Livro - Basic Laws of Electromagnetism por Irodov - Mir Publisher (1986)

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ofelectromagnetism•

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" Basic laws of electromagnetism

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I.E.IrodoYBasic laws

ofelectromagnetismTranslated from Russian

byNatasha Deineko and Ram Wadhwa

Mir Publishers IOSCOI'

~~CBS

CBS PUBLISHERS & DISTRIBUTORS4596/1-A, 11 Darya Ganj, New Delhi - 110 002 (India)

,

1,~

First pUblbhed HI86Hevisl.'d from the 1983 RUl'sian edition Preface

@> II3~aTe,lbClIlC' <d1hICIHaJI 111 KOJI3') , 1983'ij; English translation. Mir Publishers. 19aiJ

ISBN: 81-239-0306-5

For sale in India only

Copyright © English Translation, Mir Publishers, 1986

Printed at:Nazia Printers, Delhi - 110 006

The main idea behind this book is to amalgamate the description of the basic concepts of the theory and the practicalmethods of solving problems in one book. Therefore, eachchapter contains first a description of the theory of thesubject being considered (illustrated by concrete examples)and then a set of selected problems wi th solutions. The problems are closely related to the text and often complement it.Hence they should be analYfed together with the text. Inauthor's opinion, the selected problems should enable thereader to att~jn a deeper understanding of many importanttopics and /to' visualize (even without solving the problemsbut just by going through them) the wide range of applications of the ideas presented in this book.

In order to emphasize the most important laws of electromagnetism, and especially to clarify the most difficult topics, the author has endeavoured to exclude the less important topics. In an attempt to describe the main ideas conciseIy, clearly and at the same time correctly, the text has beenkept free from superfluolls mathematical formulas, and themain stress has been laid on the physical aspects of the phenomena. With the same end in view, various model representations, simplifying factors, special cases, symmetry considerations, etc. have been employed wherever possible.

SI units of IDtaSUrements are used throughout the book.However, considering that the Gaussian system of units isstill widely used, we have included in Appendices 3 and 4the tables of conversion of the most important quantitiesand formulas from SI to Gaussian units.

The most important statements and terms are giYen initalics. More complicated material and problems involvingcumbersome mathematical calculations are set in breviertype. This material can be omitted on fIrst reading withoutany loss of continuity. The brevier type is also used fot'problems and examples.

Irette'" aasie La,"First Indian Edition: 1994Reprint: 1996Reprint: 1998Reprint: 1999Reprint: 200I

This edition has been published in IndIa by arrangement withMir Publishers, Moscow

All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrievalsystem without permission, in writing, from the publisher.

Published by :Satish Kumar Jain for CBS Publishers & Distributors.4596/I-A, II Darya Ganj, New Delhi - 110 002 (India)

I. Irodov

The book is intended as a textb k fdents specializing in physics (i~ t~O f or undergraduate stu-on general physics) It' c I e bramework of the courseteachers. . . an a so e used by university

LTNhel~uthor is grateful to Prof. A.A. Detlaf

. . \.aptsov who re' d I and Readernumbet. of valuable VIewe t Ie manuscript and made a

comments and sllg'gestions.

t. Electrostatic Field in a Vaeuum

1.1. Electric Field • • • . . .t,2. The Gauss Theorem .1.3: Applications of the Gauss Theorem . .t,4. Differential Form of the Gauss Theorem1.5. Circulation of Vector E. Potential . . .1.6. Relation Between Potential and Vector E1.7. Electric Dipole . • • . . . •Problems •..........

it

itt6192325303439

5to

Contems

.. . . .. ..

. . .. .. ..Preface ••.••••List of Notations • . .

Prefacefi

2. A Condudor in an Electl'08tatic Field 45

2.1. Field in a Substance . . . . . . . • 452.2. Fields Inside and Outside a Conductor . . • 472.3. Forces Acting on the Surface of a Conductor 492.4. Properties of a Closed Conducting Shell . 5t2.5. General Problem of Electrostatics. Image Method 542.6. Capacitance. Capacitors 57problems •....•. 60

3. Eledm Field in Dielectrics • 67

3.1. Polarization of Dielectrics 673.2. Polarization 703.3. Properties of the Field of P 723.4. Vector D . . . . . . . . 76.3.5. Boundary Conditions . . . 803.6. Field in a HomogeneouS Dielectric 84Problems •.•.• 86

4. EIlt'TgY of Electric Field • . . . • • • • 94

4.1. Electric Energy of a System of Charges . . . . ., 944.2. Energies of a Charged Conductor and a Charged·

Capacitor 994.3~ Energy of Electric Fidd . . . . . tOO4.4. A System of Two Charged Bodies t054.5. Forces Acting in a Diplectric t06Problems •............ t to

8...... _~ Contents

"I,

5. Direct Current . . . • • . . . • • • . . . • .

5.1. Curr~nt Density. Continuity Equation ..5.2. Ohm s Law for a Homogeneous Conductor5.3. Generalized Ohm's Law . . • . . . . . .5:4. Branched Circuits. Kirchhoff's Laws . . .5.5. Joule'3 r qw .......•••....5.6. Transient blCesses in a Capacitor CircuitProblems •.......•..••••.•

6. Magnetic Field in a Vaeuum

6.1. Lorentz Force. Field B6.2. The Biot-Savart Law . . .. '.6.3. Basic Laws of Magnetic Field . • • • • . • . . .6.4. Applications of the Theorem on Circulation of Vec-

tor B ...........•••..•...•6.5. Diffe!e~tia! Forms of Basic Laws of Magnetic Field6.6. Ampere s Force .6.7. Torque Acting on a Current Loop •6.8. Work Done upon Displacement of Current LoopProblems .....

7. Magnetic Field in a Substance .•.•••

7.1. Magnetization of a Substance. Magnetization Vec-tor J ........•.•..•.......

7.2. Circulation of Vector J . . . • .7.3. Vector H . . .7.4. Boundary Conditions for Band H .7.5. Field in a Homogeneous Magnetic7.6. Ferromagnetism .......•Problems

8. Relative Nature of Electric and Magnetic Fields

8.1. EIElctromagnetic Field. Charge Invariance8.2. Laws of Transformation for Fields E and B8.3. Corollaries of the Laws of Field Transfo;m~tio~8.4. Electromagnetic Field InvariantsProblems '" . . . ••....

116

ft6119122126129133136

141

141145148

150154155159161163

172

172176178182185188192

198

198200206208209

9.7. Energy and Forces in Magnetic FieldProblems .

to. Muwell's Equations. Energy of Electromagnetic Field

10.1. Displacement Current . . . . . . . .10.2. Maxwell's Equations. .... . . . .10.3. Properties of Maxwell's Equations . . . .10.4. Energy and Energy Flux. Poynting's Vector10.5. Electromagnetic Field l'\foml'ntumProblems .....

11. Etedric Oscillations

11.1. Equation of an Oscillatory Circuit11.2. Free Electric Oscillations .11.3. Forced Electric Oscillations11.4. Alternating CurrentProblems "

Appendices it" •

1. Notations for Units of Measurement . . .2. Decimal Prefixes foI'" Units of Measurement .....3. Units of Measurement of Electric and Magnetic Quan-

tities in SI and Gaussian Systems .4. Basic Formulas of Electricity and Magnetism in SI

and Gaussian Systems . .5. Some Physical Constants . .

3ubject Index . . . . . . . . . .

240244

253

253251261264268271

277

277280285290293

299

299299

299

300304305-

9. Electromagnetic Induction 217

9.1. Faraday's Law of Electromagnetic Induction. Lenz'sLaw ...•.............•.•• 217

9.2. Origin of Electromagnetic Induction 2209.3. Self-induction 2269.4. Mutual Induction . " ..••. 2319.5. Magnetic Field Energy . . . . . • • . 2349.6. Magnetic Energy of Two Current Loops 238

List of Notations

Vectors are denoted by the bold-face upright letters{e.g. r, E). The same letter in the standard-type print (r, E)denotes the magnitude of the. vector.

Average quantities are denoted by angle brackets ( >,e.g. (v), (P).

The symbols in front of the quantities denote:A is the finite increment of the quantity, viz. the differencebetween its final and initial values, e.g. AE = E z - E

I,

Acp = CPt - <fl;d is the differential "(infinitesimal increment), e.g. dE, dcp;6 is the incremental value of a quantity, e.g. 6A is the elementary work.

Unit 'vectors:ex, ell' e z (or i, j, k) are the unit vectors of Cartesian coordinates x, y, z;ep , eq., e z are the unit vectors of cylindrical coordinatesp, (f'. z;n is the unit vector of the normal to a surface element;'t is the unit vector of the tangent to the contour or to aninterface.

Time derivative of an arbitrary function f is denoted by

.af/at or by the dot above the letter denoting the function, f.Integrals of any multiplicity are denoted by a single sym-

bol ) and differ only in the notation of the element of inte

gration: dll is the volume element, dS is the surface ele

ment al;d dt is the element of length. Symbol 1, denotes the

integration over a closed contour or a closed surface.Vector operator V (nabla). The operations involving this

operator are denoted as follows: vcr is the gradient ofcp (grad cp), v· E is the divergence of E (div E), and V X Eis the curl of E (curl E).

Vector product is denoted la·: b], where a and barevectors.

(~rostatic Field in a Vacuum

1.1. Electric Field

Electric Charg~. At present it is knowntl~at ~iverse ~henomena in nature are based on four fundamentallllteractl~ns

among elementary particles, viz. strong, electromag,!etIc,weak, and gravitational interactions. Ea~h. type of I~ter

action is associated with a certain charactel'lstic of a particle.For example, the gravitational interaction. depends o.n t~e

mass of particles, while the electromagnetic lllteractlOn ISdetermined by electric charges. .

The electric charge of a particle is one of its basl~ charac-teristics. It has the following fundamental properties: .

(1) electrlc charge exists in two forms, I.e. it can be pOSI-tive or negative; . .

(2) the algebraic sum of charges in any electl'lcally lllSUlated system does not change (this statement expresses ·thelaw of 'conservation of electric charge); .

(3) electric charge is a relativistic invariant: its magllltude does not depend on the reference system, in othe~ words,it does not depend on whether the charge moves or IS fixed.

I t will be shown later that these fundamental propertieshave far-reaching consequences. '. ,

Electric Field. In accordance with modern theory, lllteracti,on among charges is accomplished through a field .. Anyelectric charge q alters in a certain way the .propertIes ~f

the space surrounding it, I.e. creates an electnc fielri.. ThiSmeans that another, "test" charge placed at some pomt ofthe field experiences the action of a for~e. .

Experiments show that the force F actmg on a fixed testpoint charge q' can always be represented in the form

F == q'E, (1.1)

where vector E is called the intensity of the electric iield at .a given point. Equation (1.1) shows that vector .E can bedefined as the force acting on a positive fixed HIllt charge.Here we assume that the test charge q' is sufficiently small so

12 1. Blectrodctic Fidd in JI FUllUm 1.1. Electric Field 13

(1.3)

where ri is the distance between the charge qi and the pointunder consideration.

This statement is called the principle of superposition ofelectric fields. It expresses one of the most remarkable properties of fields and allows us to calculate fIeld intensity ofany system of charges by representing this system 8!' an aggregate of point charges whose individual contribut.ions aregiven by formula (1.2).

Charge Distribution. In order to simplify mathematicalcalculations, it is of~n convenient to ignore the fact thatcharges have a discr te structure (electrons, nudei) andassume that they are" eared" in a certain way in space. Inother words, it is conv nient to replace the acturJl distribution of discrete pointIcharges by a fictitious continuousdistribution. This makes it possible to simplify calculationsconsiderably without introducing any significant erNr.

While going over to a continuous distribution, the conceptof charge dlWsity is introduced, viz. the volume density p,surface density 0, and linear density A., By definition,

dq • dq dqP=dV' o=dS' A.=df' (1.4)

where dq is the charg'e contained in the volume dV, on thesurface dS, and in the length dl respectively.

Taking these distributions into consideration we can represent formula (1.3) in a different form. For example, if thecharge is distributed over the volume, we must replace qjby dq == p dV and summation by integTation. This gives

E=....!- r pedl' =_1_ \ prdV (1.5)41t80 J r' 4nEo J r' ,

where the integration is performed over the entire space withnonzero values of p.

Thus, knowing the distribution of charges, we can completely solve the problem of fmding the electriy field intensityby formula (1.3) if the distribution is discrete or by formula (1.5) or by a similar formula if it is continuous. In thegeneral case, the calculation involves certain difficulties(although they are not of principle nature). Indeed, in orderto find vector E, we must first calculate its projections E~,

Ell' and E z, which means that we must take three integrf!.ls

that it.s intr?du~tion does not noticeably distort the fieldunder mveStl~atlOn (as a result of possible redistribut' fcharges creatmg the field). Ion 0

. The Field of 8 Point Charge. It follows directly from ex eri.ment (~Qulomb's law) that the intensity of the field :f ~1Jdxe~ pomt charge q at a ,distance r from it ean be repre.sente 111 the form

E=_1 .!L e I41t80 r 2 T' (1,2)

:~)er~l:~o, ~s the electric constant and eT is the unit vector ofe l"ad, l,:; vector r drawn from the centre of the field h

i~Je charge q i~ located, to the point we are interest:c't ~~eormula (1.2) IS written in SI. Here the eoefficient .

1/41t1,o "'= 9 X 109 m/F,t~le r d~a.rge q is measured in coulombs (C) and the field in tenSIt) E In, volts pe~ metre (Vim). Vector E is directed alon ror ~pposItely to it depending on the sign of the charueg~o~mula (1.2) expresses Coulomb's law in the "field" t q.

It .IS nnpol'tan: that the intensity E of the field created ~rm~

dlJ?Int charge is inversely proportional to the square of tYhl~tRn("e r All e . tIl e. ~: I'd .' . xperJmen a resu ts indicate that this law

IS HI 1 tor dIstances from 10-13 cm to several kilometresa~dl thedre

fare no .grounds to expect that this law will ~

VIO ate or larger distances.It should also be noted that the force actin on a te

~harg~ in th~ field created by a fixed point charie does n~:.ete~ .on wether the test charge is at rest or moves. ThisreLer~ t.? a system of fixed charges as well.(1 Prm.clplc of Superposition. Besides the law expressed bth· 2)fi :~ 81;0 follows from experiments that the intensity J

e .e1 0 a system of fixed point charges is equal to thevectOf

dsum of the intensities of the fields that would be

create by each ('harge separately:

!IE=~

1

,1

I, I

1. Electro.tcttc Fteld In • Vacuum

,-

151.1. Electric Field

E':;" q/2l 2 I q4.n:eox Vl~-J-x2 4.n:eox Jf~+X2

In this case also E ~ qI4n.8eX"s for x » 1 as the lield of a point charge.

problem: we must lind the.component dEx of the lield create~1 by the1·lement dl of the filament, having the charge dq, and then llltel:l'ratethe result over all the elements or the tilament. In this ca~e

1 AdtdEx=dE cos ':'=-4- -,- cos ':t,. ;leo r-

where A = q/2t is the linear charf{e density. Let ~IS reduce this equationto the form convenient for lIltpgratlOn, Figure 1.2 shows thatdl cos a = rda and r = x/cos a; consequently,

1 Ar Ib, AdEx = --·--2- -=-=-4--. cos ada.

4.n:eo r lleox

This ·expressio.l can be easily integrated:

ao AE = _f.._ 2 i cos a da = -4·-- 2 sin a o_

4.n:eox J ;leoxI)

where a o is the maximum value of the angle a, sin a o= 11}tr12 + x 2 •

Thus,E

z

t:--..:;;!l--oo

E

of the t.vpe (1.5). A8 a rul~, the problem becomes muchsimpler in cases when a system of charges has a certain symmetry, Let us consider two examples.

Example 1. The field on the axis of atbin uDiformly charged ring.A charge q > 0 is uniformly distributed over a thin ring of radius a.Find the electric field intensity E on tbe axis 'Of tbe ring as a functionof the distance z from itseentre.

It ean be easily shown that vector E in this case must be directedalong the axis of the ring (Fig. 1. t). Let us isolate element dl on the

r

rl'I,'1

ring in the vicinity of pOInt A. We write the expression for the component dEL of the field created by this element at a point C:

t Adl .dEL = -z-· -1- cos a,

..n.lle r

where A. = q/2rr.a. The values of r and a will be the same for all theelements of the ring, 'Ind hence the integration of this equation is simply reduced t{) the replacement of Adl by q. As a result, we obtain

E=-q- I4n.eo (al +zl)'fl

It can be seen that for z~ a the field E ~ qf4rr.eoz2, i.e. at large dis-tances the system behaves as a point charge. '

Example 2. The field of a ..niformlyeharged straight filaDlellt. Athin straight filament of length 2l is Uniformly charged by a charge q.Find ~he ~eld intensity E at a point separated by a distance x froIl1the mIdpomt of the filament and located symmetrically with respectto its ends.

I t is clear from symmetry considerations that vector E must bedirected as shown in Fig. 1.2. This shows the way of solving this

Fig. 1.1 l"ig. 1.2 Geometrical Description of Electric Field. If we knowvector E at each point, the electric field can be visually represented with the help of field lines, or lines of E. Such aline is drawn so that a tangent to it at each point coincideswith the direction of vector E. The density of the lines,Le. the number of lines per unit area normal to the lines isproportional to the magnitude of vector E, Besides, the linesare directed like vector E'. This pattern gives the idea aboutthe configuration of a given electric field, Le. about the direction and magnitude of vector E at each point of the field.

On the General Properties of Field E. The field E definedabove has two very important properties. The knowledge ofthese properties helped to deeper understand the very concept of the field and formulate its laws, and also made it possible to solve a number of problems in a simple and elegantway. These properties, viz. the Gauss theorem and the theorem on circulation of vector E. are associated with two mostimportant mathematical characteristics of all vector fields:the flux and the circulation. It will be shown below that in

- '\

17

Fig. 1.5

1.2. The Gauss Theorem~--------

Fig. 1.'(

E through a closed surface is equal to the algebraic sum of thecharges enclosed by this surface, divided by eo.

Proof. Let us fIrst consider the field of a single point chargeq. We enclose this charge by an arbitrary closed surface S(Fig. 1:4) and find the flux of E through the area element dS:

dlP =~s ex = -4.1 ..J." dS cos ex = -4q dQ, (1.8)

-~__~ neo r· neo

where dQ is the solid angle resting on the area element dSand having the vertex at the point where the charge q is located. The integration of this expression over the entiresurface S is equivalent to the integration over the entiresolid angle, Le. to the replacement of dQ by 4n. Thus weobtain <lJ = qleo, as is defined by formula (1.7).

It should be noted that for a more complicated shape of aclosed surface, the angles a may be greater thnn n12, andhence cos a hnd dQ in (1.8) generally assume eitlICr positiveor negative vlJlues. Tlius, dQ is an algehraic quantity: ifdn rests Oft the inner l'idp of the surface S, dQ > 0, while ifit rests on the outer side, dQ < O.

The Gauss Theorem. The flux of E through an arbitraryclosed surfaco S has a remarkable property: it depends onlyon the algebraic sum of the charges embraced by this surface, i.e.

I "_ 1I A'EdS·--Qln, (1.7)I j eo

where the circle on the integral symbol indicates that theintegration is performed over a closed surface.

This expression is essentially the Gauss theurem: the flux of

(1.6)

E

Fig. 1.3

Flux of E. For the sake of clarity, we shall use the geomet··rica1 description of electric field (with the help of lines

of E). Moreover, to simplify theanalysis, we shall assume that thedensity of lines of field E is equalto the magnitude of vector E. Thenthe number of lines piercing thearea element dS, the normal n- towhich forms angle ex with vector E,is determined as E ·dS cos a. (seeFig. 1.3). This quantity is just the

flux d<1l of. E through the area element dS. In a more compaetform, this can be written as

1.2. The Gauss Theorem

terms of these two concepts not only all the laws of electricity but also all the laws of magnetism can be describefJ,. Letus go over to a systematic description of these properties.

16

(!J= I E as.s

This is an algebraic quantity. since it depends not only onthe configuration of the field E but also on the choice of thenormal. If a surface is closed it is customary to direct thenormal n outside the region enveloped by this surface, i.e.to choose the outward normal. Henceforth we shall alwaysassume that this is the case. .

Although we considered here the flux of E, the concept offlux is applicable to any \'Setor field as well.

d<1l = En dS = E.dS,

where En is the projection of vector E onto the normal n tothe area element dS, and dS is the vector whose magnitudeis equal to dS and the direction coincides with the directionof the normal. It should be noted that the choice of the direction of n (and hence of dS) is arbitrary. This vector couldbe directed oppositely.

If we have an arbitrary surface S, the flux of E throughit can be expressed as

.,I

II

2-181

191.3. Applications of the Gauss Theorem~:.--;.;...:..:-'.-::...:.~---_:..:.

1. Electrostatic Field in a Vacuum18

I

Ii

In particular, this leads to the following conclusion' ifthe charge q is located outside a closed surface S the fluxof .E t.hr?ugh th~s surface is equal to zero. In order'to provetills, It IS suffiCient to draw through the charge q a conicalsurface tangent to the closed surface S. Then the integrationo! Eq. (1.8) o~er the surface S is equivalent to tho integratIon over Q (FIg. 1.5): the outer side of the surface S will beseen from the point q at an angle Q > O. while the inner side~t an angl~ -Q (the two angles being equal in magnitude):r~~ su~ IS equal to zero, and <1> =-c; 0, which" also agreeswltn (1./). In terms of fIeld lines or lines of E, this meansthat the num~er of lines entering the volume enclosed byth~ surface S IS equal to the number of lines emerging fromthIS surface.

Let us now consider the case when the electric field iscreate~ by a system of point charges ql' Q2' ...• In thiscase, III accordance with the principle of superposition E == E1 + E 2 + ..., where E1 is the field created by thecharge qil etc. Then the flux of E can be written fh the form

~EdS=~(EI-' E2 + ... )ciS

= .k-. EidS + A; £2 dS -'.- ... =ClJ i -I- (}J2 _!-J J I I I •.••

If- accordanc~ wi~h what was sa~d above, each integral on therIght-hand SIde IS equal to qi/eo if the charge qi is insidethe closed surface S and is equal to zero -if it is outside thesurf.ace S. Thus, the right-hand side will contain the algebraIC sum of only those charges that lie inside the surface S.

To c?mplete the proof of the theorem, it remains for usto consIder the case when the charges are distributed continuousl~ with the volume density depending on coordinates.In thIS case, we may assume that each volume element dVcontains a "point" charge p dV. Then on the right-hand sideof (1.7) we have

qln = ~ P dV, (1.9)

where the integration is performed only over the volumecontained within the closed surface S.

We must pay attention to the following important circum-

stance: while the licld E itself depolld:-; Oil the mutual configuration of all the charges, the [lux of E through an arbitrary closed surface S is determined by the algebraic sumof the charges inside the surface S. This means that if wedisplace the charges, the fwld E willlJC changed everywhere,and in particular on the surface S. Generally, the flux of Ethrough the surface S will also change. However, if thedisplacement of charges did not involve their crossing ofthe surface S, the flux of E through this surface would remain lJ,nchanged, although, we stress again, the field E itselfmay change considerably. What a remarkable property ofelectric field I

1.3. Applications of the Gauss Theorem

Since the p,eld E depends on the configuration of all charges,the Gauss t'heorem generally does not allow us to determine this field. However, in certain cases the Gauss theoremproves to be a very effective analytical instrument since itgives answers to certain principle questions without solvingthe problem and allows us to determine the field E in a verysimple way. Let us consider some examples and then formulate several general conclusions pbout the cases when application of the Gauss theorem is the most expedient.

Example 1. On the impossibility 01 stable eqUilibrium 01 a chargein an electric field. Suppose that we have in vacuum a system of fixedpoint charges in equilibrium. Let us consider one of these charges,e.g. a charge q. Cani~ be stable?

In order to answer this question, let us envelop the charge q by asmall closed surface S (Fig. 1.6). For the sake of defmiteness, we assume that q > O. For the equilibrium of this charge to be stable, it isnecessary that the field E creatl'd by all the remaining charges of thesystem at all the points of the surface S he directed towards the chargeq. Only in this case any small displacement of the charge q from theeqUilibrium position will gi ve rise to a restoring force, and the equilibrium state will actually he stable. But such a configuration of thefield E around the charge q is in contradiction to the Gauss theorem:the flux of E through the surface S is nl'gative, while in accordancewith the Gauss theorem it must he l'qual to zero since it is created bycharges lying outside the surface S. On tile other hand, the fact thatE is equal to zero indicates that at S()Illl' p()ints of thl' surface S vl'clorE is directed inside it and at some other lliJints it is llirect.ed outside.

1. iUectroltatlc Field In a VaclJlJm1.3. Applications 01 the Gauss Theorem 21

Hence it follows that in any electrostatic~lield_.acharge cannot be instable equilibrium.

Example 2. The field of a unHormly charged plane. Suppose thatthe surface charge density is o. It is clear from the symmetry of theproblem that vector E can only be normal to the charged plane. Moreover, at points symmetric with respect ~o t~is pl.ane, vectors E ohviously have the same magnitude but OPPOSite directiOns. Such a configuration of the field indicates that a right cylinder should be chosen forthe closed surface as shown in Fig. 1.7, where we assume that {1 > O.

to the field from the positively charged plane, while the lower arrows,to that from the negatively charged plane. In the space between theplanes the intensities of the fields bdng added have tlw same (Hrection, hence the result (1.10) will be doubled, and the resultant fieldintensity \\ill be

E=oho' (1.11)where !J sLands for the magnitude of the surface charge density. It canbe easily seen that outside this space the field is equal to zero. Thus,

(1.12,.. - _A_ (r> a).'r - 21teoT

£=0

------

Fig. 1.8 Fig. 1.9

in the given case the field is located between the planes and is uniform.

This result is approximately valid for the plates of finite dimensions as well, if only the separation between the plates is considerablysmaller than their linear dimensions (parallel-plate capacitor). In thiscase, noticeable deviations of the field from uniformity are observed.only near the edges of the plates (these distortions are often ignored incalculations).

F..xample 4. The field vf an infinite circular cylinder uniformlycharged ov('r th~ surface 1,0 that the charge A corresponds to its unit]en~th.

In this case, as follows from symmetry considerations, the fieldis of a radial nature, i.e. vector E at eaen pomt is-perpendicular tothe cylinder axis, and its magnitude depends only on the distance rfrom the cylinder axis to the point. This indicates that a closed surface here should be taken in the form of a coaxial right cylinder(Fig. 1.9). Then the flux of E through the enrlfaces of the cylinder i~equal to zero, while the flux through the lateral surface is ET 2 mhwhere E r is the projection of vPetor E onto the rad ius vector r coinciding with the normal n to the lateral surface of the cylinder of radimr and height h. According to the Gauss theorem, 1'7 r'21trh =0 Altho fOJ> a, whence

+o

Fig. 1.7Fig. 1.6

The flux through the lateral surface of this cylinder is equal tozero, and hence the total flux through the entire cylindrical surfaceis 2E l!.S, where l!.S is the area of each endface. A charge (J AS isenclosed within the cylinder. According to the Gauss theorem, 2E AS == (J tJ.S/eo, whence E = 0/2eo' In a more exact forni, this expressionmust be written as

En = 0/2eo' (1.10)

where En is the projection of vector E onto the normal n to the chargedplane the normal n being directed away from this plane. If (J > 0then En > 0, and hence vector E is directed away from the chargedplane, as shown in Fig. 1.7. On the other hand, if {1 < 0 then En < ~,and vector E is directed towards the charged plane. The fact thiit E ISthe same at any distance from the plane indicates thatJ the corresponding electric field is uniform (both on the right and on the left of .the plane).

The obtained result is valid only for an infinite plane surface,since only in this case we can use the symmetry considerations discussed above. However, this result is approximately valid for the regionnear the middle of a finite uniformly charged plane surface far fromits ends.

Example 3. The field of two parallel planes charged uniformly withdensities (J and -(J by unlike charges.

This fIeld can be easily found as superposition of the f,elds createdby each plane separately (Fig. 1.8). Here the upper arrows correspond

22 1. Electrostatic Fteld tn a Vacuum

r

i.e. inside a uniformly charged sphere the field intensity grows linearly with the distance r from its centre. The curve representing the dependence of E on r is shown in Fig. 1. to.

General ConcllJsions. The results obtained in the above

where Er is the projection of vector E onto the radius vector r coincidingwith the normal n to the surface at each of its points. The sign of thecharge q determines the sign of the projection Er in this case !is well.Hence it determines the direction of vector E itself: either away fromthe sphere (for q > 0) or towards it (for q < 0).

If r < a, the closed surface does not contain any charge and hencewithin this reg-ion E = 0 everywhere. In other words, inside a uniformly charged spherical surface the electric field is absent. Outsidethis surface the field decreases with the distance r in accordance withthe same law as for a point charge. .

Example 6. The field of a uniformly charged sphere. Suppose thata charge g is uniformly distributed over a sphere of radius a. Obviously, the field of such a system is centrally symmetric, and hence for determining the field we must take a concentric sphere as a closed surface. It can be easily seen that for the field outside the sphere we obtainthe same result as in the previous example [see (1.13)1. However,inside the sphere the expression for the field will be different. Thesphere of radius r < a encloses the charge q' = q (rla.)S since in ourcase the ratio of charges is equal to the ratio of volumes and is proportional to the radii to the third power. Hence, in accordance withthe Gauss theorem we have

1 (r}SE r·4nri =- q - •llo a 1.4. Differential Form of the Gauss Theorem

A remarkable proper~y of electri~ r~e~~s::fedsf~da ~rff:~:n?f~~:theorem suggests thdat ~htls the~rb~ml'lt~s a~ an i~strument for analysiswhich would broa en I s POSSI 1

andI~~~~~;~~~10 (1.7~ wticg is c~~:~r:: ~hi~h~:f:bii~~:~h~l ::l~~the differential formlo t eh agU~ensity p' and the changes in the fieldtion between the vo ume c ar e. .,intensity E in the vicinity of a gIven pOInt m space.

(1.14)1 . q

Er=-4- .... r (r<a),nllo a

whence

25

(1.20)

(1.21

1.5. Circulation of Vector E. Potential

1.5. Circulation of Vector E. Potential

Theorem on Circulation of Vector E. It is known frommechanics that any stationary field of central forces is conservative, Le. the work done by the forces of this field isindependent of the path and depends only on the positionof the initial and final points. This property is inherent inthe electrostatic field, viz. the field created by a system ojfixed charges. If we take a unit positive charge for the testcharge and carry it from point 1 of a given field E to point 2.the elementary work of t he forces of the field done over thEdistance dl is equal to E· dl, and the total work of the fiekforces over the distance between points 1 and 2 is defined a1

2

~ E dI.t

The Gauss theorem in the differential form is a local theorem:the divergence of the field E at a given point depends only on theelectric charge density p at this point. This is one of the remarkableproperties of electric field. For example, the field E of a point chargeis different at different points. Generally, this refers to the spatialderivatives oE:dox, oEy/By, and oEz/o% as well. However. the Gausstheorem states that the sum of these derivatives, which determines thedivergence of Ef"turns out to be equal to zero at all points of the field(outside the charge itself).

At the points of the field where the divergence of E is positive, wehave the sources of the fieIa (positive charges), while at the pointswhere it is negative, we have Binks (negative charges). The field linesemerge from the field sources and terminate at the sinks.

only in combination with a scalar or vector function by which it issymbolically multiplied. For example, if we forin the scalar productof vector V and vector E, we obtain

° ° aV·E=VxEx+VyEy..LVzEz=- Ex+- Ey+- Ez•I ax oy 0%

It follows from (1. i 7) that this is just the divergence of E.Thus, the divergence of the field E can be written as div E or V·E

(in both cases it is read as "the divergence of E"). We shall be usingthe latter, more convenient notation. Then, for example, the Gausstheorem (1.18) will have the form

1. Electrostatic Field i- a Tr•• F aCllllm

For this purpose we first rep t henveloped by a clo~d surface S i':r: : e charge q in the volume Vis~h~ volum~ charge ?ensity, averaged ~~ '1Vr = I(p) V, Where {p)~ sthl~uhte t.hlS expressIOn into Eq. (1 7) and d' e;;o bumeh r Then We

, w IC gives . IVI e ot Its sides by

~. ~ E dS=(p)/Bo• (1.15)

We now make the volume Y Je d .point we are interested in. In this c~ to zero. by c0';1tracting it to the~iluhaneof P at the given point of the S:' I~P) wJltohvlOusly tend to the

t- d side of Eq. (1.15) will tend toe I' an ence the ratio on theP Bo•

The quantity which is the limit of the ratio of t. E dS toIs called the divergence of the fi ld Ed' 'j' Vas V--+-Ohy defuIition. e an IS denoted by div E. Thus,

div E= lim ..!.- k. E dSv...o V 'j'. (1.16)

The dive!'lfence of any other v to fi .

furiway• It· follows from definitioneC(1 16)el

tdh

ls de~rmined in a similarctlon of coordinates. . at divergence is a scalarIn order to obtain the ex ressi .

we m~t, in accordance with f116,o~ fkr the. divergence of the field Edetermine the flux of E thrO~ h 't a e an Infinitely small volume Y'volu~e. and find the ratio of tfis f1~~ closed surface enveloping thi~ohtaIned for the divergence will d :I the volume. The expressiondfflte sys)tem (in different systems ~~o~~. th:e c~oice of the coordi-

erent. For example, in Cartesian coordi~~:e S'tl~ tU!D8 out to be. s 1 IS given by

div E = ~+ oEy _L oE%0:& oy I 0;-. (i.17)

Thus. we have found that as Y --+-' '.::nd,j. to plea, while the left-hand sideO~~i1~5)d" Its nght.hand side

e Ivergence of the field E' 1 d s IV E. Consequentlylame point through the equati~~re ate to the charge density at th~

f divE=pleo' I (1.18)This equation expre -th-----The form of man e sses .e Gauss theorem in the differential f

ira~IY simplified ilw~P[:::;~d::~i:~ir:;P~i~fftions.can be co~:r:tn artesian coordinates, the operator Vehasrth~ ;:;:tIal operator V.

V=i.!-+ 3-. aox j oy +k 8i" ' (1.19)

where I, J, and k are the unit vecto f thoperator V itself does not have any~~anin~X[-i r-. and Z-axes. The

. ecomes meaniD~fu J

I

I

26 1. Electrostatic Field tn a Vacuum 1.5. Ctrculation of Vector E. PotenUal 27

Q.E.D.A field haVing property (1 22) . I

Hence, any electrostatic field . IS ca led .the potential field.The theorem on' . lS a potentzal fleld.

to draw a numbe~I~Ctl?tIon of vector E ~akes it possiblesorting to calculations ~mfortant ~oncluslOns without 1'e-

- . e us conSIder two examples.'Example f TI r Id Iclosed. • Ie Ie ines of an electrostatic field E cannot be

This integral is taken alon '.therefore called the line inte:ra~ certam hne (path) and is

We shall now show that f' h .integral (1.21) of the path hrotm t e mdependence of linethat when taken along an b\ween two points it followsgral is equal to zero. Integr:I (~r;f)Y closed path, this iute-

. . over a closed contour iscalled the cIrculation of vector Ed' d. t... an IS enoted by 'V'

Thus we state that cireul t' f ..static field is equal to ze a ~on 0 vector E in anv electro-ro, I.e. "

(1.23)

where <p 1 and <P2 are the values of the function <P at thepoints 1 and 2. The quantity <P (r) defined in this way is

Indeed, if the oppollite were true and some lines of field E?lerJ!closed, then taking the circulation of vector E along this line we wouldimmediately come to contradiction with theorem (1.22). This meansthat actually there are no closed lines of E in an electrostatic field:the lines emerge from positive chargesand terminate on negative ones (or go F!;::::jto infinity). I • 'I

Example 2. Is the configuration of an I Ielectrostatic field shown in Fig. 1.12 -4 Ipossible? . lr

No, itisnot. This immediately becomes tclear if we apply the theorem on circula- I lr Ition of vector E to the closed contour L -.Jshown in the figure by the dashed line. ----The arrows on the contour indicate the •direction of circumvention. With such aspecial choice of the contour, the contri- Fig 1t2bution to the circulation from its ver- . .tical parts is equal to zero, since inthis case E .L dl and E ·dl = O. It remains for us to consider thetwo horizontal segments of equal length:,. The figure shows that thecontributions to t.lte circulation from these regions are opposite insign, and unequal in magnitude (the contribution from the upper segment is larger since the field lines are denser, and hence the value ofE is larger). Therefore, the ci'rculation of E differs from zero, whichcontradicts to (1.22).

Potential. Till now we considered the description of electric field with the help of vector E. However, there existsanother adequate way of describing it by using potential cp(it should be noted at the very outset that there is a one-toone correspondence between the two methods). It will beshown that the second method has a number of significantadvantages.

The fact that line integral (1.21) representing the work'ofthe field forces done in the displacement of a unit positivecharge from point 1 to point 2 does not depend on the pathallows us to!state that for electric neld there exists a certainscalar function <p (r) of coordinates such that its decrease isgiven by

(1.22)~Edl=O.1248

This statement is' called th hvector E. e t eorem on circulation of

In order to prove this th. . eorem.. we break an arbitrary closedpa.th Into two parts la2 and 2bl

2 (FIg. 1.11). Since line integral (1.21)

c:? (we denote it by i ) does not

1 depend on the path be~~een'points 1(a) (b)

b and 2, we have f = i '. On the.otb h d 12 • 12er an, it is clear that

Fig. 1.11 T (b) (b)

• 112 - - \ ,where \ is theIntegral over the same segment b b ,21 • ., 21direction. Therefore ut taken In the opposite

(n) (b) (a) (b)

~ 12 -I- L1 = ~ 11 - L2 = 0,

I ~

28 1. Electrostattc Field in tJ Vacuum 2U

called the field potential. A comparison of (1.23) with theexpression for the work done by the forces of the potentialfield (the work being equal to the decrease in the potentialenergy of a particle in the field) leads to the conclusionthat the potential is the quantity numerically equal to the potential energy of a unit positive charge at a given point of thefield.

We can conditionally ascribe to an arbitrary point 0 of thefield any value CPo of the potential. Then the potentials ofall other points of the field will be unambiguously determined by formula (1.23). If we change CPo by a certain valueAcp, the potentials of all other points of the field will changeby the same value.

Thus, potential cp is determined to within an arbitraryadditive constant. The value of this constant does not playany role, since all electric phenomena depend only on theeleetric field strength. It is determined, as .will be"'shownlater, not by the potentia] at a given point but by the·potential difference between neighbouring points of the field.

The unit of potential is the volt (V).

Potential of the Field of a Point Charge. Formula (1.23)contains, in addition to the definition of potential cp, themethod of finding this function. For this purpose, it is suf-

ficient to evaluate the integral ~ E dl over any path be

tween two points and than represent the obtained result asa decrease in a certain function which is just <p (r). We canmake it even simpler. Let us use the fact that' formula(1.23) is valid not only for finite displacements butffor elementary displacements dl as well. Then, in accorda~cewiththis formula, the elementary decrease in the potential overthis displacement is

-dcp =--0 E·dl. (1.24)

In other words, if we know the field E (r), then to find cpwe must represent E·dl (With the help of appropriate transformations) as a decrease in a certain function. This functionwillbo the potential cpo

Let us apply this method for finding the potential of the

field of a lixed point charge: •1 q IJ Ii r

E.dt -- -- - (' ·die- /trrf,o ,.2 r ,l.'l:'o r:!.

= -d (_1_ _.'L _i-const)4Jlco r I ,

where we took into account that cr dt =-= 1· (dl}r= dr, silleethe projection of dl onto en ~nd hence ?ll 1', is ~qual to ~heincrement of the magnitude of vector r, Le. dr. 'I he quantltyappearing in the parentheses under the diffe~enti~l is exactly i:p (1'). Since the additive constant contamed III ~he fo~'mula- does not play any physical role, it is usually omitted ill

order to simplify the expression for (p. Thus, the potentialof the field of a point charge is given by

I~ = __1 }L II (1.2:»__4n~_'_

The absence'6f an additive constant in this expressionindicates· that we conventionally assume that the potentialis equal to zero at infinity (for r -+ 00).

Potential of the Field ola System of Charges. Let a systemconsist of fixed point charges ql' q2' .... In accordancewith the principle of superposition, the field intensity at anypoint of the field is given by E = EI + E 2 + ... , wh~reEl is the field intensity from the charge ql' etc. By llsmgformula (1.24)1 we can then write

E·dl = (El + E2 + ...)dl = EI·dl + E 2 ·dl + ...= -dcpl - dCP2 - ... = --dcpl

where cp = 2jcpf, Le. the principle of superposition t~rns outto be valid for potential as well. Thus, the potential of asystem of fixed point charges is given by

I .- _1_ '" !!i- -I (1.26)cp - 4nco L..J r i 'I1. _

where ri is the distance froID the point charge qj 10 the pointunder consideration. Here we also omitted an arbitrary constant. This is in complete agreement witll the faet that anyreal system of charges is Lounded in space, and honce itspotential can bo taken equal to zero at inflIlity.

(1.28)

3t

(1.31)

(1.30)

1.6. Relation Between Potential and Vector I!:

pression with formula (1.24) gives

Ex = - orp/vx, (1.29)

wher~ the symbol of partial.derivative emphasizes that thefunctIOn rp ~x, y, z) must he differentiated only with respectto x, assummg that y and z are constant in this case.. In a similar wa~, w.e can obtain the corresponding expres

810.ns for the prOJectIOns E y and E z. Having determinedE", Ell' and E z, we can easily fmd vector E itself:

E= _ ( ocp i +~- j +~ k)ox iJy oz .

T~e quantity in the par~Iltheses is the gradient of the potentzaZ cp (grad cp or Vcp). We shall be using the latter moreconvenient notation and will formally consider VCP 'as theproduct of a symbolic vector V and the scalar cpo ThenEq. (1.30) can be represented in the form

Let ~s derive one more useful formula. We write the righthand sl~e of (1.24) in the form E·dl c=o R I dZ, where dl =. Id1l IS an elementary displacement and E I is the projec

tIOn of vector E onto the displacement dl. Hence

I E l = -ocpI8Z, I (1.32)

Le. the projection of vector E onto the direction of the dis-

L~. the field inte~sity E'is equal to the potential gradientWith the miflUS SIgn. This is exactly the formula that canbe used for reconstructing the field E if we know thefunction <p (r).

f E~ample. Find the field intensity. E if the field potential has theorm. (1~ ~ (x, y) = -axy, where. a IS a c~nstant; (2) cp (r) = -a'r,whe~ a Is.a constant vector and r IS the radlUs vector of a poiut underconslderahon.

(1) By using formula (1.30), we obtain E = a (yi -1- xj)._ (2) Let us first represent the function ip tls cp == --ax.x --. llll!! __

izz, where ax, ay and az are cl)nstants. Then with the help of formthil! a (1.30) we find E = axi -+ ayj + a k = a. It can be seen that in

..8 case the field E is uniform. Z

(1.27)

1.6. Relation Between Potential and Vector E

It is known that electric fwld is completely described byvector function E (r). Knowing this function, we can findthe force acting on a charge under investigation at any pointof the neld, calculate the work of fwld forces for any displacement of the charge, and so on. And what do we get byintroducing potential? First of all, it turns out that if weknow the potential cp (r) of a given electric fIeld, we can reconstruct the fwld E (r) quite easily. Let us consider thisquestion in greater detail.

The relation between rp and E can be established with thohelp of Eq. (1.24). Let the displacement dl be parallel tothe X-axis; then dl = i dx, where i is the unit vector alongthe X-axis and dx is the increment of the coordinate x. Inthis case

where (J is the surface charge density and dS is the elementof the surface S. A similar expression corresponds to thecase when the charges have a linear distribution.

Thus, if we know the charge distribution (discrete or continuous), we can, in principle, fmd the potential of anysystem.

E·dl = E·i dx = Ex dx,

where Ex is the projectioll of vector E onto the ullit vector i(and not on the displacement dll). A comparison of this ex-

where the integration is performed either over the entirespace or over its part containing the charges. If the chargesare located only on the surface S, we can write

1 ~ adS<p = 4neo -r-'

30 1. Electrostatic Field in a Vacuum-----------_._---- .-'-'-'=";~---

If the charges forming the system are distributed cUlItinuously, then, as hefore, we assume that each volume element dV contains 3 "point" charge p dV, where p is the charge

. density in the volume dV. Taking this into consideration,we call write formula (1.26) in a different form:

331.6. llelation Between Potential and Vector E--------------~_. ----------~ ------

On the Advantages o~ Potential. 1l was noted earlier that. electrostatic field is completely cltaracterizeu by vectorfunction E (r). Thea what is the use of introducing potential?There are several sound reasons for doing that. The conceptof potential is indeed very useful, and it is not by chancethat this concept is widely used not only in physics but inengineering as well..it. If we know the potential <P (r), we can easily calculatethe work of field forces done in the displacement of a pointcharge q' from point 1 to point 2:

Al2~ q' «PI - <pz), (1.33)

where <PI and..<vz are the potentials at points 1 anu 2. Thismeans that the required work is equal to the decrease in thepotential energy of the charge q' upon its displacement frompoint 1 to 2. Calculation of the work of the lield forces withthe help of formula (1.33) is not just very simple, but isin some cases the only possible resort.

Example. A df~rgc q is distributed over a thin ring of radius a.Find the work of the lield forces dune in the displacement of a pointcharge q' from the centre of tpe ring to i:llinity. .

Since the distributiun o[ the charge q over the ring is unknown, wecannot say anythiug detinite about the intensity E of the field createdby this charge. This means that we cannot calculate the work by eval-

uating the integral ~ q'E dl in this case. This problem can be easily

solved with the help of potential. Indeed, since all elements of thering are at the same distance a from the centre of the ring, the potential of this point is Ipo ~= q/4Jvou. And we know that cp = 0 at intinity.Consequently, the work A = q'cpo = q'q/4nlOoa.

2. It turns out in many cases that in order to find electricfield intensity E, it is easier lirst to calculate the potential

" lp and then take its gradient than to calculate the value of Edirectly. This is a considerable advantage of potential. Indeed, for calculating <p, we must evaluate only one integral,while for calculating E we must take three integrals (sinceit is a vector). Moreover, the integrals for calculating <P areusually simpler than those for Ex, By, and E z·

Let us note here that this does not apply to a comparatively small number of problems with high symmetry, inwhich the calculation of the lielu E directly or with the helpof the Gauss theorem turns out to be much simpler.

3-0181

1. Electrostatic Feeld in a Vacuum

Fig. 1.13

g2

placement dl is equal to the d' •. .p.otential (this is emphasized b;r~~~onal telflvfative .of therivative). sym 0 0 partial de-

Equipotential Surfaces Let . dequipotential surface, viz. 'the s:;a~~t:~ :l~e t~e tco~cept.ofpotentiallp has the same value. We shall sho~o~~~ 0 whIch:~ each point o.f the surface is directed along thean::::.~r~

e eqUipotentIal surface and. towards the decrease in thepotential. Indeed, it followsfrom formula (1.32) that theprojection of vector E ontoany direction tangent to theeq,!ipo~entialsurface at a givenpomt 18 equal to zero. Thismeans that vector E is normal to the given surface. Further, let us take a displaC&ment dl along the normal to thesurface, towards decreasingp. Then Olp < 0, and accordmg to (1 32) E > 0 .

v,ector E is. directed towards decreasing ~, ~r i~ the d;;;'tIon ?pposite. to ~hat of the vector Vlp.

It IS expedIent,.to draw equipotential surfaces in such away that the potential difference between two neighbourinsurfaces b~ t~e same. Then the density of e ui otentiJs~aces WI.ll vIsually indicate the magnitudes offieYd intensttIes a~ dIfferent points. Field intensity. will be higher .t e re~lOns ~h~re equipotential surfaces are denser '"th~pot~ntial rehef IS steeper"). \

Smce vector E ~s normal to an equipotential surface eve where, the field hnes are orthogonal to these surfaces. ry

Th F~guhedf l 3 shows a two-dimensional patt~rn of an electric fieldlin:s :: h liDes eorres~o~d to equipoten~ial rlUrfaces, while the MUdI. t. e mes of E. Such a representatIOn ean be easil v· tiledi~~~:t:yd\:tiYhhows dthe hdirec.t!o.n of vector E, the regiorrs wi::e field

~ g er an were 10 13 lower as we'l's tIl . "\h~~~~r steepness of the putential relief. SU(~ri a ;ai'tern ~~?':sed'tod. t' qilal~~i~lve answers to a number of .~lU.,.stiOJl8 sU~.h as ';In'wha.t

lrec ,IOn WI "charge placed at a certain point m~veP Whe I> •

iliagiutude o! the potential gradient hi~her? At which U:in~.;t.e1e orce actmg on the charge be greater~" et.c. A

(1.36)

1.7. Electric Dipole

from the negative to the positive charge:

p = ql, (1.35)

where q > 0 and I is the vector directed as p.It can be seen from formula (1.34) that the dipole field

depends on its electric mom~nt p.. It will be shown belowthat the behaviour of the dIpole III an external field .al~o

depends on p. Consequently, p is an important characterIstIcof the dipole. ' .

It should also be noted that the potential of the dI~)Qle fielddecreases with the distance r faste~ than the

z~otentlal of the

field of a point charge (in proportIOn to 1fr Illstead of 1fr).!In order to find the dipole field, we shall use formula (1.32)

a~d calculate the projections of vector E onto two mutuallyperpendicular directions along the unit vectors er and ~(Fig. 1.14b):

E aq> 1 2pcos~"",' r = - --ar = 4nllo -,-.- ,

E aq> 1 p sin ~t) = - r'~ = 4neo --,.-

Hence, the modulus of vector E will be

E=VE~+E6= J.: -?V-1-+-3-c-o-sz-tt. (1.37)'tJ,80 r

In particular, for tt = 0 and tt = n/2 we obtain the expressions for the field intensity on the dipole axis (E u) andon the normal to it (E.d:

Ell = ~80 2~, EJ. = 4~80 ~, (1.38)

Le. for the same r the intensity E u is twice as high as EJ.'The Force Acting on a Dipole. Let us place a dipole into

a nonuniform electric field. Suppose that E+ and E_ arethe intensities of the external field at the points where thepositive and negative dipole charges are located. Then theresultant force F acting on the dipole is (Fig. 1.i5a):

F = qE+ -qE_ = q (E+ - E_).

The difference E+ - E_ is the increment LiE of vector Eon the segment equal to the dipole length 1 in the direction

(1.34)

p(b)

eb

+q

/q P

Fig. 1.14

34 1. Electrostatic Field in a Vacuum----------'--'--'-.:-:..'-.:-:..~....:..::..:..::......:::..:......:=-:..=::.::::=-------


The Field of a Dipole. The electric dipole is a system of twoequal in magnitude unlike charges + g and - g, separatedby a certain distance l. When the dipole field is considered. ,

There are some other advantages in using potential whichwill be discussed later.

it is assumed that the dipole itself is pointlike, i.e. the distance r from the dipole to the points under consideration isassumed to be much greater than l.

The dipole field is axisymmetric. Therefore, in any planepassing through the dipole axis the pattern of the field isthe same, vector E lying in this plane.L~t us first find the potential of the dipole field and then

its illtensity. According to (1.25), the potential of the dipolefield at the point P (Fig. 1.14a) is defined as

1 (lJ '1) 1 q(r_-r.)<.p ,- 4itl::o 7..- - -,:: = 4ne

or.r_ •

Since r» l, it call 1e seen fro.:n Fig. 1.14a that r__ r + == 1cos \} and r +r_ = r2

, where r is the distance from thepoint P to the dipole (it is pointlikel). Taking this intoaccoulJt, we get

where p = ql is the electric moment oj the dipole. This quantity corresponds to a vector directed along the dipole axis

rI

3*

J

1. Electro$tatlc Field In a Vacuum

Fig. 1.15

~E aE~E=E+-E_=-. l = az l.

37

Fig. 1.16

~F

q...-----tpFI

q-----~


• We take the moment with respect to the centre of mass in orderto eliminate the moment of inertial forces.

where fJE~/az is the derivative of the corresponding projection of vector E again onto the direction of vector I or p.

Let a dipole with moment p be oriented along the symmetry axis of a certain nonuniform field E. We take the positivedirection of the X-axis, for example, Fas shown in Fig. 1.17. Since the incre- qe <= p.ment of the projection E~ in the direction of vector p will be negative, F:r< O.and hence vector F is directed to theleft, i.e. towards increasing field intensity. If we rotate vector p shown inthe figure through 90° so that the dipolecentre coincides with the symmetryaxis of the field, it can be easily seenthat in this position F:II: = O.

The Moment of Forces Acting on 8

Dipole. Let<4Is consider behaviour of a dipole in an external electric field in its centre-of-mass system and find outwhether the dipole will rotate or not. For this purpose, wemust find the moment. of external forces with respect to thedipole centre of mass*.~By definition, the moment of forces F + = qE+ and F _ =

= -qE_ with respect to the centre of mass C (Fig. 1.18) isequal to

M = [r+ X F+I + [r_ X FJ=[r+ X qE+l -[r_ X qEJ,

where r + and r _ are the radius vectors of the charges + qand - q relative to the point C. For a sufficiently small dipole length, E+ ~ E_ and M = [(r+ - r_) X qEl. It remains for us to take into account that r + - r _ = I andqI = p, which gives

IM= [p X EJ.! (1.41)

This moment of force tends to rotate the dipole so thatits electric moment p is oriented along the external field E.Such a position of the dipole is stable.

Thus, in a nonuniform electric field a dipole behaves as

f

(.~ ...~:-q~~q~-

<bJ z:s.JL

of vector I. Since the length of this segment is small, we canwrite

Substituting this expression into the formula for F, wefind that the force acting on the d'ipole is equal to

I F=p"*, (1.39)

where p = ql is the dipole electric moment. The derivativeappearing in this expression is called the directional deriva

tive of the vector. The symbol ofpartial derivative indicates that itis taken with respect to a certaindirection, viz. the direction coinciding with vector I or p.

Unfortunately, the simplicity offormula (1.39) is delusive: takingthe derivative fJE/fJl is a rather complicated mathematical operation.We shall not discuss this questionin detail here but pay attention tothe esse\"~e of the obtained result.First of all, note that in a uniformfield aEliJl = 0 and F = O. Thismeans that generally the force acts

on a dipole only in a nonuniform field. Next, i::l the general case the direction of F coincides neither with vector Enor with vector p. Vector F coincides in direction onlywith the elementary increment of vector E, taken along thedirection of I or p (Fig. 1.15b).

Figure 1.16 shows the directions of the force F acting on a dipolein the field of a point charge q for three different dipole orientations.We suggest that the reader prove independently that it is really so.

lf we are interested in the projection of force F onto acertain direction Xi, it is sufficient to write equation (1.39)in terms of the projections onto this direction, and we get

Fx = P a:zx, (1.40)

I'

follo,,":s: under the action of the moment of force (1.41),the. dIpole tends to.get oriented along the field (p tt E),whIle under the action of the resultant force (1.39) it is

position, the moment of external forces will return the dipole to the equilibrium position..

38 1. Electroltattc Field in a VacuumProblem' 39

£.

Fig. 1.20Fig. Lt9

1dEz = -4- (J dO.

JtEo

1 lJdSdEz =-4- -I cos 'It. (1'

JtEo r

In our case (dS cos 'It)/rz = dO is the solid angle at which the areaelement dS is seen from the point A, and expression (1) can be written as

E

Hence, the required quantity is1

E=-4- aQ.Jtto

It should he noted that at large distances from the disc, Q = SIrS,where S is the area of the disc and E = ql4nEors just as the field of thepoint charge q = lJS. In the immediate vicinity of the point O,thesolid angle g = 2a and E = a/2eo'

Solu.io1l. It is clear from symmetry considerations th.at o~ the discaxis vector E must coincide with the direction of this aXIS (FIg. 1.19).Hence, it is safWentto find the component d!£z from the charg; ofthe area e1emeBt tiS at the point A and th~n mtegrate the o.btamedexpression over the entire surface of lthe dISC. It can be easIly seenthat'

Problems

• t.t. A very thin disc is uniformly charged with sut'fa?e char~density lJ > O. Find the electric field intensity E on the axiS of thISdisc at the point from which the disc is seen at an angle .Q.

+q

-q

Fig. 1.17 Fig. 1.18

displaced towards the region where the field E has largermagnitude. Both these motions are simultaneous.

The Energy of a Dipole in an External Field. We knowthat the energy of a point charge q in an external field isW. qcp, where cp is the fIeld potential at the point of locatIOn of the charge q. A dipole is the system of two chargesand hence its energy in an external field is '

W = q+cp+ + q_cp_ = q (cp+ _ cp_),

where CP+ and cp_ are the potentials of the external field atthe points of location of the charges +q and -q. To withina quantity of the second order of smallness, we can write

alpcp+-cp-= az 1,

,,:here acp/Ol is the derivative of the potential in the directIOn of the vector l. According to (1.32), acp/al = -E It

and henen '"+ - '"_ ~ LE ,I - Ell. from which we Il"l

IW = -p.E. (1.42)

It follows from this formula that the dipole has the minimUt;D. energy (Wmin = -pE) in the position p tt E (thepositIOn of stable equilibrium). If it is displaced from this

40 1. Electrostatic Field in a VacuumProblems

• f .2. A thin nonconducting ring of radius R is charged with alinear density A = Ao cos cp, where '-0 is a positive eoDStant and cpis the azimuth angle. Find the electric field intensity E at the centreof the ring.

Solution. The given charge distribution is shown in Fig. t,20.The symmetry of this distribution implies that vector E at the point 0is directed to the right, and its magnitude is equal to the sum of theprojections onto the direction of E of vectors dE from elementarycharges dq. The projection of vector dE onto vector B Is

1 dqdE cos cp = -4- -R' cos cp, (1)neo

where dq = AR dlp = AoR cos cp dcp. IntegTating (1) over cp between 0and 2n, we find the magnitude of the vector E:

2nE "-0 r t d "-0. = 4neoR J cos cp cp = 4eoR •

oIt should be noted that this Integral is evaluated in the most simpleway if we take into account that (cost cp) = 1/2. Then

2n

~ cos' cp dcp=(cos' cp) 2n=n.o

• 1.3. A semi-infinite straight uniformly charged filament has acharge'" per unit length. Find the magnitude and the direction of thefield Intensity at the point separated from the filament by a distanC'-t! yand lying on the normal to the filament, passing through its end.

Solution. The prohlem is reduced to finding Ex and E., viz. theprojections of vector E (Fig. L21, where it is assumed fliat '" > 0).Let us start with Ex. The contribution to Ex from the-charge elementof the segment dz Is .

1 Adx.dEx =-4- -t-sma. (1)

neo r jLet us reduce this expression to the form convenient for integration.

In our case, dx = r dalcos a, T = ylcos a. Then "

dEx = -4A sin ex da.neoy

Integrating this expression over a between 0 and n/2, we find

Ex = "'/4neoy·

In order to find the projectiou Ey , it is sufficient to recall thatdEli diffeq from dE~ in that sin a"ln (1) is simply replaced by C08 lx.

This givesdEy = (A cos a da)/4neoy and Ey = A/4.tteoy·

We have obtained an interesting result: Ex = Ey independently of y,

a

Fig. 1.21

i.e. vector E is oriented at the angle of 45° to the filament. The modulus of vector E is

E=VE~+E~=j,. Y2i4rt8oY •

• 1.4. The Gauss theorem. The intensity of an electric field depends only on the coordinates x and y as follows:

E = a (xi + yj)/(x2 + y2),

where a is a constant, and i and j are the unit vectors of the X-andY;-axes. Find the charge within a sphere of radius' R with the centreat the origin.

Solution. In accordance with the Gauss theorem, the requiredcharge is equal to the flux of E through this sphere, divided by &0'

In our case, we can determine the flux as follows. Since the field E isaxisymmetric (as the field of a uniformly charged filament), we arriveat the conclusion that the flux through the sphere of radim(R is equalto the flux through the lateral surface of a cylinder having the sameradius and the height 2R. and arranged as shown in Fig. 1.22. Then

q=80 ~ E dS=8oErS,

where Er = aIR and S = 2nR·2R = 4nR2. Finally, we getq = 4n8oaR.

• 1.5. A system consists of a uniformly charged sphere of radiusR and a surrounding medium filled by a charge with the volume density p = aIr, where a is a positive constant and r is the distance from

43

Fig. 1.25

Problem.

Fig. 1.24

the surface density CT = CTo cos {Jo, where 0"0 is a constant and {Jo isthe polar angle...'

Solution. Let us consider two spheres of the same radius, havinguniformly distributed volume charges with the densities p and -poSuppose that the centres of the spheres are sep-arated by the distance 1 (Fig. 1.26). Then, inaccordance with the solution of the previous tzproblem, the field in the region of intersectionof these spheres will be uniform:

E = (p/380 ) 1. (1)

In our case, the volume charge differs from zeroonly in the surface layer. For a very small l,we shall arrive at the concept of the surfacecharge density on the sphere. The thickness ofthe charged layer at the points determined byangle -& (Fig. 1.26) is equal to 1 cos {Jo. Hence,in this region the charge per unit area is(J = pl cos {Jo = 0"0 cos a, where °0 = pl, andexpression (1) can be represented in the form Fig. 1.26

E = -(0"0/380) k,

where k is the unit vector of the Z-axis from which the angle it ii'Jmeasured.

• 1.8. Potential. Tho potential of a certain electric field has theform Ip = ex (xy - z2). Find the projection of vector E onto thp. direction of the vector a = i + 3k at the point M (2, 1, -3).

Solution. Let us first find vector E:E = -Vip = -a (yl + xj - 2zk).

of the spheres and of the distance between their centres. In particular,it is valid when one sphere is completely within the other or, in otherwords, when there is a spherical cavity in a sphere (Fig. 1.25).

• 1.7. Using the solution of the previous problem, find the fieldintensity E inside the sphere over which a charge is distributed with

1. Electroirtattc Field in a Vacuum42

the centre of the sphere F' d th helectric field intensity E; o~~ide ~hc arghe of.th~ dsphere for which thethe value of E. e sp ere IS In ependent of r. Find

Solution. Let the sought charge of the sphere be q. Then, Using the

Fig. 1.22 Fig. 1.23Gauss theorem we can writ th f II .surf~ of radi~s r (outside :he :ph~r~~i~t ~h~r~h~ir~~ ~r a spherical

r

E ·4n 2 - q t 1 \ a~ r -_. -- . -4nr2dr.8 0 £0. r

R

After integration, we transform this equation to

E ·4nr2 = (q ~ 2naR2)leo + 4nar2/280

• •

The intensity E does t d dtheses is equal to ze~~. J!rt~~ on r when the expression in the paren-

q = 2naR2 and E = a/280

'

~ f .6. Find the electric fi ld . t· .sectIon of two spheres unifo I In enslty .Ii: In the region of inter-volume densities p and _p Wth c3~rged by unlike charges with thespheres is determined by v~ctor le(F::~af.r:3)~etween the centres of the

Solution. Using the Gauss thelectric field intensity Within eo~efm, lwe can easilY.lshow that the

a UD! orm Y charged spflere isE = (p/380 ) r,

where r is the radius vector r I t' .can consider the field in the r e.a Ive .to the c~ntre of the sphere. WesUperposition of the fields of et~on of.;n\ersecthlOn of the spheres as thean arbitrary point A (Fig 1 24)0 \lfntlh~r my.c arged spheres. Then at

" 0 IS regIOn we haveE = E+ + JL = P (r+ - r_V3en = pI/3en.

Thus, in the region of intersectio f th hform. This conclusion is valid retpardleo f etseh sp ~rebs the field is uni-

" ss 0 e ratio etween the radii

I I·, I1 lie

I Iii

IiiI

2.1. Field in a Substance 45

1. Electrodatlc Field In a Vacuam

which gives :-1 after substituting the limits of integration. As a re-sult, we obtam ~/

lp = oRfneo'

• 1.10. T.he potential of the field inside a charged sphere depends?nly on the dI.stance r from its. Cl'ntre to the point under considerationI~ the foll?wI~g 'Yay: lp = ar' + b, where a and b are constants.Fmd the dIstrIbution of the volume charge p (r) within the 'sphere.

Solution. Let us first find the field intensity. According to (1.32)we have •

Er = -oq>/ar = -2ar. (1)

T~en we US? th~ Gauss theorem: 4nr'Er = q/eo• The differential ofthiS expressIOn IS

1 i4n d(rIEr)= - dq = - p·4nr' dr,

66 eJ

2.1. Field in a Substance

Micro- and Macroscopic Fields. The real electric field inany substance (which is called the microscopic field) variesabruptly both in space and in time. It is different at different points of atoms and in the interstices. In order to findthe intensity E of a real field at a certain point at a giveninstant, we should sum up the intensities of the fields of allindividual charged particles of the substance, viz. electronsand nuclei. The solution of this problem is obviously notfeasible. In any case, the result would be so complicated

E=_1_ 2P2...' 4nEo za

Taking the derivative of this expression with respect to I and substituting it into the formull\. for fl, we obtain

F__1 _ 6PIP2- 41tEo l4'

It should be noted that the dipoles will be attracted when PI t t P2 andrepulsed when PI H P2'

2. A Conductor in an Electrostatic Field

where dq is the charge contained between the spheres of radii randr + dr. Hence

. 1,d aEr +2.E'r

__ ...£...r:dEr+2rEr dr=-pr r, _Eo ar r eo

Substituting (1) into the last equation, we obtain

p =-" -flEoa,

i.e. the charge is distributed uniformly within the sphere.

• 10ft. Dipole. Find the force of interaction between two pointdipoles with moments PI and P2, if the vectors PI anti P2 are directedalong the straight line connecting the dipoles and the distance between the dipoles is l.

Solution. According to (1.39), we have

F = PI I aElal I•where E is the field intensity from the dipole P2' determined by thefirst of formulas (1.38):

olp=- oR J~sin'6'd{;.

neon/2

We integrate by parts, denoting ~ = •and sin t}- dt}- = dv:

) 'It sin t}- dt}- = - '6- cos '6-

+ \ cos'6-dtJo=-'6-cos{}+sinil'.01

Fig. 1.27

The sought projection is

Ea=E...!..= -0: (yl-zj- 2lk)(i+3k). -0: (y-61)

. a }f1+31 JfiOAt the point M we have

E. -0:~1t18) J~o 0:.

• f.9. Find the potential lp at the edge of a thin df!e of l'Jdius Rwith a charge uniformly distributed over olle of its sides with the ....face density o.

. S~lut~on. ~y definition, the potential in the case of a surface chargedl8trl~ution IS defined by integral (1.28). In order to simplify in-

. tegratlOD, we shall choose the area element dS in the fonn of a partof the ring of radius rand width!,dr (Fig. 1.27). Then dS = 2'&r dr,r = 2R cos '6-, and dr = -2R sin '6- t:H}. After substituting these expressions into integral (1.28), we obtain the expression for cp at &bepoint 0:

I I

Ii

that i~ would ~e impossibl~ to use it. Moreover, the knowledgeof tillS field IS not reqUIred for the solution of macroscopic problems. In many cases it is sufficient to have a simplerand rougher description which we shall be using henceforth.

Under the electric field E in a substance (which is calledthe rruu:roscopic field) we shall understand the microscopicfield averaged over space (in this case time averaging issuperfluous). This averaging is performed over what is calleda physically infinitesimal volume, viz. the volume containing a large number of atoms and having the dimensionsthat are many times smaller than the dli!tances over whichthe macroscopic field noticeably changes. The averaging oversuch vo~umes smooth~ns all i.rregular and rapidly varyingfluctuatlOns of the mIcroscopIC field over the distances ofthe order of atomic ones, but retains smooth variations ofthe macroscopic field over macroscopic distances.

Thus, the field in the substance is

E=Emacro=(Emlcro). (2.f)

The Inftuence ofa Substance on a Field. If any suhstance isintroduced into an electric field, the positive and negativecharges (nuclei and electrons) are displaced, which in turnleads to a parti.al separation· of these charges. In certainregions of the suhstance, uncompensated charges of differentsigns appear. This phenomenon is called the electrostatic

. / induction, while the charges appearing as a result of separation are called induced charges.

Induced charges create an additional electric field whichin combination with the initial (external) field forms theresultant field. Knowing the external field and the distribution of induced charges, we can forget about the presence ofthe substance itself while calculating the reSultant iieldsince the role of the substance has already been taken int~account with the help of induced dlarges

Thus, the resultant field in the presence of a substance isdetermined simply as the superposition of the external Baldand the field of .wauced charges. However, in many casesthe situation is complicated by the fact that we do not kaowbeforehand how all these charges are distributed in space,and the problem turns out to be not as simple as it eooJdseem at first sight. It will be shown later that the distri-

46 2. A Co..a~ctor·tn an Electrodatic Pield

2.2. Fields Inside and Outside a Conductor 41

bution of induced charges is mainly determined by the properties of the substance, i.e. its physical nature and theshape of the bodies. We shall have to consider these questions in greater detail.

