
cha rge nea r a g rounded conduc t ing p la t e . Equipo ten t i a l s and l ines o f fo rce nea r
the cha rge cannot be shown because they ge t t oo c lose toge the r . Equipo ten t i a l
surfaces are generated by rota t ing the f igure about the axis des ignated by the
curve d arro w. The im age f ie ld to the r ight of the cond uct ing pl a te is indica ted by
broken l ines .
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3.8 Summa ry 81
I t wi l l be observed that image charges are located outside t he reg ion where
we calculate the field. In this case we require the field in the region to the left of the
plate , whereas the image is to the r ight .N ow wha t is t he e l ec t ros t a t i c force of a t t rac t ion be tween th e cha rge Q and the
gro und ed p la t e? I t i s obvio us ly the s ame as be tween two cha rge s Q a n d — Q
sepa ra t ed by a d i s t ance 2D , s ince Q cannot te l l whether i t i s in the presence of a
po in t cha rge — Q or of a grounded pla te . The force on Q i s thus Q 2/4ne0{2D)2. T h i s
is a lways t rue . The force be tween a cha rge an d a con du c tor i s a lways g iven cor rec t ly
by the Coulomb force be tween the cha rge and i t s image cha rges .
SUMMARY
T h e solid angle sub tended by a c losed su r f ace S at a point P is 4n s t e r a d i a n s
if P i s s i tuated inside S, and zero i f P i s s i tua ted ou t s id e .
Gauss's law fo l lows f rom Coulomb 's l aw. I t can be s t a t ed e i the r in
integra l or in di f ferentia l form :
(3-9)
w h e r e S' i s the sur face that encloses r ' , or
V • E = fi/e 0. (3-11)
Poisson's equation,
\2V = -p/co, (3-16)
then follows from Eq. 3-11 and from E = — W.
W he n the char ge dens i ty i s ze ro we have Laplace's equation,
\2
V = 0. (3-37)
A c c o r d i n g t o t h e uniqueness theorem, for a given p and for a given set
o f bo un da ry c ond i t ions , t he re i s on ly one poss ib le po ten t i a l sa t is fy ing
P o i s s o n ' s e q u a t i o n .
T h e m e t h o d o f images is o n e r e m a r k a b l e a p p l i c a t i o n o f th e u n i q u e n e s s
theorem. For example , t he f i e ld o f a po in t charge Q in front of an infinite
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82 Fields of Stationary Electric Charges: II
gro und ed con duc ting plate is the same as if the plate were replaced by a
charge — Q at the position of the image of Q. The method of images gives
the correct field only outside the region where the images are situated.
PROBLEMS
3-1E ANGLE SUBTENDED BY A LINE AT A POINT
W h a t is the angle 0 s u b t e n d e d at a poin t P by a l ine of length a s i tua t ed at a
di s t ance b as in Fig. 3-15?
Figure 3-1 5 Line of l eng th a at a
di s t ance b from a p o i n t P. See P r o b .
3-1.
3-2 SOLID ANGLE SUBTENDED BY A DISK AT A POINT
W h a t is the so l id angle sub tended at a p o i n t P by a circular area of r a d i u s a
at a di s t ance b as in Fig. 3-16?
Hint: Consider f i rs t th e so l id angle sub tended at P by a r ing of r a d i u s /• and
width dr. and t hen in t egra t e .
Figure 3 -1 6 Circu la r a rea of r a d i u s a
s i tua t ed at a di s t ance b from th e p o i n t
P. See P r o b . 3-2.
3-3E GAUSS'S LAW
C o u l d one use G a u s s ' s law to ca lcu la t e th e elect r ic f ie ld intens i ty near a d i p o l e ?
3-4E SURFACE DENSITY OF ELECTRONS ON A CHARGED BODY
The maximum elect r ic f ie ld intens i ty that can be m a i n t a i n e d in air is 3 x 1 0 6 vol t s
pe r me te r .
a ) Ca lcu la t e th e surface charge dens i ty that wi l l produce this f ie ld.
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Problems 83
b) Co ns ide r a me ta l w i th an in t e ra tom ic spac ing of 0 .3 nanom ete r . W ha t i s
t h e a p p r o x i m a t e n u m b e r o f a t o m s p er s q u a r e m e t e r ?
c ) How many ex t ra e l ec t rons a re requi red , pe r a tom, to g ive a sur face cha rgedens i ty o f t he abo ve m agn i tud e?
3-5 THE ELECTRIC FIELD IN A NUCLEUS
a) Calculate the e lect r ic f ie ld intens i ty in vol ts per meter a t the surface of an
iod ine nuc leus (53 pro ton s and 74 neu t rons ) .
b) Fin d the e lect r ic pote nt ia l a t the center of the nucleu s .
Assu me tha t the cha rge dens i ty is uniform and th at the radiu s of the nucle us is
1.25 x 1 0- 1 5
j 41 / 3
mete r , where A i s the tota l number of part ic les in the nucleus .
3-6E THE SPACE DERIVATIVES OF Ex, £
v, E
z
Lis t a l l the re la t ions between the various space derivat ives of the three com
ponents of E, us ing Gauss 's law and the fact that an e lect ros ta t ic f ie ld is conservat ive .
3-7 PHYSICALLY IMPOSSIBLE FIELDS
An elect r ic f ie ld points everywhere in the z di rec t ion .
a ) W ha t c an you con c lude a bo ut the va lues of t he pa r t i a l de r iva t ives o f Ez
with respect to x, to y. and to z , ( i ) i f the volume charge dens i ty p is zero, (ii) if the
volum e cha rge dens i ty i s no t ze ro ?
b) Sketch l ines of force for som e f ie lds that ar e poss ible and for som e that are not .
PROTON BEAM
A 1 .00-mic roampere beam of p ro ton s i s acce le ra t ed th ro ugh a di ffe rence of po
tent ia l of 10,000 vol ts .
a ) Ca lcu la t e the vo lume cha rge dens i ty in the beam, once the pro tons have
been acce le ra t ed , a s sumin g tha t t he cur re n t dens i ty i s un i form wi th in a d i am ete r o f
2.00 mil l imeters , and zero outs ide this diameter . (The current dens i ty in an ion
be am is in fact larges t in the center a nd fa lls off gradua l ly wi th increas in g radius .)
Solution: The cha rge pe r un i t l eng th k i s equ al to the curre nt I, div ided by the
veloci ty v o f t h e p r o t o n s :
;. = i/ v. i n
Also,
( l / 2 ) m i ' 2 = eV , (2)
w h e r e m i s t he pro ton mass , e i s t he pro ton cha rge , and V i s the accelerat ing vol tage.
T h u s ,
(2eVjm)m \2eVj
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84
Figure 3-17 Cylin dric al surface sur
rounding the ion beam of Prob. 3-8
for calculat ing the magnitude of E
us ing Gauss ' s l aw.
a n d
p = )JnR 2
= 2.300 x 1 0 "7
c o u l o m b / m e t e r3
, (4)
wh e re R is the radius of the beam.
b) Calculate the radial electric f ield in tensi ty , both inside and outs ide the beam.
Solution: Outside the beam, E is radial . Applying Gauss 's law to a cyl indrical
surface as in Fig. 3-17,
2nr(e0Et) = /., (5)
E„ = X/2ne0r, (6)
/ / m Y '
2
1H r'ne 0 \2eVj
= 1.299 x 10"2
/r volts /meter, (8)
= 12.99 volts / mete r at r = 1 0 "3
meter. (9)
Inside the beam, we find E by applying Gauss 's law to a cyl indrical surface of
radius r < R. T h e n
E i = nr2
p/2tie0r = pr/2e0, (10)
= 1.299 x 104
r vol ts /m eter, (11)
= 12.99 volts /m eter at r = 1 0 "3
meter. (12)
c) Draw a graph of the radial electric f ield in tensi ty for values of r ranging from
zero to 10 mil l imeters .
Solution: See Fig. 3-18.
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85
Figure 3-18 P o t e n t i a l V and electric field intensity E as funct ions of the dis tance r
from the axis for the pr oto n be am of Pro b. 3-8. The be am h as a radiu s of
1 mi l l ime te r .
d) Now le t the beam be s i tuated on the axis of a grounded cyl indrical conduct ing tube with an ins ide radius of 10.0 mil l imeters .
Are the above va lues o f E s t i l l val id?
Solution: The y a re s t il l val id: £„ de pe nd s sole ly on " /. and r , and £, o n p a n d r.
It is the value of V that i s affected by th e presen ce of the tu be.
e ) D raw a graph of V ins ide the tube.
Solution: O u ts id e th e b ea m , K = 0 a t r = 1 0 " 2 m e t e r a n d b e y o n d . T h e n
V„ = Jr' ° " E„ dr = j ' " " 1.299 x 1 0 " 2 dr/r, (13)
= 1.299 x 1 0 " 2 [ l n ; - ]r
1 0
"2
, (14)
= 1.299 x 1 0 " 2 [ l n 1 0 " 2 - In r] , (15)
= - (5 .9 82 + 1 .299 In r ) 1 0" 2 volt, (16)
= 2.991 x 1 0 " 2 vol t a t r = 1 0 " 3 meter . (17)
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86 Fields of Stationary Electric Char ges: II
Ins ide the beam,
11= 2.991 x 1CT
2
+ (18)
2 .991 x 1(T 2 + 6.495 x 10 3 ( 1 ( T 6 - r 2 ) vol t , (19)
= 36 .41 x 10" 3 volt at r = 0. (20)
F igure 3-18 shows V as a funct ion of r.
3-9 ION BEAM
An ion beam of un i form dens i ty p passes between a pair of paral le l pla tes , one
at x = 0, a t zero pote nt ia l , an d on e a t x = a, a t t he po ten t i a l V 0. Neglec t t he mot ion of
the ions , as wel l as edge effects . The beam complete ly f i l l s the space between the pla tes .
3-10 A UNIFORM AND A NON-UNIFORM FIELD
Con s ide r a pa i r o f pa ra l l e l condu c t ing p la t e s , one a t x = 0 tha t i s g rou nde d ,
and o ne a t x = 0.1 me ter tha t is ma int a in ed a t a po tent ia l of 100 vol ts .
a ) Expres s V as a funct ion of x when p = 0 . Show the equipo ten t i a l s a t V =
0, 10, 2 0 , . . . 100 vol ts . Draw a curve of £ as a funct ion of x.
b) Do the same for the case where p/e 0 i s un i form and equa l t o 10 4 .
3-11 VACUUM DIODE
In a vacuum d iode , e l ec t rons a re emi t t ed by a hea ted ca thode and co l l ec t edby an anode. In the s imples t case , the e lect rodes are plane, paral le l , and c lose together ,
so edge effects can be neglected.
Since the e lect rons move at a f ini te veloci ty, the space charge dens i ty is not
ze ro . It is given by
w h e r e V0 i s the vol tage a t the anode ( the vol tage a t the cathode is zero); s i s the dis tance
between the e lect rodes ; and x is the dis tance of a point to the cathode. So, a t x = 0,
V = 0, while, at x = s, V = V 0.
This pecu l i a r cha rge d i s t r ibu t ion comes f rom the fac t t ha t t he e l ec t rons a reaccelerated in the e lect r ic f ie ld, which i tse l f depends on the charge dens i ty, and hence
on the e lect ron veloci ty. We assume that the current i s l imi ted by the space charge,
and no t by e l ec t ron emis s ion a t t he hea ted ca thode .
a ) Use Poi s son ' s equa t ion to f ind V as a funct ion of x.
b) At any point , the current dens i ty J is pv , w h e r e v i s the veloci ty of the e lect ro ns
at that point .
Use the value of pv at x = s to find J.
F i n d V and the t ransverse £ as funct ions of x.
P = ~
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Problems 87
3-12 IMAGES
Tw o cha rges + Q a n d — Q a re s epa ra t ed by a d i s t ance a and are a t a dis tance b
from a large conduct ing sheet as in Fig. 3-19.F ind the x and y com pon ent s o f t he force exe rt ed on the r igh t -h and cha rge .
+ Q
Figure 3-19 Pair of charges near a
conduc t ing sur face . See P rob . 3 -12 .
3-13 IMAGES
A c h a r g e Q i s s i tuate d ne ar a cond uct ing pla te form ing a r ight angle as in Fig. 3-20.
Find the e lect r ic f ie ld intens i ty a t P .
The re a re th ree images of equa l mag ni tud es .
Figure 3-20 Po int c harg e nea r a
conduc tor fo rming a r igh t ang le . See
Prob . 3 -13 .
3-14 IMAGES
A poin t cha rge Q i s s i tuated a t a dis tance D from an infini te conduct ing pla te
connec ted to g round , a s in F ig . 3 -13 .
a) F ind th e surface charg e dens i ty on the co nd uc tor as a funct ion of the radiu s .
b ) Show tha t t he to t a l cha rge on the conduc tor i s — Q.
3-15 ANTENNA IMAGES
An tenna s a re o f ten mo un ted n ea r condu c t ing sur faces . On e example is t he wh ip
antenna shown in F ig . 18-4 .
F igure 3-21 shows s chemat i ca l ly th ree an tennas , o r por t ions of an tennas , nea r
conduc t ing sur faces . The a r rows show the d i rec t ion of cur ren t f low a t a g iven
ins t an t .
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Figure 3-21 Th ree an te nna s s i tua t ed abov e conduc t ing p lanes , and the i r images .
The a r rows show the d i rec t ion of the cur ren t s in the an tennas , a t a g iven ins t an t .
See Prob. 3-15.
D raw figures show ing the di rect io n of cu rren t flow in the ima ges .
Solution: (a) Con s ide r a pos i t ive cha rge movin g dow nw ard in the ve r t i ca l
an ten na . I t s image i s a nega t ive cha rge tha t mov es upw ard . No w a nega t ive cha rge
mo ving up war d g ives a down wa rd cu r ren t . So the cur ren t i n the image is a s show n
in Fig. 3-22a.
b) Similar ly, a pos i t ive charge moving to the r ight has a negat ive image a lso
moving to the r ight . The current in the image is in the opposi te di rect ion, as shown
in Fig. 3-22b.
c ) A pos i t ive cha rge moving to the r igh t and downward has a nega t ive image
mo ving to the r igh t and u pwa rd . The image cur ren t i s i n the d i rec t ion show n in
Fig. 3-22c.
(a) (b) ^ (<)
: : _ "
Figure 3-22 R elat io n betwee n the curr ent di rec t ion s in the ante nn as and in thei r
images .
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CHAPTER 4
FIELDS OF STATIONARY
ELECTRIC CHARGES: III
Capacitance, Energy, and Forces
4.1 CAPACITANCE OF AN ISOLATED CONDUCTOR
4.1.1 Exam ple: Isolated Spherical Conductor
4.2 CAPACITANCE BETWEEN TWO CONDUCTORS
4.2.1 Exam ple: Parallel-Plate Capacitor
4.2.2 Capacitors Connected in Parallel
4.2.3 Capacitors Connected in Series
4.3 POTENTIAL ENERGY OF A CHARGE DISTRIBUTION
4.4 ENERG Y DENSITY IN AN ELECTRIC FIELD
4.4.1 Exam ple: Isolated Spherical Conductor
4.5 FORCES ON CONDUCTORS
4.5.1 Exam ple: Electric Force for E = 3 x 106 volts per meter
4.5.2 Parallel-Plate Capacitor
4.6 SUMMARY
PROBLEMS
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In th is chapter we shal l complete our s tudy of the elect r ic f ie lds of s ta t ionary
e lec t r i c charges in a vacuum. We sha l l s t a r t wi th the concep t o f capac i t ance .
This wi l l lead us to the energy stored in an elect r ic f ie ld , and this energy,
in turn, wi l l g ive us the force exer ted on a conductor s i tuated in an elect r ic
field.
CAPACITANCE OF AN ISOLATED CONDUCTO R
Ima g ine a co nd uc to r s i tua te d a long d i s t anc e awa y f rom any o the r bod y .
As ch arg e is ad de d to i t , i t s po ten t ia l r i ses, the m ag ni tu de of th e ch an ge in
p o t e n t i a l b e in g p r o p o r t i o n a l t o th e a m o u n t o f c h a r g e a d d e d a n d d e p e n d i n g
on the geom et r i ca l conf igura t ion o f the con du c to r .
T h e r a t i o
C = Q/V ( 4 - 1 )
i s cal led the capacitance of the i so la t ed conduc to r . A capac i t ance i s pos i t ive ,
by def ini t ion.
Th e un i t of cap ac i t a nce i s t he cou lom b per vo l t o r farad.
EXAMPLE: ISOLATED SPHERICAL CONDU CTOR
A n isolated spherical conductor of rad ius R carr ies a tota l surface charge Q. T h e n
i ts capaci tance is
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91
Figure 4-1 Tw o cond uc to rs ca r ry ing equa l and oppos i t e cha rges . Th e in t egra l ofE • dl f rom one conduc tor to the o the r , a long any pa th , g ives the po ten t i a l
difference V. The capac i t ance be tween the conduc tors i s Q/V.
4.2 CAPACITANCE BETWEEN TWO CONDUCTOR S
W e hav e jus t de f ined the cap ac i t a nce o f an i so la t ed cond uc to r . W e can
proceed in a s imi l a r way to def ine the capac i t ance be tween two conduc to r s ,
as in Fig . 4- 1.
In i t i a l ly , bo th conduc to r s a r e uncharged . Le t us assume tha t one o f
the cond uc to r s i s g ro und ed , so tha t i t s po ten t i a l i s kep t equa l to ze ro . W e
then g radua l ly add smal l charges to the o ther un t i l i t ca r r i es a charge Q.
In the p rocess , charges o f opp os i t e s ign a r e a t t r a c t ed to the g ro un ded con
duc to r un t i l i t ca r r i es a charge —Q. A t t h a t p o i n t t h e g r o u n d e d c o n d u c t o r
i s s ti ll a t ze ro po ten t i a l , wh i l e the o the r co nd uc t o r i s a t a po ten t i a l V.
Th e capac i t anc e be twe en the two con du c to r s is aga in def ined as in
E q . 4 - 1 :
C = Q/V. (4-3)
Th e cap ac i t a nce be tw een tw o cond uc to r s i s a l so pos i tive , by def in i tion .
In oth er wo rds , if we establ ish a po tent ia l d i f ference V b e t w e e n t w o
c o n d u c t o r s b y g i v i n g t h e m c h a r g e s +Q a n d — Q, t h e n t h e c a p a c i t a n c e
b e t w e e n t h e c o n d u c t o r s i s Q/V.
4.2 .1 EXAMPLE: PARALLEL-PLATE CAPACITOR
Figure 4-2 shows a parallel-plate capacitor with the spacing s gros s ly exaggera t ed .
W e neglect edge effects. We hav e a unifo rm pos i t ive charg e dens i ty er on the to p
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92
Figure 4- 2 Para l le l -p la te capac i to r . The lower end of the small cyl indrical f igure is
si tuated inside th e lower p la te where E = 0.
surface of the lower plate and a uniform negative charge densi ty — a on the b o t t o m
surface of the upper p la te . The electric field intensity E p o i n t s u p wa rd .
App ly ing Gauss ' s law to a cylindrical volume as in the figure, th e outward flux
of E is Eu. where a is the cross-sect ion of the cylinder. Since the enclosed charge
is (jo. then
Ea = aa/e0, E = a'e0, (4-4)
so tha tV = Es = as/e0. (4-5)
N o w Q = a A, wh e re A is the a re a of one p la te , and
C = Q/V = e0A/s. (4-6)
T h i s is the capac i tance be tween two para l le l conduct ing p la tes of a re a A s e p a ra t e d
by a d is tance s.
4.2.2 CAPACITORS CONNECTED IN PARALLEL
Figure 4-3a shows two capacitors C, and C2
connected in parallel and
carrying charges Ql
and Q2. We wish to find the value of C in Fig. 4-3b tha t
will give the same V for the same charge Qx
+ Q2.This is simple:
Qi + Q2= c,v + c
2v = cv. (4-7)
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93
VQ
(a) (b)
Figure 4 -3 The s ing le capac i to r C has the s ame capac i t ance a s the two capac i to rs
C, and C 2 connec ted in pa ra l l e l .
T h e n
C = C, + C 2 . ( 4 - 8 )
S o , t he capac i t ance o f two capac i to r s connec ted in para l l e l i s t he sum
of the two capac i t ances . I f we have severa l capac i to r s in para l l e l , t he to t a l
capa c i t an ce i s t he sum of the ind iv idua l ca pac i t an ces .
4.2.3 CAPAC ITORS CONNE CTED IN SERIES
F i g u r e 4 - 4 a s h o w s t w o c a p a c i t o r s C , a n d C 2 con nec ted in series . W e f i rs t
c o n n e c t p o i n t s A a n d B t o g r o u n d , t e m p o r a r i l y . T h i s m a k e s t h e v o l t a g e s
V l a n d V 2 z e r o . T h e c a p a c i t o r p l a t e s a re u n c h a r g e d . W e n o w d i s c o n n e c t
A a n d B f r o m g r o u n d .
We then app ly a charge Q t o A. W e a s s u m e t h a t t h e v o l t m e t e r s d r a w
zero curre nt . N o cha rge can flow int o or ou t of B. A c h a r g e — Q leaves theupper p l a t e o f C 2 and goes to the lower p la t e o f Cy. This l eaves a charge + Q
on the top plate of C 2 . Then the lower p la t e o f C , t akes on a charge — Q,
as in the f igure.
T h e n
Vi = QIC,, V 2 = Q/C 2. ( 4 - 9 )
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94
VQV Q
(a ) (b )
Figure 4-4 T h e c a p a c i t o r C has the s ame capac i t ance a s the two capac i to rs C, a n d
C 2 connected in ser ies .
Now what should be the value of C in Fig . 4-4b to give a vol tage V
equal to K, + V 2 wi th the same charge Q'l W e m u s t h a v e
Q Q QV = - = — + — , (4-10)
c c x c 2
o r
+ ( 4 - l Dc c , c 2
C = (4-12)
I f we have three capaci tors in ser ies ,
I * . j . J_ + _L ( 4 - , 3 )
= C ^ C j
C 2 C 3 + C 3 C , + c , c2
N ow t he inverse of the capa ci ta nc e, 1 /C, i s cal led the elastance. S o ,
if cap ac i t o r s a r e co nne c ted in se r i es , t he to t a l e l as t ance i s t he sum of the
ind iv idua l e l as t ances .
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4.3 Potential Energy of a Charge Distribution 95
Note tha t , when capac i to r s a r e connec ted in para l l e l , t he vo l t ages
are the same . W he n cap ac i to r s a r e con nec ted in se r ies , t he charges a r e the
s a m e .
POTENTIAL ENERGY OF A
CHARGE DISTRIBUTION
W e can find a gen eral exp ress ion for the pote nt ia l energy of a cha rge dis
t r ibu t ion by cons ider ing the energy expended in charg ing a pa i r o f con
ductors , as in Sec. 4 .2 .
When we add a smal l charge dQ t o the l e f t -hand conduc to r , we mustexpend an energy
dW = V dQ = | dQ, (4-15)
w h e r e Q i s t he charg e a l r eady on the co nd uc to r . Then , to charg e the con du c to r
to a po ten t i a l V = QIC, we mu st supp ly an energy
W = FC %dQ = %- = 1 QV = - CV2. (4-16)
J o
c •2C 2 2
Th e reaso n for the factor of one -half shou ld be clear f rom the rea son ing
that we have used to ar r ive at i t . I t i s that , on the average, the potent ia l V
at the t ime of ar r ival of a charge dQ i s just o ne-ha lf the f inal pote nt ia l .
M ore genera l ly , if we hav e a nu m be r o f charg ed con duc t ing bod ies ,
the s to red energ y i s t he sum of s imi l a r t e rm s fo r each on e :
W = WQiV, (4-17)
^ i
F o r a c o n t i n u o u s c h a r g e d i s t r i b u t i o n o f d e n s i t y p(x',y',z') o c c u p y i n g a
v o l u m e t ' , we can r ep lace Q : by p clr' a n d t h e s u m m a t i o n b y a n i n t e g r a t i o n :
W = - { v p d f , (4-18)
w h e r e V a n d p are bo th func t ions o f the coord ina tes in the genera l case .
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96 Fields of Stationary Electric Charges: III
4.4 ENERGY DENSITY IN AN
ELECTRIC FIELD
The energy stored in a parallel-plate capacitor is therefore
\ c r ' ' - H O ' l 4 • l 9 ,
= Q e0£
2
) As, (4-20)
where As is the volume occupied by the field. Th e sto red energy may therefore
be calculated by associating with each point an energy density [e0/2)E2
joules per cubic meter.
This is a general result. The energy associated with a charge distribu
tion, that is, the energy requ ired to assemble it, sta rti ng with the cha rge
spread over an infinite volume, is
W = ^ [ E2 dr, (4-21)2 J t
where the volume t extends to all the region where the field under consid
eration differs from zero.
The energy stored in an electric field can therefore be expressed, either
as in Eq. 4-18, or as in Eq. 4-21. No te th at the terms under the integral signs
in these two equations are unrelated. For example, at a point where p = 0,
Vp is zero but E2 is usually not zero.
4.4.1 EXAMPLE: ISOLATED SPHERICAL CONDUCTOR
F o r the isolated spherical conductor of radius R and carrying a charge Q,
W = f £ 2
dr.
Us i n g a spherical shell of radius r and thickness dr as element of volume dr.
W = f2 r9 JR2 J* V / W 2
Q2
47tr2 dr.
(4-22)
(4-23)
(4-24)
T h i s energy is jQV, orjCV2, where Ci s the capaci tance of the sphere , as in Sec. 4.1.1.
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97
a
Figure 4-5 The local e lect r ic charge dens i ty a t the surface of a conductor gives r ise
to opposi te ly di rected e lect r ic f ie ld intens i t ies <r/2e 0 as shown by the two a r rows on
the lef t ; the other charges on the conductor give r ise to the f ie ld o/2e0 shown by
the arrow on the r ight . The net resul t i s <r/e 0 out s ide , and ze ro ins ide . Th e v ec tor
n i s a un i t vec tor no rm al to the con duc tor sur face , and i t po in t s o u tw ard .
An e lement o f charge a da on the su r face o f a co nd uc to r exp er i ences the
elect r ic f ie ld of a l l the other charges in the system and is therefore subject
to an elect r ic force. In a s ta t ic f ie ld , th is force must be perpendicular to the
sur f ace o f the conduc to r , f o r o therwise the charges would move a long the
surface. Since the charge a da i s bo un d to the co nd uc to r by in t e rna l fo rces ,
the force act ing on a da is t r a n s m i t t e d t o t h e c o n d u c t o r itself.
T o ca lcu la t e the ma gn i tud e o f th i s fo rce , we cons ider a con du c to r
wi th a su r f ace charge dens i ty a and a f ie ld E at the sur face. From Gauss 's
law, the elect r ic f ie ld in tensi ty just outside the conductor i s <r /e 0. This field
i s pe rp end icu la r to the su r f ace. No w th e fo rce on the e l emen t o f cha rge a dais not Eo da, since the f ie ld that acts on a da is the f ield due only to the other
charges in the sys t em.
Let us f i r s t calculate the elect r ic f ie ld in tensi ty produced by a da itself.
We can do th i s by us ing Gauss ' s l aw. The to t a l f lux emerg ing f rom a da m u s t
b e a da/e0, half of i t inw ard an d hal f out w ar d, as in Fig . 4-5. Th en th e elect r ic
f ie ld in tensi ty due to the local su r f ace charge dens i ty must be f r /2e 0 i n w a r d
a n d o/2e0 o u t w a r d .
FORCES ON CONDUCTORS
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98 Fields of Stationary Electric Charges: III
The element of charge a da itself therefore produces exactly half the
total field at a point outside, arbitrarily close to the surface. This is reasonable,
for the nearb y char ge is mo re effective th an th e rest.
Now if a da produces half the field, then all the other charges must
produce the other half, and the electric field intensity acting on a da must be
ff/2e0.
The force dF on the element of area da of the conductor is therefore
given by the product of its charge a da multiplied by the field of all the other
charges:
dF = —o da, (4-25)2e
0
and the force per unit area is
dF a2
1— = — = - e 0E
2
. (4-26)da 2e 0 2
No te tha t this force is just equa l to the ener gy den sity of Sec. 4.4 at
the surface of the conductor. We can show that this is correct by using the
method of virtual work. In this method we assume a small change in the
system and apply the princip le of cons erv ati on of energy. Let us imaginethat the conducting bodies in the field are disconnected from their power
supplies. Then, if the electric forces are allowed t o perform mech ani cal wor k,
they must do so at the expense of the electric energy stored in the field.
Imagine that a small area a of a conductor is allowed to be pulled into the
field by a small distance x. The mechanical work performed is ax times the
the force per unit area. It is also equal to the energy lost by the field, which
is ax times the energy density. Thus the force per unit area is equal to the
energy density.
If we use the same type of argument for a gas, we find that its energy
density is equal to its pressure.Note that the electric force on a conductor always tends to pull the
conductor into the field. In other words, an electric field exerts a negative
pressure on a conductor.
In order to visualize electric forces, it is useful to imagine that electric
lines of force are real and under tension—also, that they arrange themselves
in space as if they repelled each other.
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99
F o r e x a m p l e , t h e l i n e s o f f o r c e b e t w e e n t h e p l a t e s o f a c a p a c i t o r t e n d
t o p u l l t h e p l a t e s t o g e t h e r a n d b u l g e o u t a t t h e e d g e s o f t h e p l a t e s , a s i n
F i g . 4 - 6 .
4.5.1 EXAMPLE: ELECTRIC FORCE FOR E = 3 x 10 6 VOLTS PER METER
Th e m ax im um elect r ic f ie ld intens i ty th at can be sus ta ined in a i r a t no rm al tem
pera ture and pres sure i s abo ut 3 x 1 0 6 vol ts per meter . The force is then 40 new ton s
pe r squa re me te r .
EXAMPLE: PARALLEL-PLATE CAPACITOR
Let us calcula te th e forces on the p la tes of a parallel-plate capacitor of area S ca r ry inga charge dens i ty + n on one p la t e and — <r on the other , as in Fig. 4-7. We assume
tha t the dis ta nce betw een the pla tes is smal l com pa red to thei r l inear exten t , in orde r
to make edge effects negl igible .
Th e force per uni t are a on th e pla tes is equ al to the energy d ens i ty in the f ield
e 0 £ 2 / 2 , and th e force of a t t r act ion betw een the pla tes is therefore e 0E2S/2, o r
(< x 2/2 e 0)S .
We shal l calcula te this force in two other ways in Probs . 4-13 and 4-14.
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10 0
Figure 4-7 A cha rged pa ra l l e l -p l a t e
capaci tor wi th i t s pla tes insula ted.
The p la t e s a re he ld in equ i l ib r ium by
mechanical forces F m that act on each
plate in a di rect ion tending to increase
the sepa rat io n s , an d by e lect r ic forces
Fe t end ing to dec rease the s epa ra t ion .
s
SUMMARY
If Q is t he ne t charge ca r r i ed by an i so la t ed c ond uc t o r , an d if V i s i t s potent ia l ,
t he r a t io Q/V is called i ts capacitance.
T h e capacitance between two isolated conductors is Q/V, w h e r e Q is
the charge t r ansfe r r ed f rom one to the o ther , and V i s t he r esu l t ing po ten t i a l
difference.
T h e c a p a c i t a n c e o f t w o o r m o r e c a p a c i t o r s c o n n e c t e d i n parallel is
the sum of the capac i t ances .
The inver se o f a capac i t ance i s an elastance.
The e l as t ance o f two o r more capac i to r s connec ted in series is the
sum of the e l as t ances .
T h e potential energy assoc ia t ed wi th a cha rge d i s t r ibu t io n can be
wr i t t en e i the r as
In the f i r s t in tegral , T ' mu s t be chosen to inc lu de a ll t he cha rge d i s t r ib u t ion ,
and in the second , m us t in c lude a l l r eg ions o f space where E i s n o n v a n i s h i n g .
The ass ignment o f an energy density e 0Ez/2 to every point in space leads
to the co r r ec t po ten t i a l energy fo r the whole charge d i s t r ibu t ion .
F ina l ly , t he force per unit area on a cha rged co nd uc to r i s equa l to
c r 2 / 2 e 0 , or to the energy density in the electr ic f ield, e 0 £ 2 / 2 .
(4-18)
or as
(4-21)
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10 1
Figure 4-8 Para l l e l -p l a t e capac i to r
with seven pla tes . See Prob. 4-3.
PROBLEMS
4-1E Show tha t e 0 i s expresse d in farad s per m eter .
4-2E THE EARTH'S ELECTRIC FIELDa) The earth has a radius of 6.4 x 1 0 6 mete rs . Wha t i s i t s capac i t ance?
b) The earth carr ies a negat ive charge that gives a f ie ld of about 100 vol ts per
meter a t the surface.
Ca lcu la t e the to t a l cha rge .
c) Calculate the potent ia l a t the surface of the earth.
4-3E PARALLEL-PLATE CAPACITOR
Show th at the cap aci t anc e of a paral le l -p la te cap aci t or hav ing TV plates , as in
Fig. 4-8, is
C = 8.85(N - l)A/t picofa rads ,
w h e r e N i s t he number of p l a t e s , A i s the area of one s ide of one p la te , an d t is the
d i s t ance be tween the p l a t e s . One p icofa rad i s 10" 1 2 farad.
4-4E PARALLEL-PLATE CAPACITOR
Can y ou sugges t reaso nable d im ens ion s for a 1 p i cofa rad ( 1 0 "1 2
farad) air-
insul ated p aral le l -p la te capa ci to r tha t is to ope rat e a t a pote nt ia l di fference of 500 vol ts?
4-5 PARALLEL-PLATE CAPACITOR
A para l l e l -p l a t e capac i to r has p l a t e s o f a rea S sepa ra t ed by a d i s t ance s .
Ho w is the cap aci ta nce -affected by the int r od uc t ion of an ins ulated sheet of
metal of thickness s ' , paral le l to the pla tes?
4-6 CYLINDRICAL CAPACITOR
Calcu la t e the capac i t ance pe r un i t l eng th C between two coaxial cyl inders of radi i
R\ a n d R 2 as in Fig. 4-9.
C a l c u l a t e C w h e n R 2 = 2R l.
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102
Figure 4-9 Cylindrical capacitor. See
Prob. 4-6.
Solution: Let us assume that the inner conductor is positive and that it carries
a charge of k coulombs per meter of length. Then the capacitance per unit length
is k divided by the voltage V of the inner conductor with respect to the outer one,
where
V = [Rl
Ech= r ^ d r . (1)
k R
2ne0 P ,(2)
Thus
For R2 = 2Rlt
_ _2ne^
l n ( P 2 / R 1 )
C = 27T€ 0/l n 2 = 8.026 x 10"1 1
farad/meter. (4)
DROPLET GENERATOR
There are a number of devices that utilize charged droplets of liquid. Electrostatic
spray guns (Prob. 2-8), colloid thrusters for spacecraft (Prob. 2-15). and ink-jet
printers (Prob. 4-17) are examples. As we shall see, drop size is controlled by the
specific charge (coulombs per kilogram) carried by the fluid.
a) When a charged droplet splits into two parts, electrostatic repulsion drives
the droplets apart and the electrostatic energy decreases.
Calculate the loss of electrostatic energy when a droplet of radius R carrying a
charge Q splits into two equal-sized droplets of charge Q/2 and radius R'.
Assume that the droplets are repelled to a distance that is large compared to
R'. Neglect evaporation.
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Problems 103
Solution: Since
then
4 42 x -nR'
3
= ~ T I R \ (1)
R' = R/2 113. (2)
Initially, the charge Q is at the potential of the surface of the drop of radius R
and the electrostatic energy is
QV = 1 Q2
2 2 4ne0
After splitting, the electrostatic energy is
2( Q / 2 )
2
2 "3
8rce 0«
and the electrostatic energy gained is
(3)
(4)
Q 1 \ Q2
AW = —^— KTT - 1 = -0 .3 70 0 (5)
b) The energy associated with surface tension is equal to a constant that is a
characteristic of the liquid, multiplied by the surface area. The constant is called
the surface tension of the liquid. The surface tension of water is 7.275 x 10~2
joule
per square meter.
When a droplet splits, the surface area increases, and the energy associated with
the surface tension increases proportionately.
Calculate the gain in this energy when a droplet of water of radius R splits into
two equal-sized droplets.
Solution: The increase in surface energy is
AW = (2 x 4nR'2
- 4nR2
)T = - \ ]4nR2
T, (6)V 2
2
'3
= 0.2599 x 4 ; i R2
r = 0.2376K2
, (7)
where T is the surface tensi on of water.
c) A dro ple t of wate r has a radiu s of one micro mete r and carri es a specific charge
of one coulomb per kilogram.
Will the droplet split?
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104 Fields of Stationary Electric Cha rges: III
Solution: Th e dro ple t will split if the tot al AM 7 i s nega t ive . The cha rge Q is nu
merical ly equal to the mass , in this case:
4 ,
Q = 1000 x - nR coulombs . (8 )
T h u s
[(4000rc/3)J?3
]2
AW = - 0 . 3 7 0 0 - — — — + 0 .2 37 6K2
, (9)Sn e 0R
= - 2 . 9 1 9 x 1016
RS + 0 .2376K 2 . (10)
F o r R = 1 0 " 6 ,
AW = - 2 . 9 1 9 x 1 0 "1 4
+ 0.2376 x 10"1 2
joule . (11)
The droplet wi l l not spl i t .
d) W ha t is the radiu s of the larges t dro plet of wate r that i s s table wi th this specif ic
c h a r g e ?
Solution: For a spec i f i c cha rge of one coulomb pe r k i logram, a d rop le t o f wa te r
will not spl it spo nta neo us ly if i ts radius is such that
- 2 . 9 1 9 x 1 0 1 6 R 5 + 0 .2376K 2 > 0.
or for radi i smal ler than
0.2376
2.919 x 10 1 6
1 /3
= 2 .012 x 10" 6 meter . (12)
4-8E ELECTROSTATIC ENERGY
a) Two capaci tors are connected in ser ies as in Fig. 4-10a. Calculate the ra t io
of the s tored energies .
b) Calculate this ra t io for capaci tors connected in paral le l as in Fig. 4-10b.
4-9E ELECTROSTATIC ENERGY
Tw o capa c i to rs o f capac i t an ces C, and C 2 h a v e c h a r g e s Q x a n d Q 2, respect ively.
a) Ca lcula te the am ou nt of energy diss ipated wh en they are con nec ted in paral le l .
b ) Ho w i s t h i s ene rgy d i s s ipa ted ?
4-10 PROTON BOMB
a) Calculate the e lect r ic potent ia l energy of a sphere of radius R ca r ry ing a to t a l
c h a r g e Q uni formly d i s t r ibu ted th r oug hou t it s vo lum e .
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10 5
+ --O V o -
c ,
Figure 4-10 T w o c a p a c i t o r s : (a ) connected in ser ies , and (b ) connec ted in pa ra l l e l .
See P rob . 4 - 8 .
b) N ow ca lcu la t e the grav i t a t iona l po ten t i a l ene rgy of a sphe re of rad ius R' of
to t a l mass M.
c) The moon has a mass of 7.33 x 1 0 2 2 ki lograms and a radius of 1.74 x 10°
mete rs . Ca lcu la t e i ts g rav i t a t io na l po ten t i a l ene rgy .
d) Imagine tha t you can a s semble a sphe re of p ro tons wi th a dens i ty equa l t o
tha t of wate r . Wh at wo uld b e the rad ius of this sphe re if i t s e lect r ic pote nt ia l energy
were equa l t o the grav i t a t iona l po ten t i a l ene rgy of the moon?
4-11 ELECTROSTATIC MOTOR
Can you sugges t a rough des ign for an e l ec t ros t a t i c motor?
Draw a sketch, explain i t s operat ion, specify vol tages and currents , and make a
roug h es t ima te o f wha t i t s pow er would be .
4-12 ELECTROSTATIC PRESSURE
A l ight spherica l bal lo on is m ad e of con du ct in g mate ria l . I t i s sugges ted th at i t
cou ld be kep t sphe r i ca l s imply by connec t ing i t t o a h igh-vol t age source . The ba l loon
has a d i am ete r o f 100 mi l l ime te rs , and the max imu m brea kdo wn vol t age in a i r is 3 x 1 0 6
vol ts per meter .
a) W ha t m ust be the vo l tage of the sou rce i f the e lect r ic force is to be as large
as pos s ib l e?
b) Wha t gas p res sure ins ide the ba l loon would produce the s ame e f fec t ?
4-13 PARALLEL-PLATE CAPACITOR
Calcu la t e the force of a t t rac t ion F betw een the pla tes of a paral le l -pla te cap aci to r .
Assum e tha t t he capac i to r i s conne c ted to a ba t t e ry supply ing a cons tan t vo l t age V.
Use the method of vi r tual work, assuming a smal l increase ds in the spacing s
between the pla tes , and set
dWB
+ dWm
= dWe,
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106 Fields of Stationary Electric Char ges: III
w h e r e dW B i s t he work done by t he ba t t e ry , dW m = Fds i s t he mech anica l wo rk don e
on the sys tem, and dWe is the increase in the energy stored in the electric field.
You should f ind that
I K 2 1 ,F = - e„ — S =- e0E
2
S.2 . r 2
No te tha t on e ha lf of t he ene rgy suppl i ed by the ba t t e ry app ea rs a s me chanica l
work, and one half as an increase in the energy s tored in the e lect r ic f ie ld. This is a
general rule .
4-14 PARALLEL-PLATE CAPACITOR
Ca lcul ate the force of a t t ra ct io n betw een the pla tes of a paral le l -pla te capa ci tor ,
a s in the preced ing prob lem, a s sumin g now tha t t he capa c i to r i s cha rg ed and d i s
connec ted f rom the ba t t e ry .In this case ,
dW m = dW e.
You should f ind the same resul t .
4-15 OSCILLATING PARALLEL-PLATE CAPACITOR
Figure 4-11 shows a pa ra l l e l -p l a t e capac i to r whose upper p l a t e o f mass m is
suppor t ed by a spr ing insu la t ed f rom ground . A vo l t age V i s app l i ed to the upp er
plate . Ther e is zero ten s ion in the sprin g whe n the leng th of the a i r gap is x 0 .
Set m = 0.1 k i log ram , x 0 = 0.01 meter , k = 150 newtons pe r me te r , V =
100 vol ts , and A = 0.01 square meter . Assume that the mass of the spring is negl igible ,
and neglect edge effects.
a) Set the var ious po tent ia l energies of the com ple te sys tem equal to zero w hen
the tens ion in the spring is zero, a t x = x 0 . Show that the potent ia l energy is then
W = mg(x - x 0) + I k(x - x 0 )2 - \ e 0A V
2
(- - —2 2 \ x x 0
Draw a curve of W as a function of x b e t w e e n x = 1 0 ~ 6 a n d x = 1 0 ~ 2 mete r .
Use a logari thmic scale for the x-axis .
b ) Sh ow tha t , a t equ i l ib r ium ,
mg + k(x - x0) + i~e
0AV 2/x 1 = 0.
You can arr ive a t this resul t in two different ways .
The re i s s t ab le equ i l ib r ium a t x eq = 3.46 x 1 0 " 3 mete r and uns tab le equ i l ib r ium
at x = 2.93 x 1 0 " 5 m e t e r .
c) If the sys tem is br ou gh t to the poin t of s table equi l ib r ium , and then dis tu rbed
s l ight ly, i t osci l la tes . For smal l osci l la t ions , the res toring force F i s p ro por t ion a l t o the
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10 7
Figure 4-11 Para l l e l -p l a t e capac i to r , w i th the
lower pla te f ixed in pos i t ion and the upper
p la t e suspended by a spr ing insu la t ed f rom
g r o u n d . S e e P r o b . 4 - 1 5 .
d i s p l a ce m e n t a n d w e h a v e a s i m p l e h a r m o n i c m o t i o n :
(d 2x\ (dW\
where the de r iva t ives mus t be eva lua ted in the ne ighborhood of .x = x eq.
The ci rcular frequency is o j = (K/m)111, w h e r e
S h o w t h a t
'k - (e 0AV 2/x 3
eq)OJ = —
m
This gives a frequency of 6.16 hertz .
4-16 HIGH-VOLTAGE GENERATOR
Imagine the fo l lowing s imple -minded h igh-vol t age gene ra tor . A pa ra l l e l -p l a t e
capac i to r has one f ixed p la t e tha t i s pe rm anen t ly connec ted to g round , and one p la t e
that i s movable . When the pla tes are c lose together a t the dis tance s , the capaci tor i scha rged by a ba t t e ry to a vo l t age V. The n the mova ble p l a t e is d i s connec ted f rom
the ba t t e ry and moved ou t t o a d i s t ance ns . The vol tage on this pla te then increases
to nV , if we neglect edg e effects. O nc e the vo l tage h as been ra ised to nV , the pla te is
d i s cha rged th rough a load re s i s t ance .
a) Veri fy tha t there is con serv at io n of energ y.
b) C an yo u sugges t a rou gh des ign for such a high -vo l tage ge ner ator wi th a
m o r e c o n v e n i e n t g e o m e t r y ?
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10 8
Q K
mm N
Figure 4-12 Ink-je t pr inter . A nozzle N forms a f ine je t that breaks up into droplets
ins ide e lect ro de £*. A va riab le vol ta ge V appl i ed to the e l ec t rode induces a known
charge on each drop le t i n succes s ion . The drop le t s a re de f l ec t ed in the cons tan t
elect r ic f ie ld between the deflect ing pla tes and impinge on the paper form P. U n
charged drop le t s a re co l l ec t ed a t C. See P rob . 4 -17 .
4-17 INK-JET PRINTER
Th e ink-je t pr inte r i s one solut io n to the pro ble m of prin t ing da ta a t high sp eed.I t operates more or less l ike an osci l loscope, except that i t ut i l izes microscopic droplets
of ink, ins tead of e lect rons . See Prob. 4-7.
I ts pr inciple of operat ion is shown in Fig. 4-12. A nozzle produces a f ine je t of
conduc t ing ink tha t s epa ra t e s in to drop le t s i ns ide a cy l indr i ca l e l ec t rode . The po ten t i a l
on the e lect r ode is e i ther p os i t ive or zero . At a given ins tant , the charge car r ied b y a
drop le t t ha t b re aks off f rom the j e t depend s on the cha rge induc ed on the j e t an d
hence on V.
The droplets are then deflected as in an osci l loscope, except that here the de
f lecting f ie ld is con s tan t , each drop let being deflected acco rdin g to i ts charg e. U n
cha rged drop le t s a re co l l ec t ed in an ink sump. The j e t ope ra t e s con t inuous ly , an d
only a smal l f ract ion of the dro ple ts is actua l ly u sed.
Th e deflect ion is vert ical and the je t move s ho rizo nta l ly.Th e pres sure in the nozz le is mo du la t ed a t f requenc ies up to 0 .5 m egah er t z ,
p ro duc ing up to 5 x 10 ' d rop le t s pe r s econd , w i th rad i i of t he orde r o f 10 mic ro mete r s .
a) L et the rad ius of the je t be R, and th e ins ide radiu s of e lect ro de E b e R2-
W ha t is the charg e per uni t leng th on the je t , negle ct ing end effects?
b) The j e t b reaks u p in to l a rge r d rop le t s o f rad ius 2R X. W ha t length of je t i s
requi red to fo rm one drop le t ?
c ) Ca lcu la t e the cha rge Q per d rop le t .
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Problems 109
d) Ca lcu la t e the spec if ic cha rg e in cou lom bs pe r k i logram for = 20 mic rom
eters , R 2 = 5 mi l l ime te rs , V = 100 vol ts . Set the ink dens i ty equal to that of water
(1000 k i lograms pe r cub ic me te r ) .
e) Ca lcul ate the veloci ty of the drop lets i f they are pro du ced a t the ra te of
1 0 ' per s econd and if t hey a re s epa ra t ed by a d i s t ance of 100 mic ro mete rs .
f) Calcu late the vert ical deflect ion a nd the vert ical veloci ty of a dro plet up on
leaving the deflect ing pla tes , assuming a uniform t ransverse f ie ld of 10 5 vol ts per meter
ex tending ove r a hor i zonta l d i s t ance of 40 mi l l ime te rs .
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CHAPTER 5
DIRECT CURRENTS IN ELECTRIC CIRCUITS
5.1 COND UCTION OF ELECTRICITY
5.1.1 Conservation of Charge
5.1.2 Charge Density in a Hom ogeneous Conductor
5.2 OHM 'S LAW5.2.1 The Joule Effect
5.3 NON-LINEAR RESISTORS
5.3.1 Examp le: Incandescent Lamp
5.3.2 Examp le: Voltage-Dependent Resistors
5.4 RESISTORS CONNE CTED IN SERIES
5.5 RESISTORS CONNE CTED IN PARALLEL
5.6 THE PRINCIPLE OF SUPERPO SITION
5.6.1 Examp le: Simple Circuit with Two Sources
5.7 KIRCHOFF 'S LAWS
5.7.1 Examp le: The Branch Currents in a Two-M esh Circuit5.8 SUBSTITUTION THEORE M
5.8.1 Examp le: Resistor Replaced by a Battery
5.8.2 Examp le: Battery Replaced by a Resistor
5.9 MESH CURRENTS
5.9.1 Examp le: The Mesh Currents in a Two-M esh Circuit
5.10 DELTA-STAR TRANSFOR MATIONS
5.10.1 Examp le: Transformation of a Symm etrical Delta and of a
Symmetrical Star
5.11 VOLTAGE AND CURRENT SOURCES
5.12 THEVEN IN'S THEORE M5.12.1 Examp le: Thevenin's Theorem Applied to a Simple Circuit
5.12.2 Examp le and Review: The Wheatstone Bridge
5.13 SOLVING A SIMPLE DIFFEREN TIAL EQUATION
5.14 TRANSIENTS IN RC CIRCUITS
5.14.1 Examp le: Simple RC Circuit
5.15 SUMMARY
PROBLEMS
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Unti l now we have l imi ted ourselves to the study of the f ie lds of s ta t ionary
charges. We shal l now consider the f low of e lect r ic current , and we shal l
d e v e l o p m e t h o d s f o r c a l c u l a t i n g t h e c u r r e n t s i n c o m p l e x a r r a n g e m e n t s o fc o n d u c t o r s .
We sha l l no t be concerned as ye t wi th magne t i c f i e lds .
CONDUCTION OF ELECTRICITY
We have seen in Sec. 3 .3 that , under s ta t ic condi t ions, E is zero inside a
con duc to r . H ow ever , if an e l ec t r i c fi eld i s ma in ta i ned by an ex te rna l sou rce —
for exam ple , when on e conn ec t s a l eng th o f cop per wi r e be tween the t e r
minals of a bat tery—then charge car r iers dr i f t in the f ie ld and there is an
e lec t r i c cu r r en t .
W i t h i n t h e c o n d u c t o r , t h e current density vec to r J po in t s in the
dire ct io n of f low of pos i t ive charg e car r i ers ; for neg at ive car r iers , J po ints
in the d i r ec t io n o ppo s i t e to th e f low, as in F ig . 5 -1 . The ma gn i tud e o f J i s
the qua n t i ty o f char ge f lowing th rou gh an in f in i t es imal su r f ace perp end icu la r
to the di rect ion of f low, per uni t area and per uni t t ime. Current densi ty i s
expressed in cou lombs per square mete r pe r second , o r in amperes per
square mete r . I n good conductors, such as copper , t he charge ca r r i e r s a r e
the condu c t io n e l ec t rons , an d we hav e the s i tua t ion sho wn in F ig . 5 - lb .
Semiconductors con ta in e i th e r o r bo th o f two types o f mo bi l e c harges ,
n a m e l y c o n d u c t i o n e l e c t r o n s a n d h o l e s . A hole is a vaca nc y left by the r eleas e
of an ele ct ron by an at om . A ho le can mi gra te in the ma ter i al as if i t were a
pos i t ive par t i c l e .
F o r o r d i n a r y c o n d u c t o r s , t h e c u r r e n t d e n s i t y J i s p r o p o r t i o n a l t o
the elect r ic f ie ld in tensi ty E:
J = crE, ( 5 - 1 )
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11 2
^ 0
0 — ^ 0 ^ - 0
^ 0
da
^ 0
0 -
da
^ 0
Figure 5-1 Current dens i ty vector J for (a ) pos i t ive and (b ) nega t ive cha rge ca r r i e r s .
Table 5- 1
Conductor
Conductivity a
(siemens per meter)
Alu min um 3 .54 x 10 7
Br ass (65.8 Cu , 34.2 Zn ) 1.59 x 107
C h r o m i u m 3 .8 x 1 0 7
Co pp er 5 .80 x 10 7
Go ld 4 .50 x 10 7
G rap hi te 1.0 x 10 5
Ma gn et ic i ron 1.0 x 107
M um eta l (75 Ni , 2 Cr, 5 Cu, 18 Fe) 0.16 x 10 7
Nic kel 1.3 x 10
7
Sea wa ter ss 5
Silver 6.15 x 10 7
T in 0.870 x 1 07
Zi nc 1.86 x 10 7
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5.1 Conduction of Electricity 113
w h e r e a is the conductivity of the m a t e r i a l . C o n d u c t i v i t y is e x p r e s s e d in
a m p e r e s pe r vol t per m e t e r , or in Siemens11
per m e t e r . T a b l e 5-1 gives the
c o n d u c t i v i t i e s of a n u m b e r of c o m m o n m a t e r i a l s .
The dr i f t veloci ty v of c o n d u c t i o n e l e c t r o n s is easi ly calculated. It
i s su rp r i s ing ly low. If n is the n u m b e r of c o n d u c t i o n e l e c t r o n s per c u b i c
m e t e r and if e is the m a g n i t u d e of the e lec t ron charge , t hen
J = nev. (5-2)
In copper , t he re is one c o n d u c t i o n e l e c t r o n per a t o m , and 8.5 x 1 02 8 a t o m s
per cub ic mete r , so n = 8.5 x 1 02 8 e l e c t r o n s pe r c u b i c m e t e r . For a cur
r e n t of one a m p e r e in a wire hav ing a cross- sec t ion of one s q u a r e m i l l i m e t e r ,
J= 10
6
a m p e r e spe r
s q u a r e m e t e rand
1 06
x 7.4 x 105 m e t e r / s e c o n d , (5-3)
8.5 x 1 02 8 x 1.6 x 10
o r a b o u t 0.26 m e t e r per hour}
5.1.1 CONSERVATION OF CHARGE
N o w c o n s i d e r a v o l u m e T b o u n d e d by a surface S i n s i d e c o n d u c t i n g ma
ter ia l . The i n t e g r a l of J • da o v e r the surface is the charge f lowing out of Sin one s e c o n d , or the charge los t by the v o l u m e % in one s e c o n d . T h e n
l3
-Aa=
-Jtlpch
-(5
"4)
T h i s is the law of conservation of charge.
U s i n g the d i v e r g e n c e t h e o r e m (Sec. 1.10) on the lef t -hand side,
J V j * (5-5)
f Afte r Erns t Werne r von Siemens (1816-1892) . The word the re fore t akes a t e rmina l s
in th e s i n g u l a r : on e Siemens . The Siemens wa s formerly cal led a " m h o . "
{ T h e n how is it t h a t one can, say, t u rn on a l i gh t k i lomete rs away , ins t an taneous ly?
T h e m o m e n t on e closes th e swi t ch , a guided e l ec t romagne t i c wave is l a u n c h e d a l o n g
the wire . This wave t ravels at the veloci ty of l ight an d es t ab l i shes th e electric field in
a n d a r o u n d th e wire .
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114 Direct Curren ts in Electric Circuits
Since this eq ua t io n is val id for any vo lum e,
V • J = —2-. (5-6)
dt
Th is is aga in the law of co ns erv at io n of cha rge , s ta ted in di f ferentia l form.
5.1.2 CHARGE DENSITY IN A HOMOGENEOU S CONDUCTOR
Under s t eady- s t a t e cond i t ions , t he t ime der iva t ive in Eq . 5 -6 i s ze ro . Then
V • J = V • <rE = 0. (5-7)
If n o w w e h a v e a h o m o g e n e o u s c o n d u c t o r w i t h a i nde pen den t o f the co
ord ina tes , V • E is zero . Th us , f rom Eq. 3-11, the net ch arg e den si ty inside a
homogeneous c o n d u c t o r c a r r y i n g a c u r r e n t u n d e r s t e a d y - s t a t e c o n d i t i o n s i s
z e r o .
5.2 OHM 'S LA W
Figure 5 -2 show s an e l em ent o f co nd uc to r o f l eng th / and c ross - sec t ion A.
If a difference of potential V i s main ta ined be tween i t s two ends , t hen
J = I/A = oE = aV/l, (5-8)
a n d
aVA V Vl
—r~m = R> ,5" 9)
w h e r e
R = 1/oA (5-10)
is the resistance of the e l ement . Res i s t anc es a r e express ed in vo l t s pe r am per e
or in ohms.
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115
Figure 5-2 Element o f con du c tor o f
c ros s - s ec t ion A an d length / carry ing a
cur ren t / .
Figure 5 -3 The cur ren t / t h rough a
l inea r re s i s to r i s p ro por t iona l t o th e
V vol t age V across i t .
Equat ion 5 -9 shows tha t t he cu r r en t th rough a r es i s to r i s p ropor
t iona l to th e vol ta ge ac ros s i t , as in F ig . 5-3. This i s Ohms law. A res i s to r
satisfying this law is said to be ohmic, o r linear. Equat ion 5 -1 i s ano ther ,
mo re genera l fo rm of O hm 's l aw.
The cur r en t and the vo l t age a r e measured as in F ig . 5 -4 . I t i s a ssumed
tha t the cu r r en t f lowing th rough the vo l tmete r V i s neg l ig ib le co mp are d
t o t h a t t h r o u g h R. Th at is usua l ly the case .
V W V \ AR
Figure 5-4 Me asur em ent o f t he cur ren t / t h ro ug h a re s i s to r R. and of the vol tage V
across i t . A circle marked 1 represents an ammeter, and a circle marked V, a
voltmeter.
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/16 Direct Currents in Electric Circuits
5.2.1 THE JOULE EFFECT
W h e n a cur r en t f lows th rough a c o n d u c t o r , the charge ca r r i e r s ga in k ine t i c
energy by m o v i n g in the elect r ic f ie ld . Their veloci ty never becomes large,
however , because they can t ravel only a shor t d i s t ance befo re co l l id ing wi th
a n a t o m or a m o l e c u l e . The k ine t i c energy ga ined by the car r i e r s thus se rves
to raise the t e m p e r a t u r e of the c o n d u c t i n g m e d i u m . T h i s p h e n o m e n o n is
called the Joule effect.
T h e e n e r g y t h a t is d i ss ipa ted as h e a t in t h i s way is eas i ly ca lcu la t ed .
C o n s i d e r a c u b i c m e t e r of c o n d u c t o r w i t h E para l l e l to one e d g e . The c u r r e n t
dens i ty J is the charge f lowing through the c u b e in one s e c o n d . The v o l t a g e
dif ference across the c u b e is E. T h e n the k ine t i c energy ga ined by the car r i e r s
and los t to the c o n d u c t o r as h e a t in one s e c o n d is JE, or <T EZ
, or J2
/o, fromEq . 5-1. So the h e a t e n e r g y p r o d u c e d per c u b i c m e t e r in one s e c o n d is oE
2
o r J2
/a.
If one has a r es i s t ance R c a r r y i n g a c u r r e n t L t h e n the vo l t age ac ross
it is IR and the k ine t i c energy ga ined by the c h a r g e c a r r i e r s in one s e c o n d
is I(IR). T h e n the p o w e r d i s s i p a t e d as h e a t is I2
R.
5.3 NON-LINEAR RESISTORS
A l t h o u g h O h m ' s law app l i es to ord inary r es i s to r s , it is by no m e a n s g e n e r a l .
EXAMPLE: INCANDESCENT LAMPS
F i g u r e 5-5 s h o w s th e c u r r e n t as a funct ion of vol t age for a 60-wa t t incandescent
lamp. The re s i s t ance V/I is a b o u t 20 o h m s at a few vol t s , an d a b o u t 250 o h m s at
100 volts. The reason is t ha t , in the i n t e rva l , the t e m p e r a t u r e of the tungsten filament
has changed from 300 to 2400 kelvins .
If th e t e m p e r a t u r e of the f i l ament were ma in ta ined cons tan t by exte rna l means ,
sa y by i m m e r s i n g it in a c o n s t a n t - t e m p e r a t u r e oil ba th , t hen it would fo l low Ohm's
law.
EXAMPLE: VOLTAGE-DEPENDENT RESISTORS
Voltage-dependent resistors ut i l i ze ce ramic s emiconduc tors for which V = CIB
,
w h e r e C and B ar e c o n s t a n t s , and B is l e s s than un i ty . Thi s equa t ion appl i e s at
r o o m t e m p e r a t u r e . F i g u r e 5-6 s h o w s th e cur ren t -vo l t age curve for one p a r t i c u l a r
type wi th C = 100 an d B = 0.2.
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11 7
I I I I I I I I I i I0 50 100
V (volts)
Figure 5-5 Cur ren t -vo l t age re l a t ionsh ip for a 60-wa t t i ncandescen t l amp.
0 25 50 75 100
V (volts)
Figure 5-6 Cur ren t -vo l t age re l a t ionsh ip for a ce r t a in type of vo l t age -dependent
res is tor .
Vol t age -dependent re s i s to rs a re used on t e l ephone l ines fo r shor t -c i rcu i t ing
l ightning to ground, for suppress ing sparks , and so on, s ince thei r res is tance de
creases a t large vol tages .
We shal l be concerned henceforth only wi th l inear res is tors .
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11 8
0
Figure 5-7 Three res is tors connected in ser ies .
5.4 RESISTORS CONNECTED IN SERIES
Figure 5 -7 shows th ree r es i s to r s connec ted in series. Since
V = IRi + IR 2 + IRi = HR, + R 2 + R 3) = IR S. (5-11)
the r es i s to r s have a to t a l r es i s t ance
R s = R l + R 2 + R 3. (5-12)
Th e resista nce of a set of resis tors con nec ted in ser ies is the su m of th e in
d iv idua l r es i s t ances .
5.5 RESISTORS CONNECTED IN PARALLEL
In Fig . 5-8 the resistors are connected in parrallel. T h e n
_ V V V _ ( 1 1 1 \
7 = R x
+ R~ 2
+ ~R 3 ~ V + R~ 2
+ R~J
and the r es i s t ance R p is given by
1 1 1 1
Rp Ri R 2 R 3'
R _R
lR
2R
3
- r;(5
-131
(5-14)
(5-15)
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11 9
O- <Z>
- 0 -V \ A A A
W v A A
R 3
Figure 5-8 Three res is tors connected in paral le l .
I t i s useful to rem em be r that , for tw o resistance s in pa ral le l ,
R,R2
" R { + R 2
I t i s of ten co nve nien t to use conductance
(5-16)
G = \/R, (5-17)
ins t ead o f r es i s t ance . Th en
Gi + G 2 + G 3. (5-18)
Th e con du c ta nce o f a set of r es i s to r s con nec te d in para l l e l is t he sum of the
i n d i v i d u a l c o n d u c t a n c e s .
Conduc tances a r e expressed in amperes per vo l t o r in S i e me ns . W e
have a l r eady used the Siemens in Sec. 5.1.
THE PRINCIPLE O F SUPERPOSITION
Th e p r inc ip le o f superp os i t io n which w e s tud ied in Sec . 2 .3 app l i es to e l ec t r ic
c i r cu i t s compr i s ing sources and linear res isto rs: if a c i rcui t com pris es tw o
or more sources , each source ac t s independen t ly o f the o ther s and , a t any
po in t in the c i r cu i t, t he cu r r e n t i s t he a lgebra ic sum of the cu r r en t s p ro duc ed
by the ind iv idua l sources .
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120
Figure 5-9 The current flowing through
th e circuit is the algebraic sum of the
c u r r e n t due to the 6-volt battery, plus
t h a t due to the 12-volt battery.
2C1
NA A A / ^
ion
12V
EXAMPLE: SIMPLE CIRCUIT WITH TWO SOURCES
F i g u re 5-9 shows a simple circuit comprising three resistors and two sources, all
c o n n e c t e d in series. The tota l resis tance is 15 ohm s. The 12-volt sour ce prod uce s a
c o u n t e r c l o c k wi s e cur ren t of 12/15 = 0.8 ampe re, while the 6-volt source pro duc es a
c lockwise curre nt of 0.4 ampere. The net curr ent is 0.4 ampere, counterc lockwis e.
T he power dissipated as heat in the 2-ohm resistor is (0.4)2
x 2 = 0.32 watt,
and the total power dissipated in the circuit is (0.4)2
x 15 = 2.4 wat ts .
5.7 KIRCHOFF'S LAWS
Let us consider a general circuit composed of resistors and voltage sources,
for example, that of Fig. 5-10.
Junctions such as A, B, C, D, E are called nodes; connections between
nodes, such as AB, BD, DC .. ., are called branches; closed circuits, such
as ABDE, or ABCDE, or BCD, are called meshes.
Kirchoff's laws provide a method for calculating the branch currents
when the values of the resistances and of the voltages supplied by the sources
are known.
Kirchoff's current law states that, at any given node, the sum of the
incoming currents is equal to the sum of the outgoing currents. This is simply
because charge does not accumulate at the nodes. For example, at A,
h = h+ h- (5-19)
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12 1
Figure 5-10 Part of a complex circuit .
A c c o r d i n g t o Kirchoff's voltage law, the vol tage r ise , or the vol tage
d r o p , a ro un d any g iven mesh i s ze ro . For e xam ple , t he vo l t age r ise from A
t o B, to D, to £ , and bac k to A, is clearly zero, so
-V, + / , / ? , - I2R 2 + I5R 5 + I6R b = 0. (5-20)
To f ind the b ranch cur r en t s , one wr i t es down as many equa t ions o f
the above two types as the re a r e b ranches , and one so lves the r esu l t ing
s y s t e m of s i m u l t a n e o u s e q u a t i o n s .The d i r ec t ions chosen fo r the cu r r en t s a r e a rb i t r a ry . I f t he ca lcu la t ions
give a posi t ive value for , say, / , , then that current f lows in the di rect ion of
the ar ro w . I f the calc ula t ion s give a neg at ive value , then th e cur ren t f lows
in the oppos i t e d i r ec t ion .
EXAMPLE: THE BRANCH C URRENTS IN A TWO-MESH CIRCUIT
Fo r the c i rcu i t of F ig . 5 -11 , we have th ree un kno wn s , / , , I2. 13. Hence we need
three equa t ions . A t node A,
h = h + h- (5-21)
For the l e f t -hand mesh , s t a r t ing a t B, c lockwise .
V — lxR — I
3R = 0. (5-22)
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12 2
i — A A / V f M A / — it /
3
K T
Figure 5-11 One can ca l cu la t e Ilt
I2, / 3 , given .R an d K
Similar ly, for the r ight-hand mesh, s tar t ing a t B, c lockwise ,
Z 3i? - 7 2 P - I2R = 0.
Solving, we f ind that
I , = 3F /5R , / , = 1 5K. / 3 = 2 F / 5 R .
(5-23)
(5-24)
SUBSTITUTION THEOREM
Let us return to Fig . 5-10 and to Eq. 5-20. Consider the resistance R 2. T h e
vol tage drop across i t i s I2R 2. Let us assume tha t I2 i s real ly in the di rect ion
of the a r row. Then the upper end o f R 2 i s pos i t ive . Now le t us subs t i tu t e
fo r R 2 a vol tage sou rce as in Fig . 5-12. Eq ua t io n 5-20 is unaffected. Simila r ly ,
t h e c o r r e s p o n d i n g e q u a t i o n f o r t h e m e s h BCD i s unaf fec ted . The n the c u r r en t s
11, I2, ^ 3 , and so for th , are unaffected.
Thus , i f a cu r r en t I f lows th rough a r es i s t ance R, one can r ep lace the
res i s t ance by a vo l t age source IR of the sam e pol ar i ty wi tho ut af fect ingan y of the cur ren ts in the c i rcui t .
Inversely , if we ha ve a vol ta ge so urce V pass ing a cu r r en t / , with the
current entering the source at the positive terminal, one can r ep lace the source
by a r es i s t ance V/I without af fect ing any of the currents in the ci rcui t .
This i s the substitution theorem.
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5.8.1 EXAMPLE: RESISTOR REPLACED BY A BATTERY
In Fig. 5-13a,
h = h + h,
fR + I2R = V,
-I 2R + 1,R = 0,
(5-25)
(5-26)
(5-27)
R /, /,
+
1 AA/V^—*—< — 1
+ +
j V R * - K/3
<(6)
Figure 5-13 Th e curren ts / , , I2,1 3 are not affected by subs t i tut ing the bat tery F/3
for the res is tor R.
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124 Direct Curren ts in Electric Circuits
giving
7j = 2V/3R, I2
= V/3R, l3
= VJ3R. (5-28)
Us ing now the subs t i tu t ion theorem, we can rep lace the re s i s to r on the r igh t
by a source I3R = V/3, giving the c i rcui t of Fig. 5-13b. Then
h = h+ h, (5-29)
1,R + 1 2R = V, (5-30)
-I 2R + (V/3) = 0. (5-31)
Solving, we f ind the same values for J 1 , 1 2 , / 3 as before .
Inversely, one can subs t i tute the c i rcui t of Fig. 5-13a for that of Fig. 5-13b
without affect ing J j , I2, 7 3 .
EXAMPLE: BATTERY REPLACED BY A RESISTOR
In Sec. 5.6.1 we found that th e curre nt in Fig. 5-9 is cou nter cloc kw ise . Since a cur
rent of 0.4 am pe re f lows into the pos i t ive term inal of the 6-vol t sourc e, we can
rep lace th i s source by a re s i s tance of 6 /0.4 = 1 5 oh ms wi tho ut a f fec ting the cur re n t
f lowing in the c i rcui t . Let us check. We no w have a 1 2-vol t source an d a tota l res is
tanc e of 15 + 10 + 3 + 2 = 30 ohm s. Th en th e curre nt i s 12/30 = 0.4 amp ere ,
a s p rev ious ly .
5.9 MESH CURRENTS
I f we use Ki r cho f f ' s l aw s as such , the nu m be r o f un kn ow n c ur r en t s is equa l
to the number o f b r anches . I t i s poss ib le to r educe the number o f unknowns
in the fol lowing way . Ins tea d of usin g bran ch c urr en ts as in Fig . 5-10, on e
uses mesh currents as in Fig . 5-14, s ince there are fewer mesh currents than
bra nc h cur r en t s . Ki r ch of f ' s cu r r en t l aw i s then au to ma t i ca l ly sa ti s fi ed , andwe need to app ly on ly the vo l t age l aw.
Th e adva n ta ge o f us ing mesh cur r en t s com es from the f ac t t ha t t he
t ime r equ i r ed to so lve a se t o f s im ul t an eou s equa t ion s decreases r ap id ly
as the number o f equa t ions i s r educed .
In the end , one mu st find the b ra nc h cu r r en t s , bu t tha t is a s im ple
m a t t e r , o n c e t h e m e s h c u r r e n t s a r e k n o w n .
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12 5
Figure 5-14 Mesh currents in the c i rcui t of Fig. 5-10.
5.9.1 EXAMPLE: THE MESH CURRENTS IN A TWO-MESH CIRCUIT
In the circuit of Fig. 5-15 we have two meshes . S ta r t ing a t B i n bo th cases and pro
ceeding in the c lockwise di rect ion,
a n d
V — IaR — (/„ - I„)R = o,
(/„ - Ib)R - I
b2R = 0.
7 = 3 1 7 5 / ? , h=V/5R.
(5-32)
(5-33)
(5-34)
Note tha t we have on ly two unknowns , / „ and Ib. ins tead of the three we had
prev ious ly .
R A R
- W W — f — v A A A / ^ — i
Figure 5-15 Mesh currents in the c i rcui t of Fig. 5-11.
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126 Direct Curren ts in Electric Circuits
5.10 DELTA-STAR TRANSFORM ATIONS
Figure 15-6 shows th ree nodes A, B, C , delta-connected in par t a a n d star-connected in par t b. T h e n o d e s A, B, C form par t of a larger c i rcui t , and
can be connected together e i ther as in Fig . 5-16a or as in Fig . 5-16b. We shal l
see tha t , if cer ta in rela t ion s are sat isfied betw een the three resis tance s o n
the lef t and the three on the r ight , one ci rcui t may be subst i tu ted for the other
wi th ou t d i s tu rb ing the mesh cur r e n t s . I n f ac t, if one had two box es , one
con ta in ing a de l t a c i r cu i t , and the o ther con ta in ing a s t a r c i r cu i t sa t i s fy ing
E q s . 5-38 to 5-40, wi th only th e term ina ls A, B, C s h o w i n g o n e a c h b o x ,
the re would be no way o f t e l l i ng which box con ta ined the de l t a and which
con ta ined the s t a r . The two c i r cu i t s a r e thus sa id to be equivalent.
Th is t ran sfo rm at io n is of ten used for s impl i fying the calc ulat ion ofm e s h c u r r e n t s .
We could f ind R A, R B, R c in terms of R a, R b, R c, and inversely , by
a s s u m i n g t h e s a m e m e s h c u r r e n t s IA, IB, Ic in the two ci rcui ts , as in the
f igure , an d then m ak ing the vo l t ages V A — VB,V B - V c, V c - V A in Fig. 5-16a
equa l to the co r r esp on d in g vo l t ages in F ig . 5 -16b . Th i s is t he ob jec t o f P ro b .
Figure 5-16 (a ) Three -node c i rcu i t hav ing the shape of a cap i t a l de l t a , (b ) F o u r - n o d e
circui t having the shape of a s tar . Nodes A, B, C are part of a more extens ive
circu it . It is sh ow n th at, if eith er Eq s. 5-38 to 5-40 or Eq s. 5-44 to 5-46 a re
sat is f ied, the above two ci rcui ts are complete ly equivalent .
A A
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5.10 Delta-Star Transformations 127
5-19 . W e sha l l use here ano the r de r iva t io n th a t is som ew hat l ess conv inc ing ,
bu t shor t e r .
We disconnect the del ta and the star f rom the rest of the ci rcui t . I f thec i r cu i t s a r e equ iva len t , t hen the r es i s t ance be tween A a n d B must be the
s a m e i n b o t h :
RAB = „*<[V R b l = R A + R B- (5-35)
S imi la r ly ,
Ra+ R h + R c
N ote tha t , in the abo ve equ a t ion s , i s a ssoc ia t ed wi th RhRc. R B with i? c/?„ .
a n d R c with R 0R, , . Thus , s imply by inspec t ion , we have tha t
R,. + R h + R r
R B = R
f° , (5-39)
R c=
D- <
5
-4
° )
Now we a l so need the inver se r e l a t ionsh ip . We cou ld deduce i t f rom
the above th r ee equa t ions , bu t tha t would be r a ther long , and no t ve ryins t ruc t ive . I ns t ead , l e t us use the conduc tance be tween node A a n d t h e
n o d e s B a n d C shor t - c i r cu i t ed toge ther . Th i s g ives
G b + G e = r
Gf> + °i • (5 -41 .VIA + O il + (/(•
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128 Direct Curre nts in Electric Circuits
Similar ly ,
G c + G " - G , + G R + G r ' , 5 " 4 2 )
G r ( G , + G JG a + G b=
c\ A . ,5-43)G .4 + G B + O c
Again , by inspec t ion .
GBGC
G A+ G B + G c
GCGA
G A+ G B + G c
GAG
B
(5-44)
(5-45)
G c = . (5-46)G A + G B + G c
EXAMPLE: TRANSFORMATION OF A SYMMETRICAL DELTA
AND OF A SYMMETRICAL STAR
In Fig. 5-16a, suppose R a, R b, R c are a l l 100-ohm res is tances . Then, from Eqs . 5-38
to 5-40, the equiva lent s ta r-co nne cted c i rcui t is m ad e up of thre e res is tances of
(100 x 100)/300 = 33 oh m s each .
If , ins tead, we had a s tar-connected c i rcui t wi th res is tances R A, R B, R c of 100 ohms
each , t hen the equiva len t de l t a -connec ted c i rcu i t would have th ree iden t i ca l re s i s
t ances w i th condu c tanc es g iven by Eqs . 5-44 to 5 -46 : 10" 4 / ( 3 / 1 0 0 ) = 1 0 " 2 /3 Siemens .
The res is tances in the del ta would thus each have a res is tance of 300 ohms.
5.11 VOL TA GE A ND CURRENT SO URCES
The idea l vo l t age source supp l i es a cons tan t vo l t age tha t i s i ndependen t
o f the cu r r e n t d r a wn , and hence o f the load res i s t ance . A go od e l ec t ron ica l ly
vol tag e-stab i l ized po w er sup ply is c lose to ideal , up to a speci f ied cu rren t ,
beyond which the vol tage fal ls as in Fig . 5-17.
S imi la r ly , t he idea l cu r r en t source supp l i es a cons tan t cu r r en t tha t
i s i ndependen t o f the vo l t age ac ross the load . Aga in , cu r r en t - s t ab i l i zed
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12 9
O I
Figure 5-17 O ut pu t vo l tage as a funct ion of ou tpu t curre nt , for a vol tag e source .
/
0 V
Figure 5-18 O ut pu t cur ren t as a funct ion of ou tpu t vol tage, for a curr ent sourc e.
suppl ies are near ly ideal , but only up to a speci f ied vol tage beyond which
the curre nt de crea ses . Th is is sh ow n in Fig . 5-18.
5.12 THEVENIN'S THEOREM
I t i s f ound e xper im en ta l ly tha t t he vo l t age supp l i ed by a source decreases
wh en the load curre nt incre ases. Th is is t rue , even for vo l tage-s tabi l ize d
sources: the plateau in Fig . 5-17 is not real ly f la t but s lopes down sl ight ly .
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13 0
• I
- Q -
0
V
(a ) (b )
Figure 5-1 9 (a ) Any real sourc e is equiv alent to an ideal sou rce V 0 in series with a
re s i s t ance Rt. The source i s shown he re connec ted to a load re s i s t ance R d r a w i n g a
cur ren t / . The vo l t age V be tween the t e rmina l s i s V 0 — IR h o r IR. (b) T h e t e r m i n a l
vo l t age V as a funct ion of the load current I. The s lope is — Rj .
Thevenin's theorem state s th at any so ur ce — a f lashlight bat tery , an
osc i l l a to r , a vo l t age - o r cu r r en t - s t ab i l i ze d pow er supp ly , e t c .—is e qu iva len t
to an idea l vo l t age source V 0 in series with a resistance R,, as in Fig. 5-19,
where R, i s cal led the output resistance, or the source resistance. An idea l
vol tage source, by def ini t ion, has a zero R, .
In Fig . 5-19, the current I is F 0 / (R , + R) , whi le the vo l tag e V is IR .
T h e v o l t a g e V i s smal l e r than V 0 by the factor R/(R; + R) , be cau se of the
vo l t age d rop on R, . Wi th R d i sconnec ted (R -> o c ) , V is eq ua l to F 0 .
The internal resistance R, i s g iven by the slope of the curve of V as a
funct ion of I. I f on e increase s / by decrea sing R, then V decreases and R,i s equa l to lAK /A/ l . For one par t i c u la r e l ec t ron ica l ly ro / f age- s t ab i l ized
power supp ly , t he ou tpu t vo l t age d rops by one mi l l ivo l t when the ou tpu t
cu rren t increas es f rom zero to 100 mil l iam pere s. Th us R, i s 0 .01 oh m . A
smal l ba t t e ry charger has an in t e rna l r es i s t ance tha t i s o f the o rder o f one
ohm: i f t he load cur r en t inc reases by one ampere , t he ou tpu t vo l t age d rops
by abou t one vo l t .
A cur r a i r - s t ab i l i zed power supp ly has a l a rge V 0 and a large R ; . I ts
ou tpu t cu r r en t i s V0/Rhas long as R « R, , Th us i ts ou tpu t vo l t age (F 0 / R , ) R
i s p ropor t iona l to the load r es i s t ance R. I f t he load r es i s to r i s d i sconnec ted ,
then R is inf ini te and the output vol tage r ises to some l imi t ing value thatcan be much smal l e r than V 0 an d tha t is f ixed in t e rna l ly . I n one par t i c u la r
cur r en t - s t ab i l i zed supp ly , V 0 i s 50 ki lovo l ts , R ; i s 100 ki lohms, and the l imi t ing
vol tage is 50 vol ts .
Thus , i f V 0 a n d R t are known, one can p red ic t t he va lues o f V a n d I
fo r a g iven load r es i s t ance R. Theven in ' s t heorem there fo re p rov ides us wi th
a s imple mode l fo r desc r ib ing the opera t ion o f a r ea l source .
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5.12 Thevenin's Theorem 131
Th ev en in ' s th eo rem is a lso useful for calc ulat i ng cu rre nts f lowing
through ci rcui ts , as we shal l see in the fol lowing examples.
EXAMPLE: THEVENIN'S THEOREM APPLIED TO A SIMPLE CIRCUIT
Let us see how Thevenin 's theorem appl ies to the c i rcui t of Fig. 5-20a, where the
part of the c i rcui t to the lef t of the terminals is cons idered as a source feeding the
load re s i s t ance R. H e r e V x i s a vol tag e-s tabi l ized sour ce wi th an intern al res is tance
tha t i s neg l ig ib l e com par ed to R j .
Th e Th eve nin equiv alent of this sour ce is show n in Fig. 5-20b. The vol tage V 0
i s the vol tag e betwee n the term inal s of the c i rcui t of Fig. 5-20a when the res is tor R
i s removed. Also, the res is tance Rt i s t he re s i s t ance measured a t t he t e rmina l s w hen
R i s r emoved , and when V x i s reduced to zero. This is s imply R, a n d R 2 in paral le l .The cur ren t F 0 / (Ri + R) in Fig. 5-20b i s the same curre nt on e would f ind direct ly
from Fig. 5-20a using mesh currents as in Sec. 5.9.
figure 5-20 The circuit to the left of the terminals in (a ) gives the s ame cur ren t I in
the load res is tance R as the circuit to the left of the terminals in (h).
EXAMPLE AND REVIEW: THE WHEATSTONE BRIDGE
Figure 5-21a shows a Wheatstone bridge, s imilar to that of Pr ob . 5-8, except t ha t
we have shown the re s i s t ances R s a n d R d of the source and of the detector , respec
t ively, and that R d is not infinite here.
We wish to f ind the current in the detector , when the res is tance of the upper
lef t -hand branch is R( l + x) , as a funct ion of the curre nt I in tha t b ran ch .
W hen x = 0 , t he br idge is ba l ance d and the cur ren t t h r oug h the de tec tor i s
zero (Prob. 5-8) .
We could so lve th i s p rob lem by us ing e i the r b ranch or mesh cur ren t s . Ins t ead ,
we shal l solve i t in a more roundabout way that wi l l use most of the t r icks that we
have l ea rned in th i s chap te r .
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13 2
Figure 5-21 (a ) Whea t s tone br idge c i rcu i t , showing the in t e rna l re s i s t ance of the
s o u r c e Rs and the res is tance of the detector R
d. W h e n x = 0, the bridge is balanced
and no cur ren t f lows th rough R d. (b, c, d) Success ive t rans fo rma t ions of the c i rcu i t
that are used in calcula t ing the current through the detector as a funct ion of x.
a ) Le t t he cur ren t t h rough the re s i s t ance R(\ + x) be / . Then, us ing the substitu
tion theorem, we can replace the res is tance Rx by a source IRx and we have the
circuit of Fig. 5-21b.
b) We now have two sources , V s a n d IRx. Accord ing to the principle of super
position, each one ac t s i ndependent ly o f the o the r . The cur ren t s due to V s a re now
those for the ba lanced br idge . Thus the cur ren t t h rough the de tec tor i s due on ly to
the source IRx, and we may d i s rega rd V 5. We now have c i rcui t c .
c ) Res i s t ances R s, bR, abR form a del ta . I f we use the delta-star transformation,
we have c i rcui t d. We now have a f i f th node, £.
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5.12 Thevenins Theorem 133
Using the t rans forma t ion ru l e s ,
(bR)R s n (ahR)R s
(5-47)bR +.R S + abR' abR + R s + bR'
= BR, = aBR, (5-48)
w h e r e B is R2/R .
d) We can now use Thevenins theorem to f ind the current through the dectector
W e c o n s i d e r R d as the load res is tance, and a l l the res t of the c i rcui t as a source.
