Magnetostatics - ece.uccs.edu

Chapter 5Magnetostatics

Contents5.1 Magnetic Forces and Torques . . . . . . . . . . . . . . 5-4

5.1.1 Force on Current Carring Conductors . . . . . . 5-5

5.1.2 Torque on a Current-Carrying Wire . . . . . . . 5-6

5.2 The Biot-Savart Law . . . . . . . . . . . . . . . . . . 5-115.2.1 Surface and Volume Current Distributions . . . . 5-12

5.2.2 Magnetic Dipole . . . . . . . . . . . . . . . . . 5-19

5.2.3 Force Between Conductors . . . . . . . . . . . . 5-20

5.3 Maxwell’s Magnetostatic Equations . . . . . . . . . . 5-215.3.1 Gauss’s Law for Magnetisim . . . . . . . . . . . 5-21

5.3.2 Ampère’s Law . . . . . . . . . . . . . . . . . . 5-22

5.4 Vector Magnetic Potential . . . . . . . . . . . . . . . . 5-285.5 Magnetic Properties of Materials . . . . . . . . . . . . 5-29

5.5.1 Magnetic Permeability . . . . . . . . . . . . . . 5-29

5.6 Magnetic Boundary Conditions . . . . . . . . . . . . . 5-335.7 Inductance . . . . . . . . . . . . . . . . . . . . . . . . 5-35

5.7.1 Self Inductance . . . . . . . . . . . . . . . . . . 5-38

5.7.2 Mutual Inductance . . . . . . . . . . . . . . . . 5-44

5-1

CHAPTER 5. MAGNETOSTATICS

5.8 Magnetic Energy . . . . . . . . . . . . . . . . . . . . . 5-455.9 Appendix: Solenoid Exact Magnetic Field Off Axis . . 5-46

5-2

� As we move from from electrostatics to magnetostatics weleave behind two of Maxwell’s equations and begin using theremaining two equations

r � B D 0

r �H D J ;

where as a quick review B is the megnetic flux density, H is themagnetic filed intensity, and of course J is the current density

� Also recall that B D �H, but in magnetics it is not always alinear relation; ferromagnetic materials are the exception wewill see later in this chapter

� With electrostatics behind us it is good to look at what liesahead in the context of the past

Table 5.1: Moving from electrostatics to magnetostatics.Table 5-1 Attributes of electrostatics and magnetostatics.

Attribute Electrostatics Magnetostatics

Sources Stationary charges ρv Steady currents JFields and Fluxes E and D H and BConstitutive parameter(s) ε and σ µ

Governing equations• Differential form

• Integral form

∇ ·D= ρv∇×××E = 0

♥∫

SD ·ds = Q

♥∫

CE ·dl= 0

∇ ·B = 0∇×××H = J

♥∫

SB ·ds = 0

♥∫

CH ·dl = I

Potential Scalar V , with Vector A, withE= −∇V B= ∇×××A

Energy density we = 12εE

2 wm = 12µH

2

Force on charge q Fe = qE Fm = qu×××BCircuit element(s) C and R L

5-3


5.1 Magnetic Forces and Torques

� An electric field acting on a charge produces a froce, so it iswith magnetic flux acting on a moving charge

Fm D qu � B .N/

� The right-hand rule will be used frequency in this chapter

(a)

(b)

θ

+q

uB

Fm = quB sin θ

−

+

F

F

B

u

u

Figure 5-1 The direction of the magnetic force exertedon a charged particle moving in a magnetic field is (a)perpendicular to both B and u and (b) depends on thecharge polarity (positive or negative).

Figure 5.1: Right-hand rule for a moving charge with B present.

� With the dot product the projection of A � B always involvesthe cosine of the angle between, i.e., AB cos �

� With the cross product the resultant now involves the sine ofthe angle between A and B, i.e., AB sin �

� When both E and B are present the total force becomes

F D qE„ƒ‚…Fe

C qu � B„ ƒ‚ …Fm

D q.EC u � B/ .N/

� It is somewhat peculiar that magnetic force does not exist un-less the charged particle is moving, and even then no work isdone as the force is perpendicular to the motion u

5-4

5.1. MAGNETIC FORCES AND TORQUES

5.1.1 Force on Current Carring Conductors

� A charged particle in motion receives a force from B, so itfollows that B acting on current flow in a wire imparts force tothe charged particles in the wire and hence the wire itself

z

yx

I

B

B

B

I = 0

(a)

I

(b)

(c)

Figure 5-2 When a slightly flexible vertical wire isplaced in a magnetic field directed into the page (asdenoted by the crosses), it is (a) not deflected when thecurrent through it is zero, (b) deflected to the left whenI is upward, and (c) deflected to the right when I isdownward.

Figure 5.2: Force on a wire carrying current under varous conditions.