2.2. Fields Inside and Outside a Conductor

Inside a Conductor E = O. Let us place a metallic conductor into an external electrostatic field or impart a certaincharge to it. In both cases, the electric field will act on allthe ch2.l.'ges of the conductor, and as a result all the negativecharges (electrons) will be displaced in the direction againstthe field. This displacement (current) will continue until(this practically takes a small fraction of a second) a certain charge distribution sets in, at which the electric fi.eldat a~l the points inside the conductor vanishes. Thus, in thestatIC case the electric field inside a conductor is absent(E = 0).

Fu.rther, siw;e E = 0 everyWhere in the conductor, thedensIty of excess (uncompensated) charges inside the conductor is· also equal to zero at all points (p = 0). This can beeasily explained with the help of the Gauss theorem. Indeedsince inside the conductor E = 0, the flux of E through an;closed surface inside the conductor is also equal to zero.And this means that there are no excess charges inside theconductor.

Excess charges appear only on the conductor surface witha certain density (J which is generally different for differentpoints of the surface. It should he noted that the excesss~face charge is located in a very thin surface layer (whosethIckness amounts to one or two interatomic distances).

The absence of a field inside a conductor indicates inaccordance with (1.31), that potential cp in the cond~rhas the same value for all its points, i.e. any conductor inan electrostatic field is an equipotential region, its surfacebeing an equipotential surface.

The fact that the surface of a conductor is equipotentialimplies that in the immediate vicinity of this surface thefield E at each point is directed along the normal to the surface. If the opposite were true, the tangential component ofE would make the charges move over the surface of the

2. A Condllctor in an Electrostatic Field-------

conductor, Le. charge equilibrium would be impossible. 2.3. Forces Acting on the Surface of a Conductor 49

n

4'Fig. 2.2 Fig. 2.3

= 0/18/80' where En' is ..the projection of (vector E ontothe outward normal n (with respect to the conductor), /18is"9the cross-sectional area of the cylinder and 0 is the locals~face charge density of the conductor. Cancelling bothsides of this expression by /18. we get

I En = a/so· I (2.2)

If 0 > 0, then En > 0, Le. vector E is directed from theconductor surface (coincides in direction with the normal n).If 0 < 0, then En < 0, and vector E is directed towards theconductor surface.

Relation (2.2) may lead to the erroneous conclusion thatthe field E in the vicinity of a conductor depends only onthe local charge density 0'. This is not so. The intensity Eis determined by all the charges of the system under consideration as well as the value of a itself.

2.3. Forces Acting on the Surface of a Conductor

Let us consider the case when a charged region of the surface of a conductor borders on a vacuum. The force actingon a small area· ~S of the conductor surface is

~F = at.S .Eo, (2.3)

shall take a small cylinder and arrange it as is shown inFig. 2.3. Then the flux of E through this surface will beequal only to the flux through the "outer" endface of thecylinder- (the fluxes through the lateral surf~ce and theinner endface are equal to zero). Thus we obtam En /18 =

Fig. 2.1

r

Example. Find the potential of an undlarged condudiug sphereprovided that a point charge q is located at a distance r from its centre

(Fig. 2.1).PotentialljJ is the same fur all points

of the sphere. Thus we can calculateits value at the centre 0 of the sphere,because only for this point it canbe calculated in the most simple way:

1 q., (1)Ip= 4nco' r~t-fjl·

where the tJrst term is the potential ofthe charge q, while the second is thepotential of the charges induced on

the surface of the sphere.Dut since aU induced charges are at the samedistance a from the point 0 and the total induced charge is equal tozero, q;'=O as well. Thus, in this case the putential \if the sphere willbe determined only by the Erst term ill (1).

l"igure 2.2 shows the field and the charge distributions fora system consisting of two conducting spheres one of which(left) is charged. As a result of electric induction, the ch?rgesof the opposite sign appear on the surface of the rightuncharged sphere. The field _of these.eharges will in turncause a redistribution of charges on the surface of the leftsphere, and their surface distribution will become nonuniform. The solid lines in the figure are the lines of E, whilethe dashed lines show the intersection of tlquipotential surfa~s with the plane of the figure. As we move away fromthis...system, the equipotential. surfaces become closer_andcloser to spherical, and the field lines become closer to radial. The field itself in this case resembles more and more thefield of a point charge q, viz. the total charge of the givensystem.

The Field Near aConductor Surface. We shall show thatthe electric fwld intensity in the immediate vicinity of thesurface of a conductor is connected with the local chargedensity at the conductor surface through a simple relation.This relation can be established with the help of the Gausstheorem.. Suppose that the region of the conductor surface we areinterested in borders on a vacuum. The field lines are normal to the conductor surface. Hence for a closed surface we

\

2.4,., Properties of a Closed Conducting Shell 5i

where we took into account that 0 = eoEn\ and E; = E 2•

The quantity Fu is called the surface density of force. Equation (2.7) shows that regardless of the sign of 0, and henceof the direction of E, the force Fu is always directed outsidethe conductor, tending to stretch it:

Example. Find the expression for the electric force acting in avacuum on a conductor as a whole, assuming that the field intensitiesE are known at all points in the vicinity of the conductor surface.

Multiplying (2.7) by dS, we obtain the expression for the force dFacting on the surface element dS:

dF=+ eo£2 dS,

where dS = n dS. The resultant force acting on the entire conductorcan be found by integrating this equation over the entire conductorlIUliace:

2.4. Properties of a Closed Conduding Shell

It was sho.wn that in equilibrium there.are no excess chargesinside a conductor, viz. the material inside the conductor is electrically neutral. Consequently ~ if the substance isremoved from a certain volume inside a conductor (a closedcavity is created), this does not change the field anywhere,Le. does not affect the equilibrium distribution of charges.This means that the excess charge is distributed on aconductor with a cavity in the same way as on a uniformconductor, viz. on its outer surface.

Thus, in the absence of electric charges within the cavitythe electric field is equal to zero in it. External charges,including the charges on the outer surface of the conductor, donot create any electric field in the cavity inside the conductor.This forms the basis of electrostatic shielding, Le. the screeningof bodies, e.g. measuring instruments, from the influenceof external electrostatic fields. In practice, a solid conducting shell can be replaced by a sufficiently dense metallicgrating.

That there is no electric field inside an empty cavity canbe proved in a different way. Let us take a closed surface Senveloping the cavity and lying completely in the material(2.7)

AS


Fig. 2.4 :-

50

where 0 f!S ;is the charge of this element and Eo is the fieldcreated by all the other charges of the system "in the regionwhere the charge 0 f!S is located. It should b~ noted at thevery outset that Eo is not equal to the field intensity E inthe vicinity of the given surface element of the cojductor,although there exists ~a certain relation between th~m. Letus find this relation, Le. express Eo through Be'

Let Ea be the intensity of the field created by the chargeon. the area element f!S at the points that are very close tothIS element. In this region, it behaves as an infinite uniformly charged plane. Then, in accordance with (1.10), Ea == 0/280'

The resultant field both inside and outside the conductor(near the area element f!S) is the superposition of the fieldsEo and Eo. On both sides of the area element f!S the field Eis p~actically ~he same, while the field Eo has opposite di~ractIOns (see FIg. 2.4 where (for the sake of definiteness it isassumed that C1 > 0). From 'the condition E = 0 inside theconductor, it follows that Eo = Eo, and then outside theconductor, near its surface, E = Eo + Eo = 2Eo. Thus,

Eo = E/2, (2.4)and Eq. (2.3) becomes

1AF=T aAS·E. (2.5)

Dividing both sides of thisequation by f!S, we obtain theexpression for the force actingon unit surface of a conductor:

1Fu-T.aE. (2.6)

We can write this expressionin a different form since the

quantities a and E appearing in it are interconnected. Indeed,in accordance with (2.2), En = a/eo, or E = (a/eo) n, wheren is the outward normal to the surface element at a givenpoint of the conductor. Hence

4*

52 2. A Conductor in an Electrostatic Field 2.4. Properties Df a Closed Conducting Shell 53

of the conductor. Since the field E is equal to zero inside theconductor, the flux of E through the turface S is also equalto zero. Hence, in accordance with the Gauss theorem, thetotal charge inside S is equal to zero as well. This does notexclude the situation depicted in Fig 2.5, when the surf/tge

1 qqJ = 4n:eo -,:-.

An infinite conducting plane is a special case of a closedconducting shell. The -space on one side of this plane is

.electrically independent of the space on its other side.We shall repeatedly use this property of a closed con

ducting shell.

space outside the cavity, the field created by the chargeslocated inside the cavity.

Since the conducting medium is electrically neutral everywhere it does not influence the electric field in any way.Therefore, if we remove the medium, leavi.ng only a conducting shell around the cavity, the field WIll not be ~hangedanywhere and will remain equal to zero beyond thIS shell.

Thus the field of the charges surrounded by a conductingshell a~d of the charges induced on the surface of the cavity(on the inner surface of the shell) is equal to zero in the entire outer space.

We arrive at the following important conclusion: a closedconducting shell divides the entire space into the inner .andouter parts which are completely independent oj one anotherin respect of electric fields. This must be i~te:preted as follows:any arbitrary displacement of charges InSIde the shell doesnot introduce any change in the field of the outer space,and hence the ch.arge distribution on the outer surface of theshell remains unchanged. The same refers to the field insidethe cavity (if it contains cbarges) and to the ~istribution ?fcharges induced on the cavity walls. They WIll also remamunchanged upon the displacement of charges outside theshell. Naturally, the above arguments are applicable onlyin the framework of electrostatics.

Example. A point charge q is within an electrically neutral sh~llwhose outer surface has spherical shape (Fig. 2.6). Find the potentialcp at the point P lying outside the shell at a distance r from thecentre 0 of the outer surface.

The field at the point P is determined only by charges induced on. the outer spherical surface since, a~ was shown ab«;lve, the field of thepoint charge q and of the charges mdu.ced on the. mner surf~ee ?f thesphere is equal to zero everywhere outsIde the caVIty. Next, III VIew ofsymmetry, the charge on the outer surface of the shell is distributeduniformly, and hence

pr

Fig. 2.6Fig. 2.5

of the cavity itself contains equal quantities of positiveand negative charges. However, this assumption is prohibited by another theorem, viz. the theorem on circulation· ofvector E. Indeed, let the contour r cross the cavity alongone of the lines of E and be closed in the conductor material.It is clear that the line integral of vector E along this contourdiffers from zero, which is in contradiction with the theoremon circulation.

Let us now consider the case when the cavity is not emptybut contains a certain electric charge g (or several charges).Suppose also that the entire external space is filled by aconducting medium. In equilibrium, the field in this mediumis equal to zero, which means that the medium is electrically neutral and contains no excess charges.

Since E = 0 inside the conductor, the field flux through aclosed surface surrounding the cavity is also equal to zero.According to the Gauss theorem, this means that the algebraic sum of the charges within this closed surface is equal tozero as well. Thus, the algebraic sum of the charges inducedon the cavity surface is equal in magnitude and opposite insign to the algebraic sum of the charges inside the cavity.In equilibrium the charges induced on the surface of thecavity are\arranged so as to compensate compleLely, in the

2. A Conductor in an El~ctrostatic Field 2.5. General Problem of Electrostatilcs 55

static case the charge is distributed over thlle surface of aconductor in a unique way as well. Indeed, th'Iere is a one-toone correspondence (2.2) between the charges; on the conductor and the electric field in the vicinity of itts surface: (J =

Fig. 2.7

= EllEn. Hence it immediately follows that the uniquenessof the field E determines the uniqueness of the chargedistribution over the conductor surface.

The solution of Eqs. (2.8) aM (2.9) in the generail case is a complicated and laborious ·problem. The analytic solutions i of these equationswere obtained only for a few particular cases. As ffor the uniquenesstheorem, it simplifies the solution of a number of «electrostatic problems. If a solution of the problem satisfies the Lalplace (or Poisson)equation and the boundary conditions, we can stat6e that it is correctand unique regardless of the methods by which itt was obtained (ifonly by guess).

Example. Prove that in an empty cavity of a cconductor the fieldis absent.

The potential <P must satisfy the Laplace equatiwn (2.9) inside thecavity and acquire a certain value <Po at the'cavity's ,walls. The solutionof the Laplace equation satisfying this condition c~an immediately befounrl. It is d that makes itpossible to calculate in a simple way the ,electric field insome (unfortunately few) cases. Let us consiider this methodby using a simple example of a point charge ~q near an infiniteconducting plane (Fig. 2.7a)., The idea of this method lies in that we mmst find another

problem which can be easily solved and whwse solution or apart of it can be used in our problem. In ourr case such a simple problem is the problem about two changes: q and -g.

2.5. General Problem of Electrostatics.Image Method

d .~r~~ue~tlY~ we must solve problems in which the charget~:.rl ~tJon IS

d. unknown but the potentials of conductors

Jr s ape an relative arrangement are given We mustfind

dth~ pot~ntial cp (r) at any point of the field between the

~~a~ uc ors. t should be recalled that if we know the poten-d th (r),. the field E (r) itself can be easily reconstructed

an f en Its value in the immediate vicinity of the condu~tors'!r a~es .can be used for determining the surface chdlstrlbutlOll for the conductors. arge

The Poisson and I aplac t· L .equation for the functi e equa .lOns. et ~s derIve the differentialinto the left-hand sid~~(/r~t)n~hal). For t~IS p¥rpose: we substitutei.e. E = -v<p. As a result . b .e expressIon o~ E In .terms of 'P,for potential which is cali ':,(}tOh tapIn.the general dIfferentIal equation

, ell e Olsson equation:

. IV2<p= -P/eo,/ (2.8)

7thh:~ r~eifo~ Lapl~ce operator (Laplacian). In Cartesian coordinates

V2=~ ~ a2

ax 2 + ay2 + az' 'i.e. 1:\ the scalar product v·v [see (1.19)).is tran~f~:ma:d ~~tcharg~s beltween th!! cond.uctors (p = 0), Eq. (2.8)

o a SImp er equatIOn, VIZ. the Laplace equation:

IV'<P = O. I (2.9)

Eq.S:(2~h8)t~~(4~~ f~\h~t~~ii;;:s~a~~t ~:t~:e~u~h~i~~n~~~ :~~sfiesqUIres t e given values <Ph 'P2' ... on the surfaces of the conducto~~

It can ~e prove? theoretically that this problem has a unique solutIOn. ThiS statement is called the uniqueness theorerr;. Fro~ the physical point of view, this conclusion isq'!Ite obvIOUS: If there are more than one solution therewIll be .several potential "reliefs", and hence the field E ateach pom~ generally has not a single value. Thus we arriveat a phYSically absurd conclusion.Usm~ the uniqueness theorem, we can state that in a

(a) (b) (e) l

----

56 2. A Conductor in an Electrostatic Field 2.6. C.pacUance. Capacitors 57

qb---lf---<>- q

2.6. CapaCitance. Capacitors

Capacitance of an Isolated Conductor. Let I~S fconSid~r asolitarv conductor, i.e. the conductor .remove( r~m 0 101'

. 1 d' . and charge" Expenments show that theconductors, )0 Ies" ,,,. . . 1 'tcl·r.r e of this conductor is directly propor~lOJ.la to I Sp~~e~tia\ <p (we assumed that at infin.ity potent~,aI IS eJual t~zero): cp oc q. Consequently, the r~tlO g!cp ~oe.., n~t ~~~n yon the charge q and has a certain value or eac so I rconductor. The quantity

q

(2.10)

Fig. 2.8

c ~'" g!cp

is called the electrostatic capacitance of an isolated conductor(Qr simply capacitance). It is numerically.equal to th~ chargethat must be supplied to the conductor III order to lIlcre~se

its potential by unity. The capacitance depends on the SIze

and shape of the conductor.

Example. Find the capacit:mce of an isolated conductor which hasthe shape of a sphere of radIUS R.

It can he seen from formula (2.10) that for this jlurpose. we mustmentally chargre the conductor by a charge q and calculate Its potell-

charges whose action on the charge q,is eq~ivalent to the action of all

cha~~\\~~deu~~dii~3:~~~etc~;l~l~~r::'chargesfor which the eqnipoten-tial ~ul'face~ with cp = 0 wo~J!d (a) (~)coinCide With the cond~c!Illg

half-planes. One or two fI~tltIO~StL'q fcharges are insuflicient III tillS Icase' there should be three of -q Ithe~ (Fig. 2.8b). Only with such aconfiguration of the ~:ystem of fOl~r Icharges we can realize the ft- .--l---quired "trimming"', i.e. ensure 7, jthat the potential on the con- Idueling" half-plane~ ~e equal tozero. These three hctl tlOllS cl.Jarges create just the same heldwithin the "right angle" as thefield of the charges induced on

the con~ucting plah.es. n alion of point chal'gl's {anolher proble~n),HavIll\r.found I IS con 19l1b f tlH'r questions, for example, lmd

ili'e ~~~e~t~~{a~ldsfi~f/i~~~~~i~; ~n ~ny point within the "right angle"or determine the fw-ce acting Oil the charge q.

The field of this system is well known (its equipotentialsurfaces and field lines are shown in Fig. 2.7b).

Let us make the conducting plane coincide with the middle equipotential surface (its potential qJ = 0) and removethe charge --g. According to the uniqueness theorem, thefield in the upper half-space will remain unchanged. Indeed,qJ = 0 on the conducting plane and everywhere at infinity.The point charge g can be considered to be the limiting caseof a small spherical conductor whose radius tends to zeroand potential to infinity. Thus, the boundary conditions forthe potential in the upper half-space remain the same, andhence the field in this region is also the same (Fig. 2.7c).

It should he noted that we can arrive at this conclusionproceeding from the properties of a closed conducting shell[see Sec.2.1J. since hoth half-spaces separated by the conducting plane are electrically independent of one another,and the removal of the charge -g will not affect the fieldin the upper half-space.

Thus, in the case under consideration the field differs fromzero only in the upper half-space. In order to calculate thisfield, it is sufficient to introduce a fictitious image chargeg' = -g, opposite in sign to the charge q, by placing it onthe other side of the conducting plane at the same distanceas the distance from g to the plane. The fictitious charge g'creates in the upper half-space the same field as that of tliecharges induced on the plane. This is precisely what ismeant when we say that the fictitious charge produces the same"effect" as all the induced charges. We must only bear inmind that the "effect" of the fictitious charge extends only tothe half-space where the real charge q is located,. In anotherhalf-space the field is absent.

Summing up, we can say that the image method is essentially based on driving the potential to the boundary conditions, I.e. we strive to find another problem (eonfignration of charges) in which the field configuration in the regionof space we are interested in is the same. The image methodproves to be very effective if this·can be done with the helpof sufficiently simple configurations. Let liS consider onemore example.

Example. A point charge q is placed between two mutually perpendicular half-planes (Fig. 2.8a). Find the location of fictitious point

58 2. A Conductor in an Electro8tattc Field 2.6. Capacitance. Capacitors 59

tial <po In accordance with (1.23), the potential of a sphere is

00 00

J' 1 f q 1 q<p= Erdr=-- --2 dr =-- -.4nfo . r 4n8o R

R R

Substituting this result into (2.10), we find

C = 4moR. (2.U)

The unit 'ofcapacitance is the capacitance of a conductorwhose potential changes by 1 V when a charge of 1 C issupplied to it. This unit of capacitance is called the farad (F).

The farad is a very large quantity. It corresponds to thecapacitance of an isolated sphere 9 X 106 km in radius,which is 1500 times the radius of the Earth (the capacitanceof the Earth is 0.7 mF). In actual practice, we encountercapacitances between 1 ~F and 1 pF.

Capacitors. If a conductor is not isolated, its capacitancewill considerably increase as other bodies approach it. Thisis due to the fact Jhat the field of the given conductor causesa redistribution of charges on the surrounding bodies, Le.induces charges on them. Let the charge of the conductorbe g > O. Then negative induced charges will be nearer tothe conductor than the positive charges. For this reason, thepotential of the conductor, which is the algebraic sum of thepotentials of its own charge and of the charges induced onother bodies will decrease when other uncharged bodiesapproach it. This means that its capacitance increases.

This circumstance made it possible to create the systemof conductors, which has a considerably higher capacitancethan that of an isolated conductor. Moreover, the capacitanceof this system does not depend on surrounding bodies.Such a system is called a capacitor. The simplest capacitorconsists' of two eonductors (plates) separated by a smalldistance.

In order to exclude the effect of external bodies on thecapacitance of a eapacitor, its plates are arranged with respeet to one another in such a way that the field created bythe charges accumulated on them is concentrated almostcompletely inside the capaeHor. This means that the linesof E emerging on one plate must terminate on the other,

. the charges on the plates must be equal in magnitudeI.e. ( d )and opposite in sign ~ :.n f! ~apacitor is its capacitance.

The basic charactens lC 0 . I d onductor the capaci-Unlike the capacitance of an Isolate

lc t' f i'ts charge to

f 'tor is denned as t le ra 10 0 •tance 0 a. cap~cl b t n the plates (lhis difference 18the potentIal difference e weecalled the voltage):

\ \(2.12)C=qlU.

The charge g ofa capacitor is the charge of its positively

charged' plate. ., f a capacitor is also measuredNaturally, the capaCitance 0 ,

in farads. "t depends on its O'eometryThe capacitance of a capaCl or a between th~ plates,

(size and shape. of i~S pl~\~s)th~leC;p~citor. Let us deriveand the m~teqalf t ~~e capacitances of some capacitorsthe expreSSIOns or. cuum between their plates.assuming that there is a rr I I te Capacitor. This capacitor

Capacitance of a Para e -p a d b a ap of width h.consists of two parallel pl~ttes ~epar~~:n :cco~ding to (1.11),If the charge of the capaCi or IS g, . '1 . E _ ale

. h fi Id between Its pates IS - 0'the intensity of t e .e h f each plate. Consequent-where a = glS and S IS t e area 0 .

ly, the voltage between the plates ISU = Eh = ghlfoS.

Substituting this expression into (2.12), we obtainIh (2.13)C = foS .

d 'thout taking into accountThis calculation was rna e WI lates (edge effects).

field disto~tions n~ar thel e~~~e °1a~~~lor is determined ~YThe capaCItance 0 a rea p tely the smaller the gap h IIIthis formula the more. accura. nsions of the plates.compari~on with th; l~ne~r t~~acitor. Let the radii of thE

CapaCItance of a p. enca es be a and b respectively. IIinner and outer capaclt?r pl.at field intensity between thEthe charge of the capaCItor. IS g, .plates is determined by the Gauss theorem.

1 qEr = into rt·

where the first integ-ral is taken over all the charges induced on theinner surface of the layer, while the second integral, over all the

It can be easily seen that the capacitance of a sphericalcapacitor is given by

\'

(1)dF=-taEdS.. '"

Problems

It follows from symmetry considerations that t~e resultant force fis directed along the Z-axis (Fig. 2.11), and henc0 it can be represented

charges on the outer surface. It follows from this expression that

__q_ (..i..__1 +..!..).cP - 4n80 r a b

It should be noted that the potential.can. be found in such a~f~eform only at the poin~ 0 s~nce all the. like mduced charges are a esame distance from this pomt and theIrdistribution (which is unknown to us)does .not play any role.

• 2.2. A system consists of twoconcentric spheres, the inner sphere ofradius R l having a charge ql' What charge q must be placed onto the outerspher~ of radius R 2 to make the potential of the inner sphere equal to zero?What will be the dependence of potential cP on the distance r from the centre of the system? Plot schematica~lythe graph of this dependence, assummgthat ql < o. 4.' 'Fig. 2.9

Solution. We write the expressions. . hfor potentials outside the slstem (CPn) and In the regIOn between t espheres (CPI):

1 ql +q2 1 ql +CPn=-- ---, CPI=-4-- -- CPo,

4n80 r n80 r

where CPo is a c~r~ain constant. Its val~ can be easily found from theboundary conditIOn: for r = R 2 , CPn - CPl, Hence

lfu = q2/4ncoR2'From the condition <PI (RJ) = 0 we find that q2 = -ql R2IRl •

The <P (r) dependence (Fig. 2.10) will have the form:

_ ql 1-R2/R J =.3..L (_1 ~).<P II - 4n80 r ,CPr 4n80 r R1

• 2.3. The force acting on a surface charge. An. uncharged metallic sphere of radius R is placed into an external umform. field'Js aresult of which an induced charge appears on the sphere Wit~ su icedensity ° = 00 cos 'ft, where 00 is a positive consta.nt an~ ~ IS ~ po arangle. Find the magnitude of the resultant electnc force actmg onlike charges.

Solution. According to (2.5), the force acting on the area elementdS is

(2.14)C=4Jt£ ~o b-a .

"u = r E dr = _q_ ( ..!.. __1 ).J r 4n80 a ba


The voltage of the capacitor is

60

It is interesting that when the gap between the plates issmall, Le. when b - a «a (or b), this expression is reduced to (2.13), viz. the expression for the capacitance of aparallel-plate capacitor.

Capacitance of a Cylindrical Capacitor. By using the sameline of reasoning as in the case of a spherical capacitor, weobtain

- 1 (q t- .~' 0_ dS , 10+ dS )qJ---- --- \ --T' )---4rtEo r • a :. b '

c-· 2n8ol (2.15)- In (b/a) ,

where l is the capacitor's length, a and b are the radii of theinner and outer cylindrical plates. Like in the previouscase, the obtained expression is reduced to (2.13) when thegap between the plates is small.

The influence of the medium on the capacitance of a capacitor will be discussed in Sec. 3.7.

Problems

• 2.1. On the determination or potential. A point charge q is ata distance r from the centre 0 of an uncharged spherical conductinglayer, whose inner and outer radii are equal to a and b respectively.Find the potential at the point 0 if r < a.

Solution. As a result of electrostatic induction, say, negative chargeswill be induced on the inner surface of the layer and positive charges on its out{~r surface (Fig. 2.9). According to the principle of superposition, the sought potential at the point 0 can be represented inthe form

xdxx

F q

Fig. 2.13

-q

Eq

Fig. 2.12

o

I

Solution. By definition, the work of this force done upon an elementary displacement dx (F;.ig. 2.13) is given by

ql6A = F% dx = - 4neo (2%)1 dx,

where the expression for the force is obtained with the help of theimage method. Integrating this equation over x between land 00, wefind

1

Problem.

00

ql r dx qlA= - 16neo J 7= - 16neol •

IRemark. An attempt to solve this problem in a different way (through

potential) leads to an erroneous result ~h!ch differs from w~atwas obtained by us by a factor of two. ThIS IS because the re.latIOnA = q (IJ>l - <PI) is valid only for potential fields. However, In thereference system fixed to the conducting plane, the electric field ofinduced charges is not a potential field: a displacem.ent o.f the chargeq leads to a redistribution of the induced charges, and theIr field turnsout to be time-dependent.

• 2.6. A thin conducting ring of radius R, having a charge q, isarranged 80 that it is parallel to an infinite conducting plane at a ~istance I, from it. Find (1) the surface charge density at a point of theplane, which is symmetric with respect to the ring and (2) the electricfield potential at the centre of the ring. .

Solutton. It can be easily seen that in accordance with the imagemethod, a fictitious charge -q must be located on a similar ring but

where the minus sign indicates that the induced charge is oppositeto sign to the point charge q.

• 2.5. A point charge q is at a distance 1 from an infinite conducting plane. Find the work of the electric force acting on the charge qdone upon its slow removal to a very large distance from the plane.

f

(2)

dS =

Fig. 2.11

,Fig. 2.10

o

as the sum (integral) of the projections of elementary forces (1) ontothe Z-axis:

dFz = dF cos {}.

It is expedient to take for the area element dS a spherical zone

~olutton. Accord.ing to (2.2), the :surface charge:density on a conductor IS connected .wIth the electric field near its surface (in vacuum)through the relatIOn a = eoEn' Consequently, the problem is reducedto determining the field E in the vicinity of the conducting plane.

Using the image method, we find that the field at the point P(Fig. 2.12) which is at a distance r from the point 0 is

q 1E=2Eqcosa=2 ----4neox2 x'

Hence

2. A Conductor in an Electrostatic field

= 2nR sin t}·R dt}. Considering that E = a/eo, we transform (2) asfollows:

clFz = (na I R2/eo) sin t} cos {} d{} = - (na~R2/po) cos3 {} d (cos t}).

Integrating this expression over the half-sphere (i.e. with respect tocos if between 1 and 0), we obtain

F = na8R2/4eo'

• 2.4. Image method. A point charge q is at a distance 1 from aninfinite conducting plane. Find the density of surface charges inducedon the plane as a function of the distance r from the base of the perpendicular dropped from the charge q onto the plane.

2. A Conductor in an Electrostatic Field Problems 65

on the other side of the conducting plane (Fig. 2.14). Indeed, only inthis case the potential of the midplane between these rings is equal tozero, Le. it coincides with the potential of the conducting plane. Letus now use the formulas we already know.

(1) In order to find {J at the point 0, we must, according to (2.2),find the field E at this point (Fig. 2.14). The expression for E on the

R

the required force (see Fig. 2.15b)

2 V2- 1 q2F =c F2 - F1= ---'74-- 2a2 ,neo

and answer the second question.• 2.8. Capacitance of parallel wir~s. :rwo long straight wires

with the same cross section are arranged In aIr parallel to one another.

B

1 2

Fig. 2.17

A

Fig. 2.16

(b)

Fig. 2.15

(8)

Fig. 2.14

1 II

-qC---l;---~........_----

(3 )

where a is the radius of the wires' cross section and b is the separationbetween the axes of the wires.

5-0181

The distance between the wires is 11 times larger th~n the radi~s of thewires' cross section. Find the capacitance of the WIres per umt lengthprovided that 11:;;}> 1.

Solution. Let us mentally charge the two wires by charg~s of thesame magnitude and opposite signs so that t~e charge per um.t lengthis equal to "'. Then, by defmition, the reqUIred capaCitance is

Cu = ",IU, (1)

and it remains for us to find the potential difference between the wi~es.It follows from Fig. 2.16 showing the dependences of the potentl~ls

lp + and Ip_ on the distance between the plates that the sought potentialdifference is

u= ILlcp+1 + ILlCP-l =2jLl(jl+I· (2)

The intensity of the electric field created by one of the wires at a distance x from its axis can be easily found wi th ,the help of the Gausstheorem: E = "'/2neox. Then

b-a\ '" b-a

I 8(jl+ I = J E dx = 2neo In --a- ,a

{J= 21t (RI+12)'/1

(2) The potential at the centre of the ring is equal to the algebraicsum of the potentials at this point created by the charges q and -q:

1 (q q)ljl= 41t8o if yR2+411 •

• 2.7. Three unlike point charges are arranged as shown inFig. 2.15a, where AOB is the right angle formed by two conductinghalf-planes. The magnitude of each of the charges is I q I· and the distances between them are shown in the figure. Find (1) the total chargeinduced on the conducting half-planes and (2) the force acting on thecharge -q.

Solution. The half-planeD forming the angle AGE go to infinity,and hence their potential q; = O. It can be easily seen that a ·~YlltemhaVing equipotential surfaces with ljl = 0 coinciding with the conducting half-planes has the form sho1inl in Fig. 2.15b. Hence the actionof the charges induced on the conducting hali-planes Is equivalent tothe action of the fictitious charge -q placed in the lower left comer ofthe dashed sauare.

Thus we have already answered the first question: -qeBy reducing the syst0m to iour point ch:trges, we can easily find

axis of a ring was obtained in Example 1 (see p. 14). In our case, thisexpression must he doubled. As a result, we obtain

ql

tili67

x

--r------r--r---=-01" _02~

-r--r- a ;r1

B

3.1. Polarization 0/ Dielectrics

(a) (b)

Fig. 2.1st= Fig. 2.19

A

3. Electric Field in Dielectrics

of the image method, since it would require an infmite series of fIctitious charges arranged on both sides of the charge q, and to find thefield of such a system is a complicated problem.

3.1. Polarization of Dielectrics

Dielectrics. Dielectrics (or insulators) are substances thatpractically do not conduct electric current.. This ~neans thatin contrast, for example, to conductors dielectriCS do notcontain charcrcs that can move over considerable distances

t>

and create elcctric current.

difference between them is equal to zero. Consequently,

E1Xl 1 + E2x (l - 11) = 0,

where E 1x and E x are the projections of ve.ctor E onto the X-axis tothe left and to tte right of the plane P (Fig. 2.19b).

On the other hand, it is clear that

a = -(a1 + a.),

where, in accordance with (2.2), a1 = eOE1n1 = eOE1X and a2 == e E

2= -e

OE

2X(the minus sign indicates that the normal U2

o n, . f h X .)is directed oppositely to the umt vector 0 t. e ,-axIs. .Eliminating E!X and E 2x from these equatIOns, we obtam

a1 = -a (I - 11)/1, a2 = -al1/l·

The formulas for charges q1 and q2 in terms ~f q have a si~ilar form.It would be difficult, however, to solve thiS problem WIth the help

p

2. A Conductur in an Electroltatlc Field-------

It follows from (1), (2) and (3) .nat

Cu = neo/ln 1').

where we took into account that b :it> a.

• 2.9. Four identical metallic plates are arranged in air at thesame distance h from each other. The outer plates are connected by aconductor. The area of each plate is equal to S. Find the capacitanceof this system (between points 1 and 2, Fig. 2.17).

Solution. Let us charge the plates 1 and 2 by charges qo and -qo.Under the action of the dissipation field appearing between theseplates (edge effect), a charge will move in the connecting wire, afterwhich the plate A will be charged negatively while the plate B willacquire a positive charge. An electric field appears in the gaps between the plates, accompanied by the corresponding distribution ofpotential cp (Fig. 2.18). It should be noted that as follows from thesymmetry of the system, the potentials at the Imiddle of the system aswell as on its outer plates are equal to zero.

By defini tion, the capacitance of the system in this case is

C = qo/V . (1)

where U is the required potential difference between the points 1and 2. Figure 2.18 shows that the putential difference V between theinner plates is twice as large as the potential difference between theoutside pair of plates (both on the right and on the left). This alsorefers to the field intensity:

E = 2E'. (2)

And since E ex a, we can state that according to (2) the charge qo onplate 1 is divided into two parts: qo/3 on the left side of the plate 1and 2qo/3 on its right side. Hence

V = Eh = ahlEo = 2qoh/3eoS,

and the capacitance of the system (between points 1 and 2) is

C= 3eoS2h •

• 2.10. Distribution 01 an induced charge. A point charge q isplaced between two large parallel conducting plates 1 and 2 separated by a distance 1. Find the tutal charges q1 and q2 induced on eachplate, if the plates are connected by a wire and the charge q is locatedat a distance 11 from the left plate 1 (Fig. 2.19a).

Solution. Let us use the superposition principle. We mentallyplace somewhere on a plane P the same charge q. Clearly, this willdouble the surface charge on ~ach plate. If we now distribute uniformlyon the surface P a certain cnarge with surface density a, the electricfield can be easily calculated (Fig. 2.19b).

The plates are connected by the wire, and hence the potential

.,!i'I'

'I'1,1., .

'I'I ! i

!

* There exist ionic crystals polarized even in the absence of anexternal field. This property is inherent in dielectrics which arecalled electrets (they resemble permanent magnets).

When even a neutral dielectric is introduced into an external electric field, appreciable changes are observed in thefield and in the dielectric itself. This is because the dielec~ric is acted up?n. by a force, the capacitance of a capacitorIncreases when It IS filled by a dielectric, and so on.

In order to understand the nature of these phenomena wemust take into consideration that dielectrics consist eitherof neutral molecules or of charged ions located at the sites

. of a crystal lattice (ionic crystals, for example, of the NaCItype). The molecules can be either polar or nonpolar. In apolar molecule, the centre of "mass" of the negative chargeis displaced relative to the centre of "mass" of the positivecharge. As a result, the molecule acquires an intrinsic dipolemoment p. Nonpolar molecules do not have intrinsic dipolemo~ents, since the "centres of mass" of the positive and negatIve charges in them coincide.

Polarization. Under the action of an external electricfield, dielectric is polarized. This phenomenon consists inthe following. If a dielectric is made up by nonpolar molecules, the positive cll~rge. in each molecule is shifted along thefield and the negatIve, In the opposite direction. If a dielectric consists of polar molecules, then in the absence of thefield their d~pole moments are oriented at random (due tothermal motIOn). L nder the action of an external field thedipole moments acquire predominant orientation in' thedirection of the external field. Finally, in dielectric crystals?f the NaCI ty~e, an external field displaces all the positiveIOns along the field and the negative ions, against the field. *

Thus, the m~chanismof polarization depends on the structure of a dielectric. For further discussion it is only important that regardless of the polarization mechanism all thepositive ~harges duri~g this process are displaced aiong thefield, whIle the negatIve charges, against the field. It shouldbe noted that under normal conditions the displacements ofcharges are very small even in comparison with the dimensions of the molecules. This is due to the fact that the intensity of the external field acting on the dielectric is consider-

693.1. Polarization of Dielectrics

£=0.!.-

p~ ~Ip~ p~ It

7~ I , I p,+I I II I I:I I I.I I I.

X X X

(a) (b) (c)

Fig. 3.1A.~

of emergence of these charges (and especially bulk charges),let us consider the following model: Suppose that we have aplate made of a neutral inhomogeneous dielectric (Fig. 3.~a)whose density increases with the coordinate x accordIngto a certain law. We denote by p~ and p~ the magnitudes ofthe volume densities of the positive and negative charges inthe material (these charges are associated with:nuclei andelectrons).

In the absence of an external field, p~ = p~ at each pointof the dielectric, since the dielectric is electrically neutral.However, p~ as well as p~ increase with x due to inhomog~neity of the dielectric (Fig. 3.1b). This figure shows that III

, the absence of external field, these two distributions exactly coincide (the distribution of p~(x) is shown by thesolid line, while that of p~ (x), by the dashed line).

Switching on of the external field leads to a displacementof the positive charges along the field and of the negativecharges against the field, and the two distributions will beshifted relative to one another (Fig. 3.1c). As a result, uncompensated charges will appear on the dielectric surface aswell as in the bulk (in Fig. 3.1 an uncompensated negativecharge appears in the bulk). It should be noted that the re-

ably lower than the intensities of internal electric fields inthe molecules.

Bulk and Surface Bound Charges. As a result of polariza-tion, uncompensated charges appear on the dielectric surf!"ceas well as in its bulk. To understand better the mechamsm

~'

,"~i)

3. Electric Field in Dielectrics68

• ~xtraneous c~arg~s are frequently called free charges, but thisterm IS not convement In some cases since extraneous charges may benot free.

3.2. Polarization

Definition. It is natural to describe polarization of a dielectric with the help of the dipole moment of a unit volume.If an .ext~rnal field or a dielectric (or both) are nonuniform,polarl.zatlOn. turns out to be different at different points oft~e dlel~ctrlc. In order to characterize the polarization at agiven POIllt, we must mentally isolate an infinitesimal volume!'1V containing this point and then find the vector sum of the

7f

(3.2)

3.2. Polarization

Vector P defined in this way is called the polarization of adielectric. This vector is numerically equal to the dipole moment of a unit volume of the substance.

There are two more useful representations of vector P.Let a volume !'1V contain !'1N dipoles. We multiply and divide the right-hand side of (3.2) by !'1N. Then we can write

P = n (p), (3.3)

where n = !'1NI!'1V is the concentration of molecules (theirnumber in a unit volume) and (p) = ('T.PI)!!'1N is the meandipole moment of a molecule.

Another expression for P corresponds to the model ofa dielectric as a mixture of positive and negative "fluids".Let us isolate a very small volume !'1V inside the dielectric.Upon polarization, the positive charge p+!'1V contai~edin this volume will he displaced relative to the negativecharge by a distance I, and these charges will acquire thedipole moment !'1p = p~ !'1V·1. Dividing both sides of thisformula by !'1V, we obtain the expression for the dipole moment of a unit volume, i.e. vector P:

P = pi.!. (3.4)

The unit of polarization P is the coulomb per square meter

(elm').Relation Between P and E. Experiments show that for

a large number of dielectrics and a broad class of phenomena,'Polarization P linearly depends on the field E in a dielectric.For an isotropic dielectric and for not very large E, thereexists a relation

P = 'X8oEJ (3.5)

where 'X is a dimensionless quantity called the dielectricsusceptibility of a substance. This quantity is independentof E and characterizes the properties of the dielectric itself.'X is alwavs greater than zero.

Hencef~rth, if the opposite is not stipulated, we shall

dipole moments of the molecules in this volume and writethe ratio

3. Electric Field in Dielectric,70

version of the field direction changes the sign of all thesecharges. It can be easily seen that in the case of a plate madeof a homogeneous dielectric, the distributions p' (x) andp~ (x) would be IT-shaped, and only uncompensated surfacecharges would appear upon their relative displacement inthe field E.

Uncompensated charges appearing as a result of polarization of a dielectric are called polarization, or bound, charges.The latter term emphasizes that the displacements of thesecharges are limited. They can move only within electricallyn~utr~l m?lecules. We shall denote bound charges by a prime(q , p , (J).

Thus, in the general case the polarization of a dielectric!ea~s to the appearance of surface and bulk bound chargesIII It.

We shall call the charges that do not constitute dielectricmolecules the extraneous charges. * These charges may belocated both inside and outside the dielectric.

The Field in a Dielectric. The field E in a dielectric is theterm applied to the superposition of the field E of extraneous charges and the field E' of bound charges: 0

E = Eo + E', (3.1)

where Eo and E' are macroscopic fields, Le. the microscopicfields of extraneous and bound charges, averaged over aphysically infinitesimal volume. Clearly, the field E in thedielectric defined in this way is also a macroscopic field.

, I

1,1'1'1'

iii!','II

72 3. Electric Field in Dielectrics 3.3. Properties 0/ the Field 0/ P 73

3.3. Properties of the Field of P

The Gauss Theorem for th£' Field of P. We shall showthat the field of P has the following remarkable and important property. It turns out that the flux of P through an

Fig. 3.2

ar~itrary closed sur~ace S is equal to the excess bound charge(wIth the reverse sIgn) of the dielectric in the volume en

. closed by the surface S, i.e.

consider only isotropic dielectrics for which relation (3.5)is valid.H~wever, there exist dielectrics for which (3.5) is not

applIcable. These are some ionic crystals (see footnote onpage oR) and ferroeZectrics. The relation between P and Efor ferr~electr~cs .is nonlinear and depends on the historyof the dIeI?ctnc, l.e. on the previous values of E; (this'phe-nomenon IS called hysteresIs). -

(3.9)IV·P=-p',

(3.7)

where l = l+ "+ L is the relative displacement of positiveand negative bound chaJ;ges in the dielectric during polari-zation.

Next, according to (3.4), p:l = P and dq' = P dS cos at

ordq' = P n dS = P·dS. (3.8)

Integrating this expression over the entire closed surfaceS, we find the total charge that left the volume enclosedby the surface S upon polarization. This charge is equal to

~ P.dS. As a result, a certain excess bound charge ?' will

be left inside the surface S. Clearly, the charge leavlllg thevolume must be equal to the excess bound charge remainingwithin the surface S, taken with the opposite sign. Thus, wearrive at (3.6).

Differential form of Eq. (3.6). Equation (3.6), viz. the Gausstheorem for the field of vector P, can be written in the differential formas follows:

of the positive and negative bound charges as a result ofpolarization. Then it is clear that the positive chargep:Z+dS cos a inclosed in the "inner"part of the obliquecylinder will pass through the area element dS from thesurface S outwards (Fig. 3.2b). Besides, the negative chargep:tdS cos a enclosed in the "outer" part of the oblique cylinder will enter the surface S through the area element dS.But be know that the transport of a negative charge in acertain direction is equivalent to the transport of the positive charge in the opposite direction. Taking this into account,we can write the expression for the total bound charge passing through the. area element dS of the surfa~e S in the outward direction:

dq' = p:Z+ dS cos a + Ip:IL dS cos a.

Since I r: I = p~ we havedq' = P+ (l+ + L) dS cos a = p+l dS cos a,

(3.6)

p

( b)

~ PdS= -qint.

(a)

This equation expresses the Gauss theorem for vector P.Proof of the theorem. Let an arbitrary closed surface S

envelope a part of a dielectric (Fig. 3.2a, the dielectric ishatched). When an external electric field is switched onthe ~ielectric is polarized-its positive charges are displacedrelative to the negative charges. Let us find the charge whichpasses through an element dS of the closed surface S in theoutward direction (Fig. 3.2b).

Let 1+ and C be vectors characterizing the displacement

c:

i.e. the divergence of the field of vector P is equal to the volume densityof the excess bound charge at the same point, but taken with the oppOsite sign. This equation can be obtained from (3.6) in the same manneras the similar expression for vector E was obtained (see p. 24). Forthis purpose, it is sufficient to replace E by P and p by p' ..

When Is p' Equal to Zero in a Dielectric? We shall showthat the volume density of excess bound charges in a dielectric is equal to zero if two conditions are simultaneouslysatisfied: (1) the dielectric is homogeneous and (2) thereare no extraneous charges within it (p = 0).

Indeed, it follows from the main property (3.6) of thefield of vector P that in the case of a homogeneous dielectric,we can substitute xeoE for P in accordance with (3.5), takex out of the integral, and write

x &eoE·dS .. -q'.oJ

75

(3.13)

3.3. Properties of the Field of P

where Pn

is the projection of vector ~ onto th~ outward normal to the surface of a given dielectrIC. The SIgn of the pro-

such a dielectric there is no excess bound bulk1ch~rge tn~only a surface bound charge appears as a resu t 0 po arl-

zat~~·us find the relation between polarizatio? P afnd ~he

f d 'ty a' of the bound charge at the lOter ace e-sur ace ensl . 1 t s use propertytween the dielectrics. For this purpose, e u '(3.6) of the field of vector~. Wechoose the closed surface lO the D

form of a flat cylinder whose end- 2~J~~~t---faces are on different sides of the 1interface (Fig. 3.3). We shall as- Isume that the height of the cyl- • n'

inder is negligibly small and the F' 3 3area !1S of each endface is so small Ig. .that vector P is the same at all . d'

. ts of each endface (thiS also refers to the surface ensl~Y~?I~f the bound charge). Let 11 be the common normal to t einterface at jl. given point. We shall always draw vector n

froDm. dielecd~riC ~h~o g~~e~}ri~ ~hrOugh the lateral surfaceIsregar Illg d 'th (3 6)'

of the cylinder, we ca'h write, in accor ance WI .,

P 2n !1S + Pin' t1S = - a' t1S,

h P and P , are the projections of vector P in d ielec-w. er; on2to the n~~aln and in dielectric 1 onto the normalt~I(p' 3 3) Considering that the projection of. vector Pontonh

Ig. . 1 ~, is equal to the projection of thIS vector o~to. t e norma. ( m n) normal n taken with the opposItet~e o~pos~te ~m ~ we can ~ite the previous equationsIgn, l.e. tn' - - In' 11' t1S)'in the following form (after cance mg .

P2n - Pin = - a'. (3.12)

This means that at the interface bet~een ~iel~ctrics thenormal component of vector P ha~ a dlsc.ofntIDdu!ty, ;~ose

. d d d a' In partIcular, I me mID IS amagmtu e epen s on . d d't' (3 12) acquires avacuum, then P 2n = 0, an con I Ion .simpler form:


The remaining integral is just the algebraic sum of all thecharges-extraneous and bound-inside the closed surfaceS under consideration, Le. it is equal to q + q'. Hence,x (q + q') = - q', from which we obtain

q' = - 1~X q. (3.10)

This relation between the excess bound charge q' andthe extraneous charge q is valid for any volume inside thedielectric, in particular, for a physically infinitesimal volume, when q' __ dq' = p'dV and q -+- dq = pl1V. Then, aftercancelling out dV, Eq. (3.10) becomes

, x •p = - 1+x p. (J.H)

Hence it follows that in a homogeneous dielectric p' = 0when p = O.

Thus, if we place a homogeneous isotropic dielectric ofany shape into an arbitrary electric field, we can be surethat its polarization will give rise only to the surface boundcharge, while the bulk excess bound charge will be zero atall points of such a dielectric,

Boundary Conditions for Vector P. Let us consider thebehaviour of vector P at the interface between two homogeneous isotropic dielectrics.. We have just shown that in

74

jection Pn determines the sign of the surface bound chargea' at a given point. Formula (3.13) can be written in a diHerent form. In accordance with (3.5), we can write

('

77

(3.19)

(3.17)

(3.18)

a.4. Vector D

. '. ftPJ] called dielectric displacement! or electro-* The ql;'antity Dh~~llo i b u~ing this term, however, III order tostatic inductton. We s a no e ~ . D

emphasize the auxiliary nature of \~etor .

I~ DdS = qlnt.\

.- This statement is called the ~auss thearer- for fiel~e~

I t should h~..note? ~h~t ve~orDdISpth~~~~~ist~Za~~~Pit is

ly different quantities. eo a~. have an' deepinde~d an aux~liary veft~~rW~~: ;::~e~t~ of the leld ofphysICal meamng. Howe 't. (3 18) justifies the intro-vdectt~r Do'f etxhf:~:~~or~~ne~ua~~o~ase~ it ~onsiderablysimpli-

uc IOn . d' It' *fies the ~naly(s3iS107f) thedfle(idH~n) a:: ~can~sfor any dielectric,

RelatIOns . an '. ,

both isotr?pic (~nf7)anS\~~~~P;~~t the dimensions o.f vect~rD ~~~r~:~lOs~me' as those of vector P. The 2quantlty D ISmeasured in coulombs per square metre (elm ).

Differential form of Eq. (3.18) is

I V·D=p, I. f th field D is equal to the volume densi ty of anLe. the divergence 0 e oint

extran~ous ch~rge .at the obat~i~fJ fro~ (3.18) in the same way as itThis equatiOn can e, 24\ It suffices to replace E by D andwas done for the field E \see p. "

take into account only extraneous charges.

(3.15) Can he transformed as follows:,expression

~ (eoE + P) dS = qlnt. (3.1.6)

The quantity in the parentheses in t~~ integrand ~ denotedby D. Thus, we have defined an auxIlIary vector

ID=eoE+P, Ih fl through an arbitrary dosed surface is equal.to the

w ose ux 1 closed hy this sur-'algebraic sum of extraneous c larges enface:

(3.14)


where En is the projection of vector E (inside the dielectricand in the vicinity of its surface) onto the outward normal.Here, too, the sign of En determines the sign of a'.

A Remark about th(' Field of Vector P. Relations (3.6)and (3.13) may lead to the erroneous conclusion that thefield of vector P depends only on the hound charge. Actually.this is not true. The field of vector P, as well as the field of E,depends on all the charges, both hou~d and extraneous.This can he proved if only by the fact that vectors P andE are connected through the relation P = xeoE. The houndcharge determines the flux of vector P through a closed surface S rather than the field of P. Moreover, this flux is determined not hy the whole hound charge hut by its partenclosed hy the surface S.

3.4. Vedor D

The Gauss Theorem for Field D. Since the sources of anelectric field E are all the electric charges-extraneous andhound, we can write the Gauss theorem for the field E inthe following form:

~ eoE dS = (q + q')lnt, (3.15)

where q and q' are the extraneous and hound charges enclosed by the surface S. The appearance of the hound chargeq' complicates the analysis, and formula (3.15) turns outto be of little use for finding the field E in a dielectric evenin the case of a "sufficiently good" symmetry. Indeed, thisformula expresses the properties of unknown field E in termsof the bound charge q' which in turn is determined hy unknown field E.

This difficulty, however, can he overcome hy expressingthe charge q' in terms of the flux of P hy formula (3.6). Then

78 3. Electric field in Dielectrics 3.4. Vector lJ 79

At the points where the divergence of vector D is positive we havethe sources of the field D (p > 0), while at the points where the divergence is negative, the sinks of the lield D (p < 0).

Relation Between Vectors D and .E. In the case of isotropic dielectrics, polarization P = xeoE. Substituting thisexpression into (3.17), we obtain D = 8 0 (1 + x) E, or

Fig. ;LS

r

Fig. ~L4

constant e. Find"'ihe projectiun 1,', of Iield intensity E as a functiun ofthe distance r frolll the centre of this sphere.

The syIllllletry uf the system allows us to use the Gauss th~oremfor vector D for sol ving the 'problem (we cannut usc here the srn.l1lartheorem for vector E, since the bound char~e is unlwo.wn to us). l' or asphere of radius r with the centre at the POlllt of locatIOn of the chargeq we can write the following relation: 4;r.r"lJ r . = q. Hence we can lindlj, and then, usillg formula (3.20), the required quantity t r :

1 '/ E ) 1 qE r (r < a) = 4ne"; <:/:2' r (r > a =="4neo ~.

Figure 3.4 shows the curves lJ (r) and E (r).Example 2. Suppose that a system consists of apoint ~harg.e q >.0

and an arbitrary sample 01 a homogelleou~. IsotropIc dielectrrc(Fig. 3.5), where S isa certain closed surface. h~d ?ut what Will happen to the ileIds of vectors E and D (and to theIr fluxes through thesurface .) if the dielectric is removed.

The field E at any point of space is determined by the charge qand by bound charges of the polarized diele~tr.ic. Since in o.ur caseD = e eE this refers to the lield D as well: It IS also determIned bythe exrra~eous charge q and by the bound char~es of the dielectric.

The removal of the dielectric will change the held E, and hence thelield D. The flux' of vector E through the surface S will also cha~ge,since negative bound charges will vanish from inside. this s~rface.However the flux of vector D through the surface SWill remalll thesame in ;pite of the change in the lield D. . .

. Example 3. Let us consider a system C?ntallllllg no extraneo.uscharges and having only bound charges. :::;uc~ a ~ystem can, forexample, be a sphere made 01 an electret (see p. 6b). 1, I~ure 3.?a showsthe field E of this sy::;tem. What can we say about the fIeld Dt

Let us illustrate what was said above by several examples.

Example t. An extrallPUUS 'poil~t charge '1. is loc.atc(~ at th~ cenl~eof a sphere of radius a, made 01 an IsotropIC dIelectrrc WIth a dlelectrrc

(3.21)

(3.20)

where e is the dielectric constant of a substance:

8=1+x.

I D=eoeE, I

The dielectric constant e (as well as x) is the basic electriccharacteristic of a dielectric. For materials 8> 1, whilefor vacuum 8 = 1. The value of 8 depends on the nature ofthe dielectric and varies between the values slightly differing from unity (for gases) and several thousands (for someceramics). The value of e for water is rather high (8 = 81).

Formula (3.20) shows that in isotropic dielectrics vectorD is collinear to vector E. For anisotropic dielectrics, thesevectors are generally noncollinear.

The field D can be graphically represented by the linesof vector D, whose direction and density are determined inthe same way as for vector E. The lines of E may emerge andterminate on extraneous as well as bound charges. We saythat any charges may be the sources and sinks of vector E.The sIJurces and sinks of field D, however, are only extraneouscharges, since only on these charges the lines of D emerge andterminate. The lines of D pass without discontinuitiesthrough the regions of the field containing bound charges.

A Remark about the Field of Vector D. The field of vectorD generally depends on extraneous as well as bound charges(just as the field of vector E). This follows if only from therelation D = 8 oeE. .However, in certain cases the field ofvector D is determined only by extraneous charges. It isjust the cases for which vector D is especially useful. Atthe same time, this may lead to the erroneous conclusionthat vector D always depends only on extraneous chargesand to au incorrect interpretation of the laws (3.18) and(a.1!J). These laws express only a certain property of fieldD but do not determine this field proper.

81

(3.22)

3.5. Boundary Conditions

.. I '!------ ---:I c I~

2 [ ~ J~ '"'r' •

n

n

I.e. the tangential component of vector E turns out to bethe same on both sides of the interface (it does no.t have adiscontinuity). .

Boundary Condition for Vector D. ~et us ta~e a cylInderof a very small height and arrange it at the lll~erface between two dielectrics (Fig. 3.8). The cross sectIOn of thecylinder must be such that vector D is the same within eachof its endfaces. Then, in accordance'with the Gauss theoremfor vector D, we have

D 2n .!1S + DIn' ·!1S = (JAS,

where (J is the surface density of the extraneous charge atthe interface. Taking both projections of vector D onto thecommon normal n (which is directed from dielectric 1 todielectric 2), we obtain Din' = - DIn' and the previousequation can be reduced to the form

\ D2n - Din = a. (3.23)

It follows from this relation that the normal componentof vector D generally has a discontinuity when passing

6-0181

Fig. 3.7 Fig. 3.8

projection of vector E not onto the unit vector 't' bu~ ontothe common unit vector 't, then En' = - EH , and It follows from the above equation that

Ii:

where the projections of vector E are taken on the directionof the circumvention of the contour, shown by arrows inthe figure. If in the lower region of the contour we take the

Cb)

Lines of DLines ofE

(a)

Fig, 3.G

3, Electric Field in Dielectrics


Let ~s first consider the behaviour of vectors E and Dat. the mterface between two homogeneous isotropic dielectrics. Suppose that, for greater generality, an extraneoussu:face charge exists at tl16 interface between these dielectrICS. The required conditions cari be easily obtained withthe help of two theorems: the theorem on circulation ofvector E and the Gauss theorem for vector D:

~E dI = 0 and ~ n dS =--= glnt-

. Boundary Condition for Veclor E. Let the field near thelDterface be E1 in dielectric 1 and E 2 in dielectric 2. Wechoose ~ sm~ll ?l(~_ngateu rectangular contour and orient it as~hown In Fig, .L'. The sides of the contour parallel to thelllterfac~ must have slIch a length that the fwld E over this!,eng.th ,,1Tl each dielectric can be assumed constant. Theheight of th~ contour must be negligibly small. Then, in

accordance With the tllf'orem on circulation of vector Ewe IJave '

First of all, t~e abs~nee of extraneous charges means that the fieldJ) has ~o sou:(;~s, the Imes of D d.o IIOt emerge or terminate anywhE're.lJowever, the 11I~ld J) PXISts 'md IS sllown I'n I"I'g 3 "b 'fl}' l' L'_ (.. . .u. Ie Illes 0 L

and ~ coi~cid~ outsi?e the sphere, but inside the sphere they have01.PdPoslte dlfectlOns, smce here the relation D = 2 eE is no longer va-

l and D = FoE + P. 0'

RO

83

(3.26)

FieldD

Fig. 3.1U

FieldE


Fig 3.9

We see that in the case under consideration, the lines of E arerefracted and undergo discontinuities (due to the presence of boundcharges), whileihe lines of D are only refracted, WI~hou t thscon ttnuities (since there are no extraneous charges at the Illterlace).

Boundary Condition 'on the Conductor-Dielectric Interface. If medium 1 is a conductor and medium 2 is a dielectric(see Fig. 3.8), it follows from formula (3.23) that

pouents o[ vectors D are equal leads to the conclusion that LJ 2 > LJ1in magnitude, Le. the lines of D must be denser in dielectric 2.

EH /E2n

Eu /E l1l •

tan Cl2

tan Cll


throngh tllO inlt'l'[a(;('. If, however, tltel'O aro /10 extraneouscharges at the interface (u = 0), wo ohtain

I Din fl~1l" IIn this case the normal components of vector l) do not havea discontinuity and tUI'll out to he tlte same on differentsides of the interfaGo.

Thus, in the absence of extraneous charges at the interface hetween two homogeneous isotropic dielectrics, thecomponents E, anfl J)n vary continuollsly dnring a transition through this interface, while the components En andD, have discontinuities.

Hdl'action o[ E and D Lines. The boundary conditionswhich we obtained for the components of vectors E and Dat the interface between two dielectrics indicate (as willbe shown later) that these vectors have a break at this interface, Le. are refracted (Fig. 3.9). Let us flIld the relationbetween the angles a l and a 2 •

In the absence of extraneous charges at the interface,we have, in accordancp- with (3.22) and (3.24), E 2 , = E l ,

and 8 2E2n = 8 rE ln • Figure 3.9 shows that

82

Taking into aceount the above conditions, we obtain thelaw of refraction of lines E, and hence of lines D: .

This means that lines of D and E will form a larger anglewith the normal to the interface in the dielectric with alarger value of 8 (in Fig. ;3.9, 8 2 > 8 1),

0*

where n is the conductor's outward normal (we omitted thesubscript 2 :;ince it is inessential in the given case). Let usverify formula (~).26). In equilibriulll, the electric fteldinside a conductor is E =,c 0, and hence the polarizationP = O. This means, according to (;3.17), that vector D =~ 0inside the conductor, Le. in the notations of formula (:L~3)

D1 = 0 and DIn = O. IIenceD2n = a.Bound Charge at the Conductor Surface. If a homogeneous

dielectric adjoins a charged region of the surface of a COIl

ductor, bound charges of a certain r1en:;ity a' appeal' at theconductor-dielectric interface (recall that the volume density of bound charges p' = 0 for a homogeneous dielectric).Let us now apply the Gauss theorem to vector E in the sameway as it was done while deriving formula (2.2). Considering that there are both bound and extraneous charges (a

(;3.2!'i)~ tan Cl2 E2

tan (;(, E t

Example. Let us represent graphically the fields E and D at thei II ["ri'ace hetween two homogeneous dielectrics 1 and 2, assuming that:.'2> 1-'1 awl llli~re is 110 extraneous charge on this surface.

Since 1'2 > ;'n in accordance with (3.25) Cl2 > at (Fig. 3.10).Considering that the tangential component of vector E remains unchanged anrl using Fig. ;·1.9, We can easily show that F 2 < Hl inmagnitude, i.e. the lines of E ill dielectric 1 must be denser than indielectric 2, as is shown in Fig. 3.10. The fact t~at. the normal corn-

I t can be seen that the surface density G' of the boundcharge in the dielectric is unambiguously connected withthe surface density G of the extraneous charge on the conductor, the signs of these charges being opposite.

and G') at the conductor-dielectric interface we arrive atthe following expression: E" = (G + a')leo' On the otherhantl, according to (3.2G) E" = Dn/ee,o = aleeo' Combining these two equations, we obtain Gle = G + a', whence

85

(3.29)

3.6. Field in /l Homogeneous Dielectric

Multiplying both sides of this equation by eeo, we obtain

• D = Do, (3.30)

Le. the field of vector D does not change in this case.It turns out that formulas (3.29) and (3.30) are also valid

in a more general case when a homogeneous dielectric fillsthe volume enclosed between the equipotential sur1acesof the field Eo of extraneous charges (or of an external field).In this caBe also E = Eo/e and D = Do inside the dielectric.

In the eases indicated above, the intensity E of the fieldof bound charges is connected by a simple relation withthe polarization P of the dielectric, namely,

E' = -Ple{}. (3.3t)

This relation can be easily obtained from the formula E == Eo + E' ifl we take into account diat Eo = eE andP = xeoE. .

As was mentioned above, in other cases the situationis much more complicated, and forml1las~ (3.29)-(3.3t) areinapplicable.

CornU.Fie. Thus, if a homogeneous dielectric fills theentire space oeeupied by a field, the intensity E of the field

a' are unambiguously connected with the extraneous chargesa on the surface of the conductor.

As before, the,re will be no field inside the conductor (E == 0). This means that the distribution of surface charges(extraneous charges a and bound charges a') at the conductordielectric interface will be similar to the previous distribution of extraneous charges (G), and the configuration ofthe resultant field E in the dielectric will remain the sameas in the absence of the dielectric. Only the magnitude ofthe field at each point will be different.

In accordance with the Gauss theorem, a + G' = eDEn,where En = Dnleeo = a/eeo' and hence

G + a' = Gle. (3.28)

But if the charges creating the field have decreased by a factorof e everywhere at the interface, the field E itself hasbecomeless than the field Eo by the same factor:

E = Eo/e.

(3.27)I ' 8-1 I0=--8- 0 .


3.6. Field in a Homogeneous Dielectric

It was noted in Sec. 2.1 that the determination of theresultant field E in a substance is associated with considerable difficulties, since the distribution of induced charges inthe substance is not known beforehand. It is only clearthat the distribution of these charges depends on tho natureand shape of the substance as well as on the configurationof the external field Eo.

Consequently, in the general case, while solving the problem about the resultant field E in a dielectric, we encounterserious difficulties: determination of the macroscopic fieldE' of bound charges in each specific case is generally a complicated independent problem, since unfortunately thereis no universal formula for finding E'.

An exception is the case when the entire space where thereis a field Eo is filled by a homogeneous isotropic dielectric.Let us consider this case in greater detail. Suppose that wehave a charged conductor (or several conductors) in a vacuum. Normally, extraneous charges are located on conductors. As we already know, in equilibrium the field E insidethe conductor is zero, which corresponds to a certainunique distribution of the surface charge G. Let the fIeldcreated in the space surrounding the conductor be Eo.

Let us now fill the entire space of the field with a homogeneous dielectric. As a result of polarization, only surfacebound charges G will appear in this dielectric at the interface with the conductor. According to (3.27) the charges

II

III

III':1 1

I'

Problems87

(2)

E = pl/ee:o (1 ~ a),

E = paleo (l ~ a).

, 2 dr2 dP r +2rPr dr=-pr r,

DS = pSI, D = pi,

DS = pSa, D = pu,

which is just the required result.f vector D An infmitely large pla~e

• 3.2. The Gauss thedo~i t~~c with th~ dielectric constant e ISmade of a hBtnogeneous Ie ee 1uniformly charged by an. extraneouscharge with volume densIt~ P > O. Ex I{J

The thickness of the platt! IS 2a. (ll 1~h1lii-l~Find the magnitude of vector E an. l;; E?,the potential cp as functions of the

ldI-

stance 1 from the middl~ of. the p a~e(assume that the poten~Ial IS ~ero hnthe middle of the plate), ChOOSllly\ I'X-axis perpendicular to the pal'.Plot schematic curves for the pro-. . n E (x) of vector E and the po~~~t:fal ~(x). (2) Find the surface andvolume densities of the bound charge. /

Solution. (1) From symmetry co~- .siderations it is clear that ~'I= ~;ii FIg. 3.11the middle of the plate, w 1 I' a d-'other points vectors E fa~h pefP:: In order to determine E, w,e s~bIlicular to the surface 0 I' P a . D (since we know the dIstrr u'use the Gauss theorem for vector W take for the closed su~face ation of only extran~ous charges)! hose endfaces coincid~s WIth theright cylinder of helghthl, one 0 ~'onal area of this cylInder be S.midplane (x = 0). Let t I' cross-sec 1

Then

In the case under consideration we havee-l 8-1 q

Pr=xeoE r = -e Dr=-;- 4n:r2

. (2) will have the formand after certain transformation expressIOn

1 qp'=~ ?"

whence

. h' I er between the spheres withHere dq' is the bound c~arg~ ill th \ ~n, a;: p'4Jtr2 dr, we transform (1)radii rand r + dr. Considermg a qas follows:

Problems

. .3.1. Polarization of a dielectric and the bound charge. An extraneous point charge q is at the centre of a spherical layer of a heterogeneous isotropic dielectric whose dielectric constant varies only in theradial direction as 8 = air, where a is a constant and r is the distancefrom the centre of the system. Find the volume density p' of a boundcharge as a fuuction of r within the layer.

Solution. We shall use Eq. (3.6), taking a sphere of radius r asthe closed surface, the centre of the sphere coinciding with the centreof the system. Then

4Jtr2 • Pr = -q' (r),

where q' (r) is the bound charge inside the sphere. Let us take thedifferential of this expression:

4Jt d(rl·pr) = - dq'. (1)

will be lower than the intensity Eo of the field of the sameextraneous charges, but in the absence of dielectric, by afactor of e. Hence it follows that potential cp at all pointswill also decrease by a factor of e:

cp = cpo/e, (3.32)

where CPo is the field potential in the ahrnce 01. the dielectric.The same applies to the potential difference:

U = Uoh, (3.33)

where U0 is the potential difference in a vacuum, in theabsence of dielectric.

In the simplest case, when a homogeneous dielectricfills the entire space between the plates of a capacitor, thepotential difference U between its plates will he by a factorof e less than that in the absence of dielectric (naturally,at the same magnitude of the charge q on the plates). Andsince it is so, the capacitance e = qlU of the capacitorfilled by dielectric will increase e times

e' = ee, (3.34)

where e is the capacitance of the capacitor in. the absenceof dielectric. It should be noted that this formula is validwhen the entire space between the plates is filled and edgeeffects are ignored.

9. Electric Field in Dtl!lectrics

The graphs of the funtions Ex (x) and <p (x) are shown in Fig. 3.11.It is useful to verify that the graph of Ex (x) corresponds to the derivative -alp/ax.