The open-c i rcu i t vo l t age K 0 i s that between D a n d B when the branch DB is
cu t . Th e cur ren t f lowing a ro un d the ou te r b ranc hes i s
IRx
R + R 2 + Ri + aR'
a n d
IRx
(aR + R 2). (5-49)R + R 2 + K 3 + aR
IRx
R + BR + aBR + aR(aR + aBR), (5-50)
ail + B)= IRx — , (5-51)
(1 + a)(\ + B)'
IRx. (5-52)1 + a
Note tha t t h i s vo l t age i s i ndependent o f R s, of b, and of the quant i ty B of Eq. 5-48.
We now requi re the re s i s t ance of R + R t of Sec. 5.12, which is the resistance
measured a t a cu t i n the branch DB, when the source IRx i s replaced by a short -
c i rcui t . We can f ind this res is tance from Fig. 5-2Id, bu t i t i s s imp ler to re tur n to
Fig. 5-21b. We imagine a source inserted in the branch DB, wi th bo th V s a n d IR x
rep laced by shor t -c i rcu i t s . Th e br idge i s ba l anced . Then a source in the bra nch
DB gives ze ro cur ren t i n the branch AC , so we may d i s rega rd the re s i s t ance Rs. So
(R + bR)(aR + abR)R ' + R > = R< + R
+bR + aR
+abR' ^
(1 + b)(\
i(l + b
1 + a
o(l + b)
Rd+ , R (5-55)
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134 Direct Currents in Electric Circuits
and, finally,
V0
a IRx /(l + a)
h = = „ , r „ [\„ u , (5-56)d + R, R d + 0 ( 1 + fe)«/( l + a) ] '
/.V
( 1 + - ) (R d/R )a
(5-57)
If we wish to maximize the ra t io Idjl, t hen R d/R should be sma l l , b should be
smal l , and a large. These las t two condi t ions are not a t a l l cr i t ical and one can say,
a s a ru l e o f t humb, tha t bo th a a n d b should be approxima te ly equa l t o un i ty .
5.13 SOLVING A SIMPLE DIFFERENTIAL EQUATION
In the nex t sec t ion we sha l l have to so lve the fo l lowing equa t ion :
a d-X + bx = c, (5-58)at
w h e r e a, b, c a r e k n o w n c o n s t a n t s , a n d x is an un kn ow n funct ion of f.
This i s a differential equation. I ts general solution i s co mp ose d of two
p a r t s : (a) the particular solution, ob ta ined by conserv ing on ly the l as t t e rm
on the left,
x i = c/b, (5-59)
plus (b) the complementary function, whic h is the solu t ion for c = 0, or
x2 = Ae- {h,a)t, (5-60)
w h e r e A is a constant of integration.
T h u s
x = x , + x 2 = C-+ Ae-Wa)t. (5-61)
The numer ica l va lue o f the constant of integration A i s obtained, as a rule ,
f rom the known va lue o f x a t t = 0.
I t i s easy to check this solu t ion by su bst i tu t ing i t in the or igin al eq ua t io n
(Eq . 5 -58). A un iquen ess t heo rem s t a t es tha t t h i s so lu t ion is t he on ly p oss ib le
o n e .
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5.14 Transients in RC Circuits 135
5.14 TRANSIENTS IN RC CIRCUITS
When a source is connected to a circuit comprising resistors and capacitors,transient currents flow until the capacitors are charged. Such currents are
large, at first, and then decay exponentially with time.
5.14.1 EXAM PLE: SIMPLE RC CIRCUIT
F i g u r e 5-22 s h o w s a c a p a c i t o r C, i n i t i a l ly uncharged , t ha t is c o n n e c t e d at t ime t = 0
to a vol t age source V t h r o u g h a res is tor R.
W e ca n find th e vol t age ac ros s th e c a p a c i t o r by us ing Ki rchof f ' s vo l t age law.
A d d i n g up the vol t age r i s e s a round th e l oop , s t a r t ing at the lowe r lef t -hand corn er ,
c lockwise ,
V — IRQ
c
Since / is dQ/dt,
RdQ
dt
1
0.
V
and, from th e prev ious s ec t ion ,
Q = CV + Ae-'IRC.
Since, by h y p o t h e s i s , Q = 0 a t t = 0, A = — C F a n d
Q = CF(1
After a t ime r » RC, Q % CV.
The vo l t age ac ros s th e c a p a c i t o r is
Vc = F( l I.
(5-62)
(5-63)
(5-64)
(5-65)
(5-66)
F igure 5-23 shows Vc/V as a funct ion of t i m e for R = 1 m e g o h m , C = 1 mic rofa rad ,
RC = 1 s e c o n d .
Figure 5-22 An RC circui t .
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13 6
1 .0
0 2 3t (seconds)
4 5
Figure 5-23 The vol tage V c across the capaci tor C in Fig. 5-22 as a funct ion of the
t ime . I t is a s sum ed th a t V c = 0 a t t — 0 and tha t RC = 1.
T h e p r o d u c t RC is called the time constant of the c i rcui t . For t = RC, V c/V =
1 - 1/e as 2/ 3.
I t wi l l be shown in Prob. 5-23 that , i f C is discharged through R, V c = Ve~"RC.
T h e current density vector J points in the di rect ion of f low of posi t ive charge
and i s expressed in am pere s per squ are mete r .
For any surface S b o u n d i n g a v o l u m e t .
which is the law of conservation of charg e.
T h e n e t c h a r g e d e n s i t y i n s i d e a h o m o g e n e o u s c o n d u c t o r c a r r y i n g a
cur r en t under s t eady- s t a t e cond i t ions i s ze ro .
F o r o r d i n a r y c o n d u c t o r s , Ohm's law a p p l i e s :
5.75 SUMMARY
(5-4)
J = <rE, (5-1)
where J is t he cu r r en t dens i ty , a i s the conduct ivi ty , and E is the elect r ic f ie ld
in tens i ty . Th e e l ec t r i c char ge dens i ty ins ide a un i fo rm cur r en t - c a r ry ing
c o n d u c t o r is z e r o u n d e r s t e a d y - s t a t e c o n d i t i o n s .
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5.15 Summa ry 137
Ohm's l aw can a l so be s t a t ed as
/ = V/R, (5-9)
w h e r e I i s the curre nt f lowing t hr ou gh a resistance R , acro ss wh ich a vol tage
V i s m a i n t a i n e d .
The power d i ss ipa ted as hea t by the Joule effect per cubic meter i s
J2/a, and the power d i ss ipa ted in a r es i s to r R car ry ing a cu r r en t I is I2
R.
Resi s to r s for which / is p ro po r t io na l to V are said to be linear, o r ohmic.
Other r es i s to r s a r e sa id to be non-linear.
The resistance of a set of resistors connected in series i s equ al to th e
sum of the i r i nd iv idua l r es i s t ances .
The conduc tance o f a se t o f resistors connected in parallel i s equ al to
the sum of the i r i nd iv idu a l co nd uc ta nce s .
T h e principle of superposition states that , as long as a l l the resistors
in a c i r cu i t a r e l i near , each source ac t s independen t ly o f the o ther s and
produces i t s own se t o f cu r r en t s .
Kirchoff's current law s t a t es tha t t he a lgebra ic sum of the cu r r en t s
ente r ing a no de , in an elect r ic c i rcui t , is equ al to zero . His voltage law s t a t es
tha t t he sum of the vo l t age d ro ps a ro un d a mesh i s equa l to ze ro . Cu r ren t s
in the var ious b ranches o f a c i r cu i t a r e usua l ly ca lcu la t ed by us ing mesh
cur r en t s and the vo l t age l aw.
A c c o r d i n g t o t h e substitution theorem, one can r ep lace a r es i s t ance Rin a c i rcui t by a vol tage source IR of the co r r ec t po la r i ty wi thou t a l t e r ing
the cur ren ts f lowing t hr ou gh the ci rcui t or , inversely , if a cu rre nt ente rs a
v o l t a g e s o u r c e V at i t s posi t ive terminal , one can replace the source by a
r es i s t ance V/I w i t h o u t c h a n g i n g t h e c u r r e n t s .
O n e c a n t r a n s f o r m t h e delta-connected circuit of Fig. 5-16a into the
star-connected circuit of F ig . 5 -16b by means o f the fo l lowing equa t ions :
RA =RhRr
R a + R b + R c
(5-38)
RBR , R „
R„ + R„ + R ,(5-39)
R,R„R„
R„ + Rh + R r
(5-40)
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138 Direct Currents in Electric Circuits
Also,
Gb
=
G , =
GB
GC
GA
+ GB
+ Gc
GCG
A
GA
+ GB
+ Gc
GAG
B
G, + GB+ Gr
(5-44)
(5-45)
(5-46)
A voltage source supplies a voltage that is nearly independent of the
current through the load; a current source supplies a current that is nearly
independent of the voltage across the load.
Thevenins theorem states that any source may be represented by an
ideal voltage source, in series with a resistance called it s output resistance.
When a source is connected to a circuit comprising capacitors, transient
currents flow temporarily in the circuit until all the capaci tors are charged.
PROBLEMS
5-1E CONDUCTION IN A NON-UNIFORM MEDIUM
Two p lane pa ra l l e l copper e l ec t rodes are s e p a r a t e d by a pla t e of t h i cknes s s
whose conduc t iv i ty a varies l inearly from a0. n e a r the pos i t ive e l ec t rode , to u 0 + a
n e a r th e negat ive e lect rode. Neglect edge effects . The cur ren t dens i ty is J.
F i n d the electric field intensity in the c o n d u c t i n g p l a t e , as a funct ion of th e dis
t a n c e x from the pos i t ive e l ec t rode .
5-2E RESISTIVE FILM
A square f i lm of N i c h r o m e (a n al loy of nickel and c h r o m i u m ) is d e p o s i t e d on a
sheet of glass , and copper e l ec t rodes are t hen depos i t ed on two oppos i t e edges .
S h o w t h a t the re s i s t ance be tween the e lec t rodes depends on ly on the t h i cknes s
of th e film an d on its conduc t iv i ty , as l ong as the film is s q u a r e .Thi s re s i s t ance is given in ohms per square. Nichrome f i lms range approxi
mately from 40 to 40 0 o h m s pe r s q u a r e .
5-3E RESISTOJET
F i g u r e 5-24 s h o w s the pr inc ip le of o p e r a t i o n of the res is toje t , which is used as
a t h r u s t e r for cor rec t ing e i the r the t r a j ec tory or the a t t i t u d e of a sate l l i te . See a l so
Probs . 2-14, 2-15, an d 10-11.
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13 9
Figure 5-24 S che ma t ic diag ram of a res is toje t . Th e prop el la nt gas P is ad mi t ted on
the l e f t , hea ted to a h igh t empera ture by the re s i s to r , and exhaus ted th rough the
nozzle . See Prob. 5-3.
Assuming comple te convers ion of the e l ec t r i c ene rgy in to k ine t i c ene rgy , ca l cula te the thru s t for an input p ow er of 3 ki lo wa t ts and a f low of 0.6 gram of hy dro gen per
second .
5-4E JOULE LOSSES
A res is to r has a re s is t ance of 100 k i loh ms an d a pow er ra t ing of one -qu a r t e r wa t t .
Wh a t i s t he maxi mu m vo l t age tha t can be appl i ed ac ros s i t ?
5-5E VOLTAGE DIVIDER
Figure 5-25 shows a vo l t age d iv ide r o r a t t enua tor . A source suppl i e s a vo l t age
V, t o the input , and a h igh- res i s t ance load (no t shown) R » R 2 i s con nec ted acro ss P 2 .
The vo l t age on R 2 a n d R is V 0.S h o w t h a t
K = R 2
F Ri + R 2
R
Figure 5-25 Vol tag e divid er: V 0 is
R 2/(R 1+ R
2) t imes V
t. See P rob . 5 -5 .
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14 0
unknown. When the s l id ing contac t i s
adjus ted for zero current , Vb/V
ais
Figure 5-26 P ote nt i om ete r c i rcui t . Th e
vol t age V a i s known, and V b is
equal to R 2/(R! + R 2). See Prob. 5-6.
5-6E POTENTIOMETER
Figure 5-26 shows a po ten t io mete r c i rcu it .
Show tha t , when7
= 0,
Th is c i rcui t i s extens ively used for me asu ring the vol tage sup pl ied by a sour ce
V b, wit hou t draw ing any curren t from i t . Th e c ircui t is used in s t r ip- cha rt rec ord ers .
In that case the current 7 i s ampli f ied to actuate the mo to r that displa ces the pen, and
s imul taneous ly moves the tap on the res is tance in a di rect ion to decrease 7. This is one
type of servomechanism.
5-7E SIMPLE CIRCUIT
Show that , in Fig. 5-27,
V = —R2 - R
VR2
+ R
if t he vo l tme te r d raw s a neg l ig ib l e amo un t o f cur ren t .
We shal l use this resul t in Prob. 15-5.
R, Ri
I — A W f A A A / — |
+ +
Figure 5-27 See P ro b. 5-7.r i
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14 1
WHEATSTONE BRIDGE
Figure 5-28 shows a c i rcui t known as a Wheats tone bridge. As we shal l see , the
vol t age V i s zero when
R1 Rl
— = — • (1)
R$ R 4
Fhe bridge is then said to be balanced. As a rule , the bridge is used unbalanced, the
vol t age V be ing a measure of the unknown res i s t ance , s ay R{.
T he circu it is so widely used to da y th at i t is alm ost imp ossi ble to list all its
appl i ca t ions . Some of the be t t e r known a re the fo l lowing .
1. I f one of the res is tors is a temperature-sens i t ive res is tor , or thermistor, th e
bridge serves as a thermometer, V be ing a measure of the t empera ture .
2. If one of the res is tors is a temperature-sens i t ive wire , heated by the bridge
cur ren t and im mersed in a gas , V i s a me asu re of the gas veloci ty, or of i t s turbu lence .
One then has a hot-wire anemometer.
3. In the Pirani vacuum gauge the heated wire is in a part ia l vacuum and is
cooled more or less by the res idual gas , according to the pressure .
4. In certa in gas analyzers, t he hea te d wi re is exposed to the unkn ow n gas .
T h e n V i s a measure of the the rma l conduc t iv i ty o f t he gas . Such ana lyze rs a re
used , i n pa r t i cu la r , for ca rbo n d iox ide . The the rma l co nduc t iv i ty o f ca rbo n d iox ide
is roug hly one half that of a i r (16.6, again s t 26.1 mil l iw at ts per m eter-kelvin ) .
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142 Direct Curre nts in Electric Circuits
5. In flammable-gas detectors the wire is he ate d suff ic iently to igni te a sam ple
of the gas con tain ed in a smal l cell . Th e heat genera ted by the com bu sio n increases
the tem pe ratu re and ch ang es the res is tance of the wire . Th e deflect ion of the vo l t me ter poin ter is a me asu re of the f lammabi li ty.
6. I f one of the res is tors is a f ine wire whose res is tance changes when elongated,
the bridge serves as a strain gauge for measur ing mic roscopic d i sp lacement s o r
d e f o r m a t i o n s .
7 . I f t he s t ra in gauge i s a t t ached to a d i aphragm, one has a pressure gauge.
a ) Sh ow tha t , when V = 0, Eq . 1 is satisfied.
Solution : Le t us supp ose tha t node C i s g ro und ed .
The vo l t age V being zero, there is zero current in the branch DB. T h e c u r r e n t
in the left-hand side is Vs/(R
t+ R 3 ) and the vol tage a t D is
V.
Similar ly, a t B,
R , + R 3
R 2+ R 4
Since V D = V B,
Ri + R 3 R 2 + R4
R 3 R 4
and , sub t rac t ing 1 on bo t h s ides ,
R i R 2
(2)
(3)
(4)
(5 )
b) Sup po se no w tha t R, increases by the factor 1 + x as in Sec. 5.12.2. We
r e q u i r e V/V s as a funct ion of x, whe re x can vary from —0.2 to + 0. 2. Th e vo l tme terV draws a neg l ig ib l e cur ren t .
Set
R 3 = aR lt R 2 = bR u R 4 = ab R u (6)
again as in Sec. 5.12.2. W ith x = 0, the bridg e is bala nce d.
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14 3
0.2
x
0
a
Figure 5-29 Th e rati o V/V s for the Wheats tone bridge of Fig. 5-28, as a funct ion of
x and a.
Find V/V s as a funct ion of x, a, b.
No te tha t F /F tu rn s ou t t o be indep ende nt o f b.
Solution : W e now have tha t
FK 2 + K 4 £, (1 + x) + R
R4R 3
I RJ R 2 R 3/i?, (8)1 + ( R 4 / R 2 ) (1 + x) + (R 3 / R i ) '
(9)1 + a 1 + x + a '
ax(10)(1 + a)(l + a + x)
Figure 5-29 shows F/F, as a funct ion of both x and a.
c) For what value of a i s F/F m ax im um , for a given value of x? This is the
condi t ion for maximum sens i t iv i ty . F igure 5-29 shows tha t t h i s condi t ion i s = t 1.
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144 Direct Curren ts in Electric Circuits
0.05 "
i /
/
1 1 1 1 ^ i i i i1 1
- 0 . 2i i i
0.2
V/ // o . 3 /
/ 1
- 0 . 0 5 .
Figure 5-30 R a t i o V/Vs for the Wheats tone bridge of Fig. 5-28 as a funct ion of x,
fo r a eq ua l to 0 .3, 1, 3.
Solution: VjV s i s m a x i m u m w h e n d(V/V s)/da = 0. Th us we set
(1 + a) (l + a + x)x - axl(l + a + x) + (1 + a)]
(1 + a ) 2 ( l + a + x) 2 " 1 1 '
(1 + a)(l + a + x) = a(2 + 2a + x), (12)
1 + a + x + a + a 2 + ax = 2a + 2a 2 + ax , (13)
a = (1 + x )1
'2
. (14)
Since x is smal l , the con di t i on for ma xi m um sens i t ivi ty is a * 1. as we ha d
gues sed .
d) Draw curves of V/V s as a func tion of x, for x = —0.2 to x = + 0 .2 , an d for
a = 0.3, 1, and 3.
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Problems 145
No te the s l igh t non- l inea r i ty o f t he curves . The n on- l inea r i ty dec reases w i th
inc reas ing a.
Solution: See Fig. 5-30.
AMPLIFIER
F i g u r e 5-3la shows one e lement of an analog computer. Its function is to
ampli fy the input vol tage by the factor — R 2/R i '•
Th e t r iang ula r f igure is an operational amplifier whose gain is — A. Such ampli f iers
have very high gains , of the order of 10 4 t o 10 9 , and d raw a neg l ig ib l e am ou nt o f cur ren t
a t t he i r i nput t e rmina l s .
The accuracy of the gain is l imi ted only by the s tabi l i ty of the ra t io R 2/Ri- T h e
drift in R2/R i due to ag ing , t empera ture changes , and so for th , can be sma l l e r t han the
drift in A b y o r d e r s o f m a g n i t u d e .
a) Y ou can show in the following way th at the ab ov e equa t ion is corre ct . F i rs t ,
you need an expres s ion for the vo l t age V iA at the input to the ampli f ier . S ince the
ampli f ier draws essent ia l ly zero current a t i t s input terminals , you can use the c i rcui t
of Fig. 5-31b. Then, se t t ing V 0 = — AViA, you can show tha t
K = R 2 „ _ « 2
V, K , f [ ( « , + R jFVTJ * R t
The gain is — R 2/R l if A » 1 an d if A » R 2/R 1.
(b )
ri;I
VNAA- TViA
I
Rz 1v
1
Figure 5-31 (a) Mu l t ip ly ing c i rcu it . Th e
t r i ang le represen t s an ampl i f i e r whose
gain is — A. The ci rcui t has a gain of
-R 2/R!, as long as A » 1 and
A » R 2/R 1. (b ) Equivalent c i rcui t for
calculat ing the gain. See Prob. 5-9.
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146 Direct Curren ts in Electric Circuits
C
B Figure 5-32 See Prob. 5-11.
5-12E CUBE
Twelve equa l re s i s t ances R form the edges of a cube, as in Fig. 5-33.
a ) A ba t t e ry i s conn ec ted be tween A a n d G.
Ca n you f ind one set of three po ints that are a t the sam e p ote nt i a l?Can you f ind anothe r s e t ?
b) No w imag ine that you ha ve a sheet of co pp er con nec t ing the f i rs t se t , an d
an oth er sheet of cop per con nec t ing the second set . Th is doe s no t affect the cu rren ts
in the res is tors .
You can now show tha t
RAG= (5/6)R.
b) W ha t i s t he min im um va lue of A if R, = 1000 oh ms , R 2 = 2000 ohms , and
if V 0 mus t be equa l t o — (R2/R
l)V
iwithin one part in 1000?
5-10E G = el
Yo u a re g iven a l a rge num ber of 6 -ohm res i s to rs . H ow can they be connec ted
to give a c i rcui t whos e con du cta nc e is equa l to e Siemens , where e is the base of the
n a t u r a l l o g a r i t h m s ?
Hint: Write out the ser ies for e.
5-1 IE TETRAHEDRON
Six equal res is tors R form the edges of a te t rahedron, as in Fig. 5-32.
a ) A ba t t e ry i s conn ec ted be tw een A a n d B.
Show tha t , by symmet ry , po in t s C and D wil l be a t the same potent ia l . That
being so, there is zero current in branch CD, and we can remove i t .
b) W ha t is the res is tance betwe en A a n d B ?
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14 7
F
Figure 5-33 See Prob. 5-12.
F G
H Figure 5-34 See P rob . 5 -13 .
5-BE CUBE
Twelve equa l re s i s t ances R form the edges of a cube, as in Fig. 5-33. To find
the re s i s t ance be tween A and C, we first flatten the cube as in Fig. 5-34.
a ) A ba t t e ry i s conn ec ted b e tween A and C.
Which four po in t s a re a t t he s ame po ten t i a l ?
Which two branches can be removed wi thout a f fec t ing the cur ren t s ?
b ) N o w s h o w t h a t
RAC= (3/4)/?.
5-14 CUBE
Firs t , read Probs . 5-11 to 5-13.
Now show that , in Fig. 5-33,
RAD= (1.4/2.4)R.
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148
r ^ A A A - f - A A A - f -
550 V ~±r+ R
S a a a - ^ a a a >X20 km
Figure 5-35 A 550-vol t source feeds a load res is tance R T t h r o u g h a 20-k i lomete r
line. The pos i t ion of a fault RScan be f o u n d f r o m m e a s u r e m e n t s of the c u r r e n t / ,
with and w i t h o u t a shor t -c i rcu i t ac ros s R,. See P r o b . 5-15.
5-/50* LINE FAULT LOCATION
In Fig. 5-35. a 550-vol t source feeds a f ixed load res is tance RX t h r o u g h a pa i r
of copper (a = 5.8 x 107
Siemens pe r meter) wires 20 ki lomete rs long . The wires are
3 mi l l ime te rs in d i a m e t e r .
Suddenly , dur ing a s t o r m , the c u r r e n t I suppl i ed by the source inc reases and
t he vo l t age V a c r o s s R, falls. It is conc luded tha t a fault has d e v e l o p e d s o m e w h e r e
a l o n g the l ine, and t ha t t he re is a shunt re s i s t ance R S at some d i s t ance x from the source .
If the l oad re s i s t ance R T is d i s c o n n e c t e d , / is f o u n d to be 3.78 a m p e r e s . W h e n
the l ine t e rmina l s at the l oad ar e shor t -c i rcu i t ed , / is 7.20 a m p e r e s .
W h a t is the di s t ance x?
5-16E UNIFORM RESISTIVE NET
F i g u r e 5-36 s h o w s a t w o - d i m e n s i o n a l n e t w o r k c o m p o s e d of equa l re s i s t ances .
T h e o u t s i d e n o d e s are e i the r connec ted to ba t t e r i e s in s o m e a r b i t r a r y way, or left
u n c o n n e c t e d .
a) Use Kirchof f ' s cur ren t law to show tha t th e p o t e n t i a l at any i ns ide node is
e q u a l to the ave rage po ten t i a l at the four ne ighbor ing nodes . Fo r e x a m p l e .
VA+V M+V C+ VL
4
b ) W h a t is the ru l e for a s i m i la r t h r e e - d i m e n s i o n a l n e t w o r k ?
5-17 POTENTIAL DIVIDER
F i n d th e r a t i o VJV Tin Fig. 5-37.
P r o b l e m s m a r k e d D ar e relatively difficult .
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14 9
Figure 5-36 Tw o-dim ens io na l ne twor k comp osed of equa l re s i s t ances R. A r b i t r a r y
vol tages are appl ied a t the periphery. I t i s shown in Prob. 5-16 that the vol tage on
any ins ide node is equ al to the ave rage valu e of the vol tages on the four ne igh bo ring
n o d e s .
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15 0
Figure 5-38 See Prob. 5-18.
5-18 SIMPLE CIRCUIT WITH TWO SOURCES
Show that , in Fig. 5-38,
/ =R,
r(R, + R 2)
5-19 DELTA-STAR TRANSFORMATION
Find the equa t ions for e i the r the de l t a - s t a r o r t he s t a r -de l t a t r ans forma t ion by
as suming mesh cur ren t s , a s in F ig . 5 -16 , and making the vo l t ages V A - V B, V B - F c ,
V c - V A in part a the same as those in part b.
Hint: Find on e equ a t ion of the form
( • )I A + ( • )IB + ( )Ic = 0.
Then, s ince i t must be val id, whatever the values of the mesh currents , the parentheses
must a l l be ident ical ly equal to zero. This should give you one of the equat ions of one
se t ; t he o the r two equa t io ns can be found by sym met ry .
4 KQ
i K n KQ 3 K n
5 K n
-o
Figure 5-39 See P rob . 5 -20 . The symbol Q s t ands for ohm.
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Problems 151
5-20 DELTA-STAR TRANSFORMATION
Find the res is tance of the c i rcui t shown in Fig. 5-39.
5-2IE OUTPUT RESISTANCE OF A BRIDGE CIRCUIT
Show that the output res is tance of the bridge c i rcui t shown in Fig. 5-40, as
seen a t t he vo l tme te r V, is R.
Assume tha t t he source V s has a ze ro ou tpu t re s i s t ance .
5-22E OUTPUT RESISTANCE OF AN AUTOMO BILE BATTERY
I t i s found tha t the vol tage a t the term inal s of a defect ive au tom ob i le bat te ry
dr op s from 12.5 to 11.5 vol ts wh en the headl ig hts are turne d on .
W ha t i s t he appro xim a te va lue of the ou tpu t re s i s t ance?
5-23E DISCHARGING A CAPACITOR THROUG H A RESISTOR
A res is tance of 1 me go hm i s conne c ted ac ros s a 1 mic rofa rad capac i to r i n i t i a llycharged to 100 vol ts .
Ca lcu la t e the vo l t age ac ros s the capac i to r a s a func t ion of the t ime .
5-24E RAMP GENERATOR
A con s tan t - cur r en t source feeds a cur ren t / t o a capac i to r C. A t t = 0 the
capac i to r vo l t age i s ze ro .
F ind the vo l t age ac ros s C as a funct ion of the t ime.
5-25 CHARG ING A CAPACITOR THROUG H A RESISTOR
A source supply ing a vo l t age V cha rges a capac i to r C t h rough a re s i s t ance R.
Calculate the energy suppl ied by the source, that diss ipated by the res is tor ,
and that s tored in the capaci tor , af ter an inf ini te t ime.
Yo u sho uld f ind tha t one half of the energy is diss ipa ted in R, and tha t t he o the r
half is stored in C.
5-26 RC TRANSIENT
In the c i rcui t of Fig. 5-41, the cap aci tor i s ini t ia lly un cha rge d.
a) At 1 = 0 the switch is c losed. Find the vol tage V as a function of t.
b) After a long t ime the switch is op ene d. Fin d again V as a function of t, set t ing
t = 0 a t t he moment the swi t ch i s opened .
Figure 5-40 See Prob. 5-21.
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15 2
Figure 5-41 See Prob. 5-26.
Figure 5-42 RC different ia t ing c i rcui t .
See Prob. 5-27.
5-27E RC DIFFERENTIATING CIRCUIT
Figure 5-42 shows an RC different ia t ing c i rcui t . Th e loa d res is tance con nec ted
at F> is large co m pa red to R.
Show tha t , if t he vo l t age dr op ac ro s s R is neg l ig ib le com pare d t o tha t ac ros s C ,
dVt
K.RC-^.
N o t e t h a t V„ « V,. See Prob. 5-29.
5-28E DIFFERENTIATING A SQUARE WAVE
A square wave as in Fig. 5-43 is appl ied to the RC different ia t ing c i rcui t of
Fig. 5-42.
Sketch a curve of the output vol tage as a funct ion of the t ime.
Figure 5-43 Ideal square wave. In
pract ice , the s t ra ight l ines are
L
por t ions of exponent i a l curves . See
Prob . 5 -28 .
5-29 DIFFERENTIATING CIRCUIT
The ci rcui t of Fig. 5-42 is s imple and inexpens ive, but V 0 « V t.
F i g u r e 5-44a shows a much supe r ior , bu t more complex , d i f fe ren t i a t ing c i rcu i t .
Fhe t r i ang le represen t s an ope ra t iona l ampl i f i e r a s in P rob . 5 -9 .
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15 3
(b)
O-
t
I;
I
i
<> » A A A - o
I 1
Figure 5-44 (a ) Different ia t ing c i rcui t
us ing an ampli f ier , (b ) E q u i v a l e n tci rcui t used to calcula te V 0. See P rob .
5-29.
S h o w t h a t
dV t
K = -RC—1
,dt
as long as A » 1 an d |K 0 | /RC » \dVJdt\jA. N o t e t h a t RC can be much l a rge r than
uni ty, so that V0 need no t be much sma l l e r t han V
t.
Solution: From F ig . 5 -44b .
Vt=®+ViA, V„=-AViA, (1)
dV , = 1 dQ dV u _ 1 1 dV a
dt C dt dt C A dt'
N o w
so tha t
V i4 - V 0 \ 1 ,/ = — = 1 + — K, (3)
R R \ A 1
dV t 1 / 1 \ 1 dV 0
— = 1 + - IK , -. (4)
dt RC \ A I A dt1 1 / V dV"
RC " A \RC dt J'
* -VJRC, (6)
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154 Direct Currents in Electric Circuits
if A » 1 and i f |F | / i ? C » \dVJdt\/A. T h u s , w i th t h is a p p r o x i m a t i o n ,
dV ,V„=-RC-^. (7)dt
5-30 RC INTEGRATING CIRCUIT
Figure 5-45 shows an RC i n t egra t ing c i rcu it . Th e cur ren t t h rou gh the loa d
con necte d a t F„ is negl igible com pa re d to dQ/dt at C.
Show tha t , a s long as the vo l t age ac ros s C i s sma l l compared to tha t ac ros s R,
As in Prob. 5-27, V0 « F- I t i s assu me d tha t V
0= 0 a t t = 0.
5-31 E INTEGRATING CIRCUIT
A square wave as in Fig. 5-43 is appl ied to the RC integrat ing c i rcui t of Fig. 5-45.
Sketch a curve of the output vol tage as a funct ion of the t ime.
R
W V \ A
Figure 5-45 RC integrat ing c i rcui t . See
Pro b . 5 -30. ° "
5-32 INTEGRATING CIRCUIT
The in t egra t ing c i rcu i t shown in F ig . 5 -46a pe r forms in t egra t ions w i thout the
l imi t a t ion F 0 « V t that a ppl ies to the c i rcui t of Fig. 5-45. Th e t r iang le repres ents an
ampli f ier as in Probs . 5-9 and 5-29.
An alo g co mp ute rs m ak e extens ive use of the c i rcui ts of Figs . 5-31, 5-44, and 5-46.
S h o w t h a t
if A » 1 and if \V„\/RC « A\dVJdt\.
Use the c i rcui t of Fig. 5-46b and set
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15 5
Figure 5-4 6 (a ) In t egra t ing c i rcu i t us ing
an ampli f ier , (b ) Equivalent c i rcui t for
ca lcu la t ing V„. See P rob . 5 -32 .
(a )
ri;
I
V W 9 ) | —Ti
c
1y.
I(b )
5-33 PULSE-COUNTING CIRCUIT
The ci rcui t shown in Fig. 5-47 can be used e i ther for count ing pulses or for
measur ing a capac i t ance , e i the r C\ o r C 2 , t he o the r one be ing known.
T h e p u l s e g e n e r a t o r G prod uces / pos i t ive squa re pu l ses of amp l i tude V p per
second and
Cj « c 2 , / c , « c 2 .
D i o d e s D x a n d D 2 pas s cur ren t s on ly in the d i rec t ion of the a r row.
Dur ing a pu l s e , no cur ren t f lows th rough d iode D u and capac i to rs C, and C 2
charge in ser ies .
Betw een pulses , the term inal s of G are effect ively sho rted a nd ca pac i tor C t
d i s c h a r g e s t h r o u g h d i o d e D l. D i o d e D 2 i s t h e n n o n - c o n d u c t i n g .
a) Sketch a curve of V as a function of the time, for V « Vp, wi th V = 0 at
f = 0.
b) W ha t is the value of F af ter a t ime f ?
Jl v,
4 ^ D ,
T
Figure 5-47 This c i rcui t can be used e i ther for measuring the repet i t ion frequency of
the cur ren t pu l s es o r ig ina t ing in G or for measur ing e i the r C, o r C 2 . See P rob .
5-33.
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CHAPTER 6
DIELECTRICS: I
Electric Polarization P, Bound Charges,
Gauss's Law, Electric Displacement D
6.1 THE ELECTRIC POLA RIZA TION P
6.2 THE BOUND CHARG E DENSITIES pb AND a h
6.2.1 Example: Sheet of Dielectric
6.3 THE ELECTRIC FIELD OUTSIDE AND INSIDE
A DIELECTRIC
6.3.1 Example: Sheet of Dielectric
6.4 THE DIVERGENCE OF E IN DIELECTRICS:
GAUSS'S LAW
6.5 THE ELECTRIC DISPLACEM ENT D
6.6 THE ELECTRIC SUSCEPTIBILITY Xe
6.7 THE RELATIVE PERM ITTIVITY e r
6.8 RELATION BETWEEN THE FREE CHARGE DENSITIES
AND THE BOUND CHARGE DENSITIES
6.8.1 The Volume Charge Densities p f and p h
6.8.2 The Surface Charge Densities a ( and a b
6.8.3 Exam ple: Dielectric-Insulated Parallel-Plate Capacitor
6.9 POISSON'S AND LAPLACE'S EQUATION S
6.10 ELECT RETS
6.11 SUMMARY
PROBLEMS
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Dielect r ics di f fer f rom conductors in that they have no f ree charges that can
move through the mater ial under the inf luence of an elect r ic f ie ld . In di e l ec t r i cs , a l l t he e l ec t rons a r e bound ; the on ly poss ib le mot ion in an e l ec t r i c
f ie ld i s a m inu te d i sp la cem ent o f pos i t ive and neg a t ive charges in o ppo s i t e
d i r e c t i o n s . T h e d i s p l a c e m e n t i s u s u a l l y s m a ll c o m p a r e d t o a t o m i c d i m e n s i o n s .
A d ie l ec t r i c in which th i s charge d i sp lacement has t aken p lace i s sa id
to be polarized, and i t s molecu les a r e sa id to possess induced dipole mom ents.
These d ipo les p roduce the i r own f i e ld , which adds to tha t o f the ex te rna l
cha rges . Th e dipole f ield an d the externa l ly app l ied elect r ic f ie ld can be
c o m p a r a b l e i n m a g n i t u d e .
In add i t ion to d i sp lac ing the pos i t ive and nega t ive charges , an app l i ed
e lec t r i c f i e ld can a l so o r i en t molecu les tha t possess permanen t d ipo lemoments . Such molecu les exper i ence a to rque tha t t ends to a l ign them wi th
the f ie ld , but col l i s ions ar is ing f rom thermal agi ta t ion of the molecules tend
to des t ro y the a l ignm ent . An equ i l ib r ium p o la r i za t ion i s t hus es t ab l i shed
in which th ere is , on the avera ge, a net a l ig nm ent .
This i s , in fact , a s impl i f ied view of dielect r ic behavior , because many
so l ids , such as sod ium ch lo r ide , f o r example , a r e no t made up o f molecu les
bu t o f ind iv idua l ions .
THE ELECTRIC POLARIZATION P
T h e electric polarization P i s t he d ipo le moment per un i t vo lume a t a g iven
point . I f p i s t he average e l ec t r i c d ipo le moment per molecu le , and i f N is
the nu m be r of mo lecu les per un i t vo lum e, then
p = yvp. (6-1)
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Kigure 6- 1 Under the act ion of an e lect r ic f ie ld E, which is the resul tant of an
external f ie ld and of the f ie ld of the dipoles wi thin the die lect r ic , pos i t ive andnega t ive cha rges in the molecu les a re s epa ra t ed by an ave rage d i s t ance s. In the
process a net charge dQ = NQs • da' crosses the surface da', N be ing the number of
molecu les pe r un i t vo lume and Q t he pos i t ive cha rge in a molecu le . The vec tor
da' i s per pen dic ula r to the shad ed surface. Th e c i rc les indi cate the cent ers of ,
charge for the pos i t ive and for the negat ive charges in one molecule .
THE BOUND CHARGE DENSITIES p„ AND a b
W e sha l l now sh ow tha t the d i sp lac eme nt o f char ge wi th in the d ie l ec t r i c
gives r ise to net volume and surface charge densi t ies
p „ = - V - P a n d a b = P • n 1 ( (6-2)
whe re rij is t he norm al to the su rf ace , po in t ing ou tw ard .
Let us f i r s t consider the sur face densi ty a b. We imagine a smal l e l ement
of sur face da ' inside the dielect r ic , as in Fig . 6-1 . U nd er th e act io n of the
f i e ld , pos i t ive and nega t ive charges wi th in the molecu les separa te by an
a v e r a g e d i s t a n c e s. Posi t ive charges c ross the su r f ace by moving in the
d i r ec t ion o f the f i e ld ; nega t ive charges c ross i t by moving in the oppos i t e
d i r ec t ion . For the purpose o f our ca l cu la t ion , we cons ider the pos i t ive
charges to be in the form of point charges Q and the nega t ive charges in the
form of point charges — Q. F u r t h e r m o r e , w e c o n s i d e r t h e n e g a ti v e c h a r g e s
to be f ixed and the pos i t ive charges to move a d i s t ance s. T h e a m o u n t o f
c h a r g e dQ t ha t c rosses da ' i s t hen jus t t he to t a l amount o f positive c h a r g e
wi th in the imag in ary pa ra l l e l ep iped show n in F ig . 6 -1 . Th e vo l um e o f th i s
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6.2 The Bound Charge Densities p b and c b 15 9
para l l e l ep iped i s
dx ' = s • da'. (6-3)
a n d
dQ = NQs • da', (6-4)
w h e r e N i s t he number o f molecu les per un i t vo lume and Qs i s the dipole
m o m e n t p of a molecu le . Then
dQ = P • da'. (6-5)
If da ' i s on th e sur face of the dielect r ic ma ter i al , dQ a c c u m u l a t e s t h e r e
as a su r f ace d i s t r ibu t ion o f dens i ty
(7 6 = ^ = P - n 1 , (6-6)da
w h e r e n, i s a un i t vec to r , no rmal to the su r f ace and po in t ing outward. T h e
bound su r f ace charge dens i ty a h is t hus equa l to the no rm al com po ne n t o f
P at the sur face.
We can show s imi la r ly tha t — V • P r ep resen t s a vo lume dens i ty o fcharg e . Th e ne t charge th a t f lows ou t o f a vo lum e x across an e l ement da'
of its surface is P • da', as we found above. The net charge that f lows out
o f the su r f ace S ' bound ing x i s thus
Q = J*g, P • da', (6-7)
and the ne t charge tha t r emains wi th in the vo lume x must be —Q. If p h is
the vo lum e dens i ty o f the cha rge r em ain in g wi th in th i s vo lum e, then
£, Pt dx = - Q = - J s„ P • da' = - £ , (V • P) dx'. (6-8)
S ince th i s equa t ion must be t rue fo r a l l x, t he in t egrand s must be equa l a t
every po in t , and the bound vo lume charge dens i ty i s
pb
= - V P (6-9)
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160 Dielectrics: I
We refer to p b a n d a b as bound charge dens i t i e s , a s d i s t ingu i shed f rom
th e free charge dens i t i e s p f a n d a f. B o u n d c h a r g e s a r e t h o s e t h a t a c c u m u l a t e
th rough the d i sp lacement s tha t occur on a molecu la r sca le in the po la r i za t ionprocess . Th e o th er charges a r e ca ll ed fr ee charg es . Th e cond uc t ion e l ec t rons
in a conduc to r and the e l ec t rons in j ec ted in to a d i e l ec t r i c wi th a h igh-energy
e lec t ron beam are examples o f f r ee charges .
EXAMPLE: SHEET OF DIELECTRIC
If a sheet of dielectric i s pola r ized u niformly in the di rect ion no rm al to i ts surface,
then the surface dens i ty ofbound charge crfc is P. as in Fig. 6-2. Also. p h = — V • P = 0.
s ince P is i ndepen dent o f t he coo rd ina te s .
6.3 THE ELECTRIC FIELD OUTSIDE ANDINSIDE A DIELECTRIC
Figure 6 -3 show s a b lock o f po la r i zed d ie l ec t r i c ma te r i a l i n which P is a
funct ion of posi t ion. We must f ind the elect r ic f ie ld in tensi ty that the dipoles
in the d ie l ec t r i c p roduce a t some po in t A, e i the r ins ide o r ou t s ide . Once
ob t a ine d , th i s E can then be ad de d to tha t p rod uce d by o th er charges to
give the total E .
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16 1
Figure 6-3 Block of d i e l ec t r i c w i th a d ipo le moment P per un i t vo lume . The
dipoles wi thin the e lement of volume shown ins ide the block give r ise to an
e lec t r ic po ten t i a l dV a t t he po in t A.
C o u l o m b ' s l a w a p p l ie s t o a n y n e t a c c u m u l a t i o n o f c h a r g e , r e g a r d l e s s
of o th er ma t t e r tha t ma y be p resen t . Thu s the f ie ld p ro du ced by the d ie lec t r i c ,bo th ins ide and ou t s ide , i s t he same as i f t he charge dens i t i e s p b a n d a b w e r e
s i tua ted in a vac uum .
6.3.1 EXAMPLE: SHEET OF DIELECTRIC
In the po lar ized sheet of die lect r ic of F ig. 6-2. E = 0 ou t s ide and E = o b/e 0 p o i n t i n g
to the left inside, exactly as if the sheet were replaced by its surface charges.
6.4 THE DIVERGENCE OF E IN DIELECTRICS:
GA USS'S LA W
W e have jus t seen that the elect r ic f ie ld pro du ce d by the dipoles of a p olar iz ed
dielect r ic can be calc ulate d f rom th e bou nd c ha rge densi t ies as if they w ere
si tu ated in a va cu um . Let us inve st igate the imp l ica t ion s of th is fact on
Ga uss ' s l aw (Sec . 3 .2 ) .
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162 Dielectrics: I
Gauss 's law re la tes the f lux of E through a c losed surface to the charge
Q enclosed with in tha t surface:
Js
E • da = £ V • E dx = (2/e 0. (6-10)
For d ie lec tr ics , Q inc ludes bound as well as f ree charges:
Q = j(p f + p b)dx. (6-11)
If we subs t i tu t e th is va lu e of Q in to Eq . 6 -10 and equa te the in teg rands
of the volume in tegra ls , then
e 0
This is Gauss's law in i ts more genera l form. I t is one of Maxwell 's
four fundam en ta l equa t ion s of e lec t rom agne t i sm . We sha ll f ind the o the r
three in Sees . 8 .2 . 11 .4, an d 19 .2. Th is one fo l lows from (a) C ou lo m b' s law.
(b) the concept of electric field intensity, (c) the principle of superposition,
an d (d) the fact tha t the field of the dipol es situa ted in a po lari zed m ed iu m
can be ca lcula ted f rom pb a n d ab.
Note tha t we have implic i t ly assumed the exis tence of the spaceder ivat ives of E. These der iva t ives do not ex is t a t the in terface between two
media : in such cases one must rever t to the in tegra l of E • da over a c losed
surface , which is equal to the to ta l enclose d cha rge ov er e 0 , a cco rd ing to
Eq. 6-10.
THE ELECTRIC DISPLACEMENT D
Since pb = - V • P. f rom Eq. 6-9 , then
V E = -(p f - V P ) , (6-13)
o r
V • (e 0 E + P) = Pf. (6-14)
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6.7 The Relative Permittivity e r 163
The vec to r e 0 E + P is therefore such th a t i ts d ivergen ce dep end s only on
the free charge density p f. This vector is ca l led the electric displacement a n d
is des ignated by D :
D = e 0 E + P. (6-15)
T h u s
(6-16)
In in tegra l form, Gauss 's law for D b e c o m e s
f D • da = f pj-d-c, ( 6 - 1 7 )
and the f lux of the e lec tr ic d isp lacement D through a c losed surface is equal
to the f ree charge enclosed by the surface .
No te tha t the divergence of D , as well as the surface integral of D , a re
bot h unaffec ted by the bo un d charge s .
6.6 THE ELECTRIC SUSCEPTIBILITY /e
In mos t d ie lec t r ic s the mo le cu la r cha rg e sepa ra t io n i s d i rec tly p rop o r t i ona l
to , and in the same d irec t ion as , E. Die lec tr ics tha t show th is s imple depen
dence of polarization on field are said to be linear a n d isotropic In prac t ice ,
most d ie lec tr ics are homogenous, as well as l inear and iso tropic .
The d ipo le momen t pe r un i t vo lume i s then
P = Np = x ee 0E, (6-18)
w h e r e N is again th e num be r of mol ecule s per uni t volu me , an d y e is a
d imens ion le ss cons tan t known as the electric susceptibility of the dielectric.
6.7 THE RELA TIVE PERMITTIVITY e r
F or a l inear and iso tr opic d ie lec tr ic we have , f rom E qs. 6-15 and 6-18 th a t
D = € 0 (1 + yjE = € 0 e r E = eE, (6-19)
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16 4
Table 6-1 Relative Perm ittivities of
Dielectrics Near 25 C.
Frequency (hertz)
100 106
10"'
Bake l i t e 5.50 4.45 3.55
B u t y l r u b b e r 2.43 2.40 2.38
Fu sed silica 3.78 3.78 3.78
Luc i t e 3.20 2.63 2.57
N e o p r e n e 6.70 6.26 4.0
Polys tyrene 2 .56 2.56 2.54
S tea t i t e 6.55 6.53 6.51
S tyrofoam 1.03 1.03 1.03
Teflon 2.1 2.1 2.08
W a t e r 81 . 78.2 34 .
w h e r e
e, = 1 + l e = e / e 0 (6-20)
i s a d ime ns ion les s cons tan t l a rger than un i ty an d i s kn ow n as the relative
permittivity of the mate r i a l . I n a vacuum, y e = 0, e r = 1. Th e relat iv e
permi t t iv i ty was fo rmer ly ca l l ed the dielectric constant. The quan t i ty e i s
cal led the permittivity.
Th e relat ive perm it t iv i ty of dielect r ics l ies typical ly be twe en 2 an d 7 .
However , some non- l inear d i e l ec t r i cs have r e l a t ive permi t t iv i t i e s as h igh as
1 05
.
Table 6 -1 g ives the r e l a t ive permi t t iv i t i e s o f some common d ie l ec t r i cs .
I t wi l l be observed f rom the table that e r is f r equen cy-dep ende n t . W e sha ll
return to th is phenomenon br ief ly in Sec. 7 .7 .
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6.8 Relation Between the Free and Bound Charge Densities 165
6.8 RE LATION BETWEEN THE FREE
CHARGE DENSITIES AND THE BOU ND
CHARGE DENSITIES
6.8.1 THE VOLUME CHARGE DENSIHES p f AND p„
In a l inea r and is ot ro pic dielect r ic , from Eqs. 6-15 and 6-19,
D = e 0 E + P = e r e 0 E . (6-21)
a n d t h u s
P = (e r - 1 ) e ( )E = ^ — - D . (6 -2 2)
Ta ki ng no w the divergen ce of bo th sides an d using Eqs. 6-9 an d 6-16,
p b = - ^ - P f (6-23)
N o t e t h a t p b a n d p f have oppos i t e s igns , so tha t t he to t a l charge
dens i ty i s smal l e r th an p s•:
P/ + Pt = Pf/er- (6-24)
Also, if p f if zero in a l inear an d is ot ro pic dielect r ic , which is nea r ly alw ays
the case, then p h i s a lso zero.
6.8.2 THE SURFACE CHARG E DENSITIES a f AND o„
At the in t e rf ace be twee n a d i e l ec t r ic and a condu c to r , t he re is a b ou nd
sur f ace charge dens i ty a h on the dielect r ic , and a f ree sur face charge densi ty
o f on the conduc to r , a s in F ig . 6 -4 . For s t eady- s t a t e cond i t ions , t he re i s
zero elect r ic f ie ld inside the conductor and. inside the dielect r ic , f rom Gauss 's
l a w .
€ 0E = Of + o h, D = e re 0E = a r. (6-25)
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16 6
E ,D
Figure 6-4 Interface between a die lect r ic and a conductor , wi th a pos i t ive charge
dens i ty on the sur face of the con duc tor .
This eq ua t ion i s s imi l a r to Eq . 6 -24 .
6.8.3 EXAMPLE: DIELECTRIC-INSULATED PARALLEL-PLATE CAPACITOR
Figure 6-5 shows a d i e l ec t r i c - insu la t ed pa ra l l e l -p l a t e capac i to r w i th the spac ing s
exag gera ted for c lar i ty. We assum e that s i s smal l , co m pa red to the l inear exten t
of the pla tes , in order that we may neglect the fr inging f ie ld a t the edges .Th en P is unifo rm, V • P = 0, and p b = 0, which is cons is tent w i th the fact
t h a t p, = 0.
The sur face -cha rge dens i t i e s + a s on the lower pla te and — a s on the upper
p la t e p roduce a un i form e lec t r i c f i e ld d i rec t ed upward . The po la r i za t ion in the
dielect r ic gives a bound surface charge dens i ty — ah on the lower surface of the
dielect r ic and + a b on the upper sur face . These bound cha rges produce a un i form
electr ic f ie ld di rected downward that cancels part of the f ie ld of the charges s i tuated
on the pla tes . S ince the net f ie ld between the pla tes must remain equal to V/s, th e
free charge dens i t ies +a f a n d — a f on the pla tes must be larger than when the
dielect r ic i s abse nt . The presen ce of the die lect r ic thus has the effect of increa s ing
the charges on the pla tes for a given value of V, and hence of increas ing the
c a p a c i t a n c e .To ca lcu la t e the capac i t ance , we apply Gauss ' s l aw for the d i sp lacement D to a
cylinder as in Fig. 6-5. Then the only flux of D t h rough the Gauss i an sur face i s
t h r o u g h t h e to p , a n d D i s num erica l ly equa l to a f. T h e n . E = o f/e re 0, t he po ten t i a l
di fference between the pla tes is afs/e
re
0, a n d t h e c a p a c i t a n c e
T h u s
(TF
+ a h = a f/e r. (6-26)
C = = e re 0A/s, (6-27)
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16 7
+ + + + +
U t p
+ + +
Figure 6-5 Diele ct r ic- in sulated para l le l -pla te capac i tor . The smal l rectangle is the
cross-sect io n of an ima gina ry cyl inder used for calcu la t ing D.
w h e r e A i s the area of one pla te . The capa ci ta nce is therefore increased b y a factor
of e r t h rough the presence of the d i e l ec t r i c .
The me asur eme nt o f t he capac i t ance of a su i t ab le capac i to r w i th and wi thout a
d ie l ec t r i c p rov ides a convenien t me thod for measur ing a re l a t ive pe rmi t t iv i ty er.
POISSON'S AND LAPLACE'S EQUATIONS
In homogeneous , l i near , and i so t rop ic d i e l ec t r i cs , D is p ro po r t io na l to E ,
a n d e r is i nde pen den t o f the coo rd in a tes . The n , from Eqs . 6 -12 an d 6 -24 ,
V • E = = p f/e re0. (6-28)
Also, s ince E is —VI 7,
r-V = -p
- ± ± l ± = - P / / M o . (6-29)e
o
This i s Poissoris equation for dielect r ics . The expressions involving e r a reval id only in l inear and isot ropic dielect r ics .
Laplace's equation i s aga in .
V 2 K = 0 , (6 -3 0)
w h e n b o t h p s a n d p b are ze ro .
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16 8
Figure 6-6 (a ) Bar e lect re t polar ized uniformly paral le l to i t s axis , (b ) The E field of
the bar e lect re t i s the sam e as tha t of a pai r of c i rcula r pla tes carr yin g unifo rm
surface ch arg e dens i t ies of oppo s i te polar i t ies , (c) Lines of E (black) . Lines of D
are shown in whi te ins ide the e lect re t ; outs ide , they fol low the l ines of E.
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6.11 Summary 169
6.10 ELECTRETS
A n electret i s t he e l ec tr i c equ iv a len t o f a pe rm an en t ma gne t . I n mo st d i
e l ec t r i cs the po la r i za t ion d i sappear s immedia te ly when the e l ec t r i c f i e ld i s
r emoved , bu t some d ie l ec t r i cs r e t a in the i r po la r i za t ion fo r a ve ry long t ime .
Som e po lym ers have ex t r ap o la t ed l i f et imes o f a few hu nd red ye ar s a t r o om
t e m p e r a t u r e .
One way of charging a dielect r ic i s to place i t in a s t rong elect r ic f ie ld
a t h igh t empera tu re . The bound charge dens i ty on the su r f aces then bu i lds
up s lowly as the molecu les o r i en t themse lves . F ree charges a r e a l so depos i t ed
on the su r f aces when spark ing occur s be tween the e l ec t rodes and the d ie l ec
t r ic . Th e fr ee and the bo un d cha rges hav e opp os i t e s igns . Th e samp le is
then coo led to room tempera tu re wi thou t r emoving the e l ec t r i c f i e ld .
The examples in Sees. 6 .2 .1 and 6 .3 .1 apply to a sheet e lect ret wi th
zero f r ee charge dens i ty .
Figure 6-6 shows the f ie ld of a uniformly polar ized bar e lect ret .
6.11 SUMMARY
W he n a dielect r ic ma ter i al i s p lac ed in an elect r ic field, posi t ive and neg at ive
charges wi th in the molecu les a r e d i sp laced , one wi th r espec t to the o ther ,a n d t h e m a t e r i a l b e c o m e s polarized. T h e i n d u c e d d i p o l e m o m e n t p e r u n i t
v o l u m e P i s cal led the electric polarization.
This p ro duc es r ea l acc um ula t ion s o f cha rge tha t we can use to ca lcu la t e
V and E bo th ins ide and ou t s ide the d ie l ec t r i c :
Pb V • P, (6-2)
w h e r e p b i s t he vo lu me den s i ty and o h i s the sur face densi ty of bound charge.
Th e uni t vec tor i s no rm al to the sur face of the dielect r ic an d point s
oufward .Gauss's law can be expre ssed in a form that i s val id for dielect r ics :
V • EPf + Pb
e 0
(6-12)
This i s one o f Ma xwe l l ' s f our funda men ta l e qu a t ion s o f e l ec t ro ma gne t i sm.
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170 Dielectrics: I
In l inear and i so t rop ic d i e l ec t r i cs , t he d ipo le moment per molecu le p
i s p ro po r t ion a l to E , an d
P = e 0 Z , E , (6-18)
w h e r e % e i s a d im ens ion less con s tan t ca l l ed the electric susceptibility.
T h e electric displacement
D = e 0 E + P. (6-15)
= e 0 d + yjE = e0e
rE = eE , (6-19)
w h e r e e r is the relative permittivity and e i s the permittivity. A l s o ,
V • D = p y , (6-16)
o r
f D d a = [pfdx. (6-17)
Inside a dielect r ic ,
P f + P b = r V * r . (6
'2
^
whi le , a t t he in t e r face be tw een a d i e l ec t ri c and a con duc to r ,
o f + o h = o f/e r. (6-26)
Poisson's equation for dielectr ics is
R - V = _Pl±I> m _ p//€f60) (5.29)
e
o
a n d Laplace's equation i s again
V2
V = 0 , ( 6 - 50 )
w h e n p r a n d p h are bo th ze ro .
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Problems 171
As usua l , express ions invo lv ing e r are val id only in l inear and isot ropic
m e d i a .
A n electret i s a p i ece o f d i e l ec t r i c mate r i a l t ha t i s pe rmanen t ly
p o l a r i z e d .
PROBLEMS
6-1E THE DIPOLE MOMEN T p
A sample of diamond has a dens i ty of 3.5 x 1 0 3 ki lograms pe r cub ic me te r and
a p o l a r i z a t i o n o f 1 0 " 7 c o u l o m b p e r s q u a r e m e t e r .
a ) C o m p u t e t h e a v e r a g e d i p o l e m o m e n t p e r a t o m .
b) F ind the ave rage s epa ra t ion be tw een cen te r s o f pos i t ive and nega t ive cha rge .
Ca rbo n has a nuc leus w i th a cha rg e + 6e , sur roun ded by 6 e l ec t rons .
6-2E THE VOLUME AND SURFACE BOUND CHARGE DENSITIES pb
AND ab
Co ns ide r a b lock of d i e l ec t r ic w i th bou nd ch a rge dens i t i e s p b a n d <j b.
S h o w m a t h e m a t i c a l l y t h a t
w h e r e t h e x i s the volume of the die lect r ic and S is i t s surface. In other words , the tota l
ne t bound cha rge i s ze ro .
6-3 BOUN D CHARGE DENSITY AT AN INTERFACE
Show tha t t he bound cha rge dens i ty a t t he in t e r face be tween two d ie l ec t r i c s 1
and 2 that i s crossed by an e lect r ic f ie ld is (P x — P 2 ) • n. Th e po lar iza t ion in 1 is P ,
and i s d i rec t ed into the interface; the polar izat ion in 2 is P 2 a n d p o i n t s away from the
interface. The uni t vector n is normal to the interface and points in the di rect ion from
6-4E COAXIAL LINE
Sho w th at V • E = 0 in the die lect r ic of a coaxial l ine .
Hint: App ly the diver genc e the ore m to a po rt io n of the die lect r ic .
6-5 COAXIAL LINE
The capac i t ance pe r un i t l eng th C of a die lect r ic- insulated coaxial l ine is equal
to that for an a i r- insulated l ine as in Prob. 4-6, mul t ipl ied by e r:
1 to 2.
C =
2n e re 0
In (R 2 /R , )
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17 2
Table 6-2
D i a m et er
B & S Gaug e in Mi l l imeters
20 0.812
22 0.644
24 0.511
26 0.405
28 0.321
30 0.255
32 0.202
34 0.160
For a ce r t a in appl i ca t ion , one requi re s a coax ia l cab le whose ou t s ide d i amete r
mu s t no t exceed abou t 10 mi l l ime te rs . Then R2
< 5 mi l lime te rs . I t s capac i t ance pe r
uni t length must be as low as poss ible . Then R 2 mu st be 5 mil l im eters , and R, as smal l
as poss ible . The cable wi l l be operated a t up to 500 vol ts .
Th e insulat in g mat eria l wi ll be Teflon. Teflon ha s a re la t ive perm it t ivi ty of 2.1.
I t can op era te re l iably in such a cable a t e lect r ic f ie ld inten s i t ies as high as 5 x 10 6 vol t s
per meter . (The ma xi m um e lect r ic f ie ld intens i ty before bre ak do wn is cal led the di
electric strength of a dielectric. In a uniform field the dielectric strength is larger for
th inne r spec imens . A f ilm of Teflon 0.1 mil l imeter th ick ha s a die lect r ic s t reng th of
1 08 vol ts per meter . )
a ) W ha t i s t he mi n im um a l low able d i amete r for t he inne r co nd uc t or?
Remember tha t you can so lve a t ranscendenta l equa t ion by t r i a l and e r ror .
b ) Tab le 6-2 show s the d i amete r s of som e comm onl y ava i l ab le copp er w i res .
F ine r w i res would be imprac t i ca l because they would t end to b reak a t t he connec tors
f ixed to the ends of the cable .
W h a t g a u g e n u m b e r d o y o u s u g g e s t ?
c) What wi l l be the capaci tance per meter wi th the wire s ize you have chosen?
COAXIAL LINE
A coaxia l l ine is com pose d of an in t e rna l cond uc to r o f rad ius R , an d an ex te rna l
conduc tor o f rad ius R 2 , sepa rated by a die lect r ic . Th e die lect r ic should no rm al ly
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Problems 173
f il l a ll t he space be tween the cond uc to rs . Ho wever , du e to an e r ror , t he ou te r rad ius
of the dielectric is R < R2, leaving an a i r space.
a) Assu min g that the die lect r ic i s wel l center ed, what i s the capa ci tan ce w hen
Rj = 1 mm . R2 = 5 m m . R = 4.5 mm, and e
r = 2.56 (polystyrene)'?
Solution: Th e ou te r sur face of the d i e l ec t r i c i s an equip o ten t i a l sur face . Thu s ,
we can imagine that i t i s covered with a thin conduct ing layer . This does not dis turb
the f ie ld. Then we may cons ider that we have two cyl indrical capaci tors , one ins ide
the o the r .
Be tween P , and R we have a capaci tance per uni t length of
2n e re 0
MR R t)
from Prob. 6-5. Between R a n d R 2. the capaci tance per uni t length is
l n ( P 2 / P ) '
These tw o capac i to rs a re in s e r ie s , and the capac i t ance pe r un i t l eng th be tw een
the inne r and the ou te r conduc tors i s g iven by
j _ = l n ( P / P . ) + l n ( P 2 / P )
C 2ne re 0 2n e 0
12n e
0
! P , ) + l n ( P , P ) .I n f P v / P ! ) + l n ( P 2 /P ) I. (2)
C = - ^ . (3)
— l n ( P / P , ) + l n ( P 2 / P )
2n x 8 .85 x 1 0 " 1 2 , ,= = 80.2 pico farad s /me ter . (4)
In4.5 + In —
2.56 0.9
b) Calculate the percent change in capaci tance due to the a i r f i lm.
Solution: Without the a i r f i lm, the capac i t ance would be
2n x 2.56 x 8.85 x 1 0 " 1 2
C = = 88.4 pico farad s /me ter . (5)
In 5
The a i r f i lm therefore decreases the capaci tance by about 10%.
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174 Dielectrics: I
6-7 CHARGED W IRE EMBEDDED IN DIELECTRIC: THE FREE
AND BOUND CHARGES
A conduc t in g wi re ca r ry ing a cha rge / . pe r un i t l eng th i s em bed ded a long the
axis of a c i rcular cyl inder of die lect r ic . The radius of the wire is a ; the radius of the
cyl inder is b.
a) Show that the bound charge on the outer surface of the die lect r ic i s equal
to the bound charge on the inner surface, except for s ign.
b) Show that the net charge a long the axis i s /./e r per uni t length.
6-8 PARALLEL-PLATE CAPACITOR
Th e space be tween the p l a t e s o f a pa ra l l e l -p l a t e capac i to r w i th a p l a t e s epa r a t ion
.s an d a surface are a A i s part ia l ly f i l led wi th a die lect r ic pla te of thickness t < s a n d
a r e a A.
a ) Show tha t
C =
b) Call C 0 t he capac i t ance for t = 0.
Draw a curve of C/C 0 as a function of t s for e r = 3 .
6-9 SPHERE OF DIELECTRIC WITH A POINT CHARGE AT ITS CENTER
A sphere of die lect r ic of radius R conta ins a po in t cha rge Q at i ts center. Set
R = 20 mi l l ime te rs , Q = 1 0 " 9 c o u l o m b , e r = 3.
a ) Dr aw graph s of D, E, V as funct ions of the dis tance r from the center , out to
r = 100 mil l imeters .
Note the d i s cont inu i ty in E at the surface.
b) Now ca lcu la t e a b at the surface from the value of the polar izat ion P j us t
ins ide the surface.
c ) Now use Gauss ' s l aw to expla in the d i s cont inu i ty in E.
6-10 CHARGED DIELECTRIC SPHERE
A dielect r ic sphere of radius R contains a uniform dens i ty of free charge p f.
Show that the potent ia l a t the center i s
(2 e r + l)p fR2
MEASURING SURFACE CHARGE DENSITIES ON DIELECTRICS
Figu re 6-7a shows a s chem at i c d i ag ram of an ins t ru men t tha t ha s been used to
measure charge dens i t ies a t the surface of die lect r ics . See a lso Prob. 17-9,
Th e prob e P i s sup por t ed a few mi l l ime te rs abov e the surface of the s am ple
and can be d i sp laced hor i zon ta l ly and ve r t i cal ly a long th ree or tho gon a l d i rec t ions .
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17 5
Figure 6-7 (a ) Ins t ru me nt for me asu rin g surface charg e dens i t ies on die lect r ics . A
c o n d u c t i n g p r o b e P i s held c lose to the surface of the sample S. The probe i s
connec ted to an e l ec t romete r E t h rough a coaxia l cab le CC. (h) Equiva len t c i rcu i t :
C\ i s t he capac i t ance be tween the top sur face of the s ample and g roun d , C 2 i s thatbe tween the probe and the s ample sur face , and C 3 i s the capaci tance of the coaxial
cab le , p lus the input cap ac i t an ce of the e l ec t romete r . See P ro b . 6 -11 .
As we shal l see , the surface charge on the sample induces a vol tage on P. This vol tage
is read on the e le ct rom eter .
A n electrometer i s a vo l tme te r th a t has an ex t remely h igh re s i s t ance . For e xamp le ,
one type has a res is tance of 1 0 1 4 o h m s . F o r c o m p a r i s o n , c o m m o n d i g i t a l v o l t m e t e r s
have res is tances of the order of 10 6 or 10 7 o h m s . A g o o d a n a l o g v o l t m e t e r d r a w s
abo ut 100 mi cro am per es . I f i t i s ra ted a t 10 vol ts , it s res is tance is 10 10" 4 = 1 0 5 o h m s .
An e l ec t rom ete r d raw s es sen t i al ly zero cur ren t and i s t he re fore su i t ab le fo r m easur ing
the vol tage on a smal l cap aci to r , which is wh at we have h ere .
Figure 6-7b shows the equivalent c i rcui t : we have, effect ively, three capaci tors
connec ted in s e r i e s , and the e l ec t romete r s e rves to measure the vo l t age V.
a) Find the surface charge dens i ty a in terms of Clt C 2 , C 3 , V.
Solution: Let the area of the probe be A. Then the cha rge we a re conce rned
with, at the surface of the dielectric, is Aa . We neglect the fringing field at the edge
of the probe. Part of the f ie ld of Aa goes to the probe , and pa r t goes to g round . Le t
Aa = Q + Q\ w h e r e Q = Ae 0E 2, Q' = Ae^o^i, (D
as in Fig. 6-7b. The f igure a lso shows the charges induced on the conductors . S incethe probe and i ts lead to the e lect rometer are exceedingly wel l insula ted, the net
cha rge induced on them is ze ro .
T h u s V = Q C 3 , and we must f ind a, C 3 b e i n g k n o w n .
Now we have two expres s ions for the vo l t age V b e t w e e n t h e p r o b e a n d g r o u n d :
v=Q=9L.Q. ,2 ,
c 3 c, c 2
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176 Dielectrics: I
So
Subst i tut ing in Eq. 1,
e' = e ^ + ^ . (3)
Aa = 61 1 + £ i + - i ) .C , C
Since 2 = C 3 K ,
a = C, + C , +
C , C , \ K
C , A (5)
b) Expres s <r in terms of the thickness t , of the die lect r ic , the dis tance r 2 b e t w e e n
the probe and the die lect r ic , and e r .
Solution:
and
d = e.eo/4/f!, C 2 = e0/ l / t
2
= i ^ + C3 + € ^ r 2 ) i /
r,
(6)
(7)6-12E VARIABLE CAPACITOR UTILIZING A PRINTED-CIRCUIT BOARD
A printed-ci rcui t board is a sheet of plas t ic , about one mil l imeter thick, one s ide
of which is covered w ith a thin coat i ng of cop per . P art of the copp er can be e tched away ,
l eav ing conduc t ing pa ths tha t s e rve to in t e rconnec t re s i s to rs , capac i to rs , and so for th .
The t e rmina l s o f t hese component s a re so lde red d i rec t ly to the copper .
Figure 6-8 shows a variable capaci tor in which a s l iding conduct ing pla te l ies
under a p r in t ed-c i rcu i t boa rd tha t has been e t ched to g ive a p resc r ibed va r i a t ion of
capac i t ance wi th pos i t ion z .
At the pos i t ion z . the capaci tance is
t Jo
w h e r e t i s the thickness of the plas t ic sheet of re la t ive permit t ivi ty e r .
Set e r = 3, t = 1 mil l ime ter , and f ind y(z) giving the fol lowing v alues of C, wi th
C expressed in farads and x i n me te rs :
a ) l ( T 9 z , b ) K T 8 z 2 .
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17 7
QFigure 6-8 Var i ab le capac i to r made
wi th a p r in t ed-c i rcu i t boa rd PC B a n d
a conduc t ing p la t e P. T h e c a p a c i t a n c e
be tween the t e rmina l s depends on the
pos i t ion of the pla te P and on the
shape of the copper foi l F. See P rob .
•y 6-12.
In both cases , draw a curve of y(z) from z = 0 to z = 100 mi l l ime te rs , showing
the une tched reg ion on the c i rcu i t boa rd .
6-13E EQUIPOTENTIAL SURFACES
A l iquid die lect r ic is s i tua ted in an e lect r ic f ie ld. A thin c on duc t ing sheet carry ingzero net charge is int roduced into the die lect r ic so as to coincide wi th an equipotent ia l
surface.
a) Is the e lect r ic f ie ld di s tu rbe d?
b) W ha t is the value of a s on the surfaces of the sheet a t a point where the e lect r ic
d i sp lacement i s D]
6-14E NON-HOMO GENEOUS DIELECTRICS
Show tha t a non-homogeneous d i e l ec t r i c can have a vo lume dens i ty o f bound
charge in the absence of a free charge dens i ty.
6-15 FIELD OF A SHEET OF ELECTRONS TRAPPED IN LUCITE
W hen a b lock of insu la t ing ma te r i a l such a s Luc i t e is bo m bar ded wi th h igh-ene rgy e l ec t rons , t he e l ec t rons pene t ra t e in to the ma te r i a l and remain t rapped ins ide .
In one pa r t i cu la r i ns t ance , a 0 .1 mic roa mp ere b eam b om bar ded an a rea of 25 squa re
cen t ime te rs o f Luc i t e (e r = 3.2) for f second, and essent ia l ly a l l the e lect rons were
t rap ped abo ut 6 mi l l ime te rs be low the sur face in a reg ion abo ut 2 mi l l ime te rs th i ck .
Th e b lock was 12 mi l l ime te rs th i ck .
In the fol lowing calculat ion, neglect edge effects and assume a uniform dens i ty
for the t rapped elect rons . Assume also that both faces of the Luci te are in contact wi th
g r o u n d e d c o n d u c t i n g p l a t e s .
S h o w t h a t
a ) i n the reg ion where the e l ec t rons a re t rapped .
p, = —2.000 x 1 0 " 2 c o u l o m b / m e t e r 3 .
p h = 1.375 x 1 0 " 2 c o u l o m b / m e t e r3 ;
b) in the neu tra l region,
D„ = - 2 . 0 0 0 x 1 0 " 5 c o u l o m b / m e t e r 2 ,
P„ = —1.375 x 1 0 "5
c o u l o m b m e t e r2
.
E„ = - 7 . 0 6 2 x 1 0 5 vol t s /me te r ,
V„ = 7.062 x 10 5.v - 4,237 vo l ts ;
c) at the surfaces.
a b = — 1.375 x 1 0 " ? c o u l o m b / m e t e r 2 ;
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17 8 Dielectrics: I
d) in the cha rged reg ion ,
dDc
~dx~ - 2 . 0 0 0 x 102
c o u l o m b / m e t e r3
.
- 2 . 0 0 0 x 1 0 " 2. v c o u l o m b / m e t e r 2 .
- 7 . 0 6 2 x 1 0 8 v o l t s m e t e r 2 ,
Ec
=
d2
Vc
die 2
7.068 x 1 0 8 x vo l t s /me te r ,
= 7.062 x 1 0 8 v o l t s / m e t e r 2 .
V c = 3.531 x 1 08
x2
- 3884 volts .
e) Draw curves of D, E, V as funct ions of the dis tance x t o the midplane , f rom
x = —6 to x = 6 mil l im eters .
f) Show that the s tored energy is 1.88 x 1 0- 4
j o u l e .
g) Is ther e any dan ger tha t the block wil l exp lod e?
6-16 SHEET ELECTRET
A sheet e lect re t i s polar ized in the di rect ion normal to i t s surface.
Draw a f igure showing the po la r i za t ion P . t he sur face cha rge dens i t i e s +a h
and -<r fc , as wel l as the e lect r ic f ie ld intens i ty E and the e lect r ic displacement D, both
ins ide and outs ide the e lect re t .
6-17 RELATION BETWEEN R AND C FOR ANY PAIR OF ELECTRODES
W he n a para l le l -p la te cap aci to r i s f il led wi th a die lect r ic wh ose re la t ive p er
mittivity is e r , i t s capaci tance is C. If the die lect r ic i s s l ight ly conduct ing, i t s res is tance
is P.
S h o w t h a t RC = e re 0/a , w h e r e o is t he condu c t iv i ty .
Th is rule appl ie s to any pair of e lect ro des sub me rge d in a me diu m, as long as
the conduc t iv i ty o f t he e l ec trodes i s much l a rge r than tha t o f t he m ediu m.
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CHAPTER 7
DIELECTRICS: II
Continuity Conditions at an Interface, Energy Density
and Forces, Displacement Current
7.1 CONTINUITY CONDITIONS AT THE INTERFA CE
BETWEEN TWO MEDIA7.1.1 The Potential V
7.1.2 The Normal Compo nent of D
7.1.3 The Tangential Com ponent of E
7.1.4 Bending of Lines of Force
7.1.5 Examp les: Point Charge Near a Dielectric, Sphere of Dielectric
in an Electric Field
7.2 POTENTIAL ENERGY OF A CHARG E DISTRIBUTION,
ENERGY DENSITY
7.3 FORCES ON CONDUCTORS IN THE PRESENCE
OF DIELECTRICS
7.3.1 Examp le: Parallel-Plate Capacitor Immersed in a LiquidDielectric
7.4 ELECTRIC FORCE S ON DIELECTRICS
7.4.1 Examp le: Curve Tracer
7.5 THE POLARIZATION CURRE NT DENSITY dP/dt
7.6 THE DISPLACEM ENT CURRE NT DENSITY c D [ d t
7.6.1 Examp le: Dielectric-Insulated Parallel-Plate Capacitor
7.7 FREQUENCY AND TEMPERATURE DEPENDENCE,
ANISOTROPY
7.7.1 Examp les: Water, Sodium Chloride, Nitrobenzene, Compo unds
of Titanium
7.8 SUMMARY
PROBLEMS
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This chap te r wi l l comple te our s tudy o f d i e l ec t r i cs .
We sha l l f i r s t d i scuss the impor tan t con t inu i ty cond i t ions tha t app ly
a t the in t e r f ace be tween two media . These cond i t ions concern the po ten t i a l
V, t h e t a n g e n t i a l c o m p o n e n t o f E , a n d t h e n o r m a l c o m p o n e n t o f D .
Then we must re turn to the energy stored in an elect rostat ic f ie ld .
Th i s s to r ed energy wi l l g ive us the fo rces exer t ed on conduc to r s immersed
in non-conduct ing l iquids. I t wi l l a lso give us the forces act ing wi thin the
d ie l ec t r i cs themse lves .
I f the elect r ic f ie ld is a funct ion of the t ime, then the mot ion of charge
wi th in the molecu les o f the d ie l ec t r i c g ives a po la r i za t ion cur r en t . The
d i sp lac em ent cu r r en t i s equ a l to the po la r i za t io n cur r e n t p lus ano the r
current that exists whenever the elect r ic f ie ld var ies wi th t ime, even in a
v a c u u m .
7.1 CONTINUITY CONDITIONS A T THE
INTERFACE BETWEEN TWO MEDIA
T h e q u a n t i t i e s V, E, and D must sa t i s fy ce r t a in boundary cond i t ions .
7.1.1 THE POTENTIAL V
A t t h e b o u n d a r y b e t w e e n t w o m e d i a , V must be con t inuous , f o r a d i scon
t inui ty wo uld im ply an inf inite ly large elect r ic field in tensi ty , which is
phys ica l ly imposs ib le .
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7. / Continuity Co nditions at the Interface Between Two Media 181
Th e pote nt i a l i s no rm al ly set equ al to zero at inf inity if the ch arg e
d i s t r ib u t ion i s o f f in i te ex ten t . T he po te n t i a l i s con s tan t th ro ug ho u t any
conduc to r , a s long as the e l ec t r i c charges a r e a t r es t .
7.1.2 THE NORMAL COMPONENT OF D
C o n s i d e r a s h o r t G a u s s i a n c y l i n d e r d r a w n a b o u t a b o u n d a r y s u rf a ce , a s
in Fig . 7-1. Th e end faces of the cyl inde r are paral le l to the bo un da ry an d
arbi t rar i ly c lose to i t . The boundary car r ies a f ree sur face charge densi ty a f.
If the are a S i s sm al l , D a n d a s do not vary signif icant ly over i t and, according
to Gauss 's law, the f lux of D em erg ing f rom th e f la t cyl inder i s equ al to the
c h a r g e e n c l o s e d :
(D nl - D„ 2)S = OjS. (7-1)
The only f lux of D i s through the end faces, s ince the area of the cyl indr ical
su r face i s a rb i t r a r i ly smal l . T hu s
Dm - Dm = <>{• (7-2)
Figure 7-1 Gauss ian cy l inde r on the in t e r face be tween two media 1 and 2 . The
difference D nl — D n2 be tween the normal co mp on en ts of D is equ al to the surface
dens i ty of free charge a f.
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182 Dielectrics: II
At the boundary be tween two d ie l ec t r i c med ia the f r ee su r f ace charge
dens i ty o f i s genera l ly ze ro and then D„ i s co n t in uo us ac ross the b ou nd ary .
On the o ther han d , if t he bo un da ry i s be twe en a co nd uc to r a nd a d i e l ec t r ic ,and i f the elect r ic f ie ld is constant , D = 0 i n t h e c o n d u c t o r a n d D„ = a f in
the dielect r ic , a f be ing the fr ee charg e dens i ty on the su r f ace o f the con du c to r .
7.1.3 THE TANGENTIAL COMPONENT OF E
Th is bou nd ar y co nd i t io n fol lows f rom th e fact that the l ine integra l of E • dl
aro un d a ny c losed pa t h i s ze ro for e l ec t ros t a t i c fi elds. Cons id er the pa th
shown in Fig . 7-2, wi th two sides paral le l to the boundary, of length L, and
arb i t r a r i ly c lose to i t. Th e o ther two s ides a r e perp end icu la r to the bo un dar y .In calculat ing this l ine in tegral , only the f i r s t two sides of the path are impor
tant , s ince the lengths of the other two approach zero. I f the path is smal l
e n o u g h , E, does not vary signif icant ly over i t and
E nL- E,2L = 0, (7-3)
o r
E n = E t2. (7-4)
T h e t a n g e n t i a l c o m p o n e n t o f E is t h e re f o r e c o n t i n u o u s a c r o s s th e b o u n d a r y .
Figure 7-2 C losed p ath of inte grat ion cro ss ing the interface between tw o media 1
and 2 . Wha teve r be the sur face cha rge dens i ty a f th e tangential c o m p o n e n t s o f E
on ei ther s ide of the interface are equal ; £ ( 1 = £ ( 2 -
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7.1 Continuity Conditions at the Interface B etween Two Media 183
I f t he bou nd ar y l ie s be tw een a d i e l ec t r i c and a con duc to r , t hen E = 0
i n t h e c o n d u c t o r a n d E, = 0 in both media. With sta t ic charges, E is therefore
nor ma l to the su rf ace o f a co nd uc to r .
BENDING OF LINES OF FORCE
I t f o l lows f rom the boundary cond i t ions tha t t he D a n d E v e c t o r s c h a n g e
direc t ion at the bo un da ry be twe en tw o dielect r ics . In Fig . 7-3, using Eq. 7-2
wi th a* — 0.
D : cos (f = D 2 co s 0 2, (7-5)
o r
er l
e0
£
i cos 0 , = e r 2 e 0 £ 2 cos 0 2 (7-6)
and, f rom Eq. 7-4,
Ei si n 0 l = E 2 sin d2. (7-7)
Figure 7-3 Lin es of D or of E cro ssin g the interface betw een two me dia 1 an d 2.
The l ines change di rect ion in such a way that e r l t a n 8 2 = e r2 t an 0 , .
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184 Dielectrics: II
Then, from the last two equations,
tan 0!
tan 0, In (7-8)
7.1.5
The larger angle from th e nor mal is in the mediu m with the larger relative
permittivity.
EXAMPLES: POINT CHARGE NEAR A DIELECTRIC, SPHERE OF
DIELECTRIC IN AN ELECTRIC FIELD
F i g u r e 7-4 s h o w s tw o e x a m p l e s of the b e n d i n g of l ines of force at the b o u n d a r y
b e t w e e n tw o dielect r ics . The l ines of D are b r o k e n bu t cont inuous , s ince l i nes of D
1
(a)
Figure 7-4 (a) Lines of D and e q u i p o t e n t i a l s for a p o i n t c h a r g e in air n e a r a
dielect r ic . They are not s h o w n n e a r th e cha rge because they get too close together .
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18 5
Figure 7-4 (cont.) (b) Lines of D and equipo ten t i a l s nea r a d i e l ec t r i c sphe re s i tua t ed
in a i r in a uniform elect r ic f ie ld. In both cases , the l ines of D are indicated by
a r rows , and the equipo ten t i a l sur faces a re gene ra t ed by ro t a t ing the f igures a round
the hor i zonta l ax i s .
terminate only on free charges . Some l ines of E e i ther or iginate or terminate a t the
interface, according as to whether a b i s pos i t ive or negat ive .In Fig. 7-4b, the lines of D c rowd in to the sphe re , showing tha t D is larger
ins ide than ou t s ide . However , t he equipo ten t i a l s spread ou t i ns ide , and E i s weaker
ins ide than outs ide . The dens i ty of the l ines of E is lower ins ide than outs ide , s ince
some of the l ines of E coming from outs ide terminate a t the surface.
No te tha t , in the case of Fig. 7-4b, the field ins ide the sph ere is unifo rm. No te
also that the field ou ts ide is har dly dis turb ed a t dis tances larger than on e rad ius
from the surface.
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186 Dielectrics: II
7.2 POTENTIAL ENERGY O F A CHARGE
DISTRIBUTION, ENERGY DENSITY
To calculate the potent ia l energy associated wi th an elect r ic f ie ld in dielect r ic
ma ter ia l , we refer to the paral le l -p late ca pa ci t or as in Sec. 4 .3 .
W e now h av e a dielect r ic- f i l led ca pa ci to r as in Fig . 6- 5. I f we a dd
c h a r g e s dQ un t i l t he top p la t e ca r r i es a charge Q an d is a t a po ten t ia l V,
we must expend an energy
-QV = - — = i2
V 2 C 2^ ere
0/F
~QV = - ^ = , Q 2 — , (7-9)
2 \A l e r e 0 AsA, (7-10)
= \DEsA, (7-11)
wh ere Z) i s the elect r ic dis pla ce me nt (Sec. 6 .5) , E is the electr ic f ield intensity,
a n d sA i s the volume of the dielect r ic .
T h e energy density i s now DE/2 or, if the dielectr ic is l inear, as is
usua l ly the case , € re 0E2
/2 . Th e po ten t i a l energy i s t hen g iven by the in t egra l
of e re 0E2
/2 eva lua ted over the vo lume occup ied by the f i e ld ,
2
7.3 FORCES ON CONDUCTORS IN THE
PRESENCE OF DIELECTRICS
Th e fo rces be tween con du c to r s in the p resenc e o f d i e l ec t r i cs can be ca l cu la t ed b y the me tho d o f v i r tua l wo rk th a t we used in Sec . 4 .5 .
W h e n t h e c o n d u c t o r s a r e i m m e r s e d i n a liquid dielect r ic , the forces
a re a lways found to be smaller than those in ai r by the factor e r if the charges
are the same in bo th cases . They a r e larger than in ai r by the factor e r if the
electric fields ( and hence the voltages) are the same.
The case of sol id dielect r ics wi l l be i l lust rated in Probs. 7-9 and 7-10.