� A detailed description of the force on the wire can be devel-oped using differential charge traveling at velocity ue (electron

5-5


drift velocity), by writing

dFm D dQ ue � B

� Work through the details of drift velocity, cross-sectional are,and the density of the charges particles, we can write

dFm D I d l � B .N/

and then upon integrating over a closed contour (the circuit)arrive at

Fm D I

IC

d l � B .C/

� An important side note is that if B is uniform it can be factoredoutside the contour integral, leaving the integration over theclosed contour all by it self, which is simply zero, i.e., in theend no net force

5.1.2 Torque on a Current-Carrying Wire

� In physics you learn that torque arises out of a rotational force,i.e.,

T D d � F .N� m/;

where d is a moment arm to which a force is applied on oneend and the other end is connected to a rotating shaft

– Don’t act too surprised about getting into some rotationalmechanics

– Electrical engineers do deal with motors and machines

5-6


θ

F

T

d

z

x

y

Pivot axis

Figure 5-5 The force F acting on a circular disk thatcan pivot along the z axis generates a torque T = d×Fthat causes the disk to rotate.

Figure 5.3: definition of torque T D d � F .

� Drilling down a bit, Figure 5.3 shows that

T D Ozr F sin � .N� m/

where r D jdj and F D jF j

� For a CD positioner example see the Signals and Systems forDummies on-line example1

Magnetic Field Acting a Multi-Turn Loop

� Consider initially a single turn loop of wire in a uniform B field

� We assume that B D OxB0

1http://www.dummies.com/how-to/content/cddvd-electromechanical-system.html

5-7

http://www.dummies.com/how-to/content/cddvd-electromechanical-system.html

http://www.dummies.com/how-to/content/cddvd-electromechanical-system.html


(a)

(b)

I

y

xbO

a

Pivot axis

B

B

2

4

31

Loop arm 3Loop arm 1

z

xy

a/2

O

z

−z

d1

F1

B F3

d3

Figure 5-6 Rectangular loop pivoted along the y axis:(a) front view and (b) bottom view. The combination offorces F1 and F3 on the loop generates a torque that tendsto rotate the loop in a clockwise direction as shown in (b).

Figure 5.4: A single loop of wire parallel to B and the force analysisthat results in torque.

� Due to the right-hand rule only segments 1 and 3 receive anon-zero force

5-8


� The moment arms of length a=2 receive the correspondingforces F1 and F2 respectively and produce a net torque of

T D d1 � F1 C d3 � F3 D OyIabB0 D OyIAB0 .N� m/;

where I is the current in the loop, A is the area of the loop, andB0 D jBj and the current in the loop isI

� Assuming the shaft is free to turn, the nonzero T will changethe position of the loop to the general scenario of Figure 5.5

� With the loop rotated to angle � the torque is no longer at itsmaximum, but now

T D jT j D I AB0 sin � .N� m/

Multiturn Loop

� If the single turn loop is modified to contain N turns of wire,the torque is increased to

T D jT j D N I AB0 sin � .N� m/

� There is also a magnetic moment, m that lies along the normalOn to the loop

m D OnN I A D Onm .A� m2/

and the torque can be written as

T D m � B .N� m/

5-9


(a)

(b)

x

y

z

F4F3

B

B

I

F2F1

n

ab

3

Pivot axis

21

4

ˆ

θ

m (magnetic moment)

Arm 1

Arm 3

OB

F1

F3

n

a/2

(a/2) sin θ

θ θ

ˆ

Figure 5-7 Rectangular loop in a uniform magneticfield with flux density B whose direction is perpendicularto the rotation axis of the loop, but makes an angle θ withthe loop’s surface normal n̂.

Figure 5.5: A single turn loop no longer parallel to B.

5-10

5.2. THE BIOT-SAVART LAW

5.2 The Biot-Savart Law

� Up to this point no means of generating a magnetic flux or fieldintensity has been considered

� Biot and Savart studied how a compass needed was deflectedwhen current carrying conductors were brought nearby

� The result of their work is the Biot-Savart law, which beginswith a description of the field intensity, H, due to a differentialcurrent element

dH

P

P'

R

R

dl

I

(dH into the page)

dH

(dH out of the page)

θ

ˆ

Figure 5-8 Magnetic field dH generated by a currentelement I dl. The direction of the field induced at point Pis opposite to that induced at point P ′.

Figure 5.6: Finding the magnetic field dH intensity due to a currentelement.