(2) In accordance with (3.13), the surface density of the boundcharge is

, e-1fJ =Pn="X80E n=(e-1)pa/e=--. pa>O.

e

89

(b)

Problems

(a)

Fig. 3.12

\---'---!---~, 0 ao a b

. ·f I d· tributed with the volumc.3.4. Extraneous ~argerUd·lu~r:\:ad~of a homogeneous dielec

deMity p > 0 over ~ sp ~:e 0 ~~nd (1) the magnitude of vector E as atric .with the pet:nntttvI y 8. he centre of the sphere and plot thefunction of the dl!~tanceEr(f)ro~dtm (r). (2) the surface and volume den-curves of the functions raT 'sitie5 of bound charge.

d . E we shall use the Gauss thea-Solution. (1) In .order to ketcr~h~e di~tribution of only extraneous

rem for vector D smce we now

c~: D p4 , D P r E=-=- r,

r<a, 4nr3IJ=Tnr p, =7' 8100 3880

2 4, D_ Pa3 E=!!-=pa3~.r;;:' a, 4nr D=3 na p, - 3r' ' eo 3eo r

The curves for the functions E (r) and ~ (r~ are ~hown in Fig. 3.13.(2) The surface density of the boun c arge IS

e-1 pafJ'=Pn=-e- T·

d ·t of bound charge it is sufficient toIn order to find the volume enSI y f I (3 11) ~nd we getrepeat the reasoning that led us to ormu a . ,

8--1 (1)p' =c - --;;- fl·

'rh••• __It can be obtaincd ill a different way, viZ·I. bY

t 1I~(i.llgn:;:KU E d x doc~ not depend on coor( rna es m-

Eq. (3.9). Since P = "XfO an

where P is the volume density of the extraneous charge. HenceIJ p r3 -a'

E-- -=-- --,-- eEo 3808 r

The corresponding curves for E (r) and cp (r) are shown in Fig.· 3.12b.


This result is valid for Il0th sides of the plate. Thus, if the extraneouscharge p > 0, the bound charges appearing on both surfaces of theplate are also positive.

In order to find the volume density of the bound charge, we useEq. (3.9) which in our case will have a simpler form:

p'=_ apx =_.!.- ( e-1 px)=-- e-1 p.ax ax e e

It can be seen that the hound charge is uniformly distributed overthe bulk and has the sign opposite to that of the extraneous charge.

• 3.3. A homogeneous dielectric has the shape of a sphericallayer whose inner and outer radii are a and b. Plot schematically thecurves of intensity E and potential <p of the electric field as functionsof the distance r from the centre of the system, if the dielectric ischarged by a positive extraneous charge distributed uniformly (1)over the inner surface of the layer, (2) over the layer's hulk.

Solution. (1) We use the Gauss theorem for vector D, takingfor the closed surface a sphere of radius r:

4nr'D = q,

where q is the extraneous charge within this sphere. Hence it followsthat

D (r < a) = 0, D (r > a) = q/4nr2 •

The required intensity is

E (r < a) = 0, E (r > a) = D/eeo.

The curve for E (r) is shown in Fig. 3.12a. The curve for <p (r) is alsollhown in this figure. The curve <p (r) must have such a shape that thederivative 8<plar taken with the opposite sign ~orresponds to the curveof the function E (r). Besides, we must take into account the normalization condition: <p -+ 0 as r -+ 00.

It should be noted that the curve corresponding to the functioncp (r) is continuous. At the points where the function E (r) has finitediscontinuities, the function <p (r) is only broken. .

(2) In this case, in accordance with the Gauss theorem, we have

44nr2IJ=Tlt (r3 -a') p,

91

Fig. 3.16

Problems

Fig. 3.15

after an external electric field is switched off. If the sphere is. polarized uniformly, the field intensity inside it is E' = - P/SEo, where Pis the polarization. (1) Derive this formula assuming that the sphere ispolarized as a result of a small displacement of all positive charges ofthe dielectric with respect to all its negative charges. (2) Using thisformula, find the intensity Eo of the field in the spherical cavity insidean infinite homogeneous dielectric with the permittivity f if the fieldintensity in the dielectric away from the cavity is E.

Solution. (1) Let us represent this sphere as a combination of twospheres of the same radii, bearing uniformly distributed charges withvolume densities p and -po Suppose that as a result of a small shift,the centres of the spheres are displaced relative to one another by adistance I (Fig. 3.15). Then at an arbitrary point A inside the spherewe have

E' =E~+E~=-3P(r+-r_)= __ .J:!.,eo 3eo

where we used the fact that field intensity inside a uniformly chargedsphere is E = pr/3Eo, which directly follows from the Gauss theorem.It remains for us to take into account that in accordance with (3.4),pI ~ P. ,

(2) The creation of a spherical cavity in a dielectric is equivalentto the removal from a sphere of a ball made of a polarized material.

where we assume that the charge of the inner plate is q > O. Let usfind E with the help of the Gauss theorem for vector D:

D 1 q 1 q. 4nr

2D = q; E = EEo = 4nEo ~~ 4'1Eo r;;:.

Substituting the latter expression into (1) and integrating, we find

q b 4neoaU = 4neoa In a' C = In (bla) •

• 3.7. The Ga_ theorem and the principle of SU,ertlOSitiOD.Suppose that we have a dielectric sphere which retains polarization

(1)

r


a

·Fig. 3.13

90

Fig. 3.14Ileous diplectric and hav· tIstant f. Plot approximat~r:fur~:sf~~Pf radius b and the dielectric contan

lce from the centre of the sphere Y)t~nd <fJh(r), ~here r is the dis-

VI' y. ,I e sp ere IS charged positi-

Solution. Dy definition th .potential <fJ of the conducior eS~;~I~~~agnce C =11~!<fJ· Let us find ,the

, menta y a charge q to ·t.00 b - 1 ."1 00

<fJ= \ Er dr= _ r -.!L dr-' _1 \ q. 4n£o \ £r 2 T 4 - -- dr.a ~. n8.0 ., r 2

I hntcgrating this expression, we obtain

<fJ=~4q (_1_ j-~)., 4nEoEanEoE a b ,L- = 1-c,~--"'-,--_

p -HE-1) a!bIll' curves for E (r) and ()

<fJ r are shown in Fig. 3.14.h • 3.6. Capacitance of a ca ac·t

t e radii of the plates a and b \\~er~ or:-A ~pherical .capacitor withheterogeneous dielectric who' _ . a. ;- b, IS filled wIth an isotropicfrom the centre of the svste~ea;er~ttl\'!tydepends on the distance rthe capacitance of the capacitor. f

- air, where a is a constant. Find

So/ution. In accordance w'th th d ..capacltor(C= !U) th I ~ efillltlOnofthecapacitanceofpotential djffere~ce 'u f~lrob~em IS jreduced to determination of th:

a gIven c large q:b

u= JEdr,

a

side the sphere), we obtain

p' = -V'p = -XEllV.Ewhere EoV'E = P + p' H' 'mula (1). . ence p = -x (p + p'), which gives for-

• 3.5. Capacitance of a cond t .spherical conductor of radius a, su~~o~~de~lIbd the capacitance of a

y a layer of a homoge-

~.. -;;-'-'-'..-.•'.i' ,

, "

Consequently, in a('cordance with the principle of superposition, thefield E inside the dielectric can he represented as the sum E = E' ++ Eo· Hence

93Problems

n

We immediately und that ~; = ~. no bound surface charge (with theThus, in the giv~n ca~e I~re:s 'ontact with the extraneous point

exception o~ ~hc pomt~hl~ ~lh~e~le~tric licld iu the surrounding spa~echarge q). 'I hiS mea~s a , and b' depends only on the dlsis the field of the, pomt char~e ( ~eq ~harge q' is unknown, and hencetance r from thiS cha~ge'l u m for vector D. Taking fo~ the ~lose~we must use the G~uds.s t lCo:'~h the centre at the point of locatIOn o.surface a sphere of ra 1U,S r \\1the charge q, we can write

2:rtr2Do + 2rr.r2J) = q,

I II ""uit'ltll'S of H'l'tor D at the disLa,nce rwhere Do and J) are t lC lu

o. 'I I'll tJ'e ";I'eleclric reS}lech"eiy,h . th V"CUUIn all( ,. • to

from the C arge qlHI et · <. 'tv of lhe tanrrential component of vee rBesides, from t Ie con WUI J 5

E it follows that J) = d'o.

Combining these two conditions, we find

q eqDo ~ 2lt (1+e) r2 ' j)oc= 2lt <1+1;) r 2 •

0'/2eo = e (-0'/2eo).

h t cos t} - lIr As 1 -+- 0 the quantityHere we took into account. t ,a t t th;inte~face there is no surfacea' -+- 0, i.e. if the charge q IS JUS a ,

charge on the plane. 'fa I' a thin rinO' with the centre at(2) Let us imagine at the lllt~h tC the inner a;' outer radii of this

the point 0 (Fig. 3.17). Suppose ,a are r' and r' + dr'. The surfacermg . . th' . g isbound charge WI thm ,IS rIDdq'=o' • 2:rtr' dr'. It can be ~:from the figure that r2 = 12 + r •whence r dr = r' dr', an~ the exp.re~~sion for dq' combined WIth (1) glV~S

e-1 drdq'=-e'+1 qlrz·

Integrating this equatio~ over r between land 00, we obtam

e-1Fig. 3.17 q'= - e+1 q.

, , n the plane separating a va.cu~I!1• 3.~O: ~ p(),lIlt charge qdl'~I~ctric with the dil~lectric p~rmlttlvl-

from an mtimte fiomo~eneousf I D and E in the entIre space.F' d the magmtudes 0 vectorsty e. m . II from the continuity of the normal

Solution. In this case, It fa o~ eE ,Only the surface cha.r~e .a'component of vector D thatf2n n~;:t of veo;Lor E in the vlclmtywill contribute to the n?rmd a t"c~P~ence the above equality can beof the point under consl era 10 •written in the form

(1)V-R= E~-I-E~.


UsilJJ{ conditions (3.22) and (3.21) at the interface between dielectrics,we fmd

E-r; = Eo sin a o, En = IJnlceo = Eonle = Eo cos aole,

when> Eon is the normal component of the vector Eo in a vacuum.Substituting these expressions into (1), we obtain

i.e. E < Eo.

• 3.9. A point charge q is in a vacuum at a distanc\: 1 from theplane surface of a homogeneous dielectric filling the half-space belowthe plane. The dielectric's permittivity is e. Find (1) the surface density of the bound charge as a function of the distance r from the pointcharge '1 and analyse the obtained result; (2) the total hound charge onthe surface of the dielectric.

Solutioll. Let liS Use the continuity of the normal component 0'vector D at the dielectric-vacuum interface (Fig. 3.17):

D2n = DIn' E2n = eE1n

Eo = E - E' = E + P/3EO

'

Considering that P = (e:- 1) EoE, we obtain

Eo = (2 + e) E/3.

• 3.8. Boundary conditions. In the vicinity of point A (Fig. 3.16)belonging to a dielectric-vacuum interface, the electric field intensityin vacuum is equal to Eo, and the vector Eo forms the angle ao withthe normal to the interface at the given point. The dielectric permittiv~ity is e. Find the ratio E/Eo, where E is the intensity of the field in~side the dielectric in the Vicinity of point A.

Solution. The field intensity inside the dielectric is

or

1 q (J' ( 1 q (J' )---- -- cos t}+--=e -_ -cos \'t--._ ,4n:fo r

22eo 4:rteo r 2 , 2e

owhere the u'rm a' /2£0 is the component of the electric field createdDear the region of the plane under consideration, where the surfacecharge density is a'. From the last equality it follows that

0,=_e-1 ~ (1)e+ 1 2ltr3 '

ml,.

IIII',I

'.1.[II

1.

1

'/

Ii

1'1

'II

4. Energy of Electric Field

should be noled that the field n in this case is not determined by theextranC?us charge alone (otherwise it would have the form of the fieldof a pOlllt charge).

954.1. Electric Energy 0/ a System of Charges

de. Then the corresponding work of these forces is

6/1 1• z = FI·rll l + F2 ·dlz·

Considering that F 2 = - 1'\ (according to the N.ewtonthird law), we can write this expression in the form

61'11.2= F I (dll-dl l )·

The quantity in the parentheses is the dis~la~ement .ofcharge 1 relative to charge 2. In other words, It IS ,the ~IS

placement of charge 1 in the system of. referen.ce 1! ' whIchis rigidly lixell to charge 2 and accomplishes wIth It a trallSlational motion relative to the initial reference system K.Indeed the displacement dl l of charge 1 in system K canbe rep;esented as the displacement dl 2 of .system K' plusthe displacement dli of charge 1 relatIve to systemK': dl l = dl z +dlj. Hence, dl l - dlz = dli, and

I{ 6At,2==Ft.dl~.

Thus, it turns out that, the sum of the elementary worksdone by two charges in an arbitrary system of reference Kis always equal to the elementary work done by the~or~e

acting on one charge in another reference system (/\ ) III

which the other charge is at rest. In other words, the work6A does not depend on the choice of the initial system ofl.Z

reference.The force F acting on charge 1 from charge 2 is conserva

tive (as a cellt~al force). Consequently, the work of the givenforce in the displacement dli can be represented as the decrease in the potential energy of interaction between thepair of charges under consideration:

6.111. 2 - dWl2 ,

where Wl2 is the quantity depending only on the distancebetween these charges.

2. Let us now go over to a system of three point charges(the result obtained for this case can be easily g~neralizerl

for a system of an arbitrary nllmber oE charges). fhe war\..performed by all forces oE interaction during elementarydisplacements of all the charges can be represented as thesum of the works of three pairs of interactions, i.e. 6.1= l'lA1. 2 + 6A1.3 + 6.1 2 ,3' But as it has just been shown,

FieldDFig. 3.18

4. Energy 01 Electric Field

. Field E

94

4.1. Electric Energy of a System of Charges

~nergy .Approach to Interaction. The energy approachto lllteractlOn between electric charges is, as will be shownrather fruitful in respect of its applications. Besides, thi~approach makes it possible to consider the electric fieldfrom a different point of view.

First of all, let us find out how we can arrive at the concept of the energy of interaction in a system of charges.

1. Let us first consider a system of two point charges 1and 2. We shall [lOd the algebraic sum of the elementaryworks of the forces FI and F 2 of interaction between thecharges. Suppose" that ill a certain system of reference Kthe charges were displaced by dl

land dl

2during the time

and the ele&tric field intensity in the entire space is

E = .!.!J!. = 'I80 2n (t +e) for 2

It ean ~ seen that for 8 = 1 these formulas are reduced to the already famlhar expressions for D and E of the point charge in a vacuum.

The obtained results are represented graphically in Fig. 3.18. It

t4.2)

(4.3)

q

Fig. 4.1

1~II! - - - g.(D·-- - 2 I riO

7-0181

Total Energy of Interaction. If the charges are arrangedcontinuously, then, representing the system of charges. asa combination of elementary charges dq = p dV and gOlQgover in (4.:1) from summation to integration, we obtain

w· -~- ~ plpdV, I

4J. Electric Energy of a System of Charges 97

Example. Four similar point ch,~rges q ar~local('d at th: ve:tice~of a tet.rdbedron w.ith an edge a (I'lg. 4.1). I·tnd the energ) of tnterac:tinll of charges in this syst.em.

The energy of interaction fn.. each pair of charges of t.his system i~t.he same and equal to 1V I = q2 /4m oa. It. can be seen from the figurethat. the total number of interacting pairs is six, a~d hence the energyof interaction of all point charges of the system IS

W = 6 WI = 6q2/4neoa. q

Another approach to the solution of thisproblem is based on formula. (4.3). The p,0t.ential q> at the point of 10catlOII of one of thecharges. created by the Held of all the othercharges. is q> = 3q/4neoa. lienee q q

Considering that WI =c= qi(Pio where qi is the ith cha:ge oftho system and (ei is the potential create~ ~t the pomt oflocation of the ith charge by all the remaunn~ charg~s, weobtain the final expression for the energy of mteractlOn ofthe system of point charges:

obviously docs 1I0t depend on the number ~f charg?s constituting the system. Thus, the energy of mteractlOn fora system of point charges is


W = W12 + W13 + W 23•

Each term of this sum depends on the distance betwooncorresponding charges, and hence the energy W of the givensystem of charges is a function of its configuration.

Similar arguments are obviously valid for a system ofany number of charges. Consequently, we can state that toeach configuration of an arbitrary system of charges, therecorresponds a certain value of energy W, and the work ofall the forces of interaction upon a change in this configuration is equal to the decrease in the energy W

6A = - dW. (4.1)

Energy of Interaction. Let us find the expression for theenergy W. We again first consider a system of three pointcharges, for which we have found that W = W 12 + WllI ++ W 23• This sum can be transformed as follows. Werepresent each term W ik in a symmetric form: Wilt =

= ~ (Wlk + Wkl ), since Wik = Wk !. Then

W = ; (Wt2 +W21 +W 13 + W31 + W23 + W32).

Let us group the terms with similar first indices:

W =+ [(W12+ W(3) + (TV21 +W23 ) + (W:11 +W32)].

Each sum in the parentheses is the energy WI of interactionbetween the ith charge with all the remaining charges.Hence the latter expression can be written in the form

3

W = ; (Wt +W2+Ws) =+ ~ Wj'i=l

This expression can be generalized to a system of an ubitrary number of charges, since the above line of reasoning

for each pair of interactions 6A i, k = - dWik , and hence

6A = - d (W12 + W13 + W 23) = - dW,

where W is the energy of interaction for the given system ofcharges:

96

where (jJ is the potential created by all the charges of thesystem in the volume element dV. A similar expression canbe written, for example, for a surface distribution of charges.For this pmpose, we must replace in (4.4) p by 0' and dV hydS.

Expression (4.4) can be erroneously interpreted (andthis often leads to misunderstanding) as a modification ofexpression (4.3), corresponding to the replacement of theconcept of point charges by that of a continuously distributed charge. Actually, this is not so since the two expressions differ essentially. The origin of this difference is indifferent meanings of the potential (jJ appearing in these expressions. Let us explain this difference with the help ofthe following example.j

Suppose a system consists of two small balls havingcharges qi and q2' Thedistance between the balls is considerably larger than their dimensions, hence qi and t}z can heassumed to be point charges. Let us find the energy W ofthe given system by using both formulas.

According to (4.3), we have1

W ="'2 (qt(jJt + q2(jJ2) = qt(jJl = q2(jJ2,

where <PI {(jJ2) is the potential created by charge q2 (ql) atthe point where charge ql (q2) is located.

On. the other hand, according to formula (4.4) we must.split the charge of each ball into infinitely small elementsp dV and multiply each of them by the potential <P createdby not only the charge elements of another ball hut by thecharge elements of this ball as well. Clearly, the result willbe completely different:

W = WI + W 2 + W12 , (4.5)

where WI is the energy of interaction of the charge elementsof the first ball with each other, W 2 the same but for thesecond ball, and W 12 the energy of interaction between thecjJ.arge elements of the first ball and the charge elements ofthe second ball. The energies WI and W 2 are called theintrinsic energies of charges ql and q2' while W 12 is the energyof interaction between charge ql and charge q2'

Thus, we see that the energy W calculated by formula (4.3)

!lO4.2, Charged Conductor and Capacitor Energies

7*

4.2. Energies of a Charged Conductor anda Charged Capacitor

Energy of an Isolated Conduetor. Let a conductor. havea charge q and a potential (jJ. Si?ce the value of (jJ IS thesame at all points where charge IS located,w,e .can. take (jJin formula {4.4),:outof the integral. The remallllllg !ntegralis just the charge q on the conductor, and we oLtam

.W ~ .. -* = 0f = -f-. \ (1t.G)

These three expressions are written assuming that C = q/(jJ.Energy of a Capacitor. Let q and (jJ + be the charge.and

potential of the positively charged plate. o~ a capaCitor,.According to (4.4), the integral can be splIt lllto two parts(for the two plates). Then

W = +(q+(jJ+ +q-(jJ-).

Since q_ = - q+, we have1 1

W =2 q+ «(jJ+ -<p-) =2 qU,

where q = q+ is the charge of the capacitor ~nd .U is thepotential difference across its plates. Consldermg thatC = q/ U, we obtain the following expression for the energyof the capacitor:

----------\Iw~5'{-~!f- f,. (4.7)

corresponds only to the energy WI2 , ~hile c~lcllla.tion b~formula (4.4) gives the total energy o( lllteractwn: III a~dltion to W

12, it gives intrinsic energies WI and W 2 • DIsre

gard of this circumstance is frequently a cause of grosserrors.

We shall return to this question in Sec. 4.4. Now, weshall use formula (itA) for obtaining several important re-sults.

4. Energy of Electric Field98

I

II'I

IiiI!!

1004. Energy oj Electric Field 4.3. Energy of Electric Field iOt

bLl == l./' dq' = (q'IC) dq',

the potential tltfference between the plates atwhelJ the next portion of charge dq' is being

4.3. Energy of Electric Field

On Loc?lizatioll of Energy. Formula (4.4) defines electricenergy n' of any system in terms of charge and potential.It turns out, however, t?at energy W can also be expressedt}~rough another quantity characterizing the field itselfVl~. throu~h field intensity E. Let us at first demonstrat~thIS .by u~lDg the simple example of a parallel-plate capacitor, Igno.nn g field distortions near the edges of the plates(edge effect). Substituting into the formula W = CU2/2

(4~9)

(4.10)

W= ~ llo~E' dV= j E;D dV.

footE' E·DW=-2-=-2-

It should be noted that this formula is valid only in thecase of isotropic dielectrics for which the relation P = x€oEholds. l'

For anisotropic dielectrics the situation is more comp lcat-ed.

4/The integrand in this equation has the meaning of the ?nergycontained in the volume dV. This leads us to a very Important and fruitful physic'al idea about localization ot energyin the field. This assumption was confirmed in experimentswith fields varying in time. It is the domain where we encounter phenomena that can be explained with the help. ofthe notion of energy localization in the field. These vary~ng

fields mav exist independently of electric charges WhIChhave generated them and may propaga~e in space in theform of electromagnetic waves. Expenments show thatelectromagnetic waves carry energy. This circumstance confirms the idea that the field itself is a carrier of energy.

The last two formulas show that the electric energy isdistributed in space with the volume density

the expression C = eeoS/h, we obtain

W - cUt --'- tlloSU' _ £Ilo (Y )2 Sh-2- 2 --2 h •

And since U1h = E and Sh = V (the volume between thecapacitor plates), we get

W =+eoeE2V. (4.8)

This formula is valid for the uniform field which fills thevolume V.

In the general theory, it is p~oved that the en~rgy ':vcan be expressed in terms of E (m the case of an £SOtrop~c

dielectric) through the formula

between 0 and q, we

where l./' i::>the momenttransferred.

Integruting this expression over q'obtain

JI :,;hulIJd Le, I~uled he~e thai these formulas determine thelutal cllergy 01 lIltel'aCllOn, viz. not only the energy of interactIOII Lot \\ een the charges of one plate and those of theut.he~ plate, but also the energy of interaction of chargeswlthlll each plate.

And Wh~t if We Have a Dielectric? We shall show thatformula,s (4. fJ) and (4.7) are valid in the presence of a dielectric.For t!llS purpose, we consider the process of charging acapacitor a" a transport of small portions of charge (d ')from one plate to the other. q. Th~ elen~entary .\I'ork performed against the forces of the

f1eld 111 th IS case IS

A = q2/2C,

which ~oilJcides with the expression for the total energy ofa capacitor. ~onsequel~tly,the work done against the forcesof the eleclnc lleld IS completely spent for accumulatingt~Je en~rg~ W of the charged capacitor. Moreover, the expression o.DtalD~d for the work A is also valid in the case whenthere IS a dIelectric between the plates of a capacitor. Thus,w.e hav~ proved the validity of (4.7) in the presence of adlOlectl'lc.

Obviously, all this applies to (4.6) as well.

f

103

(4.11)

4.3. Energy of Electric Field

The Work of the Field During Polarization 01 a Dielectric.An analysis of formula (,'1.10) for the volume energy densityreveals that for the same value of E, the quantity w is E

times greater in the presence of a dielectric than when it isabsent. At first glance this may seem strange: field intensityin both cases is maintained the same. As a matter of fact.when a field is induced i'l a dielectric, it does an additionalwork associated with polarizatioll. Therefore, under theenergy of the field in the dielectric we must understand the-sum of the electric energy proper and an additional workwhich is accomplished during polarization of the dielectric.

In order to prove this, let lIS substitute into (4.10) thequantity EoE + P for D, which gives

l'oE2+ E .Pw=-- --2 2'

q2 (1' 1)W= 8nEoB a -T .

. Example 2. Find the work that must be done against the electricforces in order to remove a dielectric plate from a parallel-plate char~d capacitor. I t is assumed that the charge q of the capacitor remainsunchanged and that the dielectric fills the entire space between thecapacitor plates. The capacitance of the capacitor in the abseace ofthe dielectric is C.

The work against the electric forces in this system.is equal to theincrement of the electric energy of the system:

A = 6.W = W 2 - WlJ

where WI is the energy of the field between the capacitor plates inthe presence of thi!'Qielectric and W 2 is the same quantity in the absenceof the dielectric. Bearing in mind that the magnitude of vector Dwill not change as a result of the removal of the plate, 1.1'. that D 2 == D1 = (J, we can write •

A=W2-W I = \/~ __~) V = ~ (1-~),2Bo. 2BoB 2C B

where V = Sh and C= BoS/h, S being the area of each plate and hthe distance bl'tween them.

BOllEI 4 2ddW=-Z- itl" r,

where E = q2/4nBoBr2, Integrating this expression over r between aand b, we obtain

layer is given by

4. Energy 01 Electric Field102

W= i ~ DdS,

w~ere the integration is performed ovhlth one of the equipotential surf fr a closed surlace coincidingt eorem, this integral is equal to ~chs. hn accordance with the Gausswe finally get I' c arge q of the conductor, and

W = qrp/2,Q.E.D.

Fig. 4.2 D dS ~dW=-2- \ Edt=~

•. 2 cp,A

wh ere A is a point at the b . . '.It remains for us to make ~1lllDlng of t~e Imagmary tube.

fix£ression over all the tubes andeJa~t thtep , I.e. integrate the obtainedhd. Considering that pote~tial n. the energy localized in the entire

t ~ tubes (since they originate 0: ~h esrme at the endfaces of allwrite I' sur ace of the conductor), we

A~other substantiation for fo~hr:gl of an i~olated charged cond~:~~~s(~9~ It is known that theenerg;r{:;ufhe Ifieldrrect , proceeding from th;J~~2~fel·otcuaSI~hOt~ thaftL' Iza IOn 0

et us consider an arbitrar ..denbtallly isolate a tube of infinit~ .PosItIvely charged conductor. WeI' y mes of E (Fig. 4.2), and tak::'i~.cross section, which is bound

=dS dt T~·an ellementary volume dV=. IS vo ume contains the energy

dt ED DdS ."2 dSdt=~ Edt.

Let us now find thth~d in the entire isol:te~n~~?elo~al

IS purpose, .we !ntegrate the l~st I'o~pressIOn, conSldermg that th d xD dS is the sa . I I' pro uctof the tube andehm a 1. cross sectionsout of the i~tegral~nce It can be taken

Let us consider two exam l'Il .we get by using the ide~ of ~n~r 1 ulstral~Ing. the. advantages

gy oca lZatlOn In the field.Example 1. A point char I' .

!"'lade of a homogeneous diel:ctr1 IS ~t the ce?tre of a spherical layerlI~ner and outer radii of the la eCr ~~th the dielectric constant E. TheFmd the electric energy contaiKed ine thlJuadl. tlo a ~nd b respectively.

IS Ie ectrIC layer.We mentally isolate in the d' 1 .

ical layer of radius from r to Ie +ect~lc ~hvery thin concentric spher-r r. I' energy localized in this

dJ - J- I dl- .....----1=0 + +

:===~==:..: --1-1.- L

Fig. 4.3(4.14)lV -- r FoFt Il" ) Eof~ dl

'I J E E il'- 1 -.,-- (r -2- ., eo j' ~ c. . ,. -

4.4. A System of Two Charged Bodies

Suppose that we have a system of two clwrged bodies in avacuum. Let one body create in the surrounding space thefield E I , while the other body, the field E2 • The resultantfield is E = E) -I- E 2 , and the square of this quantity is

E2=E;+E~ + 2E j ·E2 •

Therefore, according ,to (4.9), the total energy W of thegiven system is equal to the sum of three integrals:

4.4. ..1 S,/stem of Tu'o Charged Bod/.-s 10;:;-------_...::......- ----._------

which coincides with formula (4.5) and reveals the fieldmeaning of the terms appearing in this sum. The first twointegrals in (4.14) are the intrinsic energies of the first andsecond chaJitred bodies (WI and W 2), while the last integralis the energy of their interaction (WI2).

The following impprtant circumstances should be mentioned in connection with formula (4.14).

1. The intrinsic energy of each charged body is an essentially positive quantity. The total energy (1.9) is also alwayspositive. This can be readily seen from the fact that the in-tegrand contains essentially positive quantities. However,the energy of interaction can be either positive or ne~ative.

2. The intrinsic energy of hod ies remains constant uponall possible displacements that do not. change the configuration of charges on each body, and consequently this energycan be assumed to he an additive constant in the expressionfor the total energy W. In such cases, the changes in Warecompletely determined only by the changes in the interaction energy W)2' In particular, this is just the mode of behaviour of the energy of a system consisting of two pointcharges upon a change in the distance between them.

3. Unlike vector E, the energy of the electric field isnot an additive quantity, Le. the energy of a field E whichis the sum of fields E) and E 2 is generally not equal to thesum of the energies of these fields in view of the presence ofthe interaction energy lV12 . In particular, if E increases ntimes everywhere, the energy of the field increases n2 times.


The first term on the ri ht-ha d' '.energy density of the fiel~ E . n side comcldes with theus to verify that the "addit'o I~" a vacuufI.l. It remains forwith polarization I na energy E·P/2 is associated

Le~ us calculate' the work done b .polarIzation of a unit volume of th YI ~hle el~tr~e .field for

e ( Ie ectnc, I.e. for the

disRlacements of char es' , .agamst the field upon g p+ and ~- I'espectIvely along andE to E + dE. Neglect~~ I~~rease m the field intensity from

g e second-order terms we write6A=p~E.dl++r'E.dJ '

where dJ + and dl are add' . -. -increase in the fi""eld b dIEtlO(nFa.1 dIsplacements due to anP" Y Ig 4 3) C 'd'- = -- P+, we obtain . . . onSI ermg that

6..1 = p: (dl+ - dL). E ~= p~ dl. E,where dl = dl + - dl is h . .~he positive charges relative\ e t~ddItlon~1 displacement ofmg to (3.4) p' dl _ dP doe negatIve charges. Aceord-

, + - an we get

6A = E.dP.Since P E (4.12)

= xeo , we have

oA =E'xeodE= d ()(e~E2) = d (E~P).

Hence, the total work funit volume of a dielectric ~~ent or the polarization of a

. . A = E·P/2, (413)wlndl coincides w'th th .

Thus, the volu~e enee:econd t~rm in formula (4.11).the energy eoE2/2 of the lY densIty w = E· D/2 includesassociated with the 'p I . eld. proper and the energy E.P/2

o aflzatlOn of the substance.

~I;:

1064. Energy oj Electric Field 4.5, Forces A cUng in a Dielectric 107

4.5. Forces ~cting in a Dielectric

Electr~striction. E~periments show that a dielectric inan electnc field expenences the action of forces (sometimesthese forces are called ponderomotive). These forces appearwhen the diel.ectric is neutral as a whole. Ponderomotiveforces appa.ar In the lo.ng run due to the action of a nonunifo:m ~lec.tnc field on dIpole molecules of the polarized dielectnc ~It IS known that a dipole in a nonuniform electricfield IS acte~l upon by a force direcled towards the increasingfield). In thIS case, the forces Ilre cansed by the nonuniformi~YI of not only ~he ~acroscopic fIeld hut the microscopic

e d as well, whIch IS created mainly by the nearest molecules of the polarized dielectric.

. Unde~ t~e action of these electric forces, the polarizeddI~le~tnc IS deformed. This phenomenon is called electrostnctwn: As a r~sult of electrostriction, mechanical stressesappear In the dIelectric.

( 0,wing to electrostriction, not only the electric forceWhlC~l depe;'1ds on the charges) acts on a conductor in a

polanzed dIelectric, but also an additional mechanicalf~rce c.aused .by the dielectric. In the general case, the effecto a dIelectrIc on. the resultant force acting on a conductorcannot be taken Into account by any simple relations andthe ~I:oblem of calcul.ating the forces with simulta~eousaJlaJ~~Is of th,e mechamsm of their appearance is, as a rule,rather complIcated. However, in many cases these forcescan .be clllcnlat:ed in a sufficienLly simple way without a~et~I1ed :malysls of their origin by using the law of conservatIOn of energy.

Energy Method for Calculating Forces. This method is~,he most general. It allows us to take into account automat~cally' all force interactions (both electric and mechanical)IgiorIng thei~ origin, and hence leads to a correct result.

I e~ us consIder the essence of the energy method for calc~ atmg forces. The simplest case corresponds to a situation when charged conductors are disconnected from thepowe~ supply. In this case, the charges on the conductorsrem?lll unchanged, and we may state that the work A ofall mternnl forces of tII:e syst~m upon slow displacementsof the conductors and dIelectncs is done completely at the

~..,;;!.~.".'.".'•.'

... :..L ...

.' .;'

expense of a decrease in the electric energy W of the system(or its field). Here we assume that these displacements donot cause the transformation of electric energy into otherkinds of energy. To be mOre precise, it is assumed that suchtransformations are negligibly small. Thus, for infinitesimaldisplacements we can w.rite

~A= -dW\q, (4.15)

where the symbol q emphasizes that the decrease in theenergy of the system must be calculated ,when charges. onthe conductors are constant.

Equation (4.15) is the initial equation for determiningthe forces acting on conductors and dielectrics in the electricfield. This can be done as follows. Suppose that we are interested in the force acting on a given body (a conductor ora dielectric). Let us displace this body by an infinitesimaldistance dx in the direction X we are interested in. Thenthe work of the required force F over the distance dx is~A = F xdx, -Where F x is the projection of the force F ontothe positive direction of the X-axis, Substituting this expression for ~A into (4. t'5) and dividing both parts of (4.15)by dx, we obtain

(4.16)

We must pay attention to the following circumstance.It is well known that the force depends only on the positionof bodies and on the distribution of charges at a given instant. It cannot depend on how the energy process will develope if the system starts to move under the action of forces.And this means that in order to calculate Fx by formula(4.16), we do not have to select conditions un~er which allthe charges of the conductor are necessarIly constant(q = const). We must simply find the increment dW underthe condition that q = const, which is a purely mathematical operation.

It should be noted that if a displacement is performed atconstant potential on the conductors, the correspondingcalculation leads to another:expression for the force: Fx == + aw/ax 11jO' However (and it is important!) the result

of the.calculation of F:c with the help of this formula or(4.16) IS the same, as should be expected. Therefore, henceforth we shall confine ourselves to the application of onlyformula (4.16) and will use it for any conditions, includingthose where q =1= const upon small displacements. We mustnot be confus.ed: the derivative aw/ox will be cakulatedat q = const 10 such cases as well.

F,,=- oW I=--~ (1)ax q 2eeoS •

The minus sign i!I this formula indicates that vector F is directed~owards the neg~tIv~ values on the X-axis, Le. the force is attractive

y nature. Consldenng that q = as = Ds = ee ES and E = Ulhwe transform (1) to 0,

F" = -GUt/2h.

Forces in a Liquid Dieleetric. Formula (1) of the lastexample shows that the force of interaction between theplates of a parallel-plate capacitor in a liquid dielectricIS smaller t~an the corresponding force in a vacuum by afactor of e (Ill vacuum e = 1). Experiment shows that thisresul~ can be gener~liz~d: if the entire space occupied by a~eld IS ?lled by a lIqUId or gaseous dielectric, the forces ofmteractlOn between charged conductors (at constant charaes

'"

1094.5. Forces Acting in a Dielectric

on them) decrease by a factor of e:

F = Fo/e. (4.17)

Hence it follows that two point charges ql and q2 separated by a distance r and placed into an infinite liquid or gaseous dielectric interact with the force

F -- _1-~, (4.18)-- 4neo er 2

which is smaller than the force in a vacuum by a factor ofe. This formula expresses the Coulomb law for point chargesin an infinite dielectric.

We must pay a special attention to the fact that pointcharges in this law are extraneous charges concentrated onmacroscopic bodies whose dimensions are small in comparisonwith the distance between them. Th'.1s, the law (4.18) has,unlike the Coulomb law in vacuum, a very narrow fieldof applicationi!<-the dielectric must be homogeneous, infinite,liquid or gaseous, while the interacting bodies must bepointlike in the macros~opic sense.

I t is interesting to note that the electric field intensityE as well as the· forceF acting on a point charge q in ah~mogeneous liquid or gaseous dielectric filling the entirespace of the field, are by a factor of e smaller than the valuesof Eo and F 0 in the absence of dielectric. This means thatthe force F acting on the point charge q in this case is determined by the same formula as in vacuum:

F = qE, (4.19)

where E is the field intensity in the dielectric at the pointwhere the extraneous charge q is placed. Only in this caseformula (4.19) makes it possible to determine the field Ein the dielectric from the known force F. It should be notedthat Gnother field differing from the field in the dielectricwill be acting on the extraneous charge itself (which islocated on some small body). Nevertheless, formula (4.19)gives the correct result, strange as it may seem..

Surface Density of a Force. We shall be speakmg of theforce acting on a unit surface area of a charged conductorin a liquid or gaseous dielectric. For this. pu.rpos?, let .usconsider a parallel-plate capacitor in a hqUld dielectnc.


x'+dx

i08

Example: Fin? the !orce acting on one of the plates of a II 1-plate ~apaCltor III a .hquid dielectric, jf the distance het~a the~lateJ :teh

, tre capac~tance. of !he capacitor under given conditions isan vo tage U IS malOtamed across its plates.

II?- this case, if we mentally move the plates ap1l.rl the volta e U

full:~~f:::h~~el~~~: ~h~c;~f~ I~ ~;i::~/tli:~i~~sh:iin::;j:~X ~e ~orce u;nder t.he assumption that q =

- const, I.e. WIth the help of formula(1..16). Here the mo.st convenient expres-q---- ----- Sion for the energy of the capacitor is

xc F _ ~ ~+q----<J--__ W - -n- CC= __ :r

Q "toG 2eeoS'

Fig. 4.4 w~ere ~ is the permittivity of the dielectriC, ~ IS the area of each plate, and :r isth~ .dlsta;llce ~etween them (x = h).

. ~ext, let us cho~se the posItIve dIrection of the X -axis as is sho\vnInf Fthlg. 4.4. :'-CCO~dlllg to (4.16), the force acting on the upper plateo e capaci tor IS

1 q2W=--__4..-1:80 'U'

• 4.2. Intrinsic, mutual, and total energies. A system consistsof two concentric metallic shells of radii R I and R

2with charges ql

and q2 respectively. Find intrinsic energies WI and It'2 of each shell,the energy Wl2 of interaction between the shells, and the total electric energy 0 the system W, if R

2> R

I•

We have obtained an interesting and important resultof a general nature (for a liquid or gaseous dielectric). Itturns out that the surface density of the force acting on aconductor is equal to the volume density of the electricenergy near the surface. This force is directed always outward along the normal to the surface of the conductor (tending to stretch it) regardless of the sign of the surface charge.

111Problems

S l t' In accordance with (4.6), the intrinsic energy of eachh It ue/qo:~l to qm/2 where <F is the potential of a shell created o!lly

s e hiS h 'r 't' I' e (p - q/4Jt£ R where f{ is the shell rad1U~.by t e c arge q on I, .. - • '0 '.Thus, the intrinsic energy of each shell IS

1 q:. 2

W 12 =; 4..-1:£0 2R I-:;'

The energy of interaction between the charged shells is equhl tt: thechar e of one shell multiplied by the po.tential <F created b~ t e c_argeof th~ Zther shell in the point of locatIOn of the charge q. W I2 .- q(p.

In our case (R 2 > R), we have

1 q2 1 q I q2W12 ~~ qt 4n£0 R

z=--= 4)1£0 ~.

The total electric energy of the system is

1 (qI ' '1~ : qlqz)W=WI+Wz-j-It'IZ= 4ne

o-2H

tt· 28

2' R z •

• 4 3 Two small metallic balls of radii R I and H2 are in vacuumat a di~t~nce cl'1hsiderably exceeding their dimensions and ha;e I acertain total charge. Find the ratio ql /q2. bet:-v~en the chafl~es o. t l~balls at which the energy of.the sys.tem ~s. mlmmaI. What IS the potential difference between the balls III thiS case?

Solutiun. The elecLric energy of this system is

1 (qI -iL1_qlqZ)W=WI-IWz-!-W IZ = 4neo 2R

I-+- 2H

2' l '

where Wand Ware the intrinsic electric energies of the b~IIs~qljJ~~~,W is th~ energ) of their interaction (ql<F2 or qz<FI),. anhd l IU h11;ta~~e between the balls. Since q2 =, q - ql' where q IS t e to a c argeof the system, we have

W __1_ [-..1L +- (q_q,)2 -+- ql (qL-qtl J._. 4neo 2R I 2H2 l

The energy W is minimal when aWlaql ,= O. IIence

R I Rzq ~q and qz""'q RI.J.Rz

'I~ Rt+Rz I

where we took into account that R I and R 2 are consideraIlly smallerthan land

q\lq2 = HI/Hzo

The potential of each hall (lIwy can he. considered isob ted) ,is (p ~ q!,R:Hence it follows (rolll tlip ahove I'dallOll that '.I' . : qe, I.e, lhe I'0ll ntial difference is ('qllal to zp['o lor SIlI·,II a d,sll'lh"llOlI. ,

(4.20)


Suppose that the capacitor is charged and then disconnectedfrom the power supply to maintain the charge and the fieldE of the capacitor constant when the plates are moved apart.

Let us consider once again Fig. 4.4. The energy of thecapacitor is the energy of the field within it. In accordancewith (4.9), this energy is W = (1/2) EDSx, where S is thesurface area of each plate and x is the distance between them(Sx is the volume occupied by the field). By formula (4.16),the force acting on the upper plate is

1Fx=c -8W/8x l q = -"2 EDS,

whence the surface density of the force is

Problems

• 4.1. Energy of interaction. A point charge q is at a distance lfrom an infinite conducting plane. Find the energy W of interactionbetween this charge and the charges induced on the plane.

Solution. Let us mentally "freeze" the charge distributed over theplane and displace under these conditions the point charge q to infinity. In this case the charge q will move in the potential field whichis 'equivalent to the field of a fixed fictitious point charge __q, locatedat a fixed distance l on the other side of the plane. We can writestraightaway

I',

'"

112 4. Energy of Electric Field Problems 113

• 4.4. Energy localization in the field. A charge q is uniformlydistrtbuCed inside a sphere of radius R. Assuming that the dielectricconstant is equal to unity everywhere, find the intrinsic electric energyof the sphere and the ratio of the energy Wt localized inside the sphereto the energy Wz in the surrounding space.

Solution. Let us first find the fields inside and outside the spherewith the help of the Gauss theorem:

As a result of integration, we obtain

..1-- q(qo+q/2) (_1 1_)-- 4JH'o R I R z .•

Remark. If we try to calculate the work in terms o[ the potentialas A = q (<PI - <P2), where <p is the potential createll by the ch.argo q(lat the point of location of the charge q, the answer would be (II [ferent

Fig. 4.6Fig. 4.5

qZ a-b,1=-- -- <n.. 8neo IIlJ

8-0181

and incorrect. T~ is due to the fact that this approac~ does not takeinto account the additional work pNformerl by electnc forces uponthe change in the configuratiOI) of the charge q located on the l'Xpand-ing shell.

• ~.6. A point charge q is at the centr~. of a sphlfical unch~rgedconducting layer whose inner .and oute~ radl.l are a and b respectl vcl~ .Find the work done by electnc forces III thiS system upon the remo\al of the charge q from its original position through a small hole(Fig. 4.6) to a very large distance from the spherical layer.

Solution. We shall proceed from the fact that the work of electricforces is equal to the decrease in. the .electric ene.rgy of the system. Asis well known, the latter is locahzed III the field.ltself. ',fhus, the problem is reduced to determination of the change III the lleld as a resultof this process. . .

It can be easily seen that the field around the charge q wtll changeonly within the spherical layer with the inner radius a and the outerradius b. Indeed, in the initial positiono.f .the charg~; there ~as nofield in this region, while in the fmal. POSI twn there IS a certalll field(since the conducting spherical layer IS far from the charge q). Consequently, the required work is

b\' EoEz V

A=O-Wl= - J -2- d .a

Considering that R = q/4nEor2 and dV = 4m2 dr, and integrating,we obtain

. q 1Ez=-4----Z (r;?R).

neo rE t = 4 q 3 r (r ~ R),

neoR

rZ

eoW t - W z= J T (Ei- En 4mz dr,

atwhere E t and Ez are the field intensities (in the hatched region at adistance r from the centre of the system) before and after the expansion of the shell. By using the Gauss theorem, we find

E =_1_ q+qo ~ __1_ kI 4neo rZ 2 - 4neo r 2 '

It is interesting to note that the ratio W/Wz does not depend onthe radios of the sphere.

• 4.5. A spherical shell is uniformly charged by a charge q.A point charge qp is at its centre. Find the work of electric forces uponthe expansion of the shell from radius R t to Rz•

Solution. The work of electric forces ia equal to the decrease inthe electric energy of the system:

A = Wt - W 2 •

1 3qzW=----

4neo 5R '

In order to find the difference WI - Wz, we note that upon the expansion of the shell (Fig. 4.5), the electric field, and hence the energylocalized in it, changed only in the hatched spherical layer. Consequently,

We can now calculate the intrinsic electric energy of the sphere:a 00

W=Wt+Wz=) e~Ef 4nrz dr +) e~E~ 4nr3 dr= 8n~:R (--}+1).o a

Hence, it follows that

8*

(1 )

H5Problems

1=/o-/',

pl~l'manent voltage U. F.inu the force /' acting on a unit surface of theplate from the dielectrIC.

Sol~tion. The resultant force / acting per unit area of each platecan be represented as

where /0 is the electric force acting per,unit area of a plate from theother plate (it is just the force per umt area when the dielectric isabsent). In our case, we have

I = lole, 10 = uE = f.12/2eo, (2)

where E is the field intensity in the region occu~ied.by one plate, created by the charges of the other plate. ConsJderwg that a c= D == eeoUlh and substituting (2) into (1), we obtain

/' = 10 (1 - lie) = 10 (10 - 1) f.oU2/2h2•

For example, for U = 500 V, h = 1.0 mm and 10 = 81 (water), we getl' = 7 kPa (0.07 atm).

• 4.9. The force ~cti~g on a dielect~ic. A cylindri~,al,layer of ahomogeneous di~lectrlc.WIth the dielectrIC c~nstant c IS wtro,ducedinto a cylindridH capaCItor so that the layer hll~ the gap of WIdth dbetween the plates. The mean radius of the plates IS Il sHch that R ~ cl.The capacitor is connect~d tp ~ s~ur~e of a perman.ent voltage U. Fwuthe force pulling the dlelectl'lc wSlde the capaCItor.

Solution. Using the formula W = q2!2C for thf) energy of a capacitor, we fmd that, in accordance with (4.10), the required force is

F x = - a;: Iq= ~ O~2()X =~~ ~~ . (1)

Since d ,,<:: R the capacitance of the given capacitor can b~ calcul~tedby the formula for a paraIJel-plate capacitor.. Therefore, .If the dielectric is introducer! to a depth x ann the capacitor length IS l, we have

Cc= f.f.ox~2nR + Eo (l-;>.2nR eo.~nR (ex-j-l-x), (2)

Substituting (2) into (1), we obtain

Fx = eo (e - 1) nRU2!d.

• 4.10. A capacitor consists of two fixed plates in the form of asemicirck of radius R and a movable plate of thiekness h made of adielectric with the dielectric constant 10, placed between them. Thelatter plate can freely rotate about. th(' axis 0 (Fig. 4.7) and practica~

ly fills the entire gap between the l1xed plat!~s. A constant voltage f! ISmaintained between the plates. Find the moment M about the aXIs 0of forces ading 011 the movable pIa te wllPn it is placed as shown in thefigure.

SolutitJrt. The work performed by the moment of forces M upon

IiA'= qE1.dx= eoSU~~2 x 2

'

It should be noted that AW = -A',~hus, by movi,ng the plates apart, we perform a positive work

(a~amst the e!eetI'lc forces), The energy of the rapacitor deaeases intillS case. III ordrl' to understand this, we musl ronsider a SOJ'rce main!?i!ling the potentwl diff?rence of the capacitor at a constant value.J hiS source also accomplIshes the work A S' According to the law ofconserva,tion ,')f energy, A + A' = AlV, whence As = AW _ A' == ---2.'1 ~__ 0,

(It !j .8. ro~ccs ~ct~!I~ betwee~conductors in a dieledric. A parall~l-pl~te capacJ~or IS Im!!H'rsed, 111 the horizontal position, into a liqUid !helectr1c WJth the flIelectric constant 8, filling the gap of width hbetween the plates. Then thf' capacitor is f'onnected to a SOur('(' of

whpl'l', A\ is th~ intensity of the l1eld createJ by one plate (E = u/2£.0).,It !S In tillS, fJeld that the charge located on the other plate moves.fillS work IS completely spent for increasing the f'lectric energy'AW-- A', .

(2) III this case, the force acting on each capacitor pJat~ will de(lpnd on the, oistanc~ between the plat!'s. Let us write the elementarywork 01 the force actlllg on a pla1.l~ during its displacement over a distane(' d.1' relative to the other plate:

where we took into account that q = eu, £1 = U!2;r, and C c= eoS/x.Aft,,!, integration, we obtain

The ill('J'eIlH'lIt of thp el"('lrie energy of the capacitor is

• 4.7. The work done upon moving capacitor plates apartA parallel-plate air capacitor has the plates of area S eacb. Find th~work A' against the electric forces, done to increase the di8tance bct,we~n the plates from x1.to ~2' if (1) the charge q of the capacitor and(2) It~ voltage U ,fire mallltallled constant. Find the increml'nt.."l of thl;elpclnc energy or the capacitor in ,he two cases.

Snlution. (1) The required work is

H4 . 4_,_Energy 01 Electric Field

5. Direct Current5.1. Current Density. ContinUity Equation

Electric Current. In this chapter we sb~ll co ~selves to l' f .,. nlme our-

d' . an a~a YSI~ 0 conduct JOn CUl'l'ent in a conductingme HIm, espeCIally m metals. I t is well known that ele t .~ur(rent iSh the transfe~ of charge through a certain sU~f:~~

say, t e cro~s sectJOn of a conductor).In a. conductmg medium, current can be c . d b

trons/Ulhmetals), ions (in electrolytes), or sOr:;~fl:the/p:~~~=~l~as~ti~ ~ \ ~bsenc~ of electric field, current carriers performf' 0. JOn all( on average the some number of carriers

~ureIther sign passes tllrough each side of any imaginal'face S. Thus, the current passing through S in this 'cas~

(5.1)i = P+U+ + p_u_,

is zero. However, when an electric field is applied, an ordered motion with a certain average veloeity u is imposed onthe chaotic motion of the carriers. and a current flowsthrough the surface S. Thus, an electric current is essentiallyan ordered transfer of electric charges.

The quantitative measure of electric current is intensityI, defined as the charge transferred across the surface S ina unit time:

1= dQ/dt.

5.1. Current Density. Continuity Equation 1 t 7

The unit of current is the ampere (A).Current Density. Electric. current may be distributed non

uniformly over the surface through which it passes. Hence,in order to characterize the current in greater detail, currentdensity 'vector j is introduced. The magnitude of this vectoris equal to the ratio of the current dI through a surfaceelement perpen,dicular to the direction of motion of chargecarriers to th€" area dS J of this element: j =ceo dI/dS J. Forthe direction of vector j we take the direction of velocityvector u of the ordered'motion of positive carriers (or thedirection opposite to that of the velocity vector of the ordered motion of negative carriers). If the current consists ofpositive and negative charges, its density is derIDed by theformula

where p+ and p_ are the volume densities of positive andnegative charge carriers respectively, and u + and u_ are thevelocities of their ordered motion. In conductors,where thecharge is carried only by electrons (p_ < 0 and" u+ = 0),the current density is given by

j = p_u_. (5.2)

The field of vector i can be graphically represented wnhthe help of lines of current (lines of vector j), which aredrawn in the same way as for vector E.

Knowing the current density vector at each point of thesurface S under consideration, we can also fLlld the C'Irrentpassing through th is fiurfflce f1S the f1nx of vector j:

I = J j·dS, (d)

5. Direct Current

Fig. 4.7

116

thc rotation of thc platc through an ". I ., I .'tit'crcnsc in the electric cnergy of the ..gtc"e ctlllcnt da IS equal to thc

. sys em a q = const [see (4.16»):_'\;/ z da= -dWl q ,

where W = q2/2C. Hence

Mz=- oW /. =L~iJa q 2 C2 • (1)

In the case under consideration, C = C + C hdr:l~~frfcap~~tances of the parts ?f the cap~citor ~ith a~d ;;\h~~~ tf:

. e area 0 a sector wl~.h an angle a is determined as S == aR2'12, and hence

C =~ EoaR2/2h + EEo (n - a) R2/2h.

Differentiating with respect to a we findac,liJa = (EoR2/2h) (1-E). SUbstitutint~IS ~xpression into formula (1) and con~sIderlllg that C = qlU, we obtain

U2 eoRIM Z =-2- 2h (1-e)

= -(8-1) 8oR2U2 < 0'4h .

The ncgative sign of M indicat th hacting clockwise (oppositelyZ to the eps 't~t t de. mOI?ent of the force issee Fig 4 7) Th' , OSI Ive IrectlOn of the angle a'pacitor: " IS moment tends to pull the dielectric inside the ca~

~n :~;i~fb;i~~~ ~~:~ ~I~ AJz fh:n~ependent of~he angl~ a .. However,IS due to the fact that for smail vaIu oment M Z - o..ThIS dIscrepancyas was done in the solut:on of tb' es of

bal

we cannot Ignore edge eflects,'IS pro em.

(5.6)

(5.7)

(5.8)JI=UiR, lwhere Jl is the electric resistance of the conductor.

The unit of resistance is the ohm (Q).Resistance R depends on the shape and size of the conduct

or, on its material and temperature, and above all on theconfIguration (distribution) of the current through the conductor. The meaning of resistance is quite clear when weare dealing with a wire. In a more general case of the volumedistribution of current, it is meaningless to speak of resistance until we know either the position of the leads attached to the eonductor or the current configuration.

In the simplest case of a homogeneous eylindrical eOllductor, resistance is given by

lR = Ps' (:i.!l)

where l is the length of the conductor, S is its cross-sectionalarea, and p is the resistivity, which depends on the materialof the conductor and its temperature. Resistivity is measuredin ohm-metres (Q·m).

Resistivity of the best conductors like copper and aluminium is oC_the order of 10-8 Q·m at room temperature.

5.2. Ohm's Law for a Homogeneous Condnctor 1t!J

5.2. Ohm's Law for a Homogeneous Conductor

Obm's law, which was discovered experimentally, states:the current passing through a lwmogeneous conductor is proportional to the potential ditJerence across its terminals (0['

to the l.101taf'!.' r T

,.

l) .ftf

Procl'edin~ in the same way as for the flux of vector E in Sec. 1.4, welind that the divergence of vector j at a certain point is equal to thedecrease in the charge density per unit time at the same point:

IV·j=o.j

This Illeam; that Ior dircct current the field of vector j does not havesources.

rV·j= -ap/at. IThis lcatls to thc steady-state condition (when iJp/DI = 0):

(.5.4)

(5.5)

5. Direct Current118

I~idL -*.)Tllis relation is called the '.pression for the law of . .contl~ulty equation. I t is an ex-

In the case of a t dcollservatI~n of charge.d · t 'b s ea v-state (du'CCt)IS n lItion in Sp'lce . 't ." current, the chargeW I' I 'C mu~ remalll unch I I. Of( S, 011 t IS right-hand side of E allge(. II other

sequently, for direct C\lr' t I q. (5.4) dq/dt = O. COIl-, " Tell we lave.

. Current I is a scalar al ebra" -from formula (5.a) that b~side;co{luanttty . It can be seenthe current is determined b th 1 I~r factors, the sign of,the normal at each point 0 YtJ e cflolce of the direction ofof tl e d" n lie Sur ace S i e b tl h'

I ll"eetlOn of vectors dS If h .": Y Ie c Olcevectors dS are reversed th . t e dIrectIOns of all the

Continuitv Equation' L etcurr~nt I changes sign.

. .J' • e us Imagine a I rlIII ? conllucting medium throu ,h wI . c.ose . sUl'face SIt IS customary to choose the ~ut nch a. cUl~ent IS passing.normal to closed surfaces I h ward dIrectIOn for vectors

, aD( ence for ve t dS .sequently, the integralh . d .,' COl'S. Con-the volume V (' I J'] S gl\ es the charge leavingA enve oped by the su f S). ccording to the law of 'conse t' I' ace per unit time.IS ?qu~l to the decrease in tlrva ~o~ of .ch?rge, this integralunIt tIme: 18 c alge IllSHle volume V per

~ j dS= O.

This means that in this .start or terminate any' Icase, the lInes of vector j tlo notto have no sources in tl:-'" ~ere. The. fteld of vector j is said

,e case of dIrect current.5 ' Differential form or the conti . t

( JI) and (5.5) in the diITt'n'ntial forn~ul~ y ell~~ation. Let us write Eq~.

ch . j' . or t liS purpose, we express theUlo'e q u.s p d V J I '

... . aUf tit' right··hand ~ide of Eq (5 4)" d J .__ JiJ ' . as --at P dl ~,-- -- P d I' J. 1ii'. "lere, we have taken the partial time derivative of

SineI' r can dpPl'llll 011 lim. , . P.C .IS wl'I1 as on coordinatp< TI

c '-0 HIS,

(r, j ciS - _ I' -.± (,Iii;Y J at .

120 5. Direct Current5.2. Ohm's Law for a Homogeneous Conductor

121

(5.10)

Differential Form of Ohl:!. 's Law. Let us establish a relation between the current density j and the field E at apoint of a conducting medium. We shall confine ourselvesto an isotropic conductor in which the directions of vectorsj and E coincide.

Le~ us me~tally iso!ate. around a certain point in a conductIng medIUm a cylIndrICal volume element with .generatrices parallel to vector j, and hence to vector E. If the cylinder has a cr?ss-sectional area dS and length dl, we can writeon the basIs of (5.8) and (5.9) the following expression forsuch cylindrical element:

'dS E dl] = pdlldS •

After cancelling out common terms we obtain the followingequation in vector form:

Ij~fE~.E, I

wh?re a = 1/p is the conductivity of the medium. The unitreClpro~al to an ohm is called a mho (Siemens). Consequently,the UOlt .for the measurement of a is mho per metre.

EquatIOn (5.10) is the differential form' of Ohm's lawIt does not .contai~ any diff~rentials (derivatives), but i~called the dIfferentIal form SInce it establishes a relationbetween quantities corresponding to the same point in aconductor. In other words, Eq. (5.10) is an expression forOhm's law at a point.

Methods for Calculating Resistance (R). There are severalmethods for measuring resistance and all of them are ultimately ba~ed on the application of relations (5.8)-(5.10).The expedIence of using a particular method is determinedby the formulation of the problem and by its symmetry.The practical applications of these methods are described inProblems 5.1-5.3 and 5.6.

.Charge in a Current-carrying Conductor. If the current isdlrect (constant) the excess charge inside a homogeneous~Qndu~tor is ~qual to zero everywhere. Indeed, Eq. (5.5)IS valId for dIrect current. Taking into account Ohm's law

in the form (5.10) we can rewrite (5.5) as follows:

l~ aEdS=O,u

where the integration is performed over any closed surfaceS inside the conductor. For a homogeneous conductor, thequantity a can be taken out of the integral:

a~ EdS=O. '

According to the Gauss theorem, the remaining i?te.gral isproportional to the algebraic sum of the charges mSlde theclosed surface S, Le. proportional tothe excess charge in this surface. How- Enever it can be directly seen from thelast' equation that this integral isequal to zero (since a =1= 0). This meansthat the excejEl charge is also equal to

. zero. Since the surface S is chosen arbitrarily, the excess char,ge 'under th~seconditions is always equal to zero m-side a conductor. ,

The excess charge can appear only Fi~. 5.1at the surface of a homogeneous con-ductor at places where it comes in C,I}lildlL with otherconductors, as well as in the regions wher8 the conductorhas inhomogeneities.

Electric Field of a Current-carryillgConductor. Thus,when current flows, an excess charge appears on the surf~ceof a conductor (inhomogeneity region). In accordance WIth(2.2), this means that the normal component of vector Eappears at the outer surface of the conductor .. Further, fromthe continuity of the tangential component ?f vector E,we conclude that a tangential component of thIS vector also.exists near the surface of the conductor. Thus, near the. surface of the conductor, vector E forms (in the ~resence ofthecurrent) a nonzero angle (J. with the vertical (FIg. 5.1). Whenthe current is absent a = O. . . .

If the current flows in a steady state, the dlstl'lbutlOnof electric eharges in the conducting medium (generallyspeaking, inhomogeneous) does not change in time, although

122 5. Direct Current 5.3. Generalized Oh!!'.m~·s~L~a~w~ -_-1_2~3-----------

the charges are in motion: at each point, new charges continuously replace the departing ones. These moving chargescreate a Coulomb field which is identieal to tbe one createdby stationary charges of the same configuration. This meansthat the electric fIeld of stationary currents is a potentialfield.

At the same time, electric fIeld in the case of stationarycurrents differs sig-nineantly from electrostatic, or Coulomb,field of fixed charges. If charges are in equilibrium, theelectrostatic fwld inside a conductor is always equal to zero.The electric field of stationary currents is also an electrostatic (Coulomb) field, although the charges inducing thisfield are in motion. Hence, the field E of steady-state currentsexists also inside a current-carrying conductor.

5.3. Generalized Ohm's Law

Exlrant'ous Forces. If all the forces acting all chargecurriers wcre reduced to electrostatic forces, the positivecarriers would move under the action of these forces fromregions with a higher potential to those with a lower potential, while the negative carriers would move in the oppositedirection. TIds would lead to the equalization of potentials,and the potentials of all mutually connected conductorswould become the same, thus terminating the current. Inother words, lmder the action of Coulomb forces alone, astationary licld must be a static field.

In order to prevent this, a direct CUl'l'cut circuit mustcontain, besides the subcircuits in whieh positive carriersmove in the direction of deereasing potential q;, subcircuitsill whieh positive carriers move ill the direction of increasing If' or against the electric field forces. In these subcircuits,the transfer of carriers is possible only through forces thatare not electrostatic. vVe shall call such forces extraneous.

Thus, a direct current can be sustained only witlt thehelp of extraneous forces whicu are applied either to certain parts of the cirellit, or to the entire eircuit. Thephysieal naturo of oxl I'illH'OliS forcos may be ql1ite diverse.Snell fOl'ecs may bo ('!"pated by physieal or chemical inhomo,gclloili(~s ill a ('olldllclor, rot· O\,llllplc, roreps l'esllltin,gfrom a cO/llact ol" rondllctors or different types (galvanic

cells, accumulators) or eondllctors at different temperatures(thermocouples). .

G I· d Ohm's Law In order to deSCrIbe extraneousIenera lze· f e fJ eld

forces quantitatively, the concepts of extrane?us orc. I .and its strength E* are introduced. The lat~er IS .n~mencallYequal to the extraneous foree acting on a umt posltl\:e C~I~~~

Let us now turn to current density. If an ~lec~r~ e itgenerates in a conductor a current of d.enslty l - oE, Eis obvious that under the combined actlOn of the, field '11and the extraneous force field E*, the current denSIty WI

be given by tj=o(E+E*). \ (;).11)

This equation generalizes the law ~5.10) to, the ~ase ~;inhomogeneolls regions of a conductlllg medI~lm, andcalled the generalized Ohm's law in the ~iffe~entIaI fOfl~l.

Ohm's La,,' for a Nonun iform Subcll'cml.. Nonumformsubcil'cuits are characterized by the presence of extraneOIlS

forces in them,' ,., t tLet us consider a special, but practICally qUIte.lmpor an

case when a current flows along a thin wire. In thIS ~ase, t~Ied'il'ec.tion of the current will coincide with the aXIs of t 10

wire and the current density j can be assum~d to be th~ sa~~at all points on the cross section of the WIre, Suppose th ..~he cross-sectional area of the wire is equal t~ 8 (8 need notbe the same over the entire length of the WIre): _1- .

Dividing- both sides of Eq. (5.1~) by ~, formIIlg a SC,l HI '.

prodnc.t of the resulting expresSIOn ":Ith the element. dlof the wire along its axis from erosS s~c.tIOn .~ to ~ross sectI?"2 (this direction is taken as the P?SItlve dIrectIOn) andr~~~tegrating over the length of the WIre between the two c c'

sections, we obtain2 2 2

\1:1=~Edl+~E*dI. (;).12)j 1 1

Let lIS transform the integrand in the, fl.rst integr,al: .we

re.p]oc(' (J hv iff) and J. ill by jidl, where JI IS the pl'OJectIOll... ~ . .' " . . II F·t! we IIOt('

of vee tor j onto the (hrcdlOII of vecLor {, III \Cr, .' . ~that jt is an algebraic quantity and depends OIl the Ollenta

124 5. Direct Current 5.3. Generalized Ohm's Law 125

(5.13)

Ra TIRi+Ra = 1+TI •

tion of vector j relative to dl: if j tt dl then j 1 > 0; if j # dI,then j 1 < O. Finally, we replace j 1 by liS, where I thecurrent, is also an algebraic quantity (like j I)' Since' I isthe same in all sections of the circuit when we are dealingwith direct current, this quantity can be taken out of theintegral. Cosequently, we get

2 2fjdl=If ~J (J .IPs·1 1

The expression P dllS defines the resistance of the subcircuit of length dl, while the integral of this expression isthe total resistance R of the circuit between cross sections1 and 2.

Let us now consider the right-hand side of (5.12). Here,'0 R the first integral is the poteo-

o---::.j~ . tial difference 'PI - 'Pz, while1 2 the second integral is the elec-

r£~tromotive force (e.mJ.) ~, act-

"'2 iog in the given subcircuit:~ 2

~t2 = ) E*dl. (5.14)121

Fig. 5.2 Like the current I, the electro-. . . .. motive force is an algebraic

quantity: If e.m.f. facIlItates the motion of positive carriersin a certain direction, then ifIll> 0; if, the e.m.f. hindersthis motion t"12 < O.

As a result of all the transformations described above(5.12) assumes the follOWing form: '

IRI='Pl-'P2+~t2' , (5.15)

This equation is the integral form of Ohm's law for a n01tuniform subcircuit [ef. Eq. (5.11) which describes the samelaw in the differential formJ. .

. Example. Co~sider the subcircuit shown in Fig. 5.2. The resistanceIS nonz?ro. only. III the .segment H. The lower part of the tigure showsthe variatIOn of potential !P over this region. Let us analyse the courseof events in this circuit. .

Since the potential decreases over the segment R fr0l!! .left t? rig~t,I > O. This means that the current flows in the posItIve dIrectIOn(from 1 to 2). In the present case! Ipl < .lp2' but th~ curre~t .f1ows f~ompoint 1 to point 2, Le. towards .Ific~easII~g P?te~tlal. ThIs IS p~ssIb~eonly becausp.an e.mJ. ,g is actIng Ifi this CIrcUIt from 1 to 2, I.e. mthe' positive direction.

Let us consider Eq. (5.15). It follows from thi~ eq.uat~on

that points 1 and 2 are identic.al for a .closed .cIrcUIt, I.e:«PI = «PI' In this case, the equatwn acqUIres a sImpler form.

RI = ~, (5.16)

where R is the total resistance of the closed cir~uit, and ~

is the algebraic sum of all the .e.m:f. 's in t?~ circuit. .Next we consider the subClrcUIt contallllllg the source

of the ~.mJ. between terminals 1 and 2. Then, R in (5.15)will be the internal resistance of the e.m.f. source in t.hesubcircuit under consideration, while 'PI - 'P2 the potentialdifference acro~ its terminals. If the source is disconnected,I '-= 0 and l= 'P2 - 'Pl' In other words, the e.m,f. ofthe source can be defmec! as the potential difference acrossits terminals in the open circuit.

The potential difference across the terminals of an e.m.f.source connected to an external resistance is always lessthan its e.m.f. and depends on the load.

Example. The external resistance of ~ circuit is TJ times higher th,anthe internal resistance of the source. Fmd the ratIO of the potentialdillerence across the terminals of the source to its e.m.f.

Let R be the internal resistance of the source and Ra the externalresistance' of the circuit. According to (5, i5), '!P2 - !PI = ,g - R,! I

while according to (5.16), (R j + Ra ) ! =,g. From these two equations, we get

1p2-Ip, = 1- Ri! = 1- RjIf r!J Rd-Ra

It follows from here that the higher the value of TI, the closer the ~o

tential difl'crence across the current terminals to its e,m.f., and VIceversa.