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7.4 Electric Forces on Dielectrics 187
EXAMPLE: PARALLEL-PLATE CAPACITOR IMMERSED
IN A LIQUID DIELECTRIC
We found in Sec. 4.5.2 that the force between the pla tes of an a i r- in sula ted paral le l -
p l a t e capac i to r i s e 0E2A/2, o r (a 2/2e 0)A , w h e r e a i s the surface charge den s i ty a nd
A i s the area of one pla te . We can perform a s imilar calcula t ion for a paral le l -pla te
cap aci to r imm ersed in a l iquid die lect r ic . Th e force per uni t area is equ al to the
energy den s i ty (Sec. 4.5) . Th en th e mech anic al force hold ing the pla tes ap art i s
F,„ = e,f E 2
A.(7-12)
For a given electric field s trength E, or for a given voltage difference between the
plates, the force is larger than in a i r .
Also.
D a 2
F m = A = A,2e re 0 2 e r e 0
(7-13)
1 Q 2
(7-14)e r 2e„A
a n d , for given charges +Q and — Q on the plates, the force is smaller than in a i r .
7.4 ELECTRIC FORCES ON DIELECTRICS
W he n a piece of dielect r ic i s p lace d in an elect r ic f ie ld , i ts d ipo les are subje cted
to e l ec t r i c fo rces and to rques .
I f the fie ld is unifo rm, a nd i f th e mo lecu les hav e a pe rm an en t dipo le
m o m e n t p. each d ipo le exper i ences a to rque .
T = p x E , (7-15)
that tends to align i t with the f ield, but the net force is zero.
I f the f ie ld is non-uniform, then the forces on the two charges of a
dipo le are un eq ua l , an d the re is a lso a net force.Le t us cons ider a d i e l ec t r i c whose molecu les do no t h a v e a p e r m a n e n t
dipole moment . We can f ind a general expression for the elect r ic force per
uni t volu me in a no n-u nif orm f ield by con side r ing the dielect r ic- f i lled
cyl indr ical capaci tor of Fig . 7-5.
Th e elect r ic field is rad ial , and i t s m ag ni tu de v ar ies as \jr. The ne t
elect r ic force on a dipole i s inward, because the inward force on — Q is
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18 8
Figure 7-5 Dielectr ic-f i l led cyl indrical capaci tor .
l a rger than the ou tward fo rce on + Q. This net inward force is
dE
f= QE r - QE r + dr = Q ^ s ,
dE
(7-16)
(7-17)
w h e r e p = Qs is t h e d i p o l e m o m e n t .
Therefore, i f there are N d ipo les per un i t vo lume, the e l ec t r i c fo rce
per un i t vo lume i s
dE dEF
>= Np
Hr-= P
*>(7-18)
clE d(e , - = (€ r - l ) e 0 - ( E 2 / 2 ) , (7-19)
1 - £ ) ^ ( e r e 0 E 2 / 2 ) . (7-20)
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7.4 Electric Forces on Dielectrics 189
Mo re generally, the electric force per unit volu me is pro po rti on al to
the gradie nt of the energy dens ity:
Th e dielectric tend s to move t o where the field is strong est. In the abo ve
expression, E is the electric field intensity inside the dielectric, of course.
7.4.1 I EXAMPLE: CURVE TRACER
A curve t race r is an e lec t ro-mechanica l dev ice tha t d raws curves ; the device is
u n d e r th e c o n t r o l of a c o m p u t e r . The curves are d r a w n on a shee t of p a p e r t h a t
must stay rigidly fixed to the pla t en . One way of h o l d i n g th e p a p e r is s h o w n in
Fig. 7-6. Para l l e l w i res are e m b e d d e d in the plas t i c p l a t en about tw o mil l ime te rs
a p a r t and c h a r g e d to plus and m i n u s 300 vol ts . This gives a h i g h l y i n h o m o g e n e o u s
field at th e surface, with lines of force converg ing toward th e cha rged wi res . Thus
V E 2 ins ide the p a p e r has a ver t i ca l component d i rec t ed in to th e pla t en .
Figure 7-6 Curve t race r . The elect r ic f ie ld produced by the cha rged wi res embedded
in the pla t en a t t rac t s the elect r ical ly neutra l sheet of p a p e r t o w a r d th e pla t en .
( 7 - 21 )
Platen
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190 Dielectrics: II
7.5 THE POLARIZA TION CURRENT DENSITY dP/dt
W he n a die lectr ic is pla ced in an electr ic f ield th at is a func tion of t im e, the
mot ion o f the bound charges g ives a polarization current. We can f ind the
po la r i za t ion cur r en t dens i ty J h as fol lows.
F o r a n y v o l u m e i bounded by a su r f ace S, t he r a t e a t which bound
charg e f lows ou t th rou gh S must be equa l to the r a t e o f decreas e o f the bo un d
charge wi th in S (Sec. 5.1.1):
£ J „ d a = --jp„dx. (7-22)
Using the d ivergence theorem on the l e f t and subs t i tu t ing the va lue o f p h
on the r ight from Eq. 6-2,
" . " I D
f V • J„ dx = - f V • P dx = f V • — dx . (7-23)
Jr |( Ji Jr (!(We have pu t the c/ct under the in t egra l s ign because i t i s immater i a l whe ther
the der iv at ive w i th respec t to the t ime is cal cula ted f ir st or last . S ince x is
any vo lum e, the in t eg rand s must be equ a l and the po la r i za t ion cur r en t
densi ty i s
(7-24)
W e cou ld have added a con s tan t o f in t eg ra t ion ind epe nde n t o f the coo r
d ina tes , bu t such a cons tan t would be o f no in t e r es t . We have used the
par t ia l der ivat ive wi th respect to t b e c a u s e P can be a funct ion both of the
t ime and o f the space coord ina tes x, y. z.
7.6 THE DISPLACEMENT CURRENT DENSITY cD/ct
T h e displacement current density is
, 1 ) d 8 cP— = - ( e 0 E + P) = - e 0 E + — , (7-25)dt dt ct ct
w h e r e dP/dt i s t he po la r i za t ion cur r e n t den s i ty o f the p rev io us sec t ion .
Note the f i r s t term: i t can exist even in a vacuum.
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7.7 Frequency anil Temperature Dependence, Anisotropy 191
7.6.1 EXAMPLE: DIELECTRIC-INSULATED PARALLEL-PLATE CAPACITOR
Let us ca lcu la t e the d i s p l a c e m e n t c u r r e n t in a die l ec t r i c - insu la t ed pa ra l l e l -p l a t ec a p a c i t o r .
C o n s i d e r a c u r r e n t / f lowing th rough a c a p a c i t o r as in Fig. 7-7. We a s s u m e
that pos i t ive charge f lows from lef t to r ight , that Q = 0 at t = 0, and t ha t edge
effects are negl igible .
O n th e lef t -hand s ide of t he capac i to r . Q increases and dQ/dt = I. On the r ight ,
a cha rge of e q u a l m a g n i t u d e bu t oppos i t e s ign bu i lds up and, aga in , dQ/dt = /.
N o w Q = Aa. D = a = Q A. w h e r e A is the a r e a of one plate , and a is the
surface charge dens i ty.
T h e d i s p l a c e m e n t c u r r e n t in the c a p a c i t o r is
dD
~dt
dQ
dt
I. (7-26)
T h e d i s p l a c e m e n t c u r r e n t A(dD/dt) is c o m p o s e d of two parts , f i rs t
d (a\ 1 dQA±(e
0E)
d DA — \ —
dt w dt dt(7-27)
a n d th e p o l a r i z a t i o n c u r r e n t
dP dA — =A-(D
dt dt
\\da dQ
dt '(7-28)
Q -Q
1 = dQ/dt I = dQ/dt
Figure 7-7 Curren t f lowing th rough a
capac i to r .
7.7 FREQ UENC Y AND TEMPERA TV REDEPENDENCE, ANISOTROPY
There are three basic polarization processes, (a) In induced or electronic
pola rizat ion, t he center of negativ e charg e in a molecul e is displaced , relative
to the center of positive charge, when an external field is applied, (b) In
orientational polarization, molecules with a perm anen t dipole mo men t
tend to be aligned by an externa l field, the magn itu de of the susceptibility
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192 Dielectrics: II
be ing inverse ly p rop or t i on a l to the t emp era tu re , (c ) F ina l ly , ionic p o l a r i z a t i o n
occurs in ionic crystals: ions of one sign may move, wi th respect to ions of
the other s ign, when an external f ie ld is appl ied.
Fo r a g iven ma gn i tud e o f E . bo th the ma gn i tud e and the phase o f
p are funct ions of the frequency, f ir s t bec aus e of the va r iou s p ola r iz at io n
processes tha t come in to p lay as the f r equency changes , and a l so because
of the ex i s t ence o f r esonances . Thus , s ince e r i s a funct ion of the f requency .
e r i s s t r ic t ly def inable only for a pure sin uso ida l wav e. See Ta ble 6-1.
In many subs tances the r e l a t ive permi t t iv i ty decreases by a l a rge
fac to r as the t em pe ra tu re i s l owered th rou gh th e f reez ing po in t .
An i so t ro py i s an o th er dep ar tu re from the idea l d i e l ec t r i c behav io r .
Cry stal l in e sol ids co m m on ly hav e di f ferent dielect r ic pro pe r t ie s in di f ferent
c rys t a l d i r ec t ions because the charges tha t cons t i tu t e the a toms o f the c rys t a la r e ab le to move more eas i ly in some d i r ec t ions than in o ther s . The po la r i
z a t i o n P i s the n paral le l to E only whe n E is a lo ng cer tain prefer red di re ct io ns.
7.7.1 EXAMPLES: WATER, SODIUM CHLORIDE, NITROBENZENE,
COMPOU NDS OF TITANIUM
Water has a re la t ive permit t ivi ty of 81 in an e lect ros ta t ic f ie ld, and of about 1.8 a t
opt ical f requencies . Th e large s ta t ic value is a t t r ib uta ble to the ori ent a t io n of the
pe rm anen t d ipo le mom ent s , bu t t he ro t a t iona l i ne r t i a o f t he molecu les is muc h too
large for any s ignif icant response a t opt ical f requencies .
Similar ly, the re la t ive permit t ivi ty of sodium chloride is 5.6 in an electrostatic
f ie ld a nd 2.3 a t opt ical f requen cies . The larger s ta t ic value is a t t r ibu ted to ionic
motion, which again is imposs ible a t high frequencies .
In nitrobenzene, e r fa lls f rom ab ou t 35 to abo ut 3 in cha ngin g from the l iqu id
to the sol id s ta te a t 279 kelvins . In the sol id s ta te , the permanent dipoles of the
ni t robenzene molecules are f ixed r igidly in the crys ta l la t t ice and cannot rota te
un de r the inf luence of an external f ie ld.
Ceramic capac i to rs u t i l i ze va r ious compounds of titanium as die lect r ics . These
co mp ou nd s a re used because of the i r l a rge re l a tive pe rmi t t iv i t i e s rang ing up to
10.000. However, e r var i e s w i th the t empera ture , w i th the appl i ed vo l t age , and wi th
the ope r a t ing f requency .
7.8 SUMMARY
A t t h e b o u n d a r y b e t w e e n t w o m e d i a , b o t h V a n d th e t a n g e n t i a l c o m p o n e n t
o f E a r e con t inuous , bu t the d i f f e r ence be tween the normal componen t s o f
D i s equ al to the sur face de nsi ty of f ree ch arg e a f.
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Problems 193
The potential energy stored in an electric field can be calculated from
th e energy density ere
0£
2
,'2. Hence the force per unit area exerted on a con
ductor submerged in a liquid dielectric is* e r e 0 £2 / 2 .
The force per unit volume on a dielectric is
1 - ^ V( ere
0£
2
/2). (7-2/)
The displacement current density at a point is
( D d
r i dP (7-25)
where dP/dt is the polarization current density.
Because of the nature of polarization processes, er
is a function of
f r equency . For most commonly used dielectrics, however, er
changes very
l i t t le, even when th e frequency chan ges by man y orde rs of magni tud e (Table
6-1). In anisotropic media. P is parallel to E only when E is along certain
preferred directions.
PROBLEMS
7-lE CONTINUITY CONDITIONS AT AN INTERFACE
Discuss the continuity conditions at the surface of the dielectric cylinder of Prob.
6-7.
7-2E CONTINUITY CONDITIONS AT AN INTERFACE
Discuss the continuity conditions at the surface of the dielectric sphere of Prob.
6-9.
7-3E ENERGY STORAGE IN CAPACITORS
A one-m icrof arad capa cito r is char ged to a poten tial of one kilovolt.
How high could you lift a one-kilogram mass with the stored energy, if you
could achieve 100°,, efficiency'?
7-4E ENERGY STORAGE IN CAPACITORS
The maximum stored energy per cubic meter in a capacitor is ere0a2
/2, where a
is the dielectric strength of the dielectric. The dielectric strength is the max imum electric
field intensity before rupture. See Prob. 6-5.
It is suggested that a small vehicle could be propelled by an electric motor fed by
charged capacitors. Comment on this suggestion, assuming that the only problem is
one of energy storage.
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194 Dielectrics: II
A good die lect r ic to use would be Mylar , which has a die lect r ic s t rength of
1.5 x 10 s vol ts per meter and a re la t ive permit t ivi ty of 3.2.
Note that the use of a die lect r ic increases the capaci tance by e r and a l so pe rmi t sthe use of a higher vol tage . Th e die lect r ic s t reng th of a i r is only 3 x 10
6
vol t s pe r m e te r .
E, D. a, AND IF , FOR THE THREE CAPACITORS OF FIG. 7-8
Figure 7-8 shows three capaci tors . In a l l three the e lect rodes have an area S and
are se par ated by a dis ta nce .s. In A, the die lect r ic i s a i r . In B, the die lect r ic ha s a
re la t ive permit t ivi ty e r. In C, we have the same die lect r ic , plus a thin f i lm of a i r .
a) Find E. D, the surface charge dens i ty a on the e l ec t rodes , and the ene rgy
dens i ty W t for A and B.
Express your resul ts for B in terms of those found for A.
Solution : Fo r capa c i to r A ,
£ A = F / s , (1)
Z)A = e 0 £ A = e 0V/s, (2)
a
= DA
=eoV/s, (3)
W 1 A = \toEl = {-e 0V*/s2. (4)
Figure 7-8 Thr ee ca pa ci t or s : A. wi th out die lect r ic ; B, wi th die le ct r ic ; C, wi th
dielect r ic and an a i r f i lm. See Prob. 7-5.
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Problems 195
For capacitor B.
£„ = Vs = £ A . . (5)
DK = e r e 0 £ B = ere0V/s = erDA. (6)
ffB = DB
= ere
0F x = e
ra
A. (7)
W 1 B= i e r e 0 E
2
= 1 e ,e 0 K2 /x 2
= e,W 1 A . (8)
b) Find £, D, H7
, for the dielectric in C, and <r on the left-hand electrode.
Express your results in terms of the quantities found for capacitor A.
Solution: Since the air film in capacitor C is thin, we m ay set
£„ = V/s = £A. (9)
Then D, a on the left-hand electrode, and W l are the same as for capacitor B:
Da= e r C A . (10)
< Ta
= erf f
A, (11)
Wia = e,W lA. (12)
c) Find £, D, W, for the ai r film of capacitor C. and a on the right-hand electrode.
Express again your results in terms of the corresponding quantities found for
capacitor A.
Solution: The electric displacement D in the air film is the same as in th e di
electric, since there are no free charges between th e plates. Then
DCa = M>A- (13)
ffc = Da, = e r D A= e
rff
A, (14)
Eca = DcJ*o = t,D A/e 0 = e ,£ A , (15)
WJ o = I e o £ i = J e 0 6 2 £ 2 = e r
2
rV 1 A . (16)
BOUND SURFACE CHARGE DENSITY
Find the bound surface charge densities on the dielectric of capacitor B of
P r o b . 7-5 in terms of V/s.
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796 Dielectrics: II
7-7E EXAMPLE OF A LARGE ELECTRIC FORCE
O n e a u t h o r c l a i m s t h a t he can attain fields of 4 x 107 vol t s per mete r ove r a
2.5 mil l imeter gap in pur i f i ed n i t robenzene (« r = 35), an d t h a t th e resul t ing e lect r icforce pe r uni t area on the e lec t rodes is t h e n m o r e t h a n tw o a t m o s p h e r e s .
Is th e force real ly that large?
O n e a t m o s p h e r e e q u a l s a b o u t 105 pasca l s .
7-8 PERPETUAL-MOTION MACHINE
In P rob . 7-5 we s h o w e d t h a t , for the c a p a c i t o r C, the ene rgy dens i ty in the
dielect r ic an d in the air film are. respect ively,
e re 0 V2 e 2e 0 V
2
— — an d — r .2 s~ 2 s
N o w t h i s is di s turb ing because it a p p e a r s to m e a n t h a t the net force on thec a p a c i t o r is not z e r o . The forces on the capac i to r p l a t e s are
e re 0 V2
62
e 0 V2
r - 5 and =- S.2 s
2
2 s2
Is there a force on the dielect r ic? Well , the E ins ide is uni form and VE2
= 0, which
m e a n s t h a t the force dens i ty is ze ro . So it seems tha t the net force on the c a p a c i t o r is
e re 0 V2
(Cr-D-f-j-S.
W i t h M y l a r (e r = 3.2). s = 0.1 mil l ime te r . S = 1 m e t e r 2 , and V = 1 ki lovol t .
the force is a b o u t 3 x 10"1 n e w t o n s ! O ne could the re fore p rope l a large vehicle in
definitely with a set of capac i to rs l i ke th i s , fed by a smal l ba t t e ry supply ing a vol t age V
a t ze ro cur ren t ! Where have we e r r e d ?
Him: ( i )The va lues g iven above for the forces on the e lec t rodes are cor rec t ,
(ii) Inside th e dielect r ic , there is no force, (iii) Th e forces on the faces of the dielect r ic
a r e not z e r o , h o w e v e r . R e m e m b e r t h a t C o u l o m b ' s law appl i e s to any net a c c u m u l a t i o n
of charge, i r respect ive of the presence of m a t t e r .
7-9 SELF-CLAMPING CAPACITOR
A ce r t a in capac i to r is formed of two a l u m i n u m p l a t e s of a rea A, sepa ra t ed by a
sheet of dielect r ic of t h i cknes s /. and c o n n e c t e d to a s o u r c e of vol t age V.
For va r ious reasons , it is imposs ib le to a p p l y a force of m o r e t h a n a few t ens of
n e w t o n s to press the two pla t e s toge the r mechanica l ly . S ince ne i the r the pla t e s nor
the die lect r ic are perfectly flat, th e pla t e s are spaced by a di s t ance t, plus a good frac
t ion of a mil l ime te r . Thi s is highly ob jec t ionable because th e c a p a c i t a n c e m u s t be as
large as poss ib l e .
Y o u are a s k e d to i nves t iga te whe the r the elect ros ta t ic force of a t t r a c t i o n on
the pla tes might not be sufficiently large to reduce the air f i lm thickness to a negl igible
va lue .
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Problems 197
Se t A = 4.38 x 1 0 " 2 squa re me te r , t = 0.762 mil l imeter . e r = 3.0. V =
60 ki lovol ts .
See a l so the nex t p rob lem.
7-10 ELECTROSTATIC CLAMPS
Electros ta t ic c lamps are somet imes used for holding work pieces whi le they
a re be ing machined . They u t i l i ze an insu la t ed conduc t ing p la t e cha rged to s eve ra l
tho usan d vo l t s and co vered w i th a th in insu la t ing shee t . The w ork p iece is p l aced on
the shee t and grou nde d .
On e pa r t i cu la r t ype ope ra t e s a t 3000 vo l ts and i s adve r t i s ed a s hav in g a ho ld ing
power of 2 x 1 0 5 pasca l s .
The insu la tor i s Myla r (e P = 3.2).
a) Sup po se f irst th at th ere is a f ilm of a i r on e i ther s ide of the M yla r .
Use the resul ts of Prob . 7-5 to show tha t the My lar f ilm is 15 mic rom eter s th ick.
b) Now calculate the electric field intensity in the air fi lm. You should find
over 6 x 1 0 s vol ts per meter , which is over 200 t imes the die lect r ic s t rength of a i r
(Prob. 7-4) . Sparking could occur in the a i r f i lm, and the die lect r ic could deter iorate .
Myla r ope ra t e s s a t i s fac tor i ly under these condi t ions , however .
c) Now assume that the a i r f i lm is replaced by a f i lm of t ransformer oi l wi th a
high die lect r ic s t reng th an d a perm it t ivi ty th at i s not mu ch different from that of the
M y l a r .
W ha t i s t he th i cknes s o f t he My la r now ?
7-11 CALCULATING AN ELECTRIC FORCE BY THE METHOD
OF VIRTUAL WORK
Figure 7-9 shows a pair of paral le l conduct ing pla tes immersed in a die lect r ic .
The lower pla te is f ixed in pos i t ion and grounded; the upper one can s l ide horizontal ly
and is m ain ta in ed a t a fixed vol tage V.
Use the me tho d of vi r tual work (Sec. 4.5) to calcula te the hori zon tal force on
the upper p l a t e when V = 1000 vol ts , s = 1 mil l ime ter . / = 100 mil l im eters . e r = 3.0.
Figure 7-9 P air of con du ct in g pla tes im me rsed in a die lect r ic . Th e lower one is
f ixed, whi le the up per o ne can s l ide horizo ntal l y. I t is poss ible to f ind the
hor i zo nta l fo rce on the upper p l a t e by the me tho d of v i r tua l work . See P r ob . 7 -11 .
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198 Dielectrics: II
7-12 ELECTRIC FORCE
S h o w t h a t V £ 2 i s in the di rect ion of VE.
7-13 CALCULATING AN ELECTRIC FORCE BY THE METHODOF VIRTUAL WORK
Use the me th od of vi r tual w ork (Sec. 4.5) to calcu la te the force on th e die lect r ic
sheet in Fig. 7-10 when F = 1000 vol ts , s = 1 mil l imete r , / = 100 mil l im eters , e r = 3.0.
Figure 7-10 Sheet of die lect r ic p art ly
inse r t ed be tween a pa i r o f conduc t ing
pla t e s . The force on the die lect r ic canagain be calcula ted from the principle
of vi r tual work. See Prob. 7-13.
HIGH-VOLTAGE TRANSMISSION LINES
Losses on h igh-vol t age t ransm is s ion l ines a re h ighe r tha n no rma l dur ing foggy o r
ra iny we ather . O ne rea son for this i s that wa ter col lects on the wires an d forms
droplets . S ince a droplet carr ies a charge of the same s ign as that of the wire ,
e l ec t ros t a t i c repu l s ion forces the drop le t t o e long a te and form a sha r p po in t where
the e lect r ic f ie ld intens i ty can be h igh en ou gh to ionize th e a i r . Th e resul t i s a corona
discharge in the surrounding a i r . in which the ions are accelerated by the e lect r ic
f ie ld and col l ide wi th neutra l molecules , forming more ions and heat ing the a i r .
Corona d i s cha rges a re ob jec t ionable because they d i s s ipa te ene rgy , mos t ly by
heat ing the a i r . Over long dis tances , they can cause heavy losses . They also cause
radio and te levis ion interference.
Dr op lets col lect on the wire in dri f t ing by. Th ey are a lso a t t rac ted by the no n
uniform electric field.
You a re a sked to eva lua te the impor t ance of the non-uni form e lec t r i c f i e ld .
T o do this , assu me tha t the a i r is s tag nan t and ca lculate wi thin w hat dis tance of a
wire the e lect ros ta t ic force wi l l be larger than the gravi ta t ional force . Clearly, i f that
d i s t ance i s mu ch l a rge r than the cond uc to r rad ius , t hen e l ec t ros t a t i c a t t rac t io n i s
i m p o r t a n t .
Th e t ransmis s ion l ine under cons ide ra t ion co ns i s t s o f a pa i r o f cond uc to rs
11.7 mil l imeters in diameter , separated by a dis tance of 2 meters and operat ing a t
100 ki lovol ts (± 50 ki lovol ts ) .
You can f ind an approximate value for the f ie ld a t a dis tance r nea r one wi re
as follows. If / . is the charge per meter.
2 ; i e n
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Problems 199
Th e expres sion on the right is the field near a single isolat ed wir e, whic h is a goo d
app rox im ati on as long as c is muc h less than th e dis tance to the other wire. No w
the capaci tance per meter for th is l ine is approximately ne 0/5^ a n d
/. = (ne0 5 )1 05
c o u l o m b s m e t e r .
The expression for the force per unit volume given in Sec. 7.4 involves e r a n d
the value of E inside a water droplet . For water, at low frequencies , e r = 81
(Table 6-1). To est imate the £ inside a droplet , one may use the formula for a d i
electric sphere in a uniform field . ' This is not a bad approximation, because even a
ra in d ro p is smal l , com pare d to the con duc to r rad ius , so tha t £ does no t c hange
much over one d rop d iameter . Then
e r + 2 / 2ne0r
Solution: Let us fi rs t calculate £ inside a water droplet s i tuated at a d is tance c
from the center of one of the wires:
(7re 0/5)10 5
= 361.4/r vol ts /m eter. (1)11 + 2) 2ne 0r
The electrostat ic force per cubic meter on a droplet is
1 \ dF = 1 —
81 dr
1 , , (361.4x 81 x 8.85 x 1 ( T 1 2 x
2 V r
12)
.d [i\ , 14.625 x 1CT
5
— \—A = - 9 . 2 5 0 x 1 0- 5
n e w t o n s / m e t e r3
. (3)dr \r2
y
The negative s ign indicates that the force is at t ract ive.
Now the gravi tat ional force per cubic meter is 1000 g newtons. so
F 9.250 x 10" 5
= 5 • (4
1 0 0 % 9 8 0 0 c3
This ra t io is equal to uni ty for c = 2 .113 mil l imeters , which is smaller tha n the
radius of the wire.
The electrostat ic force of at t ract ion on a droplet is therefore negligible.
+ Electromagnetic Fields and Wares, Eq. B-55.
* Electromagnetic Fields and Waves, Eq. 4-174.
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200 Dielectrics: II
7-15 ELECTRIC FORCE ON A DIELECTRIC
a) Calculate the e lect r ic force per cubic meter on the die lect r ic of a coaxial
cab le whose inne r cond uc to r has a rad ius o f 1 mi l l ime te r and w hose ou te r con duc tor
has an inner radius of 5 mil l imeters . The die lect r ic has a re la t ive permit t ivi ty of 2.5.
Th e ou te r co ndu c tor i s g rou nde d , and the inne r cond uc to r is ma in ta in ed a t 25 k i lovol ts .
b) Show that the e lect r ic force near the inner conductor is about 300 t imes
larger than t he grav i ta t io nal force if the die lect r ic has the dens i ty of water , nam ely
1 0 3 ki lograms pe r cub ic me te r .
7-16 DISPLACEMENT AND POLARIZATION CURRENTS
A capaci tor wi th pla tes of area A, separated by a die lect r ic of thickness s and
rela t ive permit t ivi ty e r , i s connected to a vol tage source V t h rough a re s i s t ance R.
Calcu la t e the d i sp lacement and the po la r i za t ion cur ren t s a s func t ions of , the
t ime .
7-/7 DIRECT ENERGY CONVERSION
I t is poss ible to conve rt th erm al energy in to e lect r ical energy with the c i rcui t
of Fig. 7-11.
W i t h X connec ted to Z , t he ba t t e ry B cha rges the ba r ium t i t ana te capac i to r C.
T h e n X i s disconnected from Z and the capac i to r i s hea ted . The re l a t ive pe rmi t t iv i ty
of the t i tanate decreases and the vol tage on C i nc reases . Then X i s connec ted to Y,
a n d C recha rges the ba t t e ry B, while supplying energy to R. Th e capac i to r i s t hen coo led
and the cycle repeated.
Such devices have been bui l t to feed e lect ronic c i rcui ts on board sate l l i tes . The
capaci tors are in thermal contact wi th panels on the outer surface of the sa te l l i te that
are heated periodical ly by the sun as the sa te l l i te rota tes about i t s own axis in space.
L et V B be the po ten t i a l suppl i ed by the ba t t e ry ; Q the capac i t ance a t t he temp e r a t u r e T l: C 2 a n d V 2 t he capac i t ance and vo l t age a t T 2; \Y lh t he the rma l ene rgy
requi red to hea t t he capac i to r ; and We the electric energy fed to R.
Calculate the eff ic iency WJWlh
w h e n V B = 700 vol ts , V 2 = 3500 vol ts . e r l =
8000. e r2 = 1600, area of on e cap aci t or e lect rod e 1 squ are meter , thick ness of die lect r ic
0.2 mil l imeter , specif ic heat of barium t i tanate 2.9 x 1 0 6 j ou le s pe r cub ic me te r pe r
degree, and T 2 — T, = 30 C.
>
O
R
Figure 7-1 1 Circu i t fo r t r ans forming
thermal energy into e lect r ic energy on
board sate l l i tes . See Prob. 7-17. B
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CHAPTER 8
MAGN ETIC FIELDS: I
The Magnetic Induction B and the Vector Potential A
8.1 THE MAG NETIC INDUCTION B
8.1.1 Exam ple: Long Straight Wire Carrying a Current8.1.2 Exam ple: Circular Loop, Magn etic Dipole Mom ent m
8.2 THE DIVERGENCE OF B
8.3 MAGNETIC MONO POLES
8.4 THE VECTOR POTENTIAL A
8.4.1 Exam ple: Long Straight Wire
8.5 THE VECTOR POTENTIAL A AND THE SCALAR
POTENTIAL V
8.6 THE LINE INTEGRAL OF THE VECTOR POTENTIAL A
OVER A CLOSED CURVE
8.7 SUMMARY
PROBLEMS
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Th e next e ight cha pte rs wi ll deal wi th the m ag ne t ic f ie lds of e lect r ic c urr en ts
a n d o f m a g n e t i z e d m a t t e r .W e sha l l s t a r t he re by s tudy ing the ma gne t i c induc t ion B an d the
vec to r po ten t i a l A . The se two quan t i t i e s co r r esp ond , r espec tive ly , t o the
elect r ic f ie ld in tensi ty E and to the potent ia l V.
F o r t h e m o m e n t , w e c o n s i d e r o n ly c o n s t a n t c o n d u c t i o n c u r r e n t s a n d
n o n - m a g n e t i c m a t e r i a l s .
Figure 8-1 shows a ci rcui t car rying a current I. We def ine the magnetic
induction a t the po in t P as fol lows:
where the in t egra t ion i s ca r r i ed ou t over the c losed c i r cu i t . As usua l , t he
uni t vector I-
! p o i n t s from t he source to t he po in t o f observa t ion : i t po in t s
from t he e l ement d l to t he po in t P. Magnet i c induc t ions a r e expressed in
teslas.
T h e c o n s t a n t p 0 is defined as fol lows:
THE MAGN ETIC INDUCTION B
(8-1)
fi0 = 4n x 10 7 t e s l a m e t e r / a m p e r e . (8-2)
and is cal led the permeability of free space.
I f the current I i s d i s t r ibu ted in space wi th a cu r r en t den s i ty J a m p e r e s
per square mete r , t hen / becomes J da and must be pu t under the in t egra l
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20 3
s i g n . T h e n J da dl c a n b e w r i t t e n a s J dx', w h e r e dx ' i s a n e l e m e n t o f v o l u m e ,
a n d
B ^ f , ^ , ( 8 - 3 )
a s i n F i g . 8 -2 . T h e i n t e g r a t i o n i s c a r r i e d o u t o v e r t h e v o l u m e x o c c u p i e d b y
t h e c u r r e n t s .
W e a s s u m e t h a t J i s n o t a f u n c t i o n o f t i m e a n d t h a t t h e r e a r e n o m a g
n e t i c m a t e r i a l s i n t h e f i e l d .
Figure 8-2 The magnet ic induct ion dB due to an e l ement J dx ' of a vo lume
di s t r ibu t ion of cur ren t .
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204 Magnetic Fields: I
As in e l ec t ros t a t i cs , we can descr ibe a magne t i c f i e ld by d rawing lines
of B t ha t a r e everywhere t angen t to B.
Similar ly , i t i s convenient to use the concept of f lux, the flux of the
magnetic induction B t h rough a su r f ace S, d e f in e d a s t h e n o r m a l c o m p o n e n t
of B i n t eg ra ted over S:
O = J . B • da . (8-4)
Th e flux d> is exp res sed in webers. Th us the t es la is one webe r per sq uar e
m e t e r .
8.1.1 EXAMPLE: LONG STRAIGHT WIRE CARRYING A CURRENT
An e lement dl of a long s t ra ight wire carrying a current / , as in Fig. 8-3, produces
a magne t i c induc t ion
p0I dl si n 0d B = - — q>i,
47 i r
(8-5)
where (p t i s t he un i t vec tor po in t ing in the az imu tha l d i rec t ion . The pos i t ive d i re c t ions
for <p, and / are re la ted by the r ight-hand screw rule .
Figure 8-3 The magnet ic induct ion d B produced by an e l ement / dl of the cur ren t
/ in a long s t ra ight wire .
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20 5
Figure 8-4 Lines of B in a p l a n e p e r p e n d i c u l a r to a l ong s t ra igh t w i re ca r ry ing ac u r r e n t / . The dens i ty of the l ines is i nve rse ly propor t iona l to the di s t ance to the
wire . Lines c lose to the wire are not s h o w n .
Expres s ing dl. sin 0. and r in t e r m s of i. and p.
B = - — I cos x d% ip , = <p,. (8-6)
4n p J~nl2
Inp
T h e m a g n i t u d e of B thus falls off inversely as the first power of the dis tance from
an infinitely long wire. T he l ines of B are concentr ic c i rc les lying in a plane pe rpen
d icu la r to the wire , as in Fig . 8-4.
EXAMPLE: CIRCULAR LOOP. MAGNETIC DIPOLE MOMEN T m
A c i rcu la r l oop of r a d i u s a ca r r i e s a c u r r e n t /, as in Fig. 8-5.
An e lement / dl of c u r r e n t p r o d u c e s a d B h a v i n g a c o m p o n e n t dB. on the axis , as
i nd ica t ed in the figure. By s y m m e t r y , the tota l B is a l o n g the axis , and
dB . =— -.cost). (8-7)
4n r
p 0I 2to j p0lcr
B. =r co s 0 = j - ^ . (8-8)
4n r2
There fore , on the axis , the m a g n e t i c i n d u c t i o n is m a x i m u m at the cente r of the r ing
a n d d r o p s off as r3
for z2
» a2
.
F i g u r e 8-6 shows l ines of B in a p l a n e c o n t a i n i n g th e axis of the l o o p .
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20 6
Figure 8-5 The magnet ic induct ion d B produced by an e l ement / d l at a point on
the axis of c i rcular current loop of radius a. The pro jec t ion of d B on the axis is dB..
y \
Figure 8-6 Lines of B in a plane containing the axis of the loop of Fig. 8-5. The
direct ion of the current in the loop is shown by the dot and the cross .
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8.2 The Divergence o / B 207
Far from th e l o o p , th e field is the s a m e as t h a t of an electric dipole (Sec. 2.5.1),
except that the factor 1 /4ne 0 is rep laced by p0/4n, an d t ha t t he e l ec t r i c d ipo le moment
p is rep laced by the m a g n e t i c d i p o l e m o m e n t m. The magnetic dipole moment m is a
v e c t o r w h o s e m a g n i t u d e is na 2I an d t h a t is n o r m a l to the p l a n e of t he loop , in the
di rec t ion g iven by the r ight-hand screw rule wi th respect to the c u r r e n t /.
THE DIVERGENCE OF B
Cons ider a current element / dl and a volume tk containing a point P as
in Fig. 8-7. Th e curren t element p rod uce s at P a magnetic induction
j U 0 / dl x i - !
4n r2
dB = ^ , • ' . (8-9)
As we saw in Sec. 8.1.1, the lines of B due to the element I dl are circles
situated in planes perpendicu lar to a line thro ugh dl and centered on it as
Figure 8-7 The cur ren t e l ement / d l p r o d u c e s a m a g n e t i c i n d u c t i o n B. The net
outward f lux of B t h r o u g h th e sur face of the e l e m e n t of v o l u m e dx is z e r o .
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208 Magnetic Fields: I
in Fig . 8-7. From the f igure, i t i s obvious that the net outward f lux of the
B due to I d l th rough the su r f ace o f the vo lume dx i s ze ro . N ow , any vo lu me
can be subd iv ided in to vo lume e lement s o f the same k ind as dx . Therefo re ,fo r any vo lume t bounded by a su r f ace S ,
j ;B • da = 0. (8-10)
Th is equ at i on is t rue of a l l m agn et ic f ie lds. The n, using the div ergen ce
theorem (Sec . 1.10),
V • B dx = 0. (8-11)
So
V • B = 0. (8-12)
Th e der iva t ives con ta in ed in the oper a to r V a re wi th r espec t to the fi eld
po in t where the magn e t i c indu c t io n i s B . Th i s is an o th er o f M axw el l ' s
e q u a t i o n s .
MAGNETIC MO NO POLES
Th e s t a t e me nt tha t V • B is equ a l to ze ro impl i es th a t m agn e t i c f ie lds a r e
due so le ly to e l ec t r i c cu r r en t s and tha t magne t i c "charges" do no t ex i s t .
Otherwise , one would have the magne t i c equ iva len t o f Eq . 6 -12 , and the
d i v e r g e n c e of B w o u l d b e p r o p o r t i o n a l t o t h e m a g n e t i c " c h a r g e " d e n s i ty .
E l e m e n t a r y m a g n e t i c " c h a r g e s " a r e c a l l e d magnetic monopoles. T h e i r
ex i s t ence was pos tu la t ed by Di r ac in 1931 , bu t they hav e never been obse rved
t o d a t e . T h e t h e o r e t i c a l v a l u e o f t h e m a g n e t i c m o n o p o l e is2h/e, or 8.271 17 x1 0 " 1 5 w e b e r , w h e r e h i s P lanck ' s cons tan t (6 .625 6 x 1 0 "
3 4
j o u l e s e c o n d ) ,
a n d e i s the charge of the elect ron, 1 .602 1 x 1 0 "1 9
c o u l o m b .
Th e surface in tegra l of B • da ov er a c losed surface is eq ua l to the en
c l o s e d m a g n e t i c c h a r g e .
P ro b le m s 8-9 an d 12-6 conc ern two m eth od s tha t hav e been used
for de tec t ing monopoles in bu lk mat t e r . See a l so P rob . 19-5 .
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8.4 The Vector Potential A 209
TH E VECTOR POTENTIAL A
Since V • B = 0, it is reasonable to assume that there exists a vector A such
that /
B = V x A, (8-13)
because the divergence of a curl is identically equal to zero (see Prob. 1-29).
The derivatives in the del operator are evaluated at the field point.
We can find an integral for A, starting from Eq. 8-3,
B = £° f J x %th'. (8-14)
4tt J*' r2
From Probs. 1-13 and 1-27,
J xr
i f = v f - ) x J = V xJ
- - V x J . (8-15)r
2
Now, th e last term on the right is zero, because J is a function of x', y', z',
while V contains derivatives with respect to X, y, z. Then
47T JR
' r 4n JR
' r(8-16)
and
A Mo r Jf -dx'. (8-17)4n
If the current is limited to a conducting wire,
Note that we can add to this integral any vector whose curl is zero
without affecting the value of B in any way.
Note also that B depends on the space derivatives of A, and not on
A itself. The value of B at a given point ca n thus be calculated from A, only
if A is known in the region around the point considered.
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210 Magnetic Fields: I
EXAMPLE: LONG STRAIGHT WIRE
In Sec. 8.1.1 we calculated B for a long straight wire carrying a current /. startingfrom the definition of B. We can also calculate B starting from
HqI
An
(II(8-19)
One can see immediately that A is parallel to the wire, since the dl 's are all along
the wire.
We first calculate A for a current of finite length 2L. Referring to Fig. 8-8.
Ho i rL
' Anpi dl
Jo{p> + /2)>-2
^ i n [ z + (P
j
+ r-?2
y6.
(8-20)
(8-21)
Ikd.iUL
271 p(p
2
« L2
).
In p (8-22)
(8-23)
For L -* oo, A tends to infinity logarithmically. We can still try to calculate B,
however, because a function can be infinite and still have finite derivatives. For
exa mple , if y = x, y -» x as x -> x . but dy dx remains equal to 1.
To calculate B for p2
« L2
, we use the fundamental definition of the curl given
in Eq. 1-77, using the rectangular path shown in Fig. 8-8 for the line integral:
B = lim1
(j) A • dl.A p A r
as ApA: tends to zero. Since A is parallel to the wire,
1B = lim
A p A r[Aip) - A(p + Ap)] Ar
— hm — I In2tt Af.
In2/.
In Ap
p p + Ap
p + Ap
— hm —In Ap
In 1 +
(8-24)
(8-25)
(8-26)
(8-27)
(8-28)
Now In (1 + x) = x when x is small an d. finally, B is equal to p0I/2np, as previously.
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21 1
Figure 8-8 An e lement / dl of a current / in a long s t ra ight wire produces an
e lement o f vec tor po ten t i a l d A at the point P . The sma l l rectangle is the path of
in t egra t ion used for ca l cu la t ing B .
THE VECTOR POTENTIAL A AND THE SCALARPOTENTIAL V
T h e v e c t o r p o t e n t i a l A of magnet ic f ie lds i s in many respects s imi lar to the
sca la r po ten t i a l ( o r , s imply , t he po ten t i a l ) V of electric fields (Sec. 2.5).
Le t us r e tu rn b r ie f ly to C ha p te r s 2 and 3 .
a) W ith elect r ic f ie lds, the imp or ta nt q ua nt i ty , in prac t ice , is the in
tensi ty E: i t i s E that g ives the stored energy and the forces on charged bodies
and o n po la r i zed d ie l ec t r i cs . I t is a l so E tha t cause s b rea kd ow n in d ie lec t r i cs .
b ) T h e q u a n t i t y V i s r e l a t ed to E th rough E = — VF. So one can
add to V any qua n t i ty tha t is i nde pen den t o f the coo rd in a te s wi tho u t a ff ect ingE in any way. Also, the value of E at a point can be calculated only i f V is
kno wn in the region a r o u n d t h a t p o i n t .
c) Fin al ly , the in tegra l for F( Eq . 2-18) i s s im pler to ev alu ate than tha t
for E (Eq. 2-6). So, in general, i t is easier to f ind E by first calculating V, a n d
then V F, than to ca lcu la t e E d i r ec t ly . How ever , if t he geom et ry o f the c harge
dis t r ib ut i on is par t icu lar ly s im ple, on e can use G au ss 's law (Sec. 3 .2). Th en
E is easy to calcu late di rect ly a nd th ere is no nee d to go thro ug h V.
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212 Magnetic Fields: I
No w le t us see how A com pa res w i th V.
a) Th e qua n t i ty tha t co r r esp ond s to E i s t he ma gne t i c indu c t ion B:
it is B tha t g ives the sto red ene rgy an d the ma gn et ic forces (Ch ap ter 13).
b ) The vec to r po ten t i a l A is r e l a t ed to B on ly th ro ug h B = V x A.
On e can add to A any qu an t i ty who se cur l is ze ro wi th ou t ch ang ing B, and
on e can ded uce f rom A the value of B at a poi nt only i f on e kn ow s the va lue
of A in the r eg ion a ro un d th a t po in t .
c) The integral for A (Eq. 8-18) i s a lso simpler to calculate than that
for B (Eq. 8-1). H ow ev er , wi th simp le curren t dis t r ib ut i on s, it i s mu ch easier
to use Am pe re 's c i rcui ta l law (Sec. 9 .1) to find B than to calcu late A f ir st .
So there is a grea t s imi lar i ty betw een V and A. Later on, in Sec. 11.5 ,
we shal l f ind an equat ion that re lates E to both V a n d A , i n t i m e - d e p e n d e n t
fields.
8.6 THE LINE INTEGRAL OF THE VECTOR
POTENTIAL A OVER A CLOSED CURVE
Using S tokes ' s t heorem (Sec . 1.13),
A • dl = f (V x A) • da , (8-29)
w h e r e S is any su r f ace bo un de d by the c losed curve C. But, from Eq. 8-13,
the cur l of A is B. Then
(j), A • dl = J s B • da = cp: (8-30)
the l ine integral of A • dl over a c losed curv e C i s eq ua l to the ma gn et i c f lux
t h r o u g h C.
8.7 SUMMARY
T h e magnetic induction B du e to a c i rcui t C car ry ing a curr en t / is def ined as
fo l lows:
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Problems 213
Magnetic inductions are measured in teslas, an d u0
is denned to be exactly
An x 10"7
tesla meter per ampere.
The magnetic flux through a surface S is
(D = JsB - d a . (8-4)
Mag net ic flux is meas ure d in webers, and a tesla is one weber per square
meter.
The divereence of B is zero:
V • B = 0, (8-12)
since the magne tic flux th rou gh a closed surface is alway s identically equa l
to zero, if mag net ic mo no po le s do no t exist. Thi s is one of M axwell 's
equations.
It follows from this that
B = V x A, (8-13)
where A is the vector potential
Th e line integra l of A • dl over any closed curve C is equa l to the
magnetic flux through any surface S bounded by C:
(j). A • dl = (D. (8-30)
PROBLEMS
8-1E MAGNETIC FIELD ON THE AXIS OF A CIRCULAR LOOP
P l o t a curve of B as a function of z on the axis of a circular loop of 100 turns
hav ing a mean radius of 100 millimeters and carrying a current of one ampere.
8-2 SQUARE CURRENT LOOP
C o m p u t e the magnetic induction B at the center of a square current loop of
side a carry ing a current /.
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214 Magnetic Fields: I
8-3 MAGNETIC FIELD OF A CHARGED ROTATING DISK
An insu la t ing d i sk of rad ius R carr ies a uniform free surface charge dens i ty a
on one face . I t rota tes about i t s axis a t an angular veloci ty to.a) What is the magni tude of the e lect r ic f ie ld intens i ty E, close to the disk?
b) Show that the surface current dens i ty a a t the radius r i s wra.
c) Show that , a t the center of the charged surface,
1B = - n
0wRo.
d) Ca lcu la t e E a n d B for R = 0.1 meter , a = 1 0- 6
c o u l o m b p e r s q u a r e m e t e r ,
a n d OJ = 1000 rad ia ns pe r s econd .
8-4 SUN SPOTS
See Prob. 8-3.The Zeeman e f fec t
1
observed in the spectra of sunspots reveals the exis tence
of magnet ic induct ions as large as 0.4 tes la .
Let us assume that the magnet ic f ie ld is due to a disk of e lect rons 10 7 m e t e r s
in radius rota t ing a t an angular veloci ty of 3 x 1 0 "2
rad ian pe r s econd . The th i cknes s
of the disk is smal l co m pa red to i t s rad ius .
a) Sho w that the dens i ty of e lect ro ns requi red to achieve a B of 0.4 tesla is
a b o u t 1 01 9
per squa re me te r .
b ) Show tha t t he cur ren t i s abo ut 3 x 1 0 1 2 a m p e r e s .
c ) In v i ew of the enormous s i ze o f the Coulomb forces , such cha rge dens i t i e s
a re c l ea r ly imposs ib le . Then how could such cur ren t s ex i s t ?
8-5E HELMHO LTZ COILSOne of t en wishes to ob ta in a un i form magne t i c f i e ld ove r an apprec iab le vo lume .
In such cases one uses a pair of Helmhol tz coi ls as in Fig. 8-9.
Show that the f ie ld on the axis of symmetry, a t the midpoint between the two
coils is
(0 .8) 3 2p 0NI/a,
w h e r e TV i s the nu m be r of turn s in each coi l .
I f yo u have th e cour age to calc ula te the f ield as a funct ion of z, and i ts derivat ives ,
you wil l f ind that the f i rs t , second, and thi rd derivat ives are a l l zero a t z = 0!
F igure 8-10 shows B. as a func tion of z for value s of/- up to 0.16a.
Ro ugh ly spe akin g, B. i s unifo rm, w i thin 10%, ins ide a sphe re of rad ius 0.1a .
f W hen a gas is subjected to a s t ron g ma gn et ic f ie ld, i ts spectra l l ines are spli t into several
com pon ent s . The sp l i t t i ng is a me asure of the magn e t i c induc t ion .
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21 5
Figure 8-9 Pair of Hel m ho l tz coi ls .
When the spacing between the coi ls i s
equa l t o one rad ius a, t he magne t i c
f ie ld near the center i s remarkably
uniform. See Prob. 8-5.
-0.2a
Figure 8-10 A xial co m po ne nt of B
nea r the cen te r o f a pa i r o f He lmhol t z
I coils as in Fig. 8-9, as a function of z
-0.2a and for var ious values of r.
HELMHOLTZ COILS
You are asked to des ign a pair of Helmhol tz coi ls (Prob. 8-5) that wi l l cancel
the earth 's ma gn et ic f ie ld, wi thin 10 perce nt , over a spherica l volum e having a radiu s
of 100 mil l imeters .
Th e magne t i c indu c t ion of the ea r th in the l abora to ry i s 5 x 1 0 "5
tes la andforms an ang le o f 70 degrees w i th the hor i zo nta l . The l a bor a tor y i s s i t ua t ed in the
No r th e rn Hem isphere . See the foo tno te to P ro b . 8 -11 .
a ) Ho w mu s t the co i ls be or i en ted , and in wha t d i rec t ion m us t t he cur ren t f low?
b) Specify the coi l dia me ter and sp acing .
c ) Wh a t i s t he to t a l nu m be r of am per e - tu rns requ i red?
d) T he l abo ra to ry has a goo d supply of N o. 18 enam eled copp er w i re on h and .
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216 Magnetic Fields: I
This wire has a cross-sect ion of 0.823 square mil l imeter and a res is tance of 21.7 ohms
per ki lom eter . An adjus tab le po we r supply is a lso avai lable . I t can sup ply from 0 to
50 vo l t s a t a maximum of two amperes .
How many tu rns should each co i l have?
Spec i fy the ope ra t ing cur ren t and vo l t age .
Wil l it be necessary to cool the coi ls? If so, wh at d o you sugg es t?
8-7 LINEAR DISPLACEMENT TRANSDUCER
Draw a curve of Bz as a function of z on the axis of a pair of Helmhol tz coi ls
(Prob. 8-5) wi th the currents f lowing in opposi te di rect ions and with the coi ls separated
by a d i s t ance 2a . ins tead of a. for values of z ranging from - a t o + a.
Note the l ineari ty of the curve over most of this region.
One could bu i ld a l i nea r d i sp lacement t ransduce r fo r measur ing the pos i t ion
of an object by fixing to it a Hall probe (Sec. 10.8.1) and having it move in the field
of such a pa i r o f oppos ing He lmhol t z co i l s .
8-8E THE SPACE DERIVATIVES OF B IN A STATIC FIELD
Consider the f ie ld of a c i rcular loop, as in Sec. 8.1.2.
On the r ight and somewhat above the axis , the l ines of B f lare out , so B. d e
creases wi th z a n d dBJvz i s negat ive . At the same point , B y increases with y, so dB y/dy
i s pos i t ive . S imilar ly, cBJdx is positive.
a ) W hy i s t h i s , ma th ema t i ca l ly? Ho w a re these de r iva t ives re l a t ed ?
b) Compare these derivat ive for other points on Fig. 8-6.
8-9E MAGNETIC MONO POLES
I t is p red ic t ed theore t i ca l ly th a t a magn e t i c cha rge Q* s i tua t ed in a magne t i c
f ie ld would be subjected to a force Q*B/u0.Calculate the energy acquired by a magnet ic monopole in a f ie ld of 10 tes las
ove r a d i s t ance of 0 .16 me te r . Expres s your answer in g igae lec t ron-vol t s (10 9 e lec t ron-
volts).
In one exper iment fo r de tec t ing magne t i c monopoles , va r ious s amples , such
as deep-sea sed imen ts , were subjected to such a f ie ld, as in Fig. 8-11. N on e were found.
Figure 8-11 Appara tus used in an
unsucces s fu l a t t empt to obse rve
magne t i c monopoles . The dev ice i s
inserted ins ide a coi l producing an
axial B of 10 tes las . A s lurry of deep-
sea sediment f lows into the chamber
on the r ight a t / and out a t O. I t was
h o p e d t h a t m a g n e t i c m o n o p o l e s i n
the s ample would be acce le ra t ed in
the magnet ic f ie ld and would leave
t racks in the photographic p l a t e P.
See Prob. 8-9.
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Problems 217
8-10 MAGNETIC FIELD OF A CHARGED ROTATING SPHERE
A conducting sphere of radius R is charged to a potential V and spun about a
diameter at an angular velocity w as in Fig. 8-12.
a) Show that the surface charge density a is e0V/R.
b) Now show that the surface current density is
It turns out that B is uniform inside the sphere.
d) What would be the value of B for a sphere 0.1 meter in radius, charged to
10 kilovolt s, and spinning at 1 04
revolutions per minute?
e) Show that the dipole moment is
f) What is the dipo le mom ent for the above sph ere?
g) What current flowing through a loop 0.1 meter in diameter would give the
same dipole moment?
a = e0o)V sin 0 = M sin 0,
where M is e0coV, an d 0 is the polar angle.
c) Show that the magnetic induction at the center is
B = ( 2 / 3 ) 6 o / u 0 o j F = (2/3)/i„M.
m = (4/3 )7tPv3
M .
+ ++
++
r +
++ +
Figure 8-12 Charged sphere spinning
about a diameter. See Probs. 8-10 and8-11.
THE EARTH'S MAGNETIC FIELD
The origin of the earth 's magn eti c field is still largely unkn own . According to one
model, the earth would carry a surface charge, producing an azimuthal current
because of the rotation, and hence the magnetic field, as in Prob. 8-10.
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218
Figure 8-13 Typical line of B for the
earth's magnetic field. The shape of
the field inside the earth is unknown
With the model discussed in Prob.
8-11 , the B inside would be uniform,
and the lines of B would be straight,
as in the figure.
This model is attractive, at first sight, because it gives a magnetic field that has
the correct conf igurat ion, o r nearly so, except that one has to disregard t he fact
that the magnetic and geographic poles do not coincide. As we shall see, this model
requires an impossibly large surface charge density a.
a) What mus t be the sign of <r?+
Solution: Figure 8-13 shows the earth with its North and South geographic
poles, a line of B, and the direction of the angular velocity a).
The current must be in the direction shown, and hence a must be negative.
b) With this current distribution, it can be shown that the B inside the earth
is uniform.
Use the results of Prob. 8-10 to deduce the value of M from the fact that the
vertical component of B at the poles is 6.2 x 1 0 "5
tesla. At the poles, the value of
B is the same, immediately above and immediately below the surface.
Solution: Since
2
B = -n0M = 6.2 x 10"5
, (1)
1.5
M = — 6.2 x 10 = 74 amp ere s/m ete r. (2)
/'o
c) Calculate the magnetic dipole moment of the earth.
The earth has a radi us of 6.4 x 106
meters.
Solution: Again from Prob. 8-10,
m = (4 /3 )7 tR 3
M , (3)
= (4/3)ti (6.4 x 106
)3
74 = 8.1 x 1 02 2
ampere meters2
. (4)Remember that the magnetic pole that is situated in the Northern Hemisphere is a
South" pole: the "North" pole of a magnetic needle points North.
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Problems 219
d| Calc ulate the surface charge dens i t ) required to give the surface curren l
dens i ty ca l cu la t ed above .
If there were such a surface charge, what would be the value of the vert icale lect r ic f ie ld inten s i ty?
There does exis t a negat ive charge a t the surface of the earth. I t gives an £ of
ab ou t 100 vol ts per meter . See Prob . 4-2. Also, the m ax im um elect r ic f ie ld intens i ty
tha t can be sus t a ined in a i r a t normal t e mp era tu re and p res sure is 3 x 10 6 vol ts per
me te r .
Solution: At the equa tor . I) = 0
7. = av = M = 74. (5)
w h e r e v i s the tangen t ia l veloci ty. Th us
74= 0 . 16 c o u l o m b m e t e r 2 .=
,2;r x 6.4 x 10" )/(24 x 60 x 60)(6)
This surface charge dens i ty would give an e lect r ic f ie ld intens i ty
a 0.16= 1.8 x 1 0 1 0 vol ts meter . (7 )=
- 1 28.85 x 10
from Sec. 3.2, which is absu rdly large.
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CHAPTER 9
MAGN ETIC FIELDS: II
Ampere's Circuital Law
9.1 AMPE RE'S CIRCUITAL LAW
9.1.1 Examp le: Long Cylindrical Conductor9.1.2 Example: Toroidal Coil
9.1.3 Example: Long Solenoid
9.1.4 Examp le: Refraction of Lines of H at a Current Sheet
9.1.5 Example: Short Solenoid
9.2 THE CURL OF THE MAGNETIC INDUCTION B
9.3 SUMMARY
PROBLEMS
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This cha p te r i s devo ted to A m pe re ' s c i r cu i t a l l aw and to the cu r l o f B.
Am pe re ' s l aw is used to ca lcu la t e ma gne t i c induc t io ns in mu ch the sam eway as Gauss 's law is used to calculate e lect r ic f ie ld in tensi t ies . The value
of V x B wi l l f o l low immedia te ly f rom Ampere ' s l aw by us ing S tokes ' s
t h e o r e m .
We s t i l l l imi t our se lves to cons tan t conduc t ion cur r en t s and to non
m a g n e t i c m e d i a .
We have seen in Sec. 8 .1 .1 that the magnet ic induct ion vector B near a long
s t r a igh t wi r e is az im utha l an d tha t i t s ma gn i tud e i s p.0I/2np, w h e r e p is the
distance f rom the wire to the point considered. Thus, over a c i rcle of radius p
cen tered on th e wire as in Fig . 9-1,
AMPERE 'S CIRCUITAL LA W
(9-1)
Th is is a gen eral resu l t and , for any close d curv e C,
(9-2)
wh ere / is the curre nt enclo sed by C. Th e posi t ive di rec t ion for the in teg rat io n
is re lated to the posi t ive di rect ion for / by the r ight-hand screw rule as in
Fig. 9-1.
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22 2
Figure 9-1 Posi t ive di rect ion for the
in tegra t ion pa th a rou nd a cur ren t / .
For a vo lume d i s t r ibu t ion o f cu r r en t ,
(f ) cB • dl = / t 0 J , J • da, (9-3)
w h e r e J is t he cu r r e n t dens i ty th r ou gh a ny su r face S bo un de d by the cu rve C
as in Fig. 9-2a. This is Ampere's circuital law.
In many cases the same cur r en t c rosses the su r f ace bounded by the
curve C several t imes. With a solenoid, for example, C could fol low the
ax i s an d r e tu rn ou t s ide the so leno id , a s in F ig . 9 -2b . Th e to t a l cu r r e n t c ross ing
the su rf ace i s t hen the cu r r e n t in each tu rn mu l t ip l i ed by the num be r o f
tu rns , o r the number o f ampere-turns.
c
(a)
Figure 9-2 (a ) Am pe re ' s c i rcui ta l law s ta tes that the l ine integral of B • dl a r o u n d C
is equal to p 0 t imes the cur ren t t h rough any sur face bounded by C. (b) Pa th of
in t egra t ion C for a solenoid. In this case the l ine integral of B • dl is &p0I.
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9.1 Ampere's Circuital Law 223
The circuital law can be used to calculate B when its magnitude is the
same, all alon g the path of integ ratio n. This law is thu s somew hat similar t o
Gauss's law, which is used to compute E, when E is unifo rm over a surface.
EXAMPLE: LONG CYLINDRICAL CONDUC TOR
The long cy l indr i ca l conduc tor of Fig. 9-3 ca r r i e s a c u r r e n t / uni formly d i s t r ibu ted
over its cross-sect ion with a dens i ty
J = I/nR2
. (9-4)
O u t s i d e th e c o n d u c t o r , B is a z i m u t h a l and i n d e p e n d e n t of (p, so t h a t , a c c o r d i n g
to the circui ta l law,
B = p0I/2np.
Ins ide the c o n d u c t o r , for a c i rcu la r pa th of r a d i u s p,
PaJitp 1 p0Jp p0IpB
2n p 2n R 2
(9-5)
(9-6)
T h e m a g n e t i c i n d u c t i o n B therefore increases l inearly wi th p ins ide th e c o n d u c t o r .
O u t s i d e the c o n d u c t o r , B dec reases as 1/p. The curve of B as a funct ion of p is
s h o w n in Fig . 9-4.
Figure 9 -3 Long cy l indr i ca l conduc tor ca r ry ing a c u r r e n t / .
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224 Magnetic Fields: II
'•0 2.0 3.0 4.0 5.0 6.0
p (millimeters)
Figure 9-4 The magnet ic induct ion B as a function of radius for a wire of 1 mil l ime te r
rad ius ca r ry ing a cur ren t o f 1 ampere .
EXAMPLE: TOROIDAL COIL
A close -wo und t oro ida l coi l of squ are cross-se ct ion, as in Fig. 9-5, carr ies a curren t / .
A l o n g p a t h a, the line integral of B i s zero, s ince there is no current l inking this
p a t h . T h e n t h e a z i m u t h a l B i s zero in this region. The sam e appl ies to c and to any
s imi l a r pa th ou t s ide the to ro id . Then the az imutha l B i s ze ro eve rywhere ou t s ide .
Ins ide , a long pa th b,
2npB = p 0NI , (9-7)
w h e r e N i s t he to t a l num ber of tu rns , a nd
B = u.N I 2np. (9-8)
The re ex is t non -az im utha l com pon ent s o f t he mag ne t i c induc t ion ou t s ide the
toro id . For a pa th such a s d in Fig. 9-5, the area bo un de d by the pat h is crossed
Figure 9-5 To roi da l coi l . Th e brok en l ines show p ath s of integ rat io n.
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9.1 Ampere's Circuital Law 225
once by the current in the toroidal winding and, at a distance large compared to
the outer radius of the toroid, the magnetic induction is that of a single turn along
the mean radius.Although the magnetic induction B outside the toroid is essentially zero, the
vector potential A is not. This will be evident if one remembers that A is a constant
times the integral of / dl /r, where r is the distance between the element dl and the
point where A is calculated (Sec. 8.4). For example, at a point close to the winding.
A is due mostly to the nearby turns, it is parallel to the current, and it has approxi
mately the same value inside and outside.
The vector potential A can therefore exist in a region where there is no B field.
This simply means that we can have at the same time A # 0 and V x A = 0.
For example, A = ki, where k is a constant, satisfies this condition. We are already
familiar with a similar situation in electrostatics: the electric potential can have
any uniform value in a region where E = — VV = 0.
9.1.3 EXAMPLE: LONG SOLENOID
For the long solenoid of Fig. 9-6. we select a region remote from the ends, so that
end effects will be negligible; also, we assume that the pitch of the winding is small.
Let us choose cylindrical coordinates with the r-axis coinciding with the axis of
symmetry of the solenoid, as in Fig. 9-7.
1. We first note that B has the following characteristics, both inside and outside
the solenoid.
a) By symmetry, all three components Bp, B_, Bv are functions neither of cp
nor of z.
b) Moreover, Bp = 0 for the following reason. Consider an axial cylinder of
length / and radius p. either larger than the solenoid radius, or smaller, as in Fig. 9-6.
The integral of B • da over its surface is simply InplB^ since the integrals over the
two end faces cancel. But, according to Eq. 8-10, the integral of B • da over any
closed surface is zero. Then B (, = 0.
c) Consider a rectangular path a inside the solenoid as in Fig. 9-6. By Ampere's
circuital law, the line integral of B • dl around the path is zero. Since Bp = 0, as we
have just found, the line integrals on the vertical sides must cancel, which means
that, inside the solenoid, B. is also independent of p. The same applies to a similar
path entirely outside the solenoid.
2. Outside the solenoid.
a) We have shown abo ve that, ou tside th e solenoid. B. is independent of all
three coordinates p. <p , z. However. B is zero at infinity. Therefore B. = 0 outside
the solenoid.
b) A path such as b is linked once by the current and, outside the solenoid.
B v = p0l 2np. (9-9)
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Figure 9 - 7 C o m p o n e n t s o f B a t P . in cy l indr i ca l co ord in a te s .
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9.1 Ampere's Circuital Law27
3. Inside t he so lenoid .
a) B v = 0 ins ide because the l ine integral of B • dl over a c i rc le of radius p,
say the top edge of the smal l cyl inder shown in Fig. 9-6, i s 2-itpB^, and this must be
ze ro accord ing to Eq . 9 -2 because the re i s no cur ren t enc losed by the pa th .
b) Con s ide r in g now p a th c in F ig . 9 -6 , and rem emb er ing th a t B f, = 0 b o t h
ins ide and ou t s ide , and tha t B. = 0 outs ide , we see that B.S = p 0N'IS, w h e r e N'
is t he num be r of tu rn s pe r me te r . The re fore
B r = ix0N'I. (9-10)
The magne t i c induc t ion ins ide a long so lenoid in the reg ion remote f rom the
end s is theref ore axia l , uniform, an d equa l to p 0 t i m e s t h e n u m b e r o f a m p e r e - t u r n s
p e r m e t e r NT. I t i s mu ch l a rge r th an the az im utha l B outs ide , as long as (l/N')/2np «
1, or as long as the pi tch of the win din g 1/JV', divided by i ts c i rcumference 2np, isvery smal l .
I t i s obvious , f rom Fig. 9-10, that B. o u t s i d e a finite solenoid is not zero. This
ou t s ide B : can be l a rge r o r sma l l e r t han the ou t s ide B^ , depending on the geomet ry
of the so lenoid , bu t bo th a re much sma l l e r t han B. ins ide .
EXAMPLE: REFRACTION OF LINES OF B AT A CURRENT SHEET
Imagine a th in conduc t ing shee t ca r ry ing a cur ren t dens i ty o f a amperes pe r me te r ,
as in Fig. 9-8. We can unders tand the refract ion of l ines of B at a current sheet by
pro ceed ing a s in Sec. 7.1. Since the diverg ence of B is z e r o , t h e n o r m a l c o m p o n e n t
4B
Figure 9-8 Conduc t ing shee t ca r ry ing a cur ren t dens i ty o f a amperes pe r me te r .
Since V • B = 0 , t h e n o r m a l c o m p o n e n t o f B i s the same on both s ides of the sheet .
Accord ing to the c i rcu i t a l l aw , howev er , t he t angen t i a l com po nen t i s no t conse rved
and a line of B i s deflected in the di rect io n s how n.
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228 Magnetic Fields: II
of B i s conse rved:
B„i = B, l2. (9-11)
Also, if we app ly Am pe re ' s c i rcui ta l law to a path of length L tha t i s perp end icula r
to the sheet as in the figure.
fl„L-B ,2L = ,, 0s<L, (9-12)
B, 2= B„ - / ,„* . (9-13)
Th e l ine of force is therefore rot a te d in the c lockwise di re ct ion for an obse rver
looking in the di rect ion of the vector a .
We could have a r r ived a t t h i s re su l t i n ano the r way . The magne t i c induc t ion B
i s due to the current sheet i t se l f and to other currents f lowing elsewhere . Accordingto Am pe re 's c i rcui ta l law. the curre nt sheet pr odu ces , jus t b elow i tse l f in the f igure ,
a B that i s di rected to the left an d who se ma gn i tud e is p 0
a
/ 2 (Pr ob . 9-3). S imilar ly,
th e B jus t ab ov e the sheet i s di rected to the r ight and has the same m ag ni tu de . If we
add this f ie ld to that of the other currents , we see that the tangent ia l components of
B must di ffer as above.
EXAMPLE: SHORT SOLENOID
We can ca l cu la t e B on the axis of the short solenoid of Fig. 9-9 by summing the
co ntr ibu t ion s of the indiv idual turn s , us ing Eq. 8-8. If the length of the solenoid is /
and if i ts radius is a. t he magne t i c induc t ion a t t he cen te r i s
= Po f + i2 a 2NT dz_ Po r+i/2 j
2 J " " 2 (aa 1 + zz
Y
(9-14)
u 0a2N'I
a\a2 + z
2
)l!2
= HcjN'I si n 8„ (9-15)
as in Fig. 9-9. We hav e assum ed th at the soleno id is c lose w ou nd .
For a long so lenoid . 0m -* i t /2 , and B -> fi0NT, as in Sec. 9.1.3.
At one end. again on the axis ,
B = u 0NT si n 0J2. (9-16)
Th e magne t i c ind uc t ion thus dec reases a t bo th end s of the so lenoid , and th i s
is of course due to the fact that the lines of B flare out as in Fig. 9-10.
Upon c ros s ing the winding , t he rad ia l component o f B r e m a i n s u n c h a n g e d ,
bu t t he ax ia l com pon ent ch anges bo th i t s ma gni tude a nd i t s s ign . Fo r exam ple .
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22 9
Figure 9-10 Lines of B for a solenoid wh ose leng th is twice i ts diam eter .
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23 0
1.5 -1.0 0.5
00.5 1.0 1.5
Figure 9-11 The m a g n e t i c i n d u c t i o n B as a funct ion of z on t he axi s of a so lenoid
with a l eng th equa l to 5 t imes its d i a m e t e r . N o t e how sha rp ly th e f ie ld dr op s off at
t he ends .
t h e a x i a l c o m p o n e n t in the upper lef t -hand s ide of t he so lenoid in Fig . 9-10 c h a n g e s
from, say, —0.9p0N'I to +0.\p oN'I, s ince th e surface current dens i ty /. is N'l
(Sec. 9.1.4).
F igure 9-11 shows B as a funct ion of z for a so lenoid wi th / = 10a. See P r o b . 9-5.
THE CURL OF THE MAGNETIC INDUCTION B
Apply ing Stok es's theo rem to Eq. 9-3, we find tha t
(9-17)
for any bounded surface S. Then
V x B = HqJ. ( 9 - 1 8 )
Thi s is an oth er of Ma xwell' s equat ion s, but no t in its final form yet,
bec aus e we are still limited t o static fields, and we hav e no t yet co nsi der ed
magnetic materials.
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Problems 231
9.3 SUMMARY
Ampere's circuital law states that the l i r ie in tegral of B • dl over a c losed
curve C i s equ a l to the cu r r en t f lowing th ro ug h any su r f ace S bo un de d by C:
(j) c B • dl = fi0 js J • da. (9-3)
U p o n a p p l y i n g S t o k e s ' s t h e o r e m t o th i s e q u a t i o n , o n e f in d s t h a t
V x B = /< 0J. (9-75)
PROBLEMS
9-1E DEFINITION OF u 0
Sho w tha t the perm eabi l i ty of free space fi 0 can be defined as follows: If an
infini tely long solen oid carr ie s a cur ren t dens i ty of on e amp ere per meter , then the
magnet ic induct ion in tes las ins ide the solenoid is numerical ly equal to / i 0 .
9-2 MAGNETIC FIELD OF A CURRENT-CARRYING TUBE
A length of tubing carr ies a current / in the longi tudinal di rect ion.
a) W ha t is the value of B o u t s i d e ?
b) Ho w i s A or i en ted ou t s id e?
c) W ha t is the value of B i ns ide?
d) Sho w that A is uniform ins ide.
9-3E MAGNETIC FIELD CLOSE TO A CURRENT SHEET
A conduc t ing shee t ca r r i e s a cur ren t dens i ty o f a amperes p e r m e te r .
Show tha t , ve ry c lose to the shee t , t he magne t i c induc t ion B due to the cur ren t
in the sheet is u. 0x/2 i n the d i rec t ion pe rpendicu la r t o the cur ren t and pa ra l l e l t o the
sheet .
9-4E VAN DE GRAAFF HIGH-VOLTAGE GENERATOR
In a Van de Graaff high -vol ta ge gen erat or , a cha rged in sulat in g bel t i s used tot ranspor t e l ec t r i c cha rges to the h igh-vol t age e l ec t rode .
a) Calculate the current carr ied by a bel t 0.5 meter wide driven by a pul ley
0.1 meter in diam ete r an d ro ta t in g a t 60 revolu t ions pe r second, i f the e lect r ic f ie ld
intens i ty a t the surface of the bel t i s 2 ki lovol ts per mil l imeter .
b) Calculate the magnet ic induct ion c lose to the surface of the bel t , neglect ing
edge effects. See Prob. 9-3.
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232 Magnetic Fields: II
9-5 FIELD ON THE AXIS OF A SHORT SOLENOID
A shor t so l enoid ca r r i e s a cur ren t I a n d h a s N' t u rns pe r me te r .
Show that , a t any point on the axis .
B = - / J 0 / N ' ( C O S + cos a 2 ) i
w h e r e a l and a 2 a re the angles sub tended a t t he po in t by a rad ius R at e i ther end of
the so lenoid .
Fo r exam ple , i f the coi l has a length 2 L, and if the po int i s s i tua ted a t a dis tan ce
x from the center .
L - x L + xcos a 2
1
[R2
+ (L — x )2
]1
'2
'2
[R2
+ (L + x )2
]1
'2
9-6 FIELD AT THE CENTER OF A COIL
A coi l has an inner radius R 1? an ou te r rad ius R 2, and a l eng th L .
a) Show that the magnet ic induct ion a t the center , when the coi l carr ies a
c u r r e n t 7, is
ti 0nIL a + (i 2 + /f)" 2
B = In r—T7T-,2 1 + (1 + P )
w h e r e
a n d n i s t he tu rn dens i ty , o r t he nu mb er of w i res pe r squa re me te r ,
b) Show that the length of the wire is
\ = Vn = 2nn(y.2 - l)/?7?3
,
w h e r e V i s the volume occupied by the wire .
9-7 CURRENT DISTRIBUTION GIVING A UNIFORM B
A long s t ra igh t conduc tor has a c i rcu la r c ros s - s ec t ion of rad ius R and ca r r i e s
a current 7. Ins ide the conductor , there is a cyl indrical hole of radius a whose axis i spa ra l l e l t o the ax i s o f t he conduc tor and a t a d i s t ance b from it .
Show tha t t he mag ne t i c induc t ion ins ide the ho le is un i form a nd equ a l t o
2n(R 2 - a 2)'
See a l so the nex t p rob lem.
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Problems 233
Hint: Use a un i form c ur ren t d i s t r ibu t io n th rou gh out the ci rc le o f rad ius R,
plus a current in the opposi te di rect ion in the hole .
9-8D 1 SADDLE COILS
Figure 9-12 shows the cons t ruc t ion of a pa i r o f s addle co i l s . The windings
occupy the two ou te r reg ions of a pa i r o f ove r l apping c i rc l e s .
a) In what general di rect ion is the f ie ld oriented in the region between the
c o n d u c t o r s ?
b) I t i s sa id that the mag net i c indu ct io n is uniform in the hol low . Is this t ru e?
What is the value of B?
Assume that the length of the coi l i s inf ini te .
You ca n solve this one by s t ra ight integ rat io n i f yo u wish, bu t there is a mu ch
eas ier way. See the hint for Prob. 9-7.
Such co i l s a re used for magne tohydrodynamic gene ra tors . See P rob . 10-16 .
Figure 9-12 (a) Pair of saddle coils forproduc ing a t ransve rse magne t i c f i e ld
ins ide a tubula r enc losure . Only one
turn of each coi l i s shown, (b) C r o s s -
sect ion of the windings . See Prob. 9-8.
9-9D TOROIDAL COIL
A toroidal coi l of N t u rns has a ma jor rad ius R a n d a m i n o r r a d i u s r.
a ) Ca lcu la t e the magne t i c f lux by in t egra t ing B over a cross-sect ion.
b) A t wha t rad ius does B have i ts me an va lue?
f The le t ter D indicates that the problem is re la t ively di ff icul t .
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CHAPTER 10
MAGN ETIC FIELDS: III
Transform ation of Electric and Ma gnetic Fields
10.1 THE LORENTZ FORCE
10.1.1 Example: The Crossed-Field Mass Spectrometer
10.2 EQUIVALENCE OF E AN D v x B
10.3 REFERENCE FRAMES, OBSERVERS,
AND RELATIVITY
10.4 THE GALILEAN TRANSFORMATION
10.5 THE LORENTZ TRANSFORMATION
10.6 IN VARIANCE OF THE VELOCITY OF LIGHT c
10.6.1 Example: The Velocity of Light is Unaffected by the Earth's
Orbital Velocity
10.7 INVARIANCE OF ELECTRIC CHARGE
10.7.1 Example: The Electron Charge
10.8 TRANSFORMATION OF ELECTRIC AND
MAGNETIC FIELDS
10.8.1 Example: The Hall Effect
10.9 SUMMARY
PROBLEMS
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In th i s chap te r , we sha l l s t a r t wi th the long -kno wn f ac t t ha t a mo ving charged
par t ic le i s def lected by a m ag ne t ic fie ld . F o r exam ple, the elec t ron be am ina t e lev i s ion tube i s de f lec ted b o th h or i zo n ta l ly a nd ver t i ca l ly by the m agn e t i c
f ie lds of tw o sets of coi ls car r yin g rapi dly va rying cu rren ts .
Now why should a magnet ic f ie ld exer t a force on an elect r ical ly
charge d pa r t i c l e? The ex p lan a t ion i s t ha t t he par t i c l e " sees" an electric
f ie ld . In other words, the elect rons in the te levision tube "see" an electric
f ie ld when they pass between the def lect ing coi ls .
Le t us cons ider a more genera l case . Suppose one has some conf igura
t ion of e lect r ic charg es and ele ct r ic cu rre nts that p ro du ce an elect r ic f ie ld
in tens i ty E , an d a ma gne t i c indu c t ion in a l ab ora to ry . Bo th E t a n d B ,
are func t ions o f the coor d in a tes .v, v . z . N o w im agine an observ er mo vinga t some co ns ta n t ve loc i ty v wi th r espec t to the l abor a to r y . Th i s obse rver i s
equ ippe d wi th ap pro pr i a t e ins t rum ent s for me asu r ing the e l ec t r i c f ie ld
intensi ty and the magnet ic induct ion. At every point , he wi l l f ind a f ie ld
E 2 , B 2 that i s different f rom E , , B x (except if v. E,. B , are al l paral le l ) . This
chap te r concerns the equa t ions tha t a r e used to f ind E 2 . B 2 . g iven E , . B ^
and inver se ly .
W h e n e v e r o n e e x p r e s s e s E t , B j in terms of E 2 . B 2 , or inversely , one
transforms t he e l ec t romagne t i c f i e ld f rom one r e f e r ence f r ame to ano ther .
Most o f the p rob lems a t t he end o f th i s chap te r concern magne t i c
forces exer ted on particles; macroscopic man i fes t a t ions o f ma gne t i c fo rceswill be dea l t wi th in C ha pt er s 13 an d 15.
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236 Magnetic Fields: 111
10.1 THE LORENTZ FORCE
It is obse rved exp erim entall y that a char ge Q moving at a velocity v in
a region whe re the magn etic ind ucti on is B, is subjected to a magnetic
force
F = Qy x B. (10-1)
Since this force is perpendicular to v, F • v = 0 and the powe r supp lied
to" the particle is zero. The Q\ x B force therefore chan ges the direction of
v without changing i ts magnitude.
M or e gen erall y, if the re is als o an electr ic field E,
F = <2(E + v x B). (10-2 )
This is the Lorentz force.
10.1.1 EXAMPLE: THE CROSSED-FIELD MASS SPECTROMETER
The crossed-field mass spectrometer is illustrated in Fig. 10-1. T he positive ions of
mass in and charge Q have a velocity v given by
- me2
= QV. (10-3)
In the region of the deflecting plates, they are submitted to an upward force QE and
to a downward force QiB. Fh e net force is zero for ions having a velocity
v = E/B, (10-4)
or a mass
in = 2QVB2
E2
. (10-5)
Fhe mass spectrum is obtained by observing the collector current I as a func
tion of E, for a constant B.
This type of mass spectrometer has a poor resolution, mostly because the ion
beam is defocused in the fringing electric an d magnetic fields. It is nonetheless a
simple and useful device if one is concerned only with the very lightest elements.
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23 7
9 "
A
0
Figure 10-1 Crossed- f i e ld mass spec t romete r . Pos i t ive ions produced in a source S
mainta ined a t a po ten t i a l V are focused into an ion beam b. In pas s ing th rough
the superposed e lect r ic and magnet ic f ie lds , the beam spl i ts vert ical ly into i t s
va r ious com pon ent s . Ions of the pro pe r ve loc i ty a re undef l ec t ed and a re co l l ec ted
at C. Th e spec trom eter is enclo sed in a vacu um vessel A e v a c u a t e d b y a p u m p P.
Th e Lor en tz fo rce o f Eq . 10-2 i s i n t r igu ing . W hy shou ld v x B have the
sa me effect as the elect r ic f ie ld E? Cle ar ly , f rom E q. 10-2, the pa r t ic le ca nn ot
tel l whether i t "sees" an E or a v x B term . This i s i l lust r ated in the a bo ve
exa mp le. I t i s a lso i l lust r ated in Pr ob . 10-6, wh ere two typ es of ma ss spe c
t rom ete r a r e com pa red . Th e ion t r a j ec to ry is c i r cu la r in bo t h types , bu t
the deflecting force is QE i n one , and Q\ x B in the o ther . Thus v x B is,
somehow, an elect r ic f ie ld in tensi ty .
As we shal l see in the next sect ions, the explanat ion is provided by
the theory of re lat iv i ty . I t has to do wi th reference f rames.
By observer, w e m e a n e i t h e r a h u m a n b e i n g e q u i p p e d w i t h p r o p e r i n s t r u
m e n t s , o r s o m e d e v i c e t h a t c a n m a k e m e a s u r e m e n t s , t a k e p h o t o g r a p h s , a n d
s o o n , e i t h e r a u t o m a t i c a l l y o r u n d e r r e m o t e c o n t r o l .
10.2 EQUIVALENCE OF E AND v x B
10.3 REFEREN CE FRAME S, OBSERVERS,
AND RELA TIVITY
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238 Magnetic Fields: III
An observer t akes h i s measurement s wi th r espec t to h i s reference
frame. F o r e x a m p l e , o n e n o r m a l l y m e a s u r e s a m a g n e t i c i n d u c t i o n w i t h a n
ins t rum en t tha t is a t rest wi th respec t to the ear th . Thi s par t ic ula r f rame is
usual ly cal led the laboratory reference frame.
Special relativity i s c o n c e r n e d w i t h t h e o b s e r v a t i o n s m a d e b y t w o
observer s , one o f w ho m ha s a con s tan t ve loc i ty wi th r espec t to the o th er .
10.4 THE GALILEAN TRANSFORMATION
Imag ine two r e f e r ence f r ames Si a n d S 2 as in F ig . 10-2, wi th S 2 m o v i n g a t
a veloci ty vi wi th r espec t to S , . N ow im agin e some even t , say a nu c lea r
r eac t ion , occ ur r ing a t a ce r t a in ins t a n t a t a ce r t a in p o in t in space . An obse rver
in S { no tes the coord ina tes o f the even t as x l s y u zu t. An observer in S 2
notes, l ikewise, x 2 , .v 2, z2, t, a n d
x, = x 2 + vt, (10-6)
y, = y2, (io-7)
- i = i 2 . (10-8)
Figure 10-2 Two Car t e s i an coord ina te sys t ems , one moving a t a ve loc i ty v\ wi th
respec t t o the o the r in the pos i t ive d i rec t ion of the common x-ax i s . The two
sys tems ove r l ap when the or ig ins 0 1 and 0 2 coinc ide . We shall always refer to
these two coordinate systems w henever we discuss relativistic effects.
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10.5 The Lorentz Transformation 239
We have set t = 0 at the instant w hen th e two reference frames ov erlap .
These equations constitute the Galilean transformation.
The inverse transformation, giving the coordinates x2, y2, ?i m terms
of xit
yu
zu
is given by inter cha ngi ng the subscr ipts 1 and 2, and chan ging
the sign of v.
10.5 THE LORENTZ TRANSFORMA TION
Altho ugh the Galilean tran sfor matio n is adeq uat e for everyday p heno mena ,
it is only approximate, and the correct equations are those of the Lorentz
transformation:
x , = y(x2
+ vt2), (10-9)
J? I = yi, (10-10)
*i = z2, (10-11)
ti = y ( * 2 + 4 x A (10-12)
where
1
[1 - {v/cfY11
'(10-13)
and c is the velocity of light in a vacuum, 2.997 924 58 x 10s
meters per
second.
The inverse relationship, giving the coordinates in Sl in terms of those
in S2, is again obt ained by interchang ing the subscripts 1 and 2, and changing
the sign of v.
Note that, in general, t1
/ t2. In other words, the observer on Sj does
not agree with the observer on S2
as to the time of occurrence of a given
event.
The Lo rent z tran sfor matio n forms the basis of special relativity. On e
can deduc e from it equ at ion s of tra nsf orm ati on for the length of an object.