� The Biot-Savart law states that

dH DI

4�

d l � ORR2

.A/m/

in spherical coordinates

5-11


� Note the similarity to Coulomb’s law, except for the cross prod-uct

� Over a conductor path l we obtain H as the line integral

H DI

4�

Zl

d l � ORR2

.A/m/

5.2.1 Surface and Volume Current Distributions

� The Biot-Savart law also can be applied to surface and volumecurrent distributions

H DI

4�

ZS

Js � ORR2

ds .A/m/

H DI

4�

ZV

J � ORR2

dV .A/m/

Example 5.1: Field from a Length l Conductor

� This is a classic example of a length l conductor along the zaxis carrying current I in the Oz direction

� We desire the H in the x � y plane at radial distance r

� We set the problem up using the figure below

5-12


(a)

(b)

z

I

P

θ1

θ2

r

R2

R1

l

I

P

dl

dl dθ

r

z

R

R

dH into the page

l

θ

ˆ

Figure 5-10 Linear conductor of length l carrying acurrent I. (a) The field dH at point P due to incrementalcurrent element dl. (b) Limiting angles θ1 and θ2, eachmeasured between vector I dl and the vector connectingthe end of the conductor associated with that angle topoint P (Example 5-2).

Figure 5.7: Finite length linear conductor carrying current I .

� Writing out R, OR, and d l � OR we have

R D�Orr��Ozz�

OR DOrr � Ozzpr2 C z2

d l � OR D Ozdz ��Orr � Ozzpr2 C z2

�DO� r dzpr2 C z2

since Oz � Or D O�

� We are now ready to integrate over �l=2 � z � l=2

H DI

4�

Zl

d l � ORR2

D O�I

4�r

Z l=2

�l=2

dz�r2 C z2

�3=2D O�

I l

2�rp4r2 C l2

.A/m/

5-13


� Converting to magnetic flux

B D �0H D O��0I l

2�rp4r2 C l2

.T/

� When l � r the wire can be treated as infinite in length, thusgiving the important result that

B D O��0I

2�r(T) (for infinite length wire)

� This result parallels the infinite line charge by having the fieldinversely proportional to distance

� The field however forms concentric circles around the wire

I

Magnetic field

I

B

B

B

Figure 5-11 Magnetic field surrounding a long, linearcurrent-carrying conductor.

Figure 5.8: The field of a long linear conductor encircles the con-ductor and falls off as 1=r .

� The finite length and infinite cases are compared using Pythonas a function of r below:

5-14


Figure 5.9: Python code used for the comparison.

1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

Axial Field Point r (m)

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

Norm

aliz

ed A

zim

uth

al Fl

ux D

ensi

ty

Plot of Bφ/(µ0I) versus r with l a Parameter

l→∞l= 1 (m)

l= 5 (m)

l= 10 (m)

Figure 5.10: Normalized B� comparisons versus r .

5-15


Example 5.2: Circular Loop in the x � y Plane

� Another classical magnetostatics problem is a circular loop ofradius a carrying current I , in the x � y plane

� The field point is P D .0; 0; z/

φ θ

θ

a

φ + π

P = (0, 0, z)

R

dHr

dH'r

dHz

dH'z

dH

dH'z

dl'

dlIx

y

Figure 5-12 Circular loop carrying a current I(Example 5-3).

Figure 5.11: The field of a long linear conductor encircles the con-ductor and falls off as 1=r .

5-16


� Writing out R, OR, d l � OR, and dH we have

R D�Ozz��Ora�

OR D�OraC Ozzpr2 C z2

d l � OR D O�a d� ��OraC Ozzpr2 C z2

�D a d�

OzaC Orzpr2 C z2

dH DIa

4�R2OzaC Orzpa2 C z2

d�

since O� � �Or D Oz and O� � Oz D Or

� Finally, from Figure 5.11 we see that symmtery makes the ra-dial component Hr D 0, and integrating over 0 � � � 2�

gives

H D OzIa2

4��a2 C z2

�3=2 Z 2�

0

d�

D OzIa2

2�a2 C z2

�3=2 .A/m/

� Special Cases:

– When z D 0 we have

Hˇ̌zD0D Oz

I

2a.A/m/

– When z � a we have

Hˇ̌z�aD Oz

Ia2

2jzj3.A/m/

� Solution comparisons using Python are given below:

5-17


Figure 5.12: Python code used for the comparison.

1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

Axial Field Point z (m)

0.0

0.1

0.2

0.3

0.4

0.5

Norm

aliz

ed A

zim

uth

al Fi

eld

Inte

nsi

ty

Plot of Hφ/I versus r with l a Parameter

Full a= 1 (m)

Full a= 1/2 (m)

Approx zÀ a, a= 1 (m)

Figure 5.13: Normalized Hz comparisons versus z.