Concludin a this section, let us consider a picture illustrating the 'how of direct current in a closed circuit. Figure 5.:3 shows the distribution of potential along a closed

126 5. Direct Current

Fig-. 5.;>Fig. 5.4

as shown ill lhe figure. Applying Ohm's Jaw (;j.1S) 10 cachof the three suhcircuits. we obtain

11R I = qJ2-- cps +'f l ,

12Hz= cps - CfJI -t- 'f2 ,

I sRI. = qJl - IP2 + '(s·

III order to pl'OVl' Ihi~ law. il i~ ~ul'l'J('il'ul to eOllsidl'" theC,lse when lhe circuil cOllsists of three su!lcireuils (Fig. :i.:l)_LeI liS assunH' Ihe dirl'dioll of cil'CllmVl'lIlioll to he clockwise,

Adding these equations and cancelling all potentials, wearrive at formula (5.18), i.e. Kirchhoff's second law.

Thus, Eq. (5.18) is a consequence of Ohm's law for 1l011

lIniforIll subcireuils.Setting up of a System of Equations. In each specifIC case,

1\ irchhoff's laws lead 10 a complete syslem of alg-cbrait:equaliollswhich call he used, say, for l'ilHlilig all the llll

known currents in the circllit.The number of equations of the form (:'>.17) and (5.18)

Illust he equal to the Ilumber of unknown quan! ilies. Careshould be taken to ensure that Ilone of the equiltions is iI

corollary of allY olher equalioll ill lhe system:(1) if a branchell circuit has N jllnctiolls, illdejwllt!I'nt

equations of lype (5.17) can he set up only for N 1 JUIII'

lions, and the equation for lhe last jUliet ion will he iI r'ornllary of tile preceding equations;

(:2) if a D!'i111clied cirelli! containC' seve!'al ell'S!"! Inops.tlw indepc'nl[pill eqllil1ions o[ t~'pe C,.iS) Coli! 1)(' ~e! up onlyfol' loops which eal!lIo( Ill' "htainpd hy !'u!lL'rilililosillg"the 100Jls con:,idl'l'ed heron'_ 1"01' p\amplp, "111'11 r-ljll,l

I iO/lS will Lt, independl'lil for loops 7.'24 and 234 oj' the(5.18)

circuit containing the e.m.f. source on segment AB. Forthe sake of clarity of representation, potential <p is plotted

'P along the generatrices of a cylindri-A cal surface resting on the current con

tour. Points A and B correspond tothe positive and negative terminalsof the SOUice. It can be seen from thefigure that the flow of current can bedescribed as follows: the positivecharge carriers "slide" along the in-

F· 5 3 dined "chute" from point lOA to point19. . 't'

<P B along the external suhcircuitwhile inside the source, carriers "rise" from point qJ B to qJ ,with the help of extraneous force shown by an arrow. A

5.4. Branched Circuits. Kirchhotrs Laws

Calculations of branched circuits, for example, determination of current in individual branches, can be considerablysimplified by using the following two Kirchhoff's laws.

Kirchhoff's Fircit Law pertains to the junctions, Le.branch points in a circuit, and states that the algebraicsum of the currents meeting at a junction is equal to zero:

~Ik = O. (5.17)

~fere, currents converging at a junction and diverging fromIt are supposed to have opposite signs; for example, we canassu~e the forme~ t? ~e positi~e and .the latter. to be negative(or VIce versa, tillS IS ImmaterIal): When applIed to Fig. 5.4,Eq. (5.17) assumes the form II _.- 12 + 13 0-== O.

Equation (5.17) is a consequence of the steady-state condition (5.7); otherwise, the charge at ajunction would cllilngeand the currents would not be in a steady state.

Kirchhoff's Second Law. This law is applicahle to anyclosed contour in a brandied eirenit dnd states that thealgebraic sum of the products of cwnfit and resistance in eachpart of a network is equal tv the algebraic sum of e.m.f. 's inthe circuit: '

1295.5. loule'lJ Law

5.5. Joule's Law

The passage of current through a conductor having aresistance is invariably accompanied by liberation of heat(heating of conductors). Our task is to find the quanti~yof heat liberated per unit time in a certain part of the mr-

\I-OIM I

direction of circumvention, the:corresponding Iterm IR .in(5.t8) should be taken with the positive sign; in the OPPosl~ecase, the minus sign should be used. The sam~ pr?cedure ~sapplicable to lJ: if an e.m.f. increases po.tentIaI 1Il th~.dlrection of circumvention, it should be assIgned the pOSItivesign; the negative sign should be used in the opposite case.

Example. Find the ma~nit~de and d.irec~ion of the cur~ent passingthrough resistor R in the CIrCUIt shown m Fig. 5.8. All reSIstances ande.m.f.'s are assumed to be known. R

There are three subcircuits, and hence,three unknown currents I, I).. and 12 inthis circuit. We mark (arbitrarily) the supposed directions of these currents by ar-rows (at the right jU!lction). . . I

The circuit contams N = 2 JunctIOns.This means that there is only one inde- 12pendent equation of type (5.17):

I + II + 12 = O.

Let us now ,Met up equations of type lS2(5.18). According to the number of sub-circuits there should be two of them. Let F' 58us con;ider the loop co.taining Rand Ig. .R I and the loop w~th R an~ R 2 • Taki.ngthe clockwise directIOn for clrcumventmg each of these loops, wecan write

-IR + I 1R 1 = -1!!1> -IR -+ I 2R 2 = 1!!2'

We can verify that the corresponding equation fo.r the lo~p withRand R can be obtained from these two equatIOns. Solvmg the

I 2 • bt .system of three equatIOns, we 0 am

1- -R I I!!2+ R 21!!1- R I R 2+RR I +RR2'

If we fmd that as a result of substitution of numerical .valu~s intothis equation I > 0 the current actually flows as shown III Fig. 5.8.If I < 0, the cllrre~t flows in the opposite direction.

3

1

Fig. 5.7

r----~4

5. Direct Current

Fig. 5.6

128

circuit shown in Fig. j,G. Equation fot' tho loop 1234will follow from the two preceding olles. It is possible toset up autonomous equations for two other loops, say,124 and 1234. Theil the equation for contour 284 will be a

consequence of these two equations. The number of autonomous eql..\ations of type (5.18) will be equal to the smallestnumber of discontinuities which must be created in a circuit . in order to break all the loops. This number isequal to the number of subcircuits bounded by conductorsif the circuit can be drawn in a plane without intersections.

I:or example, it is necessary to set up three equations oftype (5.17) and three of type (5.18) for a circuit (Fig. 5.7)containing four junctions. This is so because the number ofdiscontinuities (marked by crosses in the circuit) breakingall the loops is three (the number of subcircuits 'is alsoequal to three in this case). If we assume the currents tobe unknown the number of discontinuities will be six inaccordance with the number of subcircuits between junctions,which corresponds to the number of independent equations.

The following procedure should be adopted while setting up equations of type (5.17) and (5.18).

1. The directions of currents are marked hypotheticallyby arrows, without 'caring for the direction of these arrows.

. If a certain current turns out to be positive as a result ofcalculations, this means that the direction chosen for it iscorrect. If, however, the current is found to be negative, itsactual direction will be opposite to the one pointed by thearrow.

2. Having selected an arbitrary dosed loop, we circumvent its sections in one direction, say l clockwise. Ifthe assumed direction for any current coincides with the

Iii

iii

'Ii

131

(5.20)

(5.19)

This formula expresses Joule's law in the differential form:the thermal power density of current at any point is p'rop~rtional to the square of the current density and to the res£stw£tyof the medium at that point. .

Equation (5.20) is the most. general. form, of l.oule's law,applicable to all conductors -IrrespectIve of thel~ sh~pe orhomogeneity, as well as the nature of .forces WhIC~ mducethe electric current. If the charge earners are subJected toelectric forces only, we can write, on the basis of Ohm's law(5.10),

Qd=j.E=aE2. (5.21)

This equation is of a less general n~tur~ than (~.20).Nonuniform Subcircuit. If a subClrcUIt contams a source

This is the expression for the well-known Joule's law.We shall now derive an expression for the differential

form of this law, eharacterizing the liberation of heat atdifferent parts of a conducting medium. For this purpose,we isolate a volume element of this medium in the form ofa cylinder with its generatrices parallel to vector j, viz.the current density at the given point. Let dS and .dllbe thecross-sectional area and length of the small cylmder respectively. The amount of~eat liberatedin .this volu~e durin~the time dt will be given, m accordance WIth Joule s law, by

fJQ =;,RI2 dt = ~~l (j dS)2 dt = pj2 dV dt,

where dV = dS dl is the volume of the small cylinder. Dividing the last equation' by dV dt, ':Ie .obta.in a fo7mula forthe amonnt of heat liberated per umt tIme m a umt volumeof the conducting medium, or the thermal power density ofthe currtmt:

Since in accordanee with Ohm's law CPI - lJl2 = RI,we get

5.5. Joule's Law------------_. ~~:.::.::::.::-.::...:.=~---------

5. Direct Current130._-------=:.:..=..::..:..::.:...::..::::..:.;:.:::..--------

cuit. We shall consider two cases which are possible, viz.uniform and nonuniform subcircuits. The problem is considered on the basis of the law of conservation of energy andOhm's law.

Uniform Subcircuit, Suppose that we are interested inthe region between cross sections 1 and 2 of a conductor

(Fig. 5.9). Let us find the work2 done, by the field during a time

interval dt in the region 12.If the cuuent through the con

ductor is equal to I, a chargedq = I dt will pass through each

. cross 'section of the conductor. FIg, 5.9 during the time dt. In particular,

this charge dq will enter crosss~ction 1 and the tsame charge will leave cross section 2.~lDce ~harge distribution in the conductor remains unchanged~n th~s case (the current is dire~t), the whole processIS eqUIvalent to a transfer of charge dq from section 1 tosection 2 with potentials CPI and CPI respectively.

Hence the work done by the field in such a charge transfer is

6A =dq (CPI - CPI) = I (CPI - CPa) dt.

According to the law of cons9l'vation of energy, an equivalent amount of energy must be liberated in another form.If the conductor is stationary and no chemical transformations take place in it, this energy must be liberated in theform of internal (thermal) energy.. Consequently, the conductor gets heated. The mechanism of this transformationis quite si~ple: as a result of the work done by the field,charge .c~rrlers .(fo~example, electrons in metals) acquirean ~ddItI.onal.kmetIcenergy which is then spent on excitinglattIce VIbratIOns due to collisions of the carriers with thelattice sites' atoms.

Thus, in accordance with the law of conservation of ener-gy, the elementary work 6A = 'Q dt, where Q is the heatliberated per unit time (thermal power). A comparison ofthis equation with the preceding one gives

In other words, the total]oule heat liberated per unit timein the entire circuit is equal to the power developed byextraneous forces only. This means that heat is generatedonly by extraneous forces. The role of the electric fieldis reduced to redistribution of this heat over different partsof the circuit.

We shall now derive Eq. (5.22) in the differential form.For this purpose, we multiply both sides of Eq. (5.11) by

(5.24)

.5,6. Transient Processes in a Capacitor Circllit----

5.6. Transient Processes in a Capacitor Circuit

Transient Processes. These are th~ processes i?vol.vinga transition from one stationary regIme of the CircUlt toanother. An example of such a process- I

is the charging and dischar~ing of.a 0capacitor, which will be conSIdered III

detail in this section. . C q± RSo far, we have considered only (h-

rect current#It turns out, h~wev~r, that , Kmost of the laws obtained lfl thIS caseare also applic~ble to .alterna~ing c~lr- Fig. 5.10rents. This applIes to aU cases lD Wlll.chthe variation ofeurrent is not too rapId. . )IThe 'instantaneous value of current will then practlca ybe the same at all cross sections of the eircuit. SUC~I cll~Tents

I e ld . ted' W'\·tli them are called quaslstatzonaryaIll l1e s aSSOCIa ' . . d(a more exact eriterion for quasi~tationary currents anfields is given in Sec. 11.1). . I • I ws

Quasistationary currents can be deSCribed. by t 10 a ,obtained 'for constant currents by applying these laws toinstantaneous values of quan~ities... . . r f a

Let us now caIlsidet' the dlschargmg and ChalgIIlg 0. h' t in these processescapacitor, assumwg- that t e curren s '

are quasistationan. I IDischarging of ~ Capacitor. If the plates of a c larg~(

c,apacitor wilh eapacitance C are connected t~lI'ougILa r~~Hstor R a current will flow through the resistOl'. et ,qand U be the instantaneous values ~f th~ current, ~hargeof the positive plate, ana the potential lhfference bet.weenthe plates (voltage). . ' .\ '

Assuming that the CIIlTent I is p~sitIve whel~. It df;)sfrom the positive plate to the negatIve plate (F Ig. . ,

d . I tl t ,." 1/p 311d pJ'2 ,,-~ Q"d (see (5.20».. an COIlSH or 13 lJ c.c-= ••~he thermal power density of curront 1Il an mllOffiogeneousmedium can then he written in the form

(5.23)Q='f:l.

5. Direct C', rent132

of c.m.f., the charge carriers will be subjected not only toelectric forces, but to extraneous forces as wel1. In accordancewith the law of cOllservatioD of energy, the amount of heatliberated in this case will be equal to the algebraic sum ofthe works done by the electric and extraneous forces. Thesame applies to the corresponding powers: the thermal powermust be equal to the algebraic sum of the powers due toelectric and extraneous forces. This can be easily verifiedby multiplying (5.15) by /:

R/2 = (<pt - 'P2) / + W,/. (5.22)

The left-hand side of this expression is the t,hermal power

Q liberated in the subcircuit. In the presence of extraneous

forces, the value of Qis determined by the same formula(5.19) which is applied to a uniform subcircuit. The lastterm on the right-hand side is th..! power generated by extraneous forces in the subcircuit uNIHf consideration. Itshould also be noted that the last H'1'ill (ll) is an algebraicquantity and, unlike R/2, it roverSt~;~ its sign wh!m the di-rection of the current / is revt)f<;ed. .

Thus, (5.22) indicates that tho heat liberated in the region between points 1 and 2 is equnl, to the algebraic sum ofthe electric and extraneous powers. The sum of these powers,Le. the right-hand side of this equation, is called the electricpower developed in the subcircuit. It can then be statedthat for a stationary subcircuit, the thermal power evolvedis equal to the electric power developed in the subcircuitby the current.

Applying (5.22) to the entire unbranched circuit ('PI == <P2), we get •

~~

_:i-"._

,-,'o',;,-

,

5.13.

(5.30)

oFig. 5.13

Fig. 5.12

U .ccco qlC, we can rewrite

1= ~~ = loe- tlt,

where 10 = 'f,1ll. f d I on t are shown in Fig.The depemlences 0 q an

Integrating this equation under __ 0 we obtaind't' q - 0 at t -- ,the initial COli 1 1011 -

RG In(1 - ;c )= - t,

whence _ (1 _e-t1t). . (5.29)q- qm ··t 'rh'\rge

_ ",C is the limitillg value of the capHl~1 or ' (Hero qm-_ RC .(for t -+ 00), HIIlI .~ =- 'tl' time according to the followlIlg

The current vallCS WI 1.

law:

o

•dq 8-q/C

dt"= R .

Separating the variables, we get

Rdq =dt.8-Q/C

Fig. 5.11

that I = dq/j.t and 'P2 - qJI =the last equ:ation in the form

C ·t· Circuit 1355.6. Transient Processes i.n a apacl or .~----

t tl e loop 1'f,R2, we get'. 0 1

RI = qJI - qJ2 + ~, .. . of the circuit, includlllg

where R is the total reslstalnce f source. Consideringthe internal resistance of t le e.m..

'i·~

~;~:"""

._' ,;i_';'i,

,

(5.27)

(5.25)

T = RG.

5. Direct Current ---------

dq Ifdt + RC ~oc O.

After separating the variables in this differential equationand integrating, we get

q =., qoe- Itt, (5.26)

where 10 = qo/T is the current at the instant t = O.Figure 5.11 shows the dependence of the capacitor charge

q on time t. The / vs. t dependence also has the same form.Charging of a Capacitor. Consider a circuit in which

a capacitor G, a resistor R and a source of e.m.f. '0 areconnected in series (Fig. 5.12). To begin with, the capacitoris not charged (key K is open). At the instant t == 0 thekey is closed, and a current starts to flow through the circuit,thus charging the capacitor. The increasing charge on thecapacitor plates will obstruct the passage of the cunentand gradually decrease it.

The ClllTent in the circllit will now be assllmed positiveif it flows toward the positive plate of the cnpacitor: I= dqldt. Applying Ohm's law for a non\luifonll sHbcircuit

we can write I = - dqldt. According to Ohm's law, weobtain the following expression for the external subcircuithaving resistance R: .

RI = U.

where go is the initial charge on the capacitor and T is aconstant having dimensions of time:

COJisidering that I = - dqldt and U = giG, we cantransform the above equation as follows:

134

This constant is called relaxation time. It can be seen from(5.26) that T is the time during which the capacitor chargedecreases to 1Ie of its initial value.

Differentiating (5.26) with respect to time, we obtainthe law of variation of current:

dq1= -(it=/oe- tlt , (5.28)

Problems

.• 5.~. Resistance of a conducting medium. A metallic s here of~adlUs a IS surrounded by a concentric thin metallic shell of l d' bfhe space bet,,:een the~ two ele~tr?des is filled with a homoae~~~u~poorly conductIng medIUm of reSIstIVity p. Find the resistanc: of thgap between the two electrodes. e

d ~olution. Let us ~nentally isolate a thin spherical layer betweenra 11. rand r t dr. Lmes of current at all points of this layer are rpe~dlcular to It, and therefore such a layer can be treated as a cy\'fndrIca! conductor of length dr and a cross-sectional area 4 . Z U'formula (5.9), we can write nr. SIng

drdR=p -4-nrZ

137

u(J~EndS '

eeo~ E" dSU

u

Problem'

qC=(f

conductors and C is the mutual capacitance of the conductors in thepresence of the medium.

Solution. Let us mentally supply the charges +q and -q to theconductors. Since the medium between them is poorly conduc.ting.the surfaces of the conductors are equipotential, and the neld configuration is the same as in the absence of the medium.

Let us surround, for example, the positively charged conductol' bya closed surface S directly adjoining the conductor and calc:ulat.e Rand C separately:

UR=T-

where the integrals are taken over the given surface S. While calcu.luting R, we used Ohm's law in the form j = (JE, and C was calculated with the help of the Gauss theorem.

Multiplying these expressions, we obtain

",' RC = eoe/a = eoEp.

• 5.4. Conditions on the bounda~y of a conductor. A conductorwith resistivity p borders OJ! a dielectric whose dielectric coustant ise. The electric induction at a certain point A at the surface of the conductor is equal to D, vectorD being directed away from the conductorand forming angle a with the normal to the surface. Find the surfacecharge density on th.e conductor and the current density near thepoint A.

Solution. The surface charge density on the conductor is given by

(J = Dn = D cos a.

The current density can be found with the help of Ohm's law: j == E/p. It follows from the continuity equation (5.5) that the normalcomponents of vector j are equal, and since in the dielectric ill = 0(there is no current), in the conductor we also have in = O. HeDt:e,vector j in the conductor is tangent to its surface. The same appliesto vector E inside the conductor.

On the other hand, it follows from the theorem on circulationof vector E that its tangential components on.difIerent sides of tb;einterface are equal, and hence E = E t = J) SIn alee., where E", 15the tangential component of vector E in the dielectric. Taking tJieBearguments into account, we obtain

E D sinaj =p eoep

• 5.5. The gap between the plates of a parallel-plate capacitor isfilled consecutively by two dielectric layers 1 and 2 with thicknesses11 and 12 , dielectric constants CI and E2' an~ rcsi8tiviti~s PI and f,,'The capacitor is under a permanent voltage (;, the electrIC field bemg

5. Direct Current136

Integrating this expression with respect to r between a and b, we get

R =..i!..- (1.._1.)4n: a b •

• 5.2. Two identical metallic balls of radius a are placed in ahOI.nogeneous boorly conducting medium with resistiVity p F'nd th~hSlsJ~~~e of bt e medium b~tween the balls under tbe conditi~n tba~

e IS "nce etween them IS much larger than their size.. Solution. Let us mentally impart charges +q and _ to th b II

Smce ~he ba}ls are at a large distance from one anothe/ elect~c fletdnhar t e

fsh" ace of each ball is practically determined only by the

c ~rge 0 t ~ n~arest sphere, and its charge can be considered to beuIlIformly dIstrIbuted over the surface. Surrounding the ositiv I~hargedball. by a concentric sphere adjoining directly the Call'~ s~lface, we WrIte the expression for the current through this sphere;

1= 4lta2j,

where j is the current density. Using Ohm's law (/' = Elp) and theformula E = q/4neoa2, we obtain

1= qleop.

Let us now flIld the potential difference between the balls:

U = lp + - lp_ ~ 2ql4nEoa.

The sought resistance is given by

R = UII = p/2na.

:thiStreSfUtlht is vadl~d regardless of the magnitude of the dielectric con-"an 0 erne mm .

. . ., 5.3. Two conrl~ctors of arbitrary shape are placed into an in~~dtet~~od?~e~e~~s slIghtly cond.ucting medium with the resistivity pt.hi' w ec~n: cons~~nt e. FJ!Id the value of the product RC for

Ii system. \It here R lli the rClillitunce of the medium between the

(1)

139

Fig. 5.14

Problem.

Am + As = ~W,

where Am is the mechanical work accomplished by extraneous forcesagainst electric forces, A s is the work of the voltage source in thisprocess, and ~W is the corresponding increment in the energy of thecapacitor (we assume that contributions of other forms of energy tothe change in the energy of the system is negligibly small).

Let us find ~Wand A s. It follows from the formula W = CU2/2 == qU!2 for the energy of a capacitor that for U = const

~W = ~C U2/2 = t!.q U/2. (2)

Since the capacitance of the capacitor decreases upon the removal ofthe plate (~C < 0), the charge of the capacitor also decr~ases (t'..q << 0). This means that the charge has passed through the source againstthe direction of the 1i.ction of extraneous forces, and the source hasdone negative work

taining resistors Rand Ro, we can write

(R + R o) I = It - ~o,

where the positive direction is chosen clockwise. On the other hand, forthe nonuniform subcircuit aRb of the circuit we have

As = !'"q'U,

The charge on plate 1 is determined by the formula ql = C (CPI - CPI)'Therefore, the final result is

RC...' ql= R+R

o(~-t8'o)'

It can be seen that ql > 0 f~r It > Ito and vice versa .• 5.8. The work of an e.mJ. source. A glass plate completely

fills the gap between the plates of a parallel-plate capacitor whosecapacitance is equal to C. when the plate is absent. The capacitor isconnected to a source of permanent voltage U. Find the mechanicalwork which must be done against electric forces for extracting theplate out of the capacitor.

Solution. According to the law of conservation of 'energy, we canwrite

RI = CPa - CPb + It,

-while lor the suhcircuit aCb we have

~ + CPI - CPI = CPb - CPa·

The joint solution of these equationsgives

5. Direct Currentt38

directed from layer 1 to 2 F' d hcharges at the interface b~tw::n tthe sd'!rlfacte. delnsity of extraneous

S I. . e Ie ec ric ayers.

o utJon. The required f hsur ace c arge density is given bya = Ds - D - 8 8 E En , In- 02 2-e08 1 l' (1)

In order to determine E and E h Isince h = ii' it follows lhat E 7' w~sEall make use .of two conditions:= U. Solving the two ea' I PI - 2 P2' and besides EIII + E 1 _values into (t), we obtii~ tIOns, we find Eland E 2 • SUbstituting lh~;

a= B2P2- eIPI UPIll + P212 eo •

Hence it follows that a = Of,or _, 8l PI - e2P2'

• 5.6. Nonhomogeneous c d t Icular cross section of area S .on DC or. A ong conductor of a cir-dc

hepe!1ds only on the distance ~ f~~ethf a dterial w~lOse resistivity

w re a is a constant Th d e cpn uctor aXIS as p = a/r2intensity E in the co~duct~rCon uctor came~ current I. Find (1) fieldof the conductor. and (2) the resistance of the unit length

Solution. (1) In accordance w'th Oh ' '.related to current density i wh·l l

.. ml s dlaw, field lDtensity E iscan write· ,I e I IS re ate to current I. Hence we

1= Jj21tr dr= J(Elp) 2nr dr.

The field intensity E is the sam II'th!! given conductor, Le. is inde at d pOI ts of the cross section of

:::.: :,t:;;e:.:~~~ :?:~f!e~l~. :!;~u~n:nO{h:' C':;d~~~re~~ilih:t~~~and then applying to this con\th, f~h exahmple, the conductor's axisvector E. our e t eorem on circulation of

Thus, E can be taken out f th .gration we obtain 0 e lDtegral, and as a result of inte-

E = 2naI182.

(2) The resistance of a u 't I hwHhthe help of the formul~1 R e~gt of t~e .c~nductor can be foundequation by length lof the sectio;;oF/{ DlvdldlDg both sides of thisR and voltage U, We find e con uctor, haVing resistance

R u = Ell = 2na182.

• 5.7. Ohm's law for no 'fshown in Fig. 5.14 the emf' nunl drm suhcircuit. In the circuitRaland .R" and the capacita~~esc~ot~ Iffo?{ sourckes, the resistancesn resistances of sources are ne l"blcapaci or are nown. The interplate 1 of the capacitor: g 19l Y small. Find the charge on

Solution. In accordance ~itli ali 'I fm s aw or,a closed circuit con-

141

(2)respect to

dE dtT-' RC'

6.1. Lorentz Force. Field B

R~=_1 1dt C'

Integrating the last equation, we obtain

lIt -IIRC Fig. 5.16

n 1 = RC' 1=100 , (3)o

where I. is deterqrincd by condition (2) for q = qo, Le. 10 = qolRC.Substituting (3) into (1) and intewating over time, we get

2Q=...!!!... (1_0- 21 / RC).

2C

6. Magnetic Field in a Vacuum

RI = q/C.Let. us differentiate (2) witht.ime:

6. t. Lorentz Force. Field B

Lorentz Force. Experiments show that the forc.e F actingon a point charge q generally depends not only on the position of this charge but also on its velocity v. Accordingly,the force F is decomposed into two components, viz. theelectric component Fe (which does not depend on the motionof the charge) and the magnetic component F m (whichdepends on the charge veloc.ity). The direction and magn itude of the magnetic force -at any point of space dependon the velocity v of the charge, this force being alwaysperpendicular to vector v. Besides, at any point, the magnetic force is perpend icnlar to the direction specifiCd at thispoint. Finally, the magnitude of this force is proportionalto the velocity component which is perpendicular to thi~

direction.

or

SoluUlIn. The required amount of heat is given byI

Q= JllP dt, (1)

" U

Irom which it follows that first of all we must find the time dPpendenceI (t). For this purpose, we shall use Ohm's law for the part JR2 of .t.he circuit (Fig. 5.16):

RI = fill - lJl2 = 1I,

:S. Direct Current140

Comparing formulas (3) and (2), we obtain

As = 2.1 W.

Suh~t.itution of this expression into (1) gives

Am = -.1W A 1or III = 2" (e - 1) Co1I2•

f Thus, extracting the plate out of th .o~ces) do a positivework(a<Ja't 1•.e capaCitor, we (extraneousthIs case accomplishes a n~g:d~~ e ('kLrlc fdrces) .. The e.mJ. source intor decreases: wor , an the energy of the capaci-

Am > 0, As < 0, .1 W < O.

• 5.9. Transient proccSse A . .s~lUrce of e.m.f. tE, and a re~is~~r CircUIt co,!sists of a permanentnes. The internal resistance 'of th R and. c~paclto~ ~ connected in sem?ment t = 0, the capacitance' e source Is.negliglbly small. At theWIse) decreased by a factor of 11 °t,~hd r-hPacltor w~s abruptly (jump-

functi~n ~f t·t e current III the circuit as a1 Ime.

1... . Solution. We write Ohm's law1 mhomogeneous part ltER2 f th fo~ t~e

& .!.l2 (Fig. 5.15); 0 e CircUit

T~ RI = <PI - <P2 - {! = 1I - €].

RConsidering that U = qlC' where C' -_= Cit], we obtain '

Fig. 5.15 RI = t]q/C _ Ji'. . '" . (1)

~Ve dd'Ierentillte this equation with r .III our case (q decreases) dq/dt = _l;espect to tune, considering that

R dE TJ dl([t=-C I , -r=T=- R~ dt.

Integration of this equation gives

ln~=-~ I 1 -Tjl/RC10 RC ' =. oe ,

where J 0 is determined by condition (1) I d d• 11 ee , we can wri teRIo = TJqo/C - e,

where go = ,ffC is the char e f h .has changed. Therefore, got e capacitor hefore its capacitance

10 = (11 - 1) ~/R.

~ .5.10. A charge qo was su lied t . ,.' .C, which at the moment t = lP hO a cal,antor with capacitancethe amount of heat liberated in J:eas s .unted hy resistor R. Hnd

resIstor as a function ot time I.

II1I

1'['

1

1

.1

II'I;

1

1'1

I'IIIII

143

(6.3)

B

Fig. 6.1

6.1. Lorentz Force. F£eld B

• Formula (6.3) is also valid in the case when the charge moveswith an acceleration, hut only at suffIciently small distances r fromthe charge (so small that the velocity v of the charge does not noti-ceably change during the time ric).

\

B=.fQ.. q[vx r ] \411: r3 ,

If.:

wltere ~o is the magnetic constant, the coefficient

~o/41t = '10-7 HIm,

and r is the radius vector from the point charge q to thepoint of observation. The tail of the radius vector r isfixed in a given frame of refe-rence, while its tip moves withvelocity v (Fig. 6.1). Consequently, vector B in the givenreference frame depends notonly on the position of thepoint of observation but ontime as well.

In accordance with formu-la (6.3), vector B is directednormally to the plane containing vectors v and r, the rota-tion around vector v in the direction of vector B forming

of the (inertial) reference system. On the other hand, themagnetic component of the Lorontz force varies upon a transition from one frame of reference to another (because of v).Therefore, the electric component qE must also change.Hence it follows that the decomposition of the total force F(the Lorentz force) into the electric and magnetic components depends on the choice of the reference system. Withoutspecifying the reference system, such a decomposition is

meaningless.Magnetic Field of a Uniformly Moving Charge. Experi-ments shoW that magnetic field .is generated by movingC1harges (currents). As a result of the generalization of experimental results, the elementary law defining the field Bof a point charge q moving at a constant nonrelativisticvelocity v was obtained. This law is written in the form*

(6.1)BI.Fm = q [v )(

6. Magnetic Field in a Vacuum142

All tltc!o'(' propcrtic' f .by introdll;;ill r the CO~IC~ magnetIc f~lI'e? can be describedthis field by v~ctOl" B wili~~ °II ;n~f:n.etl~ fteld. CI~al"acterizilJgat each pomt of sp'lce W(e el1ll\n~s the specIfic directionmagnetic force ,in 'tl:e f~rme can WrIte the expression for

Tl.wn the. total electromagnetic forcewIll be gIven by . acting on charge q

• Several methods of measurin fi ld B hthe fmal analysis, thev all a b g Ie ave been worked out. Indescribeu by Eq. (G.2). • re ased on the phenomena that can be

IF=qE+q Iv x RI· I (6.2)

It is called the Lorentz for E· .' . .it is valid for constant as c~~ll xpr~sslOn «(~.2) IS un.iversaI:magnctic fIelds for any velo 't as orf vlarYIn

gelectric and

Th. . CI Y V 0 t lC charge

e actIOn of the Lorentz fo .ciple be used for determinin tl~~e on a. charge can in prin-of vectors E and B. Hence gth magmt.udes and directionsforce can be considered a~ e expre~s~on for the Lorentzmagnetic fields (as was do t.he t~CfimtlOn of electric and

It should be emphasizednet}~~t m~;::7iCoffi:~c~ric field). *on an electric charge at rest I } . oes not actessentially differs from el~ct;i 1~\~eS~Cl; m~gnetic fieldonly on moving charges. c e. agnetIc fIeld acts

Vector B character'· tl f .fIeld on a movin ~zes lC oree actlllg due to magnetican analog of vecro~ ~r~~, and l~e~ce in this respect it isto electric field. aracteflZIng the force acting due

A distinctive feature of m t' f .perpendicular to the veioci:;n~eI~ ore; If that it is alwaysfore, no work is done over the c or 0 t. Ie charge. Thereenergy of a charged particle ch:zrge: TIllS means that thetic field alwa ,. . movmg In a permanent magne-motion of the y;ar~~~:.ws unchanged h'respective of the

In the no 1 l' . .(6.2), like a;~~e :tl~~~s~~~c::~[~:Si~~ii~~pe\~~~~r~~zcl~~~~:

145

(6.6)

6.2. The Biot-Savart Law

The Biot-Savart Law. Let us consider the problemof determining the magnetic field created by a direct electriccurrent. We shall solve this problem on the basis of law (6.3)determining the induction B of the fwld of it uniformlymoving point charge. We substitute into (6')~) the chargepdF for q (where dV is the volume element and p is thevolume charge density) and take into account that ill accord-

10-0181


The Principle of Superposition. Experiments show thatmagnetic fields, as well as eleetric fields, obey the principleof superposition: the magnetic field created by severalmoving charges or currents is equal to the vector sum of themagnetic fields created by each charge or current separately:

First, we have to deal with beams of particles movingat velocities close to the velocity of light, for which this"correction" and the electric force become comparable (itshould be noted that relation (6.5) is also valid for relativistic velocities).

Second, during the motion, say, of electrons, along wires,their directional velocity amounts to several tens of a millimeter per second at normal densities, while the ratio(v/C)2 ~ 10-24 • It is indeed a negligible correction to theelectric force! But as a matter of fact, in this case the magnetie force is practieally the only force since electric forcesdisappeared as a result of an almost ideal balance (muchmore perfect than 10-24) of negative and positive chargesin the wires. The participation of a vast number of chargesin creating current compensates for the smallness of thisterm.

In other words, the excess charges on the wires are negligibly small in"comparison with the total charge of carriers.For this reason, magnetic forces in this ease eonsiderablyexceed the electric forcgs acting on excess charges of thewires.

,f.

(6.5)

(6.4)B =, Eo!-to [v >< E]c= [v X Ellc\

the rig-ht-hallrlcll SystClll with vector v (Fig. li.1). It shouldbe nQtHd that vector B is Hxial (pseutlovector).

The quantity B is called maKfictic induction.Mag:nctic induction is measmetl in teslas (1').The electric field of a point charge q moving at a non

relativistic velo(~ity is described by the same law (1.2).Hence expression «U~) cnn be written in the form

Even at sufficiently high velocities, e.g. u = 300 km/s, this ratiois equal to 10-a, Le. the magnetic component of the force is a millionth fraction of the electric component and constitutes a negligiblecorrection to the electric force.

This example may give rise to the following question:Are snch forces wOI,th investigating? It turns ont that theyare, and theI'e are two sound reasons hehin(l this.

6. Magnetic Field in a VOC1l1Jm~------------

where c is t.he electrollynamic constant (c -~, 1/Veo!-to), equal

L-1,1 Z

Fig. 6.2

to the velocity of light in vacuum (this coincidence is notaccidental).

EKample. Comparison of forces of magnetic and electric interactionbetween moving charg~. Let two point charges q of a sufftciently largemass move in parallel to onc another with thc same nonrelativisticvelocity v as is shown in Fig. 6.2. Find the ratio betwecn the magneticFm and electric Fe forces acting, for example, from charge 1 Oilcli8rge 2.

According to (6.2), Fm = quB and Fe = qE where v is the velocityof charge 2 and Band E are the induction of the magnetic and theintensity of the electric fielns created by charge 1 at the point of location of charge 2.

The ratio Fm/Fe -= vB/E. According to (6.4), in our case B == IJEIc'l.. and hence


B

dBz -dB

dl A

I;t!r

z

I RdB,B

b A0

I

Fig. 6.3 Fig. 6.4

right. All curre~t deIlll'nts will form thl' cone of vectors dB, and it canbe easily seen that the resultant \"I'dol' ,B at point A will be directedupwards along the Z-axis. This II1l'allS lhat in order to fmd th~ magnitude of vcdOl" H, it is sul'ricil'lIl 10 SUIH up the compollPnts 01 vectorsdH along lhe Z-axis. Each such pl'ojl'c.liol1 has the form

flo I dldR. = dB cos f-l C~ -,- -2- cos ~,

4 '1:1 r

where we took into account that the angl~ between the element dland the radius ve/'lor I' is equal to n12, and hrllce the sine is equal toIInity. lnterrrat ing this px pression over dl (whkh gives 2rrR) and takinginto' accolI~t that {'os ~ ~- nl,. and r ~ (z~ : R2)1/2, we get

Ilo 2nR21B=.Im (Z2-, R2)8;2 • (ti.12)

Hence it follows that the rnagnitudp or v('etor B at the centre of HIPenrrpnt ring (z = 0) and at a distance z » R is given by

..l':!'... 2JlI B ~ ~~ (' 13)B z ,~'o 1m R z?/H 4n z3 0 .•

'It is clear from the ligure that dl cos a = r da and r = b/cos a. Hence

'B--~ I clIsada/ -- I.n b

Integrating this expression over all current elem~nts, which ~s equiva-lent to the integration over a between -n/2 and n/2, we lind -

B=~.Y:!- (IU1)4n b

Example 2. Magnetic field at the axis of circular current. Figure 6.4 shows vector dB Ihlln the current element I dl located to the

(6.8)j dV = I dt.

6, Magnetic Field in a Vacuum146

anre with (5.2) pv = j. Then formula (6.3) becomes

IdB ~ {;;- Ii X,:I'1' .1 (6.7)

If the current I flows along a thin wire with the crosssectional area t!S, we have

j dV = j t!S dl = I dl,

where dl is an element of the length of the wire. Intl'Oducing vedor dt ill the direction of the current I, we canwrite this expression as follows:

, Vectors j dV and I dt are called volume and linear currentelemen is resj)ecti vely. Replacing in formula (6.7) the volume current element by the linear one, we obtain

I dB=* I[~~xrl. I (6.9)

Formulas (6.7) and (6.9) express the Biot-Sauart law.In accordance with the principle of superposition the

total field B is found as a result of integration of Eq. (6.7)or (6.9) over all current elements:

Bc=_,fA:.lL \' [jXrldV B=~~ I[dlxr] (6.10)4n. r8 ' 411: 'j' r 3 •

Generally, 'calculation of the ;.nagnetic induction ofa current of ,111 arbitrary configuration by these formulasis complicated. However, the calculations can be considerably simplifIed if CllI'l'ent distribution has a certain symmetry. Let us consider several simple examples of drtel'min i Ilg' the ma~netic ind uctio 11 of CllI'rent. '

EXaJpple f. Magnetic field or the line current. i.e. the c\Il"rentflowing along a thin straight wire of infinite length (Fig. 6.3).

In accordance with (u.9), vectors dB from all current rlements haveat a .poillt A the ,same direction, :,iz. are directed behind the plane ofthe hgure. There!orp, lhe summatlOlI of vectors dB can hp r(!placed bythe summation of their magnitudes dB, where

dB=c~ I dl cos a4n r 2

10*

(f).Hi)

(6.15)

/2" 0

Fig. 6.5

I = \ j dS.

6.3. Basic Laws 0/ Magnetic Field

where I ~I,,, and I,. are algebraic 4I1alltitie~. The CUI'

rent is assumed positive if its direction is eOllllecled withthe direction of the circumven-tion of the contoui· through theright-hand serew rille. The current having the opposite directiOIl is considerc9 to be negat ive.This rule is illustrated. in Fig.6.5: here currents I] and {a arepositive sinctl their directions are'connected with the direction of /1> 0cont.Olir circumvention throughthe right-hand screw rule, while .current I 2 is negative.

The theorem on circulation (f).i5) can he proved Oil thebasis of the Biot-Savart Jaw. In the general case of arbitrary currents, the proof is rather cumbersome and ~il1 notbe considered here. We shall treat statement (h.i5) asa postulate verified experimentally. .' _ .

Let us make one more remark. If CIlrrent I 1I1 (b.L)) IS

distl'ibllted over the volume where contour r is located,it can l,e represented in the form

In this expression, the integral is taken over an arbitrarysurface S stretched on contour r. Current density j inthe integrand corresponds to the point where the areaelement. dS is located, vector dS forming the right-handedsystem with the direetion of the contour circumventioll.

Thus, in the general case Eq. (6.15) can be written a~

terminate. In other words, magnetic field has no sources.in contrast to electric fwld. .

Theorem Oft Circulation of Vector B (for till' magnellcfield of a direct. current in a vacllum). Circulation oj uector Haround an arbitrary contour r is equal to the product oj ~~oby the algehraic sum oj the currel/ts enl'eZoped by the contour 1 :

(6.14)

6. Magnetic Field in a· Vacuum148

This theorem is essentially a generalization of experience.It expresses in the form of a postulate the experimentalresult that the lines of vector B have neither beginningnor end .. Therefore, the number of lines of vector B, emerging from any volume bounded by a closed surface S, isalways equal to the number of lines entering this volume.

Hence follows an important corollary which will be repeatedly used below: the flux of B through a closed surface Sbounded by a certain contour does not depend on the shapeoj the surface S. This can be easily grasped with the helpof the concept of the lines of vector B. Since· these linesare not discontinued anywhere, their number through thesurface S bounded by a given contour (i.e. the flux of B)indeed must be independent of the shape of the surface S.

Law (6.14) also expresses the fact that there are no magnetic charges in nature, on which the lines of vector B begin

6.3. Basic Laws of Magnetic Field

Like electric field, magnetic field has two very importantproperties. These properties, which are also related to theflux and circulation of a vector field, express the basic lawsof magnetic field.

Before analysing these laws, we should consider tIlegraphic representation of field B. Just as any vector fieldfi.eld B can be visually represented with the help of th~h~es of vector B. They are drawn in a conventional way,VIZ. so that the tangent to these lines at any Doint coincideswith the direction of vector B, and the ~density of thelin.es is proportional to the magnitude of vector B at' a givenpOlDt.

The geometrical pattern obtaiped in this way makes itpossibl~ to easily judge about the configuration ofa givenmagnetic field and considerably simplifies the analysis ofsome situations.. Let us now consider the basic laws of magnetic field,I.e. the Gauss theorem and the theorem on circulation.

The Gauss Theorem for Field B. The flux of B throughany closed surface is equal to zero: .

Iii,

Iii,'

(6.19)

151

ra

Fig. 6.7

B

6.4. Theorem on Circulation of,Vector B

--......./' "

1/ r

{f,\

\" /........ ..-...._/

.fitFig. 6.6

. d !Sf vector B must be the same at all pointswire axis. The magm~u e" f the wire Therefore in accordance withat a distance r from.t e ax!s 0 . 'B for a ~ircular contour r lthe theorem on nrBcu2latlO~ of/e~~:nce it follows that outside the(Fig. 6.6), we have . nr -- flO ,

wire (6.18)B = (flo/2n) Ilr (r> a).

It should be noted that a direct solution of this pr~~I.emt(d'ith the helpof the Biot-Savart law) turns out ~I bet:~:ei~~~iro~:th;t inside the

From· the same symmetry cons". era I d' to the theorem' on\,:ire the. lines of vec~ornBf~~e :Is~i;~~l~~s. ~oC~t~l:nh (see Fig. 6"6~,circulatIOn of vcchto I = l ( la)2 is the current enveloped by the gl-B '2nr = flo l , were r r "d th .'ven contour. r Whence we I'mrl that illSI e e wIre

B ~.~ (~lo/2n) lr/a2 (r <. a)

d B ( ~ is shown in Fig. 6.7. "d 't'The dep~n ence r f tube the induction B outSI e I ~sIf the wire has the s ape 0 a. .' 'de the tube magnetic field IS

determined by formula (6.1~1)' Wl~111e mS-\h the help of the theorem onabsent. This can also be easl y s own WIcirculation of vector ~. f' Id of a solenoid. Let current I flow al?naa

Example 2: Magne Ie Ie h surface of a cylinder, Such cyhn ~rconductor hehcally wound o~ tell d a solenoid Suppose that a umtstreamlined by the ~urrent !S ca e f h c~nductor. If the pitchlength of the solenOid contams n turns 0 f'the solenoid can be approxof the helix is sufficientl

ly smdall, each~r~hall also assume that the con-

imately replaced by a c ose oop. ,

. 1 t'on of vector B and then ~ee whcthc!' t II isrem Oil Circu a I

method is universal.. fi Id' f·a strai~ht current. Direct current 1

• Example t •. Ma~netlc Ie 0; ht ~'irp wi th a circular cross sedionflows along a~ mftmtely long. st~all 'nduction B outside and inside theof radius a. Fmd the magnetIc lie ( Iwire. I h I' ft of the probl~m t lat t e Illes 0

It follows ~rom the symhll1e fry f 'ircles with the centre at the

vector B in thiS case have t e orm 0 c


6.4. Applications of the Theoremon Circulation of Vector B

Let us consider several practically i~po:tant examplesillustrating' the effectiveness of the applIcatIOn of the theQ-

150

~Bdl=~IO f jdS=f.to.\ indS. (6.17)

TIIC fact that cil'c,dati ;,1 of vector B gene.r'llly differsfrom zero illdicates thalii! contr<lst to electrostatic field,field B is not a potential field. SUcll fields are calledvortex. 01' solenoidal fiel!!",;,, Since circulation of vector B is proportional to the cur

rent 1 enveloped by the contom, in general we cannotascribe to magnetic field a scalar potential wllich would berelated to vector B through an expression similar to E == -Vip. This potential would not be single-valued: uponeach circumvention of the current and return to the initialpoint, this potential would acquire an incl'ement equal to""0/. However, magnetic potential lflm can be introduced andeffectively used in the regi-on of space whel'e the currentsare absent. .

The Role of the Theorem on Circulation of Vector B.This theorem p.Jays aJinost the same role as the Gausstheorem for vectol's E and D. It is well known that field Bis detel'milled by all CUlTellts, while circulation of vector B,only by the currents enveloped by the given contour. Inspite of this, in certain cases (in the presence of a specialsymmetry) the theorem on circulation proves to be quiteeffective since it allows us to determine B in a very simpleway. K

This can be done in the cases when the calculation ofcirculatiou of ve('{or B call be reduced, by an appropriatechoice of the contour, to the product of B (or B I) by theleugth of the contoul' 01' its part. Othel'\vise, field B mw;tbe calculated hy some other methods, for example. withthe help or the Diot-Savar! law or by solving the correspondiug differeutial equations, and the caJculatiu/I be('.omes much mOl"e d iffienlt.

follows:

153

(6.22)

Fig. 6.10

B

6.4. Theorem on Circulation of Vector B

to infmity (nt a constant toroid cross section), we obtain in the limitthe expression ((i.20) for the magnetic field of an infmitely long so-lenoid. .

If the chosen circular contour passes outside the toroid, it doesnot envelop any currents, and hence B ·2J1r = 0 for such a contonr.This means that outside the toroid magnetic field is absent.

In the above cOJl!'ic1erations it was assumed that the lines of current lie iUlIleridional plaues, i.l'. the planes passing "through the ·,axis00' of the toroid. In'real toroids, the lines of cur-rent (turns) rlo not. lie' strictly in these planes, andhence therl' is a current component arounrlthe 00' axis. This component creates an additional field similar to the field of a circular current.

Example 4. Magnetie field of a current-car·r~'ingplane. Let us consider an infmite conducting plane with unc1irectional current uniformlydistribute!1 over its surface. Figure 6.10 showst.he cross section of such a plane when the current flows behind the plane of the figure (thisis marked BV crosses). Let us introduce the con-cept of lille~r clI!tTent density i directed along thelines of current. The malPlitude of this vectoris the current per unit length which plays therole of the "cToss-sectional <1l'ea". •

l\lentally dividing the current-carrying plane into. thin ~urrenttilament::;, we can easily see that the resnltant field B wlll be threctedparallel to the plane, downwards to the right of the plan.e and up~ardsto the left of it (Fif.{, 6.10) .. These directions can be easily estabhshedwith the help of thl' right-hand screw rule.

In orc1l'r to determine induetinn B of the field, we shall use thetheorem on circulation of vector n. Knowing- the arrangement of thelines of vector B, we shall choose the contour in the form o! rectaJ;gle1234 (Fig. 6.10). Then, in accordance with the theor.em on CIrculatIon,2Bl ~, !Joil, where I is the lenl!th of the contour SIde parallel to theconducting plane. This g-ives

B 1 .=~ 2~tOI.

This formula shows that the magnetic field is uniform on both sidesof the plane. This result is also valid for a bounded current-carryingplane, but only for the points lying uear the plane and f~r from itsends.

General Consideralicms. The results obtained in the aboveexamples could be found 'directly with the help of theBiot-Savart law. However, the theorem on circulationmakes it possible to obtain "these results much more simplyand quickly.

However, the simplicity with which the field was calculated in these examples must not produce an erroneous

Fig. tUJ


Fig. G.S

: ....lLw•..........................

152

B = ~lonI, (6.20)

i.e. the. Iield il.ls!d~ a long s()lenoi~1 is uniform (With the exception of~he regIOns. ildJOIll!lIg the solenoid s endfaces, which is often ignoredIII caleul~tIOns). 11le product liT is called the Ilumber 0/ ampere-turns.For !/ ~ 20()(I turns/m and 1 = 2 A, the magnetic field inside the solenOId IS equal to 5 mT.

Exnmple 3. l\fa~netie field of a torOid. A toroid is a wire woundarollnd a torus (hg. 6.9). .

. It can he easily see,n from symmetry considerations that the Jinesof v?ctor B mus~ be CIrcles whose centres lie on the axis 00' of thet?rold. Hence It IS c1t'ar that for the contour we must take one of SUcllCIrcles.

.If a contour !iI'S in~ide I,h(~ toroid, it envelops the ('urrent N1 wher!'!V IS th~ number of turns in tIle toroidal coil and 1 is the ~urrentIII t~e wIre: Let the contour radius be r; then, according to the theoremon clrculatlOn B '2:rrr 1'01\'1, whence it follow!' that inside the toroid

B = (/to/2n) N /h.. (6.21)

II compa.rison. ofyi.21) alltl (6.1H) shows that the lIIaglll'tie licit! illsidptil(' torOJ.d cOJflCldes with the magnetic lield fIT J of tl](' straight i'lll'

rent flowmg along the 00' axis. Tellding N and radius H uf the toroid

ductor cross-sectiollal "rea is so small that the CUITent in tIll' solenoidcan be considered flowing over its surface.h 1xperiments and (,;lleulations show that the longer thl' solenoid

t ~ ess the n!aWll'tic induction outside it. For an inlinitelv long sole~nOId, ~agnetlc field ontside it is absent at all. . . *: .It IS clear from symmetry considerations that the lines of Vl'ctor Bl~slde the solenoid are. directed along its axis, vector B forming the1'Jght-handed system \nth the direction of the cnrrent in the soleIloid.

'. Even the aho~'e consideratioIls abollt the magnetic fwld configuratIOn of the. sole,nOld indicate that we must choose a rectangular contouras shown In FIg. (i.H. Circulation of vector n around this contour isequal to BI, and the l'l!ntour envelops the current nlI. According t~1the theorem or~ clrculatlOlI, BI "'" ~lonlI, whence it foHows that insidea long solenuld

I,- .....--,.......•..........~ .:! I:

-: L 4_ :oJ

6.5. Differential Forms of Basic lawsof Magnetic Field

Divergence of field B. The differential form of the Gauss theorem(6.14) for field B can be written as

1556.6. Ampere"s Force

\ 'V X E=O. \ (fj.~7). I t zero everywhere is a potentIal

A vector field whose curl IS equa?d l field Consequently, eleetroIield. Otherwise it is ~aned the,~?f:n~~ganet'c field is a solenoidal J1~ld.static field is a potential field, II: I .

or (V X Bl" = l.l.oill' Hence

. \-'V-X-B-=-Il-oj-~ \ (fj.~6)

. . of the theorem on ~irculatlOn

~f~~~o~r~~i~t;i~~l:~~~ded~~~~J~F:~i~~~i~,c~~fl~~h::;:i~~d~tor J' viz. the current ensl y .

vec. 1 t Ii j E . al to zeroof V X B is equa ? rfio id circulation of vector IS equ •

In an electrostatIc e ,and hence

6.6. Am~re's Force, h 1 I' e carrier experiences the

Ampere's Law.. Eac c~~egaction of this force is tr~llS-action of a magnetIc forc~. h which charges are lIlovmg.[nitted to a conductor t roug

where Oil the right-hand side we have the projection of curl H onto

normal n. . . f Id B can be characterized by curl BlIence, ea~h POlllt of vedor I~ I' determined by the propertie.s of

whose directlOll and. IIwglll~u~e1~1 C. I'rection of curl B is detl'rllllllClIthe field !tself at a given 1}010 t' tl:

l. ~~rface element S, for whi?h t.he

by thtetl~Il('~c~z)n ~hf~l~md~ti~esothee J;laguitude of curl H, attainS Itsquail I) . ,

maximum value: .1 B .~ obtained in coordinate represen~-In mathematiCs, cur k therfact is more importallt: It

tion. However" for our purpos~sa:b~considered as the cross pl'Oductturns out that formally curl BB' 'tV X B. Wc shall he using theof the operator 'tV ~y vecto~ r 'n I~~~ce it allows us to represcllt thelatter, morc c~nven~nt. ~htth~Ohelp of the determinant:

cross product . X 'tV\: B = \ a~~x a~Zy iJ~~z \ ' (Ii .25)

B x By Bz .• r' the unit vectors of Cartesian coordinates. ThiS

where ~x' ell' eZj.j~. th curl of not only field B, but of ,IllY otherexpressIOn I~ va I 01'.1' f f 11 Evector field a./rIso, iI!-IP:lrtthIC~:::;~r~lI1 ~~ ~irc~latioll of vector B. Accorlt-

Let us now CoUSH 1'1 e I' tl f I: . (u 24) I.'q (6 17) call bc representee ill Ie ornillg to . ,c. .

, j H dt .lim ---c;- = f!oJr\'s-u .

(6.Y.)

(6.23)

6. Magnetic Field in II Vacuum154

Le. the divergence of field B is equal to zero everywhere. As was mentioned above, this means that magnetic field has no sources (magneticcharges). Magnetic field is generated by electric currents and not bymagnetic charges which do not exist in nature.

Law (6.23) is of fundamental nature: it is valid not only for constant but for varying fields as well.

Curl of field B. The important property of magnetic field expressedbr the theorem on circulation of vector B motivates the representationo this theorem in the differential form which broadens its potentialities as a tool for investigations and calculations.

For this purpose, we consider the ratio of circulation of vector Bto the area S bounded by a contour. It turns out that this ratio tendsto a certain limit as S -.. O. This limit depends on the contour orientation at a given point of space. The contour orientation is specified byvector n (the normal to the plane of the contour around which thecirculation is calculated) the direction of n being connected with tltedirection of contour circumvention through the right-hand screw rule.

The limit obtained as a result of this operation is a scalar quantitywhich behaves as the projection of a certain vector onto the directionof the normal n to the planp, of the contour around which the circulation is taken. This vector is caller! the curl of field B and is denoted ascurl B. Thus, we can write

. I BdlIHn'-'--=(curl H)n,8-0 S

impression about the potentialities of the method basedon the application of the theorem on circulation. Just asin the case of the Gauss theorem for electric field, the number of problems that can be easily solved by using the theorem on circulation of vector B is quite limited. It is sufficient to say that even for such a symmetric current configuration as the current ring, the theorem on circulation becomes helpless. D88pitean apparently high symmetry, themagnetic field configuration does not allow us, however,to find a simple contour required for calculations, and theproblem has to be solved by other, much more cumber-some, methods. .

Fu=~ 2/112

4rt b (6.30)

Naturally, the same expression . b. b' .per unit leDgth of the d car~ e 0 tallied lor the force actin/.(

con uctor With current fl'

~:s : resul~, magnetic field acts with a certain force on a cur-~ -carryIng conductor. Let us find this force

th~~;~~i:~t~~t the volume density of charges ~hich areto .)'( W ~ cliIre~t (e.g. electrons in a metal) is equalcon~~lctore m1tenta tY ' Isolate a volume element dV in the

, . . con aInS the charcre ( t .to pdF. The tl P. f .. >' «,., curren car:ner) equalf 'h ,n ,L orce actlllg on the volume element dV

o t. e conductor can be' written by formula (6.1) as follows:

dF = p [u X Bl dV.

Since i = pu, we have

IdF = [j X B) dv·1 (6.28)

If. the current flows along a thincordmg to (6.8), j dV = I dl, and conductor, then, ac-

IdF = I fdl X BI, I (6.29)

where dJ is. the vector coinciding in direction with thecUITdent and characterizing an element of length of the thOcon uctor. , 'In

,. F~rmtdas (6.28) a~d (6.29) express Ampere's law. 1nte~~af!ng th)ese expreSSIOns over the curren. t elements (vol ume

meal', we can find the rna t' ftain vo] ume ~f d gne I.C o~'ee acting on a cer-

Th fa. can uctor or on Its lmear part.called

eA(),rce~. ~ctfmg on currents in a magnetic field are

npel e s orces.

Example. The force of 'nt t· b ".Find the Arrwere's force with wh.ah l?~ .etw~en parallel currents.?urrents /1 and I. interact in IC LW? JllfillJ~ely long- wires withIS b. Calculate the force per ~n~tcluumt'hlfrtheh distance between them

eng 0 t e system.. Each current element of I is in th . ..I.e. aeeording to (6.19), the 2field B e~agn~t1C field of ~he current 1 I,

t.ween the element of current / . d I - (flI/4:r~ 2f1/b. fhe angle befrom fonnula (6.29) that the io~~e v:~.tor B1 IS 1(/2. Hence, it followsductor with current I I'S F _ laB> mg per unIt length of the con-

2 u -- 2 It or

1576.6. Amp~re's Force

Finally, it can he easily seen that currents of the samedirection are attracted, while the opposite currents repelleach other. Here we speak only about a magnetic force.It should be recalled, however, that besides magnetic forcesthere are electric forces, i.e. the forces due to excess chargeson the surface of conductors. Consequently, if we speak aho~t

the total force of interaction between the wires, it can eitherbe attractive or repulsive, depending on the ratio betweenthe magnetic and electric components of the total force(see Problem 6.7). .

The Force Acting on a Current Loop. The resultant Ampere's force acting on a current loop in a magnetic field isdermed, in accordance with (6.29), by

F = I ~ rdl X B], (H.31)

where the integration is performed along the given loopwith current [.

If the m~netic field is uniform, vector Bean be takenout of the integral, and the problem is reduced to the calcu-

lation of the integral ~~ dl. This integral is a dosed chain

of elementary vectorsdl and hence is equal to zero. Consequently, F = 0, i.e. the resultant Ampere's force in a uniform magnetic fIeld is equal to zero.

If, however, the magnetic lield is nonuniform, the resultant force (6.31) generally differs from zero and ill eachconcrete case is determined with the help of (6.31). Themost interesting case for further analysis is that of a planecurrent loop of sufficiently small size. Such a current loopis called elementary.

The behaviour of an elementary current loop can beconveniently described with the help of magnetic llwrnent Pin'By delin ition, magnetic moment Pm is given by

Pm = ISn, (6.:32)

where [ is tJw cUlTent, S is the areil hounded hy lhe loopand n is the normal to the loop whose direction is eonnectedwith the direction of the current in .tho !f;op through therig-lit-hand screw rule (Fig. 1).11). In terms of magnetic lipid,the clemenU;' \' loop is eompletely charaeteT'iz,~d by itsmagnl.'lic moment Pm'


1\n j~volved calculation with the help of formula (6.31),taklIlg 1I1to accoullt the smaJl size of the loop, y ields th~

foHowing expression for the force acting Oil all elementarycnrrent loop in a nOlluniform magnetic field: '

IF= Pm"* ' (6.33)

where Pm is the magnitude of the magnetic moment of theloop, and oBion is the derivative of vector B along the

10 0 F~~Pm

direction of normal n or along the direction of vector p .This formula is similar to expression (1.39) for the for~eacting on an electric dipole in an electric field.

It follows from forml.da (6.33) that, like in the case ofelectric dipole,

(1) in a uniform: magnetic field F = 0 since oBian = 0;(2) the direction of vector F g-enerally does not coincide

with vector B or with vector Pm; vector F.coincides indirection only with the elementary increment of vector Btaken in the direction of vector Pm at the locus of the loop.This is illustrated by Fig. 6.12, where three arrangementsof the loop in the magnetic field of straight current Toare shown. The vector of the resultant force F actillg on theloop in each case is abo shown in the flgllre (it is useflllto verify independently that it is really so).

If we are interested in the projection of force ..~ ontoa certain dit'ectioll X, it is sufficient to write expression

159

(6.34)

Fig. 6.13

iJBxFx= Pm an-

6.7. Torque Acting on a Current Loop

6.7. Torque Acting on a Current Loop

Let us consider a plane loop with a current I in a uniformmagnetic field B. It was shown above (see p. 157) th?tthe resultant force (5.31) acting on the current loop In

a uniform magnetic field is equal to zero. It is known frommechanics that if the resultant of the forces acting on anysystem is equal to zero, the resultant moment of these for?esdoes not depend on the position of point 0 relativ~ to wInchthe tot'ques are determined. Therefore, we can speak aboutthe resultant moment of Ampere's forces in our case.

ny deflllitinll, the resultant moment of Ampere's forcesis gin~n by

M = (~ !r:< dFl, (6.35)

where dF is defined by formula (5.29). If we perform calculation by formula (6.35) (it is cumbersome and hardlyinteresting, hence we omit it here), it turns out that the

where iJlJxiiJn is the derivative of 'the corresponding projection of vector B again with respect to normal n (or with respectto Pm) to the loop.

Example. An elementary current loophaving a magnetic moment Pm is arranged perpendicularly to the symmetry.axis of a nonuniform magnetic field, vector Pm being directed along vector B. Letus choose the positive direction of theX-axis as is shown in Fig. 6.13. Sincethe increment of projection Bx will benegative in the direction of vector Pm.F < O. Hence .yector F is directed totlfe left, i.e. to .fli.e side where B is greater. If we rotate the current loop (and vector Pm) through 90° so that the loop centre -coincides with the symmetry axil\- of field B, in this position Fx = 0, aIljivector F is directed perpendicularly to the X-axis and to the same'side as vector Pm·

(0.3:1) in terms of the projections onto this direction. Thisgives

10 €lfolo'---~O Pill

F=O

Fig. 6.12

10 G).----~ Pm

F~


Fig. 6.11

b 1

iSS

161

F@

B,n

I

Fig. 6.15

(6,37)6A = Id<1J,

6.8. Work Done upon Displacement of Current Loop

where d<D is the increment of themagnetic flux through the loopupon the given displacement.

We shall prove this formulain three stages.

1. Let us first consider a particular case: a circuit (Fig. 6.15)with a movable jumper of length l is in a uniform magneticfield perpendicular to the circuit plane and directed behindthe plane of the figure. In accordance with (6.29), the jumperis acted upon by Ampere's force F c;= IlB. Displacing thejumper to the right by dx, this force accomplishes positive

6.8. Work Done upon'Displacement of Current Loop

When a current ioop is in an external magnetic field (weshall assume that this field is constant), certain elementsof the loop are acted upon by Ampere's forces which hencewill do work during the displacement of the loop. We shallshow here that the work done by Ampere's forces upon anelementary displacement of theloop with current I is given by

11-0181

Considering that ab is the area bounded by the loop and Iba = Pm'we obtain

M = PmB sin a,which in the vector form is written as (6.36).

Concluding this section, we note that expression (6.36)is valid for nonuniform magnetic fields as well. The onlyrequirement is that the size of the current loop must be

.sufftciently small. Then the effect of nonuniformity on therotational moment (torque) M can be ignored. This precisely applies to the elementary current loop.

An elementary current loop behaves in a nonuniformmagneti(~ field in the same way as an electric ~dipole inan external nonuniform electric field: it will be rotatedtowards the position of stable equilibrium (for whichPm tt B) and, besides, under the action of the resultantforce F it will be pulled into the region where induc~ion Bis larger. {t'

(6.36)IM = {Pm >: B], I

6. Magnetic Field in a Vacuum16,)

where Pm is the magnetic moment of the current loop (Pm =B = ISn for a phme loop). *

I t is clear from ((j.36) that the IlWln-

F. ent M of Ampere's forces acting ona current loop in 11 uniform magnetic field is perpendicular both to vector Pm and to vector B. The mag-nitude of vector 1\1 is M' = PmB sin ct,where ct is the angle between vec-

F tors Pm and B. When Pm tt B, M ~= 0and it is not difficult to see that the

Fig. 6.14 position of the loop is stable. WhenPm H B, 1\'1 also eq uals zel'O but such

a position of the loop is unstable: t he slightest deviationfrom it leads to the appearance of the moment of force thattends to deviate the loop from the ill itial po,'.:ition stillfurther.

moment of fOl'ces for an arbitrary shape of the loop canbe represented in the form '

M = lbBa 8i:' Ct.

~hese forces tend to rota te the' loop so tl,a t vedol' Pm becomes (hreetedlr ke vee tor B: Hence, the loop is acted npon by II couple of forceswhose torque J~ t'9'ual to the p;'oduct of the arm Ii sin 1'1, of the couplrby the force P, I e.

.Example. Let liS verify the validity of formula (6.36) by usinga Simple example of a rectangular current loop (Fig. 6.14).. It ca.n be seen from the figure that the fo~oes <Jcting on sides a areperpendicular to them and to vector B. Hence these forces <JfP directedalong the hOrIzontal (they are not sh~':Vn in, the figure) and only striveto stretch (or compress) the loop. S,'1es 0 are perpendicular to Hand hence each of them is acted upon by the force '

F c= !hB.

* If till' l00p .is ::ot plane, its ma:'ne,j,: rnO'lwnt i ... II = l' ';'d<:<~- u fil .....~t

where the integnt! is LuLen OVer the slHIael' S s\.rr'tehed over th,; current loop. This intf'g'l'dl does nM depend 011 the choice' of til(> surfacl: Sand depends only on the loop over which it is stretched.

162 6. Magnettc Field In II VacuumProblem,

ft ..

1;

,~t'

I·~·

"".: .,..•.,.....•.'...••.",

,Y;;

Problems

• 6.1. Direct calculation of induction B. Current I flows alonga thfn conductor bent as is shown in Fig. 6.16. Find the magneticinduction B at point O. The required data are presented in the figure.

Solution. The required quantity B = B _ + B~, where B - is the

where <1>1 and <1>2 are magnetic fluxes through the contourin the initial and final positions. Thus, the work of Ampere'sforces in this c;ase is equal to the product of the currentby the incremei1t of the magnetic flux through the circuit.Expression (6.40) gives not only the magnitude but alsothe sign of the accompli~hed work.

Example. A plane loop with current I is rotated in magneticfield B from the position in which n H B to the position in whichn tt B, 0 being the normal to the loop (it spould be recalled that thedirection of the normal is connected with the direction of the currentthrough the right-hand screw rule). The area bounded by the loop is S.Find the work of Ampere's forces upon such a displacement, assumingthat the current I is maintained constant.

In accordance with (6.40), we haveA = I [BS - (-BS)) = 2IBS.

In the given casa, the work A > 0, while upon the reverse rotationA <0.It should be noted that work (6.40) is accomplished not at theexpense of the energy of the external magnetic field (which does notchange) but at the expense of the e.m.f. source maintaining the cu~re~tin the loop. (This question will be considered in greater detaIl III

Chap. 9.)

(6.37), where d<l> is the increment of the magnetic flux, through the entire circuit.

In order to find the work done by Ampere's forces uponthe total displacement of the current loop from the initialposition 1 to the final position 2, it is sufficient to integrateexpression (6.37):

2

A = ) 1 d<l>. (6.39)1

! If the current 1 is maintained constant during this dis-", placement we obtain

': A = 1(<1>2 - <1>1)' (6.40)

work

6A = F dx = IBZ dx = IB dS, (6.38)

where dS is the increment of the area bounded by the contour.

In order to determine the sign of magnetic flux , weshall agree to take t~e normal n to the surface bounded byt~e c~ntour always In such a way that it forms with thedl~ectlOn of the c~rrent in the circuit a right-handed system~Ig. 6.15): In thIS case, the current I will always be a positIV~ ~uantIty. On .the other hand, the flux may be eitherposItIve or negatIve. In our case, however, both anddCf> = B dS are positive quantities (if the field B weredIrected towards us or the jumper were displaced to theleft, d < 0). In any of these cases expression (6.38) canbe represented in the form (6.37).

2. The obtained result is valid for an arbitrary directionof the fi.eld B as well. To prove this, let us decompose vector B Into thr~e components: B ~o Bn + Bz + Bx . Thecompon~nt .Bz dIrected along, the jumper is parallel to the~urrent In It and does not produce any force acting on theJumper.. The component Bx (along the displacement) isres~onsible for the force perpendicular to the displacement,whIch does not accomplish any work. The only remainingcomJilonent is Bn which is normal to the plane in whichthe Jumper moves. Hence, in formula (6.38) instead of Bwe. must take only Bn • But Bn dS = d<l>, and we againal'nve at (6.37).