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240 Magnetic Fields: III
10.6 IN VARIANCE OF THE VELOCITY OF LIGHT c
Exp erim ents designed to measu re the velocity of light c in a vac uum , with
respect to reference frames moving at various velocities, always give the same
value. Thus c is said to be invariant.
10.6.1 EXAMPLE: THE VELOCITY OF LIGHT IS UNAFFECTED BY
THE EARTH'S ORBITAL VELOCITY
The orbi ta l veloci ty of the e a r t h a r o u n d the sun is 3 x 104 m e t e r s per second . Thi s
is two o r d e r s of m a g n i t u d e l a r g e r t h a n th e t angent i a l ve loc i ty due to the r o t a t i o n
of th e e a r t h a b o u t its own axis . At n o o n t i m e , th e orbi ta l veloci ty is wes tward whi l e ,
a t midnigh t , it is e a s t w a r d . If o n e m e a s u r e s c in the eas t -wes t di rect ion in the l a b o
ratory, first at n o o n t i m e and t hen at m i d n i g h t , one finds precisely the same va lue ,
wi thin the exper imenta l e r ror .
10.7 IN VA RIA NCE OF ELECTRIC CHA RGE
Electric charge is also invar iant . This is again an exper imen tal fact.
10.7.1 EXAMPLE: THE ELECTRON CHARGE
In th e Millikan oil-drop experiment, one m e a s u r e s th e veloci ty of an elect r ical ly
c h a r g e d m i c r o s c o p i c oil d r o p in an electric field. The veloci ty is of th e o r d e r of mil l i
m e t e r s pe r m i n u t e an d is a m e a s u r e of the e lec t r i c cha rge on the d r o p . The c h a r g e
ca r r i ed by a dr op i s a lways a mul t ip l e of t he e l ec t ron cha rge , 1.602 x 10_ 1 g c o u l o m b .
If now one m e a s u r e s in the l a b o r a t o r y f r a m e th e c h a r g e on an e lec t ron emerg ing
from an acce le ra tor w i th a v e lo c it y a p p r o a c h i n g th e veloci ty of l ight , by deflect ing
it in a known magnet ic f ie ld, one finds again 1.602 x 1 0- 1 9 c o u l o m b .
Electromagnetic Fields and Waves, C h a p t e r 5.
the duration of an event, a velocity, an acceleration, a force, a mass, and
so on /
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10.8 Transform ation of Electric and Magnetic Fields 241
10.8 TRANSFORMA TION OF ELECTRIC
AND MAGNE TIC FIELDS
T h e e x p r e s s i o n for the Loren tz fo rce in Eq. 10-2 give s us a clue as to ho w to
t r ansfo rm a magnet ic f ie ld . In t h i s express ion , F, (9, E, v, B are all m e a s u r e d
in the same reference f rame S,, w h i c h is n o r m a l l y the l a b o r a t o r y f r a m e :
F , = Q{El
+ v x B,) (10-14)
N o w c a l l S2 the reference f rame of the par t i c l e , moving at the veloci ty
v at a cer t a in ins t an t .
W h a t is the field in S 2 ? Let us a s s u m e for the m o m e n t t h a t v2
« c2
,
w h e r e c is the veloci ty of l ight . In t ha t case , the force is the s a m e in b o t hreference f rames. In S
2. the par t ic le veloci ty is z e r o and t h e r e can be no
magnet i c fo rce . Then the force in S2 m u s t be QE
2, w h e r e
E , = E , + v x Bp (10-15)
So the magnetic field B, in f rame 1 b e c o m e s an electric field v x B, in
f rame 2.
T h e a b o v e e q u a t i o n is val id only for r2
« c2
. W h e n v a p p r o a c h e s c,
it is s h o w n in r e l a t iv i ty theory tha t the forces in the two reference f rames
are different, and the cor r ec t va lue of E2
is as f o l l o w s : 1
E21 = 7 ( E , i + v x B,). (10-16)
E 2 | I = E 1 H , (10-17)
w h e r e the subscr ip t s r e f e r to the c o m p o n e n t s t h a t are e i t h e r p e r p e n d i c u l a r
or paral le l to the veloci ty vi of f r ame S2 wi th r espec t to S,, as in Fig. 10-3;
7 is as def ined in Eq. 10-13.
The magne t i c f i e ld in S2 is g iven by
x E ^ . (10-18)
B 2 | , = B[||. (10-19)
Eq uat ion s 10-16 to 10-19 are the equations of transformation for E and B.
* Electromag netic Fields and Waves, pages 262 and 288.
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242 Magnetic Fields: III
Thus, given a field E 1 ( B ( in frame Su one can calculate the field E 2 ,
B 2 in frame S2. The inverse transforma tion is obtai ned, as usual, by inter
cha ngin g the subscrip ts 1 an d 2, an d chan gin g the sign of v.
When v2
« c2
, y % 1 an d
E 2 = Ei + v x B,, (10-20)
1
B 2 = B , - ^ v x E , (10-21)
c
Note that the expression for the Lorentz force given in Eq. 10-2 remains
t rue, even if v x c.
10.8.1 EXAMPLE: THE HALL EFFECT
S e m i c o n d u c t o r s c o n t a i n e i t h e r one or b o t h of two types of mobi le cha rges , namely
c o n d u c t i o n e l e c t r o n s and holes (Sec. 5.1). W h e n a cur ren t f lows th rough a ba r ofs e m i c o n d u c t o r in the p r e s e n c e of a t ransverse magnet ic f ie ld B, as in Fig. 10-4, th e
mobile charges dri f t , not only in the di rec t ion of the appl ied e lect r ic f ie ld E but also
in a d i r e c ti o n p e r p e n d i c u l a r to b o t h the appl ied e lect r ic and magnet ic f ie lds , because
of the Qy x B force. This gives rise to a vol tage di fference V be tween th e u p p e r and
l ower e l ec t rodes . The drift is s imi l a r to the m o t i o n of t he ions in the m a s s - s p e c t r o
m e t e r of Sec. 10.1.1.
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243
(c) (d)
p-type material n-type material
Figure 10-4 Hall effect in semiconductors. In p-type materials conduction is due
to the drift of positive charges (holes) and the Hall voltage is as shown in (a). In
H -type materials conduction is due to the conduction electrons and the Hall
voltage has the opposite polarity as in (b). Ordinary conductors such as copper
behave as in (b). Figure (c) shows the two opposing transverse forces QEy
and
Qv x B on a positive charge drifting along the axis of the bar at a velocity v.
Figure (d) shows how these forces are reversed in «-type material.
If the voltmeter V draws a negligible current, the plates charge up until their
field Ey is sufficient t o stop the tra nsverse drift. This transve rse electric field is called
the Hall field.
We shall first calculate the magnitude and direction of the Hall field Ey by using
the Lorentz force. Then, as an exercise, we shall find the field E 2 , B2
in the reference
frame of the charge carriers, to arrive again at Ey. We assume that the material is
zi-type as in Fig. 10-4b and 10-4d: the charge carr iers are negat ive.
a) The field Ey
must be such that Eq. 10-4 is satisfied. Then its magnitude is vB.
Since v x B is in the positive direction of the y-axis (Fig. 10-4d), Ey must point in
the negative direction and
vB. (10-22)
as in Figs. 10-4b and 10-4d.
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244 Magnetic Fields: III
b) In the l abora to ry f rame Sj , the electric field intensity is Ex\ + E j , and the
magnet ic induct ion is Bk . We have omi t ted th e subscr ip t s 1 for simplici ty .
To find the field in the reference frame S 2 of the mo vin g charges, we use Eqs. 10-20
a nd 10-21. For n- type mater ia l , th e velocity is — ri, wh e re v is a pos i t ive quan t i ty .
T h e n
E 2 = £. vi + £ r j - ri x Bk . (10-231
= £ v i + (£,. + vB)i, (10-24)
B 2 = B k + ( l / c 2 ) r i x ( £ , i + £ v j ) , (10-25)
= [B + (vEt c2
)]k. (10-26)
The force in frame 2 is QE 2, and since it has no v c o m p o n e n t , £ v = —iB as
prev ious ly .
If there are n e lec t rons per cub ic meter , each car ry ing a charge e.
I = abJ = ab(nev).
vBb = Bl/nea.
(10-27)
(10-28)
Note tha t charge car r ie rs of ei ther s ign are swep t down by the magnetic f ield .
The Hall effect is c o m m o n l y u s e d for m e a s u r i n g m a g n e t i c i n d u c t i o n s and for
var ious o ther pu rposes , some of which ar e descr ibed in P r o b s . 8-7, 10-15, and 17-12.
10.9 SUMMARY
A charge Q moving at a velocity y in superposed electric and magnetic
fields E an d B is su bm itt ed to the Lorentz force
F = Q(E + v x B), (10-2)
wher e F, E, v, and B are all mea sur ed w ith respect to the same reference
frame S,.
To find the field perceived by the charge Q in its ow n reference frame
S2 moving at the velocity ti with respect to St, one mu st use special relativity.
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10.9 Summary 245
which is bas ed o n the Lorentz transformation:
x l = y(x 2 + M 2X (10-9)
V l = y 2, (10-10)
(10-11)
h= y[t 2 + ^ x 2 ) t (11-12)
wi th
y[1 - (v/cf\
T 1 / 2 '(10-13)
an d c equ al to the veloci ty of l ight in a va cu um , 2 .997 924 58 x 10 8 m e t e r s
p e r s e c o n d .
I t i s poss ib le to deduc e f rom th i s set o f equ a t io ns o the r t r ans fo rm at ion
equat ions for a mass, a veloci ty , e tc . A given elect r ic charge Q and the veloci ty
of light c a lw ays ha ve the sam e value, wha teve r the veloci ty of the reference
f r ame wi th r espec t to which they a r e measured .
F or a given f ie ld E ^ I* ! in frame S,, the f ield in S 2, moving a t a ve loc i ty
v wi th res pect t o S l, is given by
E 2 ± = y(E1±
+ v x B,) , (10-16)
E 2 N = E j | | , (10-17)
B 2 1 = ? ( b U - ^ X E A (10-18)
Ban = Bj ||. (10-19)
If r 2 « c 2, t h e n y = 1 and
E 2 = E , + v x B , , (10-20)
B 2 = B i - - T y x E 1 . (10-21)
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246 Magnetic Fields: III
PROBLEMS
I0-1E THE CYCLOTRON FREQUENCYA par t i c l e o f mass in. c h a r g e Q. and veloci ty v. describes a c i rc le of radius R in a
plan e perpe nd icula r to the di rec t ion of a uniform ma gne t ic field B.
a ) Show tha t
BQt = mv 2 R.
b) Show tha t the an gu lar veloci ty
co = BQ/m.
The frequency BQ 2nm is called the cyclotron frequency be cau se it is the frequen cy at
which an ion c i rcula tes in a cyclotron.
N o t e h o w a) is independent of the velocity r of the part ic le . This is not s t r ic t ly
t rue , however , because m is i tself a function of the velocity:
m 0
m
~ [1 - ( r 2 / r 2 ) ] " 2 '
w h e r e c i s the veloci ty of l ight in a vacuum and m 0 is the rest mass. In pract ice . a> m ay
be cons idered to be independent of c . for c2
« c2
. T h e m a s s m i s 10 percen t large r th an
m 0 w h e n v i s ab ou t 0.4c .
The above three equat ions are val id a t a l l veloci t ies .
c) Calculate the cyclotron frequency for a deuteron in a f ie ld of one tes la . for
r2
« c2
.
I0-2E MOTION OF A CHARGED PARTICLE IN A UNIFORM B
A charged part ic le moves in a region where E = 0 and B i s uniform.
Show that the part ic le describes e i ther a c i rc le or a hel ix.
10-3E MAGNETIC MIRRORS
Figu re 10-5a show s a cross-sect io n of a solen oid hav ing a unifo rm t urn dens i ty,
except nea r the ends where ex t ra tu rns a re added to ob ta in a h ighe r magne t i c induc t ion
than nea r the cen te r .
Show q ual i ta t iv ely th at , if the axia l veloci ty is not to o large, a charge d part ic le
that spi ra ls around the axis in a vacuum ins ide the solenoid wi l l be ref lected back when
i t reaches the higher magnet ic f ie ld. The regions of higher magnet ic f ie ld are cal ledmagnetic mirrors.
Magne t i c mi r rors a re used for conf in ing h igh- t empera ture p l a smas . (A plasma
i s a highly ionized gas . I t s net charg e dens i ty is ap pro xim ate ly zero.)
In the Van Allen radiation hells, cha rged pa r t i c l e s a re t rapped in the ea r th ' s
ma gn et ic f ie ld, osci l la t ing no rth an d sou th a lon g the ma gn et ic f ie ld l ines . Th ey a re
re f l ec t ed nea r the Nor th and South magne t i c po les , where the l i nes c rowd toge the r ,
forming magnet ic mirrors , as in Fig. 10-5b.
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24 7
Solar wind
Figure 10-5 (a) Cross -s ec t ion of a so lenoid wi th magne t i c mi r rors a t t he ends .
{b ) Gray l ines : magnet ic f ie ld of the earth. The earth acts as a magnet ic dipole .
Black l ines : magnet ic f ie ld near the earth, as observed by means of sa te l l i tes . The
so la r w ind* is com pos ed of p ro ton s and e l ec t rons evapora ted f rom the surface of
the sun. Since the interplanetary pressure is low, there are few col l is ions and the
conduc t iv i ty i s h igh . In moving th rough the magne t i c f i e ld o f t he ea r th , t he cha rge
part ic les are subjected to a v x B elect r ic f ie ld, and curre nts f low acc ord ing to
Lenz 's law. The black l ines show the net f ie ld.
10-4E HIGH-ENERGY ELECTRONS IN THE CRAB NEBULA
Som e as t rophys ic i s t s be l i eve tha t t he re ex i s ts , w i th in the Cra b Neb ula , e l ec t rons
with energies of ab ou t 2 x 1 01 4
elect ro n-v ol ts spi ra l l ing in a ma gne t ic field of 2 x 10 ~8
tes la . See Prob. 10-1.
a ) Ca lcu la t e the ene rgy W of such an e lect ron in joules .
Electromagnetic Fields and Waves, p. 502.
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248 Mag netic Fields: III
b) Ca lcu late i t s ma ss from the re la t io n
W= H i e
2
.
w h e r e c is the velocity of l ight in a vacuum, 3 x 1 0 8 mete rs pe r s econd .
How does th i s mass compare wi th tha t o f a low-ve loc i ty (c 2 « c 2 ) e lec t ron ,
which is9.1 x 1 0 " 3 1 k i l o g r a m ?
c) Calculate the radius of i t s orbi t , se t t ing v = c and neglect ing the veloci ty
component pa ra l l e l t o B .
d) C a lcu la t e the numb er of days requi red to comple te one tu rn .
10-5E MAGNETIC FOCUSING
There exis t many devices that ut i l ize f ine beams of charged part ic les . The cathode-
ray tube tha t i s used in te levis ion receivers an d in osci l loscop es is the be s t -k now n
exam ple . The e l ec t ron mic ros cope i s ano the r exam ple . In these dev ices the pa r t i c l ebea m is focused an d deflected in mu ch the same way as a l ight bea m in an opt ical in
s t r u m e n t .
Beam s of char ged part ic les can be focused and deflected by prop erly shap ed
electric f ields . See, for exam ple , P rob s . 2-11 and 2-12. Th e e lect ron beam in a TV t ub e
is focused with electric fields and deflected with magnetic fields. Beams can be deflected
along a c i rcular path in a magnetic f ie ld as in P ro b. 10-1. Let us see how the y can be
focused by a magnet ic f ie ld.
Figu re 10-6 sho ws an e lect ro n gun s i tuat ed ins ide a long solen oid. The e le ct ro ns
tha t emerge f rom the ho le in the anode have a sma l l t r ansve rse ve loc i ty component
and, i f there is no current in the solenoid, they spread out as in the f igure . Let us see
wha t happens when we tu rn on the magne t i c f i e ld .
Figure 10-6 Foc usin g of an e lect ron be am in a uniform ma gn et ic f ield: F is a
heated filament, A i s the anode, and the e lect ron beam is focused a t P . The
accele rat ing vo l tage is F . A f ilament sup ply of a few vol ts is con nec ted betwe en
the top two t e rmina l s . Wh en the so lenoid is no t ene rg ized , t he beam d ive rges a s
shown by the dashed l ines . See Prob. 10-5.
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Problems 249
Let the ve loc i ty component s a t t he ho le be v x a n d c v . with r 2 » v 2. T h e n
1 1
,eV = - m ( r
2
+ 17) % - mo*.
If v y = 0. the e lect ron cont inues paral le l to the axis . I f v y =F 0, i t follows a helical
path a t an ang ula r veloci ty cu as in Pro b. 10 -1. After one com plete tu rn, i t has re tu rned
to the axis. So. if B i s adjus ted correc t ly, a ll the e lect ro ns wi l l con verg e a t the same poin t
P, as in the f igure , and form an image of the hole in the anode.
a) Show that this wi l l occur i f
L
b) Ca lcu la t e the number of ampere tu rns pe r me te r IN ' in the solenoid, for an
acce le ra t ing vo l t age V of 10 ki lovo l ts and a dis tan ce L of 0.5 meter.
In actual pra ct ice , ma gn et ic focus ing is achieve d with sho rt coi ls, an d not wi th
so lenoids .
I0-6E DEMPSTER MASS-SPECTROMETER
Figure 2-10 shows an e lect ros ta t ic veloci ty analyser . I t was shown in Prob. 2-11
that a part ic le of charge Q. mass m, and veloci ty v pass ing through the f i rs t s l i t wi th the
proper o r i en ta t ion wi l l r each the de tec tor i f
v = QER/m.
Figure 10-7 shows a Dempster mass-spectrometer. Here again, the ion describes
a c i rc le , except that the centr ipeta l force is now Qy x B . ins tead of QE . T h e m a g n e t i c
f ie ld is pro vid ed by an e lect r om agn et .
a ) Show tha t
m = QR2
B2
/2V.
b) In one pa r t i cu la r expe r iment , a mass -spec t romete r o f t h i s l a t t e r t ype was
used for the hyd rogen ions H^.H^.Hi, wi th /? = 60 .0 mi l l ime te rs and V = 1000 vol ts .
Figure 10-7 Cross -s ec t ion of a mass
spec t romete r o f t he Demps te r t ype .
Ions produced a t t he sourse S a re
col lected a t C. The ions are accel
erated through a di fference of poten
tial V and describe a semicirc le of
r a d i u s R. See Prob. 10-6.
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250 Magnetic Fields: III
T h e W ,+ i on i s a p ro ton . H% i s com pose d of two pro t ons and one e l ec tron , and H^ is
com posed of th ree pro to ns and two e l ec t rons .
Find the values of B for these three ions .
c ) Draw a curve of B as a function of in under the above condi t ions , f rom
h y d r o g e n t o u r a n i u m .
Wo uld you use th is type of spec t ro mete r for heavy ions? W hy ?
10 -7 MASS SPECTROMETER
Figure 10-8 show s a mass spec t rom ete r tha t s epa ra t e s ions , bo th a ccord in g to
the i r ve loc ie s and accord ing to the i r cha rge - to-mass ra t ios .
Sho w that an ion of cha rge Q, m a s s in. and veloci ty v i s col lected a t the poin t
Figure 10-8 Mass spectrometer . Ions injected a t A fol low hel ical paths in the E a n d
B f ie lds that are both paral le l to the y-axis . Each square is a separate col lector .
See Prob. 10-7.
I0-8E HIGH-TEMPERATURE PLASMAS
One method of in j ec t ing and t rapp ing ions in a h igh- t empera ture p l a sma i s
i l l us t ra t ed in F ig . 10-9 . Mo lecu la r i ons of deu te r ium , D j ( two deu te ron s p lus one
e lec t ron , t he deu te ron b e ing compo sed of one pro ton a nd one neu t ron) , a re in jec ted
into a magnet ic f ie ld and are dissocia ted into pairs of D + i ons (deu te rons ) and e l ec t rons
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25 1
D +
in a high -inten s i ty arc . Th e radius of the t ra jectory is redu ced an d the ions are t r ap ped .See P rob . 10-1 .
a) Cal cul ate the rad ius of cu rva tur e P for DJ ions having a kinet ic energy of
600 k i loe lec t ron-vol t s i n a B of 1.00 tesla.
b ) Ca lcu la t e R for the D + ions prod uce d in the arc . Th e D + ions have one half
the kinet ic energy of the D 2 i ons .
10-9 HIGH-TEMPERATURE PLASMAS
The type of d i s cha rg e shown in F ig . 10-10 has been used to p ro duce h igh- t em
perature plasmas . The discharge has the shape of a cyl indrical shel l and is s i tuated in
the space be tween two conduc t ing coaxia l cy l inde rs tha t ac t a s re tu rn pa ths fo r t he
c u r r e n t .
a) Is there a magnet ic f ie ld outs ide the outer cyl inder? Ins ide the inner cyl inder?W h y ?
b) Draw a f igure showing l ines of B .
c ) Cons ide r the upper pa r t o f t he d i s cha rge . Suppose a pos i t ive ion , moving
tow ard the r ight , has an upw ard c om po ne nt of veloci ty. Ho w is i t s t ra jectory affected
by the magnet ic f ie ld af ter i t has emerged from the discharge?
d) Wha t happens to a pos i t ive ion tha t t ends to l eave the upper pa r t o f t he
d i s c h a r g e b y m o v i n g d o w n w a r d ?
G p
Figure 10-10 Sec t ion th rough a dev ice
that has been ut i l ized for producing a
h i g h - t e m p e r a t u r e p l a s m a P . T h e
copper enc losure , r epresen ted by
heavy l ines , i s sepa rated from the
plasma by a pair of glass cyl inders G.
See Prob. 10-9.
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252 Magnetic Fields: III
10-lOE ION-BEAM DIVERGENCE
Let us calcula te the forces act ing on a part ic le in a beam of pos i t ively charged
part ic les having a veloci ty v a n d c a r r y i n g c h a r g e s Q. We as sume tha t t he re a re no
external ly appl ied e lect r ic or magnet ic f ie lds . We shal l cons ider pos i t ive part ic les , but
i t wi l l be a s imple mat ter to apply our resul ts to negat ive ones .
Firs t , there is an o utw ard E, as in Fig. 10-11. This is s imply a case of e lect ros ta t ic
repul s ion . A l so , t he beam cur ren t p roduces an az imutha l B . an d v x B poin t s inward ,
as in the figure.
Thus , there is an outward e lect r ic force QE and an inward magne t i c fo rce Q\ x
B. Does the beam converge or d ive rge? Exper imenta l ly , e l ec t ron and ion beams a lways
diverge, when left to themse lves , but the diverg ence is s l ight when th e part ic le veloci ty
r approaches the ve loc i ty o f l i gh t c.We consider a part ic le s i tuated a t the edge of a beam of radius R. T h e c u r r e n t
is / and the part ic le veloci ty is v.
C a l c u l a t e
a) the charge / . per mete r of bea m.
b | the outward e lect r ic force QE.
c) the inward magnet ic force QvB.
d) t he net force.
You shou ld f ind that the net force points outw ard a nd is pro po rt io na l to 1 —
e 0 / ( 0 r2 , or to 1 - (v/c)
2 (see Eq. 20-24). Th en the net force tend s to zero as r - » c.
If the part ic les are negat ive . E is inward ins tead of outward, but QE is again
o u t w a r d . A l s o . B poin t s in the oppos i t e d i rec t ion and Q\ x B poin t s aga in inward .
In pract ice , a vacu um is neve r perfect . Let u s say the ions are pos i t ive . I f thei r
energy is of the order of tens of e lect ron-vol ts or more, they ionize the res idual gas .
forming low-energy pos i t ive ions and low-energy e lect rons . These pos i t ive ions dri f t
Figure 10-11 Po rt io n of a pos i t ive-ion be am an d i ts e lect r ic and m agn et ic f ie lds .
See Prob. 10-10.
e ) E lec t rons move toward the l e f t . Wha t happens to them when they l eave the
d i s c h a r g e , b y m o v i n g e i t h e r u p w a r d o r d o w n w a r d ?
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Problems 253
away f rom the pos i t ive beam. The low-ene rgy e l ec t rons , however , r ema in t rapped in
the beam and neu t ra l i ze pa r t o f i t s space cha rge , t he reby reduc ing E. T h e m a g n e t i c
force then t ends to p inch the beam . Thi s ph eno me non is ca l led the pinch effect. It is
a lso cal led gas focusing. With gas focus ing, some of part ic les in the beam are scat tered
away, in col l iding with the gas molecules .
10-11 ION THRUSTER
Figure 10-12 shows an ion thrus ter that ut i l izes the magnet ic force . Other types
of thru s ter ar e described in Pro bs . 2-14, 2-15. and 5-3. An arc A ionizes the gas , which
enters from the lef t , and the ions are blown into the crossed e lect r ic and magnet ic f ie lds .
I f t he cur ren t f lowing be tween the e l ec t rodes C a n d D i s / . then the thrus t F is Bis.
which is the force exerted on the gas ions . We assume that B an d s a r e u n i f o r m t h r o u g h
out the thrus t chamber, and we neglect the fr inging f ie lds .
If H I ' is the mass of gas flowing per unit t ime, and if v i s the exhaus t veloci ty wi th
respect to the vehicle , then F is m'v, and the kinetic power communica ted to the gas i s
1 , 1 1P = - H i ' r
2 = -Fv = - Blsr.2 2 2
If the efficiency is defined as
PG + P D
w h e r e P D i s the pow er diss ipate d as heat be tween C and D. show tha t
1 1 1~ 2 H I ~ ~ 2~T ~ IE'
1 + r ~ 1 + 1 + —aB-x (TBV Br
Figure 10-12 Ion thrus ter . See Prob. 10-11.
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254 Mag netic Fields: III
10-12E GAMMA
For wha t va lue of v is the value of y one pe rcen t l a rge r than un i ty?
10-13E REFERENCE FRAMES
Co nsid er tw o reference frames , 1 and 2, as in Fig. 10-2, wh ere frame 2 has a
veloci ty v = c/2 with respect to 1.
An event is found to occur at .v, = \\ = r , = 1 me te r . t1 = 1 s econd .
F ind the coord ina te s .v 2, y 2 . - 2 . h-
10-14 REFERENCE FRAMES
See the preced ing prob lem.
If the event occurs a t x 2 = y 2 = r , = 1 me te r , t2 = 1 s econd .
F ind v , , V i , z u tx.
10-15 HALL EFFECT
Let us inves t igate more c losely the Hal l effect described in Sec. 10.8.1. W e a s s u m e
again th at the curre nt i s carr ie d by e lect rons of cha rge —e. Their effective mass is m* .
T h e effective mass takes into account the periodic forces exerted on the e lect rons , as
they t ravel through the crys ta l la t t ice . As a rule the effect ive mass is smaller t han the
mass of an i so la t ed e l ec t ron .
The e lect rons are subjected to a force
F = -e(E + v x B ) .
where E has two component s , t he appl i ed f i e ld E x and the Hall field E y. T h e a v e r a g e
drift velocity is
Fv =
e
wh ere . / / i s thei r mo bi l i ty. T he m obi l i t y of a part ic le in a med ium is i t s aver age drif t
velocity , divided by th e elec tric field ap plie d to it . Th e law F = m*a appl i e s on ly be tween
col l is ions wi th the crys ta l la t t ice .
a ) Show tha t
v x = -J/(E X + v yB) .
v, = -.M{E y - v xB) ,
r = = 0.
b) If i! is t he num ber of con duc t ion e l ec t ron s pe r cub ic me te r , t he cur ren t dens i ty
is
J = —nev.
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Problems 255
Show tha t
E x - //E.Ji
J x = ne .M — T A r ,
E, + M E XBJ,. = nejf— i—
1 + .//'ti
ll J y = 0, as in Sec. 10.8.1,
.//E XB,
bV = -J/VJ3.
a
Note tha t t he Ha l l vo l t age V y i s p ro por t ion a l t o the product of the appl ied vol tage
V x a n d B. This fact makes the Hal l effect useful for mul t iplying one variable by another .
See Prob. 17-12.
When connected in this way, the Hal l e lement has four terminals and is cal led
a Hall generator, o r a Hall probe.
As a rule , rc- type materia ls are used because of thei r re la t ively high mobi l i ty.
However, the Hall effect exists in al l c o n d u c t i n g b o d i e s .
On e imp or t an t ad van tage of Ha ll e l ement s is t ha t t hey can be mad e sma l l. Fo r
ce r t a in appl i ca t ions they a re incorpora t ed in to in t egra t ed c i rcu i t s .
c) Calculate F v fo r b = 1 mil l im eter , a = 5 mil l ime te rs . .// = 1 mete rs squa red
per vo l t s econd ( ind ium an t imonide ) , V x = 1 volt , B = 1 0 - 4 tesla.
d ) Show tha t , i f £ , = 0 ,
AP , ,— = .//2B 2,Pn
0, an d A P is the
increase in res is tance up on a pp l icat io n of the mag net i c f ie ld.
Fhe Hall field E y i s ma de equa l t o ze ro by ma king c smal l , say a few micro
me te rs , and p la t ing conduc t ing s t r ips pa ra l l e l t o thev ax i s , a s in F ig . 10-13. The e l ement
then ha s only two term ina ls an d is cal led a magnetoresistor.
Magne tores i s to rs a re used for measur ing magne t i c induc t ions . See P robs . 15-3
and 15-5 .
Figure 10-13 Magne tores i s to r . See
Prob . 10-15 .
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256 Magnetic Fields: III
MAGNETOHYDRODYNAMIC GENERATOR
Figure 10-14 shows schemat ical ly the principle of operat ion of a magnetohyclrody-namic, o r MHD generator. Th e func t ion of an M H D gen era to r is t o t rans form the
kinet ic energy of a hot gas di rect ly into e lect r ic energy.
A very hot gas is injected on the left at a high velocity v. Th e gas i s ma de con
duct ing by inject ing a sa l t such as K 2 C 0 3 t ha t i on izes read i ly a t h igh t empera ture ,
forming pos i t ive ions and free e lect rons . Conduct ivi t ies of the order of 100 S i e m e n s
per me te r a re thus ach ieved , at t emp era tur es o f abo ut 3000 ke lv ins . (The con duc t iv i ty
of copper is 5.8 x 1 0 7
S i emen s per me te r ) . Of course r2
« c 2.
The ions and the e lect rons are deflected in the magnet ic field. With the B s h o w n ,
the pos i t ive ions a re de f l ec t ed downward and the e l ec t rons upward . The re su l t ing
curr ent f lows thr ou gh a load res is tance R. giving a vol tage di fference V. and hence
an electric field E be tween the p l a t e s. On e such M H D gen era to r is p l ann ed to have
a power ou tpu t i n the 500-megawat t r ange . I t would use a B of several teslas over aregion 20 me ters long and 3 me ters in dia met er .
F igure 10-14 Schemat i c d i agram of a magne tohydrodynamic gene ra tor . The
kinetic energy of a very hot gas. injected on the left at a velocity v, is transformed
direct ly into e lect r ic energy. The magnet ic f ie ld B i s that of a pai r of coi ls , outs ide
the chamb er . See P ro b . 9 -8. Th e movin g ions a re de f lec t ed e i the r up or do wn,
depending on the i r s ign . See P rob . 10-16 .
a) Let us sup po se that E, B . and the veloci ty v a re un i form ins ide the chamber .
These a re ra the r c rude a s sumpt ions . In pa r t i cu la r , v i s not uniform because (a) the
ions and e lect ro ns have a vert ical com po ne nt of veloci ty, an d (b) the vert ical veloci t ies
give Q\ x B forces tha t po in t backward , fo r bo th types of pa r t i c l e ; t hese brak ing
forces s low do wn th e horizo ntal dr i f t of the charge d part ic le s . I t i s thr ou gh this
dece lerat io n that pa rt of the ini t ia l kine t ic ener gy of the gas is t ran sfor me d into
elect r ic energy.
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Problems 257
The elect rodes each have a surface area A and a re s epa ra t ed by a d i s t ance b.
S h o w t h a t
V 'I =
R + R,
where R, is the internal res is tance of the generator , and V is the value of V w h e n
R i s inf ini te . Thevenin 's theorem (Sec. 5.12) therefore appl ies here .
Note tha t , because v 2 « c 2, the curre nt den s i ty is the sam e in the reference frame
of the laboratory as i t i s in the reference frame of the moving gas .
Solution : W e hav e two reference frames , tha t of the lab ora tor y and that of the
moving gas .
In the reference frame of the gas,
E 2 = E + v x B. (1)
We omit the subsc ript 1 for the quan t i t ie s measu red in the reference frame of the
l abora tory , fo r s impl i c i ty . Thus
J 2 = ffE 2 = a(E + v x B), (2)
j = j 2 = a(vB - E) = a (vB - H , (3 )
I = JA = a A \vB - — , (4)
(5)
b
vB b V
an d
rtAR b R + R,
b a A
V = vBb. R t = b:aA. (6)
N o t e t h a t V is |v x B|/?. and that R i i s the res is tance of a conductor of conduct ivi ty
CT, length h. and c ros s - s ec t ion A.
b) Find an express ion for the eff ic iency
^ pow er diss ipa ted in R
t o t a l power d i s s ipa ted '
as a funct ion of / . that does not involve the res is tance R.
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25 8
I oAiB I (JAIB
Figure 10-15 (a) Effic iency as a funct ion of load cur ren t for the m ag ne to hy dr o-
dy na mi c gene rato r of Fig. 10-14. (£>) O ut pu t v ol tage as a funct ion of load curr ent .
W ha t is the efficiency at / = 0?
W ha t is the value of the cur rent when the eff ic iency is zer o?Sketch a curve of S as a function of / .
Solution:
I2
R R
(8)r-R + r-R, R + R,
R <TAR
R + (b/aA) oAR + b
Since
oAvBb
oAR + b"
(9)
(10)
oAR _ RI
aAvBb ~ v~Bb'
V - Rjl vBb - (b/aA)I I
vBb vBb oAvB(12)
Th e effic iency is equ al to uni ty when / = 0.
Th e eff iciency is zer o wh en I = aAvB. In that case R = 0. f rom Eq. 4 . t he ou t pu tvo l t age V i s zero, an d / = V'/Ri, from Eq. 5.
Fig ure 10-15a sho ws & as a funct ion of / .
c) Find an express ion for V as a funct ion of / that does not involve R.
W ha t is the value of V when the cur ren t i s ze ro?
W ha t is the value of the curren t when th e vol tage is zero ?
Sketch a curve of V as a function of / .
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Problems 259
Solution:
V = IR= vBb - 7 - vB b ( 1 — ).
W he n th e curr ent i s zero (R -> oo) , the output vol tage is vBb. The ions then
f low through the gene ra tor hor i zonta l ly , unde f l ec t ed . See Sec . 10.1.1.
When the vo l t age i s ze ro , R = 0 and / i s equ al to aAvB as above .
F igure 10-15b shows V as a funct ion of I.
10-I7E ELECTROMAGNETIC FLOWMETERS
Electromagnetic flowmeters operate as fol lows. A f luid, which must be a t leas t
s l igh t ly con duc t ing (b lood , fo r example ) , f lows in a nonco ndu c t ing tube be tween the
poles of a mag ne t . The ions in the f luid are then sub jected t o mag net i c forces Q\ x B
i n the d i rec t ion pe rp endic u la r t o th e ve loc i ty v of the f luid and to the magnet ic induct ion
B. Electrodes placed on e i ther s ide of the tube and in contact wi th the f luid thus acquire
charg es of op po s i te s ign s ; the resul t ing vo l tage di fference is a me asu re of v.
No te that , in the e lect ro ma gn et ic f lowmeter, ions of both s igns hav e the sam e
veloci ty v and the pola r i ty of the e lect ro des is a lway s the sam e for a given direct io n
of flow and for a given direction of B. Compare wi th the Hal l effect discussed in Sec.
10.8.1.
Faraday a t t empted to measure the ve loc i ty o f t he Fhames r ive r in th i s way in
1832. Th e mag ne t ic fie ld was of cour se that of the earth .
Consider an ideal ized case where the tube has a rectangular cross-sect ion af t ,
with side b paral le l to B , and w here the ve loc i ty is t he s ame th ro ugh ou t the c ros s - s ec t ion .
W ha t i s t he vo l t age be tween the e l ec t rod es?
In fact , the veloci ty is m ax im um on th e axis an d zero a t the inner surface of
the tube . Moreover , t he non-uni form v x B f ie ld causes currents to f low within the
fluid. I t tur ns out , curious ly e nou gh, that , for a tub e of c i rcula r cross-sect ion a nd rad ius
a, t he vo l t age be tween d iamet r i ca l ly oppos i t e e l ec t rodes i s 2avB, w h e r e v i s the average
veloci ty, which is jus t w ha t one wo uld exp ect if the veloci ty were unifo rm.
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CHAPTER 11
MAGNETIC FIELDS: IV
The Faraday Induction Law
111 THE INTEGRAL OF E • dl
11.1.1 Example: The Expanding Loop11.2 THE FA RADA Y INDUCTION LA W
11.2.1 Example: Loop Rotating in a Magnetic Field
11.3 LENTS LAW
11.4 THE FARADA Y INDUCTION LA W IN
DIFFERENTIAL FORM
11.5 THE ELECTRIC FIELD INTENSITY E IN TERMS
OF V AND A
11.5.1 Example: The Electromotance Induced in a Loop by a Pair
of Long Parallel Wires Carrying a Variable Current
11.6 SUMMARY
PROBLEMS
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In th is ch ap ter w e shal l con side r the l ine integ ral of E • d l ar ou nd a closed
c i r cu i t . Th i s wi l l l ead us to s t i l l ano ther o f Maxwel l ' s equa t ions .
In Ch ap te r 2 we saw that , in e lect ro stat i c f ie lds, the l ine integ ral of
E • d l a ro un d a c losed c i r cu i t C i s zero. This i s no t a general rule . I f the
ci rcui t C, or par t of i t, m ov es in a m ag ne t ic f ie ld , the n on e mus t take int o
a c c o u n t t h e v x B te rm of Ch ap te r 10 . Th i s wi l l g ive us the Far ad ay induc
t ion l aw, acco rd ing to which th e abov e in t egra l is equa l to mi nus the t im e
der iva t ive o f the enc losed magne t i c f lux .
What i f , in a given reference f rame, we have a changing magnet ic
field? Th en ele ctr ic f ield inte ns ity is eq ua l, no t to — W as in Chap te r 2 , bu tr a ther to — V I 7 - cA/dt, where A is the vector potent ia l of Sec. 8 .4 , and
F a r a d a y ' s i n d u c t i o n l a w a g a i n a p p l i e s .
We have seen in Sec. 2 .5 that an elect rostat ic f ie ld is conservat ive, or that
Th us the work per fo rme d b y an e l ec t ros t a t i c fi eld i s ze ro when a charg e
m o v e s a r o u n d a c l o s e d p a t h .
However , t he re a r e many cases where one has a c losed , conduc t ing
circui t , wi thout sources, wi th par t , or a l l of the ci rcui t moving in a magnet ic
f i e ld . The conduc t ion e l ec t rons in the moving conduc to r s then f ee l an e l ec t r i c
THE INTEGRAL OF E • dl
(11-1)
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262 Magnetic Fields: IV
field v x B, as in Sec. 10.1. In such cases, the ab ov e integral, eva luate d over
the circuit, is not zero.
The v x B field in a conductor moving in a magnetic field is calledthe induced electric field intensity and the integral of E • dl around a closed
circuit is called th e induced electromotance.
EXAMPLE: THE EXPANDING LOOP
The expand ing loop of Fig. 11-1 has one side that ca n sl ide to the right at a veloci ty
v in a region of uniform B. Both v and B are m e a s u re d in the l abora to ry re ference
frame Sx. In the moving wi re , there is an electric field E , = v x B . The l ine in tegral
of E • dl a r o u n d th e circui t is thus eva lua ted par t ly in St and par t ly in S2- In the
coun ter -c lockwise d i rec t ion .
= 0 + f (v x B) • dl,
Jda
= vwB.
(11-2)
(11-3)
(11-4)
T h i s is the i n d u c e d e l e c t ro m o t a n c e . It acts in the coun ter -c lockwise d i rec t ion in
the figure. If, at a certain instant , the l o o p has a res i s tance R , t h e n th e c u r r e n t is
LWB/R.
We have assumed tha t R is large so as to m a k e th e magnetic f ield of the cu rren tf lowing around th e loop neg l ig ib le compared to B .
Figure 11-1 A conduct ing wi re ad sl ides at a veloci ty c on a pai r of c o n d u c t i n g
rails in a reg ion of un i fo rm magnet ic induct ion B . The magnet ic fo rce on the
e lec t rons in the wire p roduces a c u r r e n t / in the circui t .
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11.2 The Faraday Induction Law 263
11.2 THE FARAD A Y INDUCTION LA W
Th e r igh t-h an d side of Eq. 11-4 is the are a swept by the wire per uni t t ime ,mu l t ipl ied by B, or d8>/dt, where O is the magnet ic f lux l inking the ci rcui t .
N o w it is the cu sto m to ch oo se th e posi t iv e di rec t ion for <t> relat ive to the
pos i t ive d i r ec t ion fo r the l ine in t egra l , accord ing to the r igh t -hand sc rew
ru le . Then , s ince the in t egra l has been eva lua ted in the coun te r - c lockwise
d i r ec t ion whi l e O po in t s in to the paper ,
This i s the Faraday induction law. t he e l ec t romotance induced in a c i r cu i t
i s equ al to min us th e rate of ch an ge of the ma gn et i c flux l inking the ci rcui t .
The path of in tegrat ion may be chosen at wi l l and need not l ie in
c o n d u c t i n g m a t e r i a l .
I f the re are no sou rce s in a c i rcui t , the curre nt i s equ al to the indu ced
elect romotance divided by the resistance of the ci rcui t , exact ly as i f the
e lec t romotance were r ep laced by a ba t t e ry o f the same vo l t age and po la r i ty .
Th e induced e l ec t ro mo tanc e i s expressed in vo lt s , and i t add s a lge bra ica l ly
to the vo l t ages o f the o ther sources tha t may be p resen t in the c i r cu i t .
If on e has a r ig id loo p and a t im e-d ep en de nt B. Eq. 11-5 st il l appl ies .
If one has a r igid coil and a t ime -dep end en t B, then one must add the
f luxes thr ou gh e ach tu rn. The su m of these f luxes is cal led the flux linkage.
We shal l use the symbol A for f lux l inkage, keeping O for the sur face integral
of B • da ove r a s im ple loo p. Fo r exam ple , if a coi l has N t u rns wi th a l l t he
tur ns close tog eth er a s in Fig . 11-2, A is N t imes the <D t h rough one o f the
t u r n s :
T h u s , fo r c i r cu i t s o ther than s imple loops , Faraday ' s induc t ion l aw
b e c o m e s
( H - 5 )
A = /Vd). (11-6)
(11-7)
F lux l inkage i s expressed in we ber - tu rn s .
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26 4
Figure 11-2 Coil wi th several turns , all c lose toge the r .
Equation 11-7 should be used with care. It is not as general as it is
co mm on ly said to be. It is corr ect (a) if the circuit is rigid and the ma gn eti c
ind uc tio n B is tim e-d epe nde nt, or (b) if all or pa rt of the circuit m ove s in
such a way as to produce an induced electric field intensity v x B. See
Prob . 11-4. As we sha ll see in Sec. 11.5. the E in the first case is — cA/dt.
EXAMPLE: LOOP ROTATING IN A MAGNETIC FIELD
F i g u r e 11-3 s h o w s a l o o p r o t a t i n g in a c o n s t a n t and uniform magnet ic f ie ld at an
angula r ve loc i ty to. We shal l calcula te th e i nduced e l ec t romotance us ing th e v x B
field in th e wire , and then verify th e F a r a d a y i n d u c t i o n law. This wi l l i l lus t ra te th e
pr inc ip le of o p e r a t i o n of an electric generator.
T h e i n d u c e d e l e c t r o m o t a n c e is
w h e r e the integral is e v a l u a t e d a r o u n d th e l o o p .
A l o n g the top and bot tom s ides , v x B is p e r p e n d i c u l a r to dl and the t e rm
u n d e r th e i n t egra l is zero. Set t ing 0 = o>t. and r e m e m b e r i n g t h a t v = w(a!2). th e
e l e c t r o m o t a n c e i n d u c e d in the vert ical s ides is
T h i s e l e c t r o m o t a n c e is c o u n t e r c l o c k w i s e in the figure. It is pos i t ive because it is
in the pos i t ive d i rec t ion a round th e l oop , w i th re spec t to the di rec t ion of B, ac
c o r d i n g to the r igh t -hand s c rew ru le .
111-8)
/ = 2aj(a/2)(sin (ot)bB — ouibB sin cot. (11-9)
N o w
<> = abB cos 0 = abB cos tot (11-10)
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26 5
Kigure 11 - 3 L o o p r o t a t i n g in a c o n s t a n t an d uniform magnet ic f ie ld B . The vec tor
ri! is n o r m a l to the l o o p .
an d
" ~dt' (11-11)
as predicted by the F a r a d a y i n d u c t i o n law.
T h e e l e c t r o m o t a n c e is ze ro when the p l a n e of the l o o p is p e r p e n d i c u l a r to B.
since v a l o n g th e vert ical s ides is then paral le l to B, and v x B is ze ro all a r o u n d
the loop .
N o t e t h a t th e vol t age 1 of E q. 11-9 is a s inusoidal funct ion of (. Such vo l t ages
are sa id to be alternating. C h a p t e r s 16 to 18 ar e d e v o t e d to a l t e rna t ing vo l t ages and
c u r r e n t s .
11.3 LENZ'S LA W
If t h e f l u x l i n k a g e A i n c r e a s e s , dA/dt is p o s i t i v e , a n d t h e e l e c t r o m o t a n c e is
n e g a t i v e , t h a t i s , the i n d u c e d e l e c t r o m o t a n c e is in the n e g a t i v e d i r e c t i o n .
O n t h e o t h e r h a n d , if A d e c r e a s e s . dA/dt is n e g a t i v e , a n d t h e e l e c t r o m o t a n c e
is in the p o s i t i v e d i r e c t i o n .
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266 Magnetic Fields: IV
11.4 THE FARAD A Y INDUCTION LA W
IN DIFFERENTIAL FORM
The Fa ra da y law stat ed in Eqs. 11-5 an d 11-7 gives the ele ctr omo tan ce
ind uce d in a compl ete circu it when the flux lin kage is a function of the time.
It applies to both fixed and deformable paths in both constant and time-
dependent magnetic fields, with the restriction stated at the end of Sec. 11.2.
We shall now find an important equation that concerns the value of
the E induced at a given point by a time-dependent B.
If, in a given reference frame, we have a ti me- depe nden t B, then we
can state the Fa ra da y in duct ion law in differential form as follows. Usi ng
Stokes's theorem, Eq. 11-5 becomes
J s ( V x E , . d a = - ^ = - i j s . B . d a , ,11-12,
where S is any surface b ou nd ed by the int egra tio n p ath. If the pa th is fixed
in space, we may int erch ang e the ord er of differentiation an d inte grat ion
on the right-hand side and
J > x E ) d a = - J s ^ d a . (11-13)
We have used the partial derivative of B because we now require the rate
of change of B with time at a fixed point.
Since the above equa tio n is valid for arb itr ary surfaces, the in teg ran ds
must be equal at every point, and
Th e direct ion of the indu ced curr ent is alwa ys such that it prod uce s
a magnetic field that opposes, to a greater or lesser extent, the change in
flux, depending on the resistance in the circuit. Thus, if A increases, the
induc ed cur rent pr odu ces an oppo sing flux. If A decrea ses, the in duce d
cur rent pro duc es an aiding flux. This is Lenz' s law.
The currents induced by changing magnetic fields in conductors other
than wires are called eddy currents.
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11.5 The Electric Field Intensity E in Terms of V and A 26 7
This i s a genera l l aw fo r s t a t ion ary m edia . W e have here ano t he r o f
the four Maxwel l equa t ions . We have a l r eady found two o ther s , namely
E q s . 6-12 and 8-12:
V • EPf + Pb
V • B = 0. (11-15)
Eq ua t io n 11-14 is a di f ferent ia l eq ua t io n tha t re late s the sp ace
der iv at ive s of E at a pa r t ic ula r poi nt to the t ime ra te of ch an ge of B at th e
sam e po in t . Bo th E an d B a re me asu red in the same reference f rame. The
e q u a t i o n d o e s no t give the value of E , unless i t can be integrated.
11.5 THE ELECTRIC FIELD INTENSITY E
IN TERMS OF V AND A
Equat ion 11-14 wi l l g ive us an impor tant expression for E in terms of the
p o t e n t i a l s V and A. Since
B = V x A, (11-16)
V x E = —— (V x A) = —V x ~ , (11-17)
dt ct
or
E + - — ]
Th e t e rm be tw een p aren thes es must be equa l to a qua n t i ty w hose cur l is
ze ro , namely a g rad ien t . Then we can se t
3AE = -\V - — . (1 1 -1 9 )
Fo r s t eady cur r en t s , A i s a con s tan t an d th i s equ a t io n r educes to
Eq . 2 -13 . Th e qua n t i ty V i s therefore the elect r ic potent ia l of Sec. 2 .5 , which
i s a l so known as the scalar potential.
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268 Magnetic Fields: IV
Eq ua tio n 11-19 is a general expressio n for E. It states tha t an electric
field intensity can arise bo th from acc um ula tio ns of charg e, th rou gh the
— V I 7 term, and from changing magnetic fields, through the —dA/dt term.
All three quantities, E, A, and V, are measured in the same reference frame.
Th us, if we have a field E ,, Bj in reference frame 1, the elect ric field
in frame 2, mov in g at a co ns ta nt velocity v with r espec t to 1, is given by
Eqs. 10-16 an d 10-17, or by Eq. 10-20. Ho we ve r, in an y given reference
frame, E is given by Eq. 11-19 an d B is eq ua l to t he cu rl of A as in Eq. 8-13.
11.5.1 EXAMPLE: THE ELECTROMOTANCE INDUCED IN A LOOP
BY A PAIR OF LONG PARALLEL W IRES CARRYING
A VARIABLE CURRENT
A pai r of paral lel wires , as in Fig. 11-4, carr ies equal cu rren t s / in oppos i te d i rec
t ions , and / increases at the ra te dl/dt. We shall first calculate th e induced e lec t ro
motance f rom Eq. 11-5 and then from Eq. 11-19.
a) From Sec. 9.1.1, th e c u r r e n t / in wire a p ro d u c e s a magnet ic induct ion
B„ = n0I/2npa, (11-20)
a n d a similar relat ion exis ts for wire b. The flux through th e l o o p in the d i rec t ion
s h o wn in Fig. 11-4 is t h u s
(11-21)
F r o m Eq . 11-5, the e lec t romotance induced in the c lockwise d i rec t ion is r<l> ft:
( fE
. d l = - ^ l n
J 2n dt
T h u s / ' flows in the coun terc lockwise d i rec t ion . Th is is in agreemen t wi th Lenz ' s
l a w: th e c u r r e n t / ' p ro d u c e s a magnetic f ield that opposes th e increase in d>.
b) Let us now use Eq. 11-19 to ca lcu la te th i s same e lec t ro mota nce f rom th e t ime
der iva t ive of the vecto r po ten t ia l A, V being equal to zero in th i s case . From Sec.
ra{rb + w)(11-22)
(11-23)
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269
hA
A 8 A
IP,
Idt
Pb
Figure 11-4 Pair of parallel wires carrying equal currents / in opposite directions
in the plane of a closed rectangular loop of wire. When / increases, the induced
electromotance gives rise to a current T in the direction shown. The vector
potential A and the induced electric field —dA/dt are shown on the horizontal sides
of the loop. The induced current /' flows in the counterclockwise direction because
— dA/dt is larger on the lower wire than on the upper wire.
8.4.1. A is parallel to the wires and, if we choose the leftward direction as positive.
(I 1-24)
(1 1-25)
along the lower and upper sides of the loop, respectively. Thus
E (1 1-26)
(11-27)
and, in the clockwise direction.
(11-28)
as previously.
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270 Magnetic Fields: IV
W e have disreg arded the field —dA/dt along the left-hand and right-hand sides,
because it is perpendicular to the wires .
T o find the elec trom otan ce induce d in the lo op by a changing current in a
single conductor, we set rh
-» oo, and then
(j)E-dl = ^ ~ l n ( — — V (11-29)J 2n dt \ra + w)
11.6 SUMMARY
The Faraday induction law can be stated as follows: For a circuit C linked
by magnetic flux O,
E d l<7<J>
dt '
The integral on the left is called the induced electromotance. The positive
directions chosen for <J> and for the integration ar ou nd C are related accor ding
to the rig ht- hand screw rule. This law is true if either B chang es, giving
a — cA/ct field, or if the co ndu ct or moves, giving a v x B field. If the in tegral
is evalua ted over a coil, the flux <J> must be replaced by the flux linkage A.
Lenis law states that the induced current tends to oppose the change
in flux.
In differential form, the Faraday induction law is
V x E =Hi"
(11-14)
This is ano th er of Maxwell's equa tio ns. Both E and B are measu red in the
same reference frame.
The electric field intensity E can be due to accumulations of charge,
or to changing magnetic fields, or both. In general,
E = —VF7 A
"fit"'(11-19)
where E, F , and A are all mea sur ed in the same reference frame.
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Problems 271
Figure 11-5 See P rob . 11-3 .
11-4 INDUCED CURRENTS
Figure 1 l -6a sho ws a con du ct in g disk rota t i ng in front of a bar ma gne t ,
a) In wha t di rect io n doe s the cu rren t f low in the wire , c lockw ise or co unt er
c l o c k w i s e ? W h y ?
PROBLEMS
U-1E BOAT TESTING TANKA carr iage r uns on ra i ls on e i ther s ide of a long tank of water eq uip pe d for tes t ing
boa t mod e l s . Th e ra i ls a re 3 .0 me te rs apa r t and the ca r r i age has a maxim um speed of
20 me te rs pe r s econd .
a ) Ca lcu la t e the ma xim um vol t age be tween the ra i l s if t he ve r ti ca l co mp one nt
of the ear th 's mag net i c field is 2.0 x 1 0 " 5 tesla.
b ) Wh a t wou ld be the vo l t age if t he t ank were s i tua t ed a t t he magn e t i c eq ua t or?
11-2 EXPANDING LOOP
A conduct ing bar s l ides a t a cons tant veloci ty v a long conduc t ing ra i l s s epa ra t ed
by a d i s t ance s i n a reg ion of un i form mag ne t i c indu c t ion B pe rpendicu la r t o the p l ane
of the ra i ls . A res is tance R i s con nec ted betw een the ra i ls . Th e res is tance of the res t of
the circuit is negligible.
a) Calculate the current / f lowing in the c i rcui t .
b ) Ho w much po wer is requ i red to move the ba r?
c) How does this power compare wi th the power loss in the res is tance R!
11-3E INDUCED CURRENTS
A bar magnet i s pul led through a conduct ing r ing a t a cons tant veloci ty as in
Fig. 11-5.
Sketch curves of (a) the magnet ic f lux <t>. (b) the cu rren t / . (c) the pow er dis
s ipated in the r ing, as funct ions of the t ime.
Use the pos i t ive di rect ions for <> and / shown in the f igure .
Thi s phenomenon has been used to measure the ve loc i t i e s o f p ro jec t i l e s . The
project i le , wi th a t iny perm an en t m agn et inserted in to it s nose , i s ma de to pass th rou gh
tw o coi ls in success ion, sep arat ed by a dis ta nce of ab ou t 100 mil l im eters . Th e t ime delay
betwe en the pulses is a measu re of the veloci ty. Th e me tho d h as been used up to veloci t ies
of 5 k i lome te rs pe r s econd .
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27 2
Figure 11-6 (a) C o n d u c t i n g d i s k
ro ta t in g nea r the No r th po le o f a ba r
mag ne t . The wi re is conn ec ted to the
d i sk th rough s l id ing contac t s .
(£>) Solen oid w ou nd on a b ar ma gn et .
One end of the w i re i s connec ted to a
s l id ing contac t t ha t can move a long
the length of the solenoid.
b) W ha t h app ens i f bo th th e po la r i ty of t he magne t a nd the d i rec t ion of ro t a t ion
a re reve rsed?c) Sho w that there is no curren t genera ted with the se t -up of Fig. 11 -6b, despi te
the fact that th e magn et ic f lux l inking the c i rcui t is not con s tan t .
USE INDUCED ELECTROMOTANCE
A loop of wire is s i tuated in a t ime-dependent magnet ic f ie ld wi th
B = 1.00 x 1 0 "2
co s (27t x 60)f
pe rpen dicu la r t o the p l ane of the loop .
Ca lcu la t e the induced e l ec t romotance in a 100- turn squa re loop 100 mi l l ime te rs
on the s ide .
11-6 ELECTROMAGNETIC PROSPECTION
In e l ec t romagne t i c p rospec t ion , a co i l ca r ry ing an a l t e rna t ing cur ren t causes
induced cur ren t s to f low in conduc t ing ore bodies , and the a l t e rna t ing magne t i c f i e ld
of these induced currents is detected by means of a second coi l placed some dis tance
away from the f i rs t one, as in Fig. i l -7.
Le t us cons ide r a s imple example of induced cur ren t s .
A thin con du ct in g disk of thicknes s /?, radiu s R, a n d c o n d u c t i v i ty a is placed in
a uniform al ternat ing magnet ic f ie ld
B = B 0 co s cot
parallel to the axis of the disk as in Fig. 11-8.
a) Fin d the induced cu rren t dens i ty J as a funct ion of the radius . Assu me that
the conduc t iv i ty a is smal l enou gh to ren der the magn et ic f ie ld of the induc ed curr ent
neg l ig ib l e , compared to the appl i ed B.
Choose the pos i t ive d i rec t ions for B and for J in the di rect ions shown in the
figure.
b) Sketch curve s of B a n d J as funct ions of the t ime. Explain why B a n d J are so
rela ted.
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27 3
Figure 11-7 E l e c t r o m a g n e t i c p r o s p e c t i o n . C o i l A, connec ted to a source of a l t e r
na t ing cur ren t , p roduces an a l t e rna t ing magne t i c f i e ld . Coi l B is he ld pe rpe ndicu la r
t o A and is sens i t ive only to the ma gn et ic f ie ld du e to the cu rren ts indu ced by A
in the ore body C. The f igure is not drawn to scale ; the coi l diameters are of the order
of one meter or less .
Figure 11-8 See Prob. 11-6.
11-7E INDUCTION HEATING
I t i s com mo n prac t i ce to hea t con du c tor s by sub jec t ing them to an a l t e rna t in g
magnet ic f ie ld. The changing f lux induces eddy currents t ha t hea t t he conduc tor by the
Joule effect (Sec. 5.2.1). The method is called induction heating. It is used extensively
for mel t ing metals and for hardening and forging s teel .
The powers used range f rom wa t t s t o megawat t s , and the f requenc ies f rom
60 he r t z to s eve ral hun dred k i lohe r t z .
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Induc t ion hea t ing has the adv anta ge of conven ience and of no t con tam ina t in gthe me ta l w i th combus t ion gases . I t even pe rmi t s hea t ing a conduc tor enc losed in a
vacuum enc losure . Induc t ion hea t ing has anothe r ma jor advantage . A t h igh f requenc ies ,
currents f low near the surface of a conductor . This is the skin effect. See Probs . 16-17
and 18-14. So, by cho os in g the frequency c orrect ly, one can apply a brief heat t rea tme nt
dow n to a kno wn de pth . Thi s is pa r t i cu la r ly imp or t a n t because the re a re nu me rou s
purposes for which one requires s teel parts wi th a hard skin and a soft core: the hard
sk in re s i s t s abras ion and the sof t core reduces breakage . P lowshares a re hea t - t rea t ed
in this way.
Induction furnaces are used for mel t ing metals . They cons is t of large crucibles
wi th capac i ti e s rang ing up to 30 tons , t he rma l ly insu la t ed and su r rou nde d by cur ren t -
carry ing coi ls . Op er at i on is usual ly s tar ted wi th part of the load a l rea dy mo lten.
Let us con s ider F ig. 1 l -9a where a rod of radius a. l eng th L. and conduc t iv i ty a
i s placed ins ide a solenoid having N' t u rns pe r me te r and ca r ry ing an a l t e rna t ing cur ren t
/ = I0 cos 0)t.
As usual, we neglect end effects.
We a l so a s sum e tha t t he f requency is l ow eno ugh to avoid the c om pl i ca t ion s
du e to the skin effect. Th is assu mp tion is wel l sa t is f ied a t 60 her tz wi th a rod of gra phi te
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Problems 275
(a = 1.0 x 10?
S i e m e n s per mete r ) hav ing a rad ius of 60 mi l l ime te rs . In o t h e r w o r d s , th e
m a g n e t i c i n d u c t i o n of the i n d u c e d c u r r e n t s is negl igible . We a l so a s sume tha t th e
c o n d u c t o r is n o n - m a g n e t i c . .T h i s is a d m i t t e d l y a highly s impl i f ied i l lus t ra t ion of i n d u c t i o n h e a t i n g . M o r e
ove r , th e power d i s s ipa ted in the g r a p h i t e is only a few wat t s , which is absurd ly sma l l
for such a l a rge p i ece . Thi s should none the le s s be a useful exercise on i nduced cur ren t s .
C o n s i d e r a r ing of r a d i u s /•. t h i cknes s dr. and l eng th L ins ide the c o n d u c t o r ,
a s in Fig. 11 -9b.
a ) Show tha t th e e l e c t r o m o t a n c e i n d u c e d in the r ing is
u.0nr2
coNT0sin cot.
b) Show tha t , for a cur ren t f lowing in the az imutha l d i rec t ion , th e r ing has a
re s i s t ance
R = InrioLdr.
c ) Show tha t th e ave rage power d i s s ipa ted in the r ing is
^ ( / (0o ;N7
0)
2
7T ( tL/- 3 dr.
d) Show tha t th e t o t a l ave rage power d i s s ipa ted in the cyl inde r is
1 , ,— (H 0o)NT 0)-naLa .16
e) Ca lcu la t e th e power d i s s ipa ted in the g r a p h i t e rod, set t ing L = 1 mete r ,
7 0= 20 a m p e r e s , N' = 5000 tu rns per m e t e r .
USD INDUCED ELECTROMOTANCE
A magnet ic f ie ld is desc r ibed by
. 2nyB = £ 0 sin —^- (sin wt)i.
A
In this field a s q u a r e l o o p of side A/4 lies in the i . -p l an e w i th its s ides paral le l to
th e y- and r -axes . The l o o p m o v e s at a cons tan t ve loc i ty vj.
C a l c u l a t e the e l e c t r o m o t a n c e i n d u c e d i n the l o o p as a funct ion of the t ime i f th e
t ra i l ing edge of the l o o p is at y = 0 at t = 0.
Set co = 2 m l A.
BETATRON
F i g u r e 11-10 s h o w s the pr inc ip le of o p e r a t i o n of the b e t a t r o n . Th e b e t a t r o n p r o
duces e l ec t rons hav ing ene rg ie s of the o r d e r of mil l ions of e lec t ron-vol t s . E lec t rons
are held in a c i rcu la r o rb i t in a v a c u u m c h a m b e r by a magnet ic f ie ld B . Th e e lec t rons
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27 6
/J
i i m i L L J J i n j ^ ^ 11! 11
11111! 111
1LLL
Figure 11-10 Cross -s ec t ion th rou gh the cen t ra l r eg ion of a be ta t ron . Th e e l ec t rons
desc r ibe a c i rcu la r o rb i t i n a p l ane pe rpendicu la r t o the pape r a t 0—0, in the
f ie ld o f an e l ec t romag ne t and ins ide an evacua ted ce ram ic to rus T. Only the po le
p ieces P of the e l ec t romagne t a re shown. The core i s l amina ted . The po le p i eces
a re shaped so tha t t he ave rage B ins ide the orbi t i s eq ua l to twice that a t th e o rbi t .
The e l ec t rom agne t is ope ra t ed on a l t e rna t in g cur ren t , and the e l ec t rons a re
acce le ra t ed on ly dur ing tha t pa r t o f t he cyc le when bo th B a n d dB/dt have the
proper s igns . See Prob. 11-9 and Fig. 11-11.
Figure 11-11 Cir cul ar e lect ron orbi t in the be ta t ron . It i s assu me d that th e
ma gn et ic f lux is in the di rect ion sh own and th at i t increase s . Here .
e = - 1.60 x 1 0 " 1 9 c o u l o m b .
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Problems 277
are accelerated by increasing the magnetic f lux l inking the orbi t . Betatrons serve
as sources of x-rays. There are very few that are s t i l l in use today.
Sho w that the average ma gne tic induc tion over the plane of the orbi t must betwice the magn etic ind uctio n at the orbi t , if the orbi t ra dius is to remain fixed when
bo th B and the electron energy increase.
Solution: The various vectors we are concerned with are shown in Fig . 11-11.
In the reference frame of the labo rat ory , an electron is subjected t o the Lo ren tz
force
e (E + v x B ).
Fhe electric field E is due to the t ime-d epen den t m agn etic f ield and is — dA/dt.
We do not kno w the value of A. but we do know , from F ara da y 's induc tion law,
t h a t
1 <M>E = - - - . (1)
2nr tit
We assume that the magnetic f lux <t> increases. This electric field E provides the
tange ntial accele rat ion tha t increases the electro n energy. Th e centripe tal force
cv x B keeps the electron on i ts circular orbi t .
So , in the az imutha l d i rec t ion .
d e d<t>(mv) = eE= - — — , (2)
dt 2nr dt
while , in the radial d irect ion,
,m-2
r = -Bev. (3)
In these two equations, al l the quanti t ies are posi t ive except e.
Equating the two values of mv obt ain ed from Eqs. 2 and 3 ,
emv = <t = -Ber, (4)
2nr
<t> = 2 tu -2
B, (5)
wh e re B i s the magnet ic induct ion a t the rad ius r of the electron orbi t . Now, by
defini t ion,
<t> = 7t r2
B, (6)
wh e re B is the average magnetic induction within the circle of radius r. T h e n , c o m
parin g Eqs . 5 and 6 , B is eq ual to 2B .
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278 Magnetic Fields: IV
This effect wa s pred ic t ed by Maxwel l , and was o b s e r v e d for the first time in
1916 by T o l m a n and c o l l a b o r a t o r s .
The inverse effect , namely th e acce le ra t ion of a b o d y c a r r y i n g a var i ab le cur ren t ,
was a l so pred ic t ed by M a x w e l l , and was first observed by B a r n e t t and o t h e r s in 1930.
T h e U.S. Nat iona l E lec t r i ca l Code ru l e s tha t bo th conduc tors of a circui t
o p e r a t i n g on a l t e rna t ing cur ren t , if enc losed in a meta l l i c condui t , mus t be run in the
s a m e c o n d u i t .
Let us s u p p o s e t h a t we h a v e a s ing le conduc tor ca r ry ing an a l t e r n a t i n g c u r r e n t
and enc losed wi th in a c o n d u c t i n g t u b e .
Show tha t bo th A and dA/dt are l o n g i t u d i n a l in the t u b e .
So, with a s ingle wire , a l o n g i t u d i n a l c u r r e n t is i n d u c e d in the t ube . Thi s causes
a needless power loss , and may even cause spa rk ing at fau l ty jo in t s . Wi th two conduc tors
ca r ry ing equa l and oppos i t e cur ren t s , bo th A and cXIct are essent ia l ly zero in the
c o n d u i t .
11-11E ELECTRIC CONDUITS
11-12E THE POTENTIALS V AND A
We have s een tha t
E = - VI - , \ \ ft. B = V x A.
Show tha t ne i the r E no r B arc affected if t he po ten t i a l s V and A are rep laced by
V + (G dt and A - VG.
w h e r e G is any funct ion of v, r. z, / whose s econd de r iva t ives ex i s t and are c o n t i n u o u s .
11-10 THE TOLMAN AND BARNETT EFFECTS
If a c o n d u c t o r is given an acce le ra t ion a, the c o n d u c t i o n e l e c t r o n s of m a s s m
and cha rge — e are subjec ted to inertia forces —ma.Show tha t , if E' is the equivalent tota l e lect r ic f ie ld intens i ty and if B is the m a g
ne t i c induc t ion in the c o n d u c t o r ,
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CHAPTER 12
MAGNETIC FIELDS: V
Mutual Inductance M and Self-Inductance L
12.1 MUTUAL INDUCTANCE M
12.1.1Example: Mutual Inductance Between Two Coaxial Solenoids
12.2 INDUCED ELECTROMOTANCE IN TERMS
OF MUTUAL INDUCTANCE
12.2.1 Example: The Vector Potential of the Inner Solenoid
12.3 SELF-INDUCTANCE L
12.3.1 Example: Long Solenoid
12.3.2 Example: Toroidal Coil
12.4 COEFFICIENT OF COUPLING k
12.5 INDUCTORS CONN ECTED IN SERIES
12.5.1 Zero Mutual Inductance
12.5.2 Non-Zero Mutual Inductance
12.5.3 Example: Coaxial Solenoids
12.6 INDUCTORS CONNECTED IN PARALLEL
12.6.1 Zero Mutual Inductance
12.6.2 Non-Zero Mutual Inductance
12.6.3 Example: Coaxial Solenoids
12.7 TRANSIENTS IN RL CIRCUITS
12.7.1 Example: Inductor L Connected to a Voltage Source V 0
12.7.2 Example: Horizontal B eam Deflection in the Cathode-Ray
Tube of a Television Receiver
12.8 SUMMARY
PROBLEMS
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In th i s chap te r we sha l l see how one u t i l i zes the Faraday induc t ion l aw to
ca lcu la t e induced vo l t ages and cur r en t s in e l ec t r i c c i r cu i t s . I n Chap te r 5 weascr ibed to each b ranch o f a c i r cu i t a ce r t a in r es i s t ance R. I f the current
f lowing in a b r anch p roduces an apprec iab le magne t i c f i e ld , t hen the b ranch
also possesses a self-inductance L. I f t he ma gne t i c fi eld o f on e b ran ch p r o
duces a f lux l inkage in ano ther b r anch , then the re ex i s t s a mutual inductance
M b e t w e e n t h e t w o b r a n c h e s . T h u s , th e m a g n e t i c p r o p e r t i e s o f t h e b r a n c h e s
are charac te r i zed by the two quan t i t i e s L and M.
W e sha l l r e tu rn to mu tua l indu c tan ce l a t e r on , in C ha p t e r 18, which
dea l s wi th pow er t r ansfe r an d t r a nsfo rm er s .
12.1 MUTUAL INDUCTANCE M
To ca lcu la t e the e l ec t romotance induced in one c i r cu i t when the cu r r en t
changes in another c i rcui t , i t i s convenient to express the f lux l inkage in the
f irs t one in term s of th e cu rren t in the seco nd an d of a geo me tr ica l factor
invo lv ing bo th c i r cu i t s .
Co ns id er the tw o c i r cu i t s o f F ig . 12-1 . Th e cur r en t / „ in c i rcu i t a
p r o d u c e s i n b a f lux l inkage A ah t h a t i s p r o p o r t i o n a l t o / „ :
A f l i = M ahIa. (12-1)
Similar ly , i f there is a current Ib t h r o u g h c i r c u i t b, t he magne t i c f lux
of b l i nk ing a is
Ka = M haI„. (12-2)
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28 1
Figure 12-1 Two circuits a a n d b. The f lux <t> ab shown l ink ing b and or ig ina t ing in a
i s pos i t ive . This is because i t s di rect ion is re la ted by the r ight-hand screw rule to
the d i rec t ion chosen to be pos i t ive a round b.
Since bo th c i r cu i t s can be of any shape , one na tu ra l ly d oes no t expec t
to f ind a genera l r e l a t ionsh ip be tween M ab a n d M ba. On the con t r a ry , i t can
b e s h o w n t h a t M ab = M haf Th e f ac to r o f p ro po r t ion a l i ty M = M ab = M ba
is called the mutual inductance be tween the two c i r cu i t s .
Mutua l induc tance depends so le ly on the geomet ry o f the two c losedc i r cu i t s an d on the pos i t ion a nd o r i en ta t io n o f one wi th r espec t to the o the r .
When mul t ip l i ed by the cu r r en t in one c i r cu i t , M gives the f lux l inkage in
the o ther .
A s imi la r s i tua t ion ex i s t s in r e l a t ion wi th the capac i t ance be tween
t w o c o n d u c t o r s . T h e c a p a c i t a n c e C depends so le ly on the geomet ry o f the
two con du c to r s and on the pos i t ion and o r i e n ta t ion o f one wi th r espec t to
the o ther . If one o f the con du c to r s i s g ro und ed , then the charge indu ced on
i t i s equal to CV, w h e r e V i s t he vo l t age app l i ed to the o ther .
S ince the mutua l induc tance i s t he f lux l inkage in one c i r cu i t pe r un i t
o f cu r r e n t in the o ther , i ndu c tan ce i s me asu red in we ber - tu rns per am pere ,or in henrys.
T h e m u t u a l i n d u c t a n c e b e t w e e n t w o c i r c u i ts is o n e h e n r y w h e n a
cu rren t of on e am pe re in on e of the ci rcui ts pro du ce s a f lux l inkage of on e
weber - tu rn in the o ther .
* Electromagnetic Fields ami Waves, p. 343 .
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282 Magnetic Fields: V
The sign of M is chosen as follows. The mutual inductance between
two circuits a and b is positive if a current in the positive direction in a
produces in b a flux that is in the same direction as one due to a positive
current in b. Thi s is illus trate d in Fig. 12-1 . Th e positiv e directi ons for the
currents are chosen arbitrarily.
A pair of coils designed so as to possess a mutual inductance is called
a mutual inductor. Mutual inductors are also called transformers.
1.1 EXAMPLE: MUTUAL INDUCTANCE BETWEEN TWO
COAXIAL SOLENOIDS
Let us calculate the mutual inductance between two coaxial solenoids as in Fig. 12-2.We assume that both windings are long, compared to their common diameter 2R.
that they have the same number of turns per meter N', and that they are wound in
the same direction, as in the figure. We set /„ > lb.
Let us assume a current /„ in coil a. Then the flux of coil a linking each turn of
coil b is
<t>ah = Li 0nR2
NT a. (12-3)
and the mutual inductance is
M = \JIB = N
bQJI
a = ti
0nR
2
N'Nh. (12-4)
= ti0nR
2
NaN
h/l
a. (12-5)
I.
I,
Figure 12-2 Coaxial solenoids. The two radii are taken to be approximately equal.
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12.2 Induced Electromotance in Terms of Mutual Inductance 283
We can a l so ca l cu la t e the mutua l i nduc tance by a s suming a cur ren t Ib in coil b.
T h e n
<D ba = u0nR2
NTb(12-6)
This f lux l inks only lhN' = N b turns of coi l a. since B fa l ls rapidly to zero beyond
the end of a long solen oid, as we saw in Sec. 9.1.5. Th en t he mu tua l in du ctan ce is
M = AJI„ = N„<S>JIb
= n0nR 2N'N b = p 0nR 2N aN b/l a,
as prev ious ly .
(12-7)
12.2 INDUCED ELECTROMO TANCE IN TERMS
OF MUTUAL INDUCTANCE
Fr om Sec . 11.2, the e lec t r om otan ce ind uced in c ircu i t b by a change in /„ in
Fig . 12-1 is
E d l =d A , , h
= -M—.dt dt
(12-8)
In this case, E is —dA/dt, wh ere A is the va lue of the vecto r poten t ia l of
the cu r ren t Ia at a poin t on c ircui t b (Sec. 11.5).
S imi la r ly , the e lec t romotance induced in c i rcu i t a by a change in Ib is
E d ldA
bl
dtM
dh
dt(12-9)
T h e s e e q u a t i o n s a r e c o n v e n i e n t f or c o m p u t i n g t h e i n d u c e d e l e c t r o m o
t a n c e , s ince they invo lve on ly the mu tua l induc tance and dl/dt, both of
wh ich can be measu red .
We now have a second de f in i t ion o f the hen ry : the mu tua l induc tance
between two c ircui ts is one henry i f a current changing a t the ra te of one
ampere pe r second in one c i rcu i t induces an e lec t romotance o f one vo l t in
the o the r .
12.2.1 EXAMPLE: THE VECTOR POTENTIAL OF THE INNER SOLENOID
I t i s pa rad oxica l t ha t a va ry ing cur re n t i n the inne r so lenoid sho uld induce an
e lec t romotance in the ou te r one , s ince we have shown (Sec . 9.1.3) t ha t t he magne t i c
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2 8 4 M a g n e t i c F i e l d s : V
i nduc t ion ou t s ide a l ong so lenoid is ze ro . The e x p l a n a t i o n is t ha t th e i n d u c e d
electric field intensity at any given point is e q u a l to the nega t ive t ime de r iva t ive of
t he vec tor po ten t i a l at t ha t po in t (Eq. 11-19) and t h a t th e vec tor po ten t i a l A d o e s
not van i sh ou t s ide an inf ini te solenoid, despi te th e fact that B = V x A d o e s . See
also Sec. 9.1.2.
W e can ac tua l ly ca l cu la t e th e vec tor po ten t i a l j us t ou t s ide the inner solenoid
from th e m u t u a l i n d u c t a n c e . If we cons ide r the di rec t ion of the c u r r e n t /„ in the
i nne r so lenoid as posi t ive , then th e e l e c t r o m o t a n c e i n d u c e d in the o u t e r one is
- M i l J i t , or
d l ,- | ioJt* 2
JVW»-£a t
a n d th e induced e lect r ic f ie ld intens i ty — cA/ct is 2nRNh t imes sma l l e r :
PA HoRN' dl,= - — - . (12-10)
dt 2 dt
In t egra t ing , we h a v e th e vec tor po ten t i a l at any point c lose to the s o l e n o i d :
A = ^ L (12-11)
in th e a z i m u t h a l d i r e c t i o n .
12.3 SELF-INDUCTANCE L
A single circuit carrying a current / is of course linked by its own flux, as in
Fig. 12-3. Th e flux linkag e A is pro po rti on al t o the c urr ent :
A = LI, (12-12)
where L is called the self-inductance, or simply the inductance of the circuit.Self-inductance, like mutual inductance, depends solely on the geometry and
is mea sur ed in henr ys. It is always positive.
A circuit t hat is desi gned so as to ha ve a self-induc tance is called an
inductor.
An indu cto r has a self-inductance of one he nry if a curren t of one
ampere produces a flux linkage of one weber-turn.
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2 8 5
Figure 12-3 Isola ted c i rcui t carrying a
current / and i ts f lux l inkage A.
A cha ng e in the cu r r en t f lowing th rou gh a c ir cu i t p roduc es w i th in
the c i r cu i t i tse lf an induce d e l e c t ro mo tanc e
dt dt(12-13)
T h e i n d u c e d e l e c t r o m o t a n c e t e n d s t o o p p o s e t h e change i n cu r r en t , accord ing
to Lenz ' s law, and ad ds to wha teve r o the r vo l t ages a r e p resen t . If a va ry ing
cur r en t f lows th rough an induc tance L , the vol tage across i t i s L dljdt, a s
in Fig. 12-4.
An induc to r the re fo re has a se l f - induc tance o f one henry i f t he cu r r en t
f lowing th ro ug h it cha nges a t t he r a t e o f one am pe re per second when the
vol tage di f ference between i t s terminals i s one vol t .
V
O
I L
Figure 12-4 Ideal ized inductor wi th
ze ro re s i s t ance . The vo l t age V is
-±r L dljdt.
1 2 . 3 . 1 EXAMPLE: LONG SOLENOID
I t was show n in Sec. 9.1.3 tha t the mag net i c induc t ion ins ide a long soleno id,
neglect ing end effects , i s uniform, and that
B = HoN'I,
w h e r e AT i s t he num ber of tu rn s pe r me te r . T hus
(12-14)
<P = nR\ (12-15)
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28 6
5
Rll
10
Figure 12-5 Factor K for calcu la t ing the induc tanc e of a short solen oid.
w h e r e N i s the tota l n um be r of tur ns . / i s the length of the solen oid, an d R is itsr a d i u s . T h e n
A N<t> p 0N2 ,
L = — = = — — n R 1 . (12-16)I I I
Th e s e lf - induc tance of a long so lenoid is t hus pro por t iona l t o the square of the
nu mb er of tu rn s and to i t s c ros s - s ec t ion , and inve rse ly prop or t io na l t o i ts l eng th .
The induc tance of a short solenoid is smaller by a factor K, wh ich is a function
of R/l, as in Fig. 12-5.
EXAMPLE: TOROIDAL COIL
A toroidal coi l of N t u rn s is wo und on a fo rm of non -ma gne t i c ma te r i a l hav ing a
square cross-sect ion, as in Fig. 12-6.
Ac cord ing to the c i rcui ta l law, or from Eq. 9-8. the mag net i c induc t ion in the
az imutha l d i rec t ion , a t a rad ius p ins ide the toroid. i s
B = poNl/litp. (12-17)
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287
Figure 12-6 Toro ida l co i l of square c ross -sec t ion and m e a n r a d i u s R.
T h u s th e flux linkage is
p0N
2
I r-R + W/2 w dpR + WI:
j R - w / 2
L =
2n
p0N
2
w .
w In2R + w
2R - w
2R + w
2R - iv
(12-1S
(12-19)
(12-20)
The se l f - inductance of a toroidal coi l is p ro p o r t i o n a l to the s q u a re of t h e n u m b e rof turns , l ike that of a long so leno id .
12.4 COEFFICIENT OF COUPLING k
In Sec. 12.1.1 we found that the mutual inductance between the two solenoids
of Fig. 12-2 is
Now, from Eq. 12-16,
M = n07iR
2
NaNb
/la
.
L „ = n0nR2N2Jla,
L h = n0nR2N2/lb,
(12-21)
(12-22)
(12-23)
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288 Magnetic Fields: V
a n d
M = (j) iKU)m
. (12-24)
T h e m u t u a l i n d u c t a n c e i s t h u s p r o p o r t i o n a l t o t h e g e o m e t r i c a l m e a n o f
the se l f - induc tances .
M o r e g e n e r a l l y ,
M = k(L aL b)112, (12-25)
w h e r e k is the coefficient of coupling, which can be e i the r pos i t ive o r nega
t ive . Moreover ,
|fe| < 1. (12-26 )
For example , i n the p resen t case , k = (l h/l a)in < 1 , by h ypo thes i s .
W h e n \k \ % 1, the tw o ci rcui ts are said to be tightly coupled; t hey a r e
loosely coupled w h e n \k\ « 1.
12.5 INDUCTORS CONNECTED IN SERIES
12.5.1 ZERO MUTUAL INDUCTANC E
Consider two induc to r s connec ted in se r i es as in F ig . 12-7a o r 12-7b . The i r
m u t u a l i n d u c t a n c e is a p p r o x i m a t e l y z e r o . T h e n
v=L4+L4={L
>+L
>)d
Tt'
<i2
-27)
and the ef fec tive indu c tan ce i s s imply the sum of the induc tan ces . The sam e
ru le app l i es to any number o f induc to r s connec ted in se r i es , a s long as
the mutua l induc tances a r e a l l ze ro .
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289
-Tt
(a) lb)
Figure 12-7 (a) Tw o coils connected in series. They are assumed to be far from each
other, and their mutual inductance is approximately zero, (b) Two coils connected in
series, one perpendicular to the other. T h e flux linkage and the mutual inductance are
again approximately zero.
12.5.2 NON-ZERO MUTUAL INDUCTANCE
If the coupl ing coefficient bet ween the two in duc tors con nec ted in series
is not zero, as in Fig. 1 2 -8 , then the voltage on coil 1 is L-i t imes dl/dt, plus
M times dl/dt .in coil 2 , and the voltage on coil 2 is given by a similar ex
pression. Thus
V = ( L , + M)~ + (L2 + M ) ^ , ( 1 2 - 2 8 )
= ( L , + L2
+ 2 M ) — . ( 1 2 - 2 9 )
M
Figure 12-8 Tw o coils connected in
series with a non-zero mutual
inductance M .
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290 Magnetic Fields: V
The ef fect ive inductance is thus
L = Lj + L 2 + 2M . (12-30)
Remember tha t M can be e i the r pos i t ive o r nega t ive (Sec . 12 .1 ) .
EXAMPLE: COAXIAL SOLENOIDS
Figure 12-9a shows the coaxial solenoids of Fig. 12-2 connected in ser ies , wi th the
l e f t -hand wi res connec ted toge the r . Le t us ca l cu la t e the induc tance be tween a a n d b.