5-18


5.2.2 Magnetic Dipole

� For a single-turn loop we can define the magnetic dipole mo-ment Omm D OzI�a2, so

Hˇ̌z�aD Oz

m

2�jzj3.A/m/

� It can also be shown that in spherical coordinates with R� a,H.R; �; �/ becomes

H Dm

4�R3

�OR2 cos � C O� sin �

�where

� D cos�1�

apa2 C z2

�� The magnetic dipole and electric dipole have similarities

(a) Electric dipole (b) Magnetic dipole (c) Bar magnet

+

−

E

S

N

HH

I

Figure 5-13 Patterns of (a) the electric field of an electric dipole, (b) the magnetic field of a magnetic dipole, and (c) themagnetic field of a bar magnet. Far away from the sources, the field patterns are similar in all three cases.

Figure 5.14: Dipole field patterns plus bar magnet.

5-19


5.2.3 Force Between Conductors

� Consider two parallel (infinite length) conductors that pass throughthe y axis parallel to the z axis

d/2d

x

y

z

d/2

F'1 F'2

B1

I1 I2

l

Figure 5-14 Magnetic forces on parallel current-carrying conductors.

Figure 5.15: Analyzing the force between parallel infinite lengthconductors.

� In the case shown here the current in each wire flows in thesame direction so the wires attract

� Using the fact that Fm D I` � B and the azimuthal field pro-duced by an infinite linear wire, it can be shown that the force

5-20

5.3. MAXWELL’S MAGNETOSTATIC EQUATIONS

per unit length is

F 01 DF1

lD Oy

�0I1I2

2�d.N/m/

F 02 DF2

lD �Oy

�0I1I2

2�d.N/m/

Note: The force magnitudes are equal

5.3 Maxwell’s Magnetostatic Equations

� Using Maxwell’s equation two more properties can be defined

5.3.1 Gauss’s Law for Magnetisim

� For the case of the electric field (flux)

r �D,

IS

D � ds D Q

Also for a charge volume

� On the magnetic field side,

r � B,

IS

B � ds D 0

– In magnetics there is no equivalent to a point charge (nomagnetic monopole)

5-21


5.3.2 Ampère’s Law

� In electrostatics the line integral around a closed contour isalways zero, i.e.,

r � E D 0,

IC

E � d l D 0

� Ampère’s law is the counterpart to this in magnetics

r �H,

IC

H � d l D I

� Just as in the electric field case of Gauss’s law, Ampère’s lawis best applied to scenarios where symmetry is present

– Here we seek symmetric current distributions so that con-venient contours may be utilized

Example 5.3: A Long Wire

� Determine the magnetic field H at a distance r � a and r � afrom the axis of a wire having diameter a

� For convenience choose the wire to lie along the z axis

� The Ampèrian contour is a circle of radius 0 � r1 � a orr1 � a

� The symmetry lies in the fact that H D O�H�, e.g. pure az-imuthal

5-22


(b) Wire cross section

(a) Cylindrical wire

(c)

a

H(a) = I2πa

H(r)

r

H1 H2

C2

I

z

a

C1

r2

r1

Contour C2

for r2 ≥ a

Contour C1

for r1 ≤ a

C2

a

x

y

C1

φ1

r1

r2φ̂

Figure 5-17 Infinitely long wire of radius a carryinga uniform current I along the +z direction: (a) generalconfiguration showing contours C1 and C2; (b) cross-sectional view; and (c) a plot ofH versus r (Example 5-4).

Figure 5.16: Finding H for a long wire using Ampère’s law, with C1for r � a and C2 for r � a.

� For 0 � r D r1 � a we haveIC1

H1 � d l1 D

Z 2�

0

H1 . O� � O�/r1 d� D 2�r1H1 D I1

and on the right side the current fromRS

J � ds is the ratio ofthe areas times I , i.e.,

I1 D

��r21�a2

�I D

�r1a

�2I

5-23


� Finally, we solve for H1 (H1 D I1=.2�r1/)

H1 DO�I1

2�r1D O�

.r1=a/2I

2�r1D O�

r1

2�a2I .A/m/

� For r D r2 � aIC2

H2 � d l2 D 2�r2H2 D I2 D I;

so

H2 DO�H2 D

O�I

2�r2.A/m/(b) Wire cross section

(a) Cylindrical wire

(c)

a

H(a) = I2πa

H(r)

r

H1 H2

C2

I

z

a

C1

r2

r1

Contour C2

for r2 ≥ a

Contour C1

for r1 ≤ a

C2

a

x

y

C1

φ1

r1

r2φ̂

Figure 5-17 Infinitely long wire of radius a carryinga uniform current I along the +z direction: (a) generalconfiguration showing contours C1 and C2; (b) cross-sectional view; and (c) a plot ofH versus r (Example 5-4).

Figure 5.17: Magnetic field strength magnitude H.r/ D j O�H�j ver-sus r for a conductor of radius a.

Example 5.4: A Long Coax with Current I

� Consider a long coax with inner conductor current I

� What is the magnetic field outside the coax?