3.. Let. us now ~onsider any current loop which is arbitrarIly dIsplaced I? a constant non uniform magnetic field(t~e contour of tIus loop may be arbitrarily deformed inthIS process). We mentally divide the given loop into infinitel~ small current elements and consider their infinitesimal?Ispla~ements. Under these conditions, the magnetic field10 whICh .each element of current is displaced can be assumed uOIform. For such a displacement, we can apply toeach element of current the expression dA = I d'(p for theele~entary work, where .d' expresses the contribution (Ifa gIven element of current to the increment of the fluxthrough the contour. Summing up elementary. works forall the elements of the loop, we again obtain expression

I)'I,

I

1648. Magnetic Pield in a Vacuum

• Problems 165

m a(l'netic ~eld created by the rectilinby Its curvilinear part. According t th Ba.r p~rt of the loop and B ,on p. 146), we have 0 e IOt- avart law (see Examplef

B~=~ I (2n- 2ao) a _ !-tol4n a2 - 2na (n-ao).

As a result, we obtain

B = (n - ao + tg ao) !-toI /2na.It is interesting to show that feXI,ression (6.13). or ao - 0, we arrive at the familiar

!-tol cos a da4na cos ao

!-tol2rui"' tg a"

Substituting (1) and (3) into (2) and integrating the result over rbetween a and b, we obtain

B= !-toIN In.!:2 (b-a) a

(2) The magnetic moment of a turn of radius r is Pml = Irrr2•

and of all the turns, Pm = ~ Pml dN, here dN is defined by formula

(3). The integration gives

__ nIN bS- as _ nUl ( 2 + b+ b2)

Pm - b _ a -3- - 3 a a .

• 6.3. Current I flows in a long straight conductor having th~shape of a groove with a cross section in the form of a thin half-ring

(2)

Fig. 6.19Fig. 6.18

where dI ","-C (IIn) dq; (see Fig. 6.19). Substituting (2) into (1), we get

of radius R (Fig. 6.18). The current is directed from the reader behindthe plane of the drawing. Find the magnetic induction B on the axis O.

Solution. First of all, let us determine the direction of vector Bat point O. For this purpose, we mentally' divide the entire conductorinto elementary filaments with currents dI. Then it is clear that thesum of any two symmetric filaments gives vector dB directeli to the

. right (Fig. 6.19). Consequently, vector B will also be directed to theright.

Therefore, for calculating the field B at point 0 it is suffIcient tofind the sum of the projections of elementary vectors dB from eachcurrent filament onto the direction of vector B:

B=.\dBsincp (1)

In accordance with (6.11), we have

dB = /lo III /2rrR,

(1)

(2)B=JBldN,

number of turns in the interval (r, r+ drl,

dN=~dr.h-n

Fig. 6.16 Fig. 6.17

• 6.2. A thin isolated wire fo I .a large number N of closely packed ~s \[ ane spIral consisting ofrent I is flowing. The radii of th . urns I rough which a direct curand b (Fig. 6.17). Find (1) the r::.aIn er.na. and ~xternal loops are aof the spiral (point 0) and (2) th gnebc .InductIOn B at the centrethe given current.' e magnetIc moment of the spiral for

Solution. (1) According to (6 13) hradius r to the magnetic I'nduct: ~ t e colntribution of one turn ofIOn IS equa to

and of all the turns, B1 = f.toIl2r,

where dN is the

(b)

Fig. 6.23

(a)

Fig. 6.22

Problems

solenoid, the flux through the contacting endfaees will be lI> + lI> == lI>0. where ~ is the flux through the solenoid's cross section farfrom its cndface. Then we have

11> ~ $0/2 = lLon!S /2.

By the way, we must pay attention to the following propertieof the field B at the endfacc of a long solenoid.

1. The lines of B are arranged as shown in Fig. 6.22. This can beeasily shown with the help of superposition principle: if we place onthe right one more solenoid, field B outside the thus formed solenoidmust vanish, which is pcssible only with the field configuration shownin the figure.

2. From the principle of superposition, it follows that the normalcomponent Bn will he the same over the area of the endface, since whenwe form a composite solenoid, Bn + Bn = Bo, where HI, is the fieldinside the solenoid away from its endfaces. At the centre of the elldfaceB = B n , and we obtain B = Bo/2.

• 6.6. The field Gf 8 solenoid. The winding of a long solenoid ofradius a is 11 thin conducting band of width h, wound in one layer practically without gaps. Direct current I flows along the band. Findmagnetic field 8 inside and outside the solenoid as a funetion of distance r from its axis.

Solution. The vector of linear current density i can be representedas the sum of two components:

i=i 1 + ill"

The meaning of vectors i.L and ill is clear from Fig. 6.23b. In ourcase, the magnitudes of these vectors can be found with the help of

• 6.5. Principle of superposition. Current I flows through a longsolenoid whose cross-sectional area is S and the number of turns n.Find th~ magnetic flux through the endface of this solenoid.

Solution. Let the nux of 8 through the solenoid endface be <1>. Ifwe place another similar solenoid in contact with the endface of our

(1 )


Fig. 6.20

166

..• 6.4. Theorem on circulatilIOn.. ' nsid.e a homogeneous Ion ,on .of 8 ~nd pr.!ncipl(. of sUpt"rposi_jhere IS a .clrcular cylindrical c; s.~ralg~t wlre?f ~lrcularcross section

uctor aXIs ~nd displaced relati:~ i ~t bse aXI~ IS parallel to the con~~ent of deusl ty j flows along th 0.1 Y.a dIstance I. A direct curtnsldp the cavity. I.' wire. Ftnd magnetic induction B

S.olution. In accordance with h' . .required quantity can be t e prmclple of superposition the

represented as follows: 'B = 80 - B'

where 8 0 is the magnetic induction of th~ conductorwithout cavity,

Fig. 6.21

while B' is the magnetic induction f h .

beto the current flOWing through th 0 ttl' ~elt at the same point due

. cn removed in order to create th par ?f. t e conductor which has. Thus, the problem im lie fi e cavity.In~uction 8 inside the sO~id ~O:d~ 0/ all the ~alculation of magnetic~Stng the theorem on circulation w~ or at ~ distance r from its axis.

Fi; ~~~~ ~~I'jih;h~~c~~¥rf~~~~'can cb~n ;:~~~s::[~ ;ir~~~~' ~~~n~f1

B="2f!o[jxr).

Using now this formula for Bo and B' w fi d' h' ., e n t elf dJilerence (1):

8_ lLO [' ] lLo· 1L-2 J X r -2 [J X r']= 20 (j X (r-r')).

Figure 6.21 shoWs that r = I +' hr , w ence r - r' = I, and

B 1 .="2 flo [J :< I) .

Thus, in Our case the magnetic field B . .If the current is flOWing towards us . In the cavity is uniform andplane of the lIgure and is dl'r t d (Fig. 6.21), the Held B lies i~ the

ec e upwards.

Fig. 6.23a by the formulas

Fe=AE=A~ E:.-_ 2"-24m,0 l - - 4m l '

where I is the distance betwe h 0

force acting per unit length :Pt~ee a~es of tbhe wires. The magneticWIre can e found with the help

169Problems

of the theorem on cirulation of vector ~:

Fm = (f!0/4n) 212ll,

where I is the current in the wire. .It should be noted that the two forces, electric and magnetic, are

directed oppositely. The electric force is resp'onsible for the. attractionbetween the wires, while the magnetic force causes their repulsion.Let us find the ratio of these forces:

FmlFe = eOf!0/2/A2. (1)

There is a certain relation between the quantities / and t.. (seeProblem 2.8):

'),.=C1U= Ineo U, (2)nl]

where U = RI. Hence it follows from relation (2) that/1'),. = In TJineoR. (3)

Substituting (3) into (1), we obtain

Fm f!o In2 l]------ (4)

,,' Fe 80 n 2R2'

The resultant force of interaction vanishes when this ratio is equalto unity. This is possible w~en R = Ro, where

Ro= .. /f!o Inl] =360Q.V eo n

If R < RTJ then Fm > Fe, and the wires repel each other. If, onthe contrary, R > Ro, Fm < Fe, and the wires attract each other.This can be observed experimentally.

Thus, the statement that current-carrying wires attract each otheris true only in the case when the electric component of the interaction can be neglected, Le. for a sufficiently small resistance R in thecircuit shown in Fig. 6.24.

Besides, by measuring the force of interaction between the wires(which is always resultant), we generally cannot determine current I.This should be borne in mind to avoid confusion.

• 6.8. The moment of A-mpere's forces. A loop with current /i8 in the field of a long straight wire with current 1 0 (Fig. 6.25). Theplane of the loop is perpendicular to the straight wire. Find the momentof Ampere's forces acting on this loop. The required dimensions of thesystem are given in the figure.

Solution. Ampere's forces acting on curvilinear pllrts of the loopare equal to zero. On the other hand, the forces acting on the rectilinearparts form a couple of forces. We must calculate the torque of thiscouple. ,Let us isolate two small elements of the loop (Fig. 6.26). It can be

seen from the figure that the torque of the couple of forces correspond-


:

i.l =/Cosa=1 V1-sin2a=(I/h) V1-(h/2na)2.

i lt = i sin a = Il2na.

The magnetic induction B' .d haccordance with (6.20) by th mSI.e t .1.' sole~oid is determined inby I,,: ,I.' quantIty l.L' whIle outside the solen~id,

B i = f!o/.l = (f!oI/h) Vi - (h/2na)2 (r < a),

Ba = f!oilla/r = f!0I/2nr (r > a),

Where we used the theore . I . 'side the solenoid' 2nrB ~ on2cIr~u atlOn while calculating B out-

Th . a - f!b ltalW a.. us, h~~ing represented the current' h .

poslt!,on of transverse" and "Ion 't d' i~ t I.' solenoId as the superthe conclusion that onl th I gl. U I.na components, we arrive ate;Xist~ inside such a sole~oidea:dn~I~ldlilial component of the field BSIde It (as in the case of a strai~ht y I.' transverse component out-

Beside 'f' d current).d . S, I we ecrease the and 'dth . "eDSlty unchanged, 1-+0 as h 0 hI Imamtammg the current

~.nldYal~?e field inside the solenoidre:Uai~~I.h =thconstl' In. this case,I I.' ,I.e. I.' so enold becomes

~ ..6.7. Interaction of parallelneghgtble resistance are shunted- at ~hr~ntB'd Tbwo lo~g wires with

eIr en s y reSIstor R and atthe other 'Cnds are connected to asou~ce of constant voltage. Thera.dlU~ of the cross section of eachWIre IS smaller than the di~tancebe~een t~eir axes by a factor ofTJ. - 20. Fmd the value of the reSIstance !i at which the resultantfo:ce of I~teraction between the

Fig. WIres vamshes.

. ' Solution. There are excess sur-tIVe of whether or not the . face .charges on each wire (irrespec-~ence, in.addition to the mC::~~~~If flowkng through them) (Fig. 6.24).t I.' e.lee-trlC force Fe. Suppose that orce m, we must take into accounta unIt length of the wi~e Th an exces~ charge A corresponds to

Glength of the wire by the' other en. the elebctrlc force -exerted per unit

auss theorem: WIre can I.' found with the help of the

(2)

171Problems

dF = Ii X D'] dS,

~ ,,''b./ ./

08Dr B'

~ dS ~'!m

~B'l/ Bi

Fig. 6.28 Fig. 6.29

of substitution (1):

j dV = j6h·6b dl = i as.The meaning of the lJuantities appearing in this relation is clarifiedin Fig. 6.28. In the vector form, we can write

jdV=idS. (1)

The Ampere's force acting on the surface current element" in thiScase is determined by the formula obtained from (6.28) with the help

lind the relation between the surface and volume elements of the current:

where B' is the magnetic induction of the field at the point of lo<:ationof the given current element, but created by all other current elementsexcluding the given one. In order to find B', we proceed as for calculatin~ electric force (Sec. 2.3). Let B, be the magnetic induction

Since aB/az < 0, the projection of the force Fz < 0, Le. vector Fis directed towards the loop with current /. The obtained result canbe represented in the vector form as follows:

It should be noted that if Pm (and henceZ-axis as well) were directed oppositely, thenB z = -Band oBz/oz > O,and hence Fz>Oand vector Fwould be directed to the right, Fig. 6.27Le. again oppositely to the direction of Pm.HeBee, the obtained direction for F is valid for both orientations ofPm

• 6.10. Current I flows in a long thin-walled circular cylinderof radius R. Find the pressure exerted on the cylinder walls.

Solution. Let us consider a surface current element idS, wherei is the linear current density and dS is the surface element. We shall

(t)

(2)

--X

B on the distance r

I!'ig. 6.26


dM = 2z tan cp dF,

where the elementary Ampere's force is given by

dF = I diB.

The dependence of the magnetic induction

iug' to these elements is

from the ~traight \ . b fcirculation: vlre can I' ound with the help of the theorem on

B = !-Lo[/2nr. (3)Let us now sub~titute (3) . t (2) h

ing that· it = dr ;'1d x ~= r c~~ 0 .' t en (2) into (1) and, consicleover r between a and b. This gi~~smtegrate the obtained expression

M = (fto/n) II 0 (b - a) sin lp,

vector M being- directed to the left (I!'ig. 6.26).

~ 6.9. A small coil with current h . C1 •

Pm, IS placed on the axis of a circul 'I avm"'f the magnetic momentcurrent I is flowing. Find th f ar ~op 0 radius R, alo~ whichfrom the centre of the loop f t ce

dF actlllg on the coil if its' distance

in Fig. 6.27. s an vector Pm is oriented as is shown

Salution According t (6 33) h. 0 . ,t e required force is defined as

F= Pm aB/anwhere B is the magnetic indo ' . (t)at the Iscus of the coil. Let Us~~~:n ,o!. ~~~. field cr~ate~f by the looptor Pm' Then the projection of (1) t Zt atXII~ III t,he d.lrechon of the vec-

on 0 118. aXIs wIll beFz = Pm aB/az = Pm ilB/az,

~here w~ took into account that B = .(~rr~nt !II the loop. The magnet· .Z d B. for t~le given direction of(h.12), whenn' . IC III IIctlOn B I~ defmcd by formula

aB __ 3 f.LoRz/1az ----Z(lZ+RZ)6/Z.

rI, j

II

IiiI!I

III

"'I

. . B' = B/2. (4SubstItutIng this result into (2) b' )for the required pressure: ' we 0 tam the following expression

dF 2B' B2p=- = iB' = -- B' -

dS ~o - 2~ .Taking into account (3), we finally get 0

p = ~oI2/81t2R2.

It is clear from formula (2) th t h l'pression. ate cy mder experiences lateral com-

of the field created by the surface 1 'to its surface (see Fig 6 29 wh e ~f~nt Itself at a point very closeflowing from ~s). Acc~rding' to (~~~2) ~.a~u(T/ed th?t the current is

Further, USIng the theorem . i. l 2) ~Ol.considerations we can easH OUCIrcu atwn of vector B and symmetryfield outside the cYlinder n!ars~et thatf the .magnetic induction of the

I s sur ace IS equal to

. . . B = ~oI/21tR, (3)whIle InsIde the cylinder the field' b

The latter fact indicates that th fis ld s~nt.rent elements at points 1 and 2 e e B from the remaining cur-Fig. 6.29~ must be the same and ;:~fsf~l~h }O 11he ~ylinder. s!lrfac.e (seeand outsIde the cylinder surface: e 0 OWIng condItIOns mside

B' = B t and B = B' + B t = 2B'.Hence it follows that

(7.2)IfBdS~o·1

7.1. MagneUzation of a Substance. Vector J 173

The field B', like the field Eo of conduction currents,has no sources (magnetic charges). Ilence, for the resultantfield B in the presence of a magnetic, the Gaus.~ theoremis applicable:

This means that the field lines of B remain continuouseverywhere in the presence of 11 substance.

Mechanism of Magnetization. At present, it is estab-. lished that molecules of many substances have intrinsic

magnetic moments due to the motion of intrinsic chargesin them. Each magnetic moment corresponds to a circularcurrent creating a magnetic field in the surrounding space.In the absence of external magnetic field, the magneticmoments of molecules are oriented at random, and hencethe resultant' magnetic field. due to these moments is equalto zero, as well as the total magnetic moment of the snbstance. This also appHes to the substances that have nomagnetic moments in· the absence of an external field.

If a substance is placed into an external magnetic field,under the action of this field the magnetic moments of themolecules acquire a predominant orientation, and the substance is magnetized, viz. its resultant magnetic momentbecomes other than zero. In this case, the magnetic moments of individual molecules no longer c.ompensate eachother, and as a result the field B' appears.

Magnetization of a substance' whose molecules have nomagnetic moments in the absence of external fIeld proceedsdifferently. When such materials are introduced into anexternal field elementary circular currents are inducedin the molecules, and the entire substanc.e acquires a magnetic moment, which also results in the generation of thefield B'.

Most of materials are weakly magnetized when introducedinlo a' magnetic. field. Only ferromagnetic materials suchas iron, nickel, cobalt and their many alloys have clearlypronounced magnetic. properties.

Magnetization. The degree of magnetization of a magnetic is characterized by the magnetic moment of a unit

(7.1)

7. Magnetic Field in a Substance172

7. Magnetic Field in a Substance7.1. Magnetization of a Substance

Magnetization Vector J •

Field in Magnetics If at'into a magnetic field form~~r ~m sUb.stanc~ is introducedthe field will change. This is I y. c~r~ents m conductors,substance is a magnetic . eX~t~le y theJact that eacha magnetic moment) unde~·eih I, IS. magnetIzed (acquiresA magnetized substance creat e .;ctlOn of mag~Ietic field.which forms, together with t~ I s .own magnetIc field B',by conduction currents, the re=uft:~n:a~~l:eld Bo created

B = B o + B'.

H~re B~ an~ B stand for the fieldsphysIcal mfimtely small volume. averaged over a

volume. This quantity is called magnetization and denotedby J. By de~nition, .

115

x

Fig. 7.2

Y4 I'

'I___~.. l'

--

1.1. M41,.etnatlon 01 4 Sub,tance. Vector J

Fit. 7.1rent]' induces the same mac

surface of the cyl.inder-lhe c~~at of all molecular currentsroscopic magnetic fiel astaken together.. er case' a magnetized magnetic

Let us now conSider anot~ t . tile molecular currentsLet for msance, .' kis nonhomogeneous. : F' ~ 2 where the Ime thiC 'ness

be arranged as sho~n Ill. Ig~f ~oieclllar clll'rents. It followscorresponds to the mtenslty J' directed behind the planefrom the figure th.at vector. ~a nitude with the coord iof the figure and Illcrea~st Irhe ;olecular cnrrents are notnate x. It can be seen t a h eneous magnetic, aIllI ascompensated inside a I?-0n omog rna ~etization current l'a result, th~ macrosc.OPIC vO:~~ve d1rection of the Y-axi~;appears, which flows III t~ Pb t the linear l' and surface ~Accordingly, we can spea a ou I I'n Aim and j' in A/m~.

., ., being measurec h tcurrent denSItIeS, t. • M netic. It can be stated t ~

Calculation of Field B 10 a f ~g a magnetized magnetICthe contribution to field. B t~O~ would be created by theis equal to the contribution ~" a vacuum. In othersame distribution o~ curre~tsd' tl':~ution of magnetizationwords, having established t e _t~S the help of the Biotcurrents 1', we can find, WI I'ng to them and thell

h field B' corresponc I 'Savart law, t e fi Id B by formula (7.1).caIc.ulate the resultant e

- . . he oint:- of their contact have opadjacent mol~cules at t p tach othel'. The Oil Iy uncom-

. posite directiOns and comp~nsa_ etriOSe emerging 011 the lateralpensated molecular curren;1ale currents form macroscopicsurface of the. cyl.inrler. ItesJ' circulating over the lateralIUTjace magnetization curren.

(7.3)

7. Magnetic Field in a $ubdance

where L\ V is a physical infmitely small volume surrounding a given point and Pm is the magnetic moment of anindividual molecule. The summation is performed overall molecules in volume L\ V.

By anillogy with what was done for polarization P(see (3.3)1, magnetization can be represented in the form

J = n (Pm), (7.4)

where n is molecular concentration and (Pm) is the averagemllgnetic moment of a molecule. This formula shows thatvector J is collinear precisely to the average vector (Pm)'Hence for further analysis it is sufficient to know the behaviour of vector (Pm) and assume that all the moleculeswithin the volume L\ V have the same magnetic moment(Pm)' This will considerably simplify the understanding ofquestions associated with the phenomenon of magnetization. For example, an increase in magnetization J of a material implies the corresponding increase in vector (Pm): ifJ = 0, (Pal) = 0 as well.

If vector J is the same at each point of a substance, it issaid that the substance is uniformly magnetized.

Magnetization Currents 1'. As was mentioned above, magnetization of a substance is caused by the preferential orientation of magnetic moments of individual molecules in onecl irection. The same can be said about the elementary circularcmrents associated with each molecule and called "molecularcurrents. It will he shown that such a behaviour of molecularcllrrents leads to the appearance of macroscopic currents J'called magnetization currents. Ordinary currents flowing inthe conductors are associated with the motion of chargecarriers in a substance and are called conduction currents (I).

In order to understand how magnetization currents appear,let us first imagine II cylinder made of a homogeneous magnet ic whose magnetizatiou J is uniform and directed alongthe axis. Molecular cuneuts in the magnetized mag-neticlIre oriented as shown in Fig. 7.1. Molecular CIIITcnts of

114

t77

(7.6)IvxJ=r, I

7.2. Circulation of Vector J

Le. the curl of magnetization J is equal to the magnetization currentdensity at the same point of space.

A Remark about the Field of J. The properties of thefield of vector J, expressed by Eqs. (7.5) and (7.6), do notimply at all that the field J itself is determined only by thecmrents I'. The field of J (which is hounded only by the

t 2-0 181

the contour r intersect the sUl'face S only once. Such molecular currents create a macroscopic magnetization current ]'piercing the surface S.

Let each molecular current be equal to 1m and the areaembraced by it be Sm' Then, as is shown in Fig. 7.4, theelement dt of the contour r is wound by those molecularcurrents whose centres get into the oblique cylinder withthe volume dll c= Sill cos a dl, where a is the angle betweenthe contour element dl and the direction of vector J at thepoint under consideration. All these molecular currentsintersect the surface S once, and their contribution to themagnetization current is dI' = I mn dV, where n is the molecular concentration. Substituting into this formula theexpression for dV, we obtain

dI' = l rn Smn cos a dl = J cos a dl = J dl,

where we took into account that I mS 111 = Pm is the magneticmoment of 1m individual molecular current, and I mSmnis the magnetic moment of a unit volume of the material.

Integrating this expression over the entire contour r,we obtain (7.5). The theorem is proved.

It remains for us to note that if a magnetic is nonhomogeneous, magnetization current I' generally pierces theentire surface (see Fig. 7.3) and not only the region near itsboundary, adjoining the coutour 1'. This is why we can use

the expression I' c= I j' as, where integration is performed

over the entire surface S, bounded by the contour r. Inthe above proof, we managed as if to "drive" the entirecmrent I' to the boundary of the surface S. The only goalof this method is to simplify the calculation of the current I'.

Differential form of Eq.(7.5):

rI

(7.5)

7. Magnettc Fteld in a Subl1tancet76

ItJdJ~I'·1where l' - r j' dS d'. . - J an mtegration is performed Ovarhltrary surface stretched on th er an

In order to prove this theo e contour r.algebraic sum of molecular cu ,rem, we shall calculate theLet us stretch an arbl"raI'y nenfts enSveloped by contour r.(F' 7' '. t sur ace 0 tlIg. .3). It can be seen from tJ f . n Ie contour rcurrents intersect the surface S Ie /gure that, some molecularand for the second time in 'the t~ ,ce (~H1ce .In o~le directionsuch currents make no crnt 'b .pposlte directIOn). HenceI t' . ) rI lItlOn to the reo ItIe IzatlOn current throllgl 1I ,::;nant mag-

Howe. th . . d Ie surface S\er, e molecular currents that are',

wound around

Fig. 7.3 F' _19. i.4

to the algebraic sum of . '.by the contour r: magnetIzatIOn currents I' enveloped

UIifortunately, . the distribut' 0not only on the confi uratio 1 n of currents /' dependsnetic but on the field Jas lY ~nd ~~e properties of a magcase the problem of findin;~: or t IS re~son, in the generaldirectly. I t remains for us to ~n a magnetic cannot be solved~o the solution of this problem i and find a~othe.r approachIS the establ ishment of .' he first step III Hus direction..' an Important reIat . bn etIzatlOH current /' a I . IOn etween mag-

viz. its circulation. nt a certam property of vector J,

7.2. Circulation of Vector JIt tUl'1lS out that for a statio .

magnetization J around bJl?IY case the circulation of. an ar Itrary contour r is equal

&6Wdl

I II

r

I I rI

Ii

~!! u 7._ Magnetic Field in a Substance

179

(7.12)

(7.9)

7.8. Vector H

~ (:u -J) dl=I. (7.10)

Let us denote the integrand of this expression by H.Thus, we have found an auxiliary vector H,

\H~;'-J'\ (7.11)whose circulation around an arbitrary contour r is equal

, to the algebraiO'tsum of the eonduction currents I embracedby this contour:

.lated to the circulation of the magnetization:

~ J dl =1'.

Assuming that the cirenlation of vectors Band J is takenvel' the same contour r, we express l' in Eq. (7.8) thronghormula (7.9). Then

(7.7)if = J.

.Example. Fi.nd the surface magnetizationc"!ll"ent per umt length of a cylinder made?f ~ ho~ogeneous magnetic, if its magnetIzatIOn IS J, vector J being directed everywhere along the cylinder axis.

hLet us .apply Eq. (7.5) to the contour

c D.sen as I,S shown in Fig. 7.5. It can beeaSIly see? that the circulation of vector J

Fig', 7.5 around thl.S contour is equal to the product(l. In thIS case, the magnetization current

" ' _, ., IS the surface ~urrent. If we denote its linearJ.rnslty ly ! ,the,contour under consideration embraces the ma netlzatlOll current i l. It follows from the equality It = i'l that g

S}?ltial region filled by magnet,'c) depends on all,." l'jl th .. currents,',Z,]l.' 1 on ,0 mag~letlZatlOn currents l' and the conduc-

tl?ll currents I. However, ill some caseswIth a definite symmetry the situationis such as if the field of vector J weredetermined only by currents 1'.

It should be noted, by the way that vectors i' 8:111 J are mutuallyperpend iClllar: i' 1. J. '

7.3. Vector H

T~eorem on Circulation of Vector H (for magnetic fieldsof dlre~t currents). In magnetics placed into an external~agnetI~ fiel~l magnetization currents are induced. Hencecl~c~latlOn .0;' vector B will now be de.termined not onlyb) conductIOn currents but by magnetIzation currents aswell:

~Bdl=/lo(I+I'), (7.8)

where I and l' are the conduction .and magnetization CUI'

ren~s embraced by a given contour r.. SInce the .determination of currents l' in the general caseIS. a c0m.-phcated. problem, formula (7.8) becomes inapplIcable In practlcal respect. It turns out, however thatw~ can find a. certain auxiliary vector whose circuiationwIll be determIned only by the conduction currents embracfJdby the contour r. Indeed, as we know, the current l' 'is

This formula expresses the theor{!m on circulation of vector H: the circulation of vector H around an arbitrary closedcontour is equal to the algebraic sum of the conduction c'urrcntsembraced by this contour.

The sign rule for curr~nts is the same as ill the case ofdrculation of yector B (see p. 149).

It should be noted that vector H is a combination of twoquite different quantities, B//l o and J. Hence, vector His indeed an auxiliary vector which does not haye any deepphysical meaning. * However, the important property ofvector H expressed in the theorem on its circulation justifIOsthe introduction of this vector: in many cases, it considerably simplifies the inyestigation of the field in magnetics.

Relations (7.11) and (7.12) are val id for any magnetics,induding anisotropic ones.

* The quantity II is often called magnetic field intensity, hut weshall not be using tlJis term to emphasize again its auxiliary nature.

12*

('

18t

Fig. 7.6

7.3. Vector B

When r = 0 Inside a Magnetic? We shall show now thatmagonetization ellrren'ls inside a magnetic are absent if (1)tlte magnetic is homogeneolls nnd (2) there nrc no condllctionClllTellt.S in it (j - 0). I II this case. fot' any shape of a 1l111gnetic mill allY conliglll'ation of the magnetic Held, we call be

For paramagneties ~t > 1, while for diamagnetics fL < 1.It should be noted that in both cases fL differs from unityonly slightly, Le. the magnetic properties of these magneticsare manifested very weakly.

A Remark about the Field of H. Let us consider a questionwhich often involves misunderstanding: which currentsdetermine the fIeld of veetor H? Generally, the fwld of H(just as the field of B) depends on all currents, both conduction and magnetization. This can be seen even from(7.i5). However, in some cases the field H is determinedonly by conduction enrrents. Vector H is very helpful forthese cases in particular. At the same time, this makesgrounds for the erroneous conclusion that the field of Halways lkpends on conduction currents and for incorrectinterpretations of the theorem on circulation of vector Hand Eq. (7.13). This theorem expresses only a certain property of tite lIeltl,of H without deHning the ~field itself.

<It

Example. A system consists of a long straight wire with current Iand an arbitrary piece of a papmagnetic (Fig. 7.6). Let us lind thechanges in the lieilis of Vl'ctors B and Ifami in the circulation of vector II arour.da certain fixed contour r upon theremoval of the magnetic.

The field B at each point of space isdetermined by the conduction current Ias well as by the magnetization currentsin the paramagnetic. Since in our case,in accordance with (7.15), If = B/l!l!o'this also applies to the Held of vectorII, i.e. it also depends on the conductioncurrent I and the magnetization currents.

The removal of t.he piece of paramagnetic will lead to a changein the lield B, ami hence in the lield ll. The circulation of vector Baround the contour r will also change, since the surface st.retched onthe contour r will no longer be pierced by the magnetization currents.Only the conduction current will remain. However, the circulationof v(!ctor II around the contour r will remain unchanged in spite ofthe variation of the tield II itself.

(7.13)

(7.1:) )

(7.1lj)

U·l1),B/~o.

7. Magnetic Field in a Sllhstaw'(,180

It follows from formula (7 12) I .vector H has the cli' . . I, wI, lhe IIl:\<rnitude of.. menslOUS of tI, "'Hence, H is measured ill' Ie CUlrent pCl' !wIt ll'fI!!'tll.

, amperes per metre (Aim).Differential rorm or the tlIeO ' ,

rem on cIrculatwn or \'eCtor H:

IV x H=j, Ii.e. the curl of vector H . I -,current at the same pointlsofeqtuha tO

btthe density of the conduction

e su s ance.

Relation Between Vectors J dthat magnetization J d d an H. It was shown abovet · epen s on the lllag I' . da a gIven point of the ' b ne IC 111' uction B

vect J' su stance Howev"or IS related not with B . , C!, customarilyourselves to the analvsi" of flit WIth H. \\e slwll coniinethe depeudence bet~~enUJ . ond YHt~lOS~ magnetics fOl' which

cl1l IS hllecu', viz.

J = XHwhere . h . ' (7.14)

. X IS t e magnetic susce tibil't I ',,' .quantIty typical of eacl p. l y. t IS a dllnenSlOnlessf· 'I I magnetIc (tl t . J'01 ows from the fact tl 't. .' Ia X IS hlmension1ess

S· . f H la , accol'lllllg' to (7 11) I f'IOns 0 and J aI'e tl e ) . " t LC ( 1men-U 1· , S<:.me

n II~e electric sllsceptibilit' " .magnetIc susceptibility X hx ~\ Illch IS always positiveAccordingly the mag t~nay _ e .eltI~er positive hI' negative'd"d . ., ne ICS ,.,atlsf\"l f 1, ~ . •

IVI ed mtD paramagnetics ( fig . orIllU ,1 (i .14) areIn paramagnetics, J tt H X ~ .0) a.JlII d~amagndit's (z < 0).

It should be noted' that' iI~ ~~e. I.n (hamp'{!If'hl: J H H.there exist ferromagnetic~ for' 'I ~f~tlOr to tll(',' !!lag-netics,~laS a rather complicated' fol'!~ ,wi ~I~.' t le ~epeHde',cD J (H)It has a hysteresi~ J' n J d d' s not 1111ear, and besides. " ,". epen son t It I . -,netIc. (Ferromarrnetic" will b I ..e pre IIstory of it magin Sec. 7.6.) ,., ~ e (eSenbell in greatel' dptail

Relation Between Band H Fexpression (7 11) .• or magnetics olw,XilwH . acqllJres the forlll (1 I.,) II hence , Z

B = p~loH,

where ~l is lhe !Jermeahilt·t.t/ ()f I. lie medilllll.

~ = 1 .:- Z.

I'I

(7.21)

(7.20)

21

Il2~l +- Hi~,l = iNl,

where iN is the projection of vector i onto the normal Nto the ('oQutOllI' (vector N forms a right-handed systom withthe direction of eontour circumvention). Taking the twoprojections of vector H onto the comm'on unit vec.tor of tltetangent. T (in IUf\l\illlU 2), we obtain TT lt , ,== -lfl~' C·ancelling l Ollt of tlwpreviolls eqllation, we get

7,4. Boundary Conditions for 8 and B

-~-_.~----,,,'t n'

Fig. 7.7 Fig. '7.8,same on both sides of the interface. This quantity does not

have a discontinuity.Condition for Vector H. We shall assume, for higher

generality that a surface conduction current with lineardens'iy i flows over the interfaeial surface of the magnetics.Let us apply the theorem on circulation of vector H toa very small rectangular contour whose height is negligiblysmall in comparison with its length l (the arrangement ofthe contour is shown in Fig. 7.8). Neglecting the contributions from smaller sides of the contoUl' to the circulation,

we can write for the entire contour

i.e. the normal component of vector B turns out to he the

in Fig. 7.7. Then the flux of B in the outward direction (weignore the flux through the lateral surface) can be written

in the form B2n /).8 +- BIn' /).8 = O.

Taking the two projections of vector P onto the normaln to the interface, we obtain BIn' = -13]11' and after cancelling /).8, the previous equation becomes

7. Magnetic Field in a Substance

l' =X~ Hdl.

In accordance with (7.12.) h' .. .to the algebraic SUIll of th~ ~oe ~~m?mlIlg mtegral is equalby the contour r. Hence f . n ~,ctlOn currents I embracedhave 01 a lOmogeneous magnetic, we

7.4. Boundarv Conditions for Band H

We slwll analyse tli> C d"the interface between \ onhltlOns for vectors nand H atconditions will be obtai w~ .omoge!IOOIlS magnetics. Thesewith tlte Iw.lp of the C' n~~ 'Illist as III the case of uielectricscnlatio/l Wo' l'ol'"lll 11 "ltllf!:iS t leOl'ern ctlt.d the theorem on cir~

. ,. lil or vocto .. B I Hhave llw form .' I:> alit these theorems

1,BdS'O, ~Hdl~l. (7.19)

Condition for V('dol' B I· t " ..located dt tlte il t >'f" I ,f'. liS I III [I g'1ne [I very low cylinderJ et ,ICC Jt'tw('nll (wl)"lllaglletics as is shown

l' = 'X/'1'1 " ,1" . (7.17). us Ie atlOllslup lJetweell tl ,. ' '

any contour inside the llHlgll~~·Clll~ClltsI. and I is valid for~~nall ~ontour, when l' :-+ dI' ~C:, III p.artlcular, for a. veryIh01l l' d8 ~ . dS' ln d8 and I -+dI - ] d8", r: ,,- ~ Xln ,and aftet' cancellinO' dS n ..L~ __ XllI.IJus eqnation holds f). t::' .we ohtamsmall contour, Le. for an' d" (.t any orwntatIOn of thelIcllce the vectors i" au/' t~rect~oJIl,of the normal n to it.tile same equation: J temse \ es are related through

j' = xi.. (7.18)

'I'll liS it [olio ws that' Iif J' 0 P E III a lOtllogcucous maguetie J" ce_' 0, -(. '.D.

sure that volume magnetizat'and only surface magnetizat:~~ currents are e9

ualto zero,

In order to prove this we . currents remam.c,II1atioll of vector J ar~ d shall use the theorem on cir-pletely lying inside the ~~gn:~itr~itrary contour r comgeneous magnetic we can : ~ the case of a homoill Eq. (7.5) out ~f the i~teSgurbaSltItu~/ng .XH

for J, take 'Xanu WrIte .

182

t85

(7.23)tana2 ---~

tan a l III

7,/;, Firld in (7 Ilomo~eneous Magnetic

7.5. Field ill a Homogeneous ~.lgneIic

It was mentioned in Sec. 7.1 thllt determination of theresultant magnetic field in the pl'esence of arhitral'y magnetics is generally a complicated problem. Indeed, for thispurpose, in Ilccordance with (7.1), the fIeld Bo of conductioncurrents mnst be supplemented by the macroscopic fwld B'createfl by magnetization c,ul'rents. The prohlem is that wedo not know beforehllnrl the configuration of magnetizationcurrents. We can only state that the distribution of thesecnrrents depends on 'the nat1ll'e and configllration of themagnetic IlS well as on the configuration of the externllllield Bo, viz. the field of conduction C1ll'rents. And since wedo not know the distribution of magnetization currents, wecallnot calculate the fIeld B'.

The only exception is the case when the entire spaceoccnpiell by the fwld B is fdled by a homogeneolls isotropicdielectric. Let liS (",ollsider this ease in ~rp:ltel' dptail. Blltfirst of all, we shall lIlIalyse the phenomenll obsPI'vpd whellCOlHlllclion Cllrrent llows along a hOlllogenpolls conduetol'

Fignre 7.10 depicts the fields of vector's B HIIII H near theillterface between two mllgnctics (in the absellce of condut:tion currents). Here ~12 > ~ll' A compllrison of the densities of the lines shows that lJ 2 >R1• while H2 < HI'The lines of B do not hllve a ~liscontinuity upon U transitionthrough the interface, while the H lines do (due to the surfare magnetizlltion clIrrents).

The refraction of magnetic field lines is used in magnetic protection,If, for example, a closed iron shell (layer) is introduced into an external ma~n~tic tield. the lield lim's will be concentrated (condensed)mainly on th~ shell itself. Inside th(' shell (in the cavity) the ma~netic

field turns out to!d>e considerably weakened in comparison wi th the~xtern:ll lield. In other words, the iron shell ael;; like a screen. Thisis used in order to protect sensitive devices from ('xtt'rnal magneticfields. ,

:Taking these reilltio/ls into Ilecount. we obtllin the lllw of~refraction of the lines of vector B (Ilnd hence of vector H". weB) similllr to (2.25):

(7.22)

at the intcrfarecomponents B

n

7. Magnetic Field in a SUbstance'184

1~2T=HIT·1Thus, if there is no d .

between two homo COli lICt,lOn Cllnentgeneons magnetics, the

B2Tr---fI a2

21

i.e. the tangential co -a discontinuity u on mpo~ent. ?f vector H general1which is due to thPe prea tIansltlOn through the l'nteYI'f has

If } sence of cd' . Ilee,, IOwever, there are no o~ uctlOn Cllfl'ents.

f~ce between magnetics (i ~O~d) uctthlon currents Ilt the inter-o vector H turns 0 t. - , e tangentilll cointerface' u to be the same on botl 'd mponeJlt

. I Sl es of the

BIT

Fig. 7.9 FieJdB FieJdH

Fig. 7.10Ilnd H Vll .

T ry contlJl lIOllSI ( . I .1hrough this interface. 6n ~~tel~~~ Il Jump) upon tl tl'ullsition

TItnhd , Hn in this case have d' er ~lln~1 '. the components

. s ould be noted that IsContlnlllttes.1I~ ;ect~r D, while vect::ci!r

b~ behav~s at the interface

e raellon of Lines of B T . e laves lIke E.at the interface between 'twhe lInes of. B undergo refractionthe case of dielectrics W h fl mllgnetlCs (Fig. 7 9) A. .of angles (Xl and (X2=' e s Il find the ratio of th~ t~ng:IJ~~

tan a B IB----.!. - 2T 2ntanal -~/B •W I I IT In

e ~ la I confine ourselves to~u~~)on .cllrre!!t at the interf:~: c~e w~e,n there is no con-

• ,lJI thiS Cllse we lillve . CCOr IJIg to (7.22) and

B2T/1l2 = BfT/1l1 and B B2n= In'

I I

1,1

I tically coincide (the wiresof the conduction current pr~c h . d tion B' of .the.) . d I '''ce at all pomts t e In HC

are thm ,an le...... ".1 s differs from the indue-ld f tl e magnetIzatIOll CII 1 " •fI~ BOfl ·tl field of the conuuction currents only In magtIOn 0 ° 10 1 d' the same way as the·t· de the<:e vectors being 1'0 ate Innl u., . .. .corresponding currents, VIZ. (7.24)

B' =--0 XBo.

187

(i.25)

Bo -i' B' =

7.5. Field in a Homogeneous Magnetic

n ,--~ fJB o.

I ti 'e space is filled with a homoThis means that. when. t Ie en SI . times. In other words, thegeneous magnotlc, B Increase .tJ the magnetic iielrl B upon

't -lows the increase In ..quantI y ft S. I. . d 1y the f18ld with a magnetlc.fIlling the entll'o spaI· ce OCCt·\lP~e (7 )25) by tJ-f.\. we obtain

If we divide bot 1 par so. 0' f. ')'

H -- H \7.",6)"l - 0

. . f ld II tUfllS out to be(in the case :lllder 50nsHleratlOn, Ie . •

the same as 1II a vacllum). I valid when a homogeneousForm. \lIas (7.24)-(~.26) lare abso d d by thp s~rfaces formed

t' fill' the entlre /..'0 ume oun e v. . ') Imagne lC. l S . (I field of the conduction current. .nby the lmes of B o t le .. . d tio·n B inside the magneticthis case too the magnetic 1Il uc. t' larger than Bo· . 1 . 13'IS tJ- IIDes. . I 1. hove the maanetie IIll uctlOn

In the cases ~orrsl( e:~~ .at · I ~lIrrents '\s connected withof the .jiel~l ofJ ~a~:~e :;~~~~iC through the simple relationmagnetlzatlOlI 0 ., (7.27)

B = 1l0J.

. b en<:ilv obtained from the formulaThis c:xpresslOll can 0 L~. J l' t H' -- XB

and-- B'-~- B' if we take into account t Ja _ 0

B--- 0; ., _._ J/.B ~-~ tJ-tJ-oH , where H -- ~. . I . > in other cases the

As has <~.ll'oadY.1 been,~Il~ntlOpY~cl;t'~~O\~;ldforrilult;s (7.24)situatioll IS lilliC I mouo com, .

(7.27) become ,ilwppl~cablei' II" ro))';ider two c;imple exam-Concluding iilO :'('('.11011, ec , . .

ple,.

. I 'd A solenoid having nl amp,'re-Example 1: Field H In a so \~~ltOl: ; ho'lllogene(Jus magnetic of the

turn~ per unl t length is HIler! "

. ()f t.lle resultant field BTheu the inductlOll .=(1 + X) Bo 01'

I··

I

l' = \~ i' dl = ~ .r dl = XIf II dl.

186

in a vacuum. Since each conductor is a magnetic, magnetization currents also flow through it, viz. volume currentsgiven by (7.18) and surface currents. Let us take a contourembracing our current-carrying conductor. In accordancewith the theorem on circulation of vector J (7.5), the algebraic sum of magnetization (volume and surface) currents isequal to zero everywhere since J = 0 at all points of thecontour, i.e. l' = I~ + I~ = O. Hence I~ = -I~, i.e. thevolume and surface magnetization currents are equal inmagnitude and opposite in direction.

Thus, it can be stated that in an ordinary case, when currents flow along sufficiently thin wires, the magnetic fieldin the surrounding space (in a vacuum) is determined onlyby conductioH currents, since magnetization currents c.ompensate each other (except fbI', perhaps, the points lyingvery close to the wire).

Let us now fill the space surrounding the conductor bya homogeneous nonconducting magnetic (for the sake ofdefiniteness, we assume that it is a paramagnetic, X > 0).At the interface between this magnetic and the wire, surfacemagnetization current l' will appear. It can be easily seenthat this CUlTent has the same direction as the eonductionCUlTent I (when X >0).

As a result, we shall have the conduction current 1, thesurface and volume magnetization currents in the conductor(tho' magnetic fIelds of these currents compensate one anotheralld hence can be disregarded), and the surfac.e magnetizatioll current l' on the nonconducting magnetic. For sufficiently thin wires, the magnetic field B in the magnetic will bedetermined as the field of the eUI'l'ont I + I'.

Thus, the problem is reduced to finding the current 1'.For this pnrpose, we surround tho conductor by a contourarranged in the surface layer of the nonconducting magnetic.Let tlte plane of tho contour he perpendicular to the wireax i,~, i.e. to tlte magnetization currents. ThOll, taking (7.7)ant! (7.11) into consideration, we can write

Acc(mlillg" 10 (7.12), it follows that r = If.TrIO C()llji~lIl'ati()IlS of 1111' mag·llotizatioll current l' 'llld

Fig. 7.13

B

H

,,'Fig. 7.12

J

~~_ 7.6. F rf' ')m.:-a::cg"..:n~e:..::.t~is~m~ , 1_8~!)

in the absence of eKtemal magnetic field. Typical ferromagnetics are iron, eobalt, allil many of their alloys. .

Basic Magnetization Curve. A typical featlll'e of felTomagnetics is the complex nonlinear dependence J (H) orB (H). Figure 7.12 shows the magnetization curve for aferromagnetic whose magnetization for H ~= 0 is also zero.This curve is called the Imsic maKnetization curve. Even atcomparatively small values of H, magnetization .I attainssaturation (.I"). Magnetic induction B = [lo (ll + I) alsoincreases with H. After attaining saturation, lJ continuesto grow with II accorcing to the linear law B= ilo ll ++const, where canst = [loIs. Figure 7.13 represents the basicmagnetization cllrve on the lJ-H diagram.

In view of the nonlinear dependence B (II), the permeability [.t for ferromagnetics cannot be denned as a constantclwracterizing tlw magnetic properties of J specific felTomagnetic. However, as before, ills aswmed that j.l= BllloH,bllt here ~l is a function of II (Fig. 7.14), The value i1m:.xof permeability for ferromagnetics may lIe very large. Forexample, for pure iron pmax ~ 5,000. wilde for sHpennalloy

[.tmax = bOO/lOO.it should bl~ lIoted that the concept of pCl"Iueability is

magnetics) and strongly Illilgnetic (fel'l'olllagnetics). It iswell known that in the absence of magnetic lield, paraand diamagnetics are not magnetized and are characterizedby a olle-to-one cOITespol\(lence between magnetization J

and vector H.Ferrumagl/etics are the subst;lllces (solids) that Illay possess

spontaneous magnetization, Le. which arc Illagnetizell even

ra

Fig. 7.11

7. Magnetic Field in a Subs/ane,·188

7.6. FerromagnetismbeFerromagneties. In, Il.wgnetic ,respecl, all suh~tanc(>s call

divided into weakly magnetic (para magnetics and dia-

permeability J..l > 1 F' d th . .the magTletic. . In e magnetic mduction P ot' tJ\,' field in

. According to (6.20), in the b .mduction Bo

in the solenoid is a ie~ce of mag'netJc the magnetict~e entIre space where t.he fi ld e~r'W 0lonI. Since th.e magnetic fillseffects>, th!! magnet.ic induc~ion ~ ers trobffi zero. (we Ignore the edgemus e /.L tImes larger:

. B = flP-onI. (7.28)

In thIs cast', t.he field of v 1. II .', .of the magnetic, i.e. H = J~. or remallls the S,;:!)n as III the absence

The change in field B is caused b th .current~ fl~)wing over the surface of the e appearance of magnetizationmagnetJ~ IU the same direction as thec~nd!Jcholl currents in the solenoidwlDdl.ngw~en /.L > 1. If, however, P-< 1the dl~ctIons of these currents will b~opposIte.h The obtained results are also valid

w en the magnetic has the form of ave!y long rod arranged inside the soleno~d ,so th~t it is parallel to the solenOId saxIs.

Exa"!ple 2. The field of straight.CUneDt III the presenee of a magneticSu~pose that a magnetic fills a longcyl~nder of radius a along whose axisa.g.lven current I flows. The permeablhty of the magnetic 11 > 1 Find thmagnet' 'd t' .... eder aXi~~ In uc IOn B as a function of the distance r from the cylin-

sinc';;~~:;::~t~~ ~~e directly the theorem on circulation of vector Bis very helpful~ t~nc~~::;i::.s ar? ud~now!1' ~n thlis situation, vector IIrents, For a circle of radiusIOn IS he errn

2meH, on y byhconduction cur-

r, we ave nr = I, w ence

B = llP-oH = IJ.lloIl2rrr.Upon a t Tmagnetic i~'d~a~~l IOn Bthroul~ the magnetic-vacuum interface, the

(Fig, 7,11). c Jon ,un Jke H. undergoes a discontinuity

An increase in B insid th . . .tion currents Th . e e m~gn~tIc!s c~used by surJace magnetiza-in the wire o~ h. ese. ~.urrents cOIUClde ,,, d:rect.ion with the current IOn the other \l:ndxl~~is~~esYhtem a.nll hence "amplify" this current.current is liirectl'd' oppos\t:lyt eb cil~~dder the surface magnetizationon field B in th ' .' '. 11 I oes not produce ally effectof both current: cmomagpneent!Ct' OutSIde thhe magnetic, the magnetic fields

. " sa e one anot er.

I

~ ..

II

191

(7.29)

7.6. Ferromagnettsm

These peculiarities of magnetization curves are u.sed in a conven!entpractical method for demagnetizing fei:romagnetlcs. ~ magnetlze.dsample is placed into a coil through whIch an alternatIng cu~ent ISpassed, its amplitude being gradually ~educed to zero. In thIS c~se.the ferromagnetic is subjected to multIple cyclIc reverse J.llagnehzations in which hysteresis loops gradually decrease, contractIng to thepoint where magn~tization is zero. .

Experiments show that a ferromagnetic i~ he~ted uponreverse magnetization. It can be shown that til thiS case theamount of heat Qu liberated per unit volume of the ferro mal?netic is numerically equal to the "area" Sn of the hysteresIsloop: ",'

The val lies of lJ r and IIc for II ifferen t fl'lTomagnetics var.yover oroad ranges. Fot' transformel' steel, the hysteresisloop is narrow (He is small), while for fe.rro.magl.tetics use.dfor manufacturing permanent magnets it IS Wide (He ISlarge). For eXilmple, for alnico alloy, He = ;')0,000 Aimand B T = 0.9 T.

Curie Point. As the temperature increases, the ability ?fferromagnetics to get magnetized becomes weaker, and 111

particular, saturation magnetization becomes lower. At.1lcertain temperature called the Curie point, ferromagnetIcproperties of the material disappear.

At temperatures above the Curie point, a ferromagnetic. becomes a paramagnetic. .

On the Theory of Ferromagnetism. The phYSical nat1ll'~

of ferromagnetism could be explained only with the help ofquantum mechanics. The so-called exchange forces mayappear in crystals under certain condi~ions. These forcesorient parallel to each bther the magnetic moments of electrons. As a result, regions (1-10 !lm in size) of spontaneollsmagnetization, which are called domains, appear il.l c~'ystals.

Within the limits of each domain, a ferromagnetIc IS magnetized to saturation and has a certain magnetic lllomcn t.The directions of these moments are different for rlifferentdomains. Hence, in the absence of external fIeld, the resultantmagnetic moment of the sample is equal to zero,. and thes<'lmple as a whole is macroscopically TlonmagnetlZed.

When an external magnetic field is switche(l on, tllO

Fig. 7.15

__?- Magnetic Field in a Substance

Fig. 7,14

too

applieahlt) only tu till' basic lllili[lIetizatioll curve since, aswill ()I) SI'OWII, Lite n (ll) depcndcncc is nOlltllliq Ill'.

Magnetic Hysteresis. Besides the lIonlinear d~ependenceB (Il) or.J ([j), fCiTnmn!~'neticsalso exh ibit magnetic hysteresis: tltr rrlal iOIl hd,\Vef>11 f] and II or J and II turns out to

be ambiguous and is determined by the history of the ferromagnetic's. magnetization. If we magnetize an initiallynomnaglletlZed ferromagnetic by increasing II from zeroto tho valuo at which saturation sets in (point 1 in Fig. 7.15),and then reduce H from +H1 to --HI' the ma<rnetizationc~lI've will not follow path ]0, hut will take pdth 1 234.If we th?!! c!wnge II from -HI to +H1 the magnetizationclIrve wil] he below, and take path 4561.., The ohtained closed. curve is calletl the hysteresis loop.I hc ma:nmllm hysteresIs loop is ohtained whe!' ,;atlll'atiollis attaincd ilt points] awl 4. If, however, there is no satul:atioJl at extreme points, the hysteresis loops have similarform bnt smaller size, as if inscribed into tbe maximumIt yster'esis loop.

Fig-lin' 7.15 shows that for H =-, 0 magnetization does notYilllisit (poiut.2) and is characterized by the quantity lJ

r(',,/It'd the reSidual induction. It corresponds to the residualII/I/;;mtizatio,., Jr· .The existence of permanent magnets isil.'-'so.claled wl.th thiS residual magnetization. The quantity BYillllShes (polllt .'1) only under the action of the field H",hiell has the dil'Pction opposite to that of the field causingI Ill' lllagiletization. The qnantity Tle is called the coerciveJorce.

1,1

,'I

(2)

Fig. 7.i9Fig, 7.18

Problems

13-01RI

on the conductor axis: 2nrH = I (it is clear from symmetry considerations that the' lines of II lIlust be circles lying in the planes perpendicular to the conductor with the current l). ThiR gives

t ' = (~- i) J/2nr•

• 7.3. Circulation of vector H, A long thin conductor with current I lies in the plane separating the space fIlled with a nonconducting magnetic of permeability ~ from a vacuum. Find the magneticinduction B in the entire space as a function of the distance r fromthe conductor. Bear in mind that the lines of B are circles with centreson the conductor axis.

Solution. The lines of H are clearly also circles. Unlike vector B,vector H undergoes a discontinuity at the vacuum-magnetic interface.We denote by Hand Ho magnetic fIelds in the magnetic and vacuumrespectively. Then, in accordance with the theorem on circultaionof vector H around the contour in the form of a circle of radius r withthe centre on the conductor axis, we have

nrH + nrHo = I. (1)

Besides, B = Bo at the interface, or

~H = lIo,

Solving Eqs. (1) and (2) together, we obtain

II = I H = I t 1I = [!~oI(i-j-ft)nr' Ifo (i+t.t)nr·

radial direction, We shall use the theorem on circulation of magneti~ation vector J, taking as a ,con,tonr a smal.l rec.tangl? w~~se plane ISperpendicular to the magnetIzatIOn current III thiS regIOn. Ihe ~rr~ngement of this contour is shown in Fig. 7. i8, where the crosses llldIc~tethe direction of the surface magnetization current. From the equalItyJl = i'l we get i' = J. ,

Then we can write J = XH, where H can be rletermmed from thecirculation of vector II around the circle of radius r with the centre

-j'

J..:!tlh .

" ~.. }:

"', .

,,.

I

Fig'. 7.17

7, Magnetic Field in a Suhstance

J>roblclllsContlitions at the interface Ion a III (' • II the Vicinity of point A

i1gone Ic-vacuum inferfilcl', thl' magnl'tic inductjon

Fig. 7.16

If12

• 7.1.(Fig. 7,10)

dO/llailiS ol'ipliled a/olll{ II . f' f II ,'. Ie Ie t I{I'OW al tit

t O/lldIlIS. ')/'I('lllpd £lg-aillsl Ihe fipld (' ~XPCllse f!fg'l'Owlh IS ,·p\'P!'sih/(. 1,1 'I': 111 weak i1e]tls, thIs

. s rOllger' the/ds tI ' 1l'l'ol'iPlilalioll or III 'lg'lIl'l,'(' I" , Ie SI/IIII lalleOIlS

I 'lIlOJlIPII S wilhill II I' Ila,ps plaep This pl' '. . IC I'll 1['(' f omaill

I '. oepss IS IITp veJ'si hie I ' I 1Ij',slel'esis alld I'('sidll't1 'll't" I' . . ., W lie I {'xp aills

, , ',.,lIl· lzallOlI.

in the vacuum is B vector B f .to the ,interfan'. Th'e p('rmeab)iIi~~JI~fg'ti1I,1 angle a~ w,ith the normalmag-netIc illrfuction B ill tI " . ,Il nla/:rJletlc IS /I' Find Ull'

. If' m"g'nl'tlc lJl till' vidnity or 'Sullittun. The requirerl quantity , flomt A.

11= V-B~ +/Jl" 1; , (t)

Considerin/:r conditions (7.20) and (7.22) ,'tt.the inll'rfacf', we lind

Bn = Bo cos ao,IJT=-'llflolfT~llItllf{ --'til '- B .

. • 0' t 0" - II 0 sIn a o,~here H 0-, IS the tangential component o'Substituting" these expression~ . t (1) j veefo~ Uo in the vacuum.

~ lJl 0 ,We obtamB~B l·/c 2 '.- oOS a o T ft" sin2 a o,

. • 7.2. Surfacemagnelizationf'urrenl " ,JIlg conductor with current 1 " '. A lOng' thIn currenl-carr.·_vacu ' " 1~ al'ran<~ed perpI'n I' I 'j J. UJn-IIl,lgn('tH' mtprfare IF'" ., 1- ~ 1'1 I .1 JCU 111 .\ to a plaoenetIc IS II. Find the lllleal' eu'r J,.,./ 'f' 11. \I' jlr'rnleabiU I" 01 the malY-t ' , . n'll (ell-If\, i' ()' th f' "lOll current 011 thiE in(.prJ'ac' 's .- ,,' ,', -c. .' (' sur ace magtl:ctiza

rOllduetor. l' <I. a Hll.ctlUlI vi the Ihstallce r fruJIl Hie

,";olt:tiu/I Let n" Hrst ('()/ISIIII'I' the COl j'" . ,.netlZatwlI current It j. 'j ,_ . f .. I 1.,~ratIOn o[ the surfac~ HlCig-

, ' .. (Colt rOIll 1- 19, 7,!, lIwt t1li:s current has tltt'

II

Jl - (J + dJ) I = j;'l dr,

(1 )

Fig. 7.21

",.- ..,::::J

- ..--, ro~,I---~

dr

-~r

Problems195

Fig. 7.20

• h. ds j' is directed toward" uscumvenLion of the c~ntour. In ot.~rth:ofi u're, i.e. volumr JIlagnetizain the region of the rJ~hht con~our~ the le!h-hand system.tion currents lorm Wit vec or J

h th hape of a ring with a narrowi' 7.6. A (Jermanent magn~tdl:sdia~~ter of the ring is d. The ,!?ap

gap between the poles. :rh~ ~j'd . f the field in the gap is n. F111<1width is b, the magnetic wHuctHin/inside the magnet, ignoring theth magnitude" of vectors an< hdi~sipation of 'the field at the edges of t e gap.

. ; ." on circulation of vector II around th.~Solutwn. USIll~ the the~re(r 7 22) and considering that there IS

dashed eirele of dwmeter 19':tno conduction current, we can Wfl e

(nd - b) l/-r; + bB//!o = 0,., ,II onto the direclion of cont?ur

where 1I-r; is the pI:oJe~honkof veet~hat it coincides with the (liredlOncircumvention (which IS ta en so •of vector n in the gap). Hence

bB bBIl-r;= - tJ-o(rtd-b) ;::::: - tJ-ond .

. .' h h direction of vector II inside theThe minus sign mdlcates ~ at t .he \. 't'on of vector B at the same

. t·. I' OpPOSI te to t e ( I rcc I 11magnetic ma erld IS bOll __ 0 as we .point. It should be noted t~wLt'as J 7:10 'be found by formula (7.11),

The magnitude of magnetJzalon

- . I tI d ·t· width. Hencewhere I is the contour hClg It an r I S

dJ '2lLHoj;'''-~ --ar=--~ r.

• .' . ,clor 'f is direeted agai~st t!le norm.a1The minus sIgn wdlcates t~hat 'h. d J tern with the (IIrecllOn of Clfvector n which forms the flght- an sys

7. Magnetic Field in a Substance194

The configurations of ttl(: lields n anti II in this case are shown inl·'ig. 7.1!J. We advise the reader to convince himself that for ~t = 1,we arrive aL the well-known formulas for B and II in a vacuum.

• 7.4. Circulation of vectors Hand J. L\ 'direct current I flowsalong a long homogeneous cylindrical wire with a circular cross sectionof radius R. The wire is made of a paramagnetic with the susceptibility X. Find (1) fIeld B as a function of the distance r from the axisof the wire and (2) magnetization current density j' inside the wire.

Solution. (1) From circulation of vector H around the circle ofradius r with the centre at the wire axis, it follows that

r < R, 211rH = I (rIR)2 (H ex: r),

r> R, 2nrH = I (H ex: l/r).

Figure 7.20 shows the plots of the dependences H (r) and B (r).(2) Let us use the theorem on circulation of magnetization J

arouh'l the circle of radius r (see Fig. 7.20): 211rJ = r, where I' isthe magnetization current embraced by this contour. We find thedifferential of this expression (going over from r to r + dr):

2n d (rJ) = dI'.

Since dI' ~~, j'2nr dr, the above equation can bl' transforml'd asfollows:

j' = .!... +!:!... .r dr

Let us now consider that J = XH = (XI12nR2) r. This gives

j' = xI/nR2.

It can be easily seen that this current has the same direction as theconduction current (unlike the surface magnetization current whichhas the opposite direction).

• 7.5. A long solenoid is filled wi th a nonhomogeneous isotropicparamagnetic whose susceptibility depends only on the distance rfrom the s()lenoid axis as X= ar2 , where a is a constant. The magneticinduction on the solenoid axis is B o• Find (1) magnetization J and (2)magnetization current density j' as functions of r.

Solution. (1) J = XH. In our case, H is independent of r (this directly follows from circulation of vector II around the contour shownin the left part of Fig. 7.21). Hence H = Ho on the solenoid's axis,and we obtain

J = ar2Ho = ar2B o//!o.

(2) From the theorem on circulation of magnetization J aroun dan inflllitc1y narrow contour shown in Fig. 7.21 on the right, it followsthat

I'I

III

I'

13*

Hili .~ 1f' . ._. ,_.._'_'2.:' lid Ie FlCld ill II SII!Jsfanrr Problems 197

taliiug into account (1):

x dx

Solulion. (1) Let, for the sake of defmiteriess, vector B on the axisbe oirecteo upwards and coincide with the X-axis. Then, in accordancewith (6.34), Fx 0= I'm aB/ax, where we took into account that theIDagnetic mompnt Pm is directed as vector B (for paramagnetics) ant!h('nc(' replaced an by ax.

Next, since I'm 0= IV =-XllV andeIB/ax = _2aBoxe- ax', we have

F x = - Axe-2ax2 , (1)

where A = 2aB:! XV/'IllO-Having cal;ulated the derivative

dF,,/dx and equating it to zero, we obtain the following equation for determining xm: 1 - 4ax2 = 0, whence

X m = 1/2 y~. (2)

(2) Substituting (2) into (1), wefind

___ ,toFIII .. /,c Fig. 7.24X- BltI V a'

where we took into account that f1 ~ 1 for paramagnetics.

• 7.9. A long thin cylim{rical rod made of a paramagnetic withthe magnetic susceptibility X and cross-sectional area S is arranged

Fig. 7.25

B/I-Jo

J

Fig. 7.22

II/I'll II-'1'---':"'1'-,/"-n-d- ~ -,I-II

. .The relationship. IlI'tween Vt'r1ors B/IlO, II and J atInside the magnet IS shown in Fig. 7.23.' any point

Fig. 7.23

. • 7.7. i\ Winding contailliw' "or tu' ',. I .In the form of a torlls with the m~1 'II, rflls It "dOUI\( on an 11'011 coreslot of wid Ih b (Firr 7 2?) I' .. t . (te, (lallle er : A narrow transversI'I

.... . w S CII 11\ Ie core \Vh'l t I . .Ilrough Ihe winding, Ihl' magnptic illdu.t: ,I' ~hurrlen .. IS pas...o;edthe permeability of iroll under the ( IOJI 111 e sol IS B. FindsiJ)ation of tIl!' field at the edrres o l'setI CO~II(httIOIIS, neglecting the dis-

b ICso·.Sululio/!. In accordance with tI 't1 .

arollnd the circle of diaml'ter I (It ·FI~orelll.?1lCIrculation of v('Ctor n. (see 'Ig. 7.22), we have

(rrd - !J) HlblIo~' NI,

whl're JI and /10 are the magnitud f II' . .spectivPly. Blsides, the nbsence of /i~lI r ..In tl~on and In the slot n.',that I I Isst!'n 1011 at the edges lll('ans

B 0 B o•

lJsillg th",.:" two I'qll<ItiOiIS ,md tal;i",'Illloll alld ,,«: d, we ohl:1I11 ... illto accollnt that B =

It Jl dBJ!oN l---bB

,. • 7.k. The foree adin« on a g t· TI .I· Ig. 7.24 is used for 1Il('~'l1ri~" (w' I 'j3 tc Ie" It' deVice shown inwhich a small par:JIl1a~~eticgsph~~ I t I{ 1C1p 01 a b:~lanee) a force withpole of magnet .1/. The Ill~g'n r. ~ OJ v~ I1me ~. IS attracted to theI I

' e Ie IllI uctlOn on the axis of the rv ISloe (l'pends 011 Ihe height as B ~ B ,-"x~ . ,., 1',1 estants. Find (1) tlip h,.j ht ~!, whue En aud a arc conso tha I thp allr;lcliVt' fcrrce l:" at ,,:llJch the sphere TIlust be placedeeptibijily of lhe !"ll'alll,w;".t7c l\\~~~mum a.nd (2) the m~g-lletic SIlSIS F

m, ,.,. Ie lIlaXllIllI1ll forcC' ot atlr:lctioll

along the axis of a cllI'!'l'ntearrying coil. One end of the rod is at thecenlre of the (,oil, whl'!'l' IhI' luagnetic fIeld is B, while the olh~r endlies in the regioll when' IhI' lIlagnetic fwld is practically absent. Fint!the force with which the coil acts 011 the rod.

Solutiol/. Let us IIll'ntally isolate all element of the rod of length d;T

(Fi£:,_ 7.25). The force acting o!,! this elemellt is given by

d"~d aBx. ',. Pm an .

~lIflPoSC that vector B on the axis of the roil is directed to the right(see the figure), toward::; positive valu!'s of x. Then Bx = B, an = ax,

8. Relative Nature of Electricand Magnetic Fields

8. f. Electromagnetic Field Ch .So far . arge JAvallance

se ' we have considered ele .parately, without establishi . ctnc and magnetic field~

ng any clear reIat· 1IOn >etween

I ....

them. 'I'll is could be done only because the two fields werestatic. In other cases, however, it is impo~sible.

It will be shown that electric and magnetic fields mustnlways be considered together as a single total electromagneticflCld. In othel' words, it turns out that electric and mag-netiefields are in a certain sense the components of a single physical object which we call the electromagnetic field.

The division of the electromagnetic field into electricand magnetic fields is of relative nature since it depends toa very large extent on the reference system in which thephenomena are considered. The field wldch is constant illone reference frame in the general case is found to vary inanother reference frame.

Let us consider some examples.

A charge moves in an inertial system K at a constant velocity v.In this system, we shall observe both the electric and magnetic fieldsof this ch.u·ge, the two flelds varying with time. If, however, we goover to an iner~ial system K' moving together with the charge, thecharge will be'1it rest in this' system, and we shall observe only theelectric field.

. Suppose two identical chatges move in the system K towards eachother at the same velocity'v. In this system, we shall observe bothelectric and magnetic fields, both these fwlds varying. In this case itis im:rossible to find such a system K' where only one of the lieldswoul be observed.

In the system K, there exists a permanent nonuniform magneticfield (e.g., the field of a fixed permanent magnet). Then in the systemK' moving relative to the system K we ~hall ob~crve a varying magnetic field and, as will be shown below, an ele<;tric fleld as well.

8.1. Electromagnetic Field. Charge Invariance 19!1

Thus, it is clear that in different reference frames differentrelations are observed between electric and magnetic fields.

Before considering the main points of this chapter, viz.the laws of transformation of fields upon a transition fromone reference system to another, we must answer thequestion which is important for further discussion: how doan electric charge q itself and the Gauss theorem for vector Ebehave upon such transitions?

Charge Invariance. At present, there is exhaustive evidence that the total charge of an isolated system does notchange with the change in the motion of c.harge carriers.