Supp ose the cur ren t f lows from te rm ina l A t o t e r m i n a l B. Then the f lux of a p o i n t s
left , while that of b points r ight . So the f lux of b part ly cancels that of a, a n d
La + L b2 M , (12-31)
w h e r e L a, L h, M are given in Eqs . 12-22 to 12-24 and M i s a pos i t ive quant i ty, here .
If the solenoids are connected as in Fig. 12-9b,
La + L„ + 2M. (12-32)
Figure 12-9 The coaxial solenoids of
Fig. 12-2, con nec ted in ser ies , (a) wi th
M nega t ive , and (b) wi th M pos i t ive .B A
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12.6 Inductors Connected in Parallel 291
12.6 INDUCTORS CONNECTED IN PARALLEL
•
12.6.1 ZERO MUTUAL INDUCTANC E
W e hav e jus t seen tha t tw o indu c to r s connec ted in se ri es a r e t r ea t ed l ike
resistors in ser ies (Sec. 5 .4) when thei r mutual inductance is zero. One might
guess that inductors in paral le l are calculated l ike resistors in paral le l (Sec. 5 .5)
w h e n M i s zero. As we shal l see, th is i s cor rect .
We now have two induc to r s connec ted in para l l e l a s in F ig . 12-10 ,
wi th ze ro M. T h e n
T h e t o t a l c u r r e n t I is I t + I2. If L i s the sel f - inductance that i s equivalent
to L l a n d L 2 in paral le l ,
(12-34)
(12-35)
(12-36)
(12-37)
a n d
L =LXL 2
(12-38)L , + L 2
as we guessed a t t he beg inn ing .
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29 2
I t
-T l
Figure 12-10 Two co i l s connec ted in pa ra l l e l w i th approxima te ly ze ro mutua li n d u c t a n c e .
12.6.2 NON-ZERO MUTUAL INDUCTANC E
Let us now cons ider F ig . 12-11. w h i c h s h o w s t w o c o u p l e d i n d u c t o r s ,
connected in paral le l . I f L is t he equ iva len t indu c tanc e ,
V = Ldl
Jt '(12-39)
dt dt(12-40)
Also, for the left-hand coil ,
dt dt(12-41)
and, for the r ight-hand coi l ,
dt dt(12-42)
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29 3
1 A
A/ O U 2
Figure 12-11 Two coi l s connec ted in paral le l wi th a n o n - z e r o m u t u a l i n d u c t a n c e M.
H e r e , M is pos i t ive : a pos i t ive (downward) cur ren t in L, p r o d u c e s a d o w n w a r d
B in L 2 , whi le a pos i t ive (downward) cur ren t in L2 a l so produces a d o w n w a r d B
in L , .
From the last two equations,
( L { - M)d
± = ( L 2 - M )d
^ . (12-43)
The n, from Eqs. 12-40 an d 12-41,
and
L 2 - MJ dt \1
L2 - MJ dt
L = L,(L2 — M) + M( L. - M)( L 2 - M ) + (Lj - M )
L X L 2 - M2
L , + L2 - 2M'(12-46)
When M = 0, we revert to the pre viou s case.
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294 Magnetic Fields: V
12.6.3 EXAMP LE: COAXIAL SOLE NOIDS
F i g u r e 12-12 shows aga in th e coaxia l so l enoids of Fig. 12-2, t h i s t ime connec ted
in paral le l . With th e c o n n e c t i o n s s h o w n in Fig. 12-12a, M is posi t ive and the in
duc tance be tween t e rmina l s A and B is
L =M
2
La + L„ 2M(12-47)
with the L's and th e M on the r igh t as in Eqs. 12-22 to 12-24.
If th e c o n n e c t i o n s to one of the s o l e n o i d s are i n t e r c h a n g e d as in Fig. 12-12b,
L =LaL b - M
L a+ Lb+ 2 M '(12-48)
Here aga in , M is the pos i t ive quant i ty ca l cu la t ed as in Eq. 12-24
Figure 12-12 The p a i r of coaxia l so l enoids of Fig. 12-2, con nec ted in paral le l ,(a) with M posi t ive , (b) with M nega t ive .
12.7 TRANSIENTS IN RL CIRCUITS
Wh en a circuit com pris ing resistances and induct ance s is dist urbe d in some
way, whe ther by cha ngin g the appli ed voltages or curr ents , or by modifying
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12.7 Transients in RL Circuits 295
t he c i r cu i t , t he cu r r en t s t ake some t ime to ad jus t t hemse lves to the i r new
steady- s t a t e va lues . We d i scussed a s imi l a r phenomenon in r e l a t ion wi th
RC ci rcui ts in Sec. 5 .14.
EXAMPLE: INDUCTOR L CONNECTED TO A VOLTAGE SOURCE V 0
Figure 12-13 shows an induc tor L connec ted to a vo l t age source F 0 . Th e ind uc tor
and i t s connec t ing wi res a re made of a s ing le con t inuous wi re o f conduc t iv i ty a
and c ros s - s ec t ion a.
a) The current is constant.
If the wire is in air (e r « 1), i ts surface c ha rge de nsit y is e 0 £ „ , w h e r e E„ i s the
nor ma l com pon ent o f E jus t ou t s ide the wi re . Thi s fo l lows f rom G auss ' s l aw
(Sec. 3.2) because £„ = 0 inside the wire, as we shall show below. Similarly, if the
wire is submerged in a die lect r ic , the surface charge dens i ty is e r e 0 £ „ .At any given point on the surface of the wire , the f ie ld £„ is proport ional to F 0 .
I t depends on the configurat ion of the c i rcui t , as wel l as on the pos i t ion and con
f iguration of the ne igh bo rin g bodies .
Inside the wire , the volume charge dens i ty is zero, as we saw in Sec. 5.1.2. Also,
Th us £ and |VF | , l i ke J, are the same everywhere ins ide the wire . The l ines of force
inside the wire are parallel and, if the length of the wire is / ,
E = - V F = J/ff. (12-49)
/ |VK| = / - = I— = IR = V 0.a ao
(12-50)
T h u s |VF| i n s i d e t h e w i r e i s s i m p l y 1,, /.
V
+ + + + +
Figure 12-13 Inductor £ connected to a vol tage source.
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296
t (seconds)
Figure 12-14 Current through the inductor of Fig. 12-13 as a function of the time.
It is ass umed that / = 0 at r = 0 and that RjL = 1.
b) The current builds up from zero to its steady state value V0/R.
We assume that the time the field takes to propagate from the source to the
inductor is negligible. We also neglect the stray capacitance between the turns of
the inductor and between the wire and ground.
Since the wire has both a resistance R and an induct ance L, Kirchoff's voltage
law (Sec. 5.7) gives us that
dlL - + RI= V
0.
dt(12-51)
We saw in Sec 12.3 that the voltage across a pure inductance is L dljdt. From Sec.
5.13, the solution of this equation is
RKe (12-52)
where K is a constan t of inte grati on. By hypo thesi s, / = 0 at / = 0, and th us K
- V0jR. Then
(12-53)
The current 1 increases with time as in Fig. 12-14. No te the analogy with Fig. 5-23.
The current density J is again the same throughout the wire since, by hypothesis,
there is zero capacitance, and hence no accumulation of charge at the surface of
the wire. Thus
J I
E = - = —a aa
RI. (12-54)
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12.7 Transients in RL Circuits 297
and £ is uniform throughout the wire. It builds up gradually, like /. Also.
RI = El. (12-55)
From Eq. 11-19,
RI = j)^-\V - ^Y«U, (12-56)
V F - d l - - ^ ( j ) A - d l , (12-57)
where the integration is evaluated along the wire, in the direction of E, clockwise
in Fig. 12-13.
Now, from Eq. 8-54, the line integral of A • dl around a simple circuit is equal
to the enclosed flux. Mo re generally , the integral is equa l to the flux linka ge:
A d l = A (12-58)
and
r dA
RI = - c i v K - d l , (12-59)
J
dt
r L dl
= - d ) V F - d l . (12-60)J at
from Eq. 12-12. Co mp ar in g with Eq. 12-51, we find that
) V F - d l . (12-61)
This inte gral is therefore a const ant . The negative sign comes from t he fact tha t
the integral is the voltage of the left-hand terminal with respect to the right-hand
one, or — F 0 .
Inside the wires that lead from the source to the coil, the magnetic field is weak
and |V F| is app rox ima tel y eq ual to the £ of Eq. 12-54, with / a function of the tim eas in Eq. 12-53.
Inside the coiled part of the wire, A is not negligible. The — VF field points in
the same direction as the current, while —dA/dt points in the opposite direction.
At t = 0, when the so urce is con necte d, A = 0 but I^A/f'fj is large , / = 0, and the
two terms on the right in Eqs. 12-56, 12-57, 12-59, 12-60 cancel. At the beginning,
most of the voltage drop appears across the coil while, under steady-state conditions,
the potential drop is uniformly distributed along the wire.
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298 Magnetic Fields: V
EXAMPLE: HORIZONTAL BEAM DEFLECTION IN THE
CATHODE-RAY TUBE OF A TELEVISION RECEIVER
The e lec t ron beam sweeps ho r izon ta l l ines across the screen , one be low the o ther ,
unt i l the whole screen has been covered. The process then repeats i tself . The image
is ob ta ined by modu la t ing the beam in tens i ty .
T h e b e a m1
is deflected by the magnetic f ields of two pairs of coi ls , one for the
hor iz on ta l m ot ion , and one for the ver ti ca l . The bea m sweeps ho r izon ta l ly a t a
cons tan t ve loc i ty , and re tu rns a t a much h igher ve loc i ty .
To sweep hor izon ta l ly a t a cons tan t ve loc i ty , the cu rren t th rough the pa i r o f
horizontal deflect ion coi ls must increase l inearly with t ime. This is achieved by
app ly ing a cons tan t vo l tage V to the coils, for then
dlL —
dt
VI « — t.
L
(12-62)
12.8 SUMMARY
T h e mutual inductance M betw een tw o c i rcu i t s is equ al to the mag net i c f lux
l ink ing one c i rcu i t per un i t curren t f lowing in the o ther:
9
Kb = M ahIa. (12-1)
Th e mu tu a l i n d u c t an ce d e p en d s so le ly o n t h e g eo m e t ry a n d o n th e re l a ti v e
pos i t ions an d or ie n ta t ion s of the two c i rcu i t s . I t has the sam e value , which ever
c i rcu i t p roduces the magnet ic f lux l ink ing the o ther . Mutual inductance i s
m e a s u r e d i n henrys.
M u tu a l i n d u c t an ce can b e e i t h e r p o s i t i v e o r n eg a t i v e . Th e mu tu a l
i n d u c t an ce b e tw een tw o c i rcu i t s a a n d b is pos i t ive if a cu rr en t in the pos i t ive
di rec t ion in a p ro d u ces i n b a f lux tha t i s in the same d i rec t ion as tha t p ro
duced by a posi t ive curren t in b.
Th e e l ec t r o m o tan c e i n d u ced i n c i rcu i t b by a change in curren t in
c i rcu i t a is
E d l = - M ^ . (12-8 )dt
T h e self-inductance L of a circuit is i ts flux l ink age p er unit cu rre nt .
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Problems 299
Thus
E d l = - L ~ . (12-13)dt
The voltage across an inductance L carrying a current I is L dl/dt.
The coefficient of coupling k between two circuits is defined by
M = k(LaL„)
1
'2
, (12-25)
and can have values rangi ng from —1 to + 1 . It is zero if no ne of the flux
of one circuit links the other.
The effective i ndu cta nce of tw o inductors connected in series is
L = Ll
+ L2
+ 2M, (12-30)
where M is the mut ual ind uct ance , which can be either positive or nega tive.
The effective inductance of two inductors connected in parallel is
L = J ^ U L ^ .{l2-46)
Ll + L
2- 2M
Wh en a volt age is appl ied to an inductiv e circuit, the curren ts t ake
some time to adjust to their steady-state values.
PROBLEMS
12-1E Starti ng from E q. 12-5, show th at p0 is expressed in henry s per meter.
12-2E MUTUAL INDUCTANCE
A long straight wire is situated on the axis of a toroidal coil of N turns as in
Fig. 12-15.
S h o w that
u0bN* ( b\M = — In 1 + - .
2n \ a
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30 0
Figure 12-15 See Prob. 12-2,
12-3E MUTUAL INDUCTANCE
a ) Show tha t t he mutu a l i n duc tance be tw een two coaxia l co il s , s epa ra t ed by a
dis tance z as in Fig. 12-16, one of radius a a n d N a t u rns , and the o the r o f rad ius b « a
a n d N h turns is
nu 0N aN ba2b 2
2(a 2 + z 2 ) 3 ' 2 '
We shal l use this resul t in Prob. 14-7.
You can calculate this in one of two ways . You can e i ther assume a current in
coil a and calculate the f lux through coi l b, or a s sume a cur ren t i n b and ca lcu la t e the
f lux th rough a. From Sec. 12.1, M is the same, e i ther way. Now, i f the current i s in a, you
know how to calculate the f ie ld a t b, from Sec. 8.1.2, since b « a. If the current is in b,
then the calculat ion is vas t ly more di ff icul t because you have to calcula te B away from
the axis . So you sh ould calcu late the magn et ic f lux thro ug h coi l b due to a cur ren t
th rough co i l a.
b) Ho w does the mu tua l i n duc tan ce va ry when the sma l l co i l i s ro t a t ed a ro un d
the vert ical axis?
Figure 12-16 See Prob. 12-3. N a turns
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Problems 301
c) What if one ro ta ted th e l a rge co i l a round a ver t i ca l ax i s? Would the m u t u a l
inductance vary in the s a m e wa y ?
•
I2-4E A OUTSIDE A SOLENOID
S h o w t h a t the vecto r po ten t ia l at a d is tance r from the axis outside an infinite
so leno id of r a d i u s R and N' t u r n s per meter car ry ing a c u r r e n t / is
A = n0N'R
2
II2r.
The vec to r po ten t ia l is a z i m u t h a l .
I2-5E A INSIDE A SOLENOID
S h o w t h a t the vecto r po ten t ia l at a d is tance r from th e axis , inside an infinite
so leno id of r a d i u s R and N' t u r n s per meter car ry ing a c u r r e n t / is
A = n0Nir/2.
Here a l so , the vecto r po ten t ia l is a z i m u t h a l .
12-6 MAGNETIC MONOPOLES
M a g n e t i c m o n o p o l e s are the magnet ic equ iva len t of electric charges. See Sec. 8.3.
Figure 12-17 shows schemat ica l ly one device that ha s been used in the search
fo r magnet ic monopo les in bu lk mat te r . At the beg inn ing of the exper imen t . SW is
closed and t h e r e is zero cu rren t th rough the iV-turn superconducting coi l C. A sample
of ma tte r SA, with a m a s s of the o rder of one k i log ram, is m a d e to follow a closed path
P t ravers ing C.
Imag ine tha t th e s a m p l e c o n t a i n s one m o n o p o l e of charge e* weber wi th rad ia ll ines of B. As it a p p r o a c h e s a given turn of C . th e flux through that turn is e*/2 we b e r
i i
O , « . p
SA
Figure 12-17 M a g n e t i c m o n o p o l e d e t e c t o r . The sample SA is p laced on a conveyor
and fol lows path P with th e switch SW closed. After hundreds of passes , the switch
is opened and the vo l tage pu lse across R is observed on an osci l loscope. The coil C
i s superconduct ing . See P r o b . 12-6.
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302 Magnetic Fields: V
to the r ight , and, a moment la ter , i t i s e*/2 weber to the lef t . Thus , on emerging a t the
r igh t - hand end of C, t he flux l inkage th rou gh C due to the mo no po le has inc reased by
Ne*. Ho we ver, the res is tanc e of the coi l i s zero an d the indu ced cu rren t keep s the f luxl inkage ze ro .
Ca lcu la t e the cur ren t pe r mo no po le a f t er 100 pas ses th roug h a co i l o f 1200 tu rns
with a se l f- inductance of 75 mil l ihenrys .
The cur ren t i s de t ec t ed by opening the swi t ch SW and obse rv ing the vo l t age
pulse on R.
Al l the measurement s to da te have g iven nu l l r e su l t s .
ROGOWSKI COIL
Figure 12-18 shows a Rogowski co i l sur rounding a cur ren t -ca r ry ing wi re . The
coi l has N t u rns , and r » r'. Rog owsk i co il s a re used for measu r ing l a rge f luc tua t ingcur ren t s , pa r t i cu la r ly l a rge cur ren t pu l s es .
a) If the value of R i s chosen l a rge enough , V i s unaffected b y that p art of the
circui t that i s to the r ight of the vert ical dot ted l ine . Show that
Nr'2
dl
2Vdl'(1 )
Show that , i f C i s now chosen l a rge eno ugh to mak e V « V, t hen
Nr'2
I
V=
Po - 2r RC12)
Figure 12-18 Rogowski coi l for measuring a heavy current pulse in the axia l wire .
Wi th the RC c i rcu i t shown, the vo l t age V i s p r o p o r t i o n a l t o I. See Prob. 12-7.
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Problems 303
See Prob. 5-30. The product PC must be large compared to l/f, where / is the re
petitio n frequency, dl/dt being of the order of If. The current I need not be sinusoidal.
Solution: Under those conditions, V is the induced electromotance:
dcp d ( , u0I\V = N— = N — T t r '
2
— , 3
dt dt V 2nr)
_ ^ r ' 2 d l
2r dt'
Thus the mutual inductance is n0Nr'2/2r.
Al so ,
dQV = PR + V % PR = — R, (5)
dt
dV
* P C — . (6)
Comparing now Eqs. 4 and 6,
F « u 0 . 7)' ° 2r PC
We have set V equal to zero when / is zero. For example, the current / could
be in the form of short pulses and, between pulses, both / and V would be zero.
b) Show that, if the capacitor is shorted and if R is made small so that V is de
creased by a factor of, say, 20 or more, then F * I/N.
The self-inductance of the coil is
L « p0N2
r'2
/2r. (8)
Solution: The induced electromotance is equal to the sum of the voltage drops
across the induct ance L of the coil and the resistance P :
u0
Nr'
2
dl dP™- - r = L — + RI'. (9)2r dt dt
If P were very large, then RI' would be nearly equal to the induced electro
motance. On the other hand, if R is small, we may set
L' E v ^ l
d
l u o ,dt 2r dt
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304 Magnetic Fields: V
Substituting now the value of L,
Li 0N2
r'2
dl' n0Nr'2
dl
— x— . (11)2r dt 2r dt
1
'
I' x I/N. (12)
12-8E INDUCED CURRENTS
An azimuthal curren t is induced in a conduc ting tube of length /, average radius
a, and thickness b, with b « a.
a, Show that its resistance and induct ance are given by
R = 2na/(jbl, L = [i0na2j\.
b) Calculate L, R, L/R for a copper tube (a = 5.8 x 107
S i emen s per meter),
one meter long, 10 millimeters in diameter, and with a wall thickness of 1 millimeter.
12-9 COAXIAL LINES
Coaxial lines, as in Fig. 12-19, are widely used for the interconnection of elec
tronic equipment and for long-distance telephony. They can be used with either direct^
or alternating currents, up to very high frequencies, where the wavelength c/f is of
the same order of magnitude as the diameter of the line. At these frequencies, of the
order of 10 1 0 hertz, the field inside the line becomes much more complicated than that
shown in the figure and quite unmanageable.
There is zero electric field outside the line. Since the outer conductor carries the
same current as the inner one, there is also zero magnetic field, from Ampere's circuital
law (Sec. 9.1).
A coaxial line acts as a cylindrical capacitor. We found its capacitance per
meter in Prob. 6-5:
_ 2nere0
~ In (Rt/Rj
where Rx and R2 are the radii of the inner and outer conductors, respectively.
/
Figure 12-19 Coaxial line. See Prob. 12-9.
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Problems 305
A coaxial l ine is a lso indu ct ive , s ince there is an a zim uth al m agn et ic f ield in
the annula r space be tween the two conduc tors .
Show that the se l f- inductance per meter i s .
12-10 COAXIAL LINES
I t i s kno wn th a t h igh- f requency cu r ren t s do no t pene t ra t e in to a con duc tor a s
do low -frequen cy cur ren ts . Th is is cal led the skin effect. See Pro b. 11-7.
W ou ld you exp ect the se l f- inductance of a coaxial l ine to increase or to decre ase
wi th inc reas ing f requen cy?
12-11 LONG SOLENOID W ITH CENTER TAP
Cons ide r a long so lenoid wi th connec t ions a t bo th ends , A and C, and w ith a
cen te r t ap B. The two halves are in ser ies , and thus
w h e r e M i s t he i r mu tua l i ndu c tance .
N ow . if on e uses Sec. 12.3.1 to calcula te th e three L 's , one f inds tha t M = 0.
Th is is ab sur d be cause th e coeff ic ient of cou pl in g is surely not ze ro.
Can you expla in th i s s t range re su l t ?
12-12 TRANSIENT IN RLC CIRCUIT
The switch in the c i rcui t of Fig. 12-20 is ini t ia l ly open, and then c losed a t t = 0.
LAc = L AB + L BC + 2 M ,
C
Figure 12-20 See Prob. 12-12.
Draw a curve ofQ as a funct ion off for V = 100 vol ts , L = 1 henry , C = 1 m i c r o
farad, R = 8000 ohms , w i th t varying from 0 to 20 mil l iseconds .
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306 Magnetic Fields: V
Solution : F ro m Ki rchof f ' s vo l t age l aw .
dl QRI + L h —
dt C1.
or, since / = dQjdt,
d 2Q dQ QL ~ + R-±- + ~ = V.
dt 1 dt C
The pa r t i cu la r so lu t ion of th i s equa t ion i s
Q = CV = 1C T 4,
and the comp lem enta r y func t ion i s t he so lu t ion of
d 2Q dQ QL ~ + R — + —
dt2
dt C
Let us set
Q = Ae"'
Then, subs t i tu t ing in Eq . 4 and d iv id ing by Q,
0.
1Ln 2 + Rn + -
C0 .
R
2L
R2 1
4L2 ~ LC
Note tha t t he re a re two va lues o f n, say n, and n 2. T h e n
Q = Ae"" + Be" 2'.
(1)
(2)
(3)
(4)
15)
(6)
(7)
(S )
w h e r e A a n d B a r e u n k n o w n c o n s t a n t s o f i n t e g r a t i o n .
H e r e R/2L = 400 0. the br ac ke t of Eq . 7 is 1.5 x 1 0 7 , and the general solut ion is
Q = e - 4 0 0 0 1 1 ^ 3 8 7 3 !+ Be" + 10"
= Ae~ 121' + Be- lsli' + 10"
S ince (2 = 0 a t r = 0 . A + B = - 1 0 " 4 . Also,
dQ
dt\21Ae' 7873Be"
(9)
(10)
( I D
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30 7
1 . 5 f x 1 0 " 4
0 10 20 msf
Figure 12-21 The cha rge on the capac i to r o f F i g . 12-20, as a function of the time
for various values of R. Th e t ime is expre ssed in mil l i secon ds . W he n the res is tan ce
i s sma l l , t he cha rge ove rshoot s and osc i l l a t e s about i t s a sympto t i c va lue .
a n d I = 0 at r = 0, so th at
127/4 + 78 73 5 = 0, (12)
A = - 1 . 0 1 6 x 1 0 " 4 . B = 1.640 x 1 0 " 6 . (13)
Q = - 1 . 0 1 6 x 1 0 "4
e "1 2 7
' + 1.640 x l O "6
? "7 8 7 3
' + 1 0 "4
. (14)
As a check, we can show that L dljdt = 100 at f = 0. At tha t ins ta nt , 7 = 0
and the vo l t age drop on R i s zero.
F igure 12-21 shows Q as a funct ion of t for R = 500, 2000 , and 8000 oh ms . W i th
R = 2000 ohms , the bracke t i s ze ro . Th e r igh t -h and s ide of Eq . 8 then becom es
(A + Bt) e x p nt . W i t h R = 500 ohm s , the bracke t is nega t ive and we aga in ha ve
anothe r type of so lu t ion .
12-13 VOLTAGE SURGES ON INDUCTORS
I f t he cur ren t t h roug h a l a rge indu c tor i s i n t e r rup ted sudd enly , dljdt is large
and t rans i en t vo l t ages appea r ac ros s the swi t ch an d ac ros s the co il . The vo l t ages can
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30 8
Figure 12-22 Ind uc to r L and i t s
re s i s t ance R fed by a source V
t h rough a swi t ch represen ted he re by
a res is tor R s. See P rob . 12-13.
Ia L
9
be l a rge enough to damage bo th . Care must therefore be taken to reduce the current
slowly. This is pa r t i cu la r ly im por t a n t when us ing l arge e l ec t rom agne t s . No prob lem
arises , however, on connect ing the source to the inductor . See Sec. 12.7.1.
Figure 12-22 shows the induc tor L and i ts res is tance R fed by a source V. T h e
res is tance R s plays the role of the switch. At first . R s is zero and / = VjR. T h e n R s
i s quick ly increased to some value muc h larger than R.
Set L = 10 hen rys . R = 1 0 o h m s . V = 100 volts , dljdt = - 1 0 3 a m p e r e s p e r
second .
a) Show that the vol tages across the switch and across the inductor both r iseto about 10 k i lovol t s .
Of course the insulat ion of the switch and the insulat ion between turns on the
ind uct or break d ow n mu ch before the vol ta ge can r ise to such a high valu e.
b) I t i s poss ible to suppress the t rans ient by us ing a diode. A diode i s a two-
terminal device that has e i ther a very low or a very high res is tance, depending on the
polari ty of the appl ied vol tage, as in Fig. 12-23. In other words , a diode wil l pass a
curr ent in only one dir ect ion . Of course , if the inv erse vol tage is suffic ient ly high,
current wi l l pass in the forbidden direct ion and the diode wil l be damaged.
Figure 12-23 The symbol fo r a d iode
is a t r iangle an d a ba r . In (a), th ediode has a low res is tance and can
usual ly be cons idered as a short j
cir cu it. In (/>), the re sis tan ce is hig h ' = 0
an d is usual ly con s idere d to be • . I
infinite. The curve of / as a function + * ^1 *_
* ^1
of the appl ied vol tage is non-l inear ,
even for forward vol tages . UT) (/?)
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Problems 309
A given type of d iod e can be used up to a ra ted m axim um cur ren t , and up to a
ra t ed maximum inve rse vo l t age .
You are given a diode that i s ra ted a t 20 amperes and 200 vol ts peak inverse
vol t age . How should i t be used?
W ha t is the cur ren t thr ou gh the diod e as a funct ion of the t ime if the switch is
ope ned at f = 0?
W ha t hap pe ns to the energy s tor ed in the ma gn et ic f ie ld of the indu cto r af ter
the swi t ch i s opened?
12-14D TRA NSIENT IN RLC CIRCUIT
One wishes to cha rge a capac i to r a s qu ick ly a s pos s ib l e . The source suppl i e s
an open-c i rcu i t vo l t age V s and has an internal res is tance R (Sec. 5.12), as in Fig. 12-24a.
I t i s sugge s ted that the ra te of cha rge could be increased by addin g an i nd uct or
with L = R 2C/4 in ser ies wi th the capaci tor as in Fig. 12-24b.
Dr aw cu rves of the vo l t age V across the capaci tor as a funct ion of t ime for F s =100 vol ts , R = 100 ohms , C = 1 mic rofa ra d , w i th and wi tho ut the indu c tor .
Is the sugges t ion val id ?
O - i O - i
c V S - ±
3
(b)
Figure 12-24 Capaci tor C charged by a source F s having an in t e rna l re s i s t ance R.
T h e i n d u c t o r L in {b) has been added in the hope tha t t he capac i to r w i l l cha rge
more rap id ly . See P rob . 12-14 .
I2-I5D TRANSIENT IN RLC CIRCUIT
In the c i rcui t of Fig. 12-25 the switch is ini t ia l ly open. The capaci tor i s charged
to the vo l t age V and no cur ren t i s d raw n f rom the source .
Se t V = 100 volts , L = 1 henry , R = 10 ohm s , C = 1 mic rofa rad ,
a) T he switch is c losed. Show that
/ = 10(1 - e'
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31 0
Figure 12-25 See Prob. 12-15.
L R
b) After a t ime t » L!R, the switch is ope ned . Show that , f rom tha t t ime on,
Qx - 1 0 ~ V 5 ' cos 10 3 r + 1( T 4 .
I * O . l e " 5 ' sin 10 3f.
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CHAPTER 13
MAGNETIC FIELDS: VI
Magnetic Forces
13.1 FORCE ON A WIRE CARRYING A CURRENT IN A
MAGNETIC FIELD
13.1.1 Example: The Hodoscope
13.2 MAGNETIC PRESSURE
13.3 MAGNETIC ENERGY DENSITY
13.4 MAGNETIC ENERGY
13.4.1 Example: The Long Solenoid
13.5 MAGNETIC ENERGY IN TERMS O F THE CURRENT I
AND OF THE INDUCTANCE L
13.5.1 Example: Coaxial Solenoids
13.6 MAGNETIC FORCE BETWEEN TWO ELECTRIC
CURRENTS13.6.1 Example: The Force Between Two Infinitely Lon g Parallel Wires
13.7 MAGNETIC FORCES WITHIN AN ISOLATED CIRCUIT
13.8 SUMMARY
PROBLEMS
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In Ch ap te rs 4 an d 7 , we studie d the energ y sto red in an elect r ic f ie ld , an d
the r esu l t ing e l ec t r ic fo rces . The n , in Ch ap te r 10 , we s tud ied the ma gne t i c
fo rces on ind iv idua l charge d par t i c l es . N ow w e mu st s tudy mag ne t i c energy
and macroscop ic magne t i c fo rces . Magne t i c fo rces a r e , i n p rac t i ce , so much
la rger than e l ec t r i c fo rces tha t t hey l end themse lves to many more app l i ca
t ions . For example , e l ec t r i c motor s a lmost invar i ab ly use magne t i c fo rces!*
We con t inue to assume tha t t he mate r i a l s in the f i e ld a r e non
magne t i c . We sha l l dea l wi th the magne t i c fo rces exer t ed by e l ec t romagne t s
in Chap te r 15 .
13.1 FORCE ON A WIRE CARRYING A CURRENT
IN A MAG NETIC FIELD
Consider a wi r e o f c ross - sec t iona l a r ea a, carrying a current / as in Fig . 13-1.
T h e r e a r e n con du c t ion e l ec t rons per un i t vo lum e, each one ca r ry in g a charge
— e at a velocity v.
I f the wire i s s i tua ted in a m ag ne t ic f ie ld B , due to cu rre nts f lowing
e l sewhere , t he magne t i c fo rce on one e l ec t ron i s — ev x B (Sec. 10.1) and,
f Al thou gh e l ec t ros t a t i c moto rs a re exceeding ly ra re , t hey have been s tud ied ex tensive ly .
See, for example , A . D . Moore , Edi to r , Electrostatics and Its Applications, John Wi ley ,
New York , 1973 .
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31 3
Figure 13-1 Wire of cross-section a ca r ry ing a cur ren t / . The cha rges — e m o v e
at a veloci ty v. Th e m agn et ic f ield B i s du e to curre nts f lowing elsewh ere .
for a length dl c o n t a i n i n g na dl c o n d u c t i o n e l e c t r o n s ,
d F = nadl(-ev x B). (13-1)
N ow th e charg e pass ing th rou gh a g iven c ross - sec t ion per second i s
the charge on the ca r r i e r s con ta ined in a l eng th v of the wire and
I = — nae\. ( 1 3 - 2 )
T h e n
d F = Idl x B = / d l x B, (13-3)
w h e r e dl i s in the di rect ion of the current .
Fo r a s t ra i ght leng th / of wire car ryi ng a cur ren t / in a uniform ma gn et ic
field B.
F = II x B (13-4)
F igu re 13-2 shows the l ines o f B nea r a cu r r en t - ca r ry ing wi r e , pe r pe nd icu lar to a uniform ma gn et ic f ield. The force on a length / i s the n simply
IIB, i n the d i r ec t ion shown.
As for electr ic f ields, i t is useful to imagine that lines ofB really exist ,
a r e under t ens ion , and r epe l each o ther l a t e r a l ly . Th i s s imple mode l g ives
the correct d i rect ion for magnet ic forces, as can be seen f rom Fig. 13-2.
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31 4
(a) (b)
Figure 13-2 Magnet ic f ie ld near a current-carrying wire s i tuated in a uniform
magnet ic f ie ld, (a) Lines of B for a cur ren t pe rpendicu la r t o the pape r , (b) Lines of
B for a unif orm field par allel to th e pape r, (c) Su pe rp os itio n of fields in (a) an d (b)
and the resul t ing mag net ic force F . Th e wire carr ies a cur ren t of 10 am per es , the
uniform fie ld has a B of 2 x 1 0 ~4
tes la , and the point where the l ines of B are
bro ken is a t 10 mil l im eters from th e center of the wire .
EXAMPLE: THE HODOSCOPE
T h e hodoscope i s a device that s imu lates the t ra jec tory of a charge d p art ic le in a
magnet ic f ie ld. The principle involved is s imple: i f the charged part ic le , of mass m,
c h a r g e Q, and veloci ty v, is replaced by a light wire fixed at the two ends of the tra
j ec tory and ca r ry ing a cur ren t I, the wire will follow the trajectory if
mv/Q = T/L (13-5)
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315
Figure 13-3 (a) C h a r g e Q moving at a veloci ty v in a magnetic f ield B. The rad ius
of curvature of the t rajectory is R t. (b) Light wire carrying a current / in theopposi te direct ion in the same magnetic f ield . The tension in the wire is T, and
its radius of curvature is R w. It is shown that the wire has the same radius of
curvature as the t rajectory i f mv/Q is equal to T/I. The ang le dQ is infinitesimally
small .
where T is the tension in the wire. This s tatement is by no means obvious, but we
shal l demonstrate i ts val id i ty .
The advan tage o f the so -ca l led floating wire lies in the fact that it is much easier
to experiment with a wire than with an ion beam.
Let us cons ider a region where the ion beam is perp end icula r to B as in Fig. 13-3a.
Then the charge Q is subjected to a force QvB. T h u s
QvB = mv 2/R„ (13-6)
wh e re R, is the radius of curvature of the t rajectory, and
R, = mv/BQ. (13-7)
In Fig . 13-3b we have replaced the charged part icles by a l ight wire carrying a
current / f lowing in the opposite direct ion. If the element dl is in equil ibrium, the
ou tward fo rce BI dl i s com pens a ted by the inward com pon en t o f the t ens ion fo rces T:
Bl dl = IT si n (dd/2) = IT dO/2 = T dl/R w. (13-8)
Fhe radius of curvature of the wire is thus
Rw
= T/BI, (13-9)
and the two radi i of curvature wil l be the same if
mv/Q = T/l. (13-10)
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316 Magnetic Fields: VI
Fo r exam ple , if we have a p ro to n (in = 1.7 x 1 0 " 2 7 ki logram) beam wi th a
kinet ic energy of 10 6 e lec t ron-vol t s (10 6 x 1.6 x 1 0 " 1 9 j ou le ) , t hen T/I i s about
0.15 and. if / = 1 am per e , T i s 0.15 new ton.
Note that the part ic le wi l l be deflected downward i f the magnet ic force is down
ward, but the wire wi l l curve downward i f the force is upward. The magne t i c fo rces
mus t t he re fore be d i rec t ed in oppos i t e d i rec t ions .
If the magnet ic f ie ld is not uniform, and i f the beam is not perpendicular to B,
then the wire does not a lways fol low the t ra jectory. For example , magnet ic f ie lds
are often used both to deflect and focus an ion beam. In such cases the focus ing
forces on the bea m can b eco me defocu s ing forces on the wire , which is then deflected
away from the t ra jectory.
13.2 MAG NETIC PRESSURE
In Ch ap te r 10 we cons idered the mag ne t i c fo rces ac t ing on ind iv idua l
particles, and we have jus t seen how on e can ca lcu la t e the fo rce on a c u r r en t -
c a r r y i n g wire. I f , now, we have a current sheet, it is ap pro pr i a t e to th ink in
t e r m s o f m a g n e t i c pressure.
Let us imagine a f la t cur rent sheet car rying a a m p e r e s p e r m e t e r a n d
si tu ated in a uniform tan gen t ial mag ne t ic field B/2 du e to cu rre nts f lowing
elsewhere, as in Fig . 13-4a, wi th a normal to B.
W e a d j u s t a unt i l i t produces an aiding f ie ld B/2 on one side and anop po sin g field B/2 on the othe r s ide, as in Fig . 13-4b. Th en , f rom P ro b. 9-3,
a = B'p 0 and the force per uni t area, or the pressure, i s ot. t imes the f ield B 2
due to the currents f lowing elsewhere, or B 2/2p 0.
S o, whe neve r one has a cu r r e n t flowing th rou gh a con duc t ing shee t ,
w i t h a m a g n e t i c i n d u c t i o n B on one s ide and ze ro magne t i c induc t ion on
the other s ide, the sheet i s subjected to a pressure B 2/2p 0. This p ressure i s
exer ted in the same direct ion as i f the magnet ic f ie ld were replaced by a
c o m p r e s s e d g a s .
We have ar r ived at th is resul t for a f la t cur rent sheet , but the same
appl ies to any current sheet when B = 0 on only one side.Now we saw in Sec . 13 .1 tha t , qua l i t a t ive ly , one can exp la in magne t i c
forces by as sum ing th at the l ines of B (a , repel l each ot he r la teral ly , a nd
(b) a r e und er t ens ion . Th e ma gne t i c p r essu re we have here i s c l ea r ly asc r iba b le
to the l a t e r a l r epu l s ion , s ince the l ines a r e rough ly para l l e l t o the cu r r en t
shee t .
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317
Figure 13-4 (a) A current sheet carries a current density of a amperes per meter of
its width and is situated in a uniform magnetic field B/2 due to currents flowing
e l sewhere , (b) The total magnetic field is that of (a), plus that of the current sheet.
By adjusting a so that it just cancels the magnetic field on the left-hand side of the
sheet , we have a total field B on the right.
Since the lines are under tension, there should also be a force of attrac
tion in the direction of B . This is correct, as we shall see later on in Sec. 15.3. In
fact, along the lines of B, the force per unit area is again B2
/2p0.
13.3 MAGNETIC ENERGY DENSITY
In Sec. 4.5 we found that, in an electrostatic field, the force per unit area on
a conductor and the energy density in the field are both equal to e 0E2
/2.
No w we hav e just found tha t the mag net ic pressur e is B2
/2fi0. One might
therefore expect the energy density in a magnetic field to be equal to B2
/2[i0.
This turns out to be correct/
Figure 13-5 shows the magnetic pressure, or the magnetic energy
density, B2
/2ti0, as a function of B.
+
Electromagnetic Fields and Waves, p. 367.
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31 8
1 01 0
10 8
p(pascals) 10
6
'T O 4 -1 l l./"T 1 1 1
102
1 1 1 1 1 1 1
10"* 1 0 "3
1 0"2
10 "' 1 10 102
103
B (teslas)
Figure 13-5 M a g n e t i c p r e s s u r e B2/2fi0 as a funct ion of B. The magne t i c p res sure
is equal to the magne t i c ene rgy dens i ty . A tmospher i c p res sure at sea level is a b o u t
1 05 pasca l s .
13.4 MAGN ETIC ENERGY
Since the magnetic energy density is B2
/2u 0, the energy stored in a magnetic
field is
2Ho
whe re the integral is evalu ated over all space.
$yB2
ch, (13-11)
13.4.1 EXAMPLE: THE LONG SOLENOID
The magne t i c ene rgy s to red in the field of a l ong so lenoid of Af t u rns , r ad ius R, and
length /, ca r ry ing a c u r r e n t / is
2/i<A / /inR
2
l). (13-12)
lipN 2nR 2
21(13-13)
T h e m a g n e t i c p r e s s u r e is
1 [ti0NlV p 0N2
(13-14)
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13.5 Magnetic Energy in Terms of Current 1 and Inductance L 319
13.5 MAGNETIC ENERGY IN TERMS OF THE
CURRENT I AND OF THE INDUCTANCE L
I f one compares Eq. 13-13 for the magnet ic energy stored in the f ie ld of a
sole noid w i th Eq. 12-16 for the indu cta nc e of a sole noid , one f inds th at
M ^ W i L / 2 (13-15)
This i s a general resul t : the magnet ic energy stored in the f ie ld of a c i rcui t
o f induc tance L ca r ry ing a cu r r en t I is \LI2
.
Yo u wil l recal l f rom Sec. 4 .3 tha t the energy s tore d in th e elect r ic
f ie ld of a capaci tor of capaci tance C charged to a vo l t age V is \CV2
.If one ha s two c i r cu i t s wi th induc tanc es L a a n d L fc , a n d h a v i n g a m u t u a l
i n d u c t a n c e M , t h e n t h e m a g n e t i c e n e r g y i s
W m = ~ LJ2
a + ~ LbI2 + MlJh. (13-16)
Th is is sho wn in Pr ob . 13-16.
13.5.1 EXAMPLE: COAXIAL SOLENOIDS
We can eas i ly veri fy the above equat ion for coaxial solenoids .
We f i rs t evaluate the r ight-hand s ide . From Eqs . 12-16 and 12-7,
- LJ 2 + - L„I2
b + MIJ„
1 fi 0N2
nR2
1 n 0N2
bnR2
H 0N aN bnR2
ti 0nR2
fN2
I2
N2
I2 2N aN bIaf
+ —\ r2 \ l„ • l„ ' la
(13-17)
(13-1J
To ca lcu la t e W m, we use the fact tha t the energy dens i ty is B 2/2[i0. Ov er the l eng th
lb we have a tota l B of
>A 'B
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320 Magnetic Fields: VI
giv ing a magne t i c ene rgy
1 2 (N J a N„Ib\2
2p -o \ L >b JOver the length /„ — lb, we have only the f ie ld of coi l a and a magnet ic energy
2 v / . '
Then the magnet ic energy s tored in the f ie ld of the coaxial solenoids is
II,"ii 07iR
2
(N2
aI2
lb Njll N.NJJt Nil2
N1
!2
!,
2 V \l lb I. /. W. (13 m
which reduces to the express ion on the r ight-hand s ide of Eq. 13-1S
13.6 MAGNETIC FORCE BETWEEN TWO
ELECTRIC CURRENTS
I t i s w e l l k n o w n t h a t c i r c u i t s c a r r y i n g e l e c t r i c c u r r e n t s e x e r t f o r ce s o n e a c h
o t h e r . I n F i g . 1 3 - 6 , t h e f o r c e e x e r t e d by c i r c u i t a on c i r c u i t b is o b t a i n e d b y
i n t e g r a t i n g E q . 1 3- 3 o v e r c i r c u i t b:
Kb = Ib(f bd\h x Ba
(13-20)
w h e r e B i s d e f i n e d a s i n E q . 8 - 1 .
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13.7 Magnetic Forces Within an Isolated Circuit 321
EXAMPLE: THE FORCE BETWEEN TWO INFINITELY
LONG PARALLEL WIRES
The force between two inf ini te ly long paral le l wires carrying currents /„ and Ib,
s e p a r a t e d by a di s t ance p, as in Fig. 13-7, is ca lcu la t ed as follows. The force act ing
on an e lement lh d l b is
d F = / „ ( d l t x BJ,
dF = Ib_d lbii 0lJ2np,
(13-21)
(13-22)
a n d th e force per uni t length is
dF— = p 0IJ b/2np.dU
(13-23)
The force is a t t rac t ive if the c u r r e n t s are in the same d i rec t ion ; and it is repuls ive if
t hey are in oppos i t e d i rec t ions .
Figure 13-7 Two l ong pa ra l l e l w i res ca r ry ing cur ren t s in the same d i rec t ion . The
e lement of force d F ac t ing on e l e m e n t dl fc is in the di rec t ion shown.
13.7 MAGNETIC FORCES WITHIN
AN ISOLATED CIRCUIT
In an isolated circuit the current flows in its own magnetic field and is there
fore subj ected to a force. Th is is illu stra ted in Fig. 13-8 : the int era cti on of
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32 2
x
R S
v W— f —
o — b — f ti)
•i i
4Figure 13-8 Schemat ic diagram of a ra i l gun. The bat tery A c h a r g e s , t h r o u g h a
res i s t ance R, t he capac i to r C, which can be made to d i s cha rge th rou gh th e li ne by
clos ing switch S. The role of the capaci tor i s to s tore e lect r ic charge and to supply
a very large current to the loop for a very short t ime. If s ide D i s a l lowed to move,
i t moves to the r ight under the act ion of the magnet ic force F.
t he cu r r en t / wi th it s ow n magn e t i c induc t ion B p rodu ces a fo rce tha t t ends
to extend the ci rcui t to the r ight .
Magne t i c fo rces wi th in an i so la t ed c i r cu i t a r e usua l ly ca lcu la t ed by
pos tu la t ing a smal l d i sp lacement and us ing the p r inc ip le o f conserva t ion
of energy. See Prob. 13-25.
Th e force on an eleme nt of wire of length dl c a r r y i n g a c u r r e n t I in a region
where the magne t i c induc t ion i s B , i s
If a cond uc t i ng shee t s i tua ted in a magn e t i c f ie ld ca r r i es a cu r r e n t
tha t cance ls the mag ne t ic f ie ld on one side, then i t i s subje cted to a magnetic
pressure B
2
/2p 0, w h e r e B i s t he magn e t i c ind uc t ion on the o the r s ide . Th i sexpression also gives the energy density in the magnet ic f ie ld .
T h e magnetic energy sto red in a m ag ne t ic f ie ld is ob tai ne d by inte
g ra t ing the energy dens i ty :
13.8 SUMMARY
dF = / dl x B. U3-3)
(13-11)
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Problems 323
If the field is du e to an i nd uc tan ce L , the n
Wm = (1 2)LI2 (13-15)
or, if we hav e two in duc tanc es La and Lh with a mutu al in ductance M. then
Wm = (l/2)L„/ a
2
+ ll 21/ . , , / , ; + M / a / 0 . (13-16)
The magnetic force exerted by a circuit a on a circuit b is
Ff l f t
= h§hA\
hx B„. (75-20)
M agnetic forces also exist within an isolated circuit. The y are calculate d
by postulating a small displacement and using the principle of the con
servation of energy.
PROBLEMS
13-1E MAGNETIC FORCE
C a l c u l a t e th e magne t i c fo rce on an arc 50 mil l ime te rs long ca r ry ing a c u r r e n t
of 400 a m p e r e s in a d i r e c t i o n p e r p e n d i c u l a r to a uni form B of 5 x 1 0 "2 tesla.
H i g h - c u r r e n t circuit breakers often comprise coi ls that generate a magnet ic f ie ldto b low out the arc t ha t fo rms when the c o n t a c t s o p e n .
13-2E MAGNETIC FORCE
a ) F ind the cur ren t dens i ty neces sary to float a copper w i re in t he ea r th ' s magne t i c
field at th e e q u a t o r .
A s s u m e a field of 10~4 tes la . The dens i ty of c o p p e r is 8.9 x 10
3 k i l o g r a m s pe r
cubic me te r .
b) Will th e w i re b e c o m e h o t ?
T h e c o n d u c t i v i ty of c o p p e r is 5.80 x 107
S i emen s per mete r .
c) In w h a t d i r e c t i o n m u s t the cur ren t f low? See P r o b . 8 - 1 1.
d ) W h a t w o u l d h a p p e n if the exper iment were pe r formed at on e of the m a g n e t i c
poles?
13-3E MAGNETIC FORCE
C a l c u l a t e the force due to the earth 's magnet ic f ie ld on a hor i zonta l w i re
100 meters long carrying a c u r r e n t of 5 0 a m p e r e s due n o r t h .
Set B = 5 x 1 0 "5 t e s l a , p o i n t i n g d o w n w a r d at an angle of 70 degrees wi th th e
h o r i z o n t a l .
See P rob . 8 -11 .
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324 Magnetic Fields: VI
13-4E MAGNETIC FORCE
Show that the tota l force on a c losed c i rcui t carrying a current / in a uniform
magnet ic f ie ld is zero.
13-5E ELECTROMAGNETIC PUMPS
The conduc t ion cur ren t dens i ty in a l i qu id me ta l i s
J f = ff(E + v x B ) ,
w h e r e a i s t he condu c t iv i ty , E is the electric field intensity, v is the velocity of the fluid,
a n d B i s the magnet ic induct ion, a l l quant i t ies measured in the frame of reference of
t h e l a b o r a t o r y .
Show that the magnet ic force per uni t volume of f luid is
ff(E + v x B ) x B .
For example , in crossed e lect r ic and magnet ic f ie lds , a conduct ing f luid is pushed
in the di rect ion of E x B. The field v x B t h e n o p p o s e s E. This is the principle of
o p e r a t i o n o f electromagnetic pumps.
13-6E HOMOPOLAR GENERATOR AND HOMOPOLAR MOTOR
Figure 13-9 shows a homopolar generator. I t cons i st s in a cond uc t in g d i sk ro t a t ing
in an axial magnetic field.
In one pa r t i cu la r case , B = 1 tesla, R = 0.5 meter , and the angular veloci ty is
3000 revolu t ions pe r minute .
Ca lcu la t e the ou tpu t vo l t age .
Homopola r gene ra tors a re inhe ren t ly h igh-cur ren t , l ow-vol t age dev ices .Ho mo po la r gen e ra tors us ing supe rcon duc t ing co il s an d wi th power ou tp u t s of s eve ra l
me gaw at ts are used for the puri f icat ion of me tals by e lect rolys is .
I f a vo l t age is app l i ed to the ou tpu t t e rm ina l s o f a hom op ola r gen e ra tor , i t
b e c o m e s a homopolar motor. H o m o p o l a r m o t o r s c a n p r o v i d e l a r g e t o r q u e s . T h e y a r e
su i t ab le , i n pa r t i cu la r , fo r sh ip propul s ion .
F igure 13-9 Ho m op ola r g ene ra tor .
See Prob. 13-6.
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Problems 325
13 -7 HOMOPOLAR MOTOR
See Pr ob . 13-6. N ow con s ider the device i l lus t ra ted in Fig. 13-10. Wil l the wheel
turn when the switch is c losed? If so, in which direct ion and why?*
Figure 13-10 Spiral coi l . A vol tage is
appl i ed be tween th e ax i s and the
pe r iphe ry . Contac t t o the pe r iphe ry
i s ma de wi th a ca rbon brush l ike
tha t used on motors . Does the d i sk
tur n? See P r ob . 13-7. Th e d i sk c ould
be ma de wi th a p r in t ed-c i rcu i t boa rd .
13-8 MAGNETIC PRESSURE
So as to c lar i fy the concept of magnet ic pressure , le t us cons ider , f i rs t a s ingle
soleno id, and then a pair of coax ial solen oids .
a ) F igure 13- l l a shows a longi tud ina l s ec t ion th rough a so lenoid . The sma l l
e lemen t ident i f ied by vert ical bar s (1) pro du ces i t s own m agn et ic f ie ld, bo th outs id e
an d ins ide the solenoid, a nd (2) is s i tua ted in the f ield prod uc ed b y the res t of the wind ing.
fTYTYTVTYTYn TYTYTYri rTYTYlYTYTYTYT)U A 1A L AJ A IA L J J A I A I A J J U A l A J A l A i A i A L '
0 0 © 0 0 © 0 © © 0 © 0 © 0 © 0 0
(a)
(MXKDQXDQXKD0(Tj(IXJXD(D(]XD
fTYTYTYTYTYT)U A I A l A i A JA I A l A i A JT U A l A i A i A l A l A i A i /
Figure 13-11 (a) Lon gi tud ina l s ec t ion th r oug h a long so lenoid . In P ro b . 13-8 , we
consider the f ie lds c lose to the smal l e lement ident i f ied by vert ical bars , (b) Sect ion
through a pa i r o f coax ia l so l enoids .
f See the American Journal of Physics, Vo lum e 38, 1970, p. 1273.
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326 Magnetic Fields: VI
Sho w the f irst f ie ld by me ans of sol id arrow s, and th e second o ne by mea ns of dash ed
a r r o w s .
We are of course thinking here of the fields infinitely close to an infinitely thin
winding .
Now can you expla in why the magne t i c p res sure i s B 2 2/< 0? ^ —
b) Fig ure 13-1 lb show s a pair of coaxia l solen oid s carryin g equal cu rren ts in
opposi te di rect ions . In this example we have three f ie lds .
Show, on e i ther s ide of the e lement , (a) sol id arrows for the f ie ld of the e lement ,
(b) dash ed arr ow s for the field of the rest of the inne r soleno id, (c) wavy a rro ws for the
fie ld of the outer solenoid.
W ha t is t he magne t i c p res sure on the inne r so leno id?
13-9E MAGNETIC PRESSURE
a ) Sh ow tha t t he mag ne t i c p res sure is abo ut 4 B 2 a t m o s p h e r e s . O n e a t m o s p h e r e
i s abo ut 10
5
pasca l s .b) Draw a log-log plot of the e lect r ic force per uni t area je 0E
2. in pascals, as a
funct ion of E. The maximum elect r ic f ie ld intens i ty that can be mainta ined in a i r a t
norm al t emp era tu re and pres sure is 3 x 1 06
vol ts per meter .
c) Discuss the s imilar i t ies and differences between the magnet ic pressure and
the e lect r ic force per uni t area .
13-10E MAGNETIC PRESSURE
I t i s poss ible to a t ta in very high pressures by discharging a large capaci tor
th rough a ho l low wi re .
a) Show that , for a thin tube of radius R ca r ry ing a cur ren t / . t he inward
magne t i c p res sure i s
p,„ = tt0I
2 Sn2
R2
.
b) Ca lcu late the pressure for a cur rent of 30 ki lo am per es in a tube on e mil l im eter
in d i amete r .
O n e a t m o s p h e r e i s a p p r o x i m a t e l y 1 0 5 pasca l s .
13-1 IE MAGNETIC PRESSURE
Magne t i c f i e lds a re used for pe r forming va r ious mechanica l t a sks tha t requ i re a
high power level for a very short t ime.
Fo r exam ple , if a light a lum inu m tube is inserted axia l ly in a solenoid, an d i f the
so lenoid , is suddenly con nec ted to a cha rged ca pac i to r , the induced e l ec t ro mo tance
</<t>/</f prod uc es a large curr en t in the tube, which c olla pse s un de r the m ag ne tic pre ssu re.The tube can act as a shut ter to turn off a beam of l ight or of soft X-rays .
Le t us ca l cu la t e the pres sure exe r t ed on a conduc t ing tube p laced ins ide a long
solenoid. I f the current / in the solenoid is increased gradual ly from zero to some
arbi t ra r i ly large value, the curr ent ind uced in the tube is smal l and the ma gn et ic pres sure
is negl igible . Let us assume that dl/dt i s so large that th e induced c urr ent in the tub e
ma inta ins zero ma gn et ic f ield ins ide i t. Th en th ere is a ma gn et ic f ield only in the an nu lar
reg ion be tween the so lenoid and the conduc t ing tube .
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Problems 327
a) Calculate the pressure on the tube for B = 1 tes la .
bl W ha t would th e pressure be if the cond uct ing tu be were paral le l to the axis ,
bu t some d i s t ance away?
'3-12E ENERG Y STORAGE
C om pa re the energies per uni t vo lum e in (a) a ma gne t ic f ie ld of 1.0 tesla an d
(b) an elec tro stat ic field of 106
vol ts per meter .
'3-13E MAGNETIC PRESSURE
Imagine tha t t he cur ren t i n a long so lenoid i s ma in ta ined cons ta n t , whi l e the
magne t i c p res sure inc reases the rad ius f rom R to R + dR. Then the magne t i c ene rgy
mcreases by
a ) Show tha t t he m echanica l wor k d on e by the magn e t i c p res sure p„ , i s
2nRlp„, dR .
b) Sh ow tha t t he mcrease in magn e t i c ene rgy is equa l t o the mechanica l wor k
d o n e .
c ) Dur ing the expans ion , t he magne t i c induc t ion ins ide the so lenoid remains
c o n s t a n t b e c a u s e B depends on ly on the number of tu rns pe r me te r and on the cur ren t .
So, as the radius increases , the f lux l inkage a lso increases .
Show tha t t he ex t ra ene rgy suppl i ed by the source dur in g the expan s ion , l(N (!<£>),
i s j us t tw ice the mecha nica l work done .
We conclude that one half of the energy suppl ied by the source serves to perform
mechanica l work , and the o the r ha l f s e rves to inc rease the magne t i c ene rgy . See P rob .4-13.
13-14 FLUX COMPRESSION
Flux com pres s ion is one me tho d of ob ta in ing a mag ne t i c f ie ld w i th a ve ry
large B. For example , imagine a l i gh t conduc t ing tube s i tua t ed ins ide a so lenoid pro
ducing a s teady B 0 . Th e ann ul ar space be tween the tube and th e solenoid is fi lled
with an explos ive, and t he soleno id is re inforced ex ternal ly . If the tube is now imp lode d,
i t wi l l be crushed, an azimuthal current wi l l f low, and the internal magnet ic pressure
B2
/2u 0will bui ld up unt i l it is c lose to the extern al gas pr essure .
where 6 i s t he magne t i c induc t ion when the rad ius has been reduced to P .
b) Show that the surface current dens i ty in the tube must be about 10 9 a m p e r e s
per meter to achieve a field of 10 3 tes las .
c) If the initial B is 10 teslas. and if the tu be h as initially a diam ete r of 100 milli
me te rs , wha t should be the va lue of B when th e tube i s com pres sed to a d i am ete r o f
abou t 10 mi l l ime te rs ?
a) Sho w that , if the rad ius of the tube sh rink s very rap idly.
B * B0(R0!'Rf.
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328 Magnetic Fields: VI
d) Ca lcu la t e the resul t ing increase in magne t i c ene rgy . Assume tha t th e cyl in
d e r is 200 mil l ime te rs long an d t h a t the c u r r e n t in the so lenoid is cons tan t . Neglec t
end effects.
F l u x c o m p r e s s i o n ca n also be achieved by m e a n s of a c o n d u c t i n g p i s t o n s h o t
axial ly into a so lenoid . y
If the ra diu s of t he so lenoid is R0. and if t he rad ius of the pis ton is K, the m a g n e t i c
i n d u c t i o n in the annula r reg ion be tween th e pi s ton and the so lenoid , becomes
S 0
B %(R/R0r
13-15E PULSED MAGNETIC FIELDS
Ext remely h igh magne t i c f i e lds can be o b t a i n e d by d i s c h a r g i n g a c a p a c i t o r
t h r o u g h a l ow- induc tance co i l . Th e capac i to r l eads mus t of course have a low i n d u c
t ance . Such fa s t capac i to rs cos t approxima te ly tw o d o l l a r s per j o u l e of s tored-ene rgycapac i ty .
a ) Es t ima te th e cos t of a capac i to r t ha t cou ld s to re an ene rgy equa l to t h a t of a
100-tes la magnet ic f ie ld occupying a v o l u m e of one l i ter.
b ) Es t ima te th e cos t of the elect r ic i ty required to c h a r g e th e c a p a c i t o r s .
c ) Ca lcu la t e the m a g n e t i c p r e s s u r e in a t m o s p h e r e s , at 100 tes las . One a t m o
s p h e r e is a b o u t 105
pasca l s .
13-16E MAGNETIC ENERGY
It was s t a t ed in Sec. 13.5 t h a t , if we h a v e tw o i n d u c t a n c e s La
and Lb, ca r ry ing
c u r r e n t s Iaan d Ib,
Wm =l
-LJ2
a +\Lbll + MIJb.
Y o u can prove this qui te eas i ly.
Let coil a pro du ce flux l inkages
Aaain coi l a an d Aab
in coil b.
Similar ly, let coil h produce f lux l inkages
a n d Abb in coil b.
a ) F ind Wm in t e r m s of the c u r r e n t s and of the f lux l inkages .
b) Verify th e a b o v e e q u a t i o n .
13-17E ENERGY STORAGE
a ) Show tha t th e magne t i c ene rgy s to red in an i sola ted c i rcui t carrying a
c u r r e n t / and with a f lux l inkage A is
Wm
= ( l / 2 ) /A .
b) Ca lcu la t e Wmfor a l ong so lenoid us ing th i s fo rmula .
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Problems 329
13-18 ENERGY STORAGE
a ) A cons tan t -vo l t age source s e t a t a vo l t age V i s conn ec ted to an ind uc tance
L of negl igible res is tance a t t = 0.Fin d, as funct ions of the t ime , ( i) the cu rren t / . ( ii ) the energy suppl ied by t he
sourc e, and ( i ii ) the energ y s tor ed in the ma gn et ic f ie ld.
b ) A con s tan t -c ur ren t source s e t a t a cur ren t / is conn ec ted to a capac i to r C
at r = 0.
Find , as funct ions of the t ime, (i) th e vol tag e V on the capaci tor , ( i i ) the energy
supp l ied by the sour ce, and ( ii i) the ene rgy s tored in the e lect r ic f ie ld.
13-19 ENERGY STORAGE
Electr ical ut i l i t ies must be able to meet peak power demands . I t i s therefore
des i rable to s tore energy during periods of low demand and feed i t back into the grid
when needed . On e way of s to r ing ene rgy i s t o pu m p wa te r in to an e l eva ted re se rvo i r .
I t has been sugge s ted that large am ou nt s of energy cou ld a lso be s tore d in themagnet ic f ie lds of superconduct ing coi ls .
a) C om pa re the energy d ens i t ies in (i ) an e lect r ic f ie ld of 10 8 vol ts per meter in
Mylar , for which e r = 3.2. and (ii) a ma gn eti c field of 8 teslas. Fhe se fields are ab ou t a s
high as present technology wil l permit . Fhe magnet ic energy dens i ty is larger by two
o r d e r s o f m a g n i t u d e .
b) I t i s sugges ted that 10 gig aw at t - ho urs be s tored in a solen oid wi th a length -
to -d iam ete r ra t io o f 20 . Th e so lenoid wo uld be ma de in four pa r t s , j o ined en d to end
to form a quas i - toroidal coi l located in a tunnel excavated in sol id rock. Disregard the
toro ida l shape in your ca l cu la t ion . The m agne t i c indu c t ion w ould be 8 t e s la s . ( i) Ca lcu
l a t e the to t a l l eng th and the d i amete r , ( ii) Ca lcu la t e the magn e t i c p res sure in a tmo spher es ,
( ii i) Cal cula te the surface area tha t wo uld req uire cryogen ic ther ma l in sula t ion .
13-20 SUPERCONDUCTING POWER TRANSMISSION LINE
A supe rcon duc t ing po wer t ransmis s ion l ine has been pro pose d tha t wo uld
hav e the fol lowing chara cter is t ics . I t wo uld carry 1 0 " wat ts a t 200 ki lovo l ts DC o v e r
a dis tance of 10 3 ki lome te rs . Th e cond uc to rs would hav e a c i rcu la r c ros s- s ec t ion of
5 squa re cen t ime te rs and would be he ld 5 cen t ime te rs apa r t , cen te r t o cen te r .
a ) Ca lcu la t e the magne t i c fo rce pe r me te r .
I t so happens that the force of a t t ract ion is the same as i f the conductors were
replace d by f ine wires , 5 cen t im eters apa rt .
b ) Ca lcu la t e B 2 / 2 p 0 m i d - w a y b e t w e e n t h e t w o c o n d u c t o r s .
c) Ca lcu late the s tore d energy and i ts cos t a t 0.5 cent per ki low at t - ho ur.
Th e sel f- inductanc e of such a line is 0.66 mic roh en ry p er me ter .
13-21 ELECTRIC MOTORS AND MOVING-COIL METERS
Elec t r i c motors , a s we l l a s moving-co i l vo l tme te rs and ammete rs , u t i l i ze the
torque exerted on a current-carrying coi l s i tuated in a magnet ic f ie ld.
Cons ide r an N- turn rec t angula r co i l s i t ua t ed in a un i form magne t i c f i e ld a s
in Fig. 13-12.
a) C alc ul ate th e forces on sides 1, 2, 3, 4.
b ) Sho w tha t t he to rque
T = NIBab sin 0.
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33 0
Figure 13-12 Rec tangula r co i l whose
n o r m a l n forms an angle 0 with a
uniform magnet ic f ie ld. See Prob.13-21.
In an e lect r ic motor , we have essent ia l ly a number of such coi ls , offset by, sa ;
15 degrees , aro un d th e axis, an d con nec ted t o a se t of co nta cts cal led a commutator
f ixed to the axis . Connect ion to the coi ls i s made by carbon brushes that r ide on th
c o m m u t a t o r . T h e c o il s a n d t h e i r c o m m u t a t o r f or m t h e armature of the motor . Th
magnet ic f ie ld is suppl ied by the stator coils. Th e mag net ic f lux is carr ied by lamina tci
i ron , bo th in the a rma ture and in the s t a to r yoke.
Moving coi l vol tmeters and ammeters use a uniform radial magnet ic f ie ld as i i
F ig . 13-13. The to rqu e i s t hen s imply NIBab and i s i ndepe nden t o f 0. A light spiraspr ing exe r t s a re s to r ing to rque tha t i s p ropor t iona l t o 0. giving a deflection 0 p r o p o r
t ional to / .
Figure 13-13 Moving co i l and mag
net ic f ie ld in a vol tmeter or ammeter .
Th e mag net ic f ie ld is supp l ied by a
p e r m a n e n t m a g n e t . T h e p a r t s m a r k e d
/ are made of i ron.
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Problems 331
13-22 MAGNETIC TORQUE
Consider a square s ingle- turn coi l of s ide a, ca r ry ing a cur ren t / .
a) Show that i t tends to orient i t se l f in a magnet ic f ie ld in such a way that thetota l magnet ic f lux l inking the coi l i s maximum.
b) Show that the torque exerted on the coi l i s m x B, where m is the magnet ic
m om en t of the coi l (Sec. 8.1.2) , an d B is the ma gne t ic ind uct io n w hen th e curren t in
the coil is zero.
This resul t i s independent of the shape of the coi l .
13-23 ATTITUDE CONTROL FOR SATELLITES
See the prev ious prob lem.
Many sa t e l l i t e s requ i re a t t i t ude cont ro l t o keep them proper ly or i en ted . For
exam ple , com mu nica t ion s a t e ll i t es mus t keep the i r an te nn a sys t ems on t a rge t . A t t i t ude
cont ro l requ i re s a me thod for exe r ting appr op r i a t e to r ques a s they a re requi red .
At t i t ude cont ro l can be ach ieved to a ce r t a in ex ten t by means of co i l s whosema gne t ic f ie lds interact wi th that of the ear th.
a) Show that the torque exerted by such a coi l i s
NIBA sin 0.
w h e r e N i s the nu m be r of turn s , / is the cur ren t . B is t he magne t i c in duc t ion due to
the ea r th , A i s the area of the coi l , and 0 i s the angle between the earth 's magnet ic f ie ld
and the normal to the co i l .
b ) Ca lcu la t e the nu m be r of am per e - tu rns requi red for a co il wo un d a rou nd
the outs id e surface of a sa te l l ite whose dia me ter i s 1.14 me ters . Th e torq ue requ ired
at 0 = 5 degrees i s 1 ( F 3 newton-mete r . and the magne t i c induc t ion a t an a l t i t ude of
700 ki lome ters ov er the equa tor , wher e the orbi t of the sa te l l i te wi ll be s i tuated, i s
4.0 x 1 0 " 5 tes la .
13-24 M ECHANICAL FORCES ON AN ISOLATED CIRCUIT
Sho w tha t , if the geo me try of an isola ted ac t ive c i rcui t is a l tered, the e nergy
supp l ied by the sourc e is equ al to twice the increase in ma gne t ic energy. Th us the m e
chanica l work pe r form ed is equa l t o the inc rease in magn e t i c ene rgy . Assum e tha t t he
cur ren t i s kep t cons ta n t .
I t fol lows that , on this ass um pti on , the force on an e lem ent of an act ive c i rcui t
i s equa l t o the ra t e o f change of magne t i c ene rgy .
See Pr ob s . 13-13 and 13-25.
RAIL GUN. OR PLASMA GUN
Calculate the force on the movable l ink D in the circuit of Fig. 13-8.
Ins tead of being a metal l ic rod. the l ink can a lso be an e lect r ic arc . The device
then accelerates blobs of plasma and is cal led a plasma gun. State l l i te thrus ters can
be made in this way. See Probs . 2-14. 2-15. 5-3, and 10-11.
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332 Magnetic Fields: VI
w h e r e F i s the driving force and v i s the veloci ty of the l ink. Note that both L and
/ are funct ions of the t ime.
Now what is the value of VI T h e v o l t a g e V (a) increases the flux linkage in the
induc tance and (b) g ives a vo l t age drop IR ' on the res is tor R':
(1)
(LI) + IR\ (2 )
w h e r e LI is the flux linkage.
C o m b i n i n g E q s . 1 a n d 2 .
(3)
2(4)
I-1(5)
LI 2. (6)2
Solution: We cannot so lve th i s p rob lem by in t egra t ing the magne t i c fo rce
J x B over the vo lum e of the l ink, s ince nei th er J n or B are kno wn , or even eas i ly
calculated. Ins tead, we shal l f ind the force by inves t igat ing the magnet ic and mechanical energies involved.
If we set x = 0 a t the ini t ia l pos i t ion of the l ink, then th e indu ctan ce L i n t h l T '
circuit is L 0 + L'x, w h e r e L'x i s the indu ctan ce asso cia ted w ith the ma gne t ic f lux
l inking the ra i ls . L' being the sel f- inductance per meter .
Dur ing the d i s cha rge of the capac i to r C we can neg lec t t he ba t t e ry and the
res i s t ance R. s ince the current f lowing in that part of the c i rcui t i s negl igible . Let
the res is tance on the r ight-hand s ide of the c i rcui t be only that of the arc . R' .
At any ins tant , the power suppl ied by the capaci tor serves to (a) increase the
magne t i c ene rgy \LI2. (b) increase th e kinet ic energ y, and (c) diss ipate en ergy in
the res is tance R' . T h u s
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CHAPTER 14
MAGNETIC FIELDS: VII
Magnetic Materials
14.1 THE MAGNETIZATION M
14.2 THE EQUIVALENT SURFACE CURRENT DENSITY ete
14.3 THE EQUIVALENT VOLUM E CURRENT DENSITY J (,
14.4 CALCULATION O F MAGNETIC FIELDS ORIGINATING
IN MAGNETIZED MATERIAL
14.4.1 Example: Uniformly Magnetized Bar Magnet
14.5 THE MAGNETIC FIELD INTENSITY H
14.6 AMPERE'S CIRCUITAL LAW
14.6.1 Example: Straight Wire Carrying a Current I and Embed ded
in Magnetic Material
14.7 MAGNETIC SUSCEPTIBILITY /„„ PERMEABILITY p.
AND RELATIVE PERMEABILITY p r
14.8 THE MAGNETIZATION CURVE
14.9 HYSTERESIS
14.9.1 Example: Transformer Iron
14.10 BOUNDARY CONDITIONS
14.11 SUMMARY
PROBLEMS
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Thus far we have studied only those magnetic fields that are attributable
to the motion of free charges. Now, on the atomic scale, all bodies containspinning electrons that move around in orbits, and these electrons also
produce magnetic fields.
Ou r pu rpo se in this chap ter is to express the magne tic fields of these
atomic currents in macroscopic terms.
Magnetic materials are similar to dielectrics in that individual charges
or systems of charges can possess magnetic moments (Sec. 8.1.2), and these
moments, when properly oriented, produce a resultant magnetic moment
in a macroscopic body. Such a body is then said to be magnetized.
In most atoms the magnetic moments associated with the orbital and
spinning m oti ons of the electrons cancel. If the cancellat ion is not complete,the material is said to be paramagnetic. When such a substance is placed
in a magnetic field, its atoms are subjected to a torque that tends to align
them with the field, but thermal agitation tends to destroy the alignment.
This phe nom eno n is anal ogo us to the alignme nt of polar molecules in
dielectrics.
In diamagnetic materials, the element ary mome nts are not per mane nt
but are induced according to the Faraday induction law (Sec. 11.2). All mate
rials are diamagnetic, but orientational magnetization may predominate.
Magnetic devices use ferromagnetic materials, such as iron, in which
the magnetization can be orders of magnitude larger than that of eitherpara- or diamagnetic substances. This large magnetization results from
electron spin and is associated with gro up ph en ome na in which all the
elementary moments in a small region, known as a domain, are aligned. The
magnet izatio n of one dom ain may be oriented at ran do m with respect to
that of a neighb oring doma in.
The re is one impo rt an t difference betwee n dielectric and mag net ic
materials. In most dielectrics D is propor tional to E and the medium is said
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14.2 The Equivalent Surface C urrent Density a,, 33 5
to be linear. Fer romagne t i c mate r i a l s a r e no t on ly h igh ly non- l inear , bu t
the i r behav io r a l so depends on the i r p r ev ious h i s to ry . The ca lcu la t ion o f
the fi elds assoc ia t ed wi th m agn e t i c ma te r i a l s is t he re fo re l a rge ly emp i r i ca l .This wi l l be the subject of the next chapter .
14.1 THE MAGNE TIZA TION M
T h e magnetization M i s t he mag ne t i c mom en t per un i t vo lum e a t a g iven
point . I f m i s t he average mag ne t i c d ipo le mo me nt p er a tom , and i f N is the
n u m b e r o f a t o m s p e r u n i t v o l u m e ,
M = A/m, (14-1)
if t he ind iv idua l d ip o le mo m en t s in the e l ement o f vo lu me cons id ered a r e
a l l a l igned in the same d i r ec t ion .
Th e magn e t i za t ion M i s me asu red in am pere s per mete r , and it co r
r e s p o n d s t o t h e p o l a r i z a t i o n P in dielectr ics (Sec. 6.1).
14.2 THE EQUIVALENT SURFACE CURRE NT DENSITY ae
Fig ure 14-1 show s a cyl in der of ma ter i al d ivid ed in to squa re cells of cros s-
sec t iona l a r ea a2
, w h e r e a -» 0 . Imagine that each cel l car r ies a c lockwise
sur f ace cur r en t o f M amperes per mete r , a s in the f igure . Then each one-
me te r l eng th o f cel l has a ma gne t i c mo m en t o f a 2M a m p e r e m e t e r s q u a r e d
(Sec. 8.1.2) a n d t h e m a g n e t i c m o m e n t p e r c u b i c m e t e r is M a m p e r e s p e r
meter in the di rect ion shown in the f igure. Now the current in one cel l i s
cance led by the cu rre nts in the adjoi ning cel ls , excep t a t the pe r iphe ry of the
m a t e r i a l . T h u s , if t h e m a t e r i a l h a s a m a g n e t i z a t i o n M, t h e e q u i v a l e n t c u r r e n t
dens i ty ct e at the surface is M.
More genera l ly , t he equivalent, or Amperian, surface current density is
a , = M x n, , (14-2)
w h e r e n, i s a un i t vec to r no rm al to the su r f ace and po in t ing outward.
These equ iva len t cu r r en t s do no t d i ss ipa te energy because they do no t
invo lve e l ec t ron d r i f t and sca t t e r ing p rocesses l ike those assoc ia t ed wi th
c o n d u c t i o n c u r r e n t s .
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336
Figure 14-1 A m p e r e ' s m o d e l for the equ iva len t cu rren t in a cy l inder of m a g n e t i z e d
mater ia l .
14.3 THE EQUIVALENT VOLUME CURRENT DENSITY Je
Let us now supp ose that the magnet iza tio n M is a function of the x-c oor din ate
as in Fig. 14-2, with
where p is a positive con stan t. Since M is expressed in amper es per meter,
p is expressed in amp ere s per squ are meter .
At the surface of the material, the equivalent current density «e
is
again given by Eq. 14-2, with M equal to the local value of the magnetization
Thus ae
is a function of x, as in th e figure.
Inside the material, currents parallel to the x-axis cancel.
Currents parallel to the y-axis do not cancel. Along any of the vertical
planes inside, the left-hand side of a given cell carries an upward current of
M = px, (14-3)
while the right-hand side carries a downward current of
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33 7
y
X
Figure 14 -2 Mag ne t i zed cy l inde r in which M i s a funct ion of the x-c oor d in a te :
M increases l inearly wi th x as in Eq. 1 4 - 3 .
Along any ver t i ca l p l ane , we the re fo re have a ne t downward cur r en t o f
pa a m p e r e s .
N o t e t h a t t h is d o w n w a r d c u r r e n t i s i n d e p e n d e n t o f x. This i s s imply
b e c a u s e w e h a v e a s s u m e d t h a t M i s d i r ec t ly p rop or t i on a l to x .
So, a lon g every ver t i ca l d iv i s ion be tween ce l l s, we have a ne t d ow nw ard
cur r en t o f pa a m p e r e s . T h e n e t d o w n w a r d c u r r e n t d e n s i t y is t h u s p, since
w e h a v e pa am pere s for every in t e rva l a. In o th er words , we have an equ iva len t
cur r en t dens i ty
W e c a n n o w s h o w t h a t J e i s eq ua l to V x M in the fol lowing wa y.
Along the cu rve C of Fig. 14 -2, M • dl i s zer o over the lef t -hand e nd, b eca use
M i s pe rp end icu la r to dl. The same app l i es to the r igh t -hand end . Thus
J e = -pi (14-4)
(14-5)
= -pa. (14-6)
wh ere the line integral i s eva lua ted in the di re ct ion sho wn in the f igure.
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338 Magnetic Fields: VII
Since the area of the loop is a x 1, t he y -componen t o f the cu r l o f
M is — p, from Sec. 1.12, a n d
je
= (v x rv iy. (14-7^
More genera l ly , t he equ iva len t vo lume cur r en t dens i ty i s
J e = V x M. (14-8)
14.4 CALCULA TION OF MAGNE TIC FIELDS
ORIGINATING IN MAGNE TIZED MATERIAL
Let us f ir s t recal l the reas on ing we fol lowed in Sec. 6 .3 to f ind t he elect r ic
f i e ld due to po la r i zed d ie l ec t r i c mate r i a l . We found tha t t he charge d i sp lace
me nt tha t occu r s on po la r i z a t ion r esu l t s i n the acc um ula t ion o f ne t dens i t i e s
o f b o u n d c h a r g e , a b on the sur faces and p h i n s ide . We then a rgued tha t an y
net charge density gives r ise to an electr ic f ield, exactly as if i t were in a
vacuum. As a consequence , E can be ca lcu la t ed a t any po in t in space f rom
a b a n d p b.
We now have an en t i r e ly ana logous s i tua t ion in the magne t i c f i e lds
o r i g i n a t i n g i n m a g n e t i z e d m a t e r i a l s . T h e m a g n e t i z a t i o n g i v e s e q u i v a l e n t
cur r en t s , wi th dens i t i e s aL, on the su r f aces and J e i n s ide . Thes e cur r en t s
p roduce magne t i c f i e lds , bo th ins ide and ou t s ide the mate r i a l , exac t ly as i f
they were s i tua ted in a vacuum.
In pr incip le , on e can calc ulat e B at any p oint in spac e f rom the equ iv
a len t cu r r en t dens i t i e s a e a n d J e. In pract ice , the vector M is a n u n k n o w n
func t ion o f the coord ina tes ins ide the mate r i a l , and one can do l i t t l e more
th an gues s it s valu e. W e shal l re t ur n to th is subject in the next ch ap ter .
EXAMPLE: UNIFORMLY M AGNETIZED BAR MAGNET
Th e uniformly m agn et ize d b ar ma gne t of Fig. 14-1 is a go od ex am ple to use a t this
po in t .
W e have shown in Sec. 14.2 tha t the cyl indrical surface carr ies an eq uivale nt
cur ren t dens i ty x e that i s equ al to M. Ins ide the magne t , V x M = 0 , and the vo lum e
current dens i ty J f is zero. Ove r the end faces. M is paral le l to the norm al uni t vecto r
n , , a n d a, i s zero.
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14.5 The Magn etic Field Intensity H 339
The B field of a cyl inder uniformly magnetized in the direct ion of i ts axis of
symmetry is therefore identical to that of a solenoid of the same dimensions with
JV' tu rns per meter and car ry ing a cu rren t / such tha t IN' = M, as in Fig. 9-10.
In a rea l bar magnet the magnet ic moments o f the ind iv idua l a toms tend to
al ign themselves with the B field , so that the magnetizat ion M is weaker near the
ends . The end faces a l so car ry Amper ian cu rren ts , s ince M x n is zero only on the
axis . Th e net resul t is that th ere are tw o "poles ," one at each end of the ma gne t ,
from which l ines of B radiate in al l d irect ions outs ide the magnet . The poles are
mos t c onsp ic uous if the bar ma gnet i s long and th in .
14.5 THE MAG NETIC FIELD INTENSITY H
W e fou nd in Sec. 9.2 tha t , for a stea dy cu rre nt density of free ch ar ge an dfor n o n -m ag n e t i c m a te r i a l s ,
V x B = /< 0 J/- (14-9)
No w we have jus t seen tha t magn et ize d ma ter ia l can be rep laced by i t s
equiv alen t cu rre n ts for ca lcu la t ing B . Co nse que nt ly , i f we have ma gne t ized
mater ia l as wel l as s teady curren ts ,
V x B = nQ{i f + j e ) , (1 4 -1 0 ,
w h e r e J e i s the equ ivalen t vo lum e curre n t dens i ty . Th is equ at io n i s o f course
val id on ly a t po in ts wher e the deriva t ives of B wi th respect to the coo rd i nat es
x, y, z exis t . I t i s therefore no t ap pl ica b le a t the surface of m agn et ic ma ter ia l s
an d w e d i s reg a rd a (,.
Su b s t i t u t i n g V x M for J t > ,
V x B = p.0(J f + \ x M), (1 4 -1 1 ,
V x ^— — mJ = J f. (14-12 ,
Th e vector be tween pare n the ses , whos e curl i s equa l to the free c urre n t
density at the point , is the magnetic field intensity
H = — - M , (14-13 ,I h
and i s express ed in am pe res pe r m eter . N ot e tha t H and M are expressed
in the same uni t s .
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340 Magne tic Fields: VII
T h u s
B = n0(H + M). (14-14)
This equat ion is to be compared with Eq. 6-15 , which appl ies to d ie lec tr icsrf
E = — (D - P). (14-15)
e 0
14.6 AMPERE'S CIRCUITAL LA W
Rew rit ing no w Eq. 14-12, we hav e tha t , for s teady curr ents , e i ther ins ide or
ou ts ide magne t ic ma te r ia l ,
V x H = J f. (14-16)
In tegra t ing over a surface S,
J" s(V x H ) d a = JsJ / - d a (14-17)
or , us ing Stoke 's theorem on the lef t -hand s ide ,
( f c H •«« = / „ (14-18)
w h e r e C is the curve bounding the surface S, a n d I r is the current of free
cha rges l ink ing the cu rve C. N o t e t h a t I f d o e s not inc lude the equ iva len t
currents . The term on the lef t is ca l led the magnetomotance. A m a g n e t o m o -
tance i s exp ressed in amp eres , o r in a mp ere - t u rns .
This is a more genera l form of Ampere's circuital law (Sec. 9.1), in
tha t i t can be used to ca lcula te H even in the presence of magnet ic mater ia ls .
I t is r igorously val id , however , only for s teady currents ; we shal l dea l with
va r iab le cu r ren t s in Chap te r 19 .
14.6.1 EXAMP LE: STRAIGHT WIRE CARRYING A CURREN T I AND
EMBEDDED IN MAGNETIC MATERIAL
Fi gu re 14-3 shows a s t r a i gh t wi re car ry i ng a cu r r en t / and emb edde d i n m agne t i c
mat er i a l . Th e ma gne t om ot an ce a ro und t he pa t h shown c i r c li ng t he cu r r en t I is
equal to / , and H = I/2nr, exact ly as if the wire were s i tuated in a v ac uum .
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34 1
Figure 14-3 St ra igh t w i re ca r ry ing a cur r en t / and emb edd ed in ma gne t i c m a te r i a l .
I f the ma teria l i s i sot ro pic , and if i t has the shap e of a c i rcular cyl ind er wi th /
a lon g i ts axis , M is azi mu tha l , l ike H .
14.7 MAGNETIC SUSCEPTIBILITY xm, PERMEABILITY n,
AND RELATIVE PERMEAB ILITY \i r
As for dielectr ics (Sec. 6.6), i t is convenient to define a magnetic susceptibility
~/ m s u c h t h a t
N o w , s i n c e
t h e n
w h e r e
M = Z m H .
B = p 0 ( H + M),
B = jU 0(l + z J H = p 0p rH = pU,
P-R = 1 + Xm , / ' = M O / V
(14-19)
( 1 4 - 2 0 )
(14-21)
( 1 4 - 2 2 )
T h e q u a n t i t y p i s t he permeab i l i ty and p r is the relative permeability. B o t h
X m a n d p r a r e d i m e n s i o n l e s s q u a n t i t i e s .