5-24


� Visualize an Ampèrian contour outside the cable and considerthe superposition of the field due toCI in the inner conductorand �I in the return path of the outer conductor

� The net field is zero!

Example 5.5: Magnetic Field of a Toroid Coil

� Yet another classical example, is the toroid coil with inner ra-dius a and outer radius b

� We assume N turns are tightly spaced around the toroid

rI

Ib

a

Ampèrian contour

H

C

φ̂

Figure 5-18 Toroidal coil with inner radius a and outerradius b. The wire loops usually are much more closelyspaced than shown in the figure (Example 5-5).

Figure 5.18: An N turn toroid coil.

� In Figure 5.18 the windings are spread apart to add notes, butin reality the windings are tight so each loop is parallel to thecross section of the toroid

5-25


� Find H for r < a, a � r � b, and r > b

� Based on the direction of the current flow and using the right-hand rule, the flux of each loop is normal to the loop crosssection in the � O� direction

� To be clear, there is no field present for r < a and r > b, why?

� Applying Ampère’s law for a � r � b yieldsIC

H � d l D

Z 2�

0

.� O�H/ � O�r d� D �2�rH D �NI

� Finally,

H D � O�H D

(� O� NI

2�r(A/m); for a � r � b

0; otherwise

Example 5.6: An Infinite Current Sheet

� Consider an infinite sheet of current in the x � y plane withJ2 D OxJs

� Find H everywhere

� A convenient Ampèrian contour is a rectangle in the y�z planew by l

5-26


y

z

l

w

H

H

Js (out of the page)

Ampèriancontour

Figure 5-19 A thin current sheet in the x–y planecarrying a surface current density Js= x̂Js (Example 5-6).

Figure 5.19: An infinite current sheet in the x � y plane.

� Using the right-hand rule the direction of H above and belowthe sheet is a shown in Figure 5.19, i.e.,

H D

(�OyH; z > 0

OyH; z < 0

� Now apply Ampère’s law around the contourIC

H � d l D Hl C 0CHl C 0 D 2Hl D Jsl

where the zero entries come from the length w paths

� Finally solving for H we have

H D

(�OyJs

2; z > 0

OyJs2; z < 0

5-27


5.4 Vector Magnetic Potential

� With electrostatics we found that introducing the scalar poten-tial V led to a convenient way of calculating E via �rV

� A similar idea is possible with magnetic fields via the vectormagnetic potential A, where

B D r �A .T = Wb/m2/;

and webers per square meter is an alternate unit for tesla

� Working with some vector identities it can be shown that

A D�

4�

ZV 0

J

R0dV 0 .W/m/

� Note: It is also possible to write the above using a surface orline current distribution

� The vector magnetic potential often yields easier integrationsthat the Biot-Savart law when calculating B or H

Magnetic Flux ˆ Linking Surface S

� B is the magnetic flux density, but when we consider the fluxpassing through a surface S , the term magnetic flux ˆ linkinga surface, is often used, where

ˆ D

ZS

B � ds .Wb/ (5.1)

5-28

5.5. MAGNETIC PROPERTIES OF MATERIALS

� Via Stoke’s theorem with the vector magnetic potential, we canalso calculate ˆ using a contour integral

ˆ D

ZS

�r �A

�� ds D

IC

A � d l .Wb/;

where the contour C bounds the surface S

� In practice choose the integral form which is easiest to evaluate

5.5 Magnetic Properties of Materials

� Earlier we saw that a loop of areaA and current I has magneticmoment of magnitude m D IA

� A permanent magnet has similar properties

� In the permanent magnet it is due to current loops at the atomiclevel

1. Orbital motion of electrons and protons with respect tothe nucleus

2. Electron spin

� In both of the above it is electron motion that dominates

� Classifications: diamagnetic, paramagnetic, and ferromag-netic

5.5.1 Magnetic Permeability

� For the electric we saw that in a dielectric material D D �0EC

P , where P is the polarization vector

5-29


� In a magnetic material we similarly have

B D �0HC �0M D �0�HCM

�;

where M is the magnetization vector

� In linear materials (most, but not all materials of interest)

M D �mH;

where �m is the magnetic susceptability

� Since B D �H we have that

� D �0�1C �m

��r D

�

�rD 1C �m

Table 5.2: Summary of Magnetic Material Properties.Table 5-2 Properties of magnetic materials.