This maybe proved by the neutrality of a gas consistingof hydrogen molecules. Electrons in these molecules moveat much higher velocities than protons. Therefore, if the

8. Relative Natllre of El t.ec riC and Magnetic Fields

198

and since dpm = J S dx _- XHS dx, we get

dFx=XHSdx~_ XS Bax -~ dB.Having integrated h

t is equation, we obtain()

XS \'F x = - B dH- XSB2/l/lo C' - - 2/l,J'

Th . B 0e !llmus sign indicates th

rod IS attracted to the cu::e:~tor ~ is dir~cted to the left, i. tharrymg coIl ,.c. e

• 7.10. A small s h .magnetic susceptihT P ere of volume V, made ofcarrying coil from thity ~, was displaced along th par~magnetic withregion wher th e pomt where the m '. e aXIS of a C1lITent-against the emag~~i~gfetic field is practf:~i~cab~~~~t~~ dis B to the

S I . orees accomplished in th' . In the worko utlOn. Let us d' h IS case.

the elementar k Irect t eX-axis alon th .placement of ill(7~rph donbe agaiI!st the ma~net~cafis o~ the coil. Then

ere y dx IS equal to orces Upon the dis-

6A= -F dx-. iJBxx --Pm-dxWhere F . h an ' (1)Th . x IS! e projection of the .

leS~;;~:eSlih~fn~~~:~~s ~hat thr~~~kt;~ ~~~: ~~~~~s~~t? tlle X-axis:a ues of x. Then B = Bon dt e axis is directed tow ISd orce...vn = -ax' h x . an an = a (th' ar s posItIve

i~ ~h~tf~r~~'~otn~1:~I;:tfh:/:'~/:Z~~is~I~tI1?~~d o~Xth; dir~ct~~~X ,we rewrIte Eq. (1)

6A= ~XHV!!!!...d _ XVI. a x - - - H dBntegratmg this equation bet x /l/lo .

. ween Band 0 we hI .(J ,0 ,am

A=_~o j' B2Vrr BdB=-J--

B 2/l/lo .

---~\~

(8,2)

(8.1)

201

t - xvo/c 2t/~·V 1-(00/C)2

8.2. Laws of Transformation for Fields E and B

h X d X'-axes of coordinates are directedwhere it is assumed that t e - all

. ,;t blished in the special theoryof this transformatIon are e,. a}' ted way For this reason,of relativity in a rather comp IC:es ondi~g derivations andwe shall not present h,ere ~het~~r co~tent of these laws, thewill pay mam at~entI~n 0 th m and the application ofcorollaries followm~ rom e : e~ific problems,these laws for solvlJl~ s~;n §uppose we have two inertial

Formulation of the ro em., K nd the system K' moving .systems of reference: the system a 1 't v We know the

. th fi st system at a ve OCI Y 0'relatIve to e r, ld E d B at a certain point in spacemagn~tud~s of th~ fie s K a~hat are the magnitudes of theand tIme m th~ systfm ' o'nt in space and time in the.fields E' and B at t Ie same Pa~e oint in space and timesystem K'? (Recall tl~t t~e s d trme in the two referenceis a point whose COOl' lila es ::e Lorentz transformations~)systems arc related through to this question is

As was mentioned above,. t~Ie answer} . 1 't follows that. h th O'y of relatIvItv from w llC I I I

given III tee I . f fi'elds are expressed by tIethe laws of transformatwn 0

formulas __"'-"---.----:;::----;;-;----:;:;:-----\

IEll = Ell, BII - BI" \

B - [voxEl/cz

E' E.L+[voXBl B' ~.L_=.L = V1-pz' .L = V1-~z .

I i I "k the IOn'fitudinal and trans-Here the s~Jllb()ls II ~ll( J- ~nal com ol~ents of the elec.tricverse (relative to the vectol/ v o~ d I?s t,lle velocity of light

t · c. I \s R - V ,c an C Iand mag-ne IC ne ( , p --- 0' '

in a vacuum (c2 = 1/eof!o)'" th formulas have the form

Written in terms of prOJectIOns, ese •

\E~=EX' B~=B;.,:+(!ORZIC2 \

" E'I- voEz 1'>' ,E'l"" .1 J ' Ju~ 1!1_~Z \

' y 1-(IZ •llz- vOEy/c2

7' .," EZ ',l-l1olJy B'EI.-c. V1-~I' z = V1-~2

8. Relative Nature of Electric and Magnetic Fields200

charge depended on the velocity, the charges of the electrons and protons would not compensate each other and thegas would be charged. However, no charge (to 20 decimalplaces!) has been observed in hydrogen gas.

Another proof can be obtained while observing the heating of a piece of a substance. Since the electron mass isconsidembly smaller than the mass of nuclei, the velocityof electrons upon heating must increase more than that ofnuclei. And if the charge depended on the velocity, the substance would become charged upon heating, Nothing ·ofthis kind '.vas ever observed.

'Further, if the electron charge depended on its velocity,then the total charge of a substance would change as a resultof chemical reactions, since the average velocities of electronsin a substance depend on its chemical composition. Calculations show that even a weak dependance of the chargeon the velocity would result in extremely strong electricfields even in the simplest chemical reactions, But nothingof this kind was observed in this case either.

Finally, the design and operation of all modem particleaccelerators are based on the assumption that a particle'scharge is invariable when its velocity is changed.

Thus, we arrive at the conclusion that the charge of anyparticle is a relativistic invariant quantity and does notdepend on the"particle's velocity or on the choice of thereference system.

Invariance of the Gauss Theorem for Field E. It tlll'llS

out, as follows from the generalization of experimental

results, that the Gauss theorem ~ E dS = qleo is valid

not only for fIXed charges but for moving charges also. I Tl

the latter case, the s1ll'face integral must be taken for a definite instant of tirnl' in a given reference system.

Besides, since different inerti,ll reference systems ilrephysically eq uivalent (ill accordaJICo with 1110' relati vi typrinciple), we can state that the (;aIl8s theorelll is validfor all inertial systern~ of reference.

8.2. Laws of Transformation for Fields E a'nd B

\Vltile going over from olle referellce system to anotller,fields E and B are transformed in a certain .way. The laws

along the vector va, the Y'-axis is parallel to th y_,' I h "to the Z-axis. . e aXIS aUt t e Z-axIs

E' It f~II~~s.from Eqs. (8.1) alld (8.2) that each of the vectors. (I Il I. IS expressed both ill terms of E alld 8. This i~dll eVIdence of Il common nature of "],, t " 1 .r -I I 'I'I· " <cc llC 1l1l1 magnetIcIe I s. a (On separately each of tll"8e fiellls I b I. . ' . ".. taS 110 a so liternealllllg: speaking of eloctl'ic (lUll m'IO'II"tl'C !' I I .. 1'1 (,.., \.' W I S we JIlustHI( I~ate t Ie system of reference in which these '6 Jd'conSIdered. Ie s are

It sh?llld be emphasized that the properties of electromagnetic fwld expressed by the laws of its trllllsformationa~e lo~al; ~he values of E' ,alld 8' at a certain point ill ~pace:td

tire HI ihe system K are uniqnely determined only by. Ietlva IICS 0 E and 8 at the same point in space and timeIII Ie system K.

The following- features of the laws of field t.. 'f' .al'e also . IclllS OJlllatlOn. very Important.

1. Unlike the transverse components of E· I 8 I' Ich'lJlge 0 . t " , .tli( ,w IIC I

( U~.II a .ran~ItlOn hom Olle referollce system to Rn-

°t]thor, theu. longtludmal components do not change and l'emaillIe same HI all systems.

.2. Vectors E and B in different systems are COJilicctedWIth each other through highly symmetric relations TI'.can be. seen most clearly ill the form of the Jaws of 'trm;~~formatlOll ,f~r tlte pro~ections.of the fields (see (8.2)1.

3. In o.Idel' to obtalll the fOJ'llltlla~ ()f tll(" , tt' . . " , In v er<;c 1"1JI s-ormatIOn (from K' to K), it is sufficient to I'eplac~ ill 'fo~'-

/Ill/las (8.1). and (8.2) all pl'imed quantities by llnprimodones (and VIce versa) as well as t.he sian of "

S . I C b "0'peCta ase of Field Transformation (u '~c) If tl _

system K' moves relative to tl1 0 svst"ln K O a"t' . 1 'tIC, '. " C J' " ( , a ve OCI y

Lo «c, the rad lCals III tire denominators of fonn uhs (8 1')call be replaced by IInity, alld we obtain (, .

203

Fig. 8.1

Bt\'

8.2. Laws of Transformation for Fields E and n

It should be noted that while solving this problem, we coulll followanother line of reasoning, viz. from Ihe point of vil'W of the referenceframe in whieh the plate J1loves at the velocity v. III this system therewill be an electric jidd inside the plnte. It appears as a result of theaction of the magnetic cOlllponent of the Lorentz force causing the

will be observed. It is :directl'd towardsliS. Under the action of this externalfield, the charges will be displaced sothat positive charges will appear onthe front surface of the plate and negative charges, on the lJinrl surface.

The surface density a of these charges will be such that the Heldcreated by these charges inside the plate completely con!pe.nsat~s ~he

external field E', since in equilibrium the resultant electrIc held IIIsldethe plate must be equal to zeru. Taking into acconnt relation (1.11),we ubtain '

a = foE' = fovR.

It should be noted that the first formula in (8.1) can hedirectly obtained in a very simple wa.y. Let a charge qin the system K have a velocity vo at a certain instant t.The Lorentz force actin~ on this charge is F = qE I+ q [vo X 81. We go over to the inertial system K', movilll{relative to the system K at the salUe velocity as the charge qat the instant t, viz. the velocity vo' At this instant, thecharge q is at rest in the system K', and the force actiJl~

on the immobile ehurge is purely electric: F' ..co gE'. IfVo «c (as in the case under eonsiderlltioll), this {orec isillvariant (F' = F), whellce we ol)taill the Ihst fOl'lIlula ill(8.4). .

However, the transformation fOl'llIulas fol' magnetic Heldcanbe obtained only with the help of the theory of relativityas a result of rather cllmbersome calculations.

Let us consider a simple example of the application offormulas (8,1).

Example. A.Jal'ge metallic plate moVl'S at a constant nonrelativisticvelocity v in a'unifol'In magnetic Held 1J (Fig. 8.1). Fillll tlIP surfacedensity of charges iIHluced on the surfaces of the plate as a resultofits motion. •

Let us go over to the reference system fixed to the plate. In accordancewith the first of formulas (8.4), in this,system of reference a constant uniform electric field

E' = Iv X Bl

(8.4)

Bir = 8",

B~=B.l-(voxEl/c2. (8.3)

B' = B- [vo X E}/c2• IJ

R. Relative Nature of EI('ctric and Magnetic Ficlds202

Eir=E",

E~ = E.l + (vo X B},

Hence it follows that

IE' ~ E+ (va X Bl,

205

Fig. 8.2

8.2. Laws of Transformation for Fields E and n

. the fields E aud B lUust heto an.other refcrence, sys~em,. '11 the eoneept of the

! . tl 'u .. J:< 01' Iustance, eve .treaLe( Wi i C ie. . " char re IJy a muving magnetIC-force exerted. Ull a mov lU", . ~. The force is deterfield tioes not have any preCise m~a.llln~. and B at the pointmined by the vah~es of the q¥:ntIt~e:esult of motion of thewllere the charge ISElo~~t;\heira~;lue~at this point change,sources of th~ nellis a, well Otherwise the motion of thethe force Will ch,mge as e ~laO'nitude of the lorce. .

,;\' . sources does not a~ect th ob"'lem about the force actlllg~t Thus, while solvmg the EPr 1 B at the point of location,;,,:, I e must know an( 't',\> on a c large, w . 1 . 't v All these quantI lesi'· of the charge as ~~~l a~ 1~~I:~,~e~~i~l ;eference frame we are

'~.~;,..j.•..:.,._::,: ~~I~:~e~~e~t~~I.l re a lve 0 K'I'. The expression "nlOY - K B I

ing field" (if it is lised) ...t -v-o..........should be int~rpreted .as "a convenient way of the • q I

,I • t'OIl of .\verlJal ucscnp I . ~ L _

varying neh! unllor C('l

tain conditions, anll noth-ing more. .

The following silllple tthat upon a transitionexample will illllstraf.t.c thetofa~n~ther the nel(l must befrom one refcrcnc,c [amet1'oatell with care.

.. t between the poles of a mag-Example A charge(l partlde IS at res to the system K' that moves

net lixed in 'the system K. Let us gOt?~~~ic velocity Yo relative to theto the right (Fig. 8.2) at a t~n[~l~harged particle moves i~ the n~arsystem K. (t) Can we stat~~ ,~a(2) I~ind the force acting on thiS partlc enetic tielll in the system .. the system K'. . elic field. It should be notell ,m (t) Yes, the particle ~lUves In the rn~~'\ield and not relative to ~hehowever, that it moves m. the magne • tlwt a particle moves relatIvemagnetic field. It is meaIllngful to rS~iher objects. How:~ver,:~e cannotto the reference system, a magnlti'~e to the magnetic held. lhe ~at~:rstate that a particle I.novles ~e :: I All this refrrs not 'lIlly to magnl' ICstatement has no physlca ml'unmg.but to electric tields as we}l. must take into aceount that t~l

(2) In order to lind the °lrcl ! weted towards us (Fig. 8.2), WIelectric lield E' =- l vD.'>; B ~ ~recstem the charge will move to t ~Po

in the svst"1ll I\. . III thIS sy . ' 'n occur in crossL',ld<'CtrH'appear I '. , lI,l this mottOll WI' th·.t theleft at the ve O~·,!l.i ·_~o~ '\' I of llefmiteness, we suppose •and magnetic ileitis. l' 01 t Ie sa <e

\

204 fi. Rell/tive Nature of Electric and Magnetic Fields

Relativistic Nature of Magnetism. Formulas (8.1) and(8.2) for the transformation of fields lead to a remarkableconclusion: "the appearance of the magnetic field is a purelyrelativistic effect following from the existence in natureof the limiting velocity c equal to the velocity of light ina vacuum.

If this velocity (and hence the velocity of propagation ofinteractions) were infinite, no' magnetism would exist at all.Indeed, let us consider a free electric charge. Only electricfield exists in the system K where this charge is at rest. Consequently, according to (8.1), no magnetic field wouldappear in any other system K' if c tended to infinity. Thisfield appears only due to the fact that c is finite, i.e. in thelong run due to the relativistic effect.

The relativistic nature of magnetism is a universal physicalfact, and its origin is associated with the absence of magnetic charges.

Unlike most of relativistic effects, magnetism can beeasily observed in many cases, for example, the magneticfield of a current-carrying conductor. The reason behindsuch favourable circumstances is that the magnetic fieldmay be created by a very large number of moving electriccharges under the conditions of the almost complete vanishing of the electric field due to practically ideal balance ofthe numbers of electrons and protons in cond Hctors. I TJ

this case the magnetic interaction is predominant.The almost complete compensation of electric charges

made it possible for physicists to study relativistic effects(Le. magnetism) and discover correct laws. For this reaSOJl,the laws' of electromagnetism, unlike Newton's laws, havenot<been refined after the creation of the theory of relntivity.

Field Varies Rather than Moves. Since the relations hetwoon electric and magnetir. aelds vary upon a transition

displacement of all the electrons in the plate behind the plane ofFi;t. 8.1. As a result, the front surface of the plate will be char~edpositively and the hind surface, negatively, and the electric fieldappearing- in the plate will be such that the electric force qE compensates the magnetic component of the Lorentz force, q [v X Bl, whenceE = -[ y X BI. This field is connected with the surface charge densitythrough the same formula lJ = BovB., Both approaches to the· solution of the given problem are equallyjustified. .

v

207

(8.7)

(8.8)

Fig. 8.3

8.9. Field Transformation Laws. Corollaries

tions can be easily applied to the fields varying in space andtime. An illustrative example is the field of a uniformlymoving point charge.

Field of Freely Moving Relativistic Charge. The formulasfor field transformation are interesting above all by thatthey express remarkable properties of electromagnetic field.Besides, they are important from apurely practical point of view sincethey allow us to solve some problemsin a simpler way.

For example, the problem of determining the field of a uniformly moving point charge can be solved. ,as aresult of transformation of a purelyCOUlomb field which is observed in thereference system fixed to the charge.Calculations sh9w (see Problem 8.10)that the lines 61 the field E created bya freely moving charge q have the formshown in Fig. 8.3, where'v is the velocity of the charge. This pattern corresponds to the instant "picture" of the electric field configuration. At an arbitrary point P of the reference system, vector E is directed along the radius vector r drawn from thepoint of location of the charge q to point P.

The magnitude of vector E is determined by the formula1 q 1_~2

E "''' -4:rt-e-o -r-2 7":(1:-_--;;-~2;;-s-;i.:..n;;-2n{j-";;;)3'";;;/2'

where ~ = vIc, it is the angle between the radius vector rand the vector v corresponding to the velocity of the charge.

The electric field is "flattened" in the direction of motionof the charge (see Fig. 8.3), the degree of flattening beingthe larger the closer the velocity vof the diarge to the velocity c of light. It should be also borne in mind that the fieldshoWJl in the figure "moves" together with the charge. Thatis why the field E in the reference frame relative to whichthe charge moves varies with time.

Knowing the field E, we can find the field B in the samereference system:

1, 110 q[vxr] 1-~2

B =~ [v< El = 4it ,.a (1- ~2 sin2 {j-)3/2 •

8. Relative Nature of Electric and lUagnetic Fields206

8.3. Corollaries of the Laws of Field Transformation

. Some Simple Corollaries. Several simple and useful relatIOns follow in some cases from transformation formulas(8.1).

'. If the ~lectric field E alone is present in the system K(the magnetIc field B = 0), the following relation existshetween the fields E' and B' in the system K':

B' = -[Yo X E'lIc2 • (8.5)

Indeed, if B = 0, then E~= El.lV 1 - ~2 and B' = °B~ c=. -[Yo X El/c2 V1 - ~~ c-= -[vo X 'E'1/c2 , wh;re w~t~ok mto account that in the cross product we can writeelthe~ E .01' El. (the ,same applies to the primed quantities).Consldermg that B = Bj, + B~ = B~, we arrive at formula (8.5).

2•. If the magnetic fwld B alone is present in the system K(wIllIe tIte electric field E = 0), then for the system K'we have

F' =qE' +q [-vo X B]=q [vo X B]-q[vo X B]=O

which, h.~~v~,:er, dire~tly fo,nows from the invariance of the force uponnonrelatr ,Jst1C transformatIOns from one reference frame to the other.

~harge of the particle is q > O. Then the Lorentz force in the system K'IS

E' = [Yo X B'l, (8.6)

In ieed, if E = 0, then B~ = Bl.IV1- ~2 and Ell = 0Ef. = [Yo, x. BlIV 1 - V Suhstituting Bl. for B and the~rs.6~~r BJ. m the last cross product, we arJ'ive at formula

Form~Ilas (8.5) and (8.6) lead to the following importantconclusIOn: If only one at the fields (E or B) is present in thesystem K, the electric and magnetic fields in the system K'are mutually p,erpendicular (E' -L B'). It should be notedt~a.t the ~p~oslte statement is true only under certain addItiOnal lImitations imposed on the magnitudes of the vectors E and B.

Finally, since Eqs. (8.5) and (8.6) contain only the quantities referring to the same system of reference, these equa-

I I

Problems209

. . 'f E - 0 l' B = 0 in some reference system,opposite IS Ba~s? true. Ither s-yste~ (This conclusion has alreadY heenthen E'..l III any 0 .

drA~ifinS:llY 8~~;~ more remark: it should be borne in.mind

h II finilis E and B depend both on coordlllatest at genera Y ". c

I . .... t f (8 \J) refersd on time COllsequently, eac l lIlvallHl 0 . d'an . '. 'n s ace and time of the field, the COOl' 1-

~::~eas:::i~~~~~he ~oint in different systems of .referencebeing connected through the Lorentz transformations.

Problems

. Iff Id transformation. A nonrelativistic• 8.1. Specla case 0 n I~ant velocity v. Using the transforma

point charge q mo;el. at a co t field B of this charge at a point whose. tion formulas,. fin t e

hma~ne rc. defined hy the radius vector r.

position relatIve to tee arge ISt the s stem of reference K' fixed to the

Solution'h~etus go ov1r ~he coJlomh fIeld with the intensitycharge. In t IS system, on Y

1 qE'=--, r,

41tEo r

is resent where we took into account that the radius :~~tof~o~~~h~in ~he sy~tem K' (nonrelativistic case). Leltt~s ntowth:\ystem K' at

K' t th stem K that moves re a Ive 0 hsystem 0 e sy. we shall use the formula for t I'

}.~i/~lf~~~(~;), ir~~v~?~(rl fi:~~l:~:r~~~m~~JI~:~g~i:Sth;I~~I~~Ai~~by unprimed quantIties an vice '.d' ~ v and henceb -Vo (Fig. 8.4). In the case under conSI ~ratJOn, vo ~ 'K' B' = 0l = B' + [v X E'l/c2. Considering that III the systemand that c2 = 1/Eoflo, we fInd

110 q [v X rJB = 4Jt r3 •

We have obtained formula (6.3) which was earlier postulated as1· t' of experimental data.a result of the genera lza JOn

14-0181

v

Fj~ 8.5

~ B

.f(11c"1

-- I-vo I

1 ~.

----q

Fig. 8.01

K'I

r~oI rI----

q

K8.4. ElectYomagnetic Field Invariants

Since vectors E and B characterizing electromagnetic fielddepend on the system' of reference (at the same point inspace and time), a natural qu~stion arises concerning theinvariants of electromagnetic field, i.e. the quantitativecharacteristics independent of the reference system.

It can be shown that there exist two such invariants whichare the combinations of vectors E and B, viz.

jE.B=inv. and E2-c2B2=inv.1 (8.9)

The invariance of these quantities (relative to the Lorentztransformations) is a consequence of field transformationformulas (8.1) or (8.2). This question is considered in greaterdetail in Problem 8.9.

The application of these invariants allows us in somecases to find the solution quickly and simply and make thecorresponding conclusions and predictions. We shall listthe most important· conclusions.

L The invariance of E'B immediately implies that if E ..1 Bin some reference system, Le. E· B = 0, in all other inertial systemsE' ..1 B' as well.

2. It follows from the invariance of E2 - c2B2 that if E = cB(Le. E2 - c2B2 = 0), in any other inertial system E' = cB' as well.

3. If in a certain reference system the angle betwl'cn vectors Eand B is acute (ohtuse) (this means that E· B is greater (Ies:;) than zero),in any other reference system the ang-Ie between vectors E' and H'will he acute (ohtus(') as well.

4. If in a certain reference system F > cB (or E < cB). (whichmeans that, E2 - r2B2 is greater (or l!!ss) than zero), in any otherreference system E' > cB' (or E' <; clJ 0) as well.

5. If hoth invariants are equal to zero, then E 1. R ana E = cBin all rf'ference systems. It will he shown later that this is preciselyohserved in electromagnetic waves.

6. If the invariant I>B alorw is equ..1 to zero, we can fmd a reference system in which either E' = 0 or B' = O. The sign vI ihe otherinvariant determines which of these vectors is equal to zero. The

This formula is a corollary of relation (8.5) iu which we havereplaced the primed quantities by unprimed ones and vby -v.

When v ~ c (~ «1), expressions (8.7) and (8.8) are transformed to (1.2) and (6.3) respectively.

208 8. Relative Nature of Electric and Magnetic Fields

E~ = -1'0!-toII2nr (1)

where r is the distance from the wire axis.The expression for B in this formula was obtained with the help

of the theorem on circulation.On the other hand, according to the Gauss theorem (in the moving

system) we have

C'

2ftProblem•

H' = B-Iv X EI/e2•

l" 1_~2

The arrangement of vectors is shown in l<'ig. 8.6, from which it followsthat lv X EJ tt B. Consequently, considering that according to (1)

14*

induction B' of the llIagnetic lield at the same distance frolll the beamin the system K' that moves relative to the system K at the velocity Voin the direction of the proton beam.

Solution. This problem can be solved in the most simple way withthe help of formulas (8.1). But first we must find the induction B inthe syst.em K at the same distance from the beam where the intensity Eis given.

Using the theorem on circulation of vector B and the Gauss theorem for vector E, we find

R = !JIlI/2rrr, F A/2rtfor,

where r is the distance from the beam, I = A.V is the current, alI,1 'Ais the charge per unit length of the heam. It follows from these formulasthat

BIA' = fO!-tOJ/I.. = v/c2,

where c2 = lleo!-to. Substituting the expression for B fmlll this formula into the last of transformation formulas (8.1), we obtain

B' = 1~'lv-vol

.. ,,' c2V1- (vo/C)2

If in this case Vo < u, the Jines of B' form the right-handed systemwith the vector vo. If va > v, they form the left-handed system (sincethe current ]' in the system K' in this case will flow in the oppositedirection).

• 8.5. A relativistic charged particle moves in the space occupiedby. uniform mutually perpendicular electric and magnetic Helds Eand B. The particle moves rectilinearly in the direction perpendicularto the vectors E and B. Find E' and B' in the reference system movingtranslationally with the particle.

Solution. It follows from the description of motion of the particlethat its velocity must satisfy the condition

vB = E. (1)

In accordance with the transformation formulas (B.l), we have

E'= E+[vxBI =0Vl-~2 '

since in the case under consideration the Lorentz force (and hencethe quantHy E + [v X B) is equal to zero.

According to the same transformation formulas, the magneticfield is given hy

8. Relative Nature 0/ Electric and Magnetic Fields210

E~ .= [v X B).The polarization of the diP!ectrie is gin'lI by

, e-1P= xeoE =. eo-- [v X HI

e '

where we to~k into account that according to (3.29), E' = Eolt:. Thesurface denSIty of hound charges is .

, e-1Icr I =P=lIo---vB.

e

At the front surface of the plate (Fig. 8.5) cr' > 0 while on the oppositface cr' < o. " e

• 8.3. S,~ppose we have an uncharged long straight wire witha current I. ~ Illd the char~e per uni ~ length of this wire in the referencesystem moving translatlOnally WIth a nonrelativistic velocity valong the conductor in the direction of the current I. 0

Soluti?n . . In aC,cordance with the transformation formulas (B.4),the electnc held E = [va X B) will appear in the moving referencesystem, or

•. 8..2.. A large plate made of homogeneous dielectric with thepermlttlvltY.F 1I.lOves at a cOllstant nonrelativistic velocity v in a uniform m~gneLJc. held H as shown in Fig. 8.5. Find the polarization Pof the 'dlelectnc and the surface density cr' of hound charges.

SolUtion,. In the refereIl;ce system fixe~ to the plate, in addition tothe ~agne~lc field, there .wIlI he an elec,trlc field which we shall denote

hbY Eo· In ,lccord:lllce with formulas (8.4) of field transformation we

ave '

E; = 'A'/2m:or, (2)where 1..' is t~le charge per ulli t length of the wire.

A companson of (1) and (2) gives

'A' = -vOIlc2,

,where c2 = Ih0 !:l0' The o,rigin of this charge is associated with the

~orentz contractIOn expeflenced by the "chains" of positive and negative charges (they have different velocities!).

. • 8.~. Th?re is a narrow beam of protons mOVing with a relativistiC velOCity v II.! the ~ystem of reference K. At a certain distance fromthe beam, the IlltenSlty of the electric field is equal to E. Find the

213Problems

This equation is written for the instant t = 0, when the particlemoved in the system K' as is shown in Fig. 8.8. Since the Lorentzforce F is always perpendicular to thE: velocity of the particle, vo == const, and it follows from (1) that in the system K' the particlewill move in a circle of radius

R = mvo/qB.

, Thus, the particle moves uniformly with the velocity vo in a circlein the system K' which, in turn, moves uniformly to the right withthe same velocity vo = EIB. This motion can be compared with themotion of the point q at the rim of a wheel (Fig. 8.9) rolling with theangular velocity w = voiR = qBlm.

Figure 8.9 readily shows that the coordinates of the particle qat the instant t are given by

x=vot-R sinwI=a (wI-sin WI),

y=R-Rcoswt=a (1-cos wt),

where a = mElqB2 and w = qBlm.

• 8.7. There is only a uniform electric field E in an inertial systern K. Find the magnitudes and directions of vectors E' and B' inthe system K' II14Yving relative to the system K at a constant relativisticvelocity vo at an angle a to vector E.

Solution. According to transformation formulas (8.1) and takinginto account that B = 0 in the system K, we obtain

E'n=Ecosa, E~=Esinan/i-lit, Ii=volc.

Hence the magnitude of vector E' is given by

R'-~VE~~ I-Ei-El/(1-Bt cos' a)/(1-1i')

and the angle a' between vectors E' and vo can be determined fromthe formula

tan a' = E~IEll = tan al Vi-lit.

The magnitude and direction of vector B' can be found in 11 similarway: -

B"=O, B~=-[voXEJ/(c'V1-1i'), B'=B~.

This means that B' .l vo, andB' = voE sin al(c' V i-Ii')·

, • 8.8. Uniform electric and magnetic fields E and B of the samedirection exist in a system of 'reference K. Find the magnitudes ofvectors E' and B' and the angle between these vectors in the system K'moving at a constant relativistic velocity Vo in the direction perpendicular to vectors E and B.

Solution. In accordance with formulas (8.1), in the syste lll Fe(1)

x

E

y

B q

Z

Fig. 8.6

8, Relative Nature 0/ Electric and Magnetic Fields212

or in the vector form

B' = B-E'IBe'

Y1-1i'

v = EIB, we can write

B'=BV1-(ElcB)'.

We advise the reader to ver'f th t th b .both field invariants. I y a e 0 tamed expressions satisfy

~ ~.~. The lIIotion of a charge ill crossed field ].' drelatIvIstIc particle with a specific charge qlrn movessI'n' an ~. Anlona regIOn w Iere

Fig. 8.7

u~iform, mutually perpendicular fields E. and B have been created(FIg..8.7). At the moment t = 0 the partlcle was at point 0 d 'tvelocIty was equal to zero. Find the law of motion of the pant' II sx (t) and y (t). ,arIC e,

Solutton. The particle mov~s under the action of the Lorentz f~t can b.e easily seen tha.t the particle always remains in the plane~~.tSI

motIOn can b~ descn~ed most easily in a certain system K' h .on y the magnetIc field IS present. Let us find this refe n ,were

It follows from transformations (84) that E' _ 0 .re ce sfystem.t th t

. h .' - mare erencesys em - a mov~s WIt a velocIty Vo satisfying the reI t' E = -!vo XB!. It .IS more convenient to choose the s stem

a~~ h

ve~ocIty Vo. IS ~hrected towards the positive value/on the ;a~~e(FIg. 8.7), smce m such a system the particle will move perpendicula 1sto vector B' and its motion will be the simplest r yv 2h;iB·i~h t~e l~~~er:: K' that moves to the. right at the velocitya~ce with (8e4)e

ad F-=- 0 8an7d onlyhthe field B' IS observed. In accord-

. n Ig. ., we ave -

B' = B - [yO X E]/c2 = B (1 - v3Ic2).

Since for a nonrelat~vistic par~icle vo ~ c, we can assume that B' = B, In th~ syst~m K , the partlcle will move only in the rna net' fi I .Its ve~ocIty ~em!!, perpendicular to this field. The eqUatio~ of I~ :- d,for thIS partIcle m the system K' will have the form 0 IOn

rnv~/R = qvoB.

215

Fig. 8.11

Problem.

Fig. 8.10

Find the intensity E of the fle'ld created by this charge at a point whoseradius vector relative to the charge is equal to r and forms an angle ~with the vector v.

Solution. Suppose that the charge moves in the positive directionof the X-axis in the system of reference K. Let us go over to the system K' at whose origin this charge is at rest (the axes X' and X ofthese systems coincide while the 'Y'- and Y-axes are parallel). In 'thesystem K' the field E' of the charge has a simpler form:

E' 1 q ,= 4n8o ---,:;r r ,

and in the X'Y' plane we have

E~ = 4~80 ;'3 x', E~ ='= 4~80 :' y'. (1)

Let us now perform the reverse transition to the initial system K,which moves relative to the system K' with the velocity -v. At them<tment when the charge passes through the origin. of the sys.tem K,the projections x and y of vector r are connected With the proJections;1;' and y' of vector r' through the following relations:

x=r cos 1'l'=x'Y1-Wa y=r sin {}=y' (2)

where f\ = vic. Here we took into account that the longitudin.al diIJ?-ensions undergo the Lorentz contraction, while the transverse.dlm~nslonsdo not change. Besides, in accordance with the transformatiOns lDvetse·

Q.E.D.• 8.tO. Field E of a uniformly moving charge. A point charge q

moves uniformly and rectilinearly with a relativistic veloaity v.

where we used the fact that [YO X BJ.Hvo X E.L] = v~J.E.L XX cos a = v3B.L ·E.L (Fig. 8.11). The remaining two scalar productsin (2) are equal to zero since the vectors are mutually perpendicular.

1'hus, the right-hand side of (1) becomes

Erl·B{, +EJ. .BJ. =EII·B" +EJ. .B.L =E·B

Fil{. 8.!!

B' = .. ;- BI+ (voElcl)SV 1-(volc)'

and B' can be found from its tan-

8. Relative Nature 0/ Electric and Magnetic Fields

Fig. X.X

+y'_I-,

"/ I i)D

l~lli--_Vo 0 X'

21~

opposite is also tru!': if we know E and 8 in one reference system, andthe angle between these vectors is less than 90°, there exist referencesystems where the two vectors E' and B' are parallel to one anotheL

. • tl:9. Invaria~t E: B.. Using transformation formulas (8.1)show thaI the quantIty E· n IS all invariant.

Solution. In the system K', this product will be

E' B' (E' tE' , , " , ,. = II- .lHBII+BJ.)=EII·BII+EJ..BJ.' (1)

Let us write the last term with the help of formulas (8.1):

EJ. .BJ. = (EJ. + [Yo X BJ. I)· (BJ. - [Yo X EJ. I/e l )

1-~1 (2)

Considerf'hg that the vectors E J. and BJ. are perpendicular to thevector Yo, we transform the numerator of (2) to the form

E.l.B.l-(vo!e)2E.l.Bl. =E.l·B.l (1_~2), (3)

the tWo vectors E' and D' will be also perpendicular to the vector v 0(Fig. 8.10). The magnitudes of the vectors E' and B' can be found bythe form1l\as .

E' =" ;- EI + (VOB)IV i-(volc)I'

The angle between the vectors E'gent through the formula

Ian a' = tan (as+aB)= (tan as -Han aB)/(1- tan ai: tan ail).

Since tan ai: = voBIE and tan aB = voElelB (Fig. 8.10), we obtain

tan '"-= Vo (BI + Ellcl )a (1-~1) EB

This formula shows that as vo -+ e (~ -+ 1), the angle a' --+ n12. The

-- "'ll

217

Induction

9.1. Faraday's Law. Lenz's Law

9. Electromagnetic

d ' Law of Electromagnetic Induction.9.1. Fara ay S Lenz's Law .

"t R8 established that the relationIn the previous chapter,~ w

f. 'electromagnetic field, viz.

between the "components 0 ~n

I t · and tu a f7 netic fielrls, ISe ec nc ..-.• '": bv the Cmainly determmed .choice of the reference sy~tem. In

I . ds the two componentsot leI' wor ., I in .of electromRgnetic fie dare -t lated. We shall show hereerre". '11 ]osel'that there eXIsts a stl r, I. . hetween the E- alHconneetlon ., I'B-fields which is exhlllltC( I.n

, f I ctroma f7 netlchenomena 0 e e ,..,

induction.. I 1831 Fig. 9.1Faraday's DIscovery. n

I onc of the most . tl c pheno-FRraday mal e. . ,,' ,. '11 ('1('('lrod~'lIamlcs-, I,. h tfl1l((lam'~lIlal ,1Iscov( IICSt, I . duction. It eonsists m t amenon of electromaKne lC {:~ induced current) appe~rs 1I1

an electric Cl.lrrent (cRlled t hange in the maKnetlc fluxd t ' l op 11pon a c .a closed conl.lc tnR 0 ,tJ· loop

(Le. the nux of B) enclose~ ;y ce~:-~e\lrre~t indicates that theTile appearance of the ml U . . tile loop as a result' . f r· appearsmitinduced electromotwe orce. 'n It is important here t l~

of a change of the mRgne~~ h~~'the ('lInnge in the magl\etl~'{. rloes not at Rl! rleprnd . . I ()lllv by the rate o!i' . I l' dctel'llll1l(,(. .

Oil " {f-. was realIzc( alH IS n'rl '1 CII'III f7e ill the SigH-, I'.... d<J>/dt. ueSl es" . < "its vanatlOll, I.e. by ,

11 It should be noted that the ratio= vElc2, e2 = follo· I 2

F IF. = Follovl = (v e) ,. ~ . e (6 5) It can be seen that as v ---+ C,h relatiVIstIc case .. I F'just as in tenon f h force F ten( s to e' b

the magnetic componentfo. ~ eaction (~pulsive force) is given yThe resultant force 0 III er

. _ _ 1_!fl_ yl_1\2.l=Fe-·Fn1 - 4neo 12 .

(i)

(2)

... v

IIq ••--...~vL _

K

E:x:= E~, Ey=E~1 Vl-~I.

Substituting into these expressions (1) where x' and y' are replacedby the corresponding expressions from formulas (2), we get

E%=_l_ L x E __l q_ II

4neo r'a Vl-~2' 1/- 4neo ria J/I-1\2'

It should be noted that E:x:IEII= xly, Le. vector E is directedradially along the vector r. The situation is such as if the effect of lagwere absent altogether. But this is true only when v = const. If,however, the c~arge moves with an acceleration, the field E is no longer

radial.I t remains for us to find the magni

tude of vector E:

-V~ - _1_...!L / zS+ylE- Ex, Ey - 4 'S J 1 pl.neo r _

Since x 2 + y2 = r2 and

to (8.2), we obtain

Fig. 8.12

216 8. Relative Nature of Electric and Magnetic Field.

= a ( 1- ~2 sin2 {t ) 3/2r 1- ~I ,

in accordance with (2), the electric field intensity will be given by

R- _1_..!L. 1-~2• -- 4np.o r2 (1-1\2 sin2 ,'})3/2

• 8.11. Interaction between two moving charges. The relativisticparticles having the same charge q raove parallel to each other at thesame velocity v as is shown in Fig. 8.12. The separation between theparticles is [. Using expression (8.7), find the force of interaction between the particles.

Solution: In this case, the angle between the vector v of one of theparticles and the direction towarrls the other particle is {t = 900.Hence, in accordance with formula (8.7), the electric component ofthe Lorentz force is given by

1 q2Fe=qE===- 4nEo II Vi-I\I

while the magnetic component of the Lorentz force is

It ql v2Fm=qvB= _0_ ,

4n [I Vi-~I

where we took into account that in the case under consideration Bis related to E through formula (8.5) from which it follows that B =

219

(n.2)

(9.3)'L'. = _ )\T d<D iC, l' dt .

9,1. Faraday', lAw. Len.', Law

'" If these two directions were connected through the left-handscrew rule, Eq. (9.1) would not contain the minus si~n.

This quantity is called the total magnetic flux, or themagnetic-flux linkage. In this case, the e.m.f. induced inthe loop is defined, in accordance with (9.1), by the formula

regardless of the cause of the variation of the magnetic fluxembraced by the closed conducting loop. The minus signin this equation is connected with a certain sign rule. Thesign of the magnetic flux <J.> is determined by the choice ofthe normal to the surface S bounded by the loop underconsideration, while the sign of ~i depends on the choiceof the positive direction of circumvention of the loop.

As before we assume that the direction of the normal nto the surfa~e Sand the positive direction of the loop cir-,cumvention are connected through the right-hand screw rule·(Fig. 9.2). Consequently, when we choose eb

. (arbitrarily) the direction of the normal,we denne the sign of the flux <J.> as well asthe sign (and hence the "direction") of theinduced e.mJ. ll' +

With such a choice of positive directions(in accordance with the right-hand screw Fig. 9.2rule), the qua9tities ~l and d<J.>ldt haveopposite signs..

The unit of the magnetic flux is the weber (Wb). If therate of variation of the magnetic flux is 1 Wh/s, the e. m.f.induced in the loop is equal to 1 V [see (9.1)1.

Total Magnetic Flux (Magnetic-flux Linkage). If a closedloop in which an c.m.f. is illlillced contains not one but Nturns (like H coil), 'ii will be eqnal to the sum of the e.m.f.sinduced in each turn. And if the magnetic flux embraced byeach tUl'll is the same and equal to <Pi' the total flux <J.> throughthe surface stretched over such a complex loop can berepresented as

(!).1)

9. Electromagnetic Induction218

of the del'ivative d<J.>/dt I d"direction" of f

i. ea s to a change in the sign or

Faraday found out that th . dated in two different wa s e In. u~e~ current may be gener-which shows the coil C ~iihThIs IS Illustrated in Fig. 9.1,magnetic field) and the loa clrrent I (the coil creates themeter G which indicates th: ind con3ected to the galvano-

The first way consists in th ~ce current.(or its separate parts) in the ~e1~s~~acement of t~e loop L

In the second case the I . the fixed call C.field varies either due to the ::~f0 IS ¥xed b?t the magneticof a change in the current I . I.~ 0 the coIl C or as a result

In all these cases the gal In I , or due to both reasonsence of induced current in :hanolmetelr G indicates the pres~

Len' L . e oop J.

z saw. The direction of th .hence the sign of the induce i f )~ Induced. current (andlaw: the induced current is ~.e.m .. IS determIned by Lenz'scause generating it. In oth:;e~~d;o that i.t counteracts thecreates a magnetic flux h' t r s, the Induced currentmagnetic flux generatin~ ~~ 1 ~re;ents the variation of the

If, for example the 100 eL Ill( ~ced,e.rn.f.coil C, the magne'tic flux & FIg. 9.1) is.drawn to thec~rre~t induced in the loop i~o~~~ the I?o~ Illcreases. ThedirectIOn (if we look at th 1 'f .case IS In the clockwisecreates til(' magnetic fJuxe,,~?p ~od~ the right). This currentvents an increa!';e in It Irec e. ~o the left, which preinduced current'. . Ie magnetIc flux, which causes the

The same situation takes I· 'f .in the coil C, keeping fixed ihace \ we Increase the currentother hand, if we decrca~e the e COl a~d the loop L. On therent induced in the loop L 'lfu~rent Ill. the .coil C, the curnow be directed countercloc~~is~ei~6I'seIlts dl~~ctioll (it wi11

Lenz's law expresses an im or we oo~ from the right).tendency of a system to ~ tant phYSIcal fact, viz. the(electromagnetic inertia) Coun eract the change in its state

Faraday's Law of Ele~trorn .to this law, the c.m.f i d :g~ehC Ind~ction. Accordingformula . n uce III a loop IS defined by the

~i = -vBl, . (9.4)where the minus sign is tak .sign rule: the normal n t~ntf: accordance with the adoptedloop was chosen so that ·t· ~. surface stretched over theFig. 9.3 (towards the fi:ld ISB) Irect~d behind the plane ofthe right-hand screw rule ,~~ he!lce, according tovention is the clockwise l the .pOSltIV~ dlrecf.ion of circum-In this case, the extraneo~sefit~f;'E~s~s ds~own in th~ figure.

IS . Irected agaIllst thf

9.2. Origin 01 Elet..romagnetic Induction 221

positive direction of loop circumvention, and hence ~l

is a negative quantity.The product vl in (9.4) is the increment of the area bounded

by the loop per unit time (dS/dt). Hence vBl = B dS/dt == det>/dt, where d<1> is the increment of the magnetic fluxthrough the area of the loop (in our case, d<1> > 0). Thus,

~l = -d<1>/dt. (9.5)It can be proved in the general form that law (9.1) is

valid for any loop moving arbitrarily in a permanent nonuniform magnetic field (see Problem 9.2).

Thus, the excitation of induced e.m.f. during the motionof a loop in a permanent magnetic field is explained by theaction of the magnetic force proportional to [v X BI,which appears during the motion of the conductor.

It should be noted by the way that the idea of the circuitshown in Fig. 9.3 forms the basis of all induction generatorsin which a rofOr with a winding rotates in an external magnetic field.

. A Loop Is at Rest in -a Varying Magnetic Field. In thiscase also, the induced current is an evidence of the fact thatthe magnetic field varying with time creates extraneousforces in the loop. But what are these forces? What is theirorigin? Clearly, they cannot be magnetic forces proportionalto [v X HI, since these forces cannot set in motion thecharges that have been at rest (v = 0). But there are noother forces besides qE and q [v X Bl! This leaves the onlyconclusion that the induced current is due to the appearanceof an electric field E in the wire. It is this field which isresponsible for the induced e.m.f. in a fixed loop placedinto a magnetic field varying with time..

Maxwell assumed that a magnetic field varying with timeleads to the appearance in space of an electric field regardlessof the presence of a conducting loop. The latter just allowsus to reveal the existence of this electric field due to thecurrent induced in the loop.

Thus, according to Maxwell, a magnetic field varyingwith time generates an electric field. Circulation of vect9r Eof this field around any fixed loop is defined as

~ E dl = - ~; . (9.6)

9. Electromagnetic Induction220

9.2. Origin of Electromagnetic InductionLet us now consider the h . I

appearance of induced e f p rICa reaso~s behind· theinduction (9.1) from whaim~e' a~n dtry to derIve the law oftwo cases. . rea y know. Let us consider

A Loop Moves in a Per t M . .first take al' man.en agnetIc Field. Let us

oop wIth a movable Jumpe~ o~ l?ngth 1 (Fig. 9.3).. Su~pose that It IS III a uniform mag-

E=[V"B] netIc field perpendicular to the planer-----€I-__ of the loop and directed behind

(:\ the plane of the figure. We start

: ) movmg the jumper to the right withv velo.city v. The charge carriers in

tfe Jumper, viz. the electrons, willFig. 9.3 a so ~tart to move with the same

. v~loClty. As a result, each electronforce F = -e [v X BJ w~ll be acted upon by a magneticelectrons will start mov·dlrected along the jumper. Thewhich is equivalent to ~f! d?wnwards al~:mg the jumper,This is just the ind d ctrlC current dIrected upwards.over the surfaces ot~~e cur~nt. The ~harges redistributedfield which will ind con ucto~s wIll create an electricthe loop. uce a current III the remaining parts of

The magnetic force F I hThe field correspondingPt~y\t .e r~~e of an extraneous force.It should be noted that h' 1 IS . = F/(-e) = [v X Bl.with the help of field t t .IS

fexpre~sIOn can also be Obtained

Circulation of vectorr~~s ormatIOn formulas (8.4).by definition, the magnitud:r~~~~a. cldosed contour gives,case under consideratI'on h e III uced e.m.f. In the,we ave

(9.9)

223

Bj dl... all) t'<\l E dl =, -- --\. (\) [vu at . J

9.2. Origin 0/ Electromagnetic Inductton

The expression on the right-hand side of this equation isthe total derivative deDIdt. Here the first term is dUl' tothe time variation of the magnetic field, while the second

radius (see Problem 9.5). Since the electriC field is a vortexlfieldithe direction of the force acting on the electrons always coincides witlithe direction of motion; and the electronscontinuously increase their energy. During thetime when the magnetic field increases (-- 1ros), the electrons manage to make about amillion turns and acquire an energy up to 400MeV (at such energies, the electrons' velocity is almost equal to the velocity of lightc in vacuum).Thp inductioll arcelerat.or (bptatron) re-sembles a transformer in which the role ofthe secondary winding consisting of a single Fig. \J.4turn is played by an electron beam.

Conclusion. Thus, the Jaw of electromagnetic induction(9.1) is valid when the magnetic flux through a loop varieseither due to the motion of the loop or lIw tillle v,ll'iatiollof the maglletic ticl!l (01' due [0 both reasons). Uowever,we had to fcsprt to two quite different phenOinena for explaining the f<I\V in these two cases: fol' t!le moving loop weusatl the action of tlw maglletic force proportional to (v Y. Blwhile for the magneti~ field varying with time, aB/at,the concept of vortex electric field E was employed.

Since there is nO unique and profound principle whichwould combine these phenomena, we must interpret the lawof electromagnetic induction as the combined effect of twoentirely different phenomena. These phenomena are generally independent and nevertheless (which is astonishing)the e.m.f. induced in a loop is always equal to the rate of variation of the magnetic flux through this loop.

In other words, when the field B varies with time as wellas the ConfIguration or arrangement of the loop in the field,the induced e.m.f. should be calculated by formula (9.1)containing on its right-hand side the total time derivativedetl/dt that automatically takes into consideration the twofactors. In this connection, law (9.1) can be represented

in the form

9. Electromagnetic induction222

Here the symbol of parti I d' .(a/at) emphasizes the fa:t t~~~v~~velwith respect to timethe oop and the surface~ retc ed on it are fixed. Since the flux tIl = ~ B dSIntegration is performed aro . . (theed over the loop we are int~~ed t

and

a!b)ltrar

ysur ace stretch-s e In, we have

air aBat J BdS= J at dS ,

In this equalitv, we exchan d I~ith respect to'time and int:; t': Ie or(\er' of differentiationIS possible since the loop anJ

at~on ov~r the slll'face, whichEq. (9.6) can be represented in theef:~~ace are fixed. Then

,{"Edl= _ r~• ';Y J at dS. (9.7)

This equation has the same stru tbecto~ j being played by the vec~r~:~/~q· ~6.17) thl:' role of thee written in the differential form th t. on.sequently, it cane same as Eq. (6.26), i.e.

V X E = -aB/at.Th' . . ~~

IS equatIOn expresses a local relat' hfields: the time variation ot ma neUe IOn etween e~ectric and magneticthe c~rl 0/ vector E at the same ~otnt !!;~d r ~t hgwen point determineszero IS an evidence of the presence ~f the aCI t ~t V X E differs frome e ectnc field itself. .

The fact that the circulation of th I .by a varying magnetic field d'ff e e ectflc field ind uced~h~s electric field is not a pote:ti:~sfifl~mL~r;o indica~es thatIt IS a vortex field. Thus, electric fiel e . 1 e. magnetic field,field (in electrostatics) or a vorte~ c~~l:e eIther a potential

In the general case electric field E .electrostatic field a~d the fi ld . dm~y be the sum of thefield varying in time Since the e. In :lCed by a magneticfield is equal to ze;o, Eqs.· {9~~)~(~18\lOnof t?e elec~rostat~cfor the general case as well when the' Ii l~u~n. cut to ne vahdof these two ftehls.' e IS the vector sum

. Betatron. The vortex electric field' I'1D ~e induction electron accel~ratol'uavs: Ol~~ a hrilli8.~t aPillicationCODSlSts of a toroidal evacuated ~h " .z. B.atron. TblS aecelerato"0lf an electroma~et (Fig 94) ThamDer. at:rang"d between the pol<;~e ectromagnet wmding cr~at~s ; e .varlstlon of the current in ihna vortex electric field. Ttl~ latt':rv8:y\ng,magnetic field which iIlduce;taneously retains them on an eq a'~?be ~r..1.e3. the eillctmna and aimuI-

UJ .. rlUm cIrcular Drbit of a certain

225

(9.10).. ' . 6A A = 1 d<D

9.2. Origin of Electromagnetic Induction

while the induceo\,:m.f. accomplishes during the same timethe work

a permanent magnetic fIeld B is localized (it is directed perpendicularlyto the plane of the figure). This region is encircled by a fixed metallicring R. By moving the sliding contacts to the other side of the ring(see Fig. 9.6), we introduce the magnetic flux l1> into the closed contour containing a galvanometer G (i-initial position, 2-final position). Will the galvanometer show a current pulse?

Applying formally the law (9.1), we should conclude that therewill be an induced current. But it is not so. There is no current sincein this case both oB/iJt and the Lorentz force are equal to zero: themagnetic field B is constant and the closed loop moves in the regionwhere there is no magnetic field. Thus, we have none of the physicalreasons underlying the law of electromagnetic induction.

On an Apparent Paradox. We know that the force actingon an electric charge in a magnetic field is perpendicularto its velocity and hence accomplishes no work. Meanwhile,during the motion of a current-carrying conductor (movingcharges!) Ampere's forces undoubtedly accomplish somework (electric motor!). What is the matter?

This seeming contradiction disappears if we t,ake intoaccount that the*,IDotion of the current-carrying conductorin the magnetic neld is inevitably accompanied by electromagnetic induction. And since the e.m.f. induced in theconductor performs work ort the charges, the total work ofthe forces of the magnetic field (the work done by Ampere'sforces and by the induced e.m.f.) is equal to zero. Indeed,upon an elementary displacement of a current loop in themagnetic field, Ampere's forces perform the work (seeSec. 6.8)

6A; = 'tJ dt = -1 d<1> (9.11)

where we took into account that c; = -d<1>ldt. It follows. from these two formulas that the total work is

6A A + 6A; = O. (9.12)

Thus, the work of the forces of the magnetic field includesnot only the mee-hanical work (due to Ampere's forces)but also the work done by the e.m.f. indueed during themotion of the loop. These works are equal in magnitude andopposite in sign, and hence their sum is equal to zero.

15-0181

~s' L!~/"':\"'}_--J.r-

Fig. 9.5Fig. 9.6

In such cases it is necessarthe Lorentz force qE + fv t~ resort to the Lasic laws, viz.= -aBI8t. These 1 q X BI and the law - v E _1 aws express tl J' v /,-aw of electromagnetic ind 1- l~ P lyslcal meaning of the

Let us consider two . utc 1O~1 In all cases.. Ins ructlve examples.

E~ample 1. A conduct' t.. .a regIOn in which a m Illg ~ rIp IS dIsplaced at a velocit '

~~~~J~~::fj~~~~ lh:~;ebI~ n;1ddo~t;J ~fr~f~e~~::!'t::St:1~oJ~~~h~~~:~e~~~c~:n! ct;retVi~;o~~.r ~iIli ctoh~n~~\~~~~m~~~~ ~~:U~::

At first sight th .is difficult the questIOn seems not so sim I . .he "closed" ? cthoose ~he contour itself: it is n ~ e'l smce In th!s case itd' III e strIp and how tho . 0 c ear where It shouldf urmg. the motion of the stri I IS part of the contour will hehaves~:i;'~IlIhb~o:rI.eSpslimmdediatelrc1:~rh~h:~v~:;~;lecrterSoonr~ ~o tthJe Lorentz. h ace upwards h' h . S III Ie movinIn 1\ =hlui~b~o~~~~r thi~c~t,. 'd7re~~et~lo~~~~i~e an electric curren1

~~~~?~;~:y~~g,:~~~{;:i~r!N1I{I~f:Z~!1~i~fi;r:!:;~(:~~~js III t e case of the conducting r tr~spects, the situation is the same

Exam Ie 2 ., S J'Ip.P . lhe dotted circle in F'

Ig. 9.6. shows the region in Which

224 9 "-----,--,-- - '. :~Irctromagnetic Induction

is due. to tile mution of the 100 T1 "term IS explained in "Teater de~"1 ~e 0pl'lglll of the seconu

Possible Difficulties'''' s< . al III roblern 9.2.< t' J • ometllnes we cca IOn w wn tile law of ('Iectro . . Ime across a situ-(9.1) tlll'f1~ Ollt to In' if!' I~agnetlc IIldllction in the formdifficlllti('~ associated' wi;;P~1 Ica~le. (mainly Lecallse of the

I Ie, c JOIce of the contoul' itself).

'I'I

d' If the loop is rigid and there are no ferromagnet~cs~ei;:~~ighbourhood, the inductance is a constant quantltyindependent of the current /. . h

The unit of inductance is henry (H). In acco~da~ce w:~(9 14) the inductance of 1 H corresponds to t e oop em~gnetic flux through which is equal to 1 Wb at the currentof 1 A which means that 1 H = 1 Wb/A.

E 'I Find the inductance of a solenoid, neglecting edge effec~txamp e. f th olenoid be V, the number of turns p~r um

Let the voluIDde hO e s t"c permeability of it substance inside thelength n, an t e magne I "

solenoid fJ.. 0 •

d' t (9 t4) L = $/1. Consequently, the problem IS .re-Accor lllg 0 '.'. f th t tal magnetic flux $ assummg

duced to the dete~ml!1atlOn. or theecu~rent I the magnetic held in thethat t~e ~urre~ I IS ;re~il~~agnetic flux through a turn of, th~ sole~solen?ld IS B - fJ.~o . IS whole the total magnetic flux piercIng IInoid IS $1 = BS - 11I1oll , I

turns is N'" - l B(' = IlIlon2Vl<P = 'Vi - It·" rr '

where V = Sl. JIence the inductance of the solenoid isL = l1!1onIV. "(9.t5)

On Certain Difficultiel!l. ~t should be noted that the~etermination of inductance wlth the help of formula L -: wi/is fraught with certain difficulties. ~o m~tter how t~~n the

. . b its cross-sectional area IS fU11te, and we ~lmplywue ma

ky e.'h to draw in the body of the conductor a geo

do not nOw ow I I t' <D The resultmetrical contour required for ca eu a .mg. '. .- b' For a sufficiently thm Wlre thlS amblgu-~ec~m~s amatelgrl~aolusH'owever for thick wires the situation isIty IS Imm .. ., I f I I t' f L

t' ly different: in this case, the resu. toea c.u a 1O~ 0en lre t'n a gross error due to the mdetermlllacy m thema~ cO~fa:he geometrical contour. This should be always~hO~C~ 'nd We shall later shOW (see Sec. 9.5) that t~ere~~ist;~~~~he~method of determining L, which is not subJectto the indicated drawbacks. . .

E.M.F. of Self-induction: According to (9.1), a var~~t~~I~of the current in the circUlt leads to the appearancee.m.f. of self-induction ~a

'€ _" _ d$ = _ ldd (10/). (9.16)ea - dt t

. t' ofIf the inductance L remainsconstan~ up.on a vana lonthe current (the configuration of the ClrcUlt does not change

9. Electromagnetic induction

The work of Ampere's forces is done not at the expenseof the energy of the external magnetic field but at the expenseof the power supply which maintains the current in the loop.In this case, the source performs additional work againstthe induced e.m.f., equal" to 6Aad 0= -'til dt = I dW,which turns out to be the same as the work 6A A of Ampere'sforces.

The work 6A which is done during the displacement ofthe loop against the impeding Ampere's forces (whichappear due to the current induced in accordance with Lenz'slaw) is transformed into the work of the induced e.m.f.:

6A = -6A A = 6.,1 i. (9.13)

From the point of view of energy, this is the essence of theoperation' of all induction generators.

9.3. Self-indudion

. Electromagnetic induction appears in all cases when themagnetic flux through a loop changes. It is not importantat all what causes the variation of this flux. If in a certainloop there is a time-varying current the magnetic field ofthis current will also change. And this leads to the variationof the magnetic flux through the contour, and hence to theappearance of induced e.m.f.

Thus, the variation of current in a circuit leads to theappearance of an induced e.m.f; in thi~ ~;ir(',pit. This phe-nOmenon is called self-induction. ' ",~ "

Inductance. If there are no ferromagneiics in the spacewhere a loop with current I is located, field B and hencetotal magnetic flux W through the contour are proportional-to the current I, and we can write

W = LI (9.14)

where L is the coefficient called the inductance of the circuit.In accordance with the adopted sign rule for the quantitiesWand I, it turns out that <1> and I always have the samesigns. This means that the inductance L is essentially apositive quantity.

The inductance L depends on tlti~ shape and size of-tfleloop as well as on t he magnetic properties of tlw surrounding

9.3. Self-induction227

15*

(9.20)

(9.19)

(9.18)

229

Fig. 9.8o

9.3. Self-induction

(b)

L

(a)

h . hI we obtainSeparating t e varIa es,dl RT=-T dt .

I and I) and over t. rover 1 (between 0Integration of thIS ~qua {on(111'o) = -RLlt, or(between 0 and t) glves n

I=Ioe-t/l:

where 'r is a constant having the dimension "Of time:or = LIR.

( the relaxatol! time). This quantityIt is called the time constant O\n the current: it follows fro,!, «(J.t.9)characterizes the rate .of deh-ehs~he current decreases to (tIe) tIlI!es I~Sthat or is the time durIllg w IC . f 'r the slower the decrease 1':1 ~ einitial value. The larger thhvalue ° f 'the dependence I(t) descrlbIllgCUff('nt. Figure 9.8 shows ~ e c~rve ° ve 1)the decrease of current wIth tIme (cur . I f a circuit

.. t· r current upon c osure 0 • kExample 2. Stablhza Ion 0 . 1\ . turn the switch S countercloc -At the instant t = 0, we rapH Ysition (Fig. 9.7b). We thus co.n

wise from the upper to the .lo;er po coil L. The current in. the CI~nected the source ~ to ~he III If~t~d~~tion e.mJ., coun,teractlllg t~~cuit starts to grow, an a s~n I~ceordance with Ohm slaw, RI increase, will &iain appear.

Fig. 9.7. , law counteracts the decrea:re in

which, in accor,danc~ wIth L;~~ s th~ current in the circuit WIll bethe current. At"each ~nstant ~ ~~eiR, ordetermined by Ohm 5 law - s

.. dlRI=-L dt-·

. for a very short time and then dis-the key short-circUIts t~e s?urc\hout breaking the latter.connects it from

hthe hl[h~\tnd;ctance coil L starts to decreasei.w:Jld~

The current t roug f self-inducti~n e.mJ. ~s = -leads to the appearance 0(9.17)

Example 1. Current collapse upon the disconnection of a circuit.Suppose that a circuit consists of a coil of constant inductance L,

a resistor R, an ammeter A, an e.mJ. source~· and a special key K(Fig. 9.7a). Initially, the key K is in the lower position (Fig. 9.7b)and a current 10 = ~/R flows in the circuit (we assume that the resistance of the source of e.m.f. ~ is negligibly small).

At the instant t = 0, we rapidly turn key K clockwise from thelower to the upper position (Fig. 9.7a). This leads to the following:

and there are no ferromagnetic:'», then

'f = -L~ (L=const).B dt

The minus sign here indic.ates that 'fs is always direetedso that it prevents a change in the current (in accordancewith the Lenz's law). This e.mJ. tends to keep the currentconstant: it eounteracts the current when it increases andsustains it when it decreases. In self-induction phenomena,the current has "inertia". Therefore, the induction effectsstrive to retain the magnetic flux constant just in the sameway as mechanical inertia strives to preserve the velocityof a body.

Examples of Self-induction Effects. Typical manifestations of self-induction are observed at the moments of c.onnection or disconnection of an electric circuit. Tho illcreasein the current before it reaches the stead v-state value afterclosing the circuit and the dOGl'oase in "the current whenthe circuit is disconnected Dceur grJdually and not insti!ntaneously. These effects of lag are the morc pronounced thelarger the inductance of the eircuit.

Any large electromagnet has a largo inductance. If itswinding is diseonnected~from the source, tlw current rapidlydiminishes to zerO and creates during this process a hugee.mJ. This often leads to the appearance of a Voltaic arcbetween the contacts of the switch which is not only verydangerous for the eftlctromagnet winding but may even provefatal. For this reason, a bulb with the l'{)sistance of thesame order of magnitude as that of the wire is normallyconnected in parallel to the electromagnet winding. In thiscase, the current in the winding decreases slowly and is nota hazard.

Let us now consider in greater detail the mode of vanishingand establishment of the current in a circuit.

'228 9. Electromagnetic Induction

(9.25)

231

9.9

(9.24)

9.4. Mutual Induction

The proportionality factors L12

and L 21 are called mutual indue- .tances of the loops. Clearly, mutual' inductance IS nu-merically equal to the magnetic flux through one of theloops created by a unit current in the other loop. The coefficients L

21and L

12depend on the shape, size, an~ mutual

arrangement of the loops, as well as on the magnetic permeability of the medium surrounding. the loops: These coefficients are measured in the same umts as the lllducta?ce L.

Reciprocity Theorem. Calculations show (an~ expenmentsconfrrm) that in the absence of fel'romagnetlcs, the coef-ficients L 12 and L 21 are equal:

This remarkable property of mutual inductance is usuallycalled the reciprocity theorem. Owing to this theorem\ we

9.4. Mutual Induction

Mutual Inductance. Let us consider two fixed loops 1and 2 (Fig. 9.9) arranged sufficiently close to each other.If current I flows in loop 1, it creates through loop 2the total m~gnetic flux $2 proportional (in the absence offerromagnetics) to the current II:

$2 = .{i21Il' (9.23)

Similarly, if current 12 flows inloop 2, it creates through thecontour 1 the total magneticflux

of the field of the induced current and of the. ext!,!rn~l current throu~the ring are equal in magnitude and opposite m SIgn. Hence LI -= na'B, whence

1= na2B/L.

This current in accordance with (6.13), creates a field B I = 1tl1oaB/2~at the cent;e of the ring. The resultant magnetic induction at thispoint is given by

B = B- B I = B (1 - 1tl1oa/2L).res

(9.21 )dl

RI=~-Ldt

9. Electromagnettc Induction230

= ~ + ~., or

We transp?se ~ to the left-hand side of this equation and introducea new varIable u = RI -~, du = R dl. After that we transformthe equation thus obtained to '

du/u = -dtIT,

where 't = L/R is the time constant.Integration over u {between - 'jg and RI - ~) and over t (be

tween 0 and t) gives n [(RI - ~)/(-~)) = -tIT, or

1= 10 (1_e- t /'f), (9.22)

where J0 = 'jg/R is the value. of the steady-state current (for t ->- 00).EquatIOn (9.22) shows that the rate of stabilization of the currentis?etermin.ed by the same constant T. The curve I(t) characterizing theIDcrease m the current with time is shown in Fig. 9.8 (curve 2).

On the Conservation of Magnetic Flux. Let a currentloop move and be deformed in an arbitrary external magnetIC field (permanent or varying). The current induced inthe loop in this case is given by

1= 'jgt+~. = __1 d$R R dt'

If the resistance of the circuit R = 0, d$/dt must also beequal to zero, since the current I cannot be infinitely large.Hence it follows that cI> = const.

Thus, when a superconducting loop moves in a magneticfield, the magnetic flux through its contour remains constant.This conservation of the flux is ensured by induced currentswhich, according to Lenz's law, prevent any change in themagnetic flux through the contour.

The tendency to conserve the magnetic flux through11 coutour' always exists but is exhibited in the clearest formin the circll~ts of supercoilductors.

. .Examp.le. A superconducting ring of radius a and inductance L'IS m .a u~Iform magnetic field B. In the initial position, the plane ofthe rmg IS parallel to vector B, and the current in the ring is equalt~ zero. The ring is turned to the position perpendicular to vector B.Fmd the current in the ring in the final position and the magneticinduction at its centre.

. The magne~ic flux through the ring does not change upon its rotatIOn and remams equal to zero. This means that the magnetic fluxes

III.

....,...

233

d<l>~ ell t (\:I. 2tl)'tz = -at =. - L 2 \ dt'

th at the contOtlfS are fIxedt' 1 Here we assume I'respec Ive y. . ' , a netics in the SUf!'OlllJ( lUgS. .

and there are no fenom 11 . d tiOil the current appeal'1ng,Taking into account se -lIl He'n c'urrents of hoth eircuits

sav, in contour 1 upon a ch~)I~ge ,1 hw thrOllft!i the formlll:Jis'determinod according- to 1m S , eo

dl( dI2•

Rl/l=~I-LI F- L12 dt '

9,4. Mutual Induction--_.-=.=~-

, . theorem becomes inapplicable.situation, and th~ ree~I~r()cI1tyI I of the following concreteLet us verify thiS Wltl1 t. Ie 1e p

U~~. .. l' uder of volume V has two W1D~-

Example. A long ferromagnetic ?Yl contains nl turns per umtjngs (one over the other). ~ne w:n ~~gFind their mutual inductancelength while the other contams n2 ur .ignori~g the edge effects,

__ cD.)I This means that we must createAccordinjt t? (9:23), L2a -::-alculat~ the total magnetic flux throu~h

current II in Wll1~llld~ 1 a; If winding 2 contains N 2 turns, we obtamall the turns of Will lllg ,

$2 = N 2B 1S,f th c linder Considering that

where S is the cross-sectio1l1~ jreaJ~nglhe a~d B 1 ~ f!1~lonlII' whereN 2 = n2 l , where ~ is the cYb~l' te; for cll'rrent II' we writ~ <1>2 ~~

is the magnetic permea I I Yf!1 VI where)1 = lS. Hence= /l1f!onln2 l'

L 21 = f!1f!onln2 V,

SimilarlY, we"lcan fwd L 12:

L12 = f!2f!onln2 V.• " he last two expressions are generally

Since the values of f!1 and fl2 II. \ I and I in ferromagnetics), thedifferent (they depend on curren ~ ld 2

. f L nd l do not comCI e. 'l'ngvalues 0 21 a .'12 The presence of a magnetIC ~OI;IP 1 •

Mutual InductIOn. "f t d 'n that any varIat10n of. 't '. mam eself

between clrcUlS IS . . 't leads to the appearance 0

current in one of t,he ~rcll~1 s , circuit. This phenomenonan ind uced e .m.L 1I1 t ~ 0 leIis called mutual inJuctzon. 1 f 's appearing in

According to Faraday:s law, t 1e e,m..contours 1 and 2 are given by

d$ d/2ce 1 = -L1,) -dt 'CI- dt -

Example f. Two circular loops 1and 2 whose centres coincide lie in aplane (Fig. 9.10). The radii of the loopsare 4 1 and a,. Current I flows inloop 1. Find tile magnetic flux 4>2 em-braced by loop 2, if 41 <: 41'

The direct calculation of the flux 4>1is clearly a rather complicated problemsince the configuration of the fielditself is complicated. However, theapplication of the reciprocity theoremgreatly simplifies the solution of theproblem. Indeed, let us pass the samecurrent 1 through loop 2. Then themagnetic flux $1 created by this current through loop 1 can be easilyfound, provided that al «42: it is sufficient to multiply the magnetic induction B at the centre of the loop (B ==/loI12a2) by the area 1ta¥ of the circle

11>2 = $1 in accordance with the reci-

9. Electromagnetic Induction

Fig. 9.10

Fig. 9.1

232

do not have to distinguish between L 12 and L.J.l and cansimply speak of the mutual inductance ·of two circuits.