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342 Magnetic Fields: VII
Eq ua t io n 14-20 is gene ral , bu t Eqs. 14-21 an d 14-22 are base d on th e
ass um pt io n tha t t he mate r i a l is bo th i so t rop ic and l inear , i n o the r word s ,
tha t M i s p ro po r t io na l to H and in the same d i r ec t ion . Th i s ass um pt io n i su n f o r t u n a t e l y no t val id in fer romagnet ic mater ials , as we shal l see in
nex t sec t ion .
The magne t i c suscep t ib i l i t y o f paramagnetic substances is sma l l e r th an
un i ty by severa l o rder s o f magn i tude and i s p ropor t iona l to the inver se o f
t h e a b s o l u t e t e m p e r a t u r e .
In diamagnetic materials, t he mag ne t i za t i on i s i n the d i r ec t ion opposite
to the external f ie ld; the relat ive permeabi l i ty i s less t han un i ty and i s i nde
p e n d e n t of t h e t e m p e r a t u r e . If o r i e n t a t i o n a l m a g n e t i z a t i o n p r e d o m i n a t e s ,
the r esu l t an t r e l a t ive perm eab i l i ty i s g r ea te r than un i ty .
14.8 THE MAGN ETIZA TION CUR VE
O n e c a n m e a s u r e B for var ious values of H with a Rowland ring w h o s e
minor r ad ius i s much smal l e r than i t s majo r r ad ius , a s in F ig . 14-4 . The
func t ion o f win d ing a, which has N a t u rns and ca r r i es a cu r r en t / „ , i s t o
produce a known magne t i c f i e ld in t ens i ty
H = NJJlnr (14-23)
Figure 14-4 Rowland r ing for the determinat ion of B as a funct ion of H in a
fe r romagne t i c subs tance .
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34 3
H (ampere-turns meter)
Figure 14-5 Magne t i za t ion curves for va r ious magne t i c ma te r i a l s : a, P e r m a l l o y ; b,
a n n e a l e d p u r e i r o n : c, duct i le cas t i ron; d, Alnico 5 . Th e num bers sh own a re the
re l a tive p e rmeabi l i t i e s B//i0H.
i n the sample . The magne t i c induc t ion i s
B = cp/S, (14-24 )
w h e r e S i s the cross-sect ional area of the core, and O is the magnet ic f lux.
O ne can mea sur e change s in 0 , and hence chan ges in B, by chan g ing / and
i n t e g r a t i n g t h e e l e c t r o m o t a n c e i n d u c e d i n w i n d i n g b as in Prob. 14-16. Both
B a n d H are the re fo re eas i ly m eas urab le .
If we s t a r t wi th an unm agn e t i z ed s am ple and inc rease the cu r r en t in
coi l a, t he magne t i c induc t ion a l so inc reases , bu t i r r egu la r ly , a s in F ig . 14-5 .
Al l f e r romagne t i c subs tances have such S- shaped magnetization curves. N o t e
the l a rge var i a t ions in u, for a g iven mate r i a l . W hen ever o ne uses p r in relat ion
wi th f e r romagne t i c mate r i a l s , one r e f e r s to the r a t io B/u0H, for specific
values of B or of H.
A chara c te r i s t i c f ea tu re o f the mag ne t i za t io n curve i s t he saturation
i nduc t ion beyond which M increases no fu r ther . Maximum a l ignment o f
the dom ain s is t hen ach ieved , a ft e r which dB = fi0 dH . T h e s a t u r a t i o n
ind uc t ion l ies in the ran ge f rom 1 to 2 teslas , dep en din g on the m ate r ial .
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34 4
Figure 14-6 M a g n e t i z a t i o n c u r v e ab
and hys teres is loop bcdefgb.
14.9 HYSTERESIS
Co nsid er now F ig . 14-6. Cu rve ab is t h e m a g n e t i z a t i o n c u r v e . O n c e p o i n t b
has been r eached , i f t he cu r r en t in wind ing a of Fig . 14-4 is reduced to zero,
B d e c r e a s e s a l o n g be . T h e m a g n i t u d e o f t h e m a g n e t i c i n d u c t i o n a t c is the
remanence or the retentivity fo r the par t i cu la r sample o f mate r i a l . I f t he
cur r en t i s t hen r ever sed in d i r ec t ion a nd inc rease d . B r eaches a po in t d w h e r e
it is r educ ed to ze ro . The m agn i tud e o f H a t th i s po in t i s kn ow n as the coercive
force. On fu r ther inc reas ing the cu r r en t in the same d i r ec t ion a po in t e,
s y m m e t r i c a l t o p o i n t b, i s r eached . I f t he cu r r e n t i s no w reduc ed , r ever sed ,and inc reased , the po in t b i s aga in r eached . The c losed curve bcdefgb is
k n o w n a s a hysteresis loop. If, a t any po int , the cur ren t i s var ied in a sma l ler
cycle , a smal l hysteresis lo op is descr ibe d.
Energy i s r equ i r ed to desc r ibe a hys te r es i s cyc le . Th i s can be shown
by cons ider in g F ig . 14-7. W hen the cu r r en t in win d ing a of Fig. 14-4 is
inc reas ing , t he e l ec t romotance induced in the wind ing opposes the inc rease
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345
Figure 14-7 The shaded area gives the
energy per unit volume required to go
from « t o b on th e hysteresis loop.
T h e energy per unit vo lume required
to describe a complete hysteresis loop
is equal to the area enclosed by the
loop .
in current, according to Lenz's law (Sec. 11.3), and the extra power spent by
the source is
where S is the cross-sectional area of the ring, N is the number of turns in
winding a. and B is the average magnetic induction in the core. Also,
where / is the me an circumf erence of the ring, T = SI is its volume, and
IN/I = H, as in Eq. 14-23. Thus
W, = T [b
HdB (14-27)
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346 Magnetic Fields: VII
i s the energy suppl ied by the source in going f rom the point g t o the po in t b
in Fig . 14-7. Th is in tegral cor r es po nd s to the shad ed are a in the f igure a nd
i s equ a l to the energy sup p l i ed per un i t vo lum e o f the magn e t i c co re .
W hen the cu r r en t i s i n the sam e d i r ec t ion b u t i s decreas ing , t he polarivf
of the induced e l e c t ro mo tanc e i s r ever sed , acc ord in g to Lenz ' s l aw, wi th the
resu l t t ha t t he energy
W 2 = t P H dB (14-28)
i s r e tu rne d to the source .
F ina l ly , t he energy supp l i ed by the source dur ing one cyc le i s
W=tj)HdB, (14-29)
wh ere the in t egra l i s eva lua ted a ro un d the hys te r es i s loop . The a r e a o f the
hys te r es i s loop in t es l a - ampere tu rns per mete r o r in weber - ampere tu rns
per cub ic mete r i s t he re fo re the num be r o f jou le s d i ss ipa ted pe r cub ic mete r
and per cycle in the core.
Hys te r es i s losses can be min imized by se l ec t ing a mate r i a l wi th a
n a r r o w h y s t e r e s i s l o o p .
14.9.1 EXAMPLE: TRANSFORMER IRON
Fo r the t ransfor me r i ro n of Fig. 14-7. the area enclosed b y the hys teres is loo p is
appro xim a te ly 150 web er -am pere tu rn s pe r cub ic me te r o r 150 jou les pe r cub ic
me ter and per cycle , or 1.1 wa t ts per ki log ram at 60 hertz . This i ron is sui tab le for
use in a power t ransformer, but other a l loys are avai lable wi th lower hys teres is
losses .
14.10 BO UNDA R Y CONDITIONS
Let us exam ine the con t inu i ty co nd i t ions tha t B and H mu st obey a t t he
inter face betw een t wo m edia . W e shal l pro cee d as in Sec. 7 .1 .F igure 14-8a shows a shor t cy l indr i ca l vo lume whose top and bo t tom
faces are para l le l an d inf ini te ly close to the inter face. Since th ere is ze ro ne t
f lux through the cyl indr ical sur face (Sec. 8 .2) , the f lux through the top face
m u s t e q u a l t h a t t h r o u g h t h e b o t t o m , a n d
B„i = B n2. (14-30)
Th e no rm al com po ne n t o f B i s the re fo re con t i nu ou s ac ross the in te r f ace .
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34 7
(b)
Figure 14-8 (a) Gauss ian surface a t the interface between two media , (b) C l o s e d
path cross ing the interface.
Co nsi der n ow F ig . 14-8b . Th e c losed pa th has two s ides para l l e l t o
the inter face and close to i t . From the ci rcui ta l law (Sec. 14.6) ,
H d l = 7, (14-31)
whe re / is t he con duc t ion cu r r en t l i nk ing the pa th .
If the tw o sides par alle l to the inte rface a re infinitely close to i t , 7 is
equa l to ze ro ,
H n = H t2, (14-32)
a n d t h e t a n g e n t i a l c o m p o n e n t o f H i s con t inu ou s ac ross an in t e rf ace .
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348 Magnetic Fields: VII
If we can set B equal to uH in both media, the relative permeabilities
being those that correspond to the actual values of H. then
uriu
l}H
l cos 0l
= p,r2fj,
0H
2 cos 92, (14-33/f^
from the continuity of the normal component of B, and
Hx
sin 0j = H2
sin 02, (14-34)
from the continuity of the tangential component of H. Then
tan 6j nri
tan 02
Uri(14-35)
Th e lines of B, or of H. are farther awa y from th e nor mal in the med iu m
having the larger permeability.
14.11 SUMMARY
The magnetization M is the magnetic mom ent per unit volum e.
One can calculate the magnetic field of a piece of magnetic material
by assuming an equivalent surface current density
a, = M x n, . (14-2)
where n, is a unit vector nor mal to the surface and poi nti ng out war d, a nd
an equivalent volume current density
Je = V x M. (14-8)
The magnetic field intensity is
H = — - M, (14-13)
l<0
an d
V x H = JF
. (14-16)
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Problems 349
This i s Ampere's circuital law in terms of H. In in tegral form,
i cH • dl = I (14-18)
w h e r e I f i s the current of f ree charges l inking the curve C.
I f a magn e t i c ma te r i a l i s l i near and i so t rop ic , t he n
15 = /'o(l + Zm)H = Mort-H = pH . (14-21)
14-1E
14-2E
14-3E
14-4
w h e r e ym is the magnetic susceptibility, p r is the relative permeability, a n d p
is the permeability.
F e r r o m a g n e t i c m a t e r i a l s a r e h i g h l y n o n - l i n e a r , a n d o f t e n a n i s o t r o p i c .
Th e r e l a t ion be twe en B an d H de pe nd s on the p rev io us h i s to ry o f the f ie ld ,
and i s r ep resen ted by the hysteresis loop. The a r ea o f a hys te r es i s loop g ives
the energy d i ss ipa ted in the mate r i a l pe r cub ic mete r and per cyc le .
A t t h e b o u n d a r y b e t w e e n t w o m a t e r i a l s , t h e t a n g e n t i a l c o m p o n e n t
o f H a n d t h e n o r m a l c o m p o n e n t o f B a r e c o n t i n u o u s .
PROBLEMS
MAGNETIC FIELD OF THE EARTH
A sphere of rad ius R i s uniformly magnet ized in the di rect ion paral le l to a
diameter . The external f ie ld of such a sphere is c losely s imilar to that of the earth. See
Prob . 8 -11 .
Show that the equivalent surface current dens i ty is the same as i f the sphere
carr ied a uniform surface charge dens i ty a and ro ta t ed a t an angula r ve loc i ty co such
EQUIVALENT CURRENTS
An i ron to ru s whose ma jor rad ius i s mu ch l a rge r than i ts minor rad ius is ma g
net ized in the azimuthal di rect ion with M uni form.
What can you s ay about at,'?
EQUIVALENT CURRENTS
A long tube is uniformly magnet ized in the di rect ion paral le l to i t s axis . What
is the value of B ins ide the tube?
DIELECTRICS AND MAGNETIC MATERIALS COMPARED
a ) Imagin e a pa ra l l e l -p l a t e capa c i to r whose p l a t e s a re cha rge d and insu la t ed .
Ho w are E and D affected by the int ro du ct io n of a die lect r ic betwe en the pl a tes?
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350 Magnetic Fields: VII
b) Show a polar molecule of the die lect r ic or iented in the f ie ld.
c ) I s i ts ene rgy maxim um o r min im um ?
d) Now imagine a long so lenoid ca r ry ing a f ixed cur ren t .Ho w are B an d H affected by the int ro du ct io n of a cyl inde r of fe rrom agne t ic
ma te r i a l i ns ide the so leno id? /
e ) Show a sma l l cur ren t l oop represen t ing the magne t i c d ipo le moment o f an
a tom, and show the d i rec t ion of the cur ren t on the loop .
f ) I s i ts ene rgy ma xim um or m in im um ?
14-5 MAGNETIC TORQUE
Show tha t t he to rq ue exe r ted on a pe r ma nen t magne t o f d ipo le mom ent m
situ ated in a ma gn etic field B is m x B.
See Prob. 13-22.
I4-6E MEASUREMENT OF M
The fo l lowing me thod has been used to measure the magne t i za t ion M of a small
sph ere of mat eria l , induce d by an appl ied uniform ma gne t ic f ie ld B 0 as in Fig. 14-9.
In the figure. HG is a Hall gen er ato r (Sec. 10.8.1) with its cu rre nt flow ing par allel
to B 0 . S ince Hal l generators are sens i t ive only to magnet ic f ie lds perpendicular to thei r
current f low, this one measures only the dipole f ie ld originat ing in the smal l sphere .
Figure 14-9 Se t -up for measur ing the
m a g n e t i z a t i o n M induced in a smal lspherical sample of materia l by a
uniform field B 0 . Th e Ha l l gene ra to r
HG i s or iented so as to be sens i t ive
to the v-component of the f ie ld
or ig ina t ing in the sphe re , and no t t o
B„. See Prob. 14-6.
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Problems 351
If the magnetization is Af, the magnetic moment m of the sphere is ( 4 / 3 ) 7 r P3
M ,
and it turns out that the x-component of the dipole field of the sample at HG is
3p0m sin 0 cos 0
An r3
Thus, at a given ang le 0, thi s field decreases as 1/r3
.
At what angle 0 will the measured field be largest?
14 -7 MICROMETEORITE DETECTOR
Figur e 14-10 shows an instrum ent t hat has been devised to detect ferromagnetic
micrometeorites falling to the ground. The instrument detects particles as small as
10 micr omet ers in diam eter and me asure s their magnetic mome nt, which is a measu re
of their size and composition.
A particle of radius b is magnetized in the field of the solenoid and acquires a
magnetic moment in. It is then equivalent to a small coil having a magnetic moment m.
The mutual inductance between the particle and the coil C of mean radius a
shown in the figure is then the M of Prob. 12-3. with nb2
NhI
h= m.
I
Figure 14-10 Section through a micro-
meteorite detector. Air is sucked
through a funnel F. and a particle p
is magnetized in the field B of a
solenoid S. In passing through the
tube, the particle induces a voltage
in coil C. See Prob. 14-7.
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352 Magnetic Fields: VII
a) Show that the vol tage induced in the coi l i s
3Li 0N aa2
where r i s the veloci ty of the part ic le .
b) Draw a curve of z(a 2 + r 2 ) - 5 2 as a function of z for a = 10 mil l ime ters . Th e
c o o r d i n a t e z should va ry from —50 to + 5 0 mi l l ime te rs .
Since the veloci ty is cons tant , z i s p ropor t iona l t o the t ime t.
14-8E MECHANICAL DISPLACEMENT TRANSDUCER
One of t en wishes to ob ta in a vo l t age tha t i s p ropor t iona l t o the d i sp lacement
of an object from a kn ow n reference p os i t ion .
If there are no magnet ic materia ls nearby, one can f ix to the object a smal l
permanent magnet and measure i t s f ie ld wi th a Hal l generator , as in Fig. 14-11. T h e
Hal l e lement is sens i t ive only to the v- com po nen t of the ma gn et ic f ie ld and. from
P r o b . 14-6.
3 / / 0 ' " xz
* = ~4V(x2 + _ -
2
)5 2
'
a ) Draw a curve of xz / (x 2 + r 2 ) ' 2 as a function of z for v = 100 mil l im eters
and for z = —50 to + 5 0 mi l l ime te rs .
b ) Over wha t range does B x dev iate by less than 5% from the tang ent to th e
curve a t z = 0 ?
The l inear region is longer for larger values of x. but B x decreases rapidly wi th x.
x
Figure 14-11 H a l l g e n e r a t o r HG used
as a mechanica l d i sp lacement t rans
duce r . The moving pa r t P s l ides a long
the z-axis . The Hal l generator is
or i en ted so a s to moni tor the x-
component o f t he magne t i c f i e ld o f
t h e s m a l l p e r m a n e n t m a g n e t M fixedto P . See Prob. 14-8.
HG
14-9 MAGNETIZED DISK
A thin disk of i ron of radius a and th i cknes s t i s ma gne t ized in the di rect io n
parallel to its axis.
C ' a lcula te 11 and B on the axis , both ins ide and o uts ide the i ron.
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Problems 353
I4-I0E TOROIDAL COIL WITH MAGNETIC CORE
C om pa re the fields inside tw o s imilar toro idal coi ls , bot h carry ing a curre nt / ,
one wi th a non-magnet ic co re and the o ther wi th a magnet ic co re .How are the equ iva len t cu rren t s on the magnet ic co re o r ien ted wi th respec t to
those in the coi l?
EQUIVALENT CURRENTS
A long wire of radius a carries a current / and is s i tuated on the axis of a long
hollow iron cyl inder of inner radius b and ou ter rad ius c .
a) Compute the flux of B inside a sect ion of the cyl inder / me ters long.
b) Find the equivalent current densi ty on the inner and outer i ron surfaces ,
and find the direct ion of these equivalent currents relat ive to the current in the wire.
c) Find B a t d i s tances r > c from the wire.
H ow w ould th is value be affected if the i ron cyl inder were re mo ve d?
THE DIVERGENCE OF H
We have se en in Sec. 8.2 that V • B is zero . This eq uat ion is val id even in n on
l in e a r, n o n -h o m o g e n e o u s , a n d n o n - i s o t ro p i c m e d i a .
Set B = u.runM. Un de r what cond i t ions is V • H # 0?
wTHE MAGNETIZATION CURVE
Use Fig . 14-5 to find the relat ive permeabil i ty of annealed pure i ron at a magnetic
induction of one tes la.
14-14 ROWLAND RING
A ring of duct i le cast-i ron (Fig . 14-4) has a major radius of 200 mil l imeters and
a min or ra dius of 10 mil l ime ters . A 500-tu rn toro ida l coi l is wo und o ver i t.a) What is the value of B inside the ring when the current through the coi l is
2 .4 amperes? Use Fig . 14-5.
b) A 10-turn coi l is wo un d over th e first one. Calcu late the voltage ind uced in i t
i f the current in the large coi l suddenly increases by a small amount at the rate of
10 ampe res per second . A ssume tha t u r r e m a i n s c o n s t a n t .
14-15E THE WEBER AMPERE-TURN
Show tha t one weber am pere- tu rn i s one jou le .
14-11
14-12E
14-13E
14-16 ROWLAND RING
Figu re 14-12 show s how one can m eas ure B as a function of H to ob ta in the
magnet iza t ion and hys te res i s cu rves o f a r ing o f magnet ic mater ia l .
W i n d i n g a is fed by an adjustable power supply. As in Sec. 14.8.
H = NJJlnr.
Fhe vo l tage across wind ing b is in tegrated with the circui t of Prob. 5-32.
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35 4
The cross-sect ion of the r ing is S and winding b h a s N b t u r n s .
Show tha t
B = RCV/N bS
i f . a t the beginning of the experiment . B = V = 0 . Th e in t egra t ing c i rcu it d raws
es sen t i a l ly ze ro cur ren t .
Solution: The vo l t age ac ros s w inding h is
</4> dB
~dt=
~dt'(1)
s ince there is zero current in b, and hence no vo l t age drop caused by the re s i s t ance
and in duc tanc e of the w inding .
The n , f rom Pro b . 5 -32 , d i s rega rd ing the nega t ive s ign .
1 p dBv =
RC J°N
"s
Tt= N
"S B/ RC | 2 )
a n d
B = RCV/NbS.
14-17 TRANSFORMER HUM
W h y d o e s a t r a n s f o r m e r h u m ?
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35 5
2
-2\ I I I I- 4 0 - 2 0 0 2 0 4 0
H ( a m p e r e - t u r n s , m e t e r )
Figure 14-13 Hys te res i s l oop for the a l loy De l t amax. See P rob . 14-18 .
14-18E POWER LOSS DUE TO HYSTERESIS
Figure 14-13 shows the hys teres is loop for a nickel- i ron a l loy cal led Del tamax.
W ha t i s t he appro xim a te va lue of the powe r d i s s ipa t ion pe r cub ic me te r and
per cyc le when the ma te r i a l i s d r iven to s a tura t ion bo th way s?
14-19 THE FLUXGATE MAGNETOMETER AND THE PEAKING STRIP
A wire is wo uld a ro un d a s t r ip of Del t am ax (Fig. 14-13). form ing a solenoid. A
few turns o f w i re a re then wou nd ove r the so lenoid , and con nec ted to an osc i l loscope .
An a l t e rna t ing cu r ren t i s app l i ed to the so lenoid , d r iv ing the De l t a max to
sa tura t ion in bo th d i rec t ions .
a ) W ha t i s t he shape of the waveform obse rved on the osc i l loscope?
b) W ha t hap pe ns to the pa t ter n on the osci l loscop e if a s teady m agn et ic field isnow appl i ed pa ra l l e l t o the s t r ip?
It can be shown that , upon appl icat ion of the s teady f ie ld, the output vol tage
conta ins a com po nen t a t dou ble the f requency of the appl i ed a l t e rna t ing cur ren t and
pr op ort ion al to the m ag ni t ud e of the s teady f ie ld. Th is is the principle of op erat ion of
th e fluxgate magnetometer. Fluxg a te magn e tom ete rs a re o f ten used on boa rd s a t el l it e s
for measuring magnet ic f ie lds in outer space. They are useful from about 10 _ 1 0 to 1CT
tesla.
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356 Magnetic Fields: VII
T h e peaking strip a l so u t i l i zes a so lenoid wound on a ma te r i a l w i th a squa re
hys teres is loop l ike Del ta ma x (Fig. 14-13), an d a sec on dar y w inding . I t i s used differently,
however. The peaking s t r ip is placed ins ide a solenoid whose current i s adjus ted so asto cance l t he ambien t f i e ld , making the pa t t e rn on the osc i l loscope symmet r i ca l . The
peaking s t r ip i s used to measure magne t i c induc t ions in the range of about 1 ( T 6 to
1 0 "2
tesla.
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CHAPTER 15
MAGNETIC FIELDS: VIII
Magnetic Circuits
15.1 MAGNETIC CIRCUITS
15.2 MAGNETIC CIRCUIT WITH AN AIR GAP15.2.1 Example: Electromagnet
15.3 MAGNETIC FORCES EXERTED BY
ELECTROMAGNETS
15.4 SUMMARY
PROBLEMS
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In gen eral , it i s not po ss ible to calculate m ag ne t ic f ie lds accura tely , w hen
magnet i c mate r i a l s a r e invo lved . There a r e s evera l r easons fo r th i s , ( a ) The
re la t ion be tw een H a nd B for f e r romagn et i c m ate r i a l s i s non - l inear . I t even
depends on the previous his tory of the mater ial , as we saw in Sec. 14.9. Also,
f e r roma gnet i c m ate r i a l s a r e o f ten no n- i so t r op ic . (b ) Th e i ron cores tha t a r e
used to conf ine and guide the magnet ic f lux are of ten qui te ineff icient . As we
shal l see, a large par t of the f lux can be s i tua ted o uts ide the core, (c) Per m an en t
magne t s a r e no t as s imple as one would l ike them to be : the i r magne t i za t ion
M i s non -un i fo rm a nd depen ds on the p resence o f ne ig hbo r ing ma gne t i c
m a t e r i a l s .
I t i s none the les s neces sary to be ab le to m ak e app rox im ate ca lcu la t ions .
Th e ca lcu la t ion s e rves to des ign a mo del on w hich mag ne t i c f ie lds o r m agn e t i c
forces can be measured, and which can then be modif ied to give the f inal
des ign .
15.1 MAG NETIC CIRCU ITS
Figu re 15-1 show s a f e r romagn et i c core a ro un d wh ich i s wo un d a shor t co i l
of N t u rns ca r ry ing a cur r en t / . We wish to ca lcu la t e the magne t i c f lux O
t h r o u g h t h e c o r e .
In the absen ce of fer ro ma gne t ic ma ter ial , the l ines of B are a s show n
in th e figure. At f irst s ight, on e expe cts the B insid e the co re to be m uc h
large r close to the win din g tha n on the op po s i te s ide. Th is is no t the case,
how ever , and B i s o f the s am e orde r o f ma gn i tud e a t a ll po in t s wi th in the
f e r r o m a g n e t i c m a t e r i a l .
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359
Figure 15-1 Ferromagneti c toroid with concentrated winding . The lines offeree
s h o w n are in the plane of the toroid and are similar to those of Fig. 8-6. They
ap p l y only when there is no iron present.
This can be understood as follows. The magnetic induction of the
current / magnetizes the core in the region near the coil, and this magnetiza
tion gives equi valen t cur ren ts that bo th incr ease B and extend it along t he
core. This further increases and extends the magnetization, and hence B,
until the lines of B exte nd all aro un d t he core .
Of course some of the lines of B escape into the air and then return to
the core to pass again through the coil. This constitutes the leakage flux
that may, or may not, be negligible. For example, if the toroid is made up
of a long thin wire, most of the flux leaks across from one side of the ring
to the other and the flux at P is negligible compared to that near the coil.
Let us ass ume tha t the cross-sectio n of the toroi d is large enough to
render the leakage flux negligible. Then, applying Ampere's circuital law,
Eq. 14-18, to a circu lar pa th of rad ius r goin g all arou nd inside the to roi d,
(15-1)
InrB firft
oNI, (15-2)
B Hrfi
0NI/2nr. (15-3)
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3 6 0 M a g n e t i c F i e l d s : VIII
T a k i n g R { to be the rad ius co r re spond ing to the ave rage va lue o f B, a n d
R2
to be the minor radius of the toroid .
<D = n rn 0nRlNII2nR v (15-4)
Th e flux th rou gh the core is therefore t he sam e as if we had a tor o id al coi l
(Sec. 9 .1 .2 ) of the same s ize and i f the number of ampere- turns were increased
by a factor of a r. In o ther words , for each ampere- turn in the coi l there are
u r — 1 am pe re- tur ns in the core . Th e amplif ica t io n c an be as h igh as 1 0 5 .
Th is eq uat ion sh ows that the ma gne t ic flux is g iven by the m ag ne to -
m o t a n c e NI mult ip l ied by the fac tor
UrUonRl/lnRu
wh ich is called th e permeance of the magnet ic c ircui t . The inverse of the
per me anc e is ca l led the reluctance. T h u s
^ /V/ NI
(1) = — = (15 -5)
St InRJp^nRV ' '
and the re luctance is
9t=InRJurHouRl.
(15-6)
Re luc tanc e i s exp ressed in hen rys "
Th e ana logy wi th O hm 's l aw is obv ious : if an e lec t ro mo tance V w e r e
induced in the core , the current would be
r(15-7)InRxlanRj
w h e r e 2nR\lanR\ is the res is tance of the core .
Thus the co r re spond ing quan t i t i e s in e lec t r ic and magne t ic c i rcu i t sare as fo l lows:
C u r r e n t /
Cur ren t dens i ty J
C o n d u c t i v i t y a
Magnetic flux (P
M a g n e t i c i n d u c t i o n B
Permeab i l i ty p = prp0
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15.2 Magnetic Circuit with an Air Gap 361
E l e c t r o m o t a n c e ir
Electr ic f ie ld intens i ty E
C o n d u c t a n c eG = l/R
R e s i s t a n c e R
M a g n e t o m o t a n c e NI
Magnet ic f ie ld intens i ty H
P e r m e a n c e \f.
J
AR e l u c t a n c e 88 .
T h e r e is one impor tan t d i f f e r ence be tween e lec t r i c an d m a g n e t i c
c i r c u i t s : the magnet i c f lux cannot be m a d e to follow a magnet i c c i r cu i t in
t h e m a n n e r t h a t an elect r ic current fol lows a c o n d u c t i n g p a t h . I n d e e d , a
m a g n e t i c c i r c u i t b e h a v e s m u c h as an elect r ic ci rcui t would if it w e r e s u b
m e r g e d in ta p w a t e r : p a r t of the cur ren t would f low th rough the c o m p o n e n t s ,
a n d the r es t would f low th rough th e w a t e r .
If a m a g n e t i c c i r c u i t is not p r o p e r l y d e s i g n e d , the leakage f lux ca n
easily be an o r d e r of m a g n i t u d e larger t ha n tha t flowing a ro un d the circui t .
15.2 MAGNETIC CIRCUIT WITH AN AIR GAP
F i g u r e 15-2 s h o w s a circui t wi th an air gap whose c ros s - sec t ion is different
f rom tha t of the sof t - i ron yoke . Each wind ing p rov ides NI/2 a m p e r e - t u r n s .
We wish to c a l c u l a t e th e m a g n e t i c i n d u c t i o n in the air gap.
W e a s s u m e t h a t the leakage f lux is negl igible . As we shal l see, t h i sas sumpt ion wi l l r esu l t in q u i t e a l a rge e r ro r .
A p p l y i n g A m p e r e ' s c i r c u i t a l law to the circui t , we see t h a t
NI = Hik + H glg, (15-8)
w h e r e the s u b s c r i p t i refers to the i r o n y o k e and g to the air gap; /, and lg
a r e th e pa th l eng ths . The pa th l eng th /, in the i ron can be t a k e n to be the
l e n g t h m e a s u r e d a l o n g the c e n t e r of the cros s - sec t ion of the y o k e .
If we neglect leakage f lux, the flux of B m u s t be the same over any
cros s - sec t ion of the magnet i c c i r cu i t , so
BiAi = BgAg, (15-9)
w h e r e A t and A g are, r espec t ive ly , the cros s - sec t ions of the i r on yoke and of
t h e air gap.
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36 2
Soft iron yoke
• Ii
•11
Figure 15-2 Electromagnet . The coi ls have been cut out to expose the i ron yoke.
C o m b i n i n g t h e s e t w o e q u a t i o n s .
NI, ( 1 5 - 1 0 )
and the magnet ic f lux is
<I>NI
I;(15-11)
Th e ma gne t i c flux i s there fore equa l to the mag ne tom ot an ce d iv ided by
the sum of the reluctances of the i ron and of the ai r gap.
This is a general law: reluctances in ser ies in a magnet ic ci rcui t add in
the same way as res is tances in ser ies in an elect r ic ci rcui t ; permeances in
para l l e l add l ike co ndu c ta nc es in para l le l .
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15.2 Magnetic Circuit with an Air Gap 363
Since we have neglected leakage flux, the above equation can only
pr ov id e an up pe r lim it for <I>.
N ow let = IJi.irP.QAj be the reluctance of the iron, and &g = lglp 0A g
be the relucta nce of the air gap. N ot e that
/ / , / , = . ^ , < D. ( 1 5 - 1 2 )
H glg = ®g4>. ( 1 5 - 1 3 )
If 09 t « dig, as is usually the case, since pr » 1, Eqs. 15-8 and 15-11
become
NI % Hgl
g, ( 1 5 - 1 4 )
* * NI p0A
g/l
g. ( 1 5 - 1 5 )
EXAMPLE: ELECTROMAGNET
L et us ca lcula te B and the s tored energy in the e l e c t r o m a g n e t of Fig . 15-2. If t h e r e
is a t o t a l of 10,000 turns in the two w i n d i n g s , and se t t ing / = 1.00 a m p e r e . Ax =
10* square mi l l ime te rs , Ag = 5 x 103 square mi l l ime te rs , pr = 1,000,/, = 900 mil l i
mete rs , and /,, = 10 mil l imete rs , then
<1>1 0
4
0. 9 1015-16)
1 0 3
x 4;: x 1 0 "7
x 10~2
4TT X 1 0 "7
x 5 x 10"3
= 6.0 x 1 0 "3 weber, (15-17)
B g = 6.0 x 1 0 " 3 / 5 x 1 0 "3
= 1.2 teslas. (15-18)
The se l f - induc tance is
L = NQ/I = 104
x 6.0 x 1 0 ~ 3 / 1 . 0 0 = 60 h e n r y s . 15-191
T h e l e n g t h /f is m e a s u r e d a l o n g the m i d d l e of the cross-sect ion of the y o k e .
The s tored energy is
( 1 / 2 ) L / 2
= (1/2) x 60 x 1.00 = 30 joules . (15-20)
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364 Magnetic Fields: VIII
In this part icular case the leakage f lux is 70% of the f lux in the gap. In other
wo rd s, the mag ne tic ind uct ion in the gap is no t 1.2 teslas, bu t only 1.2 1.7 = 0.71 tesla.
There exis t empirical formulae for es t imat ing leakage f lux for s imple geometries .
15.3 MA GNETIC FORCES EXER TED
BY ELECTROMAGNETS
E l e c t r o m a g n e t s a r e c o m m o n l y u s e d t o a c t u a t e v a r i o u s m e c h a n i s m s s u c h a s
switch es , valves , an d so for th . A switch act iv ated by an ele ct ro ma gn et i s
cal led a relay. These e l ec t romagnet s compr i s e a co i l and an i ron core tha t
i s m ad e in two par t s , one f ixed and one m ova ble , ca l l ed the armature, wi th
an a i r gap be tween the two . The co i l and i t s core toge ther a r e usua l ly t e rmed
a solenoid. The magne t i c fo rce i s attractive.
In the s imples t cases , one has a f la t a i r gap, perpendicular to B. T h e n
the magne t i c fo rce o f a t t r ac t ion i s B2
/2p:0 t imes the cross-sect ion of the
gap (Sec. 13.2) where B i s calc ulate d as in Sec. 15.2. I f the so leno id is energized
wi th d i r ec t cur r en t , t hen B i s app rox ima te ly p rop or t i on a l to the inver se o f
the gap l eng th , and s ince the fo rce i s p ropor t iona l to B2
, the force increases
rap id ly wi th decreas ing gap l eng th .
The ai r gap is of ten des igned so that the at t ract ive force wi l l vary wi th
the gap l eng th in som e prescr ibed w ay . Th e des ign o f the mag ne t i c c i r cu i t
is largely empir ical .
15.4 SUMMARY
The concep t o f magnetic circuit i s widely used whe nev er the m ag ne t ic f lux
i s gu ided mos t ly th r ou gh m agn e t i c mate r i a l . We then have a m agn e t i c
e q u i v a l e n t of O h m ' s l a w :
O = NI/M, (15-5)
w h e r e NI is the magnetomotance of the energizing coi l and J ? is the reluctance
of the magne t i c c i r cu i t .
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Problems 365
T h e c o r r e s p o n d e n c e b e t w e en
fol lows:
e lec t r i c and magne t i c c i r cu i t s i s as
C u r r e n t /
C u r r e n t d e n s i t y J
Conduct iv i ty r r
E l e c t r o m o t a n c e f~
Electric f ield intensity E
C o n d u c t a n c e
Res i s t ance R
M ag ne t ic F lux <P
M a g n e t i c i n d u c t i o n B
Permeabi l i ty /<r/<„
M a g n e t o m o t a n c e NI
Magnet ic f ie ld intens i ty H
P e r m e a n c e
R e l u c t a n c e 'J)
The r e luc tance o f a bar o f magne t i c mate r i a l o f permeab i l i ty /u r /u 0 ,
c ros s - s ec t ion A, an d len gth / i s l/nrfi0A. Reluctances in ser ies and in paral lelare t reated l ike res is tances in ser ies and in paral lel .
Th e magn e t i c force o f a t t r ac t ion ex er t ed by an e l ec t rom agn et i s B2
/2p.0
t imes the cross-sect ion of the gap.
PROBLEMS
15-1E RELUCTANCE
Show that the energy s tored in a magnet ic c i rcui t i s
W = (1/2)<D 2 ^,
w h e r e <> i s the magnet ic f lux and J# i s the reluctan ce.
15-2E RELUCTANCE
Show tha t the induc tance Lof a co i l of N turns is given by
L = JV2
w h e r e J
A i s the reluctance of the magnet ic c i rcui t .
I5-3E CLIP-ON AMMETER
It is often useful to be ab le to m ea su re the cu rre nt flowing in a wire w ith ou t
dis tu rbin g the ci rcui t . Th is can be do ne with a c l ip-on a mm eter . As a rule , c l ip-on
am m ete rs are t ransform ers , as in Pr ob . 18-16, and can therefore be used only with
a l t e r n a t i n g c u r r e n t s .
I t i s a lso poss ible to m ak e a cl ip-on a mm eter th at will me asu re direct cu rre nts
as in Fig. 15-3. In this ins t rum ent the mag net ic flux thro ug h the yoke is me asur ed with
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366
Kigure 15-3 Clip-on ammeter for direct currents. A
hinge A permits the iron yoke to be opened at C so
that i t can be cl ipped around the current-carrying
wire. A gap G in the yoke contains ei ther a Hall
genera to r o r a magnetores i s to r . The magnet ic
induction in the gap is a measure of the current / .
See Prob. 15-3.
a Hall gene rato r (Sec. 10.8.1), or with a mag neto resi s tor ( Pro b. 10-15) s i tuated in the
gap of length lg.
a) Show that , i f lg » IJfi,, the magnet ic induct ion in the gap i s fi0I/lg-
What is the advantage of using an iron core?
b) Is this magnetic induction affected by the posi t ion of the wire inside the r ing?
15 -4 MAGNETIC CIRCUIT
The ducti le cast- iron r ing of Prob. 14-14 is cut , leaving an air gap 1 mil l imeter
long. The current in the toroidal coil is again 2.4 amperes.
C a l cu l a t e B in the air gap. using the relat ive permeabil i ty curve of Fig. 15-4.
You wil l have to solve this problem by successive approximations. Star t by
assuming a reasonable value for / i r , say 500. Find the corresponding B . Fhis wil l prob-
6 0 0
4 0 0
2 0 0
0 0.5 1.0
B (tesla)
Figure 15-4 Relat ive permeabil i ty as a function of magnetic induction for ducti le
cast i ron. See Prob. 15-4.
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Problems 367
ably not give the correct B on the curve . Then t ry ano ther va lue of p r, e tc . Draw a
table of p r and of the calculated B as a he lp in s e lec t ing your next approximat ion .
The ca lcula ted va lue of B need not agree to bet ter than 10% with the B on thecurve .
MAGNETO RESISTANCE MULTIPLIER
The m agne t i c c i rcu i t shown in F ig . 15-5 g ives an outp ut vo l t age K tha t i s pro por
t ional to the product IJ b. The vol tmete r draws a negl ig ib le cur rent .
The a i r gaps conta in magne tores i s t ances (P rob . 10-15) R x a n d R 2, c o n n e c t e d
as in Fig. 15-5b. Th is c i rcui t i s ident ica l to tha t of Pr ob . 5-7.
F r o m P r o b . 1 0 - 1 5 .
R 5 = R 0 [ l + ,J I2(B a + B„)2l
R2 = R 0 [ l + JI2
(B a - B b)2
l
where , It i s the mobil i ty of the charge carriers .
i/x
Figure 15-5 (a) Ma gne t i c c i rcu it wi th a pa i r of mag ne tores i s t an ces and R 2 for
obta in ing a vol t age V p r o p o r t i o n a l t o t h e p r o d u c t o f Ia a n d Ih. The s t ra ight
a r row s show the flux thro ugh the i ron core due to / „ . Th e wavy a r row s show the
flux due to Ib. In the left -hand gap, B = B a + B b, while on the r ight B = B„ - B b.
(b ) The electr ic c i rcui t . See Prob. 15-5.
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368 Magnetic Fields: VIII
Show tha t
v = -ue2
Bfi bv 0,
as long as ,M 2B\ « I , M 2B\ « 1.
MAGNETIC FORCE ON THE ARMATURE OF AN ELECTROMAGNET
Figu re 15-6 show s a smal l e lectrom agn et . All dim ens ion s are in mil l im eters . Eac h
coi l has 500 turns and carries a current of 2 amperes .
a ) Ca lcula te B in the ai r gaps as a funct ion of the gap length x. Set ii r = 1000.
and neglect the leakage f lux.
Solution: T h e m a g n e t o m o t a n c e i s
NI = 2 x 2 x 500 = 2000 am per e turns . (1)
The i ron yoke has a cross-sect ion of 10 x 10 squa re mi l l ime te rs , and a m ean
length of 2 x 25 mil l imeters for the vert ical parts , plus 23 mil l imeters for the
hor izo nta l p a r t . Hence i t s re luc tance i s
St y = 73 x 1 0 ~ 3 , (1000 x An x 1 0 ~ 7 x 1 0 - 4 ) = 5.8 x 1 0 s h e n r y - 1 . (2)
Figure 15-6 Elec t romagne t . See P rob . 15-6 .
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Problems 369
Th e arm at ure has a cross-sect ion of 50 squ are mil l imeters . Th e mean path length
of the f lux is about 23 mil l imeters . Hence
m a = 23 x l(r3
/ (1000 x 4n x 1 0 " 7
x 50 x 1 0 ~ 6 ) = 3.7 x 1 0 5 h e n r y - 1 . (3)
Finally , the two air gaps have a reluctance
m g = 2x/(47t x 1 0 ~7 x 1 0 ~
4
) = 1.6 x I 01 0
x h e n r y- 1
. (4)
T h u s , in the gaps.
2000B = XI if A = « = ^ 2 , (5)
(5.8 x 1 0s
+ 3.7 x 1 05
+ 1.6 x 1 01 0
x ) 1 0 ~4
200
• teslas. (6)'.5 + 1.6 x 105
.Y
b) Calcu la te the fo rce o f a t t r ac t ion exer ted on the arm atu re when x =
5 mil l imeters .
Solution: The force is equ al to the p rodu ct of the energy densi ty in the gap, by
its cross-sect ion A:
F = 2B2
A/2tt„, (7)
4 x 1 0 4 x 1 0 ~ 4
(9.3^*- 1.6 x 1 05 x 5 x 1 0
_ 3
)2
4 7 r x 10 ^ % 5 new tons . (8)
c) W ou ld it m ak e sense to use Eq. 6 to cal cul ate the force (i) at x = 0.1 mil lime ter,
(ii) at x = 20 millim ete rs?
Solution: At x = 0.1 mil l imeter , Eq. 6 gives a magnetic induction of about
12 teslas . This is mu ch mo re tha n t he sat ura t ion f ield of one o r two teslas for i ron
(Sec. 14.8). Eq ua tio n 6 is not val id at x = 1 0- 4
.
At x = 20 mil l imeters , the reluctance of the air gap is large and the leakage
f lux is mu ch larger tha n the f lux in the gaps. The a ctua l force is mu ch sm aller t ha n
that ca lculate d f rom Eq. 7 . Eq ua tio n 6 is again no t val id .
So the magnetic induction found in Eq. 6 is val id only over a l imited range of x .
15-7 RELAY
Calculate the force of at tract ion on the armature of a relay such as the one
shown in Fig. 15-7 whose magnetic circuit has the fol lowing character is t ics:
Coi l , 10,000 turns; resis tance, 1000 ohms; rated voltage, 10 volts .
G ap leng th , 2 mi l l imeter s ; gap cross - sec t ion , 1 square cen t imeter .
For s implici ty , we assume that the gap has a uniform length, that the reluctance
of the iron is negligible when the gap is open, and that there is zero leakage f lux.
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37 0
Figure 15-7. Relay. W he n the coi l is energized by clos ing the switch, the arm atu re
falls, opens the upper contac t , and c loses the lower one . Tens of contac t s can be
ac tua ted s imul taneous ly in th i s way. See P rob . 15 -7 .
15-8 MAGNETIC FLUIDS
There exis t magnet ic f luids consis t ing of f ine part ic les (*10 nanometers in
diam eter) of i ro n oxides sus pen ded in variou s f luids .*
a ) I f the mag ne t i c f lu id i s p laced in a non -ma gne t i c conta in e r on a n on-m agn e t i c
suppor t , a pe rmanent magne t p laced in i t wi l l remain suspended near the bot tom,
but wi thout touching the conta ine r , e i the r a t the bot tom or a t the s ides . Why i s th i s?b) C an you find appl ica t ion s for ma gn et ic f luids?
f Ma nufac tured by the Fer rof lu id ics Cor por a t io n , 144 Middlesex Turnp ike , Bur l ing ton ,
M a s s a c h u s e t t s 0 2 1 0 3 .
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CHAPTER 16
ALTERNATING CURRENTS: I
Complex Numbers and Phasors
16.1 ALTERNATING VOLTAGES AND CURRENTS
16.2 R M S , OR EFFECTIVE VALUES
16.2.1 Example: The Output Voltage of an Oscillator
16.3 THREE- WIRE SINGLE-PHA SE AL TERN A TING
CURRENT
16.4 THREE-PHASE ALTERNATING CURRENT
16.5 ALTERNATING CURRENTS IN CAPACITORS AND
IN INDUCTORS
16.6 COMPLEX NUMBERS
16.6.1 Examples: Using Com plex Numb ers
16.7 PHASORS16.7.1 Example: The Phasors in a Three-Phase Supply
16.7.2 Addition and Subtraction of Phasors
16.7.3 Example: The Phasors in a Three-Phase Supply
16.7.4 When Not to Use Phasors
16.8 SUMMARY
PROBLEMS
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M os t elect r ic an d mag ne t ic devices use al t ern at i ng or , a t leas t , f luctuating
cur ren t s . Th ere a r e man y r easo ns fo r th i s , bu t the two majo r one s have to
do wi th power t echno logy and wi th the t r ansmis s ion and p roces s ing o f
i n f o r m a t i o n .
Firs t , wi th alternating cur ren t s , t he e l ec t r i c power supp l i ed by a
source a t a g iven vo l t age can be made ava i l ab le a t a lmos t any o ther con
ven ien t vo l t age by m ean s of t r ans form ers . Th i s ma ke s e lec t r i c pow er ad ap t
able to a broad var iety of uses .
The power suppl ied by a <77rerf -current source can also be changed
from o ne vol tag e to an oth er . But this i s do ne by f irst swi tch ing the c urre nt
per iod ica l ly , t o ob ta in an a l t e rna t ing cur r en t , t hen f eed ing th i s to a t r ans
former , and then rect i fying and f i l ter ing the output . The operat ion is rela
tively costly and inefficient.
The s econd r eason fo r us ing a l t e rna t ing o r f luc tua t ing cur r en t s i s
tha t they can s e rve to t r ansmi t in fo rmat ion . For example , a microphone
t r ans forms the in format ion con ta ined in a spoken word in to a complex
f luctuating cu rren t . Th is cu rren t then serves to m od ul at e a ra dio-f re que ncy
cur ren t tha t is fed to an an ten na . The an te nn a l aunch es an e l ec t ro ma gne t i c
wave, and so for th .
A go od pa r t o f th i s cha p te r wi ll be dev o ted to the meth od u sed for
so lv ing the d i ff e ren t ia l e qu a t io ns as soc ia t ed wi th c i r cu i t s ca r ry ing a l t e rn a t ing
c u r r e n t s . T h i s m e t h o d r e q u i r e s th e u se o f c o m p l e x n u m b e r s . N o p r e v i o u s
knowledge o f the sub jec t i s r equ i r ed .
We sha l l as sume tha t the vo l t ages , t he cur r en t s , and the charges a r e
a l l cos ine func t ions o f t ime , wi th appropr ia t e phases . Th i s i s no t a lways the
case . T h e c u r r e n t t h r o u g h a m i c r o p h o n e is n o t n o r m a l l y s i n u s o i d a l . O r o n e
might have only the pos i t ive par t of the cos ine funct ion, or a cos ine funct ion
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16.2 R M S , or Effective Values 373
whose per iod i s a func t ion o f the t ime . However , any periodic funct ion can
be analyzed into an inf ini te ser ies of s ine and cos ine terms, forming what i s
cal led a Fourier series.
16.1 ALT ERISA TING VO L TAGES AND CURRENTS
Voltages are of ten of the form
V = V 0 cos cot, (16-1)
w h e r e V 0 is the peak voltage, to = 2n f is the circular frequency, expres sedin r ad ians per s econd , and / i s the frequency, expressed in hertz. Such vo l t ages
are said to be alternating (Sec. 11.2.1) . We have arbi t rar i ly set the phase cot
of V equa l to ze ro a t t = 0.
If such a vo l t age i s app l i ed to a c i r cu i t, t hen the b ra nc h cu r ren t s a r e
of the form
I = I0
co s (cot + tp), (16-2)
where 7 0 i s the peak current, cot + tp i s the phase, and tp i s the phase of the
branch cur ren t / wi th r espec t to the source vo l t age V.I t i s unfor tuna te ly the cus tom to use the expres s ions AC current a n d
AC voltage, w h e r e AC s t a n d s f o r A l t e r n a t i n g C u r r e n t .
16.2 R M S , OR EFFECTIVE VALUES
I f an a l t e rna t in g vo l t age V 0 cos to t i s appl ied across a res is tor R as in Fig.
16- l a ,
F 0 c os cot . , „/ = = I
0cos ojt, (16-3)
R
as in Fig. 16- lb . In that case the current i s in phase wi th the vol tage, and
the peak cur r en t 7 0 is V 0/R .
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The ins t an taneous power d i s s ipa ted in the r es i s to r i s
V 2
P i n s t = VI =0 c o s 2 cof = IQR COS
2 rat, (16-4)
Over one cyc le , t he average power d i s s ipa t ion i s
Wl 1 ,
2 ^ = 2 / a x ( i6 -5)
as in Fig. 16-2.
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37 5
Tz /a> 2n/a> 3n/a> Anjio
Figure 16-2 Ins t a n ta n e ou s a nd a ve ra ge powe r d i ss ipa t ion in a re s is to r .
Un less specified othe rwi se, on e alw ays avoid s the factor of \ by us ing
th e rms, o r effective v o l t a g e a n d c u r r e n t ,
Vvm s = V 2
1
'2
, 7 r m s = Zo/21
'2
, (16-6)
ins t ead o f V 0 a n d I0. H e r e " r m s " s t a n d s f o r root mean square, or the square
roo t o f the mean va lue o f the square o f the func t ion . See Prob . 16 -2 . T h e n
P a v = V2
rmJR = I2
msR. (16-7)
T h e p e a k v a l u e s V 0 and / 0 a re 2 1 / 2 t imes larger than the rms values , as in
Fig. 16-3, as long as V a n d I are s inusoidal funct ions of the t ime.
0.707 V 0
0
V = V 0 cos cur
\ 1 / 1 \ 1 / 1 \
' 27t/co \ 3n/o) / 47t/co \
-Figure 16-3 R e la t ion be twe e n pe a k a nd rm s valu es for a cosin e functio n.
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376 Alternating Currents: I
I t i s useful to remember that
2
1 / 2
= 1.414 a nd 1/2
1
'
2
= 2
1 / 2
/ 2 = 0.707. (16-8)
Unless stated otherwise, num erical values for voltages, currents, and/
charges are always rms values.
16.2.1 EXAMPLE: THE OUTPUT VOLTAGE OF AN OSCILLATOR
An osc i l la tor supp l ies 10 vol ts a t a pair of termin als , one of which is grou nd ed .
Then the potent ia l a t the other terminal is 1.414 x 10 cos ait. The potent i a l a t tha t
t e rmina l va r i es be tween —14.14 and +1 4.1 4 vol t s wi th respec t to grou nd.
16.3 THREE-W IRE SINGLE-PHASE
ALTERNATING CURRENT
As a rule , low-power ci rcui ts , such as those used in elect ronics , ut i l ize al ter
na t in g vo l t ages and c ur r e n t s as in Sec . 16 .1 . Fo r h ighe r pow ers , such as those
used in a house , on e can proce ed dif ferent ly .
F igure 16-4a shows two a l t e rna t ing cur r en t sources connec ted to
two res is tors f t , and R 2 by mea ns o f four wi res . Th e s igns an d the a r ro wsshow the po la r i t i es and the cur r en t d i r ec t ions a t a par t i cu la r ins t an t .
The two center wires of Fig. 16-4a can of course be replaced by a
s ingle wire as in Fig. 16-4b, where the center wire car r ies only the difference
be tween the load cur r en t s . One then has three-wire single-phase current.
Thus th ree wi res do the work o f four and , moreover , t he I2R loss in the
cen te r wi re i s low, whenever the loads a r e r easonab ly wel l ba lanced .
In N or th A me r ica , e lec t r i c u t i l it i e s ma in ta in 120 vo l t s be twe en A a n d
B, 120 vo l t s be twe en C a n d B, t hus 240 vo l t s be tween A a n d C. T h e p o t e n t i a l
o f po in t D wi th r espec t to g r ou nd i s no t qu i t e ze ro bec ause o f the cur r en t
f lowing through the center wire .I t i s t he cus to m to o per a te low -pow er dev ices , such as ligh t bu lb s ,
a t 120 vo l t s , bu t h igh -po we r dev ices , such as ba seb oa rd he a te r s , a r e op era ted
a t 240 vo l t s . For a g iven power consumpt ion , a h igh-power dev ice thus d raws
only half the curr en t i t wo uld d ra w at 120 vol ts . Th is perm its the use of l ighter
wire, and reduces the cos t of the wir ing.
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377
(a) (h)
Figure 16-4 (a) Two similar sources with oppo site phases connected to two resistors
•Rj and R2. (£>) Three-wire single -phase supply. If the two resist ances are equa l, the
co n n ec t i o n BD is unnecessary.
16.4 THREE-PHASE ALTERNATING CURRENT
Figu re 16-5a shows three sources of alt ern ati ng curren t feeding r esistors
R 2, ^ 3 - Th e sources are orien ted 120° apa rt, on the figure, so as to indicat e
their relative phases. For example, if the phase at A is zero at t = 0, then
VA
= V0
co s cot, (16-9)
VB = F
0co s (cot + 2n/3), (16-10)
Vc = V0 cos (cot + 4 tt /3) . (16-11)
If the three resistances are equal, the total current flowing in the three
grounded wires is
(F0/R)[cos cot + cos (cot + 2TI/3) + cos (cot + 4TT /3)].
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37 8
A
B(b)
Figure 16-5 (a) Thr ee s imi la r sources of a l t e rna t in g cur re nt , wi th phase d i ffe rences
of 120 degrees , conn ecte d to three res is tors R x, R 2. R^. (b) Three-phase supply . I f
the res i s t ances a re equa l , one may omi t the connec t ion DE .
I t wi l l be shown in Prob. 16-6 that the bracket i s a lways equal to zero. (At
f = 0, i t is eq ua l to 1 - 0.5 - 0.5.) Th en th e so urc es can be co nn ec te d t othe res is tances as in Fig. 16-5b and the wire DE can be d i spensed wi th .
I f the r es i s t ances a r e unequa l , t he cur r en t s do no t comple te ly cance l
in DE. We now have four wires doing the work of s ix , wi th low I2R losses
in the wire DE.
A se t o f th ree sources , s t a r - connec ted and phased as in F ig . 16-5b ,
supp l i es three-phase alternating current. T h e m a i n a d v a n t a g e o f t h r e e - p h a s e
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16.5 Alternating Currents in Capacitors and Inductors 379
current is that it can be used to produce the revolving magnetic fields that
are required for large electric motors. See Prob. 16-7.
Electric power stations generate three-phase currents. This can be
seen from the fact that high-voltage transmission lines have either three or
six wires, plus one or two light wires.
Except for a few pro bl ems at th e end of this chapt er, an d except for
the example s in Sees. 16.7.1 and 16.7.3, we shall be concerned henc efor th
solely with two-wi re single- phase cu rre nts , as in Fig. 1 -4.
(o
16.5 ALTERNA TING CURRENTS IN
CAPACITORS AND INDUCTORS
Figure 16-6a shows a source V connected to a capacitor C. Then
Q/C = V = V0
cos cot, (16-12)
I = •%•= C% = -toCV0 sin mt, (16-13)dt at
= coCV0
cos (cot + TI/2). (16-14)
Figure 16-6 (a) Capacitor C connected to a source of a l t e r n a t i n g voltage V.
(b) Voltage V and c urre nt / as functions of the time for a capac itor , with
l/a>C = 2 o h m s .
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38 0
A
- \
Figure 16-7 Power t ransfer between a source and a capaci tor , as in Fig. \6-6a, fo r
C = to = V 0 = i. A: appl i ed vol t age F 0 co s co t as a funct ion oh. B: — CwVl si n cot •
cosojf, t he power absorbed by the capac i tor . C.(j) CV 1,co s2
cot, t he energy s tored
in the capac i tor .
The cur ren t is p ro po r t io na l to the f r equency an d to C. I t leads t he vo l t age
by 7t /2 as in Fig. 16-6b. The cu rre nt i s said to be in quadrature wi th the vo l t age .
Thi s expres s ion i s used whenever the phase d i f f e r ence be tween two quan t i t i es
is n/2, l ead ing o r l agg ing .
T h e i n s t a n t a n e o u s p o w e r a b s o r b e d b y t h e c a p a c i t o r i s
Th i s quan t i ty i s shown as curve B in Fig. 16-7. I t wi l l be observed that power
f lows a l t e rna te ly in to and ou t o f the capac i to r . Curve C of the same f igure
shows the energy jCV2 s tored in the capaci tor as a funct ion of the t ime. The
s to red energy osc i l l a t es be tween zero and \CV\.
Figure 16-8a shows an idea l induc tor L wi th zero r es i s t ance connec ted
VI = -mCVl cos co t si n cot. (16-15)
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381
to a source V. From Sec. 12.3,
LiU
- = V = V0
cosojf, (16-16)dr
I = — sin ojf = — cos (cot - TI/2). (16-17)coL OJL
We have neglected the constant of integration because we are only interested
in the steady-state sine and cosine terms. The curre nt is inversely pro por tio nal
to the frequency and to L. Figure 16-8b shows V and I as functions of f. The
current is again in quadrature with V. It lags the voltage by JT/2 radians.
As in the case of the capacitor, the average energy dissipation in the
inductor is zero:
V2
VI = — cos cot sin cot, (16-18)coL
as in curve B of Fig. 16-9. The sto red energy oscillates bet ween zero and
iUyjcoL)2
, or ^LI2
0(Sec. 13.5).
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Figure 16-9 Power t ransfer between a source and an inductor as in Fig. 16-8o, for
L = to = F 0 = 1. A: appl i ed vol t age F 0 co s tot as a funct ion of t. B: (Vl/wL) sin cof
coscof, t he power absorbed by the induc tor . C:(j) L(V 0/coL)2 s in 2 cor , the energy
s tored in the induc tor .
16.6 COMPLEX NUMB ERS
Complex numbers a r e ex tens ive ly used fo r the ana lys i s o f e l ec t r i c c i r cu i t s .
Ord inar i ly , num be r s a r e used to expres s quan t i ty , s ize , e t c . Fo r
example , one s ays tha t there a r e 20 cha i r s in a room, o r tha t a bu i ld ing i s
30 .5 me te r s h igh . Such nu m be rs a r e s a id to be real numbers.
Imaginary numbers are r ea l number s mul t ip l i ed by the square roo t
of minus one .
There a r e two symbols used fo r ( — 1 )1 / 2
. In genera l , one uses the
s y m b o l i. H o w e v e r , w h e n e v e r o n e d e a l s w i t h e l e c t r i c a n d m a g n e t i c p h e
no m ena , the cus to m i s to use j ins tead of i because , in the pas t , t he symbol
for current was i. Th e SI sym bol fo r cur r en t i s no w I. T hu s 3i, or 3/, is an
i m a g i n a r y n u m b e r .
A complex number i s the sum of a r ea l num be r an d an ima gina ry
nu m be r , like 2 + 3/ , for exa mp le. It i s useful to rep rese nt com plex nu m be rs
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383
in the complex plane, as in Fig. 16-10, wi th th e real par t (2) plo t ted ho r i
zon ta l ly , and the imaginary par t (3j) ver t ical ly .
I n C a r t e s i a n c o o r d i n a t e s , t h e c o m p l e x n u m b e r z i s thus wr i t ten
z = x + jy. (16-19)
In po la r coord ina tes , f rom F ig . 16-10 ,
z = r c o s 0 + jr si n 0 = r ( cos 6 + j sin 0) , (16-20)
w h e r e r is the modulus of z, a n d 0 is its argument. H e r e
r = (x 2 + y 2 ) 1 ' 2 , (16-21)
a n d 9 i s the angle between the radius vector and the x-axis , as in the f igure.
If x is positive, z is e i the r in the f i rs t or in the four th qu ad ra nt an d
9 = a r c t a n (y/x), (16-22)
If x is negative, z i s in the s econd or in the th i rd quadran t and
0 = a r c t a n (yfx) + it, (16-23)
where the f i rs t term on the r ight i s assumed to be ei ther in the f i rs t or in the
f o u r t h q u a d r a n t .
N o w
e ie = co s 0 + j si n 6, (16-24)
if 6 i s expressed in radians. This r a ther su rpr i s ing r e l a t ion , known as Eulers
formula, can be checked by wri t ing down the ser ies for e je, for cos 6, and for
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384 Alternating Currents: I
si n 0. This will be d o n e in Pr ob . 16-12. The refore , f rom Eq. 16-20,
z = x+jy = re J0, (16-25)
w h e r e x, r, r, 0 are r e l a t ed as in E q s . 16-21, 16-22, and 16-23.
C o m p l e x n u m b e r s in Car tes i an fo rm may be a d d e d or s u b t r a c t e d
b y a d d i n g or s u b t r a c t i n g the real and i m a g i n a r y p a r t s s e p a r a t e l y .
T o a d d or s u b t r a c t c o m p l e x n u m b e r s in polar form, one f i rs t t ransforms
t h e n u m b e r s i n t o C a r t e s i a n f o rm .
C o m p l e x n u m b e r s in Car tes i an fo rm may be m u l t i p l i e d in the u s u a l
w a y , r e m e m b e r i n g t h a t j 2
= - 1:
{a + bj)(c + dj) = (tie - bd) + j(ad + be). (16-26)
In polar form.
(r 1ei9l)(r 2e
ie2) = r i r 2 e m + , h ) . (16-27)
N o t e t h a t, in th e p r o d u c t , the m o d u l u s is e q u a l to the p r o d u c t of the m o d u l i ,
whi l e the a r g u m e n t is the sum of t h e a r g u m e n t s .
To d iv ide one c o m p l e x n u m b e r by a n o t h e r , in C a r t e s i a n c o o r d i n a t e s ,
o n e p r o c e e d s as fo l lows :
(16-28)a + bj _ (a + bj)(c - dj) _ (ac + bd) + j(bc - ad)
e + dj ~ (c + dj)(c - dj) ~ c2
+ d2
Divi s ion in p o l a r c o o r d i n a t e s is s i m p l e :
- = 7 L t v < o . - » 2 i (16-29)
r 2e> e> r
In the q u o t i e n t , the m o d u l u s is e q u a l to the q u o t i e n t of the m o d u l i , and
t h e a r g u m e n t is t h a t of the n u m e r a t o r , m i n u s t h a t of the d e n o m i n a t o r .
16.6.1 EXAMPLES: USING COMPLEX NUMBERS
a) From Eq. 16-24,
^ 1 2 = j_ej« = _ i e-Jxli = _j (16-30)
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16.7 Phasors 385
b) Le t us expres s the complex nu mb er 2 + 3 / in pola r form:
2 + 3/ = (2 2 + 3 2 ) " 2 exp ./ (^arc tan ^ = 3.61 ex p (0.983/). (16-31)
R e m e m b e r t h a t t h e a n g l e 0 must be expressed in radians.
c) Th e squa re of 2 + 3/ is
(2 + 3/) 2 = (2 + 3/)(2 + 3/) = 4 - 9 + 12/ = - 5 + 12/. (16-32)
In polar form,
(2 + 3/) 2 = 3 . 6 1 2 e x p ( 2 x 0.983/) = 13 exp (1.966/). (16-33)
d) Finally, let us calculate (2 -(- 3/)/(l -(- j), f i rs t in Cartes ian and then in polarc o o r d i n a t e s .
2 + 3/ = (2 + 3/) (l - ,/) = (2 + 3) + (3 - 2) j
l + j (1 + ; )(! - j ) ' 2
= 2.5 + 0.5/. (16-35)
o r
2 + 3j 3.61 exp (0.983/)= — - = 2.55 ex p 0.197/ . 16-36
l + j 1.4 14 e x p [ (7 r / 4 ) / ]
76.7 PHASORS
W he n dea l ing wi th a l t e rna t ing cur r e n t s an d vo l t ages , one mus t d i f fe r en t ia t e
s ine and cos ine func t ions wi th r espec t to t ime over and over aga in . Now
the usua l p roce dur e for d i f f e r en ti a t ing these func t ions i s inconv en ien t . Fo r
example, to di f ferent iate cos cot, the cos ine funct ion is ch an ge d to a s ine,
the resul t i s mul t ipl ied by OJ, and t he s ign is cha nge d. To f ind the se con dder iva t ive, the s ine is cha ng ed bac k to a cos ine, and th e resul t is aga in m ul t i
pl ied by co , but th i s t ime wi thou t chang ing s ign , and so on .
I t i s poss ible to s impl i fy these o pe rat ion s in the fol lowing w ay.
F i r s t , r e m e m b e r t h a t
eJ"" = cos cot + j si n cot, ( 1 6 - 3 7 )
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38 6
Figure 16-11 The phasors V, /o;V, and — orV, and the vol t age V i n the complex p lane .
No te tha t the pha sor ro ta t es abou t the or ig in a t the angu la r ve loc i ty co. W e hav e
m a d e co equal to 1/2 . T h e q u a n t i t i e s dV/dt a n d d2
V / d t2 are the pro jec t ions of j w V
a n d — ( o2
\ l on the real axis .
f rom Eq. 16-24. Then
V = V 0 c o s co t = R e (V 0ej""l (16-38)
where the opera to r Re means " r ea l par t o f " what fo l lows .
Re me mb er a l so tha t the t ime der iva t ive o f the exp one n t i a l func t ion
r e d u c e s t o a s i m p l e m u l t i p l i c a t i o n :
j^V^'-") = )co(V QejM
). (16-39)
C o n s i d e r n o w t h e q u a n t i t y
V = V0eJu
" = V 0 cos cot + jV 0 si n cot. (16-40)
Thi s quan t i ty , which i s ca l l ed a phasor. is a vec to r tha t ro ta t es a t t he a ng u la r
veloci ty co in the complex plane, and i ts project ion on the real axis i s V, as
in Fig. 16-11. In d i f f e ren t i a t ing V, the r ea l and the im agina ry pa r t s r em ain
s e p a r a t e :
c/V— = — coK 0 sin cof + jcoV0 c o s cot, (16-41)dt
co2 V 0 c o s cot - jco
2
V 0 sin cot. (16-42)d
2
M
' d t 2
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16.7 Phasors 387
The same applies to all the derivatives. Thus
dV dy d2
V d2M— = Re — , — = Re -dt dt dt
1
dtR e - — = R e ( 1 6 - 4 3 )
One therefore replaces the real voltage
V = V0
cos cot (16-44)
by the phasor
V = V0eh
°'. (16-45)
This then adds to V the parasitic term jV0
sin at. From then on, the operator
d/dt can be replaced by the factor jw. In the end, one recovers the real part.
In this way, differential equations become simple algebraic equations.
Inversely, an inte gra tio n with respect to time is replace d by a division
by jca.
Fi gu re 16-11 also shows y'coV an d — OJ2
V . Note that multiplying a
phasor by jw increases its argument by 71/2 rad ian s a nd multipl ies its
modulus by OJ.
Ph as or s are used for F, 7, and Q. They are also widely used to represent
all kinds of sinusoidally varying quantities.
We use bold-faced sans serif letters for phasors (for example, V), but
it is mor e com mo n t o use or di nar y light-faced italic letters. For e xample ,
the letter V usually means either V0
cos ojf or F0
ex p jot, depending on the
context. Moreover V often stands for F0/ 2
1 / 2
! Surprisingly enough, this
seldom leads to confusion.1
16.7.1 EXAMPLE: THE PHASORS IN A THREE-PHASE SUPPLY
In Fig. 16-5, the voltages at A. B , C can be written either as in Eqs. 16-9 to 16-11,
o r as fol lows.
V.4 = F0
exp./cor, (16-46)
V„ = V0exp ./(cor + 2n/3), (16-47)
+
In a handwritten text, one can identify a phasor by means of a wavy line under the
s y m b o l , for e x a m p l e , ^ .
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388
jy
4
Figure 16- 12 T h e p h a s o r s V 4 . V B , V ( for the voltages at A, B, C in Fig. 16-5, at
tut = nil.
V< = V 0 exp j(an + 47 i/3).
These three phasors are shown in Fig. 16-12 at o>t = nil.
(16-48)
16.7.2 ADDITION AND SUBTRACTION OF PHASORS
The r ea l quan t i ty V. o r I, o r Q. tha t i s rep rese nte d by a ph as or is given by
the p ro jec t ion o f the phasor on the x -ax i s . Thus , two vo l t ages , o r two cur
ren t s , or two charges , can be adde d by fi rs t add in g the i r pha sor s vec to r i a l ly ,
and then t ak ing the p ro jec t ion o f the sum on th e x -ax i s. Th i s is becau se the
pro jec t ion o f the sum of tw o vec to r s i s equa l to the sum of the p ro jec t ion s .
Similar ly , the di f ference between two vol tages , or two currents , or two
charge s , can be ob ta in ed f rom the d i ff e rence be twee n the c or r e spo nd ing
p h a s o r s .
16.7.3 EXAMPLE : THE PHASORS IN A THREE-PHASE SUPPLY
Figur e 16- 13 show s the th r ee phasor s V 4 . V B . V ( - of Fig. 16-12 at r = 0. W e wish to
f ind the voltage of A with respect to B, o r V A — V„. as a function of r, kno win g tha t
VAis V 0
co s cot.
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389
Figure 16-13 Phasor diagram for calculating V 4 - V B for the star-connected sources
of Fig. 16-5.
We could, of course, find VA — VB by trigonometry, since
VA - VB = V0 cos cot - F 0 co s (cot + 2TC/3). (16-49)
Is it easier and more instructive to use phasors. On e can see immediately, from Fig.
16-13, that the phasor V,, — V f l has a modulus of 2(3 ' 2 2)K0 and an argument of
— 30 degrees, or — jr/6 radian at t = 0. Thus
VA
- VB
= 31 2
F 0 cos (cot - TC/6). (16-50)
If the rms voltage at A, B , C is 120 volts, the rms voltage between A and B is 31 , 2
x
120, or 208 volts.
Figures such as 16-13 that serve to perform calculations on phasors are known
as phasor diagrams.
16.7.4 WHEN NOT TO USE PHASORS
a) One can use phasors only if the time dep end enc e is strictly of the
form cos (cot + cp). where cp is a cons tant . Phas ors must not be used if one
has, say, a squar e wave for F , or even a da mp ed sine wave. They must not
be used if the com po ne nt s are non-lin ear, for exam ple, if one has n on- line ar
resis tors as in Sec. 5.3 .
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390 Alternating Currents: I
b) Wh ene ver p ro duc t s of pha sor s , o r p ro du c t s o f e j'0' t e rms a re en
counte red , one r ever t s to the s ine o r cos ine func t ions .
For example , suppose one wishes to ca lcu la t e the power d i s s ipa tedin a r es i s to r . The ins t an taneous power i s VI, wi th
V = V 0 cos tot, I = 7 0 c os cot. (16-51)
I f one uses the phasor s
V = V0eic
", I = I0eJm
', (16-52)
one is t em pted to conc lude tha t
VI = V 0I0e2J
<"' a n d t h a t VI = V 0I0 co s 2cot. (16-53)
This is wrong, as one can see from Sec. 16.2.
The er ror , here, comes f rom the fact that
R e (ej<
°') R e (eiu
") # R e (e2jm
'). (16-54)
O n e c a n d i v i d e a p h a s o r b y a n o t h e r p h a s o r o r b y a c o m p l e x n u m b e r ,
however , as we shal l see in the next chapter . In those cases one has s t raight
f o r w a r d o p e r a t i o n s w i t h c o m p l e x n u m b e r s .
16.8 SUMMARY
Al te rna t ing vo l t ages and cur r en t s a r e o f the fo rm
V = V 0 cos cot, (16-1)
I = I0 cos (cot + cp). (16-2)
We have as sumed here tha t the phase o f V is ze ro at f = 0. In gen era l, the
c u r r e n t I i s no t in phase wi th the ap pl ie d vol tag e. Unles s specified oth erw ise,
one does no t use V 0, b u t r a t h e r
(16-6)
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16.8 Summary 391
as a measure o f the magni tude o f a vo l t age . The s ame app l i es to a cur r en t :
one uses
7 r m s = IJ2il2
(16-6)
t o speci fy the ma gn i tud e o f a c ur r en t .
In three-wire single-phase current, one uses two sources , back to back ,
wi th a common cen te r wi re . Wi th three-phase current, one has th ree sources
with ph ase di f ferences of 120 degre es , co nne cte d to thei r loa ds wi th th ree
wi res , p lus a common cen te r wi re .
In a capac i to r C, t he cur r e n t is p ro po r t io na l to toC a n d leads t h e
app l i ed vo l t age by n/2 r ad ians . In an induc tor L , the current i s inversely
p r o p o r t i o n a l t o OJL a n d lags t he vo l t age by n/2 r ad ians . In bo th cases energy
f lows al te rna tely in an d out of the elem ent an d ther e is zer o ave rage e nergy
transfer .
Complex numbers are of the form a + bj , w h e r e a a n d b are real , and
j = ( — l ) 1 ' 2 . To p lo t complex numbers in the complex p lane , one wr i t es
: = x+jy, (16-19)
o r
z = r (co s 0 + j sin 0), (16-20)
w h e r e r is the modulus, a n d 0 is the argument of z, as in Fig. 16-10, and
r = (x 2 + y 2 ) 1 / 2 , (16-21)
0 = arc tan - or 0 = arc tan - + n. (16-22, 16-23)
x x
Also ,
e>° = cos 9 + j sin 6, (16-24)
so tha t
z = re j". (16-25)
C o m p l e x n u m b e r s m a y b e a d d e d , s u b t r a c t e d , m u l t i p l ie d , a n d d i v i d e d .
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392 Alternating Currents: I
A phasor is a com plex nu mb er wh ose real part is a cosi ne function
of the time. For example.
V = V 0eiv
" = V0 C O S cot + jV0 sin tot. (16-40)
Th e additi on of the parasitic term jV0 sin rot to t he re al vol tag e
V0 co s cot trans form s the cosine function into an exp onen tial function.
Differentiation with respect to time can then be replaced by a multip licati on
by the factor jo. Inversely, an integration with respect to time becomes a
division by jco.
PROBLEMS
16-1E RECTIFIER CIRCUITS COMPARED
Figures 16-14aand bshow. respec t ive ly , half-wave a n d full-wave rectifier circuits.
W h a t are the waveforms ac ross th e res i s t ances?
B 1
(a ) (b)
Figure 16-14 (</) Half-wave rect i f ier c i rcui t . The t r i angula r f igure at the top
represent s a rectifier, or d i o d e . C u r r e n t ca n p a s s t h r o u g h it only in the di rec t ion of
t he a r row head , (b) Full-wave rect i f ier c i rcui t . The s igns show th e pola r i t i e s at a
given ins tant : poin t s A and B have oppos i t e phases . See P r o b . 1 6 - 1.
16-2E RMS VALUE OF A SINE WAVE
S h o w t h a t th e root mean square va lue of V0 co s on is V0/2112
.
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Problems 393
BRIDGE RECTIFIER CIRCUIT
The vo l tage V A — V B
applied to the circuit of Fig. 16-15a is a symmetr ical sawtooth wave as in Fig. 16-15b. with a peak voltage of 100 volts. Fhe diodes in A C . A D .
etc.. pass current only in the direct ion of the ar row head.
Battery chargers use either this circuit or thos e of Fig. 16-14. Th e re sistan ce
shown in the f igures then represen ts the bat tery. Th e source for a bat te ry c harge r
is a t r ans former connected to A C power l ine, and V 0 is about 20 volts for a 12-volt
ba t te ry .
a) Calculate the average and rms values of V A — V B in Fig. 16-15b.
Solution: The ave rage value of V A — V B is zero.
The rms value is given by
( F , - F B ) r
2
m s = i j o2 ( ^ - VBfdt. (II
The l imits of integrat ion cover one per iod. To evaluate this integral , we spli t the
t ime interval into four equal par ts of 0 .5 second, giving four integrals:
\j:5
(VA-VBfdt+
l
-^(VA-VBfdt + - - :
(a) (<)
Figure 16-15 (a) Bridge rect if ier circuit , (b) Saw-tooth voltage applied between
poin ts A a n d B. (c) Vol tage be tween po in ts C and D. See Prob. 16-3.
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394 Alternating Currents: I
Now these in tegra l s a re a l l equa l and
(V A - VB)rm = 100 /3 1 2 = 57.7 vol ts. (4)
b) Draw the curve of V c — V D as a funct ion of the t ime.
Solution: Con s ider the in te rva l 0- 1 second in F ig . 16-15b. where A i s posi t ive
with respect to B. During this t ime the current f lows along the path ACDB a n d
V c — VDis positive as in Fig. 16-15c.Dur ing the in te rva l 1-2 seconds , A i s negat ive with respect to B and the cur rent
f lows along BCDA. T h u s V c — V D is again positive, as in Fig. 16-15c, and so on.
c) What are the average and rms values of V c — V D1
Solution: The average value is 50 vol ts .
The rms value is the same as for V A — V B, or 57.7 vol ts .
16-4 R M S VALUE
Find the rms values for the waveforms shown in Fig. 16-16.
16-5E THREE-WIRE SINGLE-PHASE CURRENT
W hat i s the max im um va lue of the vol t age d if ference b e tween poin t s A a n d C
in Fig. 16-4b, in the 1 20/240 vol t sys tem com mo nly u sed for inter ior w iring ?
16-6E THREE-PHASE CURRENT
Show that , for any t ime r ,
cos cot + cos (cot + 2n/3) + cos (cor + 4n/3) = 0.
76-7 ROTATING MAGNETIC FIELD
Th ree iden t ical coi ls, orie nted as in Fig. 16-17, pr od uc e ma gn et ic f ie lds as
fol lows:
B a = B 0 co s cot, B b = B 0 cos (cor + 2n/3), B L = B 0 cos (cor + 4TC /3).
Show that the resul t ing f ie ld has a magni tude of 1.5B 0 and ro ta tes a t an angula r
veloc ity co.
This is the principle ut i l ized to obtain rotat ing magnet ic f ie lds in large electr ic
m o t o r s .
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39 5
V
2V 0\-
Figurc 16-17 (a) Three coi l s , fed wi th a l t e rna t ing cur rent s of the same ampl i tude
and sam e f requency, prod uce m agne t i c f ie lds of equa l mag ni tud es and or i en ted as
in (ft). Th e two a rro ws in («) sho w wh ere th e coi ls tou ch. If the cur ren ts are pha sed
as in Pr ob . 16-7. the direct io n of the resul t ing m agn et ic f ield rotate s a t the
frequency / ' .
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396 Alternating Currents: I
16-8E DIRECT-CURRENT HIGH-VOLTAGE TRANSMISSION LINES
From many points of view, alternating current is much preferable to direct
current for power distribution. However, as we shall see, line losses are lower withdirect current.
Consid er a long high-voltage overhea d transmission line. Its maxi mum operat ing
voltage depends on several factors, such as corona losses (Prob. 7-14), the size of the
insulators, etc. So, for a given line, the instantaneous voltage between one conductor
and ground must never exceed a certain value, say F 0 . Otherwise, the down time and
the cost of maintenance become excessive. The cost of a line increases rapidly with its
voltage rating.
Th e curren t in the line can be ma de nearly as large as one likes with out dam agi ng
it, since the conductors are well cooled by the ambient air. However, the power loss
increases as the square of the current. So. the lower the current the better.
With direct current, one has two conductors operating at + V0and — F 0 with
respect to ground. Fhe power delivered to the load is 2V0IDC.
a) With single phase alternating current, we have two wires at + F 0 cos tot an d
— F 0 cos ojt.
Show that, for the same power at the load, the rms current /s p
is 2l/2
IDC.
The I2
R losses in the line are twice as large as those with direct current.
b) With three-phase alternating current, we have three wires at F 0 cos cot,
V0
cos (cot + 2TI/3), F0
COS (cot + 4TI/3). We assume that the three load resistances
connected between these wires and ground are equal. Then the current in the ground
wire is zero.
Show that, for the same total power delivered to the three load resistances, the
rms currents / l p a re ( 2 /3 ) 21 / 2
/ D C * 7 D C .
With three-phase alternating current, the rms currents are about the same as
with direct current, but we have three current-carrying wires instead of two, so that
the losses are 50% larger th an with direct curre nt.
16-9 ELECTROMAGNET OPERATING ON ALTERNATING CURRENT
An electromagnet with a variable gap length is opera ted on alter nating current.
Show that the rms value of the magnetic flux is independent of the gap length,
for a given applied alternating voltage, and neglecting leakage flux.
16-10E COMPLEX NUMBERS
a) Express the compl ex n umb ers 1 + 2/', — 1 + 2/', — 1 — 2/, 1 — 2/ in pol ar
form.
b) Simplify the following expressions, leaving them in Cartesian coordinates:
(1 + 2/)(l - 2j). (1 + 2/ )2
,
1/(1 + 2y)2
, (1 + 2/) /(l - 2j).
16-11E COMPLEX NUMBERS
Com plex num be rs in pol ar form are often writt en as rZ_0, where /• is the
modulus and 0 is the argument.
Expre ss 1 + 2/ in this way.
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Problems 397
16-12E COMPLEX NUMBERS
Use the series
e* =1 + x + | . \- 2!) + (x 3 /3! ) + • • •
sin x = x - (x 3 3!) + (x s / 5 ! ) - ( x 7 7!) + • • •
cos x = 1 - (x 2 2!) + (x 4 4!) - (x* 6!) + • • •
to show tha t
e>* = cos v + j sin x.
I6-I3E COMPLEX NUMBERS
Show tha t exp jn = — 1.
16-14E COMPLEX NUMBERS
W hat hap pen s to a comp lex nu mb er in the complex p lane when i t i s (a) mul t ip l i ed
by j. (b | mul t ip l i ed by j2, (c) divided by /'?
16-15 THREE-PHASE ALTERNATING CURRENT
Show, by t r igonomet ry , tha t the expres s ions on the r ight -hand s ide of Eqs . 16-49
and 16-50 a re equ a l .
16-16 CALCULATING AN AVERAGE POWER WITH PHASORS
In Sec. 16.7.4 we saw that , wi th al ternat ing currents , one must no t calculate a
power by us ing the product VI.
a) Show that , for the general case where
V = V 0 co s mt, I = 70
cos (wt + </>)
in a load, the average power is
P , v = ( l / 2 ) 7 0 / 0 cos q> = F r m 5 / r m 5 co s q> .
The t e rm cos q> is called the power factor of the load (Sec. 18.2).
b) Show tha t
P„ = ( l /2)ReVI*,
where I* is the complex conjugate of I, ob tain ed by ch ang ing th e s ign of i ts imag inary
par t .
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CHAPTER 17
ALTERNATING CURRENTS: II
Impedan ce, Kirchoff's Laws, Transform ations
17.1 IMPEDANCE
17.1.1 Example: The Impedance and the Admittance of a Coil
17.2 THE EQUATION 1 = V / Z IN THE COMPLEX PLANE
17.3 KIRCHOFF'S LA WS
17.3.1 Example: The Series LRC Circuit
17.4 ST A R-DEL TA TRANSFO RM A TIONS
17.4.1 Example: Transforming a Delta into a Star
17.5 TRANSFORMATION OF A MUTUAL INDUCTANCE
17.5.1 Example: Two Superposed Coils
17.6 SUMMARY
PROBLEMS
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This cha p te r rough ly para l l e l s Cha p te r 5 , mo s t o f which dea l t wi th d i r ec t
cur r e n t s f lowing th ro ug h r es is t ive c i r cu i ts . I ns t ead o f d i r ec t cur r en t s , we no whav e a l t e rna t i ng cur r en t s and , ins t ead o f r es i s t ances , we have im peda nces .