Diamagnetism Paramagnetism FerromagnetismPermanent magnetic No Yes, but weak Yes, and strongdipole momentPrimary magnetization Electron orbital Electron spin Magnetizedmechanism magnetic moment magnetic moment domainsDirection of induced Opposite Same Hysteresismagnetic field [see Fig. 5-22](relative to external field)Common substances Bismuth, copper, diamond, Aluminum, calcium, Iron,

gold, lead, mercury, silver, chromium, magnesium, nickel,silicon niobium, platinum, cobalt

tungstenTypical value of χm ≈ −10−5 ≈ 10−5 |χm| ≫ 1 and hystereticTypical value of µr ≈ 1 ≈ 1 |µr| ≫ 1 and hysteretic

Ferromagnetic Materials and Hysteresis

� Materials such as iron, nickel, and cobalt have magnetic mo-ments that easily align with an applied magnetic field

5-30

5.5. MAGNETIC PROPERTIES OF MATERIALS

� Permanent magnets are made of these materials and are key inthe many applications of magnetics in engineering

� Ferromagnetic materials contain magnetized domains whichcontain magnetic moments that remain permanantly alignedwith one another once magnetized

� At a high level the key result is the magnetization curve whichrelates B D jBj to H D jHj

A2

A1

A3HO

B

A4

Br

Figure 5-22 Typical hysteresis curve for a ferromag-netic material.

Figure 5.20: Ferromagnetic B �H hysteresis curve.

� An unmagnetized material starts at O and B increases as theapplied H increases (A1); the curve flattens as saturation isreached

� When H is decreased back to zero the point A2 is the resid-ual flux, or Br , that remains in the material and represents themagnetization

5-31


� Reversing the direction of the appliedH (e.g., chaning the cur-rent flow in the wire) hysteresis requires H to reach A3 beforeB again equals zero (demagnetized)

– The term hysteresis means lag behind

– The hysteresis loop shows that memory or material his-tory is involved

� Saturation in the opposite direction is reached at A4 and theprocess repeats as H is decreased and then increased in theopposite direction

� Hard materials have wide loops and soft materials have narrowloops

H

B

H

B

(a) Hard material (b) Soft material

Figure 5-23 Comparison of hysteresis curves for (a) ahard ferromagnetic material and (b) a soft ferromagneticmaterial.

Figure 5.21: Hard vs soft ferromagnetic material hysteresis curves.

� A wide loop has large Br useful for magnets in motors, etc.

� A narrow loop is easy to magnetize/demagnetize

5-32

5.6. MAGNETIC BOUNDARY CONDITIONS

5.6 Magnetic Boundary Conditions

� The boundary conditions of the B and H fields at a materialboundary �1=�2 parallels the electric field case of Chapter 4

∆h2∆h2

H1H1n

H1t

H2

H2n

H2t

}}

a

dc

b

∆l

μ1

Medium 1

μ2

Medium 2

n2

Js

ˆn̂

l̂ll1

l̂ll2

Figure 5-24 Boundary between medium 1 with µ1 and medium 2 with µ2.Figure 5.22: Establishing magnetic boundary conditions.

� Gauss’s law for magnetics says thatIC

B � ds D 0;

so from Figure 5.22 it must be that

B1n D B2n or �1H1n D �2H2n

� In words, B is continuous across a boundary

� Note: The corresponding D field is not continuous unless �s D0

� Using Ampère’s law a closed contour at the boundary (see Fig-ure 5.22)) establishes that

On2 ��H1 �H2

�D Js;

where On2 is the normal at the material interface

� In words, the tangential components of H are:

5-33


1. Continuous if they are parallel to Js

2. Discontinuous by Js if orthogonal to Js

� With that said, understand that surface current Js only existson perfect conductor, e.g., superconductors

� When finite conductivity is involved, Js D 0 then it is alwaystrue that

H1t D H2t

Example 5.7: Finding the Angle Between H1 and On2

� Suppose that H2 D�Ox3C Oz2

�(A/m), On2 D Oz, �r1 D 2, �r2 D

8, and Js D 0

� Find the angle �1 between H1 and On2

�n2

𝑧𝑧

𝑥𝑥

Medium 1 Above𝑥𝑥 − 𝑦𝑦 plane

Medium 2 Below𝑥𝑥 − 𝑦𝑦 plane

𝑦𝑦3

2

𝜇𝜇𝑟𝑟1

𝜇𝜇𝑟𝑟2

𝜃𝜃1

𝜃𝜃2H2

H1

Figure 5.23: Description of the interface and the arrival angle of H2

5-34

5.7. INDUCTANCE

� From the interface boundary conditions we know that

H1x D H1t D H2t D H2x

and

�r1H1z D �r1H1n D �r2H2n D �r2H2z

so

�2 D tan�1�H2x

H2z

�D tan�1

�3

2

�D 56:31ı

�1 D tan�1��r1�r2

H2x

H2z

�D tan�1

�2

8�3

2

�D 20:56ı

5.7 Inductance

� The inductor is to magnetic fields as the capacitor is to electricfields

� Recall that a capacitor can store the energy of and electric field,so too and inductor can store the energy of a magnetic field

� Inductors often take the form of a solenoid, which is multi-ples turns of wire wound around a cylindrical core as shown inFigure 5.24

5-35


(a) Loosely woundsolenoid

(b) Tightly woundsolenoid

N

S

B

NB

S

Figure 5-25 Magnetic field lines of (a) a loosely woundsolenoid and (b) a tightly wound solenoid.