The meaning of equality (9.25) is that in any case themagnetic flux <1>1 through loop 1, created by current Iin loop 2, is equal to the magnetic flux <1>2 through loop 2,created by the same current lin loop 1. This circumstanceoften alIows us to considerably simplify the calculation,for example, of magnetic fluxes. Here are two examples.

and take into account thatprocity theorem.

Example 2. A loop with current 1 has the shape of a rectangle.Find the magnetic flux $ through the hatched half-plane (Fig. 9.ft)whose boundary is at a given distance from the contour. Assume thatthis half-plane and the loop are in the same plane.

In this case too, the magnetic field of current I has a complex configuration, and hence it is very diffIcult to directly calculate the flux lDin which we are interested. However, the solution can be considerablysimplified by using the reciprocity theorem.

Suppose that current I flows not around the rectangular contourbut along the boundary of the half-plane, enveloping it at infinity.The magnetic field created by this current in the region of the rectangular loop hall a simple configuration-this is the field of a straightcurrent. Hence we can easily lind the magnetic flux $' through there.ctanguJar contour (with the help of simple integration). In accordanceWith the reciprocity theorem, the reqUired flux $ = <1>' and theproblem is thus solved., However! the presence of ferromagnetics changes the

I'

2359.5. Magilettc Field Erler",

e.m.f. ~o. As we a-lready know, the current in the circuitwill start increasing. This leads to the appearance of a self-·induced e.m.f. ~,. In accordance with Ohm's law, RI == ~o + ~" whence

lo = RI - ~•.

Let us find the elementary work done by extraneous forces(Le. the source of ~o) during time dt. For this purpose, wemultiply the above equality by I dt:

~oI dt = RI2 dt - ~,I dt.

Taking into account the meaning of each term and the relation ~, == -dcD/dt, we can write

6Antr = flQ + I d(]).

It can be seen that in the process of current stabilization,when the flux (]) varies so that dcD > 0 (if I > 0), the workdone by the s(Jurce of ~o turns out to be greater than theJoule heat liberated in. the circuit. A part of this work(additional work) is performed against the self-inducede.m.f. It should be noted that after the current has beenstabilized, d(]) = 0, and the entire work of the source of ~o

will be spent for the liberation of the Joule heat.Thus, the additional work accomplished by extraneous

forces against the self-induced e.m.f. in the process of current. stabilization is given by

IflA add = I d<l>·1 (9.27)

This relation is of a general nature. It is valid in thepresence of ferromagnetics as well, since while deriving it110 assumptions have been made concerning the magneticproperties of the surroundings.

Here (and below) we shall assume that there are no ferro·magnetics. Then dcD = L dI, and

fJAadd cc= LI dI. (9.28)

Having integrated this equation, we obtain A add == L[2/2. According to the law of conservation of energy,any work is equal to the increment of any kind of energy.It is clear that a part of the work done by extraneous forces

9. Electromagnetic Induction

(b)(a}

234

where ~1 is the extraneous e f'induced e.m.f.s) and L is th m: 'd In contour 1 (besidesA similar equation can I be ~t In ufctance of contour 1.contour 2. WrI en Or the current 1

2in

It should be noted that tr ffor transforming currents a dans ~rmers, devices intendedprinciple of induction. n vo tages, are based On the

On the Sign of L U l'k .' d• 12· n 1 e In uctance L h' hmentIOned earlier is essentiall . . w IC , as was

inductance L i~ an l b ! a POsItIve quantity, mutual12 a ge rate quantity (which may in

00 00particular, be equal to zero). This isdue to the fact that, for instance in(~.23) the quantities cD

2and II r~fer

;alldtfferent contours. It immediatelyo ows from Fig. 9.9 that the si no~ the. magnetic flux cD

2for a giv:n

dIrectIOn of current I will d dthe h' f 1 epen on

c OICe 0 the normal to the Sur-Fig. 9.12 fahc~ boufnhded by contour 2 (or On the

. c OICe 0 . t e positive direction of cir~

The positive directi~~~!oentIOn of this contour).contours can alwa 'S be I' curre!1ts (and e.m.f.s) in bothdirection of cont~ur ci~hosen ar~Itra~Ily (and the positivethe right-hand screw rUl~)~:le~t~n IS huniquel~ (throughnormal n to the surface bo d ~ e

bto t e dIrectIOn of the

long run with the sign ofu~~lee y th~ contour, i.e. in thethese directions are s e . magnetIc flux). As SOOn asassumed to he Positi':e C.lr~~' the quantity !-12 should be~hrough the contours a:e ~~~i~~~esfoof mu~~al inductionI.e. when they coincide in sign witl : If?Sltive. currents,

In other word:,;, L > 0 'f f. ~ ~e -InductIOn fluxes.tours "magnetize" ea~h otl I 0 o~ pOs.Itlve currents the concases, the positive di~ect:~~'s o~ e~wIse, L 12 :<- O. In specialt?UI'S can be established b f . CircumventIOn of the conSIgH of the quantity L ,e ~Iehall(~ so that the req uired

. 12 can e obtaJlIed (Fig. U. U).

9.S. Magnetic Field EnergyMagnetic Energy of C t L '

containing induction C~i~r~n ~udet us. connect a fixed circuit.J resistor R to a source of

236 9. Electromagnetic Induetlon9.5. Magnetic Field Energy 237

(t0) is spent on increasing the internal energy of conductors(it is associated with the liberation of the Joule heat); whileanother part (during current stabilization) is spent on something else. This "something" is just the magnetic field, sinceits appearance is associated with the appearance of thecurrent.

Thus, we arrive at the conclusion that in the absence offerromagnetics, the contour with the induction coil £and current [ has the energy

Iw= +Lf2 = i- [ = ;. .1 (9.29)

This energy is called the magnetic, or intrinsic, energy ofcurrent. It can be completely converted into the internalenergy of conductors if we disconnect the source of tofrom the circuit as shown in Fig. 9.7, Le. if we rapidly turnthe key K from position b to position a.

Magnetic Field Energy. Formula (9.29) expresses themagnetic energy of current in terms of inductance and current (in the absence of ferromagnetics). In this case, however,the energy like the electric energy of charged bodies, can bedirectly expressed in terms of magnetic induction B. Letus show this first for a simple case of a long solenoid, ignoringfield distortions at its ends (edge effects). Substituting theexpression L = flflon2V into (9.29) we obtain

W = (112)£[2 = (1/2)flflon2[2V.

And since nI = H = B1flflo,. we haveBI B.H

W=--V=-V. (9.30)211110 2

This formula is valid for a uniform field filling the volume V(as in the case of the solenoid under consideration).

The general theory shows that the energy W can be expressed in terms of Band H in any case (but in the absenceof ferromagnetics) through the formula

l~= JB;JI dV.! (9.31)

The integrand in this equation has the meaning of theenergy contained in the volume element dV...

Thus, as in the case of the electric field er~ergy, we ar~iveat the conclusion that the magnetic energy IS also localIzedin the space occupied by a magnetic field.

It follows from formulas (a.30) and (9.31) that the m~gnetie energy is distributed in space with the volume density

Iw=¥=-,~,\ (9.32)

It should be noted that tltis expl'e~sion call be applie.donly to the media for whieh till' B VE'. II dependence islin;ar, i.e. p in the relation B,= I-tl-toH is inrlependent.of H.In other words, expressions (tl.31) (lnr! (fl.32) are applicable.only to para magnetics and diamagnetics. III t.he ease 01fefi:olllagneties, these expression::, ,Il'l' illill'plieable*

I l sho'tdd also be Hoted thatthe llwglletic elll'rgy is an essentially positive ".quantity. Thiscan be easily seen from the lasttwo form ulas.

Another Approach to the Substantiation of Formula (9.32). Letus prove the validity of this formulaby proceeding "the ot~er way roun~", Fig. 9.13Le. let us show that If formula (9.32)is valid, the magnetic energy of thecurrent loop is W = LJ2/2. . . .

For this purpose, we conSider the magnetIc field of an. arbltrar,Yloop with current I (Fig. !U3). Let us imagine that the entire fiel? IS

divided into elementary tubes whose generatrices are the fIeld lIn~sof B. We isolate in one such tube a volume element dV = dt as.According to formula (9.32), the energy ~ BH dt dS is localized in

this volume. .Let us now find the energy dW in the volume of the entire ~lementa-

ry tube. For this purpose, we integrate the latter expressIOn .alon,gthe tube axis. The flux dcD == B dS through the tuue cross-sectlOIl IS

* This is due to the fact that in the long run expressions (9.31)and (9.32) are the consequences of the formula 6A add =, J d(fJ and ofthe fact that, in the absence of hysteresis, the work 6A add i~ spent onlyon increasing the magnetic energy dW. For a ferromagnetic medIUm,the situation is different: the work 6A add is also spent on increasingthe internal energy of the medium, i.e. on i~s }:eating.

"!'t- ·:1t

238 9. Electromagnetic Induction 9.6. Magnetic Energy of Two Current l,oop.~ 239

Let us find this work over the time dt:

c5Aadd = -(tal + til) II dt - (£82 + ei2) /2 dt = dW.

Considering that t SI = -LldIlldt, til = -L12 dI 2ldt,etc., we transform this formula as follows:

dW = L1/ 1 d/1 + L12[1 d/ 2 1- L 2/ 2 d/ 2 1- L 21/ 2 (1I 1•

Taking into account the fact that L 12 = L 21 , we representthis expression in the form

dll' = d (LJ~/2) + d (L 2 /;t'2)+ d (L 12/ 1/ 2 ), ~

wlleflce ~

IW=¥ +¥ -I LIZ/liz· \ (!I.:H) Fig. 9.14

The first two terms in this expression are called the intrinsicenergies of CUl1'ents II and /2' while the last term, the mutualenergy of the currents. Unlike the intrinsic energies of currents, the mutual energ¥ is an algebraic quantity. A changein the direction of one of the currents leads to the reversalof the sign of the last term in (9.34), viz. the mutual energy.

Example. Suppose that we have two concentric loops with currentsI and I whose directions are shown in Fig. 9.14. The mutual energyot these ~urrents (W12 = 1. 21112) depe~ds on thr~e.alge~raic.quantit~eswhose signs are determineJ by t~e chOIce of pos~tlve dIrectIOns of CIrcumvention of the two loops. It IS useful to verIfy, however, that .thesign of W (in this case W12 > 0) depends only on the mutual ?flentation of lhe currents themselves anft is independent of the chOice ofthe positive direction~ of circumventi?n of t~e loops. We recall thatthe sign of the quantity La was conSidered m Sec. 9.4.

Field Treatment of Energy (9.34). There are some moreimportant problems that can be solved by calculating themagnetic energy of two loops in a different way, viz. fromthe point of view of localization of energy in the field.

Let B1 be the magnetic field of current. /1' and ~2.thefield of current /2' Then, in accordance WIth the pnncipleof superposition, the field at each point is B~-= B1 -t- B2 ,

and according to (9.31), the field energy of this system of

currents is W = ~ (H2/2~lflo) dV. Substituting into this

formula .13 2nO B~ + B; + 2B1B 2 , we obtain

9.6. Magnetic Energy of Two Current Loops

Intrinsic and Mutual Energies. Let us consider two fixed10J}ps 1 and 2, arranged near one another at a sufficientlysmall distance (so that there is a magnetic coupling betweenthem). We assume that each loop includes its own sourceof constant e.m.f. Let us close each loop at the instantt = O. In each loop, its Own current will start being t-lsLablished, and hence self-induced c.m.f. t 8 and the c.m.£. Ciof mutual induction will appear. The additional work donein this case by the sources of eonstant e.m.£. against ~sand t i is spent, as was shown above, for cTcating the magnetic energy.

constant along the tube, and hence d<1J can be taken out of the integral:

dW = dq,{ II dl = I d<1J2 ';Y 2 •

Here we used the theorem on circulation of vector H (in our case, projection HI = H).

Finally, we sum up the energy of all elementary tubes:

W= iI .\ d¢J=I¢J/2=Lf2/2,

'"where ¢J ,= LI is the total magnetic flux enveloped by the currentloop, Q.E.D.

Determination of Inductance from the Expression forEnergy. We have intlfoduced the inductance L as the proportionality factor between the total magnetic flux (J) andellrJ'Cllt l There is, however, another possibility of ealculatin/.{ 1.1 froUl the exprcssioJl for energy. Indeed, COlli paringformulas (9.31) and (9.29), we see that iIl' the absence offerromagnetics,

l' BIL=Ji" \ -dV. (9.33)

• IJlJo

The value of L found in this way is free of indeterminacyassociated with the calculation of the magnetic flux <1>in formula (9.14) (see p. 227). The discrepancy appearingsometimes in determining L through formula (9.33) andfrom the expression (9.14) for the flux is illustrated inPrOblem 9.9 about a coaxial cable.

(9.37)

(9.41)

<1>} = L}1} + L 1212 , <1>2 = L 21 2 + L 211}. (9.38)

According to the law of conservation of energy, the workbA* done by the sources of e.m.f. included into circuits 1and 2 is converted into the heat /jQ and is spent to increasethe magnetic energy of the system by dW (due to the motionof the loops or the variation of currents in them as well asto accomplish the mechanical work bAm (as a result of adisplacement or deformation of the loops):

bA * = bQ + dW + bAm. (9.39)

This formula forms the basis for calcula.ting the mechanicalwork bAm and then the forces acting in a mllgnetic fwld.

Formula (9.41) can be used for obtaining simpler expressions for /jAm by assuming that either all the magnetic fluxesthrough the loops, Or currents flowing in them remain unchanged during a displacement. Let us consider this ingreater detail.

1. If the fluxes are constant (<t>" = const), it immediately

16-0181

9.7. Energy! and Forces in Magnetic Field 241

into account that

We assumed that the capacitance of the loops is negligiblysmall and hence the electric energy can he ignored.

Henceforth, we shall be interested not in the entire work6A * of the source of e.m.f. hut only in its part wllich is doneagainst the induced and self-induced e.m.f.s (in each loop).This work (wlW;h was called the additiOl:aJ w",l\ is equal

to bAadd = -(~i} + ~B1) I} dt - (~12 .t- (s ,) I dt.•

Considering that ~ i + ~8 = -d<t>/dt fOJ C';I, loop, we

can write the expression for the. additional wllrk ;t lhe form

6A add = I} d<1>} + 1 2 d(l>"o (9.40)

It is just this part of the work of the e.l' I 'iomces (thework against the induced and self-induc,"! c.nd.f') thatis associated with the variation of the nuxes (1\ and <1>2and spent for increasing the magnetic energy of tile systemas well as for accomplishing the mechanicnl work:

(9.36)

IW= i-')~ dV+ r2LdV+ I' -!Jl·B2 dV I (9.35)• 1__~l.lJ.lo J 2l.lft o J l.ll.lo • I

1( he correspondence between individual terms in formulas9.35) and (9.34) is beyolHl doubt.

Formulas (9.:3'1) and (9.35) lead to the following importantsequences.

1. The. magnetic energy of a system of two (or more)CUl'wnts IS an essentially positi~e quantity (W > 0). This

rOllo\~s from tile fact that tv ex: \ Jr2 dF, wllOre the integrandcontall1S positive quantities. •

2. The ~nergy of currents is a nonadditive quantity (dueto, tll,n, ex lsten~e of m 1I1llal energy). .

.3. I he last lJItogTHI .in (~).;3:l) is proportional to the pro-l!t,Ct 11/ 2 of Cllrren t,<; since H ex: 1 'IJIII I? ex: I 'I'] .t' I . . 1 1 ( ., .,. lC pI 0-POl' IOJ,Ja Ily fa~tol'. (i.e. tile rpmaining- ilJtogral) turns outto he I'ymmetnc WIlli rospeet to indices 1 and 2 and hencec({~n he denoted by L I2 01' L 21 ill accordance with formulav.34). Thus, L I2 == L 21 indeed.

. 4. Expression (9.35) leads to another defuiition of mutualI~ductance L12 . Indeed, a comparison of (9.3.')) and (934)gives .

L12

= _1_1

- r ]l1 ·B2 av.II 2 J l.lfto

~ . -=.9-,-._ Electromagnetic Induction

9.7. Energy and Forces in Magnetic Field

. The most general method for determining the forces actingI? a magnetic fwld is. the energy method, in which the expresslOn

Tfor the magnetic fwld energy is used.

":~. ~hall confine our.selves to the ease when the systemconSIsts of two loops With CULTPnts 1 ~Jld 1 'r-l t'

. . • 1 (A 2' Ie magne ICenel gy of such a system can be represented in the form

U? 1rr '" - (1 (Il -+-- 1 (I) )2 111 22'

where tVJ and (1)2 are the total magnetic fluxes piercingloop~ ~ all,d ~ respectivelr,. This expression can be easily?b~allIed from for~ula (\:1.01) by representing the last termIIllt as the sum (1i2) L 12IJ2 + (1/2) L 2 / 2/

1and then taking

Indeed, for 1k const, it follows from formula (9.37) that

dWlr = ~ (11 d<t\ + 12 d<tl 2 ),

where the subscript <D indicates that the increment of t.hemagnetic energy of the system must be calculated at constantfluxes through the loops. The obtained formula is similarto the corresponding formula (4.15) for the work in an elec-tric field. .

2. If the currents are constant (I k = const), we have

243

Fig. lUG

o x X

~-~

9.7. Energy and Forces in Magnetic Field

X_---1----__

B

-.:-

Fx = _ ao: 1;= ~22 ~: = I; ~: 'I.e. the calculations carried out with the help of the two formulasgive in accordance with (9.44), the same result.

~ample 2. Interaction between two current-carrying coil~. Twocoils 1 and 2 with currents 11 and 1 2 are mounted on a ma~nehc core(Fig. 9.16). Suppose that the mutual inductance of the ~.ods dependson their separation x according. to a known law £12 (x). lllld the forceof interaction between the coils.

The magnetic energy of the syste~ of two .coils is given by formul~(9.34). For determining the force of lllteractlOn.. we shall use expres. (9 43) Let us displace coil 2 through a distance d:1: at consta!?t

510n t' [' and 1 The correspondin<r increment of the magnetic.curren s 1 < 2' "energy of the system is

dW 11 = 1112 dL I2 (x).

1 Inecllanl'cal work 6A m = F2~ dr, according toSince the e ementary ~

(9.43) we obtai n

or

OL I2 (x)F2x =1112 ax .

Let currents 11 and 12 magnetize each other. Then L 12 > 0, t: n1fordr > 0 the increment dD 12 < U, i.e. F2x < U. Consequently,. t ~ orc~exerted on coil 2 by coil' is the attractive force: vect.or 1"2 IS dlrecte(to the left ill the ligure.

16*

Fig. \J.15

In the case under consideration, t~lC ma~netic en.ergy o~ the systemcan be represented, in accordance With (\1.29), as follows.

W = L12/2 = (P2/2L,

where <D = Ll. Let us displace the jumper, for example, to the rightby dr. Since 6A m = F", dx, we have

OW I 12 OJ,F -'-- =--,,,'- ax I 2 ax

I, t I' e L (x) is known. Find the Ampere's force acting on the( Ina e x, . . d f' ,h tjumper, in two ways: for 1 = const an or .... = cons.

(9.42)

(9.43)16Am=dWlr./

9. Electromagnetic Induction24~

follows hom (9.11) that

1=--6A-n-.=----1~1

i.e. in this case, in accordance with (9.40), the incrementof the magnetic energy of the system is equal to half of theadditional work done by the sources of e.m.f. Another halfof this work is spent for doing mechanical work. In otherwords, when the currents are constant, dWlr = 6A m,Q.E.D.

It should ·be emphasized that the two expressions (9.42)and (9.43) obtained by us define the mechanical work ofthe same force, Le. we can write

F·dl = -dWI<1> = dWlr. (9.44)

In order to calculate the force with the lielp of theseformulas, there is no need, of course, to choose a regime inwhich either magnetic fluxes or currents would necessarilyremain constant. We must simply find the increment dWof magnetic energy of the system provided that either <Ilk == const or 1k = const, which is a purely mathematicaloperation.

The value of the obtained expressions (9.42) and (9.43)lies in the.ir generality: they are applicable to systems consisting of any number of contours-one, two, or more.

Let us consider several examples illustrating the application of these formulas.

Example 1. Force in the case of one current loop. Suppose thatwe have a current loop, where AB is a movable jumper (Fig. 9.15).The inductance of this loo~ depends in a certain way on the coor-

I'

III

y

ds = [ dr" d I J

Fig. 9.18

f,

xFig. 9.17

Problema 245---_=....:....::..:..:..::.:::.:=-------_._----

Solution.. Let u~ consider an element dt of loop length, which ata given instant moves at a velocity v in the magnetic field B. Inaeoordance with formulas (8.4) of field transformation, in the referencesystem fixed to the given elemQllt the electric field E = [v X B] willbe observed. It should be noted that this expression can be also obtained with the help of the Lorentz force, as it was done at the beginning of Sec. 9.2.

Circulation of vector E around the entire contour is, by definition,the induced e.mJ.:

~i ~-= ~ Iv X BI dt. (1)

Let us now fllld the corresponding increment of the magnetic fluxthrough the loop. For this purpose, we tum to Fig. 9.18. Suppose thatthe loop was displaced from position r 1 to r s during time dt. If inthe first position the magnetic flux through the surface S1 stretchedover the loop was <1>11 the corresponding magnetic flux in the second,position can be represented as (1)1 + dl1l, Le. as the flux through thesurface S + dS. Here d<D is the required increment of the magneticnux throngh the narrow strip dS between contours r l and r 2•

Using Fig. 9.18, we can write

d<1>,= ~ B.dS=.\ B.[drXdIJ=-~[drXBldl. (2)

Here (1) the direction of normal n is matched with the direction ofcircumvention of tile contour, viz. with vector dl (right-handed system); (2) the direction of vector cL'!, viz. the area element of the strip,is matched with the choice of normals n, and (3) the following cyclictransposition is \lSi'll in the scalar triple product:

a·lb X e] = b·le X a) = e·[a X b) = -Ib X a)·e.

This formula shows that 'jgi ex: y. The minus sign indicates that ~iin the figure acts counterclockwise

• 9.2. A loop moves arbitrarily. A closed conducting loop is movedarbitrarily (even with a deformation) in a constant nonuniform mag·netic field. Show that Faraday's law (9.1) will be fulfIlled in this case.

PMlems• !.l.f. Jnduced emf A w· . th h

is in a uniform magneii~ fiel:ran e s ape of a parabola y = kz2A )umper translates without initial v~~p~tndlc~ar to the plane XY.atlOn a from the apex f th b CI yan at a constant aceelerinduced in the formed .c~ntou~ ~:r: f~la lFig. f9.1h7). Fin~ the e.m.f.

• . DC IOn 0 t e coordmate y.Solutlon. By definition' rS - dll>ld .

to the plane of the contour'in lh~-dir / Hfvlllg chosen the normal n= B dS wh dS _ .ec IOn 0 vector B, we write dl1l =obtain' ere - 2x dy. ConSidering now that x = Vulk , we

~i= -B·2 ¥ylk dYldt.During the motion w'th= '/2'- d I I a constant acceleration, the velocity dyldt =

rag, an lence

244- ~9~._.!.Electromagnetic Induction

Example 3. Magnetic pressure t·Let us mentally increase the r (/" ac leg on SOk:noid willding.

by dr:, retaining t.be current I th:o~uh ~b th~. so~en01d cross sectionAIII perl' S f~rces ~ill accomplish the ';ork 6A \\ 1~lDg constant. Thenuuder consideratIOn, we have m - dW fl' In the case

t'>A m = pS drwhere p is the required P d ~solenoid, and· ressure an S IS the lateral surface of the

dW 11 ~~ d ( BS v) ccc~ S'd2140 :!11o < r.

ITpre we took into account the f t has well. Equating these two ex"'::o ' t at when I. = COllst, B = canst

...~Ions, we obtamp = BS/2p.o.

Magnetic Pressure The .in Example 3 can be 'genera~f:ierJs;lO~hfor pressure obtainednetic field is different (B and B ) or d .effcase when the magface with current (cond 1 t' 2 on I erent sides of a sur-

uc IOn current Or mag t' t'rent). In this case the mag t' . ne Iza IOn cur-, ne IC pressure IS given by

p=/. HI·HI -. Bs·H. I2 2' (9,45)

The situation is such as if the re io' .energy density were the region ~f ~.~Ith a hIgher magnetic

Helation (9.45) is one of the ba .Ig Ielr p.ressure.hydrody . I . h SIC re atlOns in magnetoing liq~f(~lC(i;l~~ec~~~~;fs:~e .beha."iour of electroconduct=

gllleel'lng and astrophysics).

[I

24fl .'J. Eleclromfl!:netic Inrlnclion Problems247

(5)

(1)

a

b

o---+---,I

2 IIL.-__~ __ --l

O'

Fig. 9.21

10 a

dp .. ' dB o(it =: ei o -rit .

Fig. ..(l.20

l,2m---ccevBo·

r o

The last equation can be written, after cancelling V, in the form

[J = eroBo,

. ''I resllecL to time, consideringWe differentiate this equatIOn WI. I

that ro,= const:

E ~c ~ :t (B). (3)

. k' 'nto account (2) and (3), in termsLet us now wnte Eq'I(1), ta IlIf I d the normal to the trajectury:

of the projections on to t Ie tangen an

dp-eE.-e!'!!- ~.(R), (4)7- ,- 2 dt

. hIt' momentlllli as P c.= p'f and lind its time derivativewe wrl te tee ec ron -..

d d [,2

dp _ft'f+p-!-=.L'f!m-n, (2)dt"- dt dt dt '0

. t that p - nlV In being the relativistich t k into accoun -, F' 9 20)were we 00

1t __ (vlro)n (this can be easily seen fr~m Ig. ' •

mass, and d'f d - __ ( dtl )n and the rest is ObVIOUS,Indeed, d'f = dcp'dn, -- tV Fro aday's law 2rrroE =~ Id<I)/dt I, where

Besides, accor mg 0 ar 'etJ = nrfi (B). Hence

dp/dt = eE + e Iv X Do),. . Id 't of the projections onto

where E is the vortex electnc he 's: ~rlJ.1s t ry 'For this purpose,the tangent 't and the normal n to e raJec 0 .

Let us wrl·te the relativistic eqnation for motion of anSolution.

electron,(3)

Fig. 9.19

Dividing expression (2) by rll, we find

df!>/dt = - ~ Iv X DI dl,

~j=.\ e;(r)dN.

Having integrated this expression, we obtain the following for themaximum value of the induced e.m.f.:

~im = (1/3)rraWBow.

• 9.4. A coil of N turns wi th the cross-sectional area S is placedinside a long ;;olenoid. The coil is rotated at a constant angular velocityw around the axis coinciding with its diameter and perpendicularto the axis of the solenoid. The magnetic fIeld in the solenoid variesaccording to the law B =, B o sin wt. Find the e.m.L induced in thecoil, if at the instant t = 0 the coil axis coincided with the axis ofthe solenoid. .

Sdlution. At the instant t, the total magnetic flux through thecoil is

etJ = NBS cos wt = NBoS sin wt·cos wt = (1/2)NB oS sin 2wt.

In accordance wi th Faraday'slaw, we have

0j = -detJ/dt = -(1/2)NBoS·2w cos 2wt = -NBoSw eos 2wt.

• 9.5. The betatron condition. Show that electrons in a betatronwill move in an orbit of a constant radius ro prOVided that the magneticfield Bo on the orbit is equal to half of the value (B> of the magnetic

fIeld averaged over the area inside the orbit, i.e. Eo = <:> .

where v = dr/dt. It remains for us to compare (3) with (1), whichgives ej = -detJ/dt.

. .• 9.3 .. A flat coil ,,:,ith a large number N of tightly wound turnsIS In a umform magnetic field perpendicular to the plane of the coil(Fig. 9.19). The external radius of the coil is equal to a. The magneticfield varies in time according to the law B = B 0 sin wt.' Find themaximum [value of the e.m.f. induced in the coil.

Solution. Since each turn of the coilpractically does not differ from a circle. thee.m. f. induced in it is given by

I'j = - df!>/dt = -rrr2Bow cos wt,

where r is the radius of the turn under COll

sideration. The number of turns correspondingto the interval dr of the values of the radiusis dN = (N/a) dr. The turns are connected inseries, hence the total e.m.f. induced in thecoil is

2489. mectromagnclic Induclion

Problems 249

A comparison of Eqs. (5) and (4) gives

~ IJ _ 1 ddt 0 -"2 dt (8).

In particular, the last condition will be satislied when

1lJo""" 2 ((I).

Practically, it is altai/It'd b Ill' f.. .form (in the form of 'I hluntYI <tn,lI aclllrJIlg pole pieces in a speciul. ,-lusel cones).

• !1.6. Induced CUrrent A . .a long straight conductl)I' w·I't'h I~qnare wire frallle with side a and(F' 9 2 a I Ireet eurr' t/ I' ', Ig. . 1). The inductance of the f' . . en 0 "! 1/1 the same planefhe frame has been rUlated l' "rame loS" and Its resistance is R.then stopped. Find the arnoun\ I~fu; '~ ,I ~(~. aroulHI Ihe axis or)' and~he frame. The distancl' b bet ' t~cl,rI~I,ty l/}at has 111lssell throlll{hIS assumed to be known. ween e aXIS 00 and the straight wire

Sulutiun. According to Ohm's la\ Ifrullle duril1" its rutali . I .v, t Ie current / appearirl" in the

,., . un IS eetermllled by the formula ,.,

'R1= _ dl1> _ L~dt dt •

Hence the requiredarnollnt of electricity (charge) is

_ r 1 •q-- I I dt= --- \ (dIll I I dl) 1

• R.' -=-7f(M>[-!.M).

~ince the frame has been sto I, f . .valllshes, and hence A/- 0 aPpel ,! tcr rotatIOn, the current in itof the flux Art> through 'tlll: fra~~n(r~~lJ f~r ~s to find the increment

et us choose the normal n to th I 2 - 11>1)'rl that in the lin.al Positior; n is dire~t~/b~h~f ~he frame, for instance;'ah?tg .B). Then It can be easily seen that' the ihelPlane of the figure

II' I e III the initial positio~ <1>' ,JII e ma position 11>2 > 0and A$ turns out to be' I I < 0 (the normal is opposite to B)'boumled by the jj~al aneti~~·~'leqU<~I.to the flux through the surIac~

I Ia 'pOSI tlOns of the frame: ,.

b+a

All> = $2 +1$11= \ Ba dr,

b:a

Where B is a fllnction I Ihelp. of the the~~em ~n "'ci\;c~)lset.rurm cau be easily found with the

FJIlall . t . a lOll., y, oml trng the minus sl'gn bt ., , we 0 aln

q='~ ~lon/o I b+ an ~ n -,;::::-a .

Il can be seen that the obtai:1ed quantity is independent of the induct·ance of the circuit (the situation would be oifferent if the circuit weresupereonriueting).

• 9.7. A jumper 12 of muss m slides without friction along two longcondncting rails separated by a distance l (Fig. 9.22), The sys~em,isin a uniform magnetic fIeld perpendicular to the plane of the CIrCUIt,

L1

1

R~ 1L2

y

®B .. •X

0"-':2

Fig. \).22 Fig. \.l.2:i

Tile Idl ends.~t the rails are shunted through resistor ll. At the instantt ''= 0, the jllinper received the initial velocity Vo directed to the right.Pintl the velocity of the jumper as a function of time, ignoring theresistances of the jumper and of the rails as well as self-inductance ofthe circuit.

• Solution. Let us choose the positive direction of the normal nto the plane of the circuit away from us. This means that the positivedirection of circumvention of the circuit (for induced e.m.r. and current) is chosen clockwise, in accordance with the right-hand screwmle. It follows from Ohm's law that

dtlJ dSRl = --F== -II d't= -Hlv, (1)

where we took into account that when the jumper moves to the right,d«1> > O. According to Lenz'" law, the induced eurrent I causes theAmpere force counteracting the motion, directed to the left.

Having chosen the X-axis to the right, we write the equation ofmotion of the jumper

m dvldt = /lB, (2)

where the right-hand side is the projection of the Ampere force ontothe X-axis (this quantity is negative, but we omit the minus signsina, as can be seen from (1), current I < 0).

Elinlinating' I from l':qs. (1) allli (2), we obtain

dvlv = -a dt. a = IJ2/2/mR.

The integration of this expression, taking into account the initialcondition, gives

In (171170) = -at, v = vee-t

251Problems

.• 9.8. The role of t .Fig. 9.23 the f ranslent processes In th . .inductan~es L e.:.. I!f of the source, its iniernal e .clrcUlt shown incurrents establish:d I;z ~1 sup~rconducting coils :::Iktance R ~nd theS w e cOlIs after key K h b nown. Fwd the

and ~L:~on. Let us use Kirchhoff's laws for:~ :~n closed.. ec rIC circuits ~Ll

Fig. U.25Fi~ U.24

where Bl

is determined with the help of the theorem on circulationof vecto'r n. Taking' the time derivative of <I)] and multiplying theobtained resnlt by N, we fllld the following expression for the amplitude

valm' of induced e.m.£.:'0int =~(Jllm~In!!....

2il a

• 9.11. Calculation of mutual inductance. Two solenoids of thesame length and of practically the same cross section arc completelyinserted one into the other. The inductances of solenoids are L] and L

z·

Neglecting ellge cUeds, fmd their mutual inductance (modulo).

Solution. By definition, the mutual inductance isLIZ = <lIP?; (1)

where <!ll is the total magnetic nux through all the turns of solenoid 1provitlell tbat current I"z flows in solenoid 2. The flux <D

I

= NI!J2

S

,where N l i8 the number of turns in ~o\enoitl 1. 5 is \he eross-sedlOnul

B

diownsiolls

,Ire given in Fig. H.25. The number of tllrns on the coilis N, allti the permeability of tbe surrounding medium is unity. Findthe amplitude of tlte e.Ill.£' induced in this coil if the current I == 1

mcos wt flows along the straight wire.

Solution. The required e.mJ. 8; = _det>ldt, where et> = Net>land <1)1 is the magnetic flux through the cross section o[ the coil:

Ii

\

' '\' flo !.lohI 1 b(111

0

Bn d::> 0_ -')- Ih dr = .0...----2. n -,• • ~nr n a

n

(erroneous) result, namely, insteal\ of 1J1i, 1/2 is obtained in the parentheses. The thinner the central wire, "i.e. the larger the ratio bla,the smaller the relative diflerence in the results of calculation by thesetwo methods, viz. through the energy and through the flux.

• 9.10. Mutual indudion. A long straight wire is arranged alongthe symmetry axis of a toroidal coil of rectangular crosS section, whose

(1 )

(1)

(2)

while

9. Electromagnetic Induc~ion

B ~ !.loIr<a - 2~ z r, B ,- ltoI 1"a a<rb - . (2)

e orm of these de(2), integral (1) is r~dences is shown in F'obtain sp It mto two parts'\ 19. 9.24. On account of.. s a result of integration, we

L u = .&.. (..!:-- -+- In ~ )2n 4 a'

It should he noted thatterms of magnetic flux by thenf~;~:;I~tLto~etermine this quantity in• II - ¢\/J leads to a different

FJ10 + 120 = 10 = ~ IR

rom these eq t' .ua IOns, we find

'fi, L110=- Z 6 LR L +L ' J20 = - 11 2 R L LL .

. • 9.9. Calculati f'. I', 2mternal solid conduc on 0 ~n~uetanee. A coaxial· .tube. of radius b. Fi~dr l! r~d1Us a and extemal thi c~b~e C?nSlsts. ofconsIdering that th t e lllductance of a unit I n iliaU

canductmginternal conductore.curr~nt distribution over the eng. of ~he cable,everywhere IS umform. The permeabTt c~oss sectIOn of the. I I Y IS equal to unity

Solution. In the Cnot thin and h ase under considerati tl'

(t~e3~)nergy rath::~ha~et~~o~:~athceshould ~bi d:~e:~~~~dc~~dt~~tor iSr1 ,we can write e magnetIc nux. In acco d ms.or ance WIth

b1 r BZ

L u = J2 .\ -;;- 2rtr drii"'o '

where r is the distintegral, we mu fince from the cable axis In d-circulation, we hstavend the dependence B(·r). or er to evaluate thisUsing the theorem on

RI ='f,-L dl l RI d/1 dt' = 'fb - L __z

A

2 dt

comparison of the .for stabilized curre~texpresshlOns shows that L d1 = Ls we ave t 1 zdJz,

Besides, Lt/

10 = LzJzO'

250

where lD is the magnetic flux through the coil at the beginning ot theprocess (we omitted the minus sign since it is immaterial).

Thus, the problem is reduced to the calculation of the flux $ throughthe coil. This quantity cannot be determined directly. This difficulty,however, can be overcome by using the reciprocity theorem. We mentally replace the magnet by a small current loop creating in the surrounding space the same magnetic field as the magnet. If the area ofthe loop is 8 and the current is /, their product will give the magneticmoment I'm of the magnet: Pm e 18. According to the reciprocitytheorem, f'l?/= LnT and the problem is reduced 1.0 determining the'magnetic nux through the area 8 of the loop, which creates the samecurrent /. but flowing in the coil. ARsnming the field to be uniformwithin the loop, we obtain

<JI = BS = ItoN/ S/2a. (2)

(10.1)

S' S'~........ ,-,

- (+ \ S (\.

~------J I~-L~I-.n ~nIl!:J'\. \_./ r

(a) (b)

Fig. 10.1

~ Hdl= ) jdS.

10. Maxwell's Equations.

Energy of E~edromagnetic Field

10.1 Displacement Current-------~~

Let us apply this theorem to the case when a preliminarilycharged parallel plate capacitor is being discharged througha certain external resistance (Fig. 1O.1a). For the contour rwe take a curve embracing the wire. We can stretch varioussurfaces on this contour, for example, S and Sf. ,These twosurfaces have "equal rights". However, current I flows through:mrface S, while through surfaco S' there is nO current!

It turns out Lhat the circulation of vector ijdepends onthe surface stretched over a given contour (?!), which i.s

.O'i....e.enf Current

Maxwell's Discovery. The theory of electromagnetic fieldfounded by Faraday was mathematically completed byMaxwell. One of themost important newideas put forward byMaxwell was the ideaof symmet,fy in themutual dependenceof electric and magnetic fields. Name-'ly, since a time-varying magnetic field(aR/at) creates an electric field, it should be expected thata time-varying electric field (aE/at) creates a magnetic field.

This idea necessitating the existence of a new in principlephenomenon of inductioD. can be approached, for example,by the following line of reasoning. We know that accordingto the theorem on circulation of vector H,

It remains for us to substitute (2) into (1) and recal! that /8 = Pm'Then q = floNPm/2aR, and

pm = 2aRq/lloN.

(1)

fJ. EleclrtJlnll!:netic Induction-------

Fig. 9.26

area of the solenoid, and B 2 = f-tJ.1onS/2' Hence formula (1) can bewritten (after cancelling /2) as follows:

I L I2 1=f-tf-ton2NtS=llllonln2V, (2)

where we took into account that N 1 = ntl, where l is the solenoidlength :md lS ,= V i~ its volume. Expression (2) can be representedill terms of L1 and ['2 as follows:

I L12 1= r'" f-tflon(V YflflonW= yrL1L2 •

It shonld be noted that this C'xpres~ion definC's the limiting (maximum)value of I L12 I· In general,! L12 1< YL 1L2•

• 9.12. Reciprocity them·em. A small cylindrical magnet M isplaced at the centre of a thin coil of rarlius a, containing N tnrns(Fig. 9.26). The coil is connect.ed to a ballistic galvanometer. TheresistalKeof the circuit is R. After the magnet had been rapidly removed from the coil, a chaq;e q passed through the galvanometer.Find the magnetic moment of the magnet.

Solution. In the process of removal of the magnet, the total magnetic flux through the coil was changing, which resulted in the emergence of induced current defined by the following equation:

R / = __ d(D _ L d/dt dt .

We multiply both sides of thisequation by dt and take intoacconnt that / dt = dq. Thisgives

R dq = -d<D - L d/.

Having integrated this expression, we obtain Rq = -t.<D - L t./.Considering now that !1/ = 0 (the current was equal to zero at thebeginning as well as at the end of the process), we obtain

q= !1<D/R = $/R,

252

'[I

254 10. Ma.xweU's Equations. Electromagnetic Field EnerlfY 10.1. Displacement CUr/ent 255

On the other hand, according to the continuity equation(5.4), we have

(10.7)

(10.8)

it is sufficient to introduce on the right~hand side of Eq.(10.1) the total current instead of conduction current, viz.the quantity

Indeed, the right-hand side of (10.7) is the sum of conduction current I and displacement current I d :1 t = I -+ I d'

Let us show that the total current It will be the same forsurfaces 8 and 8' stretched over the same contour r. Forthis purpose, we apply formula (10.4) to the closed surfacecomposed of surfaces 8 and 8' (Fig. to.1b). Taking into account the fact that for a closed surface the normal n is directed outwards, we write

It (8') + It (8) = O.

Now, if we direct the normal 0' fo:-, surface 8' to the sameside .as for surface 8, the first term in this expression willchange the ~gn, and we obtain

It (8') = It (8),

Q.E.D.Thus, the theorem on circulation of vector H, which was

established for direct currents, can be generalized for an arbitrary case in the following form:

I~Hdl= J(j-+{¥-)dS.In this form, the theorem on circulation of vector H is always valid, which is confirmed by the agreement of thisequation with the results of experiments in all cases withoutany exception.

(10.2)

(10.6)

(10.4)

(10.3)

sides of Eqs. (10.2)

r\'~ dS = !.!L'j' fJt at .

,h .. dS = _.!!!L,'j' J at .

Summing up the left- and right-handand (10.3) separately, we obtain

~ (j+ ~~) dS=O.

This equation is similar to the continuity equation fordirect current. It can be seen that in addition to the density jof the conduction current, there is one mOre term aD/atwhose dimensions are the same as for current density. Maxwell termed this addend the density of displacement current:

jd = aD/at. (10.5)

obviously ruled out (and never happened in the case ofdirect currents).

Is it possible to change somehow the right-hand side of(10.1) to avoid this inconvenience? Fortu nately, th is canbe done in the following way.

First, we note that surface 8' "cuts" only the electritfield. In accordance with the Gauss theorem, the flux of

vector D through a closed surface is ~ D dS = qj whence

The sum of the conduction and displacement currents iscalled the total current. Its density is given by

. .+ aDJt=] 7ft.

Several Remarks about Displacement Current. It should

where the curl of vector II is determined by the density j of conductioncurrent and an/at of displacement current at the same point.

According to· (10.4), the lines of the total current are continuous in contrast to the lines of conduction current. Ifconduction currents are not closed, displacement currentsclose them.

We shall show now that the introduction of the concept oftotal current eliminates the difficulty associated with thedependence of circulation of vector H on the choice of thesurface stretched over contour r. It turns out that for this

Ditlerential form of Eg. (10.8):

VXJI=j+ ~~ , (10.9)

256 In. Maxwell's Equation.~. Electromagnetic Field Energy 10.2. Maxwell's Equations 257

(Fig. 10.2) would differ from zero, which is in cuntradi,',tiun withEq. (7.2). Consequently, vector B Olust be perpendicular to the radialdirection at the point P. But this is also impossibh>, sincl' all directionsperpendicular to the radial dirrclion arc absolutely I'quivull'nt. Itremains for us to conclude that magnetic flCld is equal to zero every-where.

The absence of magnetic field in the presence of electric currentof density j indicates that in addition to conduction current j, displacement current jd is also present in the system. This current is such thatthe total current is equal to zero eYcrywhere, Le. jd = - j at eachpoint. This means that

. . I q iJDJd = J= 4nrz c= 4nrz = 7ft '

where we took into account that J) = q/4nrz in accordance with theGauss theorem.

10.1. Maxwell's Equations

Maxwell's Equations in the Integral Form. The introductionof displacemtilUt current brilliantly completed the macroscopic theory of electromagnetic field. The discovery of displacement current (aD/at) allowed Maxwell to create aunified theory of electric and magnetic. phenomena. Maxwell'stheory not only explained all individual phenomena of electricity and magnetism from a single point of view, but alsopredicted a number of new phenomena whose existence wassubsequently confmned. . ,

Hitherto, we considered separate parts of this theory.Now we can represent the entire theory in the form of a system of fundamental equations in electrodynamic.s, calledl'vfaxwell's equations for stationary media. In all, theN arefour Maxwell's equations (we are already acquainted witheach of these equations separately from previous sections;here we simply gather them together). In the integral form,the system of Maxwell's equations is written as follows:

(10.10)

(10.11)

and j

~ DdS= ~ pdV,

t BdS=O,"

r ~Edl= -) ~~ dS,

I'~H dl = ~ (j +- ~~ ) dS,1 _

where p is the volume density of extraneous chargesis the density of conduction ClllTent.

be kept.in mind that displacement current is equivalent toconductl~m current only from the point of view of its abilityof creating a magnetic field.~ispla~eme~t currents exist only when an electric field

varIes with time. In dielectrics, displacement cnrrent consists of two essentially differentcomponents. Since vector D == BoE+ P, it follows that the density iJD/iJt of displacement CUf

rent is t.he sum of the densitiesof the "true" displacement current,BoiJE/iJt, and of the polariztllioncurrent ap/iJt. The latter quantityappears due to the motion ofbound charges. There is nothing un-

Fig. 10.2 expected in the fact that polari-. zation currents excite a magnetic

field~ SlIlce these currents do not differ in nature from conductIOn currents. A new in principle statement consists inthat the other component of the displacement currentBoiJE~at, which is not connected with any motion of ch.and. IS only due to the variation of the electric field alsoeXCites a magnetic ~eld. Even. in ~ vacuum, any tem'poralchange of .an electnc field eXCites In the surrounding spacea magnetiC field.

The d~s~overy of this phenomenon was the most essentialand decIsIve s.tep made by Maxwell while constructing theelectrom.agnetIc field theory. This discovery was as valuableas ~he dlscov.ery of elec~romagnetic induction, according towhICh a varymg magnetic field excites a vortex electric field.It should also be noted that the discovery of displ.arementcu~rent b.y Maxwell was a purely theoretical discovery ofprImary Importance.

Let liS consider an example in which displaooment currents are manifested.

Example. A metallic positively charged sphere i8 in aniBUaitehomo~~neo~sco~duclingmedium (Fig. to.2). Electric currents flowing11'! ra~lal 11IrectIOns should excIte II magnetic field. Let us find thedirectIOn of vector B at an arbitrary point P. •

First of al~, it is clear that vector B cannot have a radial component. OtherwISC, the flux or B through the surface S of the "'re

17-0181

•

258 _-.!(): J1!a.r'Nll's Equalion.~. EleclromaKnclic Field Enerr:y

In this case, the electric and magnetic fIelds are independent of one another, which allowed us to study first a constant electric fwld and then, independently, eonstant magnetic fIeld.

It should be emphasize!! that the line of reasoning whichhas led us to Maxwell's equations can by no means be considered their proof. These equa tions C:l nnot be "derived" sincethey are the fundamental axioms, or postulates of electrodynamics, ohtained willi t.he help of generalization of expeJ'-

These et)uations express in .. concise f01'l1l all our knowlOlL:,,> <l bout electrom:lgnl'tic fIeld. The content of these equaI ;OIIS ('onsis[s in the following".

1. Circulation of vector E al'ound any closetl contonr isequal, wit h tlteminussign, to the time derivative of the magnetic llux through an arbitrary surface bounded by the givencontour, lIere E is not only vortex electric Jield but also electrostatie lipid (whose circulation is known to be equal tozero).

2. The fj ux of D through any closed surface is equal tothe algebraic sum of extraneous charges embraced by thissurface,

3. Circulation of vector H around an arbitrary closed Contour is equal to the total current (conduction current plusdbplacement current) through an arbitrary surface boundedby this contonr.

4•. The flux of B through flU arbitrary closed surface isalways equal to zero.

It follows from Max\vell's eq uations for circulations of vectors E and H that l'lectric and magnetic fields cannot betreated as ipdependent: a change ill time in One of these fieldsleads to the ilmergence of the other. Hence it is only a combination of these fields, describing a unique electromagneticfield, that has a sense.

If the fields are stationary (E = constand B ,= const),Maxwell's equations split into two groups of independentequations

(10.13)

(10.11)

v >~ E = - ~~, V· D = p,

II . aD ..,. B 0V>< =J+aT' y' = .

t 1 Its 'fllese postulates play in electrodynamicsimen a l'esu , . I 'h' thethe same role as Newt.on's laws in classlca mel, alliCs or

laws i~l ther.mlodYFnamicosr· Maxwell's Equations. Equatiol1sDillerenba OJ'lU 1. ., If,

(to 10) and (11.11) can be represented il~ dlrreJ'en~la oll.n,Le.·in the form of a system of differential equatIOns, VIZ.

__~10:.:.2::.--:M~ax~w::.'d:..:.:...I'.:...s..::E:,2q..:.:.u_a__ti_on_s ._._.__25_9

E nations (10.13) show that there may b~ two causes forf t ic field First, its sources are electl'lc charges, b?th:~t:a~::us a~d' bound (which follows f~om. th~ ~q~a~~~V . D ,= p if we take into account that = eo - .

- ~ , th~' V·E ex: p + p'). Second, the field ~alwa~sV·p- p, n . 'I. . . t'me (willch ex-emerges when a magnetic he d vanes 1~1 . I ')

F d 's law of electroffingnetlc mductlOn .pr~se~1 <1I'~h:I\and Eqs, (10.14) indicate t~lHt the magnet-. fin I :e; can be e~cited by moving electrrc charges (elec-IC 10 ( , ". I . j' Ids or by both factorst.ric cUl'rents), by varYlllg e ectl'le 10 , " V., H _'"simultaneously (this foll~ws from the

leqt~t~~J~/ A J and

= .1- DD/ilt, if we take mto acco~nt ~,1U t .. . flo. iJE/DtV ~ J ..-c j'; then V X B ex: J + J + ~P/ iJt I- eo . . :

} ., i the magnetization current denSIty and iJP/dt IS

~1~e~~I~ri:atiOncurrent density. The flf~~ tl~~~e f~:[e~~:r:~~associated \V.ith moti?n ~fel~r~gex,s~~~o:'s from the eql~ais due to a time-vary lIlg I " . fi Id in nation V· B = 0, there are no sources. of magnetIc Ie Id betlll'e (cn-lled, by analogy, magnetic charges) til at wou

similar .1.0 electric clfla~g1:~wel1's equations in differentialThe Importance 0 n I. b . I s

. t 1 'n that they express tie aSlc aw,form consists no on Y I I fi 1 Is E and Bof electromagnetic freid, but abo in that tie. re)l I 'themselves can be found by solving (integratlllg t lCseeqll<1-

tiO;~~ether with the equation of mot,ion of charged particlesnder the action of the Lorentz fOlce,

u . )dp/dt = qE + q [v X Bl, (10,1;)

(10.12)

~ DdS=q,

~ B dS=O.

~ Edl =0,

~ Hdl~I,

17*

..

Maxwell's equations for,m a fundamental system of equations. In principle, this system is sufficient for describing allelectromagnetic phenomena in which quantum effects arenot exhibitecl. '

Boundary Conditions. Maxwell's equations in integralform are of a more general nature than their differentia)counterpart, since they remain valid in the presence of asurjace oj discontinuity, viz. the surface on which the properties of the medium or fields change abruptly. On the otherhand, Maxwell's equations in the differential form presumethat all~ quantities vary continuously in space and time.

However, the differential form of Maxwell's equationscan be imparted the same generality by supplementing themwith the boundary conditions which must be observed for anelectromagnetic field at the interface between two media.These conditions are contained in the integral form ofMaxwell's equations and are already familiar to us:

DIn = D2n , Eh = E 2'f' BIn = B 2n , Hl'f = H2'f

(iO.i6)(here the fIrst and last conditions refer to the cases when thereare neither extraneous charges nor conduction currents on theinterface). It should also he noted that these boundary conditions are valid both for constant and for varying fields.

Material Equations. Maxwell's fundamental equationsstill do not form a complete system of equations for the electromagnetic field. These equations are insufficient for determining the fields from given distributions of charges andcurrents.

Maxwell's equations should be supplemented with relations which would contain the quantities characterizing individual properties of a medium. These relations are CAlledmaterial equations. Generally, these equations are rather complicated and are not as general and fundamental as Maxwell's equations.

Material equations have the simplest form for sufficientlyweak electric fields which vary in time and space at a com-'paratively low rate. In this case, for isotropic mediacontaining no ferroelectrics and ferromagnetics, the materialequations have the following (already familiar to us) form:

D = EEoE, B = /-l/-loH, j = a (E + E*), (iO.f7J

10.3. Properties of Maxwell's Equations. Tl contain only tlIe first

Maxwell's Equations Are LlDea~th ::: ect to time and spaderivatives of 1ields E and B WI P f densities P and jtial coordinates, and the first powers 0

261

t~· t ~ebcbE H

--,/" ,s, \I ~r( I\ )"'....._--"",

~ (j + ~~ ) dS = O.

Hence it follows thatiJ t iJq

~ jdS= -.7ft'Y DdS=-~ -at' .. . (4) This equatIOn

which is just the continUlt~ eqfatlOl~ :~lu~e V t.hrough ast.aWS ~hflt. the current flOWIng rom

Fig. 10.4Fig. 10.3

ts '1'11e linearity of Maxwell's. 1 es and curren . .. fof clectl'lC c Ul~g . t i .' th the princIple 0 super-equations is dIrectly cOrnnledc et aWtiisfy Maxwell's equations,

. . 'f any two 10 SSt'pOSItIOn: 1 Id'n also satisfy these equa IOns.the sum o~ these !Ie ~ W{ude the Continuity Equation. This

Maxwell s EquatIons nc ation of electric charge. Inexpresses the law off ~~?se~;t us take an infinitely smallorder to make sure 0 l~t n arbitrary finite surface Scontour r, stredtclhl over ltracat this contour to a point, so(F. 10 3) an t en COIl

19. .<, . 'fi 'te In the limit, k.H dlvanishes,that surface S remams III . '.)

surface S becomes closed, and the first of Eqs. (10.11) be-

comes

10.3. Properties 0/ Mazwell'. Equations

I the well-known constants characteriz:where e, ~, alH a are . erties of the mediuming the electric and ~~gne~~l~~~ctroconductivity), and(permittivity, permeabIlity, force field due to chemiE* is the strength of the extraneouscalor thermal processes

10. Mazwelf. Equation•. Electromagnetic "eld Ener"260

I, II,

dosed surface S is equal to the decrease in charge in thisvolume per unit time.

The same law (viz. the continuity equation) can also beobtained from Maxwell's differential equations. For thispurpose, it is sufficient to take the divergence of both sidesof the first of Eqs. (10.14) and make use of the second of Eqs.(10.13). This gives V·j = - ap/iJt.

Maxwel's Equations Are Satisfied in All Inertial Systems ofReference. They are relativistic invariants. This is a consequence of the relativity principle, according to which allinertial systems of reference are equivalent from the physical point of view. The invariance of Maxwell's equations(with respect to Lorentz' transformations) is confirmed bynumerous experimental data. The form of Maxwell's equations remains unchanged upon a transition from one inertialreference system to another, but the quantities contained inthem are transformed in accordance with certain rules. Thetransformation of vectors E and B was considered in Sec. 8.

Thus, unlike, for example, Newton's equations in mechanics, Maxwell's equations are correct relativistic equations.

On the Symmetry of Maxwell's Equations. Maxwell'sequations are not symmetric relative to electric and magnetic fields. This is again due to the fact that there are electriccharges in nature, but as far as it is known at present, magnetic charges do not.exist. However, for aneutralhomogeneous nonconducting medium, where p = 0 and j - 0,Maxwell's equations acquire a symmetric form, since Eis related to aB/at in the same way as B to aE/at:

V X E = - aB/at, V . D = 0,V X H = iJD/at, V • B = O. (10.18)

The symmetry of equations relative to electric and magnetic fields does not involve only the sign of the derivativesaB/at and aD/at. The difference in signs of these derivativesindicates that the lines of the vortex electric field inducedby a variation of field B form a left-hand screw system withvector iJB/at, while the lines of the magnetic field inducedby a variation of D form a right-hand screw system withvector aD/at (Fig. 10.4). .

Oil Electromagnetic Waves. Maxwell's equations lead tothe following important conclusion about the existence of a

. . 'pIe physical phenomenon: electro.magnetic fieldnew ill pnnci 1 . h 1 curmay exist independently, ~itho~t e ectnc c a~g~s all( -rents 'A change in its state ill thiS case necessanl~ has awave' nature. Such fields are called electromagnetIc wal~es.In a vacuum, these waves always propagate at a velOCityequal to the velocity of light c. ( D/;::> )

It also turned out that displacement cl.lrr~mt a vtplays in this phenomenon a primary role. It IS Its presence,

E U

263

B

Fig. 10.6

o

10-3. Properties of Maxwell's Equations

El _

kr:n

0<\'Fig. 10.5

along with the quantity iJB/rJt, that mak.es the ~~~earalte

of electromagnetic waves possible. Any tll~e vana~lO~ () amagnetic field induces an electric fwld, wIllIe a. va.l'latl~r! ~fan electric field, in its turn, induces a. magnetic held., 1 IllS

continuous~interconversionor interactIOn of the, fwlds pr~

serves them, and an electromagnetic perturbatIOn propa-gates in space. . f I

Maxwell's theory not only predicted the e.Hstence 0 e ~c-

tromagnetic waves, but also made it possil~le ~o ~stabh~l~all their basic properties. Any electromagnetic. \\ ave, reg~r dless of its specific form (it can be a harm~mc wave 01, anelectromagnetic p~rturbation of any form) IS charactenzedby the following properties: . . .

(1) the velocity of its propa~atlO.n m a noneonductmgneuttal nonferromagnetic medIUm IS equal to .

v=c/Ve!Jo, where c=1/Veo!Joo; (10.19)(2) vectors E, B, and v (wave velocity) are ITW~l~lllly per

pendicular and form a right-hand scre~) s#stem (F Ig. 10.5)This riaht-handed relation is an intrtnSlC property of ~'!1

electrOl~agnetic wave, independcnt of the choicc of COfll'( 1

nate system;

10. Maxwell's Equations. Electromagnetic Field Energy262

(10.23)IS= [E X H]·I

10.4. Energy and linergy Flux. Poynting's Vector 265

This equation expresses Poynting's theorem: a decreasein energy per unit time in a given volume is equal to the energyflux through the surface bounding this volume plus the workper unit time (Le. power P) which is accomplished by the fieldover the charges of the substance inside the given volume.

In Eq. (10.21), W = ) w dV, where w is the field

energy density, and P= Jj.E dV, where j is the current

density and E is the electric field strength. The above expression for P can be obtained as follows. During the timedt, field E will do the work ~A = qE.u dtoverapointchargeq, where u is the charge velocity. Hence the power of forceqE is P = qu·E. Going over to the volume distribution ofcharges, we replace q by P dV, where p is the volume chargedensity. Then dP = pu· E dV = j.E dV. It remains for Usto integrate dP over the volume under consideration.

It should ~ noted that power P in (10.21) can be eltherpositive or negative. The latter takes place when positivecharges of the substance move against field E, or negativecharges move in the opposite direction. For example, sucha situation is observed at the points of a medium where inaddition to an electric field E, the field E* of extraneousforces is also acting. At such points, j = (j (E + E*),and if E* tt E, and E* > E in magnitude, the productj.E in the expression for P turns out to be negative.

Poynting obtained the expressions for the energy densitywand vector S by using Maxwell's equations (we shall notpresent this derivation here). If a medium contains no ferro~

electrics and ferromagnetics (Le. in the absence of hysteresis),the energy density of an electromagnetic field is given by

IE·D B·B

w=-Z--\--2-' (10.22)

It should be noted that individual terms of this expression have been obtained by us !'larHer [see (4.10) and (9.32)1.

The uensity of the electromagnetic energy flux, which iscalled Poynting vector, is defined as(10.21)

264 10. Maxwell's E t·---- qlla IOns. Electromagnetic F' ld EIe nergy

. (3) vectors E and B in an electr' -cIlIate in phase (see F' 10 6 oma~netlc wave always os-~ wave), the instantan:;~ls V~ll sh~wlDg a "photograph" oflllg connected t1uough the reI t~ es EE and B at any point be-

a IOn = vB or

. VeEoE=Vf.1f.1oH.' (10.20)ThIs means that E and H ( B) .maximum values vanish or

tsImultaneously attain theil'

MIl' b' , e c.axwe s rilliant succes . tl dtromagnetic theory of'light s m d Ie eve.lopment of the elec-the ,possibility of existence ~asl ~~ to hIS u~derstanding of10wlllg from differential equat~o~csI<:~~f;)~tJC waves, fol-

f~.4. ,Energy and Energy Flux. Poynting's VectorPO~lltmg s Theorem. Proceedin f '

l?cahzation in the field itseif an g rom the ~dea of energyclple of energy conservation d On the basIs of the prillenergy decreases in a certai ' :ve. Shou}~ conclude that ifdue to its "flowing out" tllrounhl~~~on, t 11S ~ay only OCCurunder consideration (tl e l .boundal'les of the regiontionary), I me !lun IS assumed to be sta-

In this respect thCle is a fof conservation of charge . .ormall an~logy with "the lawing f tl ' 'I -ex pI eSEe( by Eq (" 4) 'I'ho 11S aw consists in tl t d . J. . e mean-time in a given volume" 'I{J a ecrease in charge pel' uni.tth . IS equa to the flux f •

e surface enclosing this ' I 0 vector l throughI "I . \0 ume.

. n.a SimI ar manner, for the law f .\\e should assume that in Id't' 0 conservatIOn of energyin" a( . I IOn to the en d'. a gIven regIOn, there exists,' ergy enslty wzlllg the energy {lux density a eel tam vector S characteri-. Ifwe speak only of the en .~ts total energy in a given vergy of ~n electromagnetic field,lllg out of this volume as ~~llre W~1l change due to its flowfield transfers its ener t as ue to the fact that thei.e. accomplishes wor:Yov~ at~ubst~nce (charged particles),form, this statement can bel' . ?ttSU stance. In macroscopic

wrl en as follows:

r--~-=---=-,yS ,fA+p. iwb~r~ til\. is Jl surface element.

Fig. 10.9

267

Fig. 10.7

__--a.._--- f{J,

I

I,;,.-_--r_--- 'P 2

10.4. Energy and Energy Flux. Poynting's Vector

. 'd the wire normally to its lateral surPoynting vector is directed ~lsl the electromagnetic energy flows intoface (Fig: 10.7) COllseque? y, 1 But does it agree with the amount

~r~~?g~'~;';~ff*~;:~;' Ithe wire of lpngtll l.

EH .2nal = 2naH ·El = I'U = Rl2,

where U is the potential differen~e acr?sts. r d R is Its reslS the ends of thIS se? lO:t ~he conclusion thatance. Thus, we arr.lve into thet~e electromag~edtlc ~de~~Yco~;i~telY eonWIre from outsl eat greeverted into Joule's heat. We mus a luthat this is a rather unexpected cone

sionit should be .not~d t~~a; ~ins~ c~~~~~source, vecto~ ~hs ~~~~~ity lr the sourceI, anpd hent?nge10 ve~tor is directed outside:the oyn 1 . gy flows. th's region ~ctromagnetlc ener d!n \ urrdunding space. In other wor s,!nto t eSt th t the energy from t~e source Fig. 10.8It turns ou a 1 the WIres butis transmitted not a ong ! ductorthrough the space surroundlllg a con t·c energy viz. the flux of vec-in the form of the flux of electromagne 1 ,

tor S. l' of a balanced (twin) line.Example 3. Figure 10.8.shows ~.sec ~~~ known as well as the fact

The directions of curre.nts III the ::'Ires Canthat the wires' potent~als are <PI h rit) the Ewe find where (to the rlgh~ or to tee

Power source (generator) IS? . b ob-Th nswer to this questiOIl can e

tainedewith the help of. the ~oyntin~::~t~h~In the case under conSIderatiOn, bet d h'l

. E '. directed downwar S W 1 ewIres, vec~or . I~t d behind the flane of thevector H IS dlrec e ~ -IE X II is direc1 n dfigure Hence, wctor S - . thto the' right, i.e. the power sou~ce IS on eleft and a consumer is on the rIght.

. L t take a parallehplateExample 4. Charging of a capacl~or. i n~rin edge effects (field

capacitor with circular plates of ra~lU~:~rg~ fiux 1hrcugh the lateraldissipation), fwd the. elect~omagnll? thO region the Poynting vector"surface" of the capaCitor, Sl?Ce on.y I~O 9)sis directed inside the capaCItor (FIg. ., d t"c

. . electric fIeld E au a magne 1On this surface, there l~ ;~ v.arYl'1 rding to the theorem on cir-

field H generated by it.s vrarlllaLJOnih< tCZnaH = na2 DD/8t. where theculation or vector H, It 0 OWS li

I

dW = EoE2c dt = VEo/Ilo E2 dt.

Let us now see what we shall obtain with the help of Poyntingvector. The same quantity dW can be represented in terms of the magnitude of vector S as follows:

dW = S dt= Ell dt= lfeo/ Ilo E2 dt.

Thus, both expressions (for wand for S) lead to the same result(the last two formulas).

Example 2. Evolution of heat in a coDductor. Let a current I nowthrough a straight circular wire of radius a (Fig. 10.7). Since the wirehas a re!listance, a certain electric field E is acting along it. The samevalue of E will be at the wire surface in a vacuum. Besides, the presence of current generates a magnetic field. According to the theoremon circulation of vector H, near the surIace of the wire we have2naH = I, H = 112n4. Vectors E andll are manged so that the

Then

dW = we dt,

where w is the energy density, w = eo£2/2 + ,""oH2/2. In accordancewith (10.20), for an electromagnetic wave we have

t oE2 = lloH2 .

This means that the electric energy density in the electromagneticwave at any instant is equal to the magnetic energy density at thesame point, so we can write for the energy density

w = eo£2.

266 10. Maxwell's Equations. Electromagnetic Field Energy

Strictly speaking, Maxwell's equations cannot give unambiguous ex'pressions for the two quantities wand S.The expressions presented above are the simplest from an infinite number of possible equations. Therefore, we shouldtreat these expressions as postulates whose validity shouldbe confirmed by the agreement with experiments of theircorollaries.

Several following examples will demonstrate that althoughthe results obtained with the help of formulas (10.22)and (10.23) may sometimes seem strange, we will not beable to find in them anything incredible, any disagreementwith experiment. And this is just the evidence that theseexpressions are correct.

Example 1. EDergy flux in an electromagnetic wavc (in vacuum),Let us calculate energy dW passing during time dt through a unit areaperpendicular to the direction of propagation of the wave.

If we know the values of E and B in the region of location of theunit area, then

E

c

(10.25)

(10.24)

Fig. 10.11

Po "I·~---po;-

IG=wle. \and momentum is inhe

The same. relation ?et~~e~h~~~~g~f relativity) in particlesre~t (as IS sho~n lUTh .s is quite natural since, according toWith zero rest mass. 1

'\ G=S/c2, \

_ [E X HI is Poynting's vector..Just a~ vectors~:'~;eIrtumdensity G is generally a functIOn of time and

coordinates. 'tl (10 20) for an electromagnetic waveIn accordance WI I . , d

~ I -. _ V- H Hence the energy en-in a vacuum we have.v ~oEs- f ~;niing's vector are re-sity wand the magllltu e 0

spectively equal tof.T2/2 E2 S = EH = lr80/11'0 E2.

w=BoE2/2+p.~.. =Bo • Y r

lr- Ad' V~ - ileIt follows that S = wi Y Bol-lo·. n slUceS _ wOe °T--;king'

I' f l' ht lU vacuum - .~here e is th

te(1vOc 2°4)It~~ o~fain that for a~ electromagnetic

lUto accoun . ,wave in a vacuum

10.5. Electromagnetic Field Momentum

BFig. to.l0

t f momentum density G ofLet us introduce the concep o. . 11 equal to

electromagnetic :re~~ea~lC\~ei~u:n~~;~ ~~~:~~aCIteulati~ns~~i~o~ft~~~mittedhere show that the momentum denSIty

is given by

t m of such a system can be conserved onlyThe momen u . ld ( ) I essesvided that the eleetromagnetlc fie wave a so poss .

pro t. the substance acquires the momentum due~o ~~~e~~~~ierred to it by the electromagnetic fIeld.