As we sha l l s ee , bo th Ki rchof f ' s l aw s and the s t a r -de l t a t r ans form at ion s
remain va l id . We sha l l a l so l ea rn how to t r ans form a mutua l induc tance
in to a s t a r ; t h i s t r an s form at ion doe s no t have a d i r ec t - cur r e n t equ iva len t .
Let us retu rn t o the s im ple ci rcu i ts of Figs . 16-1, 16-6, an d 16-8, wh ere ana l t e rn a t ing vo l t age i s app l i e d succes s ive ly to a r es i s to r , t o an induc tor , and
to a capac i to r .
I f an a l te rna t in g vol t age V is app l ied to a re s is tor ,
17.1 IMPEDANCE
I = MIR. (17-1)
I f V is app l ied t o an ind uc tor , then, rew ri t ing E q. 16-16,
V = jcoLl (17-2)
a n d
I = V / j c o L . (17-3)
T h e q u a n t i t y jtvL plays the s ame ro le as R and is cal led the impedance of
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400 Alternating Currents: II
17.1.1
t he induc tor . The symbol fo r impedance i s Z. I m p e d a n c e s a r e m e a s u r e d
in ohms, l ike res is tances .
W h e n V i s app l i ed to a capa c i to r ,
I = Qa>V = , (17-4)1/jcoC
f rom Eq. 16-13. H e r e 1/jcoC = —j/ojC i s the imp eda nce o f the capa c i to r .
M or e genera l ly , an imp eda nce i s com plex an d is wr i t t en as
Z = R+jX = \Z\ej", (17-5)
w h e r e R is the resistance, X is th e reactance, |Z | is the modulus or the magnitudeof the impedance , and 9 is its phase angle. A pos i t ive reac tan ce is said to be
inductive, while a negat ive reactance is said to be capacitive.
T h u s , i n genera l , t he vo l t age V a c r o s s a n i m p e d a n c e Z c a r r y i n g a
cur ren t I i s
V = Zl , (17-6)
a n d
I = V/Z , Z = V/l. (17-7)
Th e inver se o f an im ped anc e i s an admittance:
Y = 1 /Z . (17-8)
Impedances in s e r i es and in para l l e l a r e ca lcu la t ed in the s ame way
as resistances in series and in parallel (Sees. 5.4 and 5.5).
EXAMPLE: THE IMPEDANCE AND THE ADMITTANCE OF A COIL
A coi l has a res is tance of 1000 ohms and an inductance of 3.000 henrys . I ts impedance
at a frequency of 100.0 hertz is
R + jcoL = 1000 + 2n x 10 0 x 3/ = 1000 + 18 85 /oh ms , (17-9)
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40 1
Figure 17-1 (a) The impedance of the coi l of Sec. 17.1.1 in the complex plane.
(h) V an d I = V Z for th is coil at t = 0. Th e mo du li of V and I are n ot to scale .
o r
1 8 8 5 \Z = (1000 2 + 1 8 8 5 2 ) 1 2 e x p ( / a r c t a n — — 1 .
= 2134 exp 1 .083/ohm s . (17-10)
Figure 17-1 shows Z in the complex plane.
If one appl ies across this coi l a vol tage V 0 cos co t with an rm s value of 10 vol ts ,
the current is
2134 ex p 1.083/ 2134exp./(ojt - 1.083). (17-11)
2 1 2 x 10/ = „ „ , co s (cut - 1.083)
2134(17-12)
The n the cur ren t / ha s an rms va lue of 10/2134. or 4 .686 mi l l iamperes . The cur ren t
lags the vol tage by 1.083 radians .
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402 Alternating Currents: II
T h e a d m i t t a n c e o f th e c o i l is
y _ I _ __ J 1000 - 1885/ _ 1000 - 1885/
~ Z ~ 1 0 0 0 + 1 8 85 / ~ 1 0 0 0 2 + 1 8 8 5 2 ~ (2134) 2 '
= (2 .196 - 4 .1 39 / I1 0" 4
Siemens.
Y = e x p ( - 1 . 0 8 3 / ) ,
= 4.686 x 1 0 _ 4 e x p ( - 1 . 0 8 3 / ) S i e m e n s .
(17-13)
(17-14)
(17-15)
(17-16)
17.2 THE EQUA TION I = V / Z IN THE COMPLEX PLANE
I t i s ins t ruct ive to cons ider the equat ion I = V/Z i n the complex p lane .
W r i t i n g
\Z\eJe, (17-17)
then
\z \e* i z r(17-18)
Th us the mo du lus o f the phas or I i s the m od ulu s o f V div ided by |Z | , whi le
i t s a rgument i s the a rgument o f V minus tha t o f Z. as in Fig. 17-1.
17.3 KIRCHOFF'S LAWS
Kirchoff 's laws of Sec. 5 .7 apply to al ternat ing currents jus t as wel l as to
d i r ec t cur r en t s .
The cur r en t l aw s t a t es tha t the sum of the cur r en t s en te r ing a node
is zero. As we saw in Sec. 5 .9 , one normal ly uses mesh currents , ins tead of
branch currents , so as to avoid us ing this law expl ici t ly .
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40 3
L RC
+ Q
V
Figure 17-2 Inductor L, res is tor R, a n d c a p a c i t o r C, connected in series across a
source V.
The vo l t age l aw s t a t es tha t the sum of the vo l t age d rops a round a
me sh is zer o. T o use this law, one show s plus an d m inu s s igns for the so urce sand for the charges , and ar rows for the currents , as in Fig. 17-2.
The s igns and the d i r ec t ions o f the a r rows a re a rb i t a ry , s ave fo r one
excep t ion . If a cur r en t I f lows in to a capa c i to r e l ec t rode ca r ry ing a charg e
Q , and if on e wishes to wr i te th at I = dQ/dt, t hen one mus t have the cur r en t
f lowing int o the pos i t ive Q , as in Fig. 17-2. W ith th e ar ro w s in the op po s i te
d i r ec t ion , one would have to wr i t e I = —dQ/dt.
Note tha t these s igns and a r rows a re no t meant to ind ica te the po la r i t i es
at a pa r t ic ula r ins ta nt , as they we re in Fig. 16-4. Th ey are s imply used as a n
a id in wr i t ing down the mesh equa t ions cor r ec t ly .
17.3.1 EXAMP LE: THE SERIES LRC CIRCUIT
Figu re 17-2 sho ws a ci rcui t wi th an ind uc tor L, a res is tor R, and a capac i tor C .
connected in series across a source V.
We shal l (a) wri te down the different ia l equat ion for the mesh, (b) rewri te i t in
ph aso r no tat ion to f ind Q and I . (c) wri te dow n the valu e of I direct ly from the relat ion
I = V /Z , (d) ded uce / as a funct ion of the t im e, and (e) discuss the value of Z as a
funct ion of the frequency.
a ) F r om Ki rchof f ' s vol t age l aw. s t a r t ing a t the lower l e ft -hand c orner .
dl u
-L RI - -- + V = 0.t C(17-19)
or
dQ Q+ R — + - = V.
dt C(17-20)
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404 Alternating Currents: II
b) In pha sor nota t io n , th i s equ a t ion b ecom es
(jwfL + jmR + CQ = V,
so tha t
{jtofL + jcoR + (1/C)
Mul t ip ly ing both s ides by jco,
jwQ = IR + jmL + (1/jwC)
(17-21)
(17-22)
(17-23)
c) Since we have impedances in series , the total impedance is the sum of thei m p e d a n c e s :
Z = R + jtoL
T h e n
1
jcoC
Z R + jtoL + (1/jcoC)
(17-24)
(17-25)
as above .
d) To find I as a funct ion of the t ime, we f i rs t express the impedance in the polar
form |Z | exp (j9):
Z = R + j[coL - (1/coC)],
= \R2
+ [coL - ( l / c » C ) ] 2 } 1 / 2 e x p j arc t ancoL - (1/coC)
T h u s
| Z | = [R 2 + [coL - ( 1 / c o C ) ] 2 } 1 ' 2 ,
a n d
0 = arc t an
_ K0e"
0
'
1
" W '
Vo
IZ l
wL - (1/coC)
R
(17-26)
(17-27)
(17-28)
(17-29)
(17-30)
(17-31)
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405
150 r
f (hertz)
Figure 17-3 Reactance as a function of the frequency for the circuit of Fig. 17-2, for
L = 1 millihenry and C = 1 microfara d. F he reactan ce X is (oL — 1/coC.
The current I is the real part of I:
/ =Z
cos (cot — 0). (17-32)
e) Note that
Since
| Z | an d 0 are both functions of the frequency.
Z = R+jX = R + jcoL 11
o) 2LC(17-33)
the reactance X is zero when co2LC = 1. At tha t frequency the voltage dro p on L
exactly cancels that on C and the circuit is said to be resonant. With series resonance,the voltages across L and across C can be larger than the applied voltage.
Fig ure 17-3 shows the reac tance as a function of the frequency for L = 1 milli
henry and C = 1 microfar ad.
This is a case of series resonance because the capacitor and the inductor are
connect ed in series with the source. Proble m 17-15 concerns parallel resonance.
In the case of parallel resonance, the currents through the capacitor and through
the inductor partly cancel, and each one can be larger than the total current.
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406 Alternating Currents: II
17.4 STAR-DELTA TRANSFO RMAT IONS
In Sec. 5.10 we saw how one can t r a n s f o r m a res is t ive s tar c i rcui t in to an
equ iva len t de l t a , and vice versa.
If one has com plex im ped ance s , as in Fig. 17-4a, ins tead of r es i s t ances ,
t h e t r a n s f o r m a t i o n r u l es a r e s imi la r . Th i s comes f rom th e fact that th e r e a
s o n i n g we used in Sec. 5.10 is in n o way l imited to r e a l i m p e d a n c e s . T h u s
ZA= ^
Z
f ' , „ , (17-34)z
hz
c
za
+ zh
+ zc
Z 7
z a + z„ + z e
z„zb
za + z
b+ z
c
YBYC
YA+ YB+ YC
YCYA
YA + YB+ YC
YAYB
(17-35)
(17-36)
(17-37)
(17-38)
F. = ^ J * . (17-39)
' YA+YB+YC
[ U
" >
i A
Figure 17-4 F h r e e n o d e s A, B, C. {a) D e l t a - c o n n e c t e d , (b) S t a r - c o n n e c t e d . T h e w a v y
l ines represent impe dance s .
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17.4 Star-Delta Transforma tions 407
17.4.1 EXAMPLE: TRANSFORMING A DELTA INTO A STAR
As an example, le t us t ransform the del ta of Fig. t7-5a into a s tar a t a frequency of
one ki lohertz . In this case,
Z a = Z h= ifjcoC = - If)-'/ : ; : oh ms , (17-40)
Z f = 100 oh m s. (17-41)
Fro m Eqs . 17-34 to 17-36. and rem emb er ing tha t Z = 1/7 ,
Z A = Z B = 45.51 - 14 .30 /oh ms , (17-42)
Z c = - 2 2 . 7 5 - 7 2 . 4 3 / o h m s . ( 17 -4 3 )
T h u s Z A a n d Z B are each composed of a res is tance of 45.51 ohms, in series with
a capac i tor whose impedance i s —J/OJC, w h e r e
o>C = 1/14.30. (17-44)
C = 1/(14.30 x 2TT x 1000) = 11.13 mic rofa rads . (17-45)
Similarly, Z c i s composed of a res is tance of —22.75 ohms, in series with a capaci tor
of 2.197 microfarads .
Th e equivale nt s tar is sho wn in Fig. 17-5b.
Note that the res is tances of the equivalent c i rcui ts can have negat ive values .
A real c i rcui t can therefore be equiva lent to a f ic ti tious on e that is imp ossible tobui ld .
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408 Alternating Currents: II
Note a l so tha t the res i s t ances and capac i t ances in the s t a r depend on the
op era t ing frequency. Fo r exam ple, in Pr ob . 17-16, it i s sho wn th at , a t one me gah ertz ,
the del ta of Fig. 17-5a is equ ivale nt to a com pletely different s tar .
17.5 TRANSFORMA TION OF A MUTUAL INDUCTAISCE
Figure 17-6a shows a mutua l induc tance M wi th i t s tw o co i ls co nnec ted
a t C. We can t r ans form th i s mutua l induc tance in to the s t a r o f F ig . 17-6b
in the fol lowing way.
T h e m e s h c u r r e n t s p, q, r mus t be the s ame in the two c i r cu i t s . For
s impl ici ty , le t us assume that C i s a t gr ou nd po ten t ial . This wi ll not res t r ic t
the genera l i ty o f the t r ans f orm at ion .In Fig. 17-6a,
V A = (p - q ) Z a + jcoM (r - q). (17-46)
Th e sec ond te rm on the r ight is the vol tag e induc ed on the lef t -hand s ide,
o r M t imes the rate of change of the current on the r ight (Sec. 12.1) . Here,
M i s pos i t iv e if the tw o coi ls are w ou nd in such a way t ha t a cu rren t in th e
pos i t ive di rect ion in one coi l produces in the other coi l a f lux l inkage that
is in the sa m e direc t ion as that d ue t o a pos i t ive cu rren t in that co i l (Sec. 12.1) .
O t h e r w i s e , M is nega t ive . W e have as sume d here tha t c lockwise cur r e n t s
are pos i t ive.
In 17-6b.
V 4 = (p - q)Z A + (p - r ) Z c . (17-47)
Similarly, in 17-6a,
V B = (q - r)Z b -jcoM(p - q), (17-48)
while, in 17-6b.
V B = (q - r)Z B + (p - r ) Z c . (17-49)
E q u a t i n g n o w t h e t w o v a l u e s o f V^, and then the two va lues o f V B
and s impl i fying.
p(Z„ - Z A - Z c) + q(Z A - Z a - jcoM) + r ( Z c + jcoM) = 0, (17-50)
p ( Z c + jcoM) + q(Z B - Z„ - jcoM) + r(Z„ - Z B - Z c ) = 0. (17-51)
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40 9
Since these two e q u a t i o n s are va l id , wha tever the va lues of p, q, r. the six
coef f i c i en t s be tween paren theses mus t be zero , and
ZA
= Z„ + jcoM, (17-52)
ZB
= Zb
+ jcoM, (17-53)
Zc
= -jcoM. (17-54)
N o t e t h a t , if M is pos i t ive , the r e a c t a n c e —jcoM is c a p a c i t i v e . On the
o t h e r h a n d , if M is nega t ive , then — joM is the r e a c t a n c e of a pure i n d u c t a n c e
\M\. A g a i n , we h a v e a c i r cu i t t ha t is m a t h e m a t i c a l l y e q u i v a l e n t to one t h a t
c a n n o t be real ized in pract ice, s ince any i nduc tor , un les s it is s u p e r c o n d u c t i n g ,
pos ses ses a r es i s t ance .
17.5.1 EXAMPLE: TWO SUPERPOSED COILS
F i g u r e 17-7a s h o w s two superposed coils w o u n d in the same d i rec t ion , on e over the
other. Each coi l has a res i s t ance R and an i n d u c t a n c e L . Since the two coi ls are
s u p e r p o s e d . \ M\ = L .
In this example, with th e cur rent d i rec t ions shown in the figure. M is nega t ive an d
the equivalent c i rcui t is s h o w n in Fig. 17-7c; the i m p e d a n c e is R/2 A- JOJL. If we
c h a n g e d th e di rec t ion of one of the c u r r e n t s , we would s t i l l arr ive at the same resul t .
T h e i m p e d a n c e R/2 + j<oL is eas i ly expla ined: in effect, we h a v e one coi l made
u p of larger wire , so t h a t th e i n d u c t a n c e is L. and the res i s t ance is R/2.
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41 0
(c)
W V \ AR
V \ A AR
rrr\21.
1
L
2L R
\ A A AR
Figure 17-7 Tw o superposed coi l s , (a) w o u n d in the same d i rec t ion , and (h) w o u n d
in oppos i t e d i rec t ions : (c ) and (d), equiva lent c i rcu i t s for (a) and (h). respect ively.
F igure 17-7b shows a s imi la r pa i r of coi ls , except that now the coi ls are w o u n d
in oppos i t e d i rec t ions . For t he cur rent d i rec t ions shown. M is posi t ive, th e equiva lent
circui t is s h o w n in 17-7d. and the t o t a l i m p e d a n c e is R/2.
T h i s is correct s ince, in t h i s example , the net magnet ic f ie ld is zero , and the two
res i s t ances are in paral le l .
Th e curre nt in a circuit elem ent is given by the volta ge acro ss it, divided by
th e impedance of the element:
The impedance of a resistor is R, th at of an in du ct or is /VoL, an d t ha t of a
capaci tor \/jcoC. An impe danc e is expressed as a complex nu mber , either
in Cartesian or in polar form:
77.6 SUMMARY
I = V/Z. (17-1)
Z = R+ jX (17-5)
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Problems 411
where R is its resistance, X is its reactance, \Z\is the modulus or magnitude
of the impedance, and 0 is its phase angle.
The inverse of an impedance is an admittance:
Y = 1/Z. (17-8)
Series and parallel combinations of impedances are calculated in
the same way as with resistances.
Kirchoff's laws apply to alternating currents as well as to direct
currents.
The star-delta transformations of Cha pte r 5 apply to i mpedances.
A mutual inductance can be transformed into a star circuit as in
Fig. 1 7-6 and Eqs. 17-52 to 17-54 .
PROBLEMS
17-1 IMPEDANCE
a) Calcu la te the i m p e d a n c e Z of the circuit shown in Fig. 17-8.
W h a t is th e value of Z (i) for / = 0, (ii) for /' -» oo?
b ) Wh a t is the m a g n i t u d e and the phase ang le of Z at 1 k i l o h e r t z?
c ) Wh a t is the m a g n i t u d e and the phase ang le of Y = 1/Z at that f requency?
d) Calcu la te the power d i ss ipa t ion when th e cu r r en t is 100 mil l iamperes , againat 1 ki loher tz .
e) Can the real par t of the impedance become negat ive?
f) For what f requency range is the circuit equivalent to (i) a resis tor in ser ies
with an i n d u c t o r , (ii) a resis tor in series with a cap ac i t o r ?
g) At what f requency is the circuit equivalent to a pure r es i s tance?
i o n
io n
o — A A A
5 LIF
Figure 17-8 See P r o b . 17-1.
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412 Alternating Currents: II
17-2 REAL INDUCTORS
A real inductor has not only an inductance but also a resistance and a capaci
tance. Its equivalent circuit is shown in Fig. 17-9.Show that the impedance between the terminals is
_ R + jco[L(l - OJ2
LC) - R2
C]
(1 - to2
LC)2
+ R2
to2
C2
I — v W A
Figure 17-9 Equivalent circuit of a real
inductor. Fhe stray capacitance is C.
See Prob. 17-2.
L
- I t -
17-3E COMPENSATED VOLTAGE DIVIDER
Th e voltage divide r of Prob . 5-5 canno t be used as such at high frequencies for
the following reas on. There are stray cap aci tanc es due to the wiring, etc.. in parallel
with P, and R2. If the frequency is high enou gh, these stray capac itan ces car ry an
appreciable current and V,,/Vj is a funct ion of the frequency.
Show that, at any frequency
2*1 * 2
V, P , + R2
if P , C , = R2C2 in Fig. 17-10. Capac itance s C, a nd C , are made large comp ared to
the stray capacitances. The voltage divider is then said to be compensated.
F'igure 17-10 Compensated voltage
divider. See Prob. 17-3.
17-4 RC FILTER
Four impedances are connected as in Fig. 17-11.
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413
A C
o —n n n
— • -r r r
^ - m o
Figure 17-11 See Prob. 17- 4. O • • O
a) Show that
V„ _ BD
Vf ~ AB + (A + B)(C + D)'
b) Draw the curve of |V„/Vi| as a function of RwC for the circuit of Fig. 17-12for RmC = 0.1 to 10. Use a log scale for RcoC.
R C
o - W V A - f 1(—f o
V, c ^ < /< v„
Figure 17-12 See Prob. 17- 4. O • • O
17-5 MEASURING AN IMPEDANCE WITH A PHASE-SENSITIVE VOLTMETER
An unk now n imp edance Z is connecte d in series with a known resistance R'
as in Fig. 1 7 - 1 3 .
Figure 17-13 See Prob . 17-5.
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414 Alternating Currents: II
Show tha t the imped ance Z can be found by me asur in g the vol tage ra t io
r = a + bj = V 2 / V ! ,
in o the r words , by measur in g the in-phase and the qua dra tur e com pon ent s of V 2 , V ,
b e i n g k n o w n .
17-6E IMPEDANCE BRIDGES. THE WIEN BRIDGE
Figure 17-14 shows an impe danc e br idge . I t i s the a l t e rna t ing-cu r rent equiva lent
of the Wheats tone bridge of Prob. 5-8. In this case, V = 0 when
Z 1/Z 2 = Zi/ZA.
In a l t e rna t ing-cur rent br idges the component s mus t s a t i s fy tw o i n d e p e n d e n t
equa t ions to s a t i s fy both the rea l and the imaginary pa r t s of th i s equa t ion .The re ex i st man y types of impe danc e br idge .* One co m mo n type is the W ien
bridg e sho wn in Fig. 17-15. As a rule , on e sets
Ri = R-i — R 3 = R*, C 3 = C 4 .
I
Figure 17-14 Impedance br idge . See
P r o b . 17-6.
Figure 17-15 Wien br idge . See P rob .
17-6.
* See. for ex am ple. Reference Data for Radio Eng ineers. Fi fth Edi t ion . Ch apt e r 11
(Howard W. Sams and Co. . Inc . . Indianapol i s . 1968) .
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Problems 415
Show tha t , under these condi t ions , P 3 < a C 3 = 1.
The Wien bridge is used in tuned amplif iers and in osci l la tors , as wel l as for mea
sur ing or moni tor ing a f requency. To measure a f requency, one changes P 3 a n d P 4
s imul taneous ly unt i l R3coC } i s equ al to uni ty.
See also Prob. 17-18.
17-7E LOW-PASS RC FILTER
The circui t of Fig. 17-16 is commonly used to reduce the A C c o m p o n e n t a t t h e
output of a power supply .
As a rule , the term power supply appl ies to a c i rcui t that t ransforms 120-vol t
60-her t z a l t e rna t ing cur rent in to d i rec t cur rent . One s imple type i s the ba t t e ry charger
of P rob . 16-3 .
A power supply usua l ly compr i ses three pa r t s : a t rans former , a rec t i fy ing
circui t , an d a f i lter . The o utp ut of the rect i fying circui t is a pulsat in g direct cur ren t
tha t can be cons ide red to be a s t eady d i rec t cur rent , p lus an a l t e rna t ing cur rent . Thefunct ion of the f i lter is to a t te nu ate the al tern at in g part . Fi l ters are usual ly d ispense d
wi th in ba t t e ry chargers .
The f i l ter is often fol lowed, or even replaced, by an electronic ci rcui t that main
tains ei ther the output vol tage or the output current a t a set value (Sec. 5.11).
The circui t shown is a s imple form of low-pass f i l ter . The load res is tance R L
i s a s sum ed to be l a rge comp ared to 1 /mC,
a) Sh ow tha t the a l t e rna t ing com po nen t i s a t t en ua te d by the fac tor
|V 0/V,. * IJRcoC
if RcoC » 1. Th e vol t age dr op ac ross the res is t ance P should be smal l com pare d to
tha t ac ros s R L, so
RL » P » 1/mC.
b) Plot a curve of |V 0 / V , | as a fun ction of frequ ency from 0 to 150 he rtz for P =
1 04 ohm s and C = 100 mic ro fa rads .
c ) P lo t the cor responding curve of the a t t enua t ion in decibels as a function of
frequency, where the number of decibels is 20 log |V 0 /V f | .
R
Figure 17-16 Simp le low-p ass P C filterfor a t t enua t ing the a l t e rna t ing
com pon ent a t the outp ut of a power
supply, with P L » R » \fcoC. See
P r o b . 17-7.
17-8 PHASE SHIFTER
I t i s often necessary to shif t the phase of a s ignal . Figure 17-17 shows a s imple
circui t for do ing this . Th e res is tances are adjus table , but are kept eq ual .
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41 6
Figure 17-17 Phase shif ter . See Prob.
17-8.
Use the polari t ie s show n. These me an th at V, is the vol tag e of the top term inal
wi th respect to the bo t to m one , and V„ i s the vol t age of the r igh t -hand t e rmina l wi threspect to the left -hand one.
a ) Show tha t
V„/V, = exp [2/ arc tan (1/RcoC)].
b) Draw a graph of the phase of V„ with respe ct to V, in the range RO JC = 0.1
to 10. Use a log scale for RcoC.
17-9 MEASURING SURFACE POTENTIALS AND SURFACE CHARGE
DENSITIES ON DIELECTRICS WITH A GENERATING
VOLTMETER, OR FIELD MILLIn P r ob . 6-11 , we saw how one can mea sure the sur face potent i a l V a n d t h e
surface charge densi ty a on a d ie l ec t r i c , wi th a probe connec ted to an e l ec t romete r .
Here is a s imilar method, shown in Fig. 17-18a, in which the charge induced on the
prob e i s mo dula ted to produ ce an a l t e rna t ing cur rent . The a l t e rna t ing cur ren t pas ses
thr ou gh a res is tance, giving an al ter nat i ng vo l tage that is then am plif ied to give an
o u t p u t v o l t a g e p r o p o r t i o n a l t o V, or to a. Ins tru me nts tha t mea sur e electr ic f ie ld in
tensities in this way are called generating voltmeters, o r field mills.
One such ins t rument can measure potent i a l s up to 3000 vol t s and has a f i e ld
of view only about 2 mil l imeters in diameter . I t serves to tes t the photoconduct ive
mate r i a l s used for xe rography.
F igure 17-19 shows another one tha t measures the t e rmina l vol t age on a Van
de Graaf f h igh-vol t age gen era tor .a) Use Fig. 17-18b to find E a, as a funct ion of a, d a, dD, e r, wi thout the vane of
F ig . 17-18a . The cur rent through R i s then zero.
b) No w wh at is the value of c ,?
No te tha t cr, depe nds on bot h dD and on e r.
c) What is the value of V in terms of cr. d a, dD, efl
N o t e t h a t V d e p e n d s o n d a, and is therefo re affected by the presen ce of the
i n s t r u m e n t .
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41'
p A / V S
V X A aI -
T V h i
P
J_
JZZZZZ- VA
R 0D
£
Figure 17-18 (a) Ap pa ra tu s for me asu rin g the pot ent ia l a t the surface of a dielectr ic
s lab D. The pro be P is grou nde d th rou gh the res i st ance R. a n d t h e g r o u n d e d
v a n e VA i s made to pas s be tween D a n d P in such a way that P i s a l ternately
sc reened and exposed to D. An a l t e rna t in g cur rent Hows to P a n d p r o d u c e s a
vol t age drop on R. The vol t age ac ross R i s amplif ied (Prob. 5-9). giving a vol tageV„ propor t iona l to the sur face charge dens i ty . The vol t age ac ross R i s orders of
magni tude smal l e r than the vol t age a t the sur face of D. (b) Probe P near the
dielectr ic carrying a surface charge densi ty a.
Figure 17-19 Genera t ing vol tmete r for measur ing a vol t age of the order of mi l l ions
of vol ts , from a dis tance. A motor M rotates a shield S in front of a probe P in
view of the h igh-vol t age t e rmina l T enclosed in a pressure tank PT . Sec tors a re
cut out of the shield. An al ter nat ing c urr ent f lows to P throu gh P. Th e vol tage V
is propor t iona l to the vol t age on the t e rmina l . See P rob . 17-9 .
P T
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418 Alternating Currents: II
d) No w ca lcula te the v ol t age ac ross R w h e n R = 1 0 m e g o h m s a n d V —
1000 vol t s , and when the capac i t ance C be tween P and the dielectr ic surface varies
s inusoidal ly between 0 and 0.1 picofarad. The vane osci l la tes a t 100 hertz and screensP once per cycle .
17-10E REFERENCE TEMPERATURES NEAR ABSOLUTE ZERO
Th e circui t sho wn in Fig. 17-20 is used to define f ixed poin ts on the tem pe rat ur e
scale near absolute zero. I t ut i l izes the fact that several e lements exhibi t sharp and
reproduc ib le t rans i t ions to and f rom the superconduc t ing s t a t e a t de f in i t e t empera tures .
A s lug of one of these elements , S , is inserted in the mutual inductance M 2.
T h e n R L/R 2 i s adjus ted to m ak e V = 0 . Whe n S becom es supe rcon duc t in g , l a rge eddy
currents f low through i t . cancel ing part of the magnet ic f ie ld, M 2 i s decreased, and
V # 0.
Show tha t , when V = 0,
M , / M 2 = RJ R 2.
Figure 17-20 See Prob. 17-10.
17-11 REMOTE-READING MERCURY THERMOMETER
A good mercu ry th e rm om ete r i s a simple , inexpens ive device for makin g accu ra tem e a s u r e m e n t s o f t e m p e r a t u r e . T h e r e a r e m a n y c i r c u m s t a n c e s , h o w e v e r, w h e r e o r d i n a r y
mercury the rmomete rs a re imprac t i ca l . For example , one may wish to moni tor the
tempera ture a t many poin t s nea r the bot tom of a r ive r , every minute , 24 hours a day .
Or one may wish to use a t empera ture reading to cont ro l a manufac tur ing process
a u t o m a t i c a l l y .
I t is poss ib le to read a me rcury the r mo me te r e l ec t ronica l ly , and hen ce au to
matical ly and remotely, with the ci rcui t shown in Fig. 17-21.
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41 9
Figure 17-21 (<v) Mer cury the r mo me te r a dap ted for rem ote reading . An osc i l l a tor
appl i es abo ut 10 vol t s a t 10 meg aher tz to an e l ec t rode sur rou ndin g the bulb an d
f o rm i n g a c a p a c i t a n c e C j . T h e c a p a c i t a n c e C 2 be tween the mercury column and a
coaxia l tube va r i es l inea r ly wi th t empera ture . The vol t age on R is amplified
(Prob. 5-9) and measured at V. (r>) Equiva lent e l ec t r i c c i rcu i t . See P rob . 17-11.
a) F ind the vol t age V as a function of to, t he source vol t age V s, C 2 , and P .
» c 2 .
b) Under wha t condi t ion wi l l V vary l inea r ly wi th t empera ture?
c) Est imate the value of C2. and suggest a value for P.
d) F hen wha t is the order of ma gni tude of V if V„ is 10 volts?
17-12 WATTMETER
Figure 17-22 shows a Hal l generator (Sec. 10.8.1) used as a wat tmeter to measure
the power suppl ied by a source to a load Z.
The coi l of a smal l e lectromagnet £ has an inductance £, with toL « |Z|. It
suppl ies the magnet ic f ie ld for the Hal l generator . The res is tances Pj and P 2 are chosen
so that P, + P 2 » |Z| . In this way, the powe r supp l ied by the sour ce is appro xim atel y
equal to that diss ipated in the load.
The vol t age and the cur rent a t the source a re
V = F 0 co s cot,
I = I0 co s {cot + cp).
Uti l ize the resul t of Sec. 10.8.1 to show that the average value of the Hal l
g e n e r a t o r o u t p u t v o l t a g e V i s pro por t ion a l to F m s ' r m s c o s V-
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420
Figure 17-22 Hal l - type wa t tmete r formeasur ing the power suppl i ed to a
load imped ance Z . Th e cur rent to the
load passes through the coi l of an
e l e c t r o m a g n e t E that appl ies a
magnet ic f ie ld to the Hal l generator
HG. A small fract ion of the vol tag e
across the load is appl ied to HG a n d
the vol t age V i s pro por t ion a l to the
power d i s s ipa ted in Z . See P rob .
17-12.
17-13 TRANSIENT SUPPRESSOR FOR INDUCTOR
In P ro b . 12-13 we saw tha t , if one d i sconn ec t s a l a rge cur ren t -ca r ry in g in duc tor f rom
a d i rec t -cur rent source , h igh t rans ient vol t ages deve lop and the induc tor can be
des t royed. We a l so saw tha t the overvol t ages can be e l imina ted by connec t ing a
diode in pa ra l l e l wi th the induc tor .
I f the indu c tor i s fed by an a l t e rna t ing- cur ren t so urce , one can use the p ro tec t ive
c i rcui t shown in F igure 17-23. H e r e , R i s the res is tance and L is the inductance of
the induc tor , and C = L/R2
.
Show th a t the reac tanc e be tween A and B is ze ro . Th en th e c i rcu i t be tween A
and B mu s t ac t l ike a pure res i s tance , and the vol t age ac ross the induc tor m us t becom e
zero when the switch is opened.
—. AI O ^ C ^ - f .
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Problems 421
Solution:
(R + jcoDVR + ( l / j o C ) ]
2R + jwL + (1/jtoC) '
R2 + (L /C) + j[toLR - (R/coC)~]
2R + j[wL - (1/coC)] '
{R2 + (L/C) + j[toLR - (P/coC )]j {2P - j[_a>L - (1/ toC)]}
4 P2 + [wL - (1 /coC )] 2
Th e num era tor of th i s expres s ion is ze ro , Hence the reac tance be tw een A and B is
zero at a l l frequencies!
17-14E SERIES RESONA NCE
W ha t i s the locus of the impe danc e Z in the complex p lane when the f requency
app l ied to the series-re son ant c i rcu i t of Fig. 17-2 chan ges from zer o to infini ty?
17-15 PARALLEL RESONANCE
An induc tor of induc tance L and res i s t ance R i s connected in paral le l wi th acapac i tor C .
Se t R = 10 oh ms , L = 1.0 mil l ihe nry, and C = 1.0 m icrofa rad.
a ) P lo t curves of the rea l pa r t , t he imaginary pa r t , t he magni tude , and the phase
of the impedance as funct ions of the frequency, from zero to 10 ki lohertz .
N o t e t h a t , w h e n Z i s m a x i m u m , w2
LC w 1.
b) A t wh a t f requency i s the reac tance z e ro?
c) F ind the impedance a t 8 k i loher t z .
d) At that frequency, the ci rcui t has the same impedance as a capaci tor C a n d
a res is tor P ' in series . W ha t are the values of P ' and of C ?
This is a case of parallel resonance, L and C being in paral le l .
T h e bandwidth i s defined as the difference between the frequencies where the
magni tude of the impedance i s 1/21
'2
, or 0.707 t imes i ts maximum value.T h e Q (for Qu al i ty ) of the ci rcui t is defined to be the reson anc e frequency, divided
b y t h e b a n d w i d t h . T h e h i g h e r t h e Q, t he sharper the resonance peak . I t can be shown
The reac tan ce is the imagina ry pa r t of Z :
X =[ P
2 + (L/ C)] [ca L - (1/coC)] + 2 P [ O J L P - (R/toC)]
4R 2
+ [ c o L - ( 1 / w C ) ] 2
(4)
t h a t
T h e Q of the above circui t is 3.2.
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422 Alternating Currents: II
Para l l e l re sonance has innumerable appl i ca t ions . Here i s one example . For
high-speed , h igh-prec i s ion coi l winding , i t i s imp or ta nt to mo ni to r the wi re d iam ete r .
For s evera l reasons , mechanica l devices a re unacceptable . The moni tor ing can be done
in the fol lowing way: a smal l coi l of a few turns is wound around a quartz tube a few
mil l im eters in diam eter thro ug h which the wire passes . Th e coi l of cou rse has an in
du ctan ce. I t a lso has a capa ci ta nc e; as in any coi l , there is a vol tag e difference, and
hence a capac i t ance , be tween turns . Thi s stray capacitance i s increased by the presence
of the wire . So the coi l acts as a paral le l -resonant c i rcui t , whose resonance frequency
depends on the diameter of the wire . If the wire diameter increases , the resonant
frequency decreases and vice versa. Th e curve of Z as a funct ion of frequency is l ike that
ca lcula ted under pa r t a above , the peak moving to the right when the wi re d iamete r
decreases , and to the left when the d iamete r inc reases .
Th e coil is des ign ed so that i ts reso nan ce frequen cy is a lwa ys s l ight ly abo ve 300
mega her tz , and an osc i l l a tor opera t in g a t 300 meg aher tz i s conn ec ted to the coi l thro ugh
a res is tance. If the wire diameter increases , the resonance peak moves to the left , Z
increases , and the vol tage on the coi l increases . Measurements are accurate within 0.5" , , .
I7-16E STAR-DELTA TRANSFO RMATION
Transform the del ta of Fig. 17-5a into a s tar a t one megahertz .
I7-17E STAR-DELTA TRANSFO RMATION
Transform the s tar of Fig. 17-24 into a del ta a t one ki lohertz .
Figure 17-24 See Prob. 17-17.
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Problems 423
17-18 T-CIRCUITS
Figu res 17-25a an d b show two ty pes of T-circui ts . the twin (or paral le l ) T and the
br idged T . Th e com po nen t s a re s e lec ted in such a way tha t the outp ut vol t age V 0
becomes zero at a specif ied frequency. Circui ts that have this characteris t ic are
called notch filters. T-circui ts are used in osci l la tors and in tuned amplif iers . Some
are a l so used for measur ing impedances , in much the same way as the impedance
br idges of P r ob . 17-6 . No te tha t the input an d the ou tpu t have a com m on te rmina l ,
which i s grou nde d. Th i s is a grea t adv antag e , s ince mos t osc i l l a tors , vol tmete rs , and
osc i l loscopes have one t e rmina l grounded.
_ n n n _
.crrx. -o o »
_ p n n .
JTCX . _ n n n _
4 - 4(a )
-O O(b)
o-m—*
_ n n r i _
Yi_ n m _
Yi
J T Y l _ n m _
Yi >4
' 2 ^
- O O -
Figure 17-25 (a) Tw in, o r pa ralle l, T, an d {b) shunted T c i rcu i t s ; (c) a n d ill).
equivalent c i rcui ts . See Prob. 17-18.
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424 Alternating Currents: II
To find the conditions under which the output voltage of the twin T is zero, we
transform the star circuits into deltas, giving the circuit of Fig. 17-25c, and we set
F, + Y2
= 0.
For the bridged T. we transform the delta formed by the three impedances at
the top into a star. The bridged T then becomes a simpleT as in Fig. 17-25d. and the
out put volt age is zero when Z , + Z 2 0.
Figure 17-26 shows one common type of twin T.
4
0 * - V \ A AR
V: 2C
Figure 17-26 See Prob. 17-18. O
R
R/2 K
4 *
a) Fin d the F, and Y2 of Fig. 17-25c for the twin T of Fig. 17-26, setting RcoC = x,and show that V„ = 0 when x = 1.
Solution: In the circuit of Fig. 17-25c, let us call F, the admit tan ce ob tai ned by
transforming the R-R-2C star of Fig. 17-26. Fhen
(1/PH1/R) 1_
(1/P) + (1/P) + 2jwC ~ 2R + 2jwCR2
Similarly, for the C-C-R/2 star.
_ J'oCjcoC _ jviCjcoCR
2
~ jcoC + jo)C + (2jR) ~ 2jtoCR + 2'
Fh en V„ = 0 when
1 icoCjx 1 — x2
Fi + Y2
= + — — = = 0. (3)2P(1 + ./x) 2(./'x + 1) 2P (1 +jx)
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Problems 425
or when
1 - x1
= 0, (4)
x = 1. (5)
b) Find V„/V, as a function of x.
Assume that the load impedance connected to the right-hand terminals is in
finite.
Solution: Considering first the star C-C-R/2, K3 of Fig. 17-25c is given by
= jcoC(2/R)=
jcoC
3
jioC + (2/R) + jcoC \+jx
For the star R-R-2C.
_ (l/R)2jo)C = jcoC
4
(1/R) + 2iwC + (\/R) 1 + ; x'
Now
1
v„ y3
+ r4
n + Ji
v, i _ j _ y, + y2
+ y3
y> + y2
y3
+ y4
where
2i?(l
Thus
(1 - .x
2
)/(2iv) 1 - A'
2
V, [(1 - A -2
) /2 R ] 4- 2jcoC 1 — x + 4jx
(8)
(1 - A2
) . (9)
y 3 + n = no)1 + /A
(I D
c) Draw a curve of |V 0 / V , | as a function of A between A = 0.1 and x = 10. Use
a logarithmic scale for x.
Solution: See Fig. 17-27.
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42 6
IORC
Figure 17-27 See Prob. 17-18.
17-19 BRIDGED T
Show tha t , for the br idged T of Fig. 17-28, V 0 = 0 when
co2
C2
rR = 1, o)2
LC = 2.
w w — T r a f f i c
Figure 17-28 See Prob. 17-19.
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Problems 427
R
Li o ! M g ) L 2
3 3
Figure 17-29 See Prob. 17-20.
17-20 MUTUAL INDUCTANCE
Find the rms va lues of the cur rent s F and l 2 in the inductances of the ci rcui t of
Fig. 17-29 when
V = 10 vol ts , R = 5 o h m s , R 1 = R 2 = 10 ohms ,
L j = L 2 = 1 henry , M = +0 .9 henry , / = 1 k i loher t z .
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CHAPTER 18
ALTERNATING CURRENTS: III
Power Transfer and Transformers
18.1 POWER DISSIPATION
18.1.1 Examples: A Light Bulb, A Radio-Frequency Antenna,
an Electric Motor
18.2 THE POWER FACTOR
18.2.1 Examples: Resistors, Capacitors, and Inductors
18.3 POWER TRANSFER FROM SOURCE TO LOAD
18.3.1 Example: Whip Antenna
18.4 TRANSFORMERS
18.5 MAGNETIC-CORE TRANSFORMERS
18.5.1 The Ideal Transformer
18.5.2 The Ratio V , / V 2
18.5.3 The Ratio LJL 2
18.5.4 The Input Impedance Z i n
18.5.5 The Ratio \J\2
18.6 POWER TRANSFER FROM SOURCE TO LOAD
THROUGH A TRANSFORMER
18.7 SUMMARY
PROBLEMS
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In th i s l as t chap te r on a l t e rna t ing cur r en t s we sha l l be concerned wi th thet ransfer of power f rom a source to a load, f i r s t d i rect ly , and then through
a t r ans former .
Tra nsm is s io n l ines tha t ca r ry l a rge amo un t s o f pow er over l a rge d is
tan ces m ust o pe rat e , for rea son s of ef ficiency, a t vol ta ges of a few hu nd re d
tho usa nd v o l t s ; som e even ope ra te a t c lose to one mi l l ion vo l t s . Bu t insu la t ion
p r o b l e m s p r e v e n t g e n e r a t o r s i n p o w e r p l a n t s f r o m b e i n g o p e r a t e d a b o v e
about 20 ,000 vo l t s . At the o ther end o f the l ines , consumers r equ i r e e l ec t r i c
power at vol tages ranging f rom a f ract ion of a vol t for solder ing guns to 6000
vol t s fo r e l ec t r i c motor s o f a f ew hundred k i lowat t s .
I t i s the funct ion of t ran sfo rm ers t o chan ge the rat io V/I with li t t lelos s o f power . In th i s way , e l ec t r i c power can be p roduced , t r ansmi t t ed , and
u t i l ized a t the mo s t conven ien t vo l t age s , wi th in l imi t s. T ran s form ers a r e
also used for impedance matching, as we shal l see in Sec. 18.6.
Trans formers can ac t on ly on a l t e rna t ing cur r en t s .
18.1 POW ER DISSIPATION
Let us cons ider the p roces ses whereby the e l ec t r i c and magne t i c energy
f lowing in a ci rcui t is diss ip ated . Unt i l now , we have cons id ere d only the
I2
R loss in a res is tan ce R car ry in g a cur r e n t / . Of cour se , ma ny o th er p roces ses
are pos s ib le . For example , e l ec t r i c motor s p roduce mechan ica l energy ,
s to rage ba t t e r i es can p roduce chemica l energy , l i gh t bu lbs and an tennas
p r o d u c e e l e c t r o m a g n e t i c w a v e s , l o u d s p e a k e r s p r o d u c e a c o u s t i c w a v e s , a n d
so for th .
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430 Alternating Currents: III
Any on e of these devices, when connected to a source, draws a current,
an d the ratio V/l is its impedance. How is this impedance related to the
power that is withdrawn from the circuit?
Let us think of a current I flowing through an impedance R + jX.
If the applied voltage V is V0
ex p jcot,
V
R + jX R2
+ X2 s
"J
"' |ZT= 7 7 — 7 7 ; = 7 7 2 — w ( * - J*) =wr2
(R- m (18-1)
V0
exp jcot
= j ^ p lR
+x e x
P (-W2)] (18-2)
and, since I is th e real part of I,
I = [R cos cot + X cos [cot - 7t/2)] . (18-3)
The instantaneous power absorbed by the impedance is the product
of this quantity by V0
cos cot:
V2
^ins t = T^k [R cos 2 cot + X co s [cot - 7t /2) cos cot]. (18-4)
Now the average of co s 2 cot over one cycle is \, while the average value of
the second term between the brackets is zero. Thus the average power
absorbed by the impedance R + jX is
\_K _ vL2\Z\-
av - ^ 7^12 & - T^iTR
- lfmsR
- (18-5)
Note that P a v is / r
2
m s R, as for direct currents, but that it is V2
ms/R
only if X = 0.
If the impedance is that of some device that draws power from a
circuit, then I2
m!iR must account for all the electric power that is absorbed.
EXAMPLES: A LIGHT BULB.
AN ELECTRIC MOTOR
A RADIO-FREQUENCY ANTENNA,
In a light bulb, I2
m!iR is equal to the heat energy that is lost by conduction and
convection, plus th e electromagnetic energy that is radiated in one second.
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18.2 The Power Factor 431
A radio-frequency antenna is usually tuned to have a purely resistive impedance
P, which is called it s radiation resistance. If the current fed to th e antenna is /, then
I2
miR is the radiated power.In an alternating-current electric motor, I
2
msR includes: Joule losses (Sec. 5.2.1)
in th e copper conductors; hysteresis losses (Sec. 14.9) in the iron; eddy-current
losses (Sec. 11.3) both in the copper and in the iron; friction losses in the bearings;
windage losses in the air; and, finally, useful mechanical power. The resistance R
of an electric motor in operation is thus the su m of many terms. It is much larger
than the resistance measured with an ohmmeter when the motor is stopped. See
Prob. 18-1.
18.2 THE POWER FACTOR
The rat io R/\Z\ of an im ped anc e is called its power factor and is usually
expressed as a perce ntage . If the impe dan ce Z is repre sente d in the comp lex
plane, as in Fig. 18 - 1 , the po wer factor is cos cp, tp being the phase angle
(Sec. 1 7 . 1 ) .
Whe neve r large am ou nt s of power must be trans mitte d from a source
to a load ov er a line of app rec iab le resistance , care must be taken to adjust
the powe r factor close to unity. Th e reason for this is tha t the c om po ne nt
of I that is in qua dr at ur e with V pro du ces no useful po wer in the load, but
nonetheless gives rise to a power loss and to a voltage drop in the line
connecting the source to the load.
/.v
R
Figure 18-1 Impedance Z represented in the complex plane. F h e cosine of the
phase angle <l> is called the power factor.
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432 Alternating Currents: HI
Elec t r i c motor s a r e induc t ive loads , and . in l a rge ins t a l l a t ions , the
pow er f ac to r i s cor r ec te d by con nec t in g ca pac i to r s ac ro s s the line , as c lose
as poss ible to the motors . See Probs . 18-2 and 18-3.
EXAMPLES: RESISTORS, CAPACITORS, AND INDUCTORS
Th e po wer factor of a res is tor is 100%, and tha t of a capa ci to r is 0%.
Real inductors have both res is tance and inductance.* If R = coL, t h e n <p
45 degrees , and the power factor is 70.7%.
18.3 POWER TRANSFER FROM SOURCE TO LOAD
In Fig. 18-2. we have re place d th e sourc e by an ideal vol ta ge so urc e V, in
ser i es wi th an impedance R s + jX s. This is the general form of Thevenin 's
theorem (Sec. 5 .12) .
Source Load
Figure 18-2 Source feeding an impedance Z L = R L + jX,.
* If the frequency is qu i te high th ey be com e capac i t ive. See Pro b. 17-2.
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18.3 Powe r Transfer from Source to Load 433
The average power dissipated in the load is
V2
Pr = ILSP-L = 5 Ri- (18-6)
Under what conditions is PL maximum, for a given U r m s ? First,
XL+ X
s should be zero:
XL
= -Xs. (18-7)
In other words, the reactance of the load should be of the same magnitude
as the reactance of the source, but of the opposite sign.
If tha t co ndi tio n is satisfied, or if the rea ctanc es are bo th zero,
P =R
*<V
™ ,18-8)
(RL + Rs)2
We can now find the optimum value of RL
by setting
dPL
(RL + Ks)
2
~ RL x 2(RL
+ Rs)
2
dR~r Krms = a (18"9)
Then
RL
= Rs. (18-10)
Thus, for maximum power transfer to the load,
XL
= -Xs, R
L= R
s. (18-11)
The load impedance is then said to be matched to that of the source.
The maximum power that can be dissipated in the load, for a given
open-circuit voltage K r m s is
p
- - - ( M * • - % »8
-1 2 1
Also, from Eqs . 18-8 an d 18-12.
P L RLV2
ms 4RS MRJRs)
PL**, (RL + Rs)2 v
2
msl(R,JRs) + i ]
2
(18-13)
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43 4
J i ' i i i i i j i i i i i i i
0.1 10
R, R,
Figure 18-3 A: PJP Lmd% as a function of RJR S. The power d i s s ipa ted in the loadis m ax im um when t he load res is tan ce is equa l to the res is tan ce of the sou rce.
B: Efficiency as a function of R/R x. The eff ic iency tends to uni ty as R -> x . It is
only 50% a t ma xim um pow er t rans fe r .
Figure 18-3 shows P L ! P L M ^ as a function of R L/R S. I t wi l l be observed that
t h e c o n d i t i o n R L = R^ for max im um pow er t rans fe r i s no t c ri t i ca l . For
example , wi th R L = 2 K S , t he power P L diss ipate d in the loa d is s t i ll 89%
of /% m a x -
T h e efficiency may be denned as the r a t io PL/(PL + P S), w h e r e P S
is the pow er d i s s ipa ted in the so urc e :
PL ILSRL RL ( 1 8 . 1 4 )
PL + PS IU PL + RS) RL + R,'
= RJ R S
(RJRS) + 1 '
(18-15)
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18.3 Pow er Transfer from Source to Load 435
Figure 18-3 also shows how the efficiency varies with the ratio RL/RS.
Th e efficiency is equ al to unit y whe n RL/RS -» oo, but the n the pow er
dissipated in the load tends to zero. For RL = Rs, the power dissipated in
the load is ma xi mu m a nd t he efficiency is only 50 %.
18.3.1 EXAMPLE: WHIP ANTENNA
A radio- f requency t ra nsm i t t e r has an output res i s t ance of 50 o h m s . It is to be c o n
nec ted to the w h i p - a n t e n n a of Fig. 18-4, which has a rad ia t ion res i s t ance of 3 7 o h m s .
What wi l l be the efficiency of power t rans fe r?
In this case,
Rs = 50, Xs = 0, RL = 37, XL = 0. (18-16)
Fhe efficiency will be 37/ (50 + 37), or a b o u t 4 3 % . T h i s is sa t i s fac tory , compared
with th e o p t i m u m v a l u e of 5 0 % . f
Figure 18-4 Q u a r t e r - w a v e w h i p a n t e n n a A m o u n t e d on a m e t a l car t o p . The w h i p
is one -qu ar te r wave length long an d is s u p p o r t e d by a feed- through insula tor F. It
is fed by a t ransmi t t e r s i tua ted at the b a c k of th e ca r t h r o u g h a coaxial l ine C. The
w h i p , t oge ther wi th its i m a g e / ( P r o b . 3-15) forms a ha l f -wave antenna . Charges
flowing in the antenna form l ines of force that detach themselves and escape in to
space at the veloci ty of l ight . This type of a n t e n n a is used at frequencies ranging
from 150 to 450 m e g a h e r t z .
1
O ne could achieve 50% effic iency by us ing a matching c i rcu i t . See, for e x a m p l e , The
Radio Amateur s Handbook, Am erican Ra dio Re lay League , New ingto n , Conn . , U .S .A.
(publ ished every other year) .
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436 Alternating Currents: III
18.4 TRANSFORMERS
A t r ans fo rme r i s a mutu a l indu c tor th a t i s used to chan ge the vo l t age l eve l
of an elect r ic cu rren t , as exp laine d brief ly in the int ro du ct i on t o this ch apt er .
T h e m u t u a l i n d u c t o r c a n h a v e e i t h e r a m a g n e t i c o r a n o n - m a g n e t i c
co re . Th e fo rmer type i s co mm on ly r e fe rr ed to as a magnetic-core transformer,
and the la t ter as an air-core transformer. Air -core t r ans formers a r e used a t
h igh f r equenc ies where eddy-cur ren t and hys te res i s los ses in a magne t i c
co re wo uld be excessive (Sees. 11.3 an d 14.9).
The equivalent s tar c i rcui t (Sec. 17.5) wi l l g ive us some general resul ts
tha t a r e exac t fo r a i r co res , and on ly approx imate fo r magne t i c cores .
F igure 18-5a shows a t r ans former inser t ed be tween a source supp ly ing
a vo l t age V, , and a load impedance Z L . T h e i m p e d a n c e R x + jcoLx of the
pr imary wind ing i s Z u a n d t h e i m p e d a n c e R 2 + jcoL 2 of the s econdary
wind ing p lus Z L , is Z 2 .
T h e m u t u a l i n d u c t a n c e i s
M = fc(L1L 2)1/2. (18-17)
Since the ci rcui t of Fig. 18-5b is equ iva len t to th at of Fig. 18-5a,
f rom Sec. 17.5, the im pe da nc e "see n" by the sourc e, or the input impedance
Z i n of the t r ans former i s
v v j.i AA i + JcoM)(-ja>M) M f i i mZ i n = Z , + JcoM + , — — , (18-18)Z 2 + JCOM — JCOM
co2
M2
= Z 1 + ^ — . (18-19)
Th e inpu t impe dan ce i s there fore Z u plus the t e rm co 2M 2jZ 2, which is
cal led the reflected impedance of the s e cond ary .
T h e c u r r e n t 11 in the pr imary is of course
l i = V , / Z i n . (18-20)
Th e cur re n t in the s econd ary can be found by app ly ing Ki rchof f ' s
vol t age law to th e ci rcui t of Fig. 18-5b, us ing th e value of I given in Eqs . 18-19
and 18-20 :
(18-21)
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43 7
Figure 18-5 [a) Trans former fed by a source and feeding a load impedance Z L.
(b ) Equiva lent c i rcu i t .
18.5 MAGNETIC-CORE TRANSFORMERS
Ma gne t i c - co re t r ans form ers as in F ig . 18-6 a re used a t fr equenc ies r ang in g
f rom a f ew her t z to abou t one megaher t z .
A magne t i c - core t r ans former has two es sen t i a l f ea tu res . F i r s t , f o r a
g iven core c ros s - sec t ion , the magne t i c f lux per ampere- tu rn in the p r imary
i s l a rger than wi th an a i r core by s evera l o rder s o f magni tude . Second , the
ma gne t i c flux th ro ug h the s eco nda ry i s near ly equa l to tha t th ro ug h the
p r i m a r y .
T h u s , t he magn e t i c core per mi t s the des ign of t r ans fo rme rs tha t a r e
smal l e r and have f ewer tu rns , bo th in the p r imary and in the s econdary .
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43 8
S
(a) (b)
Figure 18-6 («) Co m m on type of magn e t i c -core t rans fo rmer . Th e core has the
shape of a f igure e ight . The pr imary and secondary windings P a n d S a r e w o u n d
on the center leg. The core is made of two types of laminat ion, one having the
shape of an £, and the other the shape of an / , placed one next to the other. The
lamina t ions a re as sembled a f t e r the co i l has been wound, (b) S c h e m a t i c d i a g r a m
of a mag ne t i c -core t rans former .
Also , with a mag ne t ic core, there is very l it t le s t ray m ag ne t ic f ie ld . F or
example , one can opera te two such t r ans formers s ide by s ide wi th neg l ig ib le
i n t e r a c t i o n .Cores made o f i ron a l loys a r e usua l ly fo rmed of para l l e l i nsu la t ed
sheets cal led laminations. This r educes the los ses due to the eddy cur ren t s
induced by the chang ing magne t i c f i e ld . The l amina t ions have a th i cknes s
of a f ract ion of a mil l im eter . At aud io f requenc ies th e al loy is som et im es in
the form of a fine pow der m old ed in a bon d in g agen t . O ne the n has a powdered
iron core .
Ferrites are used a t aud io f r equenc ies and above . These a re ce ramic-
type mate r i a l s tha t a r e molded f rom ox ides o f i ron and var ious o ther meta l s .
The i r main a dv an t ag e i s tha t the i r cond uc t iv i t i es a r e low, o f the o rde r o f
1 S iemens per mete r , which i s ab ou t 1 0 ~
6
t imes tha t o f t r ans form er i ron .The eddy current losses in fer r i tes are thus low.
Th e ana lys i s o f ma gne t i c - core t r an s form ers is r end ered d i f fi cu lt by
severa l f ac to rs . F i r s t, t he r e l a t ion sh ip be twe en B and H in f e r rom agne t i c
m ate r ials i s no t l ine ar (Sec. 14.9). F or exa mp le, if the f lux l ink age in a ci rcui t
is A when i t car r ies a current I, t hen the s e l f - induc tance L is A// , as in
Sec. 12.3. If the re are no m ag ne t ic m ate r ial s in the f ie ld . A is pr op or t io na l
to / and A/ I depends so le ly on the geomet ry o f the c i r cu i t . However , i n
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18.5 Magnetic-Core Transformers 439
t he p resence o f ma gne t i c mate r i a l s , A is no t p rop or t i on a l t o / and the self-
i n d u c t a n c e L r equ i r ed fo r the ca lcu la t ions can on ly be approx imate .
M o r e o v e r , t h e l o s s e s i n a m a g n e t i c - c o r e t r a n s f o r m e r a r e c o m p l e x :the re are ed dy -cu rre nt losses (Sec. 11.3) bo th in the i ron core an d in the c op pe r
windings , hys teres is losses in the i ron core (Sec. 14.9) , and Joule losses
resul t ing f rom the currents f lowing in the windings . Al l these losses can be
expres sed as an I2
R loss in the pr imary, but , for a given t ransformer , R d e
pends on the vo l t age app l i ed to the p r imary , on the cur r en t d rawn f rom the
secondary , and on the f r equency .
18.5.1 THE IDEAL TRANSFORMER
W e sha ll mak e the fo llowing th ree as sum pt io ns , ( a) Th ere a r e no Jou le o r
eddy-current losses , (b) The hys teres is loop for the core is a s t raight l ine
th ro ug h the o r ig in . Th en B i s p ro po r t io na l to H and there a r e no hys te res i s
losses ei ther , (c) Th ere is zer o leak age f lux. The n th e coeff icient of co upl ing
k is equ a l to un i ty , an d the f lux th rou gh the p r im ary is equ a l to tha t th r ou gh
t h e s e c o n d a r y .
As a consequence o f the f i r s t two as sumpt ions , the t r ans former i s
lossless, and its efficiency is 100%.
A f ict i t ious t ransf orm er h avi ng these cha racte r is t ics i s cal led an ideal
transformer.
T h e a s s u m p t i o n t h a t B is p ro po r t io na l to H, and henc e to / , is mos t
undes i rable , but di f f icul t to avoid.
As to the eff iciency, i t is c lose to 100% for large ind us tr i al t ran sform ers ,
bu t on ly ab ou t 70 o r 80% for a smal l pow er t r ans form er supp ly ing , s ay ,
10 wat t s .
F ina l ly , we sha l l cons ider the usua l s i tua t ion , in which the load
imp eda nce i s r ea l an d mu ch sm al l e r than the r eac ta nce o f the s e cond ary
w i n d i n g :
Z L = R L « o)L 2. (18-22)
18.5.2 TH E RATIO \IJ\I2
With these as sumpt ions , the vo l t age app l i ed to the p r imary i s
d<P
V , = N , ~ = N , jco4>, (18-23)
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440 Alternating Currents: III
and the vo l t age acros s the s econdary i s
(10V 2 = N 2-- = N 2jc»& = \2R L, (18-24)
at
w h e r e TV, a n d N 2 are the num be rs o f tu r ns in the p r im ary a nd in the
secondary wind ings , r espec t ive ly , and where 4> i s the magnet ic f lux in the
c o r e :
0 = 0 , + $ 2 , (18-25)
= ^ + ^ (18-2 6,
w h e r e 31 i s the reluctance of the core, f rom Sec. 15.1.
T h u s
V , / V 2 = NJN 2 (18-27)
and . for an ideal t ran sfor m er and for a given V 2 is inde pen den t o f the
load cur r en t . Al so , V 2 i s e i th er in ph ase w i th V u o r n r ad ians ou t o f phase .
Of cour se , the phase o f V 2 can be changed by n radians at wi l l by inter
chan g ing the wi res com ing ou t o f the s econd ary wind ing .
N o t e t h a t V t i s equal to iV, jt»4>. Thus , fo r a g iven app l i ed vo l t age ,
the total f lux 4>, + <J> 2 is a p p r o x i m a t e l y i n d e p e n d e n t o f t h e c u r r e n t d r a w n
f rom the s econdary .
Also , O is B t imes the c ros s - sec t ion o f the core . S ince the maximum
poss ible value for B depends on the core mate r i a l , Eq . 18-23 shows tha t an
increase in f requency permits the use of a core having a smal ler cross-sect ion.
18.5.3 THE RATIO LU'L 2
N o t e t h a t
L , = . 1 . I, = Nj fp j / | j = Nj/.M. (18-28)
S imi la r ly ,
l 2 = Ni/a, (18-29)
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18.5 Magnetic-Core Transformers 441
co L^ L 2 . / jo)L= > L , ( 1 - „ ; , ) . ( 18 -3 3)
(18-34)
RL + jtoL 2 R L + jcoL2
RJoLi
R L + JOJL 2
'1
R L = ( ^ A2 R L - [R L«COL 2) (18-35)
L2 \N
18.5.5 THE RATIO l , / l 2
From Sec. 18.4,
1 , = ^ = ^ , , (18-36)2m R LJo)L {
l 2 =jC
°M
2 2 V 1 = ^ . V „ (18-37)
j w M
V ,, (18-38)R LjojL i
so that
LJL 2 = Nl/Nl (18-30)
18.5.4 THE INPUT IMPEDANCE Z i n
Since the coupl ing coeff icient k is eq ua l to unity , the n, from Sec. 12.4,
M = ( L , L 2 ) 1 / 2 . (18-31)
Also ,
Z{ =jcoLu Z 2 = R, + jwL 2, (18-32)
w h e r e R L is the load r es i s t ance con nec ted be twe en the t e rmin a l s of the
s e c o n d a r y .
T h u s , f rom Eq. 18-19,
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442 Alternating Currents: III
1 2 jcoM
Disregarding the sign,
Ij/lj = L2IM = ( L
2/ L i )
1 / 2
= N2/Nl. (RL
« wL2)
18.6 POWER TRANSFER FROM SOURCE TO
LOAD THROUGH A TRANSFORMER
As we saw in Sec. 18.3, the po wer di ssipat ed in a loa d is ma xi mu m when its
resistance R, is equa l to the inter nal re sistance of the sour ce Rv, and when
XL = — Xs. As a rule, the reactances are zero. If it is impossible to vary
either RL or R
s, then one can still achieve optimum power transfer by in
serting a transformer between source and load as in Fig. 18-7a. Let us see
how this comes about.
Let us assume that the transformer has a magnetic core, that the ap
proximations used in Sec. 18.5.1 are satisfactory, and that Xs = X, = 0.
The n, from Eq. 18-35, the tra nsfo rmer has an inpu t imp eda nce of (TV i/N 2)2
RL
ohms. In other words, the current and the power supplied by the source
are precisely the sam e as if the transf orme r and its load r esistanc e RL
were
replaced by the resistance (Ar
1/A
7
2)
2
i7
\L, as in Fig. 18-7b. Remember that,
with our approximation, the transformer has an efficiency of 100%.
(18-39)
(18-40)
Figure 18-7 (a) Source feeding a load resistance RLthrough a magnetic-core
t r a n s f o r m e r . (/>) Equivalent circuit.
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18.7 Summary 443
Then, with a transformer as in Fig. 18-7a, the power transfer is optimum
when
N1
.
RS
= ^ RL
, (18-41)
or when the turns ratio is
NJN2
= (RJRL)
12
. (18-42)
Wh en thi s con dit ion is satisfied, the power dissip ation in the loa d is max i
mu m, bu t the efficiency is aga in on ly 5 0 ° o . The trans for mer is then said
to be used for impedance matching.
18.7 SUMMARY
Any device that transforms the energy flowing in an electric circuit into
some other form has a resistance R such that I2
R accounts for all the energy
withdrawn from the circuit.
The power factor of the load is the ratio R/\Z\, or the cosine of the
phase of I with respect to V. and is usually expressed as a percentage.
For maximum power dissipation in a load, the following two con
ditions should be satisfied:
XL
= -Xs, R
L= R
s. (18-11)
where the subscripts L and s refer, respectively, to the load and to the source.
The impedances are then said to be matched, and the efficiency is
5 0 % . The efficiency is defined as the pow er delivered to the load, divided by
the power produced by the source.
The input impedance of a transformer is equal to the impedanc e Z ,of the primary winding, plus the reflected impedance of the second ary circuit,
which is w2
M2
/Z2, where Z
2is the imp edan ce of the second ary windin g,
plus the load impedance ZL.
For an ideal magnetic-core transformer,
V2/ V, = l , / l
2= N
2/N
l
(18-27, 18-40)
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444 Alternating Currents: III
and the input impedance is
An = ^ R L = ~ R L . (18-35)
L2i \
2
A trans form er can serve to match the impe dan ce of a load to tha t of a source,
by making
NJN2 = (RS/RL)12
. (18-42)
The efficiency und er these cond itio ns is again 50%.
PROBLEMS
18-1E DIRECT-CURRENT MOTORS
Let us see how a d i r e c t - c u r r e n t m o t o r o p e r a t e s .
F igure 18-8 shows a highly schemat ic d iagram of a series motor. The t e rm "se r i es "
refers to the fact that th e f ie ld winding and the rotat ing coi l , cal led th e armature, are
i A Xi
Figure 18-8 S c h e m a t i c d i a g r a m of a di rec t -cur rent s e r i es motor . The a r m a t u r e A
r o t a t e s a b o u t th e vert ical axis AX. It is c o n n e c t e d in series with th e coi l s produc ing
the magnet ic f ie ld B t h r o u g h a c o m m u t a t o r C. The c o m m u t a t o r is a spl i t copper
r ing . Contac t to the r ing is made wi th ca rbon brushes . The di rec t ion of t he cur rent
t h r o u g h th e a r m a t u r e is inverted twice every turn. The m a g n e t i c c o r e is not
s h o w n . See Prob. 18-1 .
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44 5
Figure 18-9 Equiva lent c i rcu i t V,
R of the electr ic motor of Fig. 18-8.
in series. In a shunt motor they are in paral le l . Compound motors obtain part of theirf ie ld with a series winding, and part wi th a shunt winding.
W hen the m oto r i s s topp ed, i t ac t s a s a resi s t ance . Fh e res i s t ance of the a rm atu re
is mu ch sm aller tha n that of the f ie ld wind ings .
When the motor i s in opera t ion , the mot ion of the a rmature in the magne t i c
fie ld generates a counterelectrom otive force V that opposes the f low of current as in
Fig. 18-9. Here R i s the res i s t ance as soc ia ted wi th the va r ious los ses enu mer a ted in
Sec. 18.1.1.
a) Use the subs t i tu t ion theorem to show tha t the mechanica l power i s equa l
t o IV .
b) How would you define the eff ic iency?
c) If a direct-cu rren t series m ot or is con necte d to a sou rce of po we r an d run
wi thout a load , i t wi l l ga in more and more speed and, ve ry probably , the a rmature wi l lburs t . A d i rec t -cur rent series motor that is not connected direct ly to i ts load is therefore
dangerous.
Why is this?
R e m e m b e r t h a t V i s propor t iona l to the speed of the motor , mul t ip l i ed by the
m a g n e t i c i n d u c t i o n a t t h e a r m a t u r e w i n d i n g .
d) What happens when the mechanica l power i s inc reased?
18-2 POWER-FACTOR CORRECTION
A load has a power factor of 65° „ and draws a current of 100 amperes at 600 vol ts .
a ) Ca lcula te the magni tude of Z , i t s phase angle , and i t s rea l and imaginary
pa r t s .
b ) C a l c u l a t e th e in - p h a s e a n d q u a d r a t u r e c o m p o n e n t s of t h e c u r r e n t .c) What s ize capaci tor should be placed in paral le l wi th the load to cancel the
reac t ive cur rent a t 60 he r t z?
18-3 POWER-FACTOR CORRECTION W ITH FLUORESCENT LAMPS
A fluorescent lamp cons i s t s of an evacua ted tub e conta in ing merc ury vap or an d
coa ted on the ins ide with a f luorescent pow der. A disch arge is es tabl ish ed ins ide the
t ube , between electrodes s i tuated at each end. The discharge emits most of i ts energy
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446 Alternating Currents: III
a t 253.7 nanomete rs , in the u l t rav io le t . The f luorescent coa t ing absorbs th i s rad ia t ion
and re-emits vis ible l ight .
The discharge wil l operate correct ly only when connected in series with an
impedance . A res i s tor could be used , but would d i s s ipa te an excess ive amount of
energy, so i t i s the custom to use a series inductor. There exis t many types of c i rcui t .
A par t icu lar f luorescent f ixture op era t ing at 120 vol ts has a pow er d iss ipa t ion
of 80 wat ts . I ts power factor is 50%.
a) F ind the reac t ive cur rent .
b) Wh at is the s ize of the cap aci t or con nec ted in paral le l wi th the disch arge
tube and i ts inductor that wil l make the power factor equal to 100", ,?
18-4 ENERGY TRANSFER TO A LOAD
Figu re 18-10 sho ws a ci rcui t that ha s been used to me asu re the energy abs orb ed
by a load Z when the source V produces a pul se of cur rent . The cur rent s / ' and I" are
negl ig ib le compared to / . and the vol t age drop ac ross r i s negl ig ib le compared to V.As we shal l see, this energy is proport ional to the area enclosed by the curve on the osci l
loscope screen. The method is val id is val id even i f Z i s non- l inea r .
We sha l l demons t ra te th i s for one cyc le of an a l t e rna t ing cur rent , bu t the same
appl ies to an individu al curre nt pulse of any sh ape .
Q i
v
Q
RR
O r
O A
C
Figure 18-10 Circui t for measuring the energy t ransfer to a load Z d u r i n g a c u r r e n t
p u l s e ; r « Z , r « R, R » 1/coC. T h e c h a r g e Q f lows out of the top terminal of the
source and in to the lower one . The t e rmina l s x and y are the input s to the
ho rizo nta l and vert ical amplif iers of the osci l loscope. See Pr ob . 18-4.
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Problems 447
On the sc reen ,
Q' « 2x x — , v x —
1 V.
C • P , + R 2
Set V = V 0 cos cot.
a) Show that , i f R » 1/coC,
r
x x — Q.RC
b) Show th a t the energy t rans fe r red pe r cyc le is pro por t ion a l to the a rea enc losed
by the curve on the osci l loscope screen.
c) W ha t is the sha pe of the curve wh en th e load is ( i) a res is tance, ( ii ) a ca pac i tanc e,
( ii i) a pur e ind ucta nce , (iv) a res is tance in series with an ind ucta nce , with R = w L ?d) What is the shape of the curve i f the load draws current only when V is close
to i ts ma xim um va lue?
IH-5E ENERGY TRANSFER TO A LOAD
Figure 18-11 shows another c i rcu i t for measur ing the energy d i s s ipa ted when
a pulse of current originat ing in G passes through a load Z. Fhe res is tor R' serves to
di scharge the capac i tor C s lowly between pu lses : a negl igible am ou nt of cha rge f lows
through i t during a pulse. Also, C is large and the vol tage drop across i t i s negl igible
compared to tha t ac ros s Z .
Show tha t the a rea under the curve obse rved on the osc i l loscope i s propor t iona l
to the energy diss ipated in Z, whether Z is l inear or not .
Figure 18-11 See Prob. 18-5 .
18-6E REFLECTED IMPEDANCE
Show tha t a pos i t ive ( induc t ive ) reac tance in the secondary of a t rans former
is equiv alent t o a neg at ive (capaci t ive) reacta nce in the prim ary , and vice versa.
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448 Alternating Currents: III
18-7E MEASUREMENT OF THE COEFFICIENT OF COUPLING k
A t rans forme r has a pr imary induc tance L {. a second ary indu c tance L 2 , and a
mutua l induc tance M. The winding res i s t ances a re negl ig ib le .Show tha t
Z0/Z
x, = 1 - k\
w h e r e Z 0 a n d Z x are the impedances measured a t the t e rmina l s of the pr imary , when
the secondary i s shor t -c i rcu i t ed and when it i s open-c i rcui t ed .
18-8E REFLECTED IMPEDANCE
How does the impedance a t the t e rmina l s of the pr imary of a t rans former va ry
with the res is tance R 2 i n the secondary c i rcu i t ?
Cons ider a s imple case where a = L x = L 2 = k = 1, R } = 0 .
a ) Show tha t
R 1R 2
R\ + 1 R 2
2 + 1
(Note tha t , d imens iona l ly . th i s equ a t ion i s absu rd . The rea son is tha t we have as s igned
num er ica l va lues to som e of the pa ram ete rs . For exam ple , the 1 in the den om ina t ors
is really io 2Lj.)
b) Draw curves of R, X, \Z\ as funct ions of R 2, for values of R 2 ranging f rom
0.1 to 10. Use a lo g scale for R 2.
I8-9E MEASURING THE AREA UNDER A CURVE
You a re asked to measure the a rea under a curve drawn on a shee t of paper .
You a re given a pair of He lm ho ltz coi ls , as in Pr ob . 8-5, tha t give a uniform ma g
net ic f ie ld ov er a suffic ient ly large area, a lso a sour ce of a l ter nat i ng c urre nt , an electro nic
vol tmete r , conduc t ing ink , e t c .
How wi l l you proceed?
I8-I0E SOLDERING GUNS
A solder ing gun cons i s t s in a s t ep-down t rans former tha t pas ses a l a rge cur rent
through a l ength of copper wi re .
O ne sold ering gu n diss ipa tes 100 wat ts in a piece of cop per w ire (a = 5.8 x 10 7
S i emen s per meter) havi ng a cross-se ct ion of 4 squ are m il l imete rs and a length of 100
mi l l ime te rs .a) Find V and / in the secondary .
b) Fin d the cu rre nt in the pr im ar y if i t is fed at 120 volts.
18-1 IE CURRENT TRANSFORMER
Figure 18-12 shows a s ide - looking cur rent t rans former tha t i s used for measur ing
large curren t pulses . I t i s m or e conven ient th an the R ogo wsk i coil of Pr ob . 12-7 in that
the cur rent -ca r ry ing wi re need not be threaded through the measur ing coi l .
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449
I
2a
2a
Figure 18-12 Side-looking transformer.
See Prob. 18-11.bV
Show that, for a single-turn coil.
18-12E INDUCED CURRENTS
A thin-walled conducting tube has a length /. a radius a. a wall thickness £> an d
a conductivity a. When placed in a uniform axial magnetic field of the form
an induced current flows in the azimuthal direction.
Alternating currents tend to flow near the outer surface of a conductor. This
phen omeno n is called the skin effect. See Prob. 11-7. The .sAm depth 6 is (2 prp
uc>rt\
l
We are concerned here with a brass tube having a wall thickness of 0.5 millimeter
and w ith a frequency of 60 hertz. Since the skin depth is 16.3 millimete rs under those
conditions, we may disregard the skin effect and assume that the induced current is
uniformly distributed throughout the thickness of the tube.
Show that, for currents flowing in the azimuthal direction, the resistance and
self-inductance of the tube are given by
B = B 0 cos tor,
R L, =Ho™ ,
so that
coL,/P, = p0awcih/2.
* Electromagnetic Fields and Waves. Sec. 11.5.
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450 Alternating Currents: III
Fo r a bra ss tub e with / = 200 mil l imete rs , a = 5 mil l im eters , h = 0.5 mil l imeter ,
a n d (7 = 1.60 x 1 0 7
S i emen s p er meter, this ra t io is only a bo ut 1°„ at 60 hertz .
18-13 EDDY-CURRENT LOSSES
Why do eddy-cur rent los ses inc rease as the square of the f requency?
18-14 EDDY-CURRENT LOSSES
Eddy-cur rent los ses in magne t i c cores a re minimized by as sembl ing them f rom
l a m i n a t i o n s .
Co nsid er a cor e of rect ang ular cro ss-sect ion as in Fig. 18-13. T h e e d d y - c u r r e n t
los s i s propor t iona l to V2
/R , w h e r e V i s the e l ec t ro mo tance induced a ro un d a typica l
cur ren t pa th such as the one show n by a das hed curve. The res is tan ce is a lso diff icul t
to define, but i t i s of the order of twice the res is tance of the upper half, o r
o(b/2)L
Sho w th at, if th e core is split int o n laminat ions , the loss wil l be reduced by a
factor of n 2.
18-15 HYSTERESIS LOSSES
Hys te res i s los ses a re prop or t io na l to the ope ra t ing f requency / , whi l e eddy-
cur ren t los ses inc rease a s / 2 (Prob. 18-13).
You a re g iven a number of t rans former l amina t ions . Can you devi se an exper i
me nt tha t wil l perm it you to eva luat e the relat ive impo rta nc e of the two types of loss?
18-16 CLIP-ON AMMETERFigure 18-14 shows a type of c l ip-on amm ete r th a t i s mo re com mo n than tha t
desc r ibed in P rob . 15-3 . The on e shown here is s imply an i ron-core t rans fo rmer whose
prim ary is the curren t-ca rry ing wire . I t is s imilar to the Rog ow ski coi l of Pr ob . 12-7,
except for the i ron core, and except for the fact that the secondary does not surround the
wire. This type of t ransformer is often cal led a current transformer.
Ins t ruments of th i s type a re ava i l ab le for measur ing a l t e rna t ing or pul sed
cur rent s ranging f rom mi l l i amperes to hundreds of amperes .
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45 1
Figure 18-14 C l i p - o n a m m e t e r . T h e r i n g - s h a p e d m a g n e t i c
yoke i s made in two ha lves tha t can ro ta te about the
hinge A. Th e yoke can be c l ipped a r ou nd the wi re by
open ing the j aws a t C . Th e vol t age V i nduced in winding
S is pro por t ion a l to the cur ren t f lowing th rou gh the
wire. See Prob. 18-16.
a ) Ca lcula te V when the measu red c ur rent i s one amp ere a t 60 he r t z . The c ore
has a mean diameter of 30 mil l imeters , a cross-sect ion of 64 square mil l imeters , and a
perm eabi l i ty of 10,000. Th e seco nda ry has 1000 turn s .
b) How could you inc rease the vol t age V for a given current / and for a given
i n s t r u m e n t ?
18-17 AUTO-TRANSFORMERS
Figure 18-15a shows a s chem at ic d iagram of an auto- t ran s form er . I t s s inglewinding i s t app ed an d se rves both as a pr imary and as a s econd ary . A uto- t r ans
formers are widely used, part icularly for obtaining an adjus table vol tage at 60 hertz .
Th e left -hand te rm inal s are conne cted to the 120-vol t l ine, an d the r ight- han d
Figure 18-15 (a) Auto-transformer. ( / )) Equivalent c i rcui t . See Prob. 18-17.
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452 Alternating Currents: III
terminals are connected to the load. The tap can slide along the winding. The power
ratings of commercial models vary from a few tens of watts to several kilowatts.
Calculate the ratio of output voltage to input voltage VJVi. The self-inductances
L t and L 2 are those of the winding, on either side of the tap. Set
Lj = AN\, L2
= AN\, M = ( L , L 2 ) "2
= ANXN2, (1)
where M is the mutua l induc tance between Lj and L 2 , and neglect the resistance of
the winding. These assumptions are reasonable, in practice.
You will find that V„ is directly pro por tio nal to the numbe r of tur ns in L 2 ,
and is independent of Z, under the above assumptions.
Solution: First let us transform the circuit of Fig. 18-15a into that of Fig. 18-15b.
Then
jcoM
Similarly,
(Z - jo)M)jw(L2
+ M)
V _ Z - jwM + jco(L2 + M)
V;~ (Z -jcoM)jco(L2
+ M) '(
'
' jco(Li + M)Z - jo.)M + jo)(L2
+ M)
Then
V
° _V
»V
' _ Zjw(L2
+ M)
Vi ~ ~ Z[jw(L2
+ M)+jco{Li + M)] + m2
M(L2
+ M) - w2
L 2 ( L l + M)
Substituting the values of L,, L 2 , M,
V; = ZjcoA(N\ + NXN
2)
V,- ZjcoA(N\ + N\ + 2 / V , N 2 ) + m2
A2
(N\N2
2 - / V f / V2
) '
N7
JV, + N2
161
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CHAPTER 19
MAXWELL'S EQUATIONS
19.1 THE TOTAL CURRENT DENSITY J ,
19.2 THE CURL OF B
19.2.1 Example: Dielectric-Filled Parallel-Plate Capacitor
19.3 MAXWELL'S EQUATIONS
19.4 MAXWELLS EQUATIONS IN INTEGRAL FORM
19.5 SUMMARY
PROBLEMS
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At th i s s t age we have found on ly th ree o f Maxwel l ' s four equa t ions . These
were Eqs . 6-12, 8-12, an d 11-14. O u r first ob jec tive in thi s ch ap te r is todedu ce the four th one , namely the eq ua t ion for the cur l of B . The n we sha l l
r e - examine a l l f our equa t ions as a g roup .
19.1 THE TOTAL CURRENT DENSITY J,
There are several types of elect r ic current .
a) Th ere are f i rs t the usu al conduction currents through good con
ductors such as copper , which a re as soc ia t ed wi th the d r i f t o f conduc t ionelectrons (Sec. 5.1).
b ) There a r e a l so conduction currents in semiconductors in which bo t h
conduct ion elect rons and holes (Sec. 5 .1) can dr i f t .
c) Electrolytic currents are due to the mot ion o f ions th rough l iqu ids .
d ) The m ot i on o f ions o r e l ec t rons in a vac uum — for exam ple , in an
i o n b e a m o r i n a v a c u u m d i o d e — g i v e s convection currents.
e) The motion of macroscopic charged bodies a l so p roduces e l ec t r i c
cur ren t s . A good exam ple i s the be l t o f a Van de Graaf f h igh-vo l t age
g e n e r a t o r .
Al l these cur r en t s a r e as soc ia t ed wi th the mot ion o f f r ee charges . Weshal l use the symbol J f for the current dens i ty associated wi th f ree charges .
f ) The displacement current. T h e d i s p l a c e m e n t c u r r e n t d e n s i t y dD/dt
h a s t w o c o m p o n e n t s , t h e polarization current dens i ty dP/dt, a n d e 08E/dt
(Sec. 7.6).
g ) The equivalent currents i n magne t i c mate r i a l s , wi th a vo lume
dens i ty V x M and a surface dens i ty M x n, (Sees. 14.2 a n d 14.3).
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19.2 The Curl of B 455
T h u s , in the general case, the total volume current density is
j , = J / + ^ D - + V x M , ( 19 -1 )
dt
J r + e 0 ? + ? + V x M , ( 19 - 2)
' dt ct
= J,„ + C o ^ . (19-3)ct
where J„, i s the volume current density in matter, wi th
r D r E r P
" 6 0 J i+
at'
J„ , = J | + ^ + V x M . (19-5)
( 1 9 - 4 )
79.2 THE CURL OF B
In Sec. 9.2 we foun d t ha t, for sta tic f ields.
V x B = n0(J
f+ V x M ) . ( 1 9 - 6 )
For t ime-dependen t f i e lds , we mus t r ep lace the paren thes i s by the to t a l
cur r en t dens i ty . Then
V x B = /<„ (j, + e 0 ™ + ~ + V x ( 1 9 - 7 )
o r
V x B - wo ™ = /< 0 ( j / + ^ + V x M ) . (19-8)
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456 Maxwell's Equations
This is one of Maxwell's equations. In more concise form.
rEV x B - €
0/-<o -^ r = MoJ
(19-9)
Since
B = u0(H + M), (19-10)
we also have that
V x H = J,- + ~di' (19-11)
19 .2 .1 EXAMPLE: DIELECTRIC-FILLED PARALLEL-PLATE CAPACITOR
Figure 19-1 shows a parallel-plate capacitor connected across a source of alter
nating voltage V. The cap acitor co ntains a slightly conductin g dielectric that has a
permittivity e r e 0
a r
>d a conductivity rj. We neglect edge effects. Then the field inside
the capacitor is uniform and both Jf
and cD/dt are independent of r.