Figure 5.24: Loosely and tightly wound solenoids.

� The tightly wound solenoid makes a nice case study, since ithas a near uniform B field at its interior (in the limit a cylindri-cal current sheet), which simplifies the inductance calculation

� The first step is to establish the field inside the solenoid startingfrom the axial field of a single circular loop

� Recall that

H D OzI 0a2

2�a2 C z2

�3=2 ;where here I 0 is the current in a single loop and a is the loopradius

5-36

5.7. INDUCTANCE

� If we stackN such loops a tightly wound solenoid of n D N=lturns per unit length, as shown in Figure 5.25, is obtained

θ

dθ

θ2

θ1

dz

z

l

z

xP

I (out) I (in)

a

B

Figure 5-26 Solenoid cross section showing geometryfor calculating H at a point P on the solenoid axis.

Figure 5.25: Cross section of a tightly wound solenoid of length l onthe z axis, straddling the x � y plane.

� A dz length of the solenoid contains n dz turns and we can letI 0 D In dz, so that

dB D �dH D Oz�nIa2

2�a2 C z2

�3=2 dz5-37


� Integrating over the length of the solenoid yields the total field

B D OzZ l=2

�l=2

�nIa2

2�a2 C z2

�3=2 dzD Oz

�nIa2

2

Z l=2

�l=2

dz

2�a2 C z2

�3=2D Oz

�nIa2

2�

2l

a2p4a2 C l2

D Oz�NI �1

p4a2 C l2

.Wb/m2/;

where the last line comes by replacing n with N=l

� As a special case of interest, consider l � a

B ' Oz�NI

l.Wb/m2/

� It is also true that under l � a the field inside is constantindependent of the radial distance r from the center, as long asz is not near the ends

5.7.1 Self Inductance

� The inductance in Henries (H) is defined as the ratio of totalmagnetic flux, ƒ linking the circuit or conducting structuredivided by the current I flowing through the structure

L Dƒ

I.H/;

where total flux linkage ƒ is related to flux linkage ˆ definedearlier in (5.1), the vector magnetic potential section

5-38

5.7. INDUCTANCE

– Note: For an N turn solenoid we simply have ƒ D Nˆ,where ˆ is the flux linkage in just one turn

� For a structure composed of two conductors, e.g., a coax ortwo-wire line, the flux linkage is computed over the area be-tween the conductors (see examples coming up shortly)

Example 5.8: N -Turn Tightly Wound Solenoid

� Using the fact that the interior field of the solenoid is approxi-mately uniform, we have for single loop

ˆ D

ZS

B � ds

D

ZS

Oz��N

lI

�� Oz ds D �

N

lI S .Wb/

� It is tempting to say S � 2�a2, but in reality the flux linkagefor a real solenoid is more involved

� The total flux isN times this value sinceN loops are involved,i.e.,

ƒ D Nˆ D �N 2

lI S .Wb/

� Finally the inductance is found to be

L D �N 2

lS � �

N 2

l� �a2 .H/

� In reality winding factors and pitch factors come into play infind the inductance

5-39


� Empirical formulas and inductance calculators can be found onthe Web2

Table 5.3: Two of many inductance formulas from Wikipedia.

� From Table 5.3 we see that the first entry is the same as the thetightly wound solenoid with the inclusion of the scale factorK

� Use Python to compare the tightly wound solenoid withK D 1to the short air-core formula

Figure 5.26: Python code for inductance calculation.2https://en.wikipedia.org/wiki/Inductor

5-40

https://en.wikipedia.org/wiki/Inductor

5.7. INDUCTANCE

0.5 1.0 1.5 2.0 2.5 3.0

Length (cm)

0

2

4

6

8

10

12

14

16L

(µH

)Cylindrical Coils: a = 1.0 (cm)

Tightly Wound Solenoid

Short Air Core

Figure 5.27: Comparison plot of the two inductance formulas.

� For longer lengths the two formulas are close, but at shortlengths, where the inductance is larger, the difference is overone order of magnitude

� The short air-core result is likely the more accurate expressionof the two

Example 5.9: Coaxial Transmission Line

� Consider the coax geometry in Figure 5.28, which also in-cludes the appropriate flux linking surface S

5-41

CHAPTER 5. MAGNETOSTATICS(a) Parallel-wire transmission line

(b) Coaxial transmission line

Radius a

z

x

y

IS

I

d

l

l ab

IS

I

c

Figure 5-27 To compute the inductance per unitlength of a two-conductor transmission line, we need todetermine the magnetic flux through the area S betweenthe conductors.