~\ .

right-hand side contains the displacement current through the contourshown in Fig. 10.9 by the dashed line. Hence H = (1/2) a aD/at. Ifthe distance between the plates is h, the /lux of vector S through thelateral surface is

a aD aD ,EH2nah= E 2: at 2nah=E at l, (1)

where V = na2h is the volume of the capacitor. We shall assume thatthis /lux is completely spent for increasing the capacitor's energy.Then, multiplying Eq. (1) by dt, we obtain the increment of the capacitor's energy during the time dt:

dW=EdD.V=d( ee~E2 V)=d( E: V)Having integrated this equation, we find the formula for the energy Wof a charged capacitor. Thus, everything is all right in this case too.

10.5. Electromagnetic Field Momentum

Pressure of Electromagnetic Wave. Maxwell showed theoretically that electromagnetic waves, being reflected or ahsorbed by bodies on which they are incident, exert pressureon these bodies. This pressure appears as a result of the actionof the magnetic field of· a . wave on the electric currentsinduced by the electric field of the same wave.

Suppose an electromagnetic wave propagates in a homogeneous medium capable of absorption. The presence of absorption means that Joule's heat with the volume densityaE2 will be liherated in the medium. Hence a =1= 0, Le.the absorbing medium is conductive.

The electric field of tIle wave excites in this medium an electric current with a density j = aE. As a result, Ampere'sforce Fu =- !j X B) =-= a [E X B] will act on a unit volumeof the medium. This force is directed towards the propagation of the wave (Fig. 10.10) and causes the pressure of thee,leetromagnetic wave.

In the absence of absorption, the conductivity a = 0and Fu == 0, i.e. in this case the electromagnetic wave doesnot exert any pressure on the medium.

Electromagnetic Field Momentum. Since an electromagnetic wave exerts a pressure on a substance, the latter acquires a certain momentum. However, if in a closed systemconsisting of a substance and an electromagnetic wave onlythe substance possessed a momenLum, the Jaw of conservution of momep.t'llm would be yjolated.

268 10. Maxwell's Equations. Electromagnetic Field Energy

.'

r

271

Fig. 10.13

o

Problems

D dOP,

10.12

v

dD

. t 'current density jd = DD/at. Hence,Solution. The lhsplaceme~, reducPlI to determining vector D at

the solution of the problem 1;;. . time deriv'ltive. In both cases,the indicated points nTll~ toI fmdI,nt

g It~ of r Let ~IS find the derivativeD --- e /4~r2 where e IS t Ie lITH vee or .

n q r ••. , r

iJD/Dt . P (I" to '2 where it is assumerl that q > 0), we(1) At pomt I 'lg. .1,have

If q > 0, id tI v, and vice .vers~. '. ht solenoid with the• 10.2. A current n?wm~ III a long str~~{1netic field inside the

radius R of cross sec~ion I~ varIed sOrtha\~h~he l~w B = ~t2, where ~solenoid increaFs~ IW~~~ lt~pla~~~;~:~~rrent (Iensity as a function ofis a constant. 'Illl ..the distance r from the solenoId aXIS.

aD 2q or _ 2qv-=--43;;-t er- 4nr3 ,at nr u

h f oint P the derivative arMt = -v.where we took into a~cfun\t fitt~rc~arge(i~ the direction of its motioiJIf point PI were not m. ron Old b directed in the same way and woubut behind it, vector ~d wou e .have the same magmtude.. . and vice versa.

Thus, for. q > 0, :,ec~ort~d) ttl ~b I/D = v dt/r, and hence we can(2) At POint P 2 (FIg. •. ,write / _ -"

aD/at = -qv 4nr.

Problems

A int charge q moves uniformly• 10 1. Displacement currenlt· t · pot ·c velocity v Find the vector

" 1- 'th a nonre a IVIS 1 . . d' 'Ialong n ~I.ralght Inl' WI . at a oint P lymg at a 1~ ,ance ~of the di~placement curren.t ~rnl~~: (1) c~nciding with its traJectory,from the charge on a. stral~ t ry and passing through the charge.(2) perpendicular t~ ItS traJec 0

(10.20)P = (1 +p) (w).

Here the quantity P is just the pressure of the electromagnetic wave on the body. In the case of total reflection, r ._== 1 and pressure p = 2 (w) while for total absorption,p = 0 and p = (w).

It should be added that the pressure of f'lectromagneticradiation is usually very low (the exception is the pressureof high-power laser radiation beams, especially after beamfocusing, and the pressure of radiation inside hot stars).For example, the pressure of solar radiation on the Earthamounts to about 10-6 Pa, which is smaller than the atmospheric pressure by a factor of 1010. In spite of insignifIcantvalues of these quantities, experimental proof of theexistence of electromagnetic waves, viz. the pressure oflight, was obtained by Lebedev. The results of these experiments were in agreement with the electromagneticJheoryoflight.

quantum-mechanical notions, an electromagnetic wave isequivalent to the flow of photons, viz. the particles with zerorest mass.

Some More Remarks on the Pressure of EJcctl'OmagnelieWaves. Let us calculate, llsing forll1111a (10.2;"»), tlte pressure exerted by an electromagnetic wave on a body when thewave is incident normal to its surface and is partially reflected in the opposite direction. In accordance with the law ofconservation of momentum, Po = p~ -+- p, where Po, p~ arethe momenta of the incideilt and reflected waves, wltile pis the momentum transferred to the hody (Fig. 10.11). Having projected this equality onto the direction of the incident wave and referring all the quantities to unit time andunit cross-sectional area, we obtain

210 10. Maxwell's Equations. Electromar(_fI:!!.~cField Ener~.__

P = Po + P; = (G) c + (G') c,

where (G) and (G') are the average values of the moment limdensity for the incident and r.eflected waves. It remains forus to take into account relation (10.25) between (G) and(w) as well as the fact that (w') = P (TV), where {> is the

ref!ectioncoefficienl. As a resu)t, the previous expression becomes

Solution. In order to find the displacement current density, wemust, in accordance with (to.5), first find the electric fiehl strength(here it will be a vortex field). Using Maxwell's equation for circulation of vector' E, we write

2nrE = nr2 aBlat, E = r~t (r < R);2nrE = nR2 aBlat, E = R2~tlr (r> R).

Now, using the formula id = eo iJElat, we can find the displacementcurrent density:

id = eo~r (r < R); id = eo~R2/r (r > R).

The plot of the dependence ;d (r) is shown in Fig. to.13•

• to.3. A parallel-plate capacitor is formed by two discs the spacebetween which is filled with a homogeneous, poorly conducting medium.The capacitor was charged and then disconnected from a power source.Ignoring edge effects, show that the magnetic fIeld inside the capacitoris absent.

Solution. Magnetic field will be absent since the total current (conduction current plus displacement current) is equal to zero. Let usprove this. We consider the current density. Suppose that at a certaininstant the density of conduction current is j. Obviously,j ex D andD = an, where a is the surface charge density on the positively chargedplate and n is the normal (Fig. 10.14).

The presence of conduction current leads to a decrease in the surfacecharge density a, and hence in D as well. This means that conductioncurrent will be accompanied by the displacement current whose density is

jd = aD/at = (aalat) n = -jn = -j.

Hence it follows that, indeed

it = j + jd = O.

• 10.4. The space between the plates of a parallel-plate cafacitorin the form of circular discs is filled with a homogeneous poor y conducting medium with a conductivity a and permittivity e. Ignoringedge effects, find the maci'litude of vector H between the plates ata distance r from their axeb, if the electric field strength between theplates varies with time in accordance with the law E == Em cos wt.

Solution. From Maxwell's equation for circulation of vector H.it follows that

2nrH=(in +eeo 8it) nr l .

Taking into account Ohm's law jn = aEn (t). we obtain

H r (E eeo aEn ) rEnt .="2 n+oiJ't =-r(acoswt-eeows!nwt).(3}

(1)

273

q~~_V

Fig. 10.15

Problem'

8\ q. cia.8t J D dS = 2: SlD a dt .

ot ! ~Do

Fig. 10.14

• . o'f t H obtain the expression 'for Hequati0!l for thhe CIrcuI~tt~~~ rel::i~~rto the charge is characterized byat a pOlUt P w ose POSI Iradius vector r (Fig. 10.15).

-Solution. It iShcl:~r f~om ls~~~e;?v~~~g:d~t:~~~dthba:~~~e~~o:;tour arohund whi~cle wel'~hcc~:t:~ 0 (its trace is'shown in Fig. 10.16must c oose a clrby the dashed line). Then

2nRH=..!- r Dn dS,at J

wheLre~tR iSfit~e tr:ed~~xo~fte:c~~r~l~. through a surface boun~ed by thisreus n f' 1"t e shall take a spherIcal surfacecircle. For t~e safke 0 Stlmp IC(£.lg' fO 16) Then the flux of D throughwith the radIUS 0 curva ure r • • • .an elementary ring of this spherical surface IS

q ., :1_' q. , da.'D ds=--2nr SlD a ·r ..... ='2 SlUa ,

4nrll

while the total flux through the selected surface is

~ DdS=~ (i-cos a). (2)

Now, in accordance with (1), we differentiate (2) with respect to

time:

Let us transform the expression in the parentheses to cosine. For thi~

~~1~h:~7:t:d~~~I:n~1:t~~~~;~i~h~xF;::~I:SbJi/:: c~62

, te~~iiwL=sin 6. This gives

H = ~ rEm va2+(eeoco)2 cos (cot+6).

A point charge q moves in a vacuum l;lniformly ~l,drec~i~:a'~y with a nonrelalivistic velocity v. Usmg Maxwe 8

10. Maxwell', Equation,. Electromagnetic Field Ener,,,272

18-0181

274 10. MGZlIIelt, Equation•. Electromagnetic Field Energy Problem. 275

'k-\ SH I \

C;;2:'$Z;::::~=Z&-~~\ /\J

l"ig. 10.19Fig. 10.18

18*

solenoid is being increased. Show that the rate of increase in the energyof the magnetic field in the solenoid is equal to the flux of Poynting'svector through its lateral surface.

Solution. As the current increases, the magnetic field in the !'olenoidalso increases, and hence a vortex electric field appears, Suppose thatthe radius of the solenoid cross section is equal to a. Then the strengthof the vortex electric field near the lateral surface of the solenoidcan be determined with the help of Maxwell's equation that expressesthe law of electromagnetic induction:

aB a aB2rtaE = rta8 at ' E =2" at·

The energy flux through the lateral surface of the solenoid can be represented as follows:

a ( B8 )cD=EH·2nal=rta2

/ at 2f.1o '

where I is the solenoid length and rta2l is its volume.Thus. we see that the energy flux through the lateral surface of the

solenoid (the flux of vector S) is equal to the rate of variation of themagnetic energy inside the solenoid:

cD = S ·2rtal = ow/at.

SolutiorL l"igure 10.19 shows that S tt v. Let us find the magnitude of S: S = EH, where E and H depend on r. According to theGaUS8 theonm, we have

2rtrE = 'AJeo,where A is the charge per unit length of the beam. Besides, it followsfrom the theorem on circulation of vector H that

2rtrH = f.

Having deWnnined E and H from the last two equations and takinginto aee:oant. that f = 1.1.1, we obtain

S = EH = f 8/4n8eol.lr2.

e 10.8. A current flowing through the winding of a long straight

dB

p

o1 2

Fig. 10,17Fig. 10.16

angular velocity roo Find V X E in this region as a function of vectorsfI} and B.

Solution. It follows from the equation V X E = - aB/at thatvector V X E is directed oppositely to vector dB. The magnitude ofthis vector can be calculated with the help of Fig. 10.18:

I dB I = Bro dt, I dB/dt I = Bro.

For the displacement of the charge from point 1 to point 2 (Fig. 10.17)over the distanee v dt, we have 1.1 dt ·sin a. = r da., whence

da. 1.1 sin a.Cit r (4)

SubstitutiDg(4) into (3) and then (3) into (1), we obtain

H = qvr sin a./4m,a, (5)

where we took into account that R = r sin a.. Relation (5) in vectorform can be written as follows:

H=_LJvXr)4n r3

Thus we see that expression (6.3) Which we have postulated earlieris a corollary of Maxwell's equations.

el0.6.CuriofE.Acertain :re2ion ofan inertial system of referencecontains a magnetic field of m'1g11itude B = const, rotating at an

l'

Henee

V X E = -fro X .BI.e 10.7. Poynting's vector. Protons haVing the same velocity v

form a beam of a circular cross section with current f. Find the direction and magnitude of Poynting's vector S outside the beam at adistanee r from its axia.

11.1. Equation of an Olctllatory Circuit 277

216 10. Mazwtll's Equations. Eltctromagnttic Fldtl Elltr"

(3)1 rCllUm •

B=2'llolto-h-\ SlDCIlt I·

H.t. Equation of an Oscillatory Circuit

Quasi-steady Conditions. When electric oscillations occur,the current in a circuit varieswith time and, generally speaking, turns out to be different at each instant of time in different sections of the circuit (due to the fact that electromagnetic perturbations propagate although at a very high butstill finite velocity). There are, however, many cases wheninstantaneous values of current prove to be practically thesame in all sections of the circuit (such a current is calledquasistationary). In this case, all time variations should occurSO slowly that the propagation of electromagnetic pefturba-

II. Electric Oscillations

plates be h. Then the electric energy of the capacitor is equal toII EI II 'lal

We=+'la2h=~ UfucoslCllt. (1)

The maguetic energy can be determined through the formula

Wm = \ :1 dV. (2)• 110

The quantity B required for ~valuating this integral can be foundfrom the theorem on the circulation of vector H: 2'lrH = 'lr2 aDlat.Hence, considering that H = BII10 and aDlat = -'-eo (Umlh) CIl sin CIlt,we obtain

It remains for us to substitute (3) into (2), where for dV we musttake an elemeutary volume in the form of a ring for which dV == 2nr dr·". All a result of integration, we obtain

W 'l J.LollifalSa6Ufu '1 (4)m=16 h sIn CIlt.

The ratio of th.e maximum values of magnetic energy (4) and electric energy (f) is given by

WmU:-ax 1 1 1Wemax 8 l1011oiJ cIl •

,For example, for a = 6 cm and cIl = toDD S-I, this ratio is equal to5 X to-D.

(4)

(5)

Fig. to.20where S=EH is the nux density,

. 2'lr dr is the elementary ring ofwidth dr within which the value of S is constant, and a and b are theradii of the internal wire and of the outer shell of the cable (Fig. 10.20).In order to evaluate this integral, we must know the dependenceS (r), or E (r) and H (r).

Using the Gauss theorem, we obtain

2'lrE = '}Jllo, (2)

where Ais the charge per unit length of the wire. Further, by the theorem on the circulation' we have

2'lrH = f. (3)

After substituting E and H from formulas (2) and (3) into expression(1) and integrating, we get ,

AI b=--ln -

2'lllo a

The values of A, a, and b are not given in the problem. Instead,we know U. Let us find the relation between these quantities:

b

)A b

U= E dr=--ln-.2'lllo a

a

• 10.9. The energy from a source of constant voltage U is transmitted to a consumer via a long coaxial cable with a negligibly smallresistance. The curr~nt in the c!lble is f. Find the energy nux throughthe cable cross sectIOn, assumIng that the outer conducting shell of

the cable has thin walls.Et Solution. The required ener-~ gy llux is defined by the formula

dfr-:=-H= S e---o :=:; j S·2nr dr, (1)

a

A comparison of (4) and (5) gives

III = Uf.

This coincides with the value of power liberated in the load.

• 10.tO. A parallel-plate air capacitor whose plates are madein the form of disks of radius a are connected to a source of varyingharmonic voltage of frequency CIl. Find the ratio of the maximum valuesof magnetic and electric energy inside the capacitor.

Solution. Let the voltage across the capacitor vary in accordancewith the law U = Urn cos rot and the distance between the capacitor

278 11. Electric Oscillations11.1. Equation of an Oscillatory Circuit 279

R

't = lIc~T,

where T is the period of variations.For e:ample, for a circuit of length 1 = 3 m, the time

't = 10 8 s, and the current can be assumed to be quasista-

(11.2)

(11.1)I = dqldt.

RI = CPl - CP2 + ~8 + 'IS,where 'fe s is the self-induced e.m.f. In the case under consideration,

Consequently, if I > 0, then dq > 0 as well, and vice versa(the sign of I coincides with that of dq).

In accordance with Ohm's law for section lRL2 of thecircuit, we have

t s = - L dIldt and CP2 - CPl.= giG

in the circuit attains its maximum value (Fig. 11.1b).Starting from this moment, the current begins to decrease,retaining its direction. It will not, however, cease immediately since it will he sustained by self-induced e.m.f. The current recharges the capacitor, and the appearing electric fieldwill tend to reduce the current. Finally, the current ceases,while the charge on the capacitor attains its maximum value.From this moment, the capacitor starts to discharge again,the current flows in the opposite direction, and the process

. is repeated.If the conductors constituting the oscillatory circuit have

no resistance, strictly periodic oscillations will be observedin the circuit. In the course of the process, the charge on thecapacitor plates, the voltage across the capacitor and thecurrent in the induction coil vary periodically. The oscillations are accompanied by mutual conversion of the energyof electric and magnetic fields.

If, however.... the resistance of conductors R =1= 0, then, inaddition to the process described above, electromagneticenergy will be transfor~ed into Joule's heat.

Equation of an Oscillatory Ci~uit. Let us derive theequation describing oscillations in a circuit containing seriesconnected capacitor G, induction coil L, resistor R, andvarying external e.m.f. liJ (Fig. 11.2).

First, we choose the positive direction of circumvention,e.g. clockwise. We denote by q the charge on the capacitorplate the ~iirection from which to the other plate coincideswith the chosen direction of circumvention. Then current inthe circuit is defined as

L

(a) (b) 6

Fig. 11.1 Fig. 11.2

tionary down to frequencies of 106 Hz (which corresponds toT = 10-6 s).

In this chapter, we shall assume everywhere that in thecases under consideration quasi-steady conditions are observed and currents are quasistationary. This will allow us touse formulas obtained for static fields. In particular we shallb.e using the fact that instantaneous values of 'quasistatIOnary currents obey Ohm's law.

Oscillatory Circuit. In a circuit including a coil of induc~ance L and a capacitor of capacitance G, electric oscillatIOns may appear. For this reason, such a circuit is called anoscillatory circuit. Let us find out how electric oscillationsemerge and are sustained in an oscillatory circuit.. Suppose tha~ .initially, the upper plate of the capacitorIS charged p~sltlvely and t~e lower plate, negatively (Fig.11:1a). In tIus case, the entIre energy of the oscillatory cirCUlt IS co~centrated in the capacitor. Let us close key K.T~e capacItor star!s to discharge, and a current flows throughcOlI L. The electrIc energy of the capacitor is converted into the magnetic energy of the,coil. This process terminateswhen the capacitor is discharged completely, while current

tio~s c~uld be c.onsidere~ instantaneous. If 1 is the length ofa c~rcUlt, the time reqUlred for an electromagnetic perturbatIOn to cover the distance 1 is of the order of 1: = lie.F~r a periodically varying current, quasi-steady conditionWIll be observed if

1:........ "

11.2. Free Electric OscillationsFree Undamped Oscill t· If . .IIal e.m.f. ~ and if 'ta lon~. a CIrCUit contains no exter-. I S resistance R -. 0 the '11 t'111 such a circuit will be free d d-, OSCI a Ionsdescribin th '11" an un amped. The equation

I <£' g Oese OSCI atlOns IS a particular case of Eq (11 5)w len e; = and R = 0: . .

281

(11.10).. .q+ 2pq +ro:q = O.

11.2. Free Electric OscUlatlo,..--------The solution of this equation is the function

q = qm cos (root + a), (11.8)

where qm is the maximum value of the charge on capacitorplates, (J)o is the natural frequency of the oscillatory circuit, anda is the initial phase. The value of roo is determinedonly by the properties of the circuit itself, while the valuesof qm and a depend on the initial conditions. For these conditions we can take, for example, the value of charge q

and current I = qat the moment t = O.According to (11.6), roo = lIYLC; hence the period of

free undamped oscillations is given by~.

To=2~YLC (H.9)(Thomson's formula).

Having found current I [by differentiating (11.8) withrespect to time] and bearing in mind that the voltage acrossthe eapacitor~plates is in phase with charge q, Vie can easilysee that in 1ree undamped oscillations current I leads inphase the voltage across the capacitor plates by n/2.

While solving certam problems, energy approach can also be used.

Exat~. In an oscillatory circuit, free' undamped oscillationsoccur with energy W. The capacitor plates are slowly moved apartso that the frequency of oscillations decreases by a factor of 'I. Whichwork against electric forces is done in this case?

The required work can be represented as an increment of the energyof the circuit:

A= W' - W = 9; (~, - ~ )= W ( g, -1) .On the other hand, 000 ex: HVC, and hence 'I = CilMCilI = VC/C',and consequeutly

A = W ('II - 1).

Free Damped Oscillations. Every real oscillatory circuithas a resistance, and the energy stored in the circuit is gradually spent. on heating. Free oscillations will be damped.

We can obtain the equation for a given oscillatory circuitby putting ~ = 0 in (H.5). This gives

(11.5)

(11.7)

11. Electric Oscillations280

(the sign of q must coincide with th .difference (JJ2 - (JJ since G >0) He sIgnE,of the potentialwritten in the f~rm . ence q. (H.2) can be

L dI q7t+RI+c=~, (H.3)

or, taking into account (H.t),

L aSq dq 1 I(iii"+R 7t +c q=~. (11.4)

This is the equation of an 'llat . .near second-order nonho OSCl o,:y clrcU.lt, which is a H-constant coefficients U~i~ggenteho.us differ~ntiafl equation with

(t) . IS equatIOn or calculat'q ,we can easily obtain the voltage across th .mgas Uc = (JJ2 - (JJI = qlG and current I b e capaCitoren;~~r,::::uation of an oscillatory circuit ca:b~o;~~~aa~i~:L

l·q+2Pq+ro:q=~/L,where the following notation is introduced:

. 2p = RIL, ro: = lILG. (11.6)TIle quantity ro' II d hand p is the dar:;, I~nca e t e natural f!equency of the circuitwill be explaine~ b~/actor. The meanmg of these quantities

If _ . ow:tions ~The~' t~ll ~cllladatlOns are usually called free oscilla-R' WI e un mped for R = 0 and dam ed fI

=F O. Let us consider consecutively all these casts. or

283

(11.19)

(11.18)a (t)

A. = In a (t+Tl = ~T,

11.2. Free Electric Oscillations

las

where a is the amplitude oi the corresponding quantity(q, U, or I). In a different form, we can write

- ~/roo = cos l'l, ro!Ulo = sin 15. (1.1.15)

After this, the exp~ession for I takes the following form:

I = Ulqme-flt cos (Ult + a + 15). (1.1.16)

It follows from (11.15) that angle 15 lies in the second quadrant (n/2 < l) < n). This means that in the case of a nonzero resistance, the current in the circuit leads voltage inphase(1.1.14) across the capacitor by more than n/2. It shouldbe noted that for R = 0 this advance is l) = n/2.

The plots of the dependences Uc (t) and I (t) have the formsimilar to that shown in Fig. 11.3 for q (t).

Example. An oscillatory circuit contains a capacitor of c:apIICi&anceC and a coil with resistance R and inductance L. Find the ratio between the energies of the magnetic and electric fields in the circuit atthe moment wh:Jl the current is at a maximum.

According to equation (11.3) for an oscillatory circuit,

L ~; +RI+ ~ =0.

When the current is at a maximum, dIldt = 0, and RI = -q/C.Hence the required ratio is

Wm/We = L/CR2.

Quantities Characterizing Damping.1. Damping factor ~ and relaxation time "t, viz. the time

during which the amplitude of oscillations decreases by afactor of e. It can be easily seen from formula (11.11) that

"t = 1/~. (11.17)

2. Logarithmic decrement A. of damping. It is defined as theNaperian logarithm of two successive values of amplitudesmeasured in a period T of oscillations:

wheret


I t can be shown (we shall not d· .ested in another as ct of 0 It here SlDce we are interthe solution of this~ the pro~lem) that for ~2 < (J)I,

omogeneous dIfferential equation ha"sthe form

f fJ~o.ft-.../' q= qme- flt cos (oot+a),

(H.H)

(J)=V~

Fig. 11.3 /1 (R)2= l LC - 27' (11.12)

?n.~. qr and. ~ are arbitrary constants determined from th~lll~~ c

101nd

31tlOns. The plot of the function (11.11) is sho e

III I~. :. It can be seen that this is not . ,,:nfu~tlOn SInce it determines damped oscillations

aperiodiC

. "'devjedrtheless, t~e quantity T = 2n/oo is call~d the _no 0 amped oscLllations: pe

T- 2ft To- V&>B-W' Vt-(~/CiIo)2' (H.13)

wI~~e r0 is the ~er~od of free undamped oscillations.d e act?r q"!e flt III (11.11) is called the amplitude 0

1a1m3Pebd osch~llatwns. Its ?ependence on time is shown in Fig1

. y t e dashed hne. .

C. Vo~ttagKe acr~s a Capacitor and Current in an OscillatoryIrCUI. nOWIllg q (t) we fi d tl Ip·t d ,can III Ie vo tage across a ca-

c:~~C~~O:niS ~~~e~u~~nt in a circuit. The voltage across a

Uc = ~ = q~ e-lltcos(rol+a). (1.1.14)

The current in a circuit is

1= !:!L. ~c q e-Ilt [ Il (t +. dt -- m - .... cos 00 a)-rosin(oot+a)j.

i~? transform the expression in the brackets to cosine For

V~2 ~urp~s~ we multiply. and. divide this expressio~ by1- ~ "- Wo and then mtroduce angle 0 by the formu-

i I

11. Electric Oscillations

where W is the energy stored in the circuit and IIW is the decrease in this energy during the period T of oscillations. Indeed, the energy W is proportional to the square of the amplitude value of the capacitor charge, i.e. W ex:. e-2pt.Hence the relative decrease in the energy over the period is6WIW = 2~T = 21... It remains for us to take into accountthe fact that according to (11.21), A. = n/Q.C~ncluding the section, it should be noted that when

Ill> (O~, an aperiodic discharge of the capacitor will occurinstead of oscillations. The resistance of the circuit for whichthe aperiodic process'sets in is called the critical resistance:

R cr =2YL/C. (11.24)

Let us consider two examples.

Example t. An oscillatory circuit has a capacitance Cl

inductance L,and resistance R. Find the number of oscillations a ter which theamplitude of cunent decreases to fie of its initial value.

The eunent amplitude (/m ex: e -Pt) decreases to fie of its initial (11.27)or

11.3. Forced electric Oscillations. . Let us return to equations (11.3)

Steady-state OscJlla~lOns. . 't and consider the cased (11 4) for an OSCillatory Clf~Ul t al emf ~ whosean . .' 1 d a varymg ex ern .,.

when the circUlt ,mc ~ es reased by the harmonic law:dependence on tlUle IS dec. (11.25)

~ = Om cos (Ot.t' of. .al lace owing to the proper les

This law OCCUPI~S a .Sp~Cl If lo retain a harmonic form of osthe oscillatory CIrCUlt ItS~. of external harmonic c.m.!.dIlations under the aCt' lonfor a~ oscillatory circuit IS

In this case, the equa Ionwritten in the form (11 26)

L ~+RI+...L=~m·cos(Ot, .dt C

11.8. Forced Electric O,cUlations

D ' g tbl'S time Ne oscillations will, - 11~ urm 'h

value during the time : -d f damped oscillations, t enIf T is the perlo 0

occur. 1 V (W )2't 1.{~ =_ _0 -1.

Nc=-= r~ 2n ~T 2n/V ro6-P

2 = 1./LC and l> = R {2L, we obtainConsidering tllat roo ---c._-

_1._ ~ / .!£-1.N e= 2n VCR'

. . hi Ii tbe amylitude of currentExample 2. Fin~ t~e ti?1e duni~~nwQ-factor wH. de~rea~ to 1~illations in a CllCUlt WIth a g of damped OSCIllatIOns IS equ

osc. . 't'a1 value if the frequencyof 1ts lUl 1 ~ , • • hto roo . . . ex: e- IlI , the time to durmg wl1!-c

Since the current amphtu~e ;mrof Tj is determined by the equatIOnthe amplitude decreases by a ac 0

1\ = ept•• Hence Ill.to = (In 1'\) p. ,

bftDd the Q-factor is also related to ~:On the other 'Q = Ttl;, = n/~T = ro/2~.

• • A f m the laa\ two equations, we obtainElimmatmg I' ro

2Qto =c;) In Tj.

(11.23)WQ~2n 6W'

where Ne is the number of oscillations during the time 't,

i.e. the time required for the amplitude of oscillations todecrease to 1/e of its initial value. This expression can beeasily obtained from formulas (11.17) and (11.18).

If damping is small (/32 ~ (O~), then (0 ~ (00 = vyLCand, in accordance with (11.18), we have

A. ~ ~. 2n/Ulo= nRYC/L. (11.20)

3. Q-/actor of an oscillatory circuit. By definition,

Q = nlA. = nNe' (B.21)

where ). is the logarithmic decrement. The smaller the damping, the higher the Q-factor. For a weak damping (/32 ~~ (0:), in accordance with (11.20), we have

Q~ ~ V ~ . (11.22)

Here is one more useful formula for the Q-factor in the case ofa weak damping:

11. ItZectrlc OBctli4ttoIU

(11.36)

(11.34)

(11.32)

Fig. 11.5

o

Fig. H.4

leads I by n/2. ~his candbe. v:~ugal;~er:::;f~~~~w;jh vt~~help of a vector dtagram, epIC mtages

U - Rf Uem = 1m/roC, ULm = roLl rnam - m,

and their vector sum which, according to (11.31), is equalto vector ~m (Fig. 11 .4). . d' can

Considering the right triangle of. thIS Iagram,;,eeasily obtain the following expressIons for I m and cpo

~m (11.35)1m = l!R'+«(a}L-1/wC)Z'

wL-1/(a}(:tancp= R

11.3. Forced Itkctrlc O.ctUatio..

Th the problem is solved. . d bIt :~uld be noted that the vector diagram obtame a ove

UR = RI = RIm cos (rot - cp),

Uq --.J!!!. cos (rot _ '\jJ) = I Tn cos (rot - cp - ~ ) , (11.33)e=c- c we

U L = L ~~ = - roLlrn sin (rot - cp)

• 11: )= ro LIm cos trot - cp +T .

V tor Diagnun The last three formulas show that UR

is i:;hase with cu~rent I, Ue lags behind I by n/2, and U l.

I m

·1· f '£! Taking into account relation (11.30),totheexterna e.m.. ",.we write

(11.30)I rn = roqrn, cp = 'Ii' - n/2.

UL + Un + Ue = ~rn cos rot, (11.31)

where the left-hand side is the sum of voltages across induction coil L, capacitor C and resistor R. Thus, we see thatat each instant of time, the sum of these voltages is equal

As is known from mathematics, the solution of this equation is the sum of the general solution of the homogeneousequation (wiih the zero right-hand side) and a particularsolution of the nonhomogeneous equation.

We shall be interested only in steady-state oscillations,i.e. in the particular solution of this equation (the generalsolution of the homogeneous equation exponentially attenuates and after the elapse of a certain time it virtually vanishes). It can be easily seen that this solution has the form

q = qrn cos (rot -'Ii')~ (11.28)

where qrn is the amplitude of charge on the capacitor and 'Ii'is the phase shift -between oscillations of the charge and ofthe external e.m.f. ~ (11.25).• It will be shown that qrnand", depend only on the properties of the circuit itself andthe driving e.m.f. ~. It turns out that 'Ii' > 0, and hence qalways lags behind ~ in phase.

In order to determine the constants qrn and 'Ii', we must substitme (11.28) into the initial equation (11.27) and transform the result. However, for the sake of simplicity, we shallpr~ in a different way: first find current I and then substitute its expression into (11.26). By the way, we shall solve the problem of determining the constants qrn and 'Ii'.

Differentiation of (11.28) with respect to time t gives

1= - roqrn sin (rot -'\jJ) = roqrn cos (rot -'\jJ + n/2).

Let us write this expression as follows:

I = I rn cos (rot - cp), (11.29)

where I rn is the current amplitude and cp is the phase shiftbetween the current and external e.m.f. ~,

We aim at finding I rn and cpo For this purpose, we proceed asfollows. Let us represent the initial equation (11.26) in thefonn

289

(11.40)

11.3. Forced Electric Oscillattons

where CUo is the resonance frequency and fJw is the width ofthe resonance curve at a "height" equal to 0.7 of the peakheight, I.e. at resonance.

Resonance. Resonance in the case under consideration is1{2 19-0181

(J}e res = WoY1- 2 (~/(l)O)2,

(J}L res = WoY1-2 (~/WO)2.

The smaller the value of ~, the closer are the resonance frequencies for all quantities to the value Woo

Res6n3nce UII'Ves and Q-factor. The shape of resonancecurves is connected in a certain way with the Q-factor of anoscillatory circuit. This ~onnection has the simplest formfor the case of weak damping, I.e. for ~2 «w~. In this case,

Ueres/cem = Q (11.41)

(Fig. 11.7). Indeed, for ~2« cu~, the quantity wres ~ Woand, according to (11.33) and (11.35), Uem res = I m/cuoC =

= tm/woCR, Or Uemres!'tm = YLCICR = (1IR) YLICwhich is, in accordance with formula (11.22), just the Qfactor.

Thus, the Q-factor of a circuit (for ~2« w~) shows howmany times the maximum value of the voltage amplitudeacross the capacitor (and induction coil) exceeds the amplitude of the external e.m.f.

The Q-factor of a circuit is also connected with anotherimportant characteristic of the resonance curve, viz. itswidth. It turns out that for ~2« w~

Q = cuolfJw, (11.42)

The maximum of this function or, which is the same, theminimum of the radicand can be found by equating to zerOthe derivative of the radicand with respect to cu. This givesthe resonance frequency (11.38).

Let us now see how the amplitudes of voltages U R' Ue,and UL are redistributed depending on the frequency W

of the external e.m.f. This pattern is depicted in Fig. 11.7.The resonance frequencies for UR' Ue and UL are deter

mined by the following formulas:

11. Electric Oscillations--'-'-------

288

proves to be vel'y con . fproblems. It permits a vi:~~lent 01' solving .many specificvarious situations. ' easy, and rapId analysis of

.Resonance Curves. This is the .dependences of the folloWing q n:-~e gIven to the plots ofof external e.m.f. ~: current I uahn lIes on the fre~uency (J)

, c arge q on a capaCitor, and

c6

Fig. 11.6Fig. H.7

voltages UR Ue and U d fi .(11.34). " L e ned by formulas (11.32)-

Resonance curves for current I ( ) '.U.5. Expression (11.35) shows th m cu are shown m Fig.has the maximum value for cuL _ ~t the~urrent amplitudethe resonance frequency for current lcur: -: O. C?nsequentlystural frequency of the oscl'll t .col~cldes with the na-

a ory CIrCUIt:

. WI reB = Wo= lIVLC. (11.37)The maXImum at resonance i th h'

the smaller the damping facto: A. ~ ~~~er and the sharperResonance curves for the cha p L.

.shown in Fig. 11.6 (resonance rge qm (cu) On a capacitor areacross the capacitor have the curves for voltage Uemcharge amptitude is attaine~a~~ slhlape). The maximum of

. t e resonance frequency. Wqre8=.Vw~-2~2, (11.38)

which comes closer and closer to .In order to obtain (11 38) W o WIth decreasing p.ance with (11.30), i~ the~~rmust represent qm, in accord-defined by (11.35). Then m qm = I mlw where 1m is

qm = - _ ~=/L==-_V(W3-WI)2+4~IWI (11.39)

1111

III!11:1

I!II

I - Um roL-1/rocm-R2+(roL_1/roC)2' tancp= R (11.45)

The problem is reduced to determining the current amplitude and the phase shift of the current relative to U.

The obtained ~xpression for the current amplitude 1m

(ro)can be formally Interpreted as Ohm's law for the amplitudevalues of current and voltage. The quantity in the denominatOr of this expression, which has the dimension of resistance, is denoted by Z and is called the total resistance Or,

291

(11.50)

(11.48)

11.4, Alternating Current

impedance:Z = V' R2 +(roL -1/coC)2. (11.46)

It can be seen that for ro = roo = 1/V'LC, the impedancehas the minimum value and is equal to the resistance R.The quantity appearing in the parentheses in formula(11.46) is denoted by X and is called the reactance:

X = ro[, - 1/roC. (11.47)

Here the quantity roL is called the inductive reactance,while the quantity VroC is called the capacitive reactance.They are denoted XL and Xc respectively. Thus,

XL=roL, Xc =1/roC, X=XL-Xc,

z= VR2+X~".

It should be noted that the inductive reactance grows withthe frequency",w, while t~e. cap~citive rea~tan.ce decreaseswith increasing ro, When It IS saId that a CircUlt .has no capacitance, this must be t;nderstood so that there IS no c~pac

itive reactance which is equal to 1/roC and hence valllsh.esif C -lo- 00, (when a capacitor is replaced by a short-CIr-cuited section), . ,

Finally, although the reac~anc~ is I:Ue~suredII~ t~e sameunits as the resistance, they dIffer 1Il prmclple. ThIS differenceconsists in that only the resistance determines irreversi~le

processes in a circuit such as, for example, the converSIonof electromagnetic energy into Joule's heat,

Power Liberated in an A.C. Circuit. The instantaneousvalue of power is equal to the product of instantaneousvalues of voltage and current:

P {t) = VI =-c= Vml m cos rot cos (rot - cp). (11.49)

Using the formula cos (rot - (f') = cos rot cos cp ++ sin rot sin if, we transform (11.49) as follows:

P (t) = Umlm (C082 wt cos cp + sin rot cos rot sin cp),

Of praeticnl importance is the value of power averagedover a period of oscillation. ConsJdering that (cos2 cut) ==--= 1/2 and (sin wt cos wt) = 0, we obtain

U I(P) = m2

m cos cpo


11.4. Alternating Current

~otal. Hesistance (Impedance). Steady-state forced electricoscillatIOns can he treated as an altel'llatinrr current nowing in ~.circuit having a capacitance, inducta~ce,and resistance. Lnder the action of external voltage (which playsthe role of external e.m. f.)

U =" U m cos tl)t, (11.43)

the current in the circuit varies according to the law

I = 1m cos (cut '- cp), (11.44)

the excitation of strong oscillations at the frequency of external e.m.f. Or voltage. This frequency is equal to the natura~ fre~\lency of an oscillatory circuit. Resonance is usedfor slug-hng out a required component from a composite voltage. The entire radio reception technique is based on resonan~e. In order to receive with a given radio receiver thestation we are interested in, the receiver must be tuned. Inother words, by varying C and L of the oscillatory circuitwe must attain the cOincid~ncebetween its natural {requeuc;and. the frequency of radIo wayes emitted bv the radiostatlop. .

:rhe phenomenon of resonance is also associated with a certam danger: the external e.m.f. Or voltage may be small butt:le,v?ltages.acros~in(li,:i~Ju(d elem~nts of the circuit'(thecapaCItor or lllductlOn coIl} may attalll the values utlnaerousfor people. This should always be remembered. "

where

IiiIII

II

19*

29211. Electric Oscillations Problems 293

Th~s expl'ession can be written' l"ff .take lIlto account that 'IS foIl It a {I erent form If we(see Fig. 11.'1) U 0" _ nOIws ram the vector diagram

, m C S lp - m' Hence,

{P)--21

RI:". (11.51)

TTI,lis power.i~ equal to that of the direct current I' _, ' Ie quantities - I m/V2".

I = I mlV2, V = V mlV2 (11.52)are called effective (root-mea 'voltaO'e All a t IHsquare) values of current alld

'"' . mme ers 'md It tr.m.s. values of curl"ellt' . dVO m

leel'S are graduated for

'('1 an vo tage.... Ie expression for the 'Ivera .

r.m.s. \'alues of CUlTent a<nd glet

pow1er (11.50) in tenns of

vo age las the form

. . (P) = VI cos £P, (11.53)whele the factor cos ( is u IIThus, the power developed ~~a y called. th~ power factor.only on the voltage and C'lrrent an a.c. Circuit depends notbetween them. . but also on the phase angle

When <r c~ n/2, the value (P) - 0nes of V and I In th' - regardless of the val-quarter of a per:iod fr~s case, the energy transmitted over ais exactly equal to the ~ea generator. to an external circuitcircuit to the generatol' dU;~O't~~nsm~tted from the externalso that the entire energy "os""11 e

tn~;xt quarter of the period

generator and the external ~i~~u~tes uselessly between theThe dependence of the pOwer 0 .

account while desianing t ~ C?S 'P should be taken into-current. If loads b;ing fedr~I~~~lssrn lines for alternatingmay be considerably less than a ligh reactance X, cos £Pto transmit the required 0 one. n snch cases, in ordervoltage of the O'enel'ator) ~t ~'er to a COnsumer (for a givenI, which leads to an incr~~sel~tecessaryto increase currenting wires. Hence loads ." useless energy losses in feedbe always distribut~d "oln{~UCttancesand capacitances should

. s as ° make cos en a Ias possible. For this purpose it· ffi' 'Y sease to OneX ilS small as possible i 't IS su IClent to make reactanceductivc and capaciti:vc' r~~l'ct~n~~:u(~ tl~ equality of in-

When the conductors fanning an a.c. cir~lif~~~ in motion ,

the concept of electric resistance becomes wider since in addition to the conversion of electric. energy into Joule'sheat, other types of energy transformations are also possible. For example, a part of electric energy can be convertedinto mechanical work (electric motors).

Problems

• 11.1. Free undamped oscillations. Free undamped oscillationsoccur in an oscillatory circuit consisting of a capacitor with capacitanceC and an induction coil with inductance L. The voltage amplitudeon the capacitor is Urn- Find the e.m.f. of self-induction in the coil atthe moments when its magnetic energy is equal to the electric energyof the capacitor•

Solution. According to Ohm's law,

RI = U + ~s

where U is the voltage across the capacitor (U = CPl -CP2)' In ourcase, R = 0 and ,hence ~s = -U.

It remains for-Us to find voltage U at the moments when the electricenergy of the capacitor is equal to the magnetic energy of the coil.Under this condition, we Clln write

CU2 C[]2. L12 CU2~=-2--;'-2-=2 -2-'

whence I U 1= Um/Vi:"As a result, we have I ~s I = Um/ ,(2."8 11.2. An oscillatory circuit consists of an induction coil with

inductance L and an uncharged capacitor of capacitance C. The resistance of the circuit R = 0. The coil is in 11 permanent magnetic field.The total magnetic nux piercing all the turns of the coil is <1>. At themoment t = 0, the magnetic field was abruptly switched off. Findthe current in the circuit as a function of time t.

Solution. Upon an abrupt switching off of the external magneticfield at the moment t = 0, all' induced current appears, but the capacitor still remains uncharged. In accordance with Ohm's law, we have

drD dIRI=-dt-L dt

In the given case R = 0, and hence <1> + LI = 0. This gives <1> = LIo,where lois the initial current (immediately after switching off \hefield).

After the external fleld has been switched off, the process is described by the following equation:

q dlO=---L- (1)

C dt

u

(1\

295

Fig. 11.9

Problems

oFig. lUi

. h' 'h the current varies withtance L damped oscil!ations lCcu~, (T :: ;C e-Ilt sin oot, Find the volt-time in accordance With the aw ,t f 1-age across the capacitor as a function 0 Ime.. h 'f

h I kw'se directIOn as t e POSI IveSolution. Let us ch?ose (}. e c1~c9) IAccording to Ohm's law for

direction of circumven~iOn. Ig. I . . RI = cp _ qJ + 0S' In our·ction lRL2 of the CirCUit, we lave 1 2

se • _ Ie = Uc where q is the charge on~ = -"-LI and !P2 - !PI - q . '.

calset' 2' Hence Ohm's law can he written asp a e . •Uc=-RI-LI,

Having substituted i,:to this formula the expression for J (t) andits derivative, we obtam

U.- RIme-ill (_R sin oot-oo cos uH).

c- 2~ f'

. . tl parentheses to sine. ForLet us transform the expreSSIOn III lC

d 1 'd 't b 1fw2 +~ = Wo and inthis purpose, we multiply an (iVI elYtroduce angle II through the formulas. .

-~/ooo = cos 0, c,)/c,)o = Sill 6.

, d quate the obtained expres-dirIerentiate (1) with respect to t,lIne ~2 e _ 00 sin a =, 0 whencesion to zero at t = O. We obt:un, -r cos a 'tan a = -IVoo. The required ratIO IS

U (O) _ cos a = 1 ,- 1. " (2)--u;;;- - 11-;- tan2 a 1 l-r(~/W)-

.. U d (. (0) are shown in Fig, 11.8.The quantIt~es hm an.,.!-. 2 _ P.2 we transform (2) as follows:

Considerlllg t at w-- Wo p ,

U(O}/U m = 1rl-(~/OOo)2= jll-H2CI4.L,

h A - RI2l and w2 = lILC.where we took into account t at p- J 0

't C and induc-• 11 .5. In an oscillatory circui t wi th a capacl ance

11. ElectrIc O,Ciliatton,

I = 1m cos (OOot + a).

The constants 1m and a can be found from the initial conditions

1(0).==/0 , j (0)=0

(the ·second condition follows from Eq. (1), since at the initial momentt = 0 the capacitor was uncharged). From these conditions we finda = 0 and 1m = 10. As a result, we obtain

I = locos OOot = (11)IL) cos OOot,

.where 000 = 1/YLC.

• 11 .3. Q-Iactor 01 a circuit. An oscillatory circuit with a lowdamping has a capacitance C and inductance L. In order to sustainin it un(lamped harmonic oscillations with the voltage amplitude UmaeroSll the capacitor, it is necessary to supply the average power (P).Find the Q-factor of the circuit.

Solution. Since damping is low, we can make use of formula (11.23):

Q = 2nW16W, (1)

where W = CUfn/2 and 6W = (P) T, T being the period of dampedoscillations. In our case, T ~ To = 2n l" LC. HaVing substitutedthese expreSllions into (1), we obtain

Ufn .. /0Q= 2{P) V y.

• 11.4. Damped oscillations. Ali oscillatory circuit includes acapacitor of capacitance C, an induction coil of inductance L, a resistor01 resistance R, and a key. The capacitor was charged with the keyoren. When the key was closed, a current began to flow. Find the ratioo the voltage across the capacitor at time t to the voltage at the initialmoment (immediately after closing the key).

SolutIon. The voltage across the capacitor depends on time in thesame way as the charge does. Hence we can write

U=Ume-f\t cos (oot+a). (1)t-" ..,.

M tbe ltdtial Diomt'nt t= 0, thevoI~ge. f/(O) =V",eoa~, -hereU". ia tI» ..mpUfu~eat thi8 moment. We must find U (O)/Um• i.e.

..:~ t.hia rapose, we sb~l use anothe~ initial conditioll: at the

mo_' t = 0, current I = fJ = O. Since q = CU, it is sufficient to

Pifferentiation of this equation with respect to time gives•• 11+ LC 1=0.

This is ihe equation of harmonic oscillations. We seek its solutionin the form

Uc = Rlmwo -131 .211 I' I'Jn(Wi-O)=1 VLiC-l3 t .

, m e sm(wt-15),\\here angle 15 is, in accordanc " h .assumes the values :r/2 15 I' "It (1), In the second u d 'lags behind the curren~' <h n. Thus, the voltage acrossqtha rant, .1.1'.

m p ase. I' capacItor• 11.6. Steady-state os "II· t·' .

L and resistance R. . CJ a IOns. An mduction '1 f'external voltage U ,::slJonne~ted ll.!. the moment t =coJ t~ mductancea function of time m cos Wi. Fmd the current in th a ~our~e of

. e ClfCUl t asSolution. In our case, RI = .

U-LJ or. ,I-~(R/L) I =c Um cos wt

The s?IUtion of this equation' th. .equatIOn plus the particular ;~lut~ genefral solution of the homog

IOn 0 the nonhom eneousI (i) = Ac-(R/L)I _ Urn ogeneous equation:

}/R2-i- w2L2 cos (wt-<r),where A is an arbitrar .(11.36): tan fJJ = wLii constant and angle fJJ is defined b' d"

The constant A . 'f ~ con ItlOnH IS ount! from th "'a1

ence A = -(Um/yrR2 --'-. w2L2 I' Inlh condition 1(0) = 0C' ) cos fJJ· This gives .

I (t) == 'mVr

R2 -;- w2L2 [cos (Wi - ff) _e-(R/L)t cos fJJJ.

For a sufficiently lar e I

~g~~i~~ ~m;~l, an! \\~~ ~bta~~c~h~ ~~~:I~~t;~: :ic~~ts becomes. u Ion I (t) ex:

•.11.7. Forced oscillation'" A .

~f~~:;i~~n~~l~~~ec~i~~t~h a~d ri~~~~o~,o~sac~:::e~\~do:::isting of

~~:a~K~~~~e~~d~~~:~:t~gA~;r.~~~7:Ffn'd't:eh:h::P;~~l~t~t:~Solution In th

. I' case under consideration

U = Urn cos wt, I = 1m cos (~t _Where fJJ is defined b f 'P),

The unknown valu~Ofocrampul~t(11.36):. tan fJJ = -t/WCR. f aCI ance C b f, •

SlOn or the current amplitude' I _ c~n .~ ound from the expres-• m - {;ml J' R2 + (1/wC)! h, w enee

. C=1/w V(L'm1lm)2_R2.

Havmg substituted this expression' t fIn 0 ormula f~r tan ff, we obtain

tan cp= -I'I"Wm/Rlm)2_1.

20-0181

297

u~

Fig. 11.12

Problems

Fig. 11.11

1m Um-we

wLlm

Fig. 11.10

R I",

Im~·11'<0.1we U I

'" I-----

In our case cp < 0, which means that the: .current leads the externalvoltage (Fig. 11.10).

• t t .8. An a.c. circuit, containing a series-connected capacitorand induction coil with a certain resistance, is connected to the source

Wo= Y W IW 2'

• t t .9. Vector diagram. A circuit consisting of a series-connectedcapacitor of capacitance C and induction coil with resistance Randinductance L is connected to an external voltage with the amplitudeUm and frequency w. Assuming that current in the circuit leads inphase the external voltage, construct the vector diagram and use itfor determining the amplitude of voltage across the coil.

Solution. The vector diagram for the case under consideration isshown in Fig. 11.11. It is readily seen from this diagram that the

IWtL-_1_1,=1 w2L - _1_1. (1)WtG w,C

The maximum of the resonance curve for current corresponds to thefrequency equal to the natural frequency Wo = 1/yLC. Further•let Wt < Wo < Wt (the opposite inequality is also possible, it willnot affect the final result). Equality (1) can then be written in a moregeneral form: wG/wt - WI = Wt - W~/Wh or

w2-Wt = w5 (..!._..!-) .WI w2

Cancelling out (1)2 - WI on both sides of this equality, we obtain1= W5/WIW2' whence

of external alterpating \o"oltage whoSl'> frequency can be altered withoutchanging its arffplitude. At frequencies Wt and Wt the current amplitudes in the circuit proved to be the same. Find the resonance frequencyof the current.

Solution. According to (11.35), the amplitudes are equal under thecondition


Then

1'1

I

29811. Electric Oscillations Appendkes

-

10-'10-11

n, nano,p, pico-,

minutemaxwellnewtonoerstedohmradiansecondsiemensteslavoltwattweber

minMxNOeQradsSTVWWb

3. Units of Measurement dofGEIU8Seetr~ ~~~etiCQ titles in 81 an a J

t. Notations tor Units of Measurement

2. Decimal Prefixes for Units of Measurement

amperecoulombelectronvoltfaradgaussgramhenryhourhertzjoulekelvin It

meter

therli-, 1011 d, deci-, 10-1T,108 e, centi-, 10-1G, giga-,10e m, milli-, 10-3M, mega-,

micro-, 10-'&. kilo-, 102 f.L'h. hecto-, 101

de. deC8-, t01I

ACeVFG

LhrHzJKm

nan

Unit or mea-SI unitNota- 6urement

RatioCGS unitQuantity tton

I CGSSI

I .10eF N dyne1Q7

ForeeA,W J e~ 3 X lQt

Work. energyq C C SE unit

-Charge

amplitude of voltage across the coil is

ULRm=Im YRI+COILI,

where 1m = Um/Y RZ + (coL - 1/COC)I. In the presence of resistance

the voltage across the coil leads the current by less than 1(/2.

• f f .10. Power in an a.c. circuit. A circuit consisting of a series.connected resistor R with no inductance and an induction coil witha certain resistance is connected to the main system with the r.m.s.voltage U. Find the thermal power developed in the coil if the r.m.s.values of voltage across the resistor R and the coil are equal to U

Iand Uz respectively.

Solution. Let us use the vector diagram shown in Fig. 11.12. Inaccordance with the law of cosines we obtain from this diagram

UZ=Ur+U~-2UIUZcos CPL' (f)The power developed in the coil is

Pz = IUz cos CPL, (2)Where I = UI/R.

Combining Egs. (1) and (2), we obtain

Pz= (Uz_ Ui -U~)/2R.

eSc= 4nh

D=8E

~ DdS=4nq

o'=pn=')(,En

D=E+4T£P

e=1.+4T£')(,

C=q/U

Table 4. ConI.

Gaussian system

~Edl=O

p=ql

81

a'=Pn=uoEn

D=eoE+P

s=1.+k

D=880E

8E "'-{pXEl W=-p·EF=PBr,lU- •

p=KE

Appendicll8

Relation

Cireulation of vectorE in an eleetros\8de tield

Electric ,moment of adipole' .

Electric dipole p 1D

field ERelation between pG:

ladsation and fieldstrength

Relation between cr',pOOE

Definition of vector DRelation between e

and "Relat.ion between D

and EGaUll8 t.heorem for

vectorDcapacit.anC8 of a ca

pacitorCapacitanee of a par

allel-plate capaci-t.or

Energy of system ofcharges

To\81 energy of interact.ion

Energy 01 capacitorElect.ric tield energy

densit.y

Continuity equation

Ohln's lawJoule-Lens law

LoleDU fo~

E=~II

lp=..!Lr

2

E= - V .cp, CPl-lp,= ) E dl1

81

E=_G_80B

1 qcp----- 4neo r

t qE=--

4neo r 2

Te6le 3. COllI.

4. Basic Formulas of Electricity and Magnetismin 81 and Gaussian Systems

Field E of a pointcharge

Field E in a parallelplate capacitor andat the surface of aconductor

Potential of the fieldof a point charge

Relation

Relation between Eand rp

300

Unit of _-Quantity Nota- nnllDellt ~'I_'tton

ICOS ..,

81 eGS

Electric field strength E VIm CGSE unit tJ(3X{O')Potential, voltage cp,U V CGSE unit tPJX)Electric ID~ent p C·m CGSE unit 3X IOUPolarization P ClIO' CGSE unit 3XI06Vector D D ClIO' CGSE unit 12n xtOiCapacitance C F cm 9xtOUCurrent I A CG8E unit 3 X tQllCurrent density j Aim' CGSE unit 3xtOlResistance R Q CG8E unit t/(9X IOU)Resistivity

~Q·m CGSE unit t/(9X {O')

Conductance 8 CGSE unit 9X1OUConductivity cr 81m CG8E unit 9xtOtMagnetic induction B T G 101Magnetic flux, magnetic-

nux linkage <D,W Wb Mx 1(18Magnetic moment Pm A·m' CG8E unit UPMagnetization J A/m CG8E unit 10"4Vector H H A/m Oe 4axto-aInductance L H cm tOt

Appendices303

302 Appendices Table 4. Cont.

Table 4. Cont.Relation

81Gaussian system

EL_B2= inv1 dl.1l

't5i=-7CftL=cl.1ljI

L=4nltn2V1 dI

'6>8= -(2" L dt1 LP

W=-c--r-

B·Bw=--gn-

i aDid = 4it ---at

1 ~ .~ Edl= -7 ~ BdS

~ Das=4n ~ pdV

~ Hdl=

=4: ~(i+~lt)dS

.~ ndS=O1 •

~ X E=-7B

V·D =4np

VX H = ~ ( j + ~lt)\ 'V.B=O _o

~ BdS=O

vxE=-B

V·D=P

vxH=j+D

E2_c2B2= invdl.1l

~i=-dt

L=l.1ljI

L=jJ.jJ.on2VdI

~s= -LatL[2

W=~

B·Hw=~

aDjd='at

~Edl= - ~ BdS

~ D dS= ~ pdV

~ Hdl= ~ (i+D)dS

E'=E+IvoY~l \l'~'=E++lVOXBlB'=B-~l\'oXEl B'=B-+lVo>~El

cE·B=inv

Maxwell's e';\uation~\in differential form \

\

Electromagnetic fieldinvariants

Induced e.m.f.

InductanceInductanee of a so-

lenoidE.m.f. of seH-induc

tionEnergy of the'!' mag

netic field of a current

Magnetic field energydensity

Displacement currentdensity

Maxwell's equationsin integral forlD

Laws of transformationof fields E and Bfor Vo « c

1f1=1+4nXB=[tH

1A = - I (11)2-<1'1)

c .

JJ dl~>~ l'

II = B--411J

~ II dl = 4n I'j t~

B=_1_ q[vXr)c r3

dB=_1 [j Xr) dVc rll

dB,=~ l[dlXr)c r3

Gaussian system

Fu= 211[2

b

1Pm=-lS

c

B=~~c b

B="~ 2Jtlc Ii

4JtB=--nl

c

dI~= !-[dl X B)c •

1dF=- [j X B) dV

c

IJ =;<:11,

Pm =IS

aBF=Pm~'

'~J dl=I'

H = B/[lo-J

~ II dl=IoJ

81

B=flonI

B=J:L -E.4n b

B=~ 2n14n R

dF=I [dJ X B]

dF=[j X B] dV

Fll

= ~o 21112

'in b

B=A q[vXr]4n r3

dB=~ [j X rjdV4n r 3

dB= flo 1 [dJ >< r]4n r 3

Relation

Ampere's law

Force of interactionbetween parallelcurrents

Magnetic moment ofa current loop

Magnetic dipole Poiin field B

Work done in displacement of a currentloop

Circulation of mag-\netization

Definition of ve,~tor IICirculation of vector II

II in a stationaryfield

Relations between' IJ and H IfJ. and Xi ,Band H ,

Field B(a) of straight

current

(b) at the centre ofa loop

(c) in solenoid

Field B of a movingcharge

Biot-Savart law

Appendices

5. Some Physical Constants

Magnetic constant

Damping factor, 280, 283diamagnetism, 180dielcctric(s).

anisotropic, 101isotropic, 101liquid, W8fpolarization of, 67

electrodynamic, 144magnetic, 143

coulomb, 12current, I

alternating, 290ffconduction, 174direct, 178displacement, 253f, 271induced, 248magnetization, 174

surface, 175, 178, 192volume, 175

molecular, 174, 177polarization, 256

density of, 259quasistationary, 278straight, 188wt~l, 254

current element(s),linear, 146volume, 146

curve,magnetization, 189resonance. 288f

Subject Index

Capacitance, 57ff, 86, 90of cylindrical capacitor, 60of isolated conductor, 57of parallel-p\pte capacitor, 59of parallel wires, 65of spherical capacitor,.59unit of, 58

capacitor, 57ft, 86, 90charging, 134discharging, 133energy of. 110parallel-plate, 103, 108f. 1.1.4f,

137charge invariance, 198£circulation,. of vector B, 166

of vector E, 245of vector H, 179, 193of vector J, 176ff, 193

condition, betatron, 246, 260quasi-steady, 277

conductor in electrostatic field. 45homogeneous, 120nonhomogeneous, 138

constant,dielectric (electric), 12, 78£. 86,

102

Betatron, 222boundary conditions, 80, 92, 192

for Band H. 182ff

Ampere, 117

)I

Gaussian system

Tabl. 4. Cont.

cS="41tIEXHj

1G=-4- IEXnnc

c= 2.998 X 108 m/sG= { 6.67 X 10-11 m3/(kg·s2)

6.67 X 10-8 cmS/(g.s2)g=9.807 m/s2

N A = 6.022 X 1023 mole-l

{1.602 X 10-19 C

e = 4.80 X 10-10 CGSE unitsme =0.9H X 10-30 kg

e { 1.76 X 1011 C/kg--;n;- = 5.27 X 1017 CGSE units/gmp= 1.672 X 10-27 kg80=0.885 X 10-11 F/m1/4n80 = 9 X 109 m/FIto = 1..257 X 10-8 H/ml-lo/4n= 10-7 H/m

c=1/~

81

S";"[EXHJ

G=_1 [E XII)'c2

E Y80e=H~Relation between E

and H in electromagnetic wave

Poynting's vector

Density of electromagnetic field mo-mentum

Relation

Velocity of light in vacuum

Gravitational constantl

Acceleration of free fallAvogadro constant

Charge of electron or proton

Rest mass of electron

Specific charge of electron

Rest mass of protonElectric constant

Relation between velocityof light and 80 and Ito

aq

Subject Inckz

Generator, MHD, 22!t

Henry, 227hysteresis, 72, 180

magnetic, 190

moment of, 160, lG9work of, 161

coerei ve, 1gOcoulomb, 122in dielectric, 106Helectric, 157extraneous, 122f, 235induced electromotive, 217

force(s) ,of interaction,

electric, 144£magnetic, 144£

Lorentz, 141H, ~03, 212fIl1agn('tic, 157in -ma~netic, 19Gsurface density of, 109£

nux,magnetic,

conservation of, 230total, 212

of vector D, 273formula, Thomson, 281

frequency,natural, 280rl'Sonance, 289

potential, 2G, 28, 155sinks of, 25solenoidal, 155soul'ces of, 25vector,

circulation of, 15, 25divergence of, 24flux of, 15

force(s),Ampere, 155,157,159,169,171,

Farad, 58fferroelectric(s), 72ferromagnetic(s), 180, 188f, 232ferromagnetism, 188,I!H

theory of, 191field,

coulomb, 122electric, 11, 198, 206

in charge-carrying conductor,121

in dielectrics, 61intensity (strength) of, lfff,

28macroscopic, 45ffmicroscopic, 45of point charge, 12

eleclromagn,etic, 198fmomentSm of, 268deIlsi ty of, 269

electrostatic, 155in vacuum, 11

of freely moving relativisticcharge, 207

in homogeneouS magnetic, 185magnetic, 142, 1115, 155, 198,

206on the axis of circular cur-

rent, 147of current-carrying plane, 153energy of, 2/iOforces in, 240in magnetic, 175permanent, 220of solenoid, 151, 167,181of straight current, 151in substance, 172ffof toroid, 152of uniformly moving eharge,

143in vacuum, 141ffvarying, 221vortex, 222, 275

Subject Ind,:::e.:::..c -------307

--_._-------'-electrosta tic shielding, 51e.m.f.,

induced, 244of self-induction, 227

energy,of charged capacitor, 99f, 139of charged conductor, 99fof current,

intrinsic, 238magnetic, 236mutual, 238

of electric field, 94, 100, 103,105, 107, 237, 276

of electromagnetic field, 253localization, iOOff, 112magnetic, 243

of two current loops, 238of interaction, 96ff

for system of point charges,97, 105, 110f

total, 97, 99f, 110intrinsic, 98, 105, 110fof magnetic field, 234 236f

275f "of current, 234

energy flux, 266density of, 264

equation(s),continuity, 116, 118, 261

in differential form, 118Laplace, 54fmaterial, 260Maxwell, 253ff, 257f, 272, 274

in differential form, 259in integral form, 257properties of, 261fsymmetry of, 262

of oscillatory circuit, 277, 279fPoisson, 5H

equipotential surfaces 32, 47.57 '

Edge effects, 59, 66, 8G, 100electret(s), 68, 79electric charge(s), 11, 83, 88f, 92

bound (polarization), 69f, 73fbulk, 69fat conductor surface, 83surface, 69£

in current-calTying conductor120 '

extraneous, 78, 88f, 109, 138fictitious, 63induced, 46, 66surface, density of, 63

electric circuit,a.c., 291, 297branched, 126££oscillatory, 293

electri'c current, 116, 119density of, 117direct, 116, 120linear, density of, 153quasis ta ti 0 nary , 133steady-state, 118

electric oscillations, 277££electric resistance, 119

methods of calcnlation 120electromagnetic inductio~ 217ff

225 "electromagnetic wave(s), 101, 262f

pressure of, 268, 270in vacuum, 269

dielectric susceptibility, 71dipole,

electric, 34, 44energy of, 38moment of, 34ff

dipole moment, 71intrinsic, G8

domain(s), 191

306

IIIi

Subject IndexSubject Index

308

Image charge, 56impedance, 290finductance, 226, 238

calculation of, 250fmutual, 231, 240, 251

induction,electrosta tic, 46magnetic, 143ff

direct calculation of, 163mutual, 233, 251residual, 190

interaction(s), 11electromagnetic, 11,gravitational, 11of parallel currents, 168strong, 11weak, 11

invariants of electromagneticfield, 208, 214

Joule heat, 235

Law(s),Ampere, 155ffBiot-Savart, 145ff, 149f, 153,

164of conservation,

of electric charge, 11, 118of energy, 106, 114, 130, 139

235, 241 'Coulomb, 109'

in field form, 12louie, 129ff

in differential form, 131Kirchhoff, 126, 256

first, 126second, 126f

Lenz, 217ff, 243of magnetio field, in differen

tial form, -194

Newton's third, 95Ohm, 119, 127, 131, 134, 136ff,

140f, 233, 249, 293, 295In differential form, 120generalized, 123in integral form, 124for nonuniform 8ubclreult

123, 138 'of transformation for fields E

and B, 200ft, 206, 213logarithmic decrement of

damping, 283

Magnet, permanent, 68, 195magnetlc(s), 172, f 74, 181

homogeneous, 174nonhomogeneous, 175, 177

magnetic constant, 143magnetic field intensity, 179magnetic flux, 167

linkage of, 219magnetic induction, 143f, 146,

163, 166, 168, 187magnetic prot~ction, 185magnetism, relative nature of, 204magnetization, 172ft, 195

mechanism of, 177residual, 190

magnetohydrodynamics, 244method,

energy, 106, 240image, 54ft, 62f, 67

molecule(s),nonpolar, 68polar, 68

moment, magnetic, 157ft

Ohm, 119oscillation(s),

damped, 281f, 294

oscillation(s) ,forced,294free, 280steady-state, 285, 296undamped, 280

Paramagnetic(s), 180, 194, 197permeability, 180, 189point, Curie, 191polarization of dielectric, 68, 73,

86, 103polarization, 70fI

unit of, 71potential, 25, 27, 30, 33, 43f,

47f, 53f, 57, 61, 64, 66,86f, 8&r 97f, 102, 124

of point charge, 28of sphere, 58 •of system of charges, 29

potential difference, 86, 90, 99power liberated in a.c. circuit,

291f, 298power factor, 292pressure, magnetic, 244principle of superposition, 12f,

18, 29, 60, 66, 146, 166fprocess, transient, 133, 140

Q-factor, 284, 289, 294

Reflection coefficient, 270refraction of B-lines, 184relaxation time, 134, 283resistance of conducting medium,

136resistivity, 119resonance, 289f

Self-induction, 226solenoid, 151superconductor(s), 230susceptibility, magnetic, 180

Tesla, 144theorem,

on circulation of field vectors,15, 25, 52, 80f, 137f, 148f,151ft, 155, 166, 178f, 193f,195f, 211, 248, 255

Gauss, 15ft, 19, 21ft, 25, 33,41, 44, 47f, 59, 65, 72,76f, 79ft, 83, 85, 871f, 9f,93, 102, 112, 121, 137,148, 150, 157, 168, 173,200, 211

application of, 19in dHIerential form, 23ff, 73

Poynting's 264of reciprocity, 231, 252uniqueness, 54ff

theory of relativity, 269thermal power, 132transformation(s), Lorentz, 20f

Vector,D, 76ftB, 178Poynting's, 265, 269, '274f

• vector diagram, 287, 297volt, 28voltage, 59

Work,of displacement of current

loop, 161ftof electric field, 103fof e.m.f. source, 139

.......

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Livro - Basic Laws of Electromagnetism por Irodov - Mir Publisher (1986)

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Transcript of Livro - Basic Laws of Electromagnetism por Irodov - Mir Publisher (1986)