We can find the magnetic induction inside the dielectric by transforming Eq. 19-8
into its integral form, setting M = 0. Integrating over an area S,
J s V x B d a da.
and, using Stokes's theorem on the left.
H i l l(D
da.
(19-12)
(19-13)
where C is the curve bounding the surface S. By symmetry, the vector B is azimuthal.
Not e that B, J / 5 and D are at the same time vectors and phasors, here.
If C is now a circle of radius r inside the capacitor and parallel to the plates,
as in the figure, we have the following phasor equation:
IKI-B
3D
77
(19-14)
(19-15)
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45 7
lT =====V ' * ^ „
•
3Dct/3Dct
Figure 19-1 Para l l e l -p la t e capac i tor connec ted to a source of a l t e rna t ing vol t age .
T h e c u r r e n t 3 f + c D/ct gives an azimuthal magnet ic f ie ld B.
N o w ,
J f = FFE = <TV/S,
5 E
I Te r e 0 — = ; w e r e 0 - ,
dt s
(19-16)
(19-17)
a n d
B = *^(o + jcoe^r.2s
(19-18)
Thus the vec tor B i s az im utha l and i t s ma gni tu de i s pro por t ion a l to r. Figure 19-2
shows l ines of B in a plane paral le l to the plates .
Figure 19-2 Lines of B in a plane
paral le l to the plates ins ide the
capaci tor of Fig. 19-1.
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458 Maxwell's Equations
The magn e t ic ind uc t ion i s the sum of tw o te r ms . The one tha t is p r o por t ion a l
to a i s a s soc ia ted w i th the cond uc t io n cur r en t . Th a t B is in phas e with V. like the
cond uc t io n c ur r en t cE . The o the r t e r m, w hich i s p r o por t ion a l to e r e 0 , is associated
w i th 80 /dt and is thu s pro po r t io nal to the t ime rate of cha nge of V. I t is in q ua d
rature with V.
19.3 MAXW ELL'S EQUATIONS
Let us g roup toge ther the four Maxwel l equa t ions , which we found suc
cessively as Eqs. 6-12, 8-12, 11-14, 19-9:
V - E = p , / e 0 , (19-19)
V - B = 0 . (19-20)
dB
V x E + — = 0, (19-21)
V x B - € 0fi 0
cE
~d t(19-22)
As usua l .
E is the elec tric f ield inten sity in volts pe r m et er ;
Pi — Pf + Pb i s i n e t o t a l e l ec t r i c charge dens i ty in cou lombs per
c u b i c m e t e r ;
Pf i s the f ree charg e d en s i ty ;
pb i s the bou nd charg e dens i ty — V • P, w h e r e P is the electric
p o l a r i z a t i o n i n c o u l o m b s p e r s q u a r e m e t e r ;
B i s the ma gne t i c indu c t ion in t es l as ;
J„, = if + dP/dt + V x M is the cu rre nt den s i ty resul t in g f rom
the f low of charges in mat t e r , i n amperes per square mete r ;
3f
i s the current dens i ty of f ree charges ;dP/dt i s the po la r i za t io n cur r en t den s i ty ;
V x M is the equ iv a len t cur r en t dens i ty in ma gne t i zed ma t t e r ;
M i s the ma gne t i za t io n in am pere s per m ete r ;
e 0 is the permittivity of free space, 8.854 187 82 x 1 0 " 1 2 farad per
m e t e r ;
p.0
i s the permeabi l i ty of f ree space, 4n x 1 0 " 7 henry per mete r .
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19.4 Maxwell's Equations in Integral Form 459
In conduc tor s obey ing Ohm's l aw (Sec . 5 .2 ) ,
J f = c t E , (19-23)
where r j i s the conduct ivi ty in Siemens per meter .
Maxwel l ' s equa t ions a r e par t i a l d i f f e r en t i a l equa t ions invo lv ing space
and t ime der ivat ives of the f ie ld vectors E an d B , the to t a l cha rge dens i ty p„
and the cur r en t dens i ty J m . They do no t yield the values of E and B direct ly ,
bu t on ly a f t e r in t egra t ion and a f t e r t ak ing in to account the p roper boundary
c o n d i t i o n s .
T h e s e a r e t h e f o ur f u n d a m e n t a l e q u a t i o n s o f e l e c t r o m a g n e t i s m . T h e y
apply to a l l e l ec t romagnet i c phenomena in media tha t a r e a t r es t wi th
respec t to the coord ina te sys tem used fo r the de l opera to r s .
19.4 MAXW ELL'S EQUA TIONS IN INTEGRAL FORM
We ha ve s t a t ed M axw el l ' s equa t io ns in d i ff e ren t ia l fo rm; l e t us now s ta t e
the m in in t egra l fo rm, in o rde r to a r r ive a t a be t t e r und er s t an d in g o f the i r
p h y s i c a l m e a n i n g .
The in tegra l fo rm i s r ea l ly more in te res t ing and genera l . I t i s more
in te res t ing because one can v i sua l i ze in t egra l s more eas i ly than der iva t ives .
I t i s a l so more genera l because i t i s app l i cab le even when the der iva t ivesdo no t ex i s t , f o r example a t the in t e r f ace be tween two media .
In Sec. 6 .4 , we dedu ce d E q. 19-19 f rom G au ss ' s law in integra l form,
whe re S i s the su rf ace bo un din g the vo lum e p t i s the to t a l charge dens i ty
pf
+ p b, a n d Q, i s the total net charge Q f + Q b c o n t a i n e d w i t h i n x. T h e
outward f lux o f E t h rough any c losed sur f ace S i s the re fore equa l to l / e 0
t imes the total net charge ins ide, as i l lus t rated in Fig. 19-3.
We proceeded s imilar ly in Sec. 8 .2 , where we deduced Eq. 19-20 f rom
the fact that
w h e r e S i s any closed surface. Th us th e net ou tgo ing f lux of B thr ou gh any
closed surface S i s zero. This i s shown in Fig. 19-4.
(19-24)
(19-25)
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Figure 19-4 Lines of B through a closed surface S. The net outward f lux of B is
equa l to ze ro .
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46 1
Figure 19-5 Th e d i rec t ion of the e l ec t ro mo tance induced a rou nd C i s indicated by
an a r row for the case where the magne t i c induc t ion B i s in the direct ion shown
and inc reases . Th e e lec t ro mo tance is in the same d i rec t ion if B i s upw ard and
decreases .
I f we now integrate Eq. 19-21 over a surface S b o u n d e d b y a c u r v e C,
Or , if we use S tokes ' s the ore m on th e le ft an d inver t the ope ra t io ns on the
r ight , the surface S being f ixed in space,
T h e n t h e e l e c t r o m o t a n c e i n d u c e d a r o u n d a c l o s e d c u r v e C is equ a l to m inus
th e ra te of cha ng e of the m ag ne tic f lux <I> link ing C, as in Fig. 19-5. T he
pos i t ive d i r ec t ion s fo r B an d a ro un d C a re r e l a t ed accord ing to the r igh t -h and
screw rule (Sec. 1.12). I f t he curve C has more than one tu rn , then the su r f ace
S beco me s comp l ica te d a nd the su r f ace in tegra l i s ca l l ed the f lux l inkage .
Fina l ly , if we also integ rate Eq . 19-22 ove r an area S b o u n d e d b y a
curve C ,
(19-26)
(19-27)
(19-28)
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462
Figure 19-6 The direction of the magnetomotance around C is indicated by an
a r r o w for the case where the current Jf
+ (<~D dt) is in the direction shown. The
d i s p l acem en t curr ent is down war d if D is do wnw ard and increases, or if it is
u p w a r d and decreases.
If we perform the same ope rat ion on the cor re spo nd ing e qu ati on for H, as
we did in Sec. 19.2,
(19-29)
The mag ne to mo ta nc e ar ou nd C is equal to the sum of the free cur rent plus
the displacement current linking C. This is illustrated in Fig. 19-6. The
positive directions are again related by the right-hand screw rule.
19.5 SUMMARY
The total volume current density is the sum of four terms:
r E r P
J, = Jf + e0
— + — + V x M. (19-2)ct ct
where the first term is the free current density, the second and third together
form the displacement current density BD /dt, and the fourth term is the
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19.5 Summary 463
equivalent volume current density i n m a g n e t i c m a t e r i a l s . T h u s
dBBE
. f3P
Also ,
ct dt
J,„ = if + ? + V x M (79-5)
is the current density in matter.
The fo l lowing four equa t ions a r e known as Maxwell's equations:
V - E = p , / e 0 , (19-19)
V • B = 0. (19-20)
V x E + ^ = 0, (19-21)ct
V X B - e oi U o ^ = Mm- (/9
"22
>f' f
I n c o n d u c t o r s o b e y i n g Ohms law,
J f = ffE.(19-23)
I n i n t e g r a l f o r m , M a x w e l l ' s e q u a t i o n s b e c o m e
js E • da = Q , /€ 0 , (/9-2^)
J* B • da = 0. (79-25)
d) E d l = - i f B - d a = . (19-27)
J f d t J s dt
(j), B • dl = ^ J ( J m + e 0 ^ j • d a = /<„/,. (79-28)
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464 Maxwell's Equations
PROBLEMS
19-1 PARALLEL-PLATE CAPACITOR SUBMERGED IN
AN INFINITE MEDIUM
Consider a paral le l -plate capaci tor s i tuated in an infini te medium as in Fig. 19-7.
Fo r s implici ty, we assu me th at the plates are ci rcular . At r = 0, the plates carry
c h a r g e s +Q a n d —Q. The capac i t ance be tween the p la tes i s C . Fhe medium i s
s l ight ly conduct ing and the res is tance between the plates is R.
Fr om Sec. 5.14, the vol tag e difference F betw een the plate s decreases ex
ponent i a l ly wi th t:
V = V0e-"RC
. (1)
Fherefore, a t any point in the dielectr ic , the electr ic f ie ld intensi ty decreases in the
s a m e m a n n e r :
E = E 0e-,IRC
, (2)
where £ 0 var ies f rom one poin t to another .
a) Use the resul t of Prob. 6-17 to show that the total current densi ty J f + dD/dt
in the dielectr ic is zero. The dielectr ic has a conduct ivi ty o and a re l at ive pe rm i t t iv i ty
Solution: At any point , the total current densi ty is
3D
Jf + y t = Iff + e re 0-IE, (3)
P C
since P C = e re 0/a, from Prob. 6-17.
b) What does this resul t te l l you about the magnet ic f ie ld ' :
Solution: According to Eq. 19-13, for any closed curve C,
B • dl = 0. (5)
Now, by symmetry, i f there is a magnet ic f ie ld, i t must be azimuthal . So consider
a ci rcle C as in Fig. 19-7. By symmetry again, s ince the plates are ci rcular by hypo
thes is , B mus t be the same a l l a round the c i rc l e . Because of the above equa t ion , B
is zero everywhere on the ci rcle . But the ci rcle can be s i tuated anywhere, as long
as i t i s paral le l to the plates and centered on the axis of symmetry, so B is zero every
w h e r e .
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Figure 19-7 See Prob. 19-1.
19-2E MAXWELLS EQUATIONS
Show that Maxwell's equations can also be written under the form
V • D = pf,
V - B = 0,
3E
V x E + — = 0,
<3f
cDV x H = Jf.
ct
I9-3E MAXWELL'S EQUATIONS
Show that, for linear and isotropic media, and for fields that are sinusoidal
functions of the time.
V • e ,e0E = Pf
V • u ru 0H = 0,
V x E + jwfirfi0H = o,
V x H - jmurri0E =h
Here again, we have quantities that are both vectors and phasors.
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466 Maxwell's Equations
19-4E MAXWELLS EQUATIONS
T h e eq u a t i o n for the conservat ion of charge of Sec. 5.1.1 is buil t into Maxwell 's
eq u a t i o n s .
Show tha t it follows from Eq. 19-22.
T h e J used in C h a p t e r 5 is the cur ren t dens i ty of free charges Jj.
19-5E MAGNETIC MONOPOLES AND MAXWELL'S EQUATIONS
W h e n it is as s u m ed t h a t m ag n e t i c m o n o p o l e s do exist, it is the cu s t o m to write
Maxwel l ' s equat ions in the fol lowing form:
dBV • E = p,/e0, V x E = - — - J*.
ct
( dE
V • B = p*, V x B = / t o k - + J ,
w h e r e p* is the volume dens i ty of magnet ic charge in w eb er s per cub ic meter , and J*
is the magnet ic cur ren t dens i ty , expressed in weber s per second per square meter ,
a) Show that
dp*V • J* = — — .
dt
The modi f ied Maxwel l equat ions therefo re pos tu la te the co n s e r v a t i o n of m ag n e t i c
c h a r g e .
By analogy wi th Coulomb 's law, n ea r a magnet ic charge .
Q*
4m - 2 1
In a v a c u u m ,
Q*
4np 0r
an d th e force between tw o magnet ic charges is
0*0*F
f» = e « * H
6= f ^ r
1.
4np0r
b) Show that , for any closed path C, and if cB/ct is zero ,
E d l = - / * .
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Problems 467
w h e r e / * is the m a g n e t i c c u r r e n t . Note th e n e g a t i v e s i g n ; th e i n d u c e d e l e c t r o m o t a n c e
in a c o i l is e q u a l t o m i n u s th e m a g n e t i c c u r r e n t linking it . See Prob. 12-7.
19-6 Express the f o l l o w i n g q u a n t i t i e s in t e r m s of k i l o g r a m s , meters, s e c o n d s , an d a m p e r e s :
j o u l e , w a t t ,
coulomb, v o l t , ohm, S i e m e n s , f a r a d ,
weber, t e s l a , h e n r y .
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CHAPTER 20
ELECTROMAGNETIC WA VES
20.1 WAVES
20.1.1 Plane Sinusoidal Waves. Propagation of a Scalar Quantity20.1.2 Plane Sinusoidal Waves. Propagation of a Vector Quantity
20.2 THE WA VE EQUATION
20.3 THE ELECTROMAG NETIC SPECTRUM
20.4 PLANE ELECTROMAG NETIC WA VES IN FREE
SPACE: VELOCITY OF PROPAGATION c
20.5 PLANE ELECTROMAG NETIC WAVES IN FREE
SPACE: THE E AN D H VECTORS
20.6 THE POYNTING VECTOR if IN FREE SPACE
20.6.1 Example: Laser Beam
20.7 PLANE ELECTROMAG NETIC WAVES IN
NON-CONDUCTORS
20.7.1 Example: Laser Beam
20.8 SUMMARY
PROBLEMS
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In th i s l as t chap te r we sha l l have no more than a g l impse o f e l ec t romagnet i c
w a v e s . T h e s e w a v e s a r e e v e r y w h e r e — l i g h t w a v e s a r e e l e c t r o m a g n e t i c — a n dthe i r app l i ca t ions a r e innumerab le , some of the bes t known be ing r ad io ,
te levis ion, and radar . We shal l s tar t wi th a br ief review of plane waves , and
then discuss plane elect romagnet ic waves , f i r s t in f ree space, and then in
dielect r ics . This wi l l require al l four of Maxwel l ' s equat ions .
Unfor tuna te ly , the methods tha t a r e used fo r l aunch ing o r fo r de
t ec t ing these waves a r e bey on d the s cope o f th i s bo ok . H ow ever , i f yo u
wish to go fu r ther , you wi l l f ind many books on e l ec t romagnet i c waves . f
20.1 WA VES
The essent ial nature of waves is i l lus t rated by a s t retched s t r ing, f ixed at
one end and moved r ap id ly in a ver t i ca l p l ane in some a rb i t r a ry f ash ion a t
the oth er end , as in Fig. 20 -1.
If we call y(t) t he he igh t o f the moving end , then the he igh t y a t a
d i s t a n c e z along the s t r ing is y[t — (z/v)], assuming no losses , a perfect ly
f lexible s t r ing, and no ref lected w ave . Th e height at z i s thu s equ al to t he
he igh t o f the moving end a t a p rev ious t ime t — {z/v), t h e q u a n t i t y z/v
be ing the t ime r equ i r ed fo r the d i s tu rbance to t r ave l th rough the d i s t ance z
at the veloci ty v.
f Almost All About Waves by Jo hn R. Pierce, M IT Press , 1974, is a delightful l i t t le
book by an authori ty on the subject . See also Electromagnetic Fields and Waves,
Ch apte rs 11 to 14 .
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47 0
y
Figure 20-1 Wave on a s t re t ched s t r ing .
We can genera l i ze f rom th i s s imple example and cons ider waves
propaga t ing in an ex tended r eg ion , such as acous t i c waves in a i r o r l igh t
waves in space.
The quan t i ty p ropaga ted can be e i ther a s ca la r o r a vec to r quan t i ty .
Fo r examp le , in an aco us t i c wave , we hav e the p rop ag a t i on o f a p res sure ,
which is a s ca la r . I n an e l ec t rom agn et i c wave , we have the p r op ag a t io n o f
a vector E and of a vector H .
20.1.1 PLANE SINUSOIDAL WAVES. PROPAGATIONOF A SCALAR QUANTITY
If a ce r t a in s ca la r quan t i ty a p ro pag a t in g wi th a ve loc i ty v is define d at
- = 0 by
This expres s ion descr ibes a plane wave t r ave l ing a long the z - ax i s ,
s ince a is independent of x a n d y. The wave i s a l so unattenuated, s ince the
a m p l i t u d e a 0 i s con s tan t . Th e quan t i ty be tw een b ra cke t s is the phase of the
w a v e . T h e frequency f is w/2n, a n d t h e period T is iff.
y. = 7. 0 cos ojf. (20-1)
then , fo r any po s i t ion z in the d i r ec t ion o f the p ro pa ga t ion .
(20-2)
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471
Figure 20-2 The quantity i — a0c o s t y [ f
function of l.
•)] as a function of z and as a
F o r a given pos i t io n z , we ha ve a s inus oida l var iat i on of J. wi th t ime(Fig. 20-2a) ; for a given t ime f , we have a s inusoidal var iat ion of a wi th z
(Fig. 20-2b) .
The sur f aces o f cons tan t phase , ca l l ed wave-fronts, are normal to the
z-axis .
T h e q u a n t i t y v is called the phase velocity of the wave, s ince i t is the
ve loc i ty wi th which a wavef ron t i s p ropaga ted in space . For example , i f t he
bra cke t i s zero , the po s i t io n of the wa vefron t at the t ime f i s given by
z = vt . (20-3)
T h e wavelength X i s the dis tance over which coz/v changes by 2K
r ad ians , as shown in F ig . 20-2 :
coX/v = 2n,
X = 2nv/co = v'f.
(20-4)
(20-5)
Also .
A = A/27i = v/co, (20-6)
w h e r e A ( r ead " l ambda bar " ) i s the radian length. I t i s the dis ta nce o ver wh ich
the phase o f the wave cha nges by on e r ad ian ; th is is ab ou t A /6. I t turn s o ut
tha t the quan t i ty tha t appear s in near ly a l l wave ca lcu la t ions i s A, a n d n o t
A. We are al ready famil iar wi th the fact that in near ly al l osci l la t ion calcu
lat ions i t i s the ci rcular f requency <o th at is used ins tea d of the frequen cy / '.
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472 Electromagnetic Waves
The intuitively simple quantity, namely the wavelength or the frequency,
is not the one that is "na tur al" from a math emat ical stand point .
In phasor notation (Sec. 16.7), a sinusoidal wave traveling in the
positive direction along the z-axis is described by
Similarly, a wave traveling in the negative direction along the z-axis
is describ ed by
20.1.2 PLANE SINUSOIDAL WAVES. PROPAGATION
OF A VECTOR QUANTITY
In the above equations, a is a scalar. Now, as we noted at the beginning of
this chap ter, the quant it y tha t is pr op aga te d can also be a vector , say the
electric field intensity E. The n, the above equa tio ns app ly to each one of
the components Ex, E
y, E.. Or, if we multip ly th e equ ati on for E
xby i, that
for Ey. by j , and that for E
:by k, and take the sum, we have an equation like
20-7 with a replaced by E.
Not e that the quanti ty E is then both a phasor and a vector. It is avector because it is oriented in space, with the usual three components, and
it is also a pha sor b ecau se its time depe nde nce is writ ten in the form
Let us calculate the second derivatives of a (Eq. 20-7) with respect to t
and to z:
y. = <x0expj[cof - (z/X)].* (20-7)
a = a0
ex p j [cot + (z/?.)]. (20-8)
exp/(cot + cp).
20.2 THE WA VE EQUA TION
7 T = - t o a,dt
2
32a
a. (20-9)
+
In the last ch ap t e r , we shall follow the usual cus tom of not us ing a specia l no ta t ion
for p h a s o r s : vecto r phasor s wi l l be writ ten l ike vec tors , an d sca lar phasor s l ike scalars .
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20.3 The Electromagnetic Spectrum 473
T h u s
r 2 a(20-10)
or
1 T 2 a( 2 0
-u )
w h e r e v i s the phase veloci ty . This i s the wave equation for a plane wave
pro pag a t in g a lon g the z - ax i s. Th i s eq ua t io n i s va l id even fo r no n- s inu so ida l
w a v e s .
More genera l ly , fo r an una t t enua ted wave in space ,
1 r 2 a
V 2 a = ~ ^ . I2 0
"1 2
)
v dt
If we had a vec to r qu an t i ty ins t ead o f the s ca la r x , we would have the s am e
equ a t io n wi th a r ep laced by , s ay , E . Fo r an una t t en ua ted s inuso ida l w ave ,
2
V 2 a = (20-13)v
20.3 THE ELECTROM AGNETIC SPECTRUM
W e are now ready to s tudy the p rop ag a t io n o f e l ec t ro ma gnet i c waves in
f ree space and in dielect r ics .
Maxwel l ' s equa t ions impose no l imi t on the f r equency o f e l ec t ro
ma gne t i c waves . To da te , t he spec t ru m th a t has been inves t iga ted exp er i
men ta l ly i s sho wn in F ig . 20-3 . I t ex te nds con t inuo us ly f rom the long r ad io
waves to the very h igh energy gamma rays observed in cosmic r ad ia t ion .
In the former , the f requenc ies are ab ou t 10 her tz , and th e wa vele ngth s are
a b o u t 3 x 1 0 7 meters ; in the la t ter , the f requencies are of the order of
1 0 2 4 her t z , and the wave leng th s of the o rde r o f 3 x 1 0 " 1 6 m e t e r . T h e k n o w n
spec t rum thus cover s a r ange o f 23 o rde r s o f ma gni t ude . Rad io , l i gh t, an d
h e a t w a v e s . X - r a y s a n d g a m m a r a y s — a l l a r e e l e c t r o m a g n e t i c , a l t h o u g h t h e
sources and the de tec to r s , as wel l as the modes o f in t e r ac t ion wi th mat t e r ,
vary widely as the f requency c hang es by o rde r s of ma gni tud e . The n om en
clature of radio waves is given in Table 20-1.
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20.3 The Electromagnetic Spectrum 475
Tabic 20-1
Band Frequency Metric
Num bers* Range Subdivi sion Abbreviation
1 3--30 Hz
2 30 -300 Hz M e g a m e t r i c w a v e s elf
3 300--3000 Hz vf
4 3- - 3 0 k H z My riam et r i c waves v lf
5 30 -300 kH z Ki lom et r i c waves If
6 3 0 0 - -3000 kH z Hec to met r i c waves mf
7 3- - 3 0 M H z D e c a m e t r i c w a v e s hf
8 30 - 3 0 0 M H z M e t r i c w a v e s vh f
9 300--3000 M H z Dec im et r i c waves uhf
10 3- - 3 0 G H z C e n t i m e t r i c w a v e s sh f
11 30 -300 G H z Mi l l imet r i c waves eh f
12 300- 3 0 0 0 G H z Dec imi l l ime t r i c waves
* B a n d n u m b e r N ex te nds f rom 0 .3 x 10v
to 3 x 10 'v
h e r t z .
Th e funda me nta l iden t i ty o f a l l t hese types o f wav es i s dem on s t r a t ed
by many exper iment s cover ing over l app ing par t s o f the spec t rum. I t i s a l so
shown by the fact that in f ree space they are al l t ransverse waves wi th a
c o m m o n v e l o c it y o f p r o p a g a t i o n . F o r e x a m p l e , s i m u l t a n e o u s r a d i o a n d
opt i ca l obse rva t io ns o f f la r e s t a r s hav e shown tha t the ve loc i ty o f p rop ag a t io n
i s the s ame, wi th in expe r ime nta l e r ro r , fo r wave leng ths d if fe r ing by m or e
than s ix o rder s o f magni tude .W e shal l fol low the cus tom of us ing H rat he r th an B in discuss in g
elect romagnet ic waves , in spi te of the fact that , unt i l now, we have used H
only in r e l a t ion wi th magne t i c mate r i a l s . The main r eason fo r us ing H in
s t ead o f B in dea l ing wi th e l ec t romagnet i c waves i s tha t E x H gives the
energy f lux dens i ty , as we shal l see.
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476 Electromagnetic Waves
20.4 PLANE ELECTROMAGNETIC WA VES IN FREE
SPACE: VELOCITY OF PROPAGATION c
Let us start with the relatively simple case of a plane sinusoidal wave pro
pagating in a vacuum in a region infinitely remote from matter. Then
Maxwell 's equ ati on s 19-19 to 19-22 beco me
V • E = 0, (20-14)
V • H = 0, (20-15)
V x E + ,/oj/i0H = 0. (20-16)
V x H - ;<B€ 0E = 0. (20-17)
Taking the curls of Eqs. 20-16 and 20-17,
V x V x E + jcono V x H = 0, (20-18)
V x V x H — j f coe 0 V x E = 0, (20-19)
or, usin g the identi ty of Pr ob . 1-32,
V(V • E) - V2
E + jcofio V x H = 0, (20-20)
V(V • H) - V2
H - jco€0 V x E = 0. (20-21)
Finally, substituting Maxwell's equations,
V2
E = - e0/ (
0w
2
E , (20-22)
V2
H = -e0p0co2
H. (20-23)
Comparing now with Eq. 2 0 - 1 3 , we find th at these differential equa
tions are those of a wave. Also, it follows that the field vec tor s E an d H
can propagate as waves in free space at the velocity
c = l/feo^o)1 (20-24)
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20.5 Plane Electromagnetic Waves in Free Space 477
E q u a t i o n 2 0 - 2 4 l i nk s t h r e e b a s i c c o n s t a n t s o f e l e c t r o m a g n e t i s m : t h e
velocity of an electromagnetic wave c, the permit t ivi ty of f ree space e 0 , and
the permeabi l i ty of f ree space p 0. I t wi l l be rem em be red f rom Sec. 8 .1 th att h e c o n s t a n t p 0 was def ined a rb i t r a r i ly to be exactly An x 1 0 " 7 henry per
m e t e r . T h e c o n s t a n t e 0 can thus be dedu ced f rom the me asu red va lue for
the ve loc i ty o f e l ec t romagnet i c waves ,
c = 2.997 924 58 x 1 0 8 meter s / s econd: (20-25)
e 0 = l/ c
2p 0 = 8.854 187 82 x 1 0 " 1 2 f a r ad /mete r . (20-26)
We sha l l use the approx imate va lue o f c, n a m e l y 3 x 1 0 8 meter s per
second, which is accurate wi thin 7 par ts in 10,000.The permit t ivi ty of f ree space e 0 can a l so be de te rmined d i r ec t ly
f r o m m e a s u r e m e n t s i n v o l v i n g e l e c t r o s t a t i c p h e n o m e n a . T h e m e a s u r e m e n t s
l ead to the above va lue wi th in exper imenta l e r ro r , t he reby conf i rming the
t h e o r y .
20.5 PLANE ELECTROM AGNETIC WA VES IN FREE
SPACE: THE E AND H VECTORS
For a p lane e l ec t romagnet i c wave p ropaga t ing in the pos i t ive d i r ec t ion
of the z-axis , E is independent of x a n d y, a n d
V • E = <^f-£ = 0. (20-27)
CZ
The refore the z co m po ne nt of E can no t be a funct ion of z . W e shal l set
Ez = 0, (20-28)
s ince we are interes ted in waves , and not in uniform f ields .
The s ame a rgument app l i es to the H vec to r , and we can s e t
H z = 0. (20-29)
A p lane e l ec t romagnet i c wave p ropaga t ing in f r ee space i s there fore
transverse, s i n c e i t h a s n o l o n g i t u d i n a l c o m p o n e n t s .
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47 8
y
Figure 20-4 Decomposi t ion of the vector E into two vectors E t a n d E , .
W e no w as sum e tha t the wave i s p l an e-po la r i zed wi th i ts E v e c t o r
a lways po in t ing in the d i r ec t ion o f the x -ax i s . Th i s does no t invo lve any
los s o f genera l i ty s ince any p lane-po la r i zed wav e can be cons idered to be
the sum of two waves tha t a r e p lane -po la r i ze d in perp end icu la r d i r ec t ion sand in phase. For example, in Fig. 20-4, the vector E can be r eso lved in to
t w o m u t u a l l y p e r p e n d i c u l a r v e c t o r s E i a n d E2.
For a p lane-po la r i zed wave hav ing i t s E vector in the di rect ion of
the x-axis (Fig. 20-5) ,
(20-30)
Let us now subs t i tu t e th i s va lue o f E in to Eq . 20-17 :
J k
0 0 (20-31)
H x Ey 0
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47 9
(a)
H
E x H
E x H
• XXX
x M
(b)
Figure 20-5 Fh e E and H ve ctors for a pla ne electro ma gne t ic wave t ravel ing in the
posi t ive direct ion along the z-axis . (a) Fhe variat ion of E and H with r a t a
par t i cu la r moment . The two vec tors a re in phase , but pe rpendicula r to each o the r .(b) The cor responding l ines of force as s een when looking down on the xz -p lahe .
The l ines represent the electr ic f ie ld. The dots represent magnet ic l ines of force
coming out of the paper, and the crosses represent magnet ic l ines of force going
in to the paper . The vec tor E x H gives the d i rec t ion of propaga t ion .
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480 Electromagnetic Waves
On the left-hand side we have set the derivatives with respect to x and to
v equal to zero because we have a plan e wave prop agati ng a lon g the z-axis.
We have also set H2 equal to zero, from Eq. 20-29. Thus
-— Hy = jco€0E0 expjco I t - - J , (20-32)
j~zHx= 0. (20-33)
Replacing the operator d/dz by the factor —jco/c,
Hx = 0, (20-34)
H = ce0E0 expjco (t - ^ j. (20-35)
= (e0/n0)ll2Ei. (20-36)
Therefore H is perpendicular to E, and
E/H = l / e0c = u0c = (P-o/eoV2 = 377 ohms, (20-37)
E/B = c = 3.00 x 10 8 meters/second. (20-38)
The E and H vectors of the wave are perpendicular and oriented in
such a way that their vector product E x H points in the direction of
propa gat ion, as in Fig. 20-5. The E and H vectors are in phase, since E/H
is real, and they have the same relative magnitudes at all points at all times.
The electric and magnetic energy densities are equal and in phase,since
( l / 2 ) e 0 £2 = (\/2Ui
0H
2
. (20-39)
At any instant, the total energy density is therefore distributed as in Fig. 20-6.
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48 1
€ 0 £ 2
Figure 20-6 The energy dens i ty e0E2 = p 0H
2 as a function of z for a plane
e lec t romagne t i c wave t rave l ing a long the r -ax i s in f ree space .
20.6 THE POYNTING VECTOR <f IN FREE SPACE
We have found tha t a p lane e l ec t romagnet i c wave in f r ee space p ropaga tes
in the di rect ion of the vector E x H. Let us calculate the divergence of this
vector for an y elect romagnet ic f ie ld in f ree space. From Prob. 1-28.
V • (E x H ) = — E • (V x H ) + H • (V x E). (20-40)
Now, us ing the Maxwel l equa t ions fo r V x H and for V x E,
V - ( E x H ) = - E - e o - ^ - H - M o ^ , (20-41)ct ct
=
" ^ ( !f
°£ 2 + 2 ' ' o / / j < 2 ( M 2 )
I n t e g r a t i n g o v e r a v o l u m e t bo un de d by a su r f ace S , and u s ing the d ivergence
theorem on the l e f t -hand s ide ,
J , (E x H ) • d a = - l- I e 0E2
+ ~ p0H2j ch . (20-43)
Th e in tegra l on the r igh t i s the sum of the e l ec t r ic and ma gne t i cenergies , acc ord ing t o Sees . 4 .4 an d 13.3. Th e r igh t-h an d s ide is thu s th e
energy los t per uni t t ime by the volume r , and, s ince there must be con
serv at io n of energ y, the lef t -hand s ide m ust be the tota l out w ar d f lux of
e n e r g y t h r o u g h t h e s u r f a c e 5 b o u n d i n g t .
T h e q u a n t i t y
! f = E x H (20-44)
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482 Electromagnetic Waves
EXAMPLE: LASER BEAM
A laser beam carr ies a power of 20 gigawatts and has a diameter of 2 mil l imeters .
Let us calculate the peak values of E and of B.
From Eq . 20-47 ,
21 / 2
£ _ = 21
10 3 20 x 10'
,2 .66 n x 1 0 ~6
2.2 x 10 9 vol t s /meter .
(20-49)
(20-50)
Thi s is an e no rm ou s electr ic f ield; it corres pon ds to a voltage dif ference of ab ou t a
qu ar t er of a volt over a distan ce of 1 0 " 1 0 meter , which i s abou t the d iameter o f
an a t o m .We can now find B 0 f rom Eq. 20-38:
2.2 x 10 "
3 x 1 0s
7.3 teslas. (20-51)
Thi s is abo ut ten t imes larger tha n the ma gne tic induc tion betw een the pole pieces
of a power fu l pe rma nen t magn et .
is called the Poynting vector. W he n in te gra te d over a c losed sur face , it gives
the to ta l ou tw ar d f low of energy p er un i t t ime .
I n a p l an e e l ec t r o m ag n e t i c wav e , t h e Po y n t in g v ec to r p o in t s i n th ed i r ec t io n o f p r o p ag a t io n o f th e wav e . It s i n s t an ta n eo u s v a lu e a t a g iv en
po int in spa ce is E x H an d , acco r d in g to Eq . 2 0 - 3 7 ,
Sf = — £ 2 k = c e 0 £ 2 k . (20-45)
The t ime average of ff is
•^av = (^ 0E; ms)ck = [ ( l / 2 ) e 0 E 2 ] c k = (l/2)(€0/n
0)"
2
E2
0k, (20-46)
= 2.66 x 1 0 ~ 3 £ r
2
m s k watts /meter2
. (20-47)
Th e energy can therefore be con s ide red to t ravel wi th an ave rage dens i ty
e0£
r
2
m s= ( l / 2 ) e 0 £ 2 = ( l / 2 ) e 0 £ r
2
m s + (l/2)u0H2
ms (20-48)
a t t h e v e lo c i ty o f p r o p ag a t io n ck .
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20.7 Plane Electromagnetic Waves in Non-Conductors 483
20.7 PLANE ELECTROMA GNETIC WA VES
IN NON-CONDUCTORS
I n a n o n - c o n d u c t o r w i t h a p e r m i t t i v i t y e r e 0 a n d a p e r m e a b i l i t y nrpt
0,
Maxwel l ' s equa t ions a r e s imi la r to Eqs . 20-14 to 20-17 , excep t tha t e 0 a n d
p.0
are r ep laced by e r e 0 a n d fi rfJ.0- T h e w a v e e q u a t i o n s a r e t h e r e f o r e
V 2 E = - e , € 0 ^ 0 m 2 E , (20-52)
V 2 H = - e r e 0 / ( n u 0 c o 2 H , (20-53)
and al l the resul ts of Sees . 20.4 to 20.6 apply, i f one replaces e 0 b y e r e 0 a n d
Ih by n rHo •The phase ve loc i ty i s now
v = 1 ™ = C
- m . (20-54)
Th e phase ve loc i ty in no n-c on du c to r s i s the re fore l es s tha n in fr ee space ,
a n d t h e index of refraction is
n = c/v = (e rfi r)112
. (20-55)
I n a n o n - m a g n e t i c m e d i u m , p r = 1 and the inde x of ref ract ion isequa l to the square roo t o f the r e l a t ive permi t t iv i ty :
n = el12
. (20-56)
N o t e , h o w e v e r , t h a t n and e r are bo th func t ions o f the f r equency . We have
discussed br ief ly the var ia t io n of e r with f requency in Sec. 7 .7 . Since tables of
n are usua l ly compi led a t op t i ca l f r equenc ies , whereas e r i s usua l ly me asure d
a t much lower f r equenc ies , such pa i r s o f va lues cannot be expec ted to
c o r r e s p o n d .
Also , as in Sec. 20.5,
H y = (e r€ 0/p rfi 0)ll2E x. (20-57)
In non-conduc tor s , t he E and H vec to r s a r e in phase and the e l ec t r i c
a n d m a g n e t i c e n e r g y d e n s i t i e s a r e e q u a l :
( l / 2 ) e r e 0 £ 2 = ( l / 2 ) ^ / i 0 H 2 . (20-58)
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484 Electromagnetic Waves
The to ta l ins t an taneous energy dens i ty i s thus e r e 0 £ 2 o r p rpQH2, a n d t h e
average total energy dens i ty is e r e 0 £ r
2
m s o r p rp 0H2
mi. The average va lue o f
the Poynt ing vec to r i s
^ = \ ( ^ - T Eok = (^ f E2
mX (20-59)2 W ' o /
= ( e r e 0 £ r
2
m s ) rk . (20-60)
= 2.66 x 1 0 " 3 ( e r / i 0 1 / 2 £ r
2
m s k w a t t s / m e t e r 2 . (20-61)
Th e average va lue o f the Poy nt in g vec to r i s aga in eq ua l to the phase ve loc i ty
mul t ip l i ed by the average energy dens i ty .
20.7.1 EXAMPLE: LASER BEAM
If the 20 gigawat t laser beam of Sec. 20.6.1 is in a glass whos e in dex of refra ction is
1.6, then e r "2 is 1.6, u r i s uni ty , and £„ is smaller than in air by a factor of 1.6 1 2 o r
1.26. Then E 0 in th e glas s is 1.7 x 10 9 vol ts per m eter .
To c a lc u la t e B 0 we can use the fact tha t the P oy nt in g vec tor {j)E0H0 i s the sam e
in the glass as in the air; s ince £ 0 is sm aller in the glass by a factor of 1.26. B 0 m u s t
be larger by the sam e fac tor .
You can check tha t th is ma kes the e lec tr ic energy de ns i ty equ al to the ma gne t ic
energy dens i ty (Eq. 20-58) .
20.8 SUMMARY
For any p lane wave p ropaga t ing in the pos i t ive d i r ec t ion o f the r - ax i s a t a
veloci ty v, t he d i s tu rb anc e a t z , a t t he t ime t. is th e sa m e as th at at r = 0.
at a previous t ime f — (z/v). I t is as sum ed th a t there i s no a t t en ua t ion . T he
q u a n t i t y v is the phase velocity.
F o r a p l a n e , u n a t t e n u a t e d s i n u s o i d a l w a v e ,
x 0 cos to / (20-2)
where a 0 is the amplitude, an d the qua n t i ty be tween b rack e t s is the phase.
Sur faces o f con s tan t ph ase a r e wavefronts. The wave leng th i s
/. = V I. (20-5)
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T h e wave equation is
o r , i n th ree d imens ions .
20.8 Summary 485
(20-11)dz 2 v 2 dt 2
Th e kn ow n spec t rum of e l ec t ro ma gnet i c waves ex ten ds from 10 to
1 02 4 her t z .
In f ree space, the wave equations for electromagnetic fields a re
V 2 E = -e 0p 0«)2
E, (20-22)
V 2 H = -e0p0oj2
H. (20-23)
T h e phase velocity of electromagnetic waves in free space is
c = l / ( e 0 / ( 0 ) 1 / 2 = 2.997 924 58 x 1 0 s m e t e r s / s e c o n d . (20-25)
Plane e l ec t romagnet i c waves in f r ee space a re t r ansver se . The i r E
a n dH
vec tor s a r e o r thogona l and o r i en ted so tha t Ex H
points in thed i r e c t i o n o f p r o p a g a t i o n . T h e m a g n i t u d e s o f E a n d H are related to give
equa l e l ec t r i c and magne t i c energy dens i t i es :
4 = W e 0 )1 / 2 = 377 ohm s . (20-37)
H
T h e q u a n t i t y
ff = E x H (20-44)
i s cal led the Poynting vector. I t i s exp ressed in wa t ts per squa re meter , an d,wh en i ts no rm al ou tw ard c om po ne nt i s in t egra ted ove r a c losed sur face ,
i t g ives the e l ec t romagnet i c power f lowing ou t th rough the su r f ace .
F o r a p l a n e w a v e , Sf i s eq ua l to the energy dens i ty m ul t ipl i ed by the
phase ve loc i ty .
In non-conductors, the phase veloci ty is
v = c/(erp r)l
>2 (20-54)
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486 Electromagnetic Waves
PROBLEMS
20-1 PLANE WAVE IN FREE SPACE
A plane e lec t romagne t i c wave of circular frequency co p r o p a g a t e s in free space
in the di rec t ion of th e z-axis .
Show tha t , f rom Maxwel l ' s equa t ions ,
k x E = n0cH,
k x H = - e 0 c E ,
w h e r e k is the uni t vec tor in the di rec t ion of th e z-axis , and c is the veloci ty of l i gh t in
a v a c u u m .
LOOP ANTENNA
A 3 0 - m e g a h e r t z p l a n e e l e c t r o m a g n e t i c w a v e p r o p a g a t e s in free space, and its
peak electric field intensity is 100 mil l ivol ts pe r m e t e r .
C a l c u l a t e the peak vol t age induced in a 1.00 square mete r , 10- turn rece iv ing
l o o p o r i e n t e d so t h a t it s p l a n e c o n t a i n s the n o r m a l to a wave front and forms an a n g l e
of 30 degrees with the e lec t ri c vec to r .
POYNTING VECTOR
a) Ca lcula te the electric field intensity of t h e r a d i a t i o n at the surface of t h e sun
from the fo l lowing da ta : power rad ia ted by the sun, 3.8 x 1 02 6 w a t t s ; r a d i u s of the
s u n , 7.0 x 10s meters .
b ) W h a t is the electric field intensity of s o l a r r a d i a t i o n at the surface of the
e a r t h ? The average d i s t ance be tween the sun an d the e a r t h is 1.5 x 1 01 1 m e t e r s .
c ) Ca lcula te the va lue of •'/'at th e surface of t he ea r th , neglec t ing absorpt ion in
t h e a t m o s p h e r e .
20-4E SOLAR ENERGY
C a l c u l a t e the area requi red for p r o d u c i n g one m e g a w a t t of electr ic po we r from
sola r energy at the surface of the ear th , a s suming tha t the average eff ic iency over one
year is 2 % . The efficiency of solar cells is present ly about 15°, , at n o r m a l i n c i d e n c e .
See P rob . 20-3 .
20-2
20-3E
a n d , in n o n - m a g n e t i c m e d i a fji r = 1 ) , t h e index of r e f r ac t ion n is r e la ted
to the r e la t ive permi t t iv i ty by the r e la t ion
n = e r
1 / 2
. (20-56)
T h e v e c t o r s E and H are in p h a s e , and the electr ic and magnet ic energy
dens i t i es are e q u a l .
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Problems 48 7
20-5
20-6
20-7D
20-8E
20-9
POYNTING VECTOR
An electromagnet ic wave in ai r has an electr ic f ie ld intensi ty of 20 vol ts rms per
meter . I t i s absorbed by a sheet having a mass of 10~ 2 ki logram per square mete r and
a specif ic heat capaci ty of 400 jou les per ki log ram kelvin.
Assuming tha t no hea t i s los t , ca lcula te the ra t e a t which the t empera ture r i s es .
POYNTING VECTOR
A long superconduc t ing so lenoid ca r r i e s a cur rent tha t inc reases wi th t ime .
a ) Dr aw a ske tch show ing, on a longi tud ina l c ros s -sec t ion of the so leno id ,
A, -dA/dt, H , and E x H .
b) Can yo u expla in the ex i s t ence and the or i en ta t io n of the Poy nt ing vec tor?
POYNTING VECTOR
A c i rcula r ly pola r i zed wav e resul ts from the superp os i t ion of two waves tha t
a re (a ) of the same f requency a nd a mp l i tude , (b) p lane-po la r i zed in pe rpe ndicu la r
direct io ns , an d (c) 90 out of pha se.
Show that the average value of the Poynt ing vector for such a wave is the sum
of the average va lues of the Poynt ing vec tors for the two p lane-pola r i zed waves .
Hint: Proceed as in P rob . 16-16b.
POYNTING VECTOR
A wire of radius a has a res is tance of R' ohms per meter . I t carr ies a current I.
Sho w that the Po yn t ing v ector a t the surface is directed in wa rd and th at i t gives
the correct Joule power loss of I2R' wat t s pe r mete r .
COAXIAL LINE
A s t ructure des igned to guide a wave along a prescribed path is cal led a waveguide.
Th e s imples t type is the coaxial l ine i l lus t rated in Fig. 20-7. An electr om agn et ic
wave propaga tes in the annula r reg ion be tween the two coaxia l conduc tors , and the re
is zero f ield outs ide. Fh e me diu m of pr op ag at i on is a low-loss dielectr ic . (See Pr ob s .
6-5 and 6-6.)
We as su me tha t a wave pro pag a tes in the pos i t ive d i rec t ion of the z -axi s and
that there is no reflected wave in the opposi te direct ion. We also neglect the res is tance
o f t h e c o n d u c t o r s .
As in a cyl indrical capaci tor , E is radial and inversely proport ional to the radius .Since we have a wave.
w h e r e K is a cons tan t . We sha l l cons ide r an E poin t ing ou twa rd to be pos i t ive .
F h e r a d i a n l e n g t h X i s the same as for a plane wave in the same dielectr ic .
exp /' ojt —
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488
Figure 20-7 Coa xial l ine. See Pro b. 20-9.
a ) Draw a longi tudina l s ec t ion through the l ine and show vec tors E a t a g iven
ins tant over a t l eas t one wave length , va ry ing the l engths of the a r rows according
to the s t rength of the f ie ld.
Solution: See Fig. 20-8.
b) F ind the vol t age of the inner conduc tor wi th respec t to the oute r one .
Solution:
{ ; ; E - d r = K e x p . / U - l j j ; ; ^ ,
K i n — e x p 7 wf -
I l l
(2)
t 1 1 t t I 1 t
I t " ' " t 1 1 t t 1
Figure 20-8 Sect ion through a coaxial l ine showing the electr ic f ie ld intensi ty
(ver t i ca l a r rows) and the cur rent (hor i zonta l a r rows) over 1.5 wa vele ngth s at a
given ins tant . The wave t ravels from left to r ight . The surface charge densi ty
i s pos i t ive where the cur rent a r ro ws poin t r ight , and n ega t ive where they
point left.
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Problems 489
c) The v e c t o r s E and H are re l a t ed as in a plane wave , so t h a t
The magnet ic f ie ld is a z i m u t h a l .
F i n d the c u r r e n t on the i n n e r c o n d u c t o r . An equal current f lows in the o p p o s i t e
d i rec t ion a long th e o u t e r c o n d u c t o r .
N o t e t h a t V an d / are i n t e r re la t ed .
Solution: F r o m A m p e r e ' s c i r c u i t a l law (Sec. 9.1),
/ = 2nR,H = 2n R t ^ — j ^ exp. / (at - ^Y (4)
Kef2
( z\
= exp / {cot . (5)
60 P
V * /
Figure 20-8 shows how the cur rent va r i es a long the l i ne . Note tha t £, H, and /
a re all in p h a s e .
d) Ca lcula te the t ime averaged t ransmit ted power, f i rs t from the p r o d u c t VI, and
tf^en by i n t e g r a t i n g the Poynt ing vec tor over the annula r reg ion .
Solution:
K / = K l n ^ ^ c o s
2
( c o f - ~ Y (6)R, 60 V;
-/
( ) a v =
1 2 0 " RW a t t S '
EH = -=- co s- cot — - , (8)
' \ /'o } \ *>
rR 22nrclr K2
e]2
CR 2 dr.., CR 2 Inrclr ts.-e, C-R, dr(EHV, = -K
2
ef2
— = - = ^ - I - . ( 9)(
'a l
2 J*> r2
120 JR
' r
£•21/2 n
= ^ — ^ l n — w a t t s (10)12 0
COAXIAL LINE
A coaxial l ine has an i n n e r d i a m e t e r of 5 mil l imete rs and an o u t e r d i a m e t e r of
20 mil l imeters . The o u t e r c o n d u c t o r is g r o u n d e d ; th e i n n e r c o n d u c t o r is he ld at + 2 2 0
vol ts . The c u r r e n t is 10.0 amperes .
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490 Electromagnetic Waves
Integrate the Poynting vector over the annular region between the conductors,
and compare with the power 220 x 10.0 watts.
20-1 ID REFLECTION AND REFRACTION. FRESNEL'S EQUATIONS.
When light or, for that matter, any electromagnetic wave, passes from one
medium to another, the original wave is separated into two parts, a reflected wave and
a transmi tted wave. There are two ex ceptions: in total reflection there is only a reflected
wave, and a wave incident upon an interface at the Brewster angle gives only a trans
mitted wave.
The relative amplitudes of the incident, reflected, and transmitted waves are
given by Fresne/'s equations.
Let us find Fresnel's equations for an incident wave polarized with its E vector
parallel t o the pla ne of incidence as in Fig. 20-9. The plane of incidence is the plane
containing the incident ray and the normal to the interface.
Of course the angle of incidence is equal to the angle of reflection and, according
to SnelTs law.
iij sin 0, = sin 0,.
Figure 20-9 Th e incident, reflected, and tran smit ted waves when the incident
wave is polarized with its E vector parallel to the plane of incidence. The
arrows for E and H indicate the positive directions for the vectors at the
interface. See Prob. 2 0 - 1 1 .
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Problems 491
T h e q u a n t i t y n si n 0 i s therefo re con serve d on cross ing the interface. To tal reflect ion
occurs when this equat ion gives values of s in 0, that are larger than uni ty,
a ) F ind Fresne l ' s equa t ions
by us ing the condi t ions of cont inui ty for the t angent i a l component s of E and of H a t
the interface (Sees. 7.1 and 14.10).
b) Draw curves of E 0r/E oi a n d o f - £ 0 , / £ 0 , as funct ion s of the angle of inciden ce
for a wa ve in air incid ent o n a glas s surface w ith », = 1.5.
c) Sh ow th at there is no reflected ray w hen
The angle of incidence is then cal led Brewster ' s angle . The reflected ray disappears a t
Brewster ' s an gle only if the incid ent ray is polar ized w ith i ts E vector in the p lane of
incidence as in Figure 20-9.
d) Sh ow tha t , for l ight pr op ag at i ng in glass with an index of refract ion of 1.6,the Brewster angle is 32.0 degrees .
Bi + 0, = re/2.
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APPENDIX A
VECTOR DETINITIONS, IDENTITIES,AND THEOREMS
DEFINITIONS
1) v / =dx
+ — jcy
+ 7 T k
2) V - A =6A
X
dx+ ^
dy
3) V x A =(dA
\ dy
dA
' ~ 7 z
4) V 2 / =d 2
f
dx 2 + dy 2
c . 1, i M \ . , <".-l.. <"••!
c z dx / l e x f v
5) V 2 A = V2
/4,i + V 2 / 4 y j + V2
/ lzk
IDENTITIES
6) a x (b x c) = b(a • c) — c(a • b)
7) V( /g ) = / V 0 + a V f
8) V • (./A) = (V/) • A + / (V • A),
9) V • (A x B) - B • (V x A) - A • (V x B),
10) V x ( /A) = (V/) x A + ./'(V x A),
11) V x V x A = V(V • A) - V 2 A
/ 1 \ r.
12) V I - I = - j , where the g rad ien t i s ca lcu la ted a t the source po in t (x\ y', z') and r ,
is the unit vector from the source po in t (x 1, y', z') to the field point (x, y, z).
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Theorems 493
13) V j J = — " 2 -, w h e r e t h e g r a d i e n t i s c a l c u l a t e d a t t h e f ield p o i n t , w i t h t h e s a m e
u n i t v e c t o r .
THEOREMS
16 ) D i v e r g e n c e t h e o r e m : f A * da = f V • Xdx,
w h e r e S i s t h e c l o s e d s u r f a c e t h a t b o u n d s t h e v o l u m e i.
1 7) S t o k e s ' s t h e o r e m : ( £ A • dl = J* s (V x A) • d a .
w h e r e C i s t h e c l o s e d c u r v e t h a t b o u n d s t h e s u r f a c e S .
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APPENDIX B
SI UNITS AND THEIR SYMBOLS
Quantity Unit Symbol Dimensions
Length meter m
Mass kilogram k g
Time second s
Temperature kelvin K
Current ampere A
Frequency hertz 11/ 1 s
Force newton \ k g - m / s 2
Pressure pascal P a N / m 2
Energy joule J N - m
Power watt W .1 s
Electric charge coulomb c A -s
Potential volt V J /C
Conductance Siemens s A /V
Resistance o h m < > V /A
Capacitance farad 1 c / v
Magnetic flux weber W b V s
Magnetic induction tesla T W b / m 2
Inductance henry 11 W b / A
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APPENDIX C
SI PREFIXES AND THEIR SYMBOLS
Multiple Prefix Symbol
1 0 1 8 ex a E
10 1 5
peta P
1 0 1 2 tera T
10" giga G
10 6 mega M
10 3 kilo k
10 2 hecto h
10 d e k af da
I O - 1 deci d
1 0 " 2 centi c
1 0 " 3 milli m
1 0 " 6 micro
I O " 9 nano n
I O " 1 2 pico P
1 0 " 1 5 femto f
1 0 " 1 8 at to a
* This prefix is spelled "deca" in
French.
C a u t i o n : T h e symbol for the
prefix is written next to that fo r
the unit without a do t. Fo r ex
ample. m N stands for millinew-
ton, and m - N is a meter newton,
or a joule
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APPENDIX D
CONVERSION TABLE
Examples: On e meter equals 100 centimeters. One volt equals 10s
electromagnetic units of
potential.
CGS Systems
Quantity SI esu emu
Length meter 102
centimeters 102
centimeters
Mass kilogram 103
grams 103
grams
Time second 1 second 1 second
Force newton 105
dynes 10s
dynes
Pressure pascal 10 dynes/centimeter2
10 dynes/centimeter2
107
ergsnergy joule 107
ergs
10 dynes/centimeter2
107
ergs
Power watt 107
ergs/second
3 x 10"
107
ergs/second
Charge coulomb
107
ergs/second
3 x 10" 10 "1
Electric potenti al volt 1/300 108
Electric field intensity volt /met er 1/(3 x 104
) 106
Electric displacement coul omb/ mete r2
12tt x 105
4tc x 1 0 "5
Displace ment flux cou lom b 12tt x 10Q
4n x 10 "1
Electric polarization coulomb/meter2
3 x 105
1 0"5
Electric current ampere 3 x 10° 10 "1
Conductivity siemens/meter 9 x 109
1 0 " "
Resistance ohm 1/(9 x 101 1
) 109
Conductance Siemens 9 x 1 01 1
1 0 "9
Capacitance farad 9 x 1 01 1
1 0 "9
Magnetic flux weber 1/300 108
maxwells
Magnetic induction tesla 1/(3 x 106
) 104
gausses
Magnetic field intensity ampere/meter \2n x 107
47i x 10"3
oersted
Magnetomotance ampere 12ji x 10" (47t/10) gilberts
Magnetic polarization ampere/meter 1/(3 x 101 3
) 1 0"3
Inductance henry 1/(9 x 101 1
) 109
Reluctance ampere/weber 36 ti x 10" 4n x 10~9
N o T t : We have set c = 3 x 108
meters second.
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APPENDIX E
PHYSICAL CONSTANTS
Constant Symbol Value
G r a v i t a t i o n a l c o n s t a n t G 6.6720 x 1 0 " 1 1 N m 2 / k g 2
A v o g r a d o ' s c o n s t a n t N A6.022045 x 1 0 2 3 m o l " 1
Proton res t mass m p1.6726485 x 1 0 " 2 7 k g
Elec t ron res t mass1 1 1 e
9.109534 x 1 0 " 3 1 kg
Elem enta ry charge e 1.6021892 x 1 0 " 1 9 C
Permi t t iv i ty of vacuum «o 8.85418782 x 10" 1 2 F / m
Permeabi l i ty of vacuum /'n 4n x 1 0 " 7 Ft/fn
Speed of l ight in va cuu m c 2.99792458 x 10 8 m /s
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APPENDIX F
GREEK-ALPHABET
Letter Lowercase Uppercas e
Alpha a A
Beta $ B
Gamma y 1
Delta 5 A
Epsilon e i:
Zeta c /
E ta '/ IITheta eIota i i
Kappa K K
Lambda "/. A
M u /' M
N u V N
Xi { —
Omicron o ( )
Pi n 11
R h o P 1'
Sigma a
T a u T TUpsilon U r
Phi0 V
<i>
C h i X \
Psi <l>Omega w
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INDEX
Acce le ra t ion , 9
A d m i t t a n c e , 400Alte rna t ing cu r ren t , 265 , Ch aps . 16 -18 , 3 7 2 ,
3 7 3 , 449
A l t e rn a t in g - c u rr e n t m o t o r s , 3 2 9 , 3 9 4 , 4 3 1 , 4 3 2
Alternat ing vol tage. See Alternat ing current
A m m e t e r , 115
c l ip -on , 365, 450
A m p e r e , 111
Ampere ' s c i r cu i t a l law, 212 , 2 2 1 , 34 0
A m p e r e - t u r n , 22 2
Ampl i f i e r , ope ra t iona l , 145, 152, 154
Ampl i tude of a w a v e , 47 0
Analog compute r , 145
Analyze r :
cyl indrical e lectrostat ic , 58 , 61
g a s , 141
para l l e l -p la te , 58, 61
A n e m o m e t e r , h o t - w i r e , 141
Angle , so l id , 67
Aniso t ropy : in d ie lec t r i c s , 192
in magnetic mater ia ls , 342, 358
Antenna , 43 1
h a l f - w a v e , 435
l o o p , 486
w h i p , 435
Antenna image . 87
Area :
under a c u r v e , 44 8
as a vec to r . 7, 18
A r g u m e n t , 38 3Attenua to r . See Voltage divider
B a n d w i d t h , 42 1
Bamett effect , 27 8
Bat te ry cha rge r s , 39 3
Beta t ron , 37, 275
Boat test ing tank, 27 1
Boundary cond i t ions . See Cont inu i ty cond i
t i ons ; Uniqueness theorem
B r a n c h , 120
Br idge :i m p e d a n c e , 41 4
Wh e a t s t o n e , 131, 14 1, 151
W i e n , 414
Capac i t ance , 90, 91 , 281
stray, 422
Capac i t ive r eac tance , 400
Capac i to r :
a l ternat ing current in, 379, 400
cy l ind r ica l , 101
se l f - c lamping . 196
See also Paral le l-plate capaci tor
Capac i to r s :
in paral le l . 92
in ser ies , 93
Car te s ian coord ina te sys tem, 3
Cathode- ray tube , 55 , 248, 298
Charge :
conse rva t ion of, 1 1 3 , 466
invariance of, 240
m a g n e t i c . See Mo n o p o l e , m a g n e t i c
nea r an inf ini te grounded conducting plate .
79
sheet of, 71
sphe r ica l , 71
Charge densi ty
b o u n d , 159f, 165, 171
f ree , 160, 165, 182
in a h o m o g e n e o u s c o n d u c t o r , 114on the surface of a d ie lec t r i c , measure of,
174, 416
See also G a u s s ' s law
Charge dis tr ibut ion, potent ia l energy of, 95,
186
Ci rcu i t , e l ec t r i c , 120, 361
de l t a -connec ted , 126
different ia t ing, 152
equ iva len t , 126
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500 Index
Circui t , e lectr ic (continued)
in teg ra t ing , 154
magnetic force within an isolated, 321
pu lse -coun t ing , 155Q of, 421
RC , 135, 151
rectifier, 392, 393
RL , 294
RLC, 3 0 5 , 3 0 9 , 40 3
s ta r -connec ted , 126
T - , 4 2 3 , 42 7
Circu i t , magne t i c . See Magne t ic c i r cu i t
Circui t breaker , 32 3
Circular f requency, 37 3
Clamp, e lec t ros ta t i c . 197
Coaxial l ine , 171, 172, 304 , 3 0 5 , 48 7
capac i t ance of, 171
Coefficient of coup l ing , 287, 448
Coil :
field at center of, 232
H e l m h o l t z , 215, 216, 448
induc tance of, 365
moving-co i l me te r s , 32 9
R o g o w s k i , 302 , 4 4 8 , 450
s a d d l e , 23 3
to ro ida l , 224, 286
Coi l w ind ing , 42 2
Colloid thruster , 62
Compensa ted vo l t age d iv ide r , 41 2
Complemen ta ry func t ion , 134
Complex con juga te , 39 7
C o m p l e x n u m b e r s , 38 2
add i t ion , sub t rac t ion , mul t ip l i ca t ion , d iv i s ion of. 384
Complex p lane , 38 3
C o m p u t e r , a n a l o g , 145
C o n d u c t a n c e , 119, 361
Conduc t ing p la te , 74
charge near a, 79
Conduc t ing p la te s , pa i r of, 74
Conduc t ion cu r ren t , 454
Conduc t ion e lec t rons , 111
drift velocity of. 113
Conduc t iv i ty . 113 , 360
C o n d u c t o r , 73
electric field at surface of, 73
force on , 97 , 105, 186, 196f
g o o d , Ml, 454
h o l l o w , 74
h o m o g e n e o u s , 114
magnetic field of long cyl indrical , 22 3
sem i- , 11 1, 242ff
sphe r ica l , 96
See also Interface
Con juga te , 39 7
Conse rva t ion of cha rge , 113 , 466
Conservat ive f ie ld , 22, 29, 37, 45 , 47
Cons tan t of in teg ra t ion , 134
Cons tan t s , t ab le of p h y s i c a l , 49 7
Cont inu i ty cond i t ions (at an in terface):B, 2 2 7 , 34 6
D , 1 8 1 , 185
E , 182, 185
H , 34 7
V, 180
Convec t ion cu r ren t , 454
Corona d i scha rge , 198
C o u l o m b , 41
C o u l o m b f o r c e . C o u l o m b ' s law, 4 1 , 1 6 1 , 162
Counte re lec t romot ive fo rce , 44 5
Coupling, coeff ic ient of, 287. 448
Crab nebu la , 24 7
C u r l , 23. 25
of B, 2 3 0 , 455
of E, 266
of H , 4 5 6 , 46 5
Curren t , 36 0
c o n d u c t i o n , 454
convec t ion , 454
d i s p l a c e m e n t , 190, 454
e d d y , 266 , 2 7 3 , 4 3 8 , 450
e lec t ro ly t i c , 454
equ iva len t , 454
m e s h , 124
po la r i za t ion , 190, 454
t h r e e - p h a s e , 377, 387, 388
th ree -wi re s ing le -phase , 37 6
See also Alternat ing current ; Direct current
Cur ren t dens i ty , 111 , 36 0d isp lacemen t , 190 . 454
equ iva len t , 335, 336, 338
to ta l , 454
v o l u m e , in m a t t e r . 45 5
Current sheet , magnetic f ie ld c lose to, 231
Curren t source , 128
Curren t - s t ab i l i zed power supp ly , 130
Curren t t r ans fo rmer 448 , 450
Curve t r ace r , 189
Cyc lo t ron f requency , 24 6
Del ope ra to r , lOf
Del ta -connec ted c i r cu i t , 126
Delta-star t ransformation, 126, 406
Der iva t ive , t ime , 8
Diamagne t i c ma te r i a l s , 334, 342
Die lec t r i c , 157, 334
anisotropy in, 192
electr ic force on , 187
l inear and i so t rop ic , 163 , 165, 167
l iqu id , 187
n o n - h o m o g e n e o u s , 177
polar izat ion processes in, 191f
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Index 501
po la r i zed , 157
sheet of, 161
See also Interface
Die lec t r i c cons tan t , 164Die lec t r i c s t r eng th , 172, 193f
Different ia l equat ion, 134, 387
solut ions of, 134
Different ia t ing circui t , 152
D i o d e , 308f
v a c u u m , 86
Dipo le :
e lec t r i c , 49
m a g n e t i c , 20 7
Dipo le moment :
e lec t r i c , 51
induced electr ic , 157
m a g n e t i c , 20 7
Direct current , 37 2
Direc t -cu r ren t moto r , 44 4
D i s c h a r g e , c o r o n a , 198
Disp lacemen t cu r ren t dens i ty , 190 . 454
Disp lacemen t cu r ren t in paral le l-plate
capac i to r . 191
Disp lacemen t t r ansduce r , 216, 352
D i v e r g e n c e , 18, 20, 21
of B, 20 7
of D , 163
of E. See G a u s s ' s law
of H , 35 3
Divergence theorem, 21
D o m a i n , m a g n e t i c , 33 4
Drift velocity of conduc t ion e lec t rons , 113
Drop le t gene ra to r , 102D W I O H T , H. B., 16
Eddy cu r ren t , 266, 273
Eddy-cur ren t lo sses , 438 . 450
Effect ive value, 373
Efficiency of power t r ansmiss ion , 434
Elas tance , 94
Elec t re t , 169
shee t , 178
Electr ic c ircui t . See Circui t , e lectr ic
E lec t r i c condu i t s , 27 8
Elec t ri c d i sp lacem en t D , 162
d ive rgence of, 163
Electric field E, 4 3 , 44, 235, 267, 361
curl of, 266
d ive rgence of. See G a u s s ' s la w
energy densi ty in, 96 , 98 , 186, 189, 193,
194f
flux of. See G a u s s ' s law
induced , 262 . See also E l e c t r o m o t a n c e
line integral of, 2 6 1 . See also E l e c t r o m o
tance
m a x i m u m , in air, 82 , 99
t r ans fo rma t ion of, 241
Electric field intensity v x B, 23 7
Electr ic force:on conduc to r , 97 , 105, 186, 196f
on d ie lec t r i c , 187
and magne t i c fo rce , 31 2
See also Coulomb fo rce
Electr ic polar izat ion P, 157, 192
Electr ic susceptibi l i ty , . , 163
Electrolyt ic current , 454
E l e c t r o m a g n e t , 36 4
Electromagnetic f ie ld:
t ransformation of, 241
wave equa t ion for, 47 6, 483
See also Maxwel l ' s equa t ions
E lec t romagne t i c f lowmete r , 25 9
Elec t romagne t i c p rospec t ion , 27 2
Elec t romagne t i c pump, 32 4
Elec t romagne t i c spec t rum, 47 3
Elec t romagne t i c wave , 473
ene rgy dens i ty in , 4 8 1 , 4 8 2 , 4 8 3 , 48 4
energy f low in , 4 8 1 , 484
phase veloci ty of, in non-conduc to r s , 483
p l a n e , in free space, 476ff
p l a n e , in n o n - c o n d u c t o r s , 483
veloci ty of, in f re e s p a c e , 2 3 9 , 2 4 0 , 4 7 5 , 4 7 6
wave equa t ion for, 4 7 6 , 48 3
Elec t romete r , 175
E l e c t r o m o t a n c e , 3 6 1 , 46 1
i n d u c e d , 262 , 2 8 3 , 467
See also Counte re lec t romot ive fo rce
E lec t ron :c h a r g e of, 53
conduc t ion , 111
drift velocity of, 113
m a s s of, 53
Elec t ron-vo l t , 56
Elec t ros ta t i c c l amp , 197
Electrostat ic cyl indrical analyzer , 58, 61
Electrostatic field. See Electrostat ics
Electrostat ic force:
on conduc to r , 97 , 105, 186, 196f
on dielectr ic , 187
Electrostat ic paral le l-plate analyzer , 58, 61
Electrostat ic seed sort ing, 57
Electrostat ic separat ion of ore , 53
Elec t ros tat i c sp ray ing , 56
Elec t ros ta t i c s , Chaps . 2-4 and 6- 7, 45f, 73 ,
182 , 183 , 230 , 26 1
Energy.Se e Magnetic energy; Potent ia l : Poten
tial energy
Energy , so la r , 486
Energy convers ion , the rma l to e lec t r i ca l , 200
Energy dens i ty :
in electric field, 96, 98, 186, 189, 193 , 194f
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502 Index
Energy densi ty (continued)
in e l e c tr o m a g n e t ic w a v e , 4 8 1 . 4 8 2 , 4 8 3 , 4 8 4
in gas, 98
in magnetic field, 31 7Energy flow in e l e c tr o m a g n e ti c w a v e , 4 8 1 , 4 8 4
Energy s to rage , 193, 328, 329
Energy s tored in magnetic field. See Magne t ic
ene rgy
Equipotent ia l surface, 48, 49
Eule r ' s fo rmula , 38 3
Farad , 90
FA R A D A Y , M., 259
Faraday induct ion law, 263, 266, 334
Fer r i t e s , 438
Fer romagne t i c ma te r i a l s , 334 , 342, 343 , 358
Fie ld . 2
central force, 36, 45
conse rva t ive vec to r , 22, 29, 37. 45 . 47
sca la r , 3
vec to r , 3
See also Electric field E; Elec t romagne t i c
field; Electrostatics; Magnetic field
Field mil l , 41 6
Filter.
low-pass RC. 41 5
n o t c h , 423
RC . 412
Flammable -gas de tec to r , 142
Floa t ing wi re , 31 5
Flow of ene rgy . See Poynting vector
F l o w m e t e r , 25 9
Fluorescen t l amp , 445F l u x , 18, 20 , 21
of E. See G a u s s ' s law
l e a k a g e , 359, 361
m a g n e t i c , 204, 212, 2 6 3 , 36 0
Flux compress ion , 32 7
Fluxga te magne tomete r , 35 5
Flux l inkage, 26 3
Focus ing :
g a s , 253
m a g n e t i c , 24 8
Force :
cen t ra l , 36. 45
coe rc ive , 34 4
C o u l o m b , 4 1, 161, 162
coun te re lec t romot ive , 445
grav i t a t iona l , 37, 42f
Loren tz , 236, 2 4 1 , 24 2
See also Electr ic force; Lines of force;
Magne t i c fo rce ; Torque
Fourier ser ies , 37 3
F r e q u e n c y , 470
c i rcu la r , 37 3
cyc lo t ron , 24 6
Fresne l ' s equa t ions , 49 0
Gal i l ean t r ans fo rma t ion , 23 8
G a m m a r a y s , 4 7 3 , 474
Gas ana lyze r , 141
Gas focus ing , 25 3Gas p ressu re , 98
G a u g e s , 141 f
G a u s s ' s law, 69, 16 1, 1 6 3 , 2 1 1 , 2 2 3
Genera t ing vo l tme te r , 416
Genera to r :
d rop le t , 102
e lec t r i c . 26 4
h igh-vo l t age , 107
h o m o p o l a r , 324
m a g n e t o h y d r o d y n a m i c , 2 3 3 , 25 6
r a m p , 151
V an de Graaff, 2 3 1 , 416
See also Hal l gene ra to r
G r a d i e n t , 10
Grav i ta t iona l fo rce , 37, 42f
Greek a lphabe t , 498
Gun:
pa r t i c l e , 55
p l a s m a , 33 1
r a i l , 322, 331
Hall effect, 242 , 254, 259
Hall field, 24 3
Hall generator (Hall probe) , 216, 255, 350,
3 5 2 , 3 6 6 , 419
Hea t ing , induc t ion , 27 3
Hea t waves , 473
Helmhol tz co i l s , 215, 216, 448
H e n r y , 2 8 1 , 2 8 3 , 285
H e r t z , 37 3
High voltage. See Genera to r ; T ransmiss ion
l i ne ; Van de Graaff generator
H o d o s c o p e , 31 4
H o l e , 111
Homopola r gene ra to r , 324
H o m o p o l a r m o t o r , 324, 325
Hys te res i s , 344
lo s ses , 346, 450
I m a g e , 79
a n t e n n a , 87
Imag ina ry numbers , 38 2
I m p e d a n c e , 39 9
inpu t , 436, 441
m a t c h e d , 4 3 3 , 443
ref lected, 436, 447, 448
Impedance b r idge , 414
I m p e d a n c e m a t c h i n g , 4 3 3 , 443
Incandescen t l amp . 116
Induced e lec t r i c d ipo le moment , 157
Induced electric field intensity, 26 2
Induced e lec t romotance , 262 , 2 8 3 , 467
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Index 503
Induc tance , 284, 319
m u t u a l , 28 0
self-, 284
t r ans fo rma t ion of mutua l , in to s t a r , 408
Induct ion:
Fa raday law, 263 , 266, 334
See also Magne t ic induc t ion B
Induct ion furnace, 27 4
Induct ion heat ing, 27 3
Induc t ive r eac tance , 400
Induc to r , 284 , 295 , 412
al ternat ing current in, 379, 399
m u t u a l , 28 2
t ransient suppressor for, 308, 420
Induc to r s :
in paral le l , 29 1
in ser ies . 28 8
Ink-je t pr inter , 108
Integral :d o u b l e . 13
l i ne , 22
su r face , 12
t r ip le , 16
v o l u m e , 16
See also Line integral
Integrat ing circui t . 154
Interface, 180, 346
d ie lec t r i c -conduc to r , 165, 182, 183
d ie lec t r i c -d ie lec t r i c , 171 , 182, 183, 185
Inva r iance :
of e lectr ic charge. 24 0
of veloci ty of light in v a c u u m , 24 0
Ion beam, 75
pro ton beam, 83
Ion -beam d ive rgence , 25 2
Ion thruster , 61, 253
I ron , 36 6
p o w d e r e d , 438
t r ans fo rmer , 34 6
See also Fer romagne t i c ma te r i a l s
Joule effect , 116
Kirchof f ' s l aws , 120, 124, 402
Labora to ry re fe rence f r ame , 23 8
L a m b d a bar, 471Lamina t ions , 438 , 450
L a m p :
fluorescent, 445
incandescen t , 116
Lap lace ope ra to r , 29
L a p l a c e ' s e q u a t i o n , 78 , 167
Lap lac ian , 29
Leakage flux, 359, 361
L e n z ' s law, 265
Ligh t :
inva r iance of, 240
veloci ty of, 239, 475
Ligh t waves , 4 7 3 , 474, 475
Line faul t locat ion, 148
Line integral , 22
of A, 21 2
of B, 221
of E , 261. See also Elec t romotance
of H. See Ma g n e t o m o t a n c e
Lines of B, 204 , 216 , 313 , 316f
refraction of, 227, 348
Lines of fo rce , 49
bend ing of electr ic , 183. 185
e lec t r i c , 49, 98f
Lines of H , refraction of, 348
Loop , c i r cu la r , 205
Loop an tenna , 48 6
Loren tz fo rce , 236 , 2 4 1 , 24 2Loren tz t r ans fo rma t ion , 23 9
Losses :
eddy-cur ren t , 438, 450
hys te res i s , 346, 450
J o u l e , 116
Magne t :
ba r , 338
e lec t ro - , 364
p e r m a n e n t , 169, 358
Magne t ic cha rge . See Monopo le , magne t i c
Magne t i c c i r cu i t , 35 8
with ai r gap, 361
energy s tored in, 365
Ma g n e t i c d o m a i n , 33 4
Magne t ic ene rgy :
dens i ty , 31 7
stored in isolated circui t , 32 8
stored in long so leno id , 318, 319
stored in magne t i c c i r cu i t , 36 5
stored in magnetic field, 212, 318
Magnetic f ie ld , 23 5
Ear th ' s , 217 , 246f, 34 9
energy s tored in, 212, 318
of long cyl indrical conductor , 22 3
originat ing in magne t i zed ma te r i a l , 33 8
ro ta t ing , 394
Magnetic f ie ld intensi ty H, 339, 361curl of, 4 5 6 , 46 5
d ive rgence of, 353
line integral of. See Ma g n e t o m o t a n c e
l ines of, 348
Magnetic f luids , 37 0
Magnetic f lux, 204, 212, 263, 360
Magne t ic focus ing , 24 8
Magne t ic fo rce , 212, 312
exe r ted by an e lec t romagne t , 36 4
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504 Index
Magnetic force (continued)
within an isolated circuit, 32 1
between two two electric currents, 320
on a wire, 312See also Lorentz force
Magnetic induction B, 202, 360
curl of, 230, 455
divergence of, 207
flux of. See Magnetic flux
line integral of, 221
saturation, 34 3
See also Lines of B
Magnetic materials, 334, 342, 358
anisotropy in, 342, 358
isotropic an d linear, 342
magnetic field originating in, 338
Magnetic mirrors, 24 6
Magnetic pressure, 316, 325, 326
Magnetization M, 335, 350
Magnetization curve, 342
See also Hysteresis
Magnetized body, 33 4
Magnetohydrodynamic generator, 233, 256
Magnetometer, 355
Magnetomotance, 340, 360, 361, 462
Magnetoresistance multiplier, 36 7
Magnetoresistor, 255, 366 , 367
Mass, 246
effective, 25 4
rest, 246
Mass spectrometer, 25 0
crossed-field, 236f
Dempster, 249
M A X W E L L , J . C , 278
Maxwell's equations, 162, 208 , 230, 267 ,
Chap. 19, 456 , 458 , 465, 466
in integral form, 459
Mesh, 120
Mesh current, 124
Meter:
moving-coil, 329
See also Ammeter; Electrometer; Flow
meter; Gauges; Magnetometer;
Mass spectrometer; Potentiometer;
Thermometer; Voltmeter: Wattmeter
Micrometeorite detector, 35 1
Millikan oil-drop experiment, 24 0
Mobility, 254Modulus, 38 3
Monopole, magnetic, 208 , 216 , 301, 466
M O O R E , A. D., 312
Motor (electric), 31 2
alternating-current, 329, 394, 431 , 432
compound, 445
direct-current, 444
homopolar, 324, 325
series, 444
shunt, 445
Moving-coil meter, 329
Mylar, 194, 197
/7 -Type material, 243
Nitrobenzine, 192
Node, 120
Non-linear component in a circuit, 38 9
Non-linear resistor, 116
Notch filter, 423
Observer, 237
Ohm. 114
Ohm's law , 114
Output resistance, 130
p-Type material, 243
Parallel, connection in . See Capacitors;
Inductors; Resistors
Parallel-plate analyzer, 58, 61
Parallel-plate capacitor, 91,99,105,106, 174,
194, 200
dielectric-insulated, 166 , 17 8, 191
displacement current in, 191
energy stored in, 96
in infinite medium. 464
in liquid dielectric, 187
magnetic field in circular, 456
Paramagnetic material, 334, 342
Particle gun , 55
Peak current, 373Peaking strip, 355
Peak voltage, 37 3
Period, 470
Permeability, 341, 360
of free space, 202, 231 . 477
relative, 341, 342, 366
Permeance, 360, 361
Permeances in parallel, 362
Permittivity, 16 4
of free space, 42, 477
frequency-dependence of, 164, 19 If
relative, 163
temperature-dependence of, 192
Perpetual motion machine, 55 , 196
Phase, 373, 470
Phase angle of an impedance, 400
Phase-sensitive voltmeter, 413
Phase shifter, 415
Phase velocity, 47 1
Phasors, 38 5
addition an d subtraction, 38 8
diagrams, 389
when not to use, 389
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Index 505
Phosphate , separat ion from quartz , 53
P h o t o n , 474
Physical constants , table , 497
PIERCE, J .R., 469
Pinch effect , 25 3
Pi ran i vacuum gauge , 141
Planck ' s cons tan t , 208 , 474
P l a s m a , 246 , 250 , 251
Plasma gun. 331
Poisson ' s equa t ion , 75, 7 8, 167
Pola r i za t ion , e l ec t r i c , P, 157, 192
Polarizat ion current , 190, 454
Polarizat ion processes in d ie lec t r i c s , 19If
Polarized dielectr ic , 157
Polarized wave:
c i rcu la r ly , 487
p lane - , 478
Poles:
m a g n e t i c , 33 9Nor th an d S o u t h , 21 8
See also Mo n o p o l e , m a g n e t i c
Potent ia l :
e lectr ic (scalar) , V, 4 5 , 2 1 1 , 26 7
line integral of vec to r , 212
vec to r , A, 209, 2 1 1 , 267, 301
Potent ia l divider . See Voltage divider
Potent ia l energy (of a charge dis tr ibu
t i on) , 95 , 186
Poten t iomete r , 140
Powdered i ron , 43 8
P o w e r , 39 7
dissipated in a r e s i s to r , 116, 390
wat tme te r , 419
Power d i s s ipa t ion , 42 9
Power factor , 397, 431
cor rec t ion , 445
Power supp ly , 130 , 415
Power t ransfer :
from source to load , 432
th rough t r ans fo rmer , 44 2
Power t r ansmiss ion :
efficiency of, 434
See also Transmiss ion l ine
Poyn t ing vec to r , 481 , 484, 487
Pressure:
g a s , 98
m a g n e t i c , 316, 325 , 326
Pressu re gauge , 142Printed-circui t board, 176
Printer , ink-je t , 108
Product of two vec to r s :
scalar , or dot, 4
vec to r , or c r o s s , 6
Prospec t ion , e l ec t romagne t i c , 27 2
Pro ton beam, 8 3 . See also Ion beam
Pro ton bomb, 104
Pulse -coun t ing c i r cu i t , 155
P u m p , e l e c t r o m a g n e ti c , 32 4
Q of ac i rcu i t ,
42 1
Q u a d r a t u r e , 38 0
Q u a n t u m of r ad ia t ion , 474
Radian l eng th , 47 1
Radia t ion :
quan tum of, 474
Van Allen bel ts , 246
See also Cathode- ray tube ; Gamma rays ;
X- rays
Rad ia t ion re s i s t ance , 43 1
R a d i o w a v e s , 47 3
Rail gun, 322 , 331
R a y s . See Cathode- ray tube ; Gamma rays ;
X-rays
Reac tance , 400capac i t ive , induc t ive , 40 0
Rea l numbers , 38 2
Rectif ier c ircui ts , 392, 393
Refe rence f r ame , 23 8
l abora to ry , 238
Reflect ion, 490
Ref rac t ion , 490
index of, 483
of lines of B, 227, 348
of lines of D and E , 183, 185
Rela t iv i ty , 237, 238, 239, 241
Relay , 364, 369
Reluc tance , 360, 361
Reluc tances in se r i e s , 36 2
R e m a n e n c e , 34 4
Remote - read ing mercury the rmomete r ,
418
Res i s t ance , 114, 122, 3 6 1 , 400
ou tpu t , 130
radiat ion, 43 1
source , 130
Resistive film, 138
Resist ive net, un i fo rm, 148
Resistojet , 138
Res i s to r , 115, 122
l inear , 115
non- l inea r , 116
o h m i c , 115
power dissipated in, 116, 390vo l tage -dependen t , 116
See also Magne to res i s to r
Resis tors:
in paral le l , 118
in series, 118
R e s o n a n c e , 405, 421
ser ies , 40 5
Reten t iv i ty , 34 4
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506 Index
Right-hand screw rule . 3, 6, 25
RMS v a l u e , 37 3
Rogowsk i co i l , 302 , 4 4 8 , 450
Root mean square,37 5
Rowland r ing , 342, 353
Rutherford experiment , 56
Sadd le co i l s , 23 3
Satel l i te , a l t i tude control lor, 331
Satel l i te thruster . See Thrus te r
Saturat ion magnetic induct ion, 34 3
Scalar potent ia l . See Potent ia l
Seed sort ing. 57
Semiconduc to r , 1 11 , 242f f
Se r ie s , connec t ion in . See Capac i to r s ;
Inductors; Resis tors
Series LRC c i rcu i t , 305 , 309, 403
Series motor , 444
Ser ie s r e sonance , 405S e r v o m e c h a n i s m , 140
Surface tension, 103
Surface as a vec to r , 7, 18
Susceptibi l i ty:
e lec t r i c , 163m a g n e t i c , 3 4 1 , 342
Sys teme in te rna t iona l d ' un i t e s . See SI sys
te m of units
T-c i rcu i t s , 423 , 427
Tef lon , 172
Telev i s ion tube , 248, 298
T e s l a , 20 2
T h e r m i s t o r , 141
T h e r m o m e t e r , 141
r emote - read ing mercury , 41 8
T h e v e n i n ' s t h e o r e m , 129, 432
Three -phase cu r ren t , 37 7
Three -phase supp ly , 387, 388
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