Figure 5.28: Coax transmission line including the linking surface S .

� From Ampère’s law the flux generated by the inner conductor(assumed infinite length) is

B D O��I

2�r

– No flux inside due to the outer conductor

� The flux per unit length linking the area S inside the coax is

ˆ D

Z l

0

Z b

a

�I

2�rdr dz

D�I l

2�ln.r/

ˇ̌̌̌ba

D�I l

2�ln�b

a

�.Wb/

so

L0 DL

lDˆ

lID

�

2�ln�b

a

�.H/m/

Example 5.10: Parallel Wire Line

� The geometry of the parallel wire line and the appropriate link-ing surface S are shown in Figure 5.29

5-42

5.7. INDUCTANCE

(a) Parallel-wire transmission line

(b) Coaxial transmission line

Radius a

z

x

y

IS

I

d

l

l ab

IS

I

c

Figure 5-27 To compute the inductance per unitlength of a two-conductor transmission line, we need todetermine the magnetic flux through the area S betweenthe conductors.

Figure 5.29: Parallel wire transmission line including the linkingsurface S .

� The B from each line (assumed infinite) is

B D O��0I

2�rrelative to the line axis

� Since the current is flowing in opposite directions the flux isdoubled passing through S and has direction Oy (check usingthe right hand rule)

� Solving for ˆ we have

ˆ D 2

Z l

0

Z d�a

a

�0I

2�xdx dz

D2�0I l

2�ln�d � a

a

�.Wb/

� Finally,

L0 DL

lD�0

�ln�d

a� 1

�.H/m/

5-43


5.7.2 Mutual Inductance

� When magnetic coupling between two structures exists, a mu-tual inductance exists between them

� This is the basis for a transformer as you have seen in circuittheory (and perhaps power systems)

– Inductive coupling shows up in many applications

� To study mutual inductance in more detail consider two loopswith axes aligned as shown in Figure 5.30

I1

B1

S1

S2

C1

C2

N1 turns

N2 turns

Figure 5-29 Magnetic field lines generated by current I1in loop 1 linking surface S2 of loop 2.

Figure 5.30: Two multi-turn loops configured to explore mutual in-ductance.

� The flux linking loop 2 due to the flux generated by loop 1 is

ˆ12 D

ZS2

B1 � ds .Wb/

5-44

5.8. MAGNETIC ENERGY

� Likewise the total magnetic flux linkage between the loops is

ƒ12 D N2ˆ12 D Ns

ZS2

B1 � ds

� Finally, the mutual inductance is defined as

L12 Dƒ12

I1DN2

I1

ZS2

B1 � ds .H/

5.8 Magnetic Energy

� When an inductor is connected to a current source, i increas-ing/charging from zero to I , the total work performed is

Wm D1

2LI 2 .J/

– The dual of the capacitor charging up with voltage V

� Wm represents the magnetic energy stored in the inductor

Example 5.11: Energy Stored in Solenoidal Inductor

� Here L D �N 2S=l and B D �NI=l , so

Wm D1

2LI 2 D

1

2

��N 2

lS

��Bl

�N

�2D1

2

B2

�.lS/ D

1

2�H 2V .J/;

where V D lS is the volume of the solenoid interior

5-45


� The above example motivates the definition of magnetic energydensity in general as

wm DWm

V D1

2�H 2 .J/m3/

� A general way of calculating Wm is

Wm D1

2

ZV�H 2 dV .J/

5.9 Appendix: Solenoid Exact MagneticField Off Axis

� In this appendix we consider the magnetic field intensity of theN turn tightly wound solenoid along a radial starting at the zaxis

� Assume that the solenoid has radius a and length l

� Setting up the Biot-Savart law in cartesian coordinates for asingle loop and the field point on the x axis we have

R D Ox.r � a cos.�//C Oy.�a sin.�//C Ozzd l D O�a d� D

�� Ox sin� C Oy cos�

�a d�

d l � OR DOxz cos� C Oyz sin� C Oz

�a � r cos�

�p.r2 C a2/ � 2ra cos� C z2

a d�

DOrz C Oz

�a � r cos�

�p.r2 C a2/ � 2ra cos� C z2

a d�;

� Due to symmtery, the field point initially on the x axis can bemoved to any point in the x � y plane at radial distance r

5-46

5.9. APPENDIX: SOLENOID EXACT MAGNETIC FIELD OFF AXIS

� H for the single loop is

H D OzI 0a2

4�

Z 2�

0

1 � r=a cos��.r2 C a2/ � 2ra cos� C z2

�3=2 d�C Or

I 0a2z

4�

Z 2�

0

1�.r2 C a2/ � 2ra cos� C z2

�3=2 d�� Assuming a continuous distribution of current loops running

from z D �l=2 to z D l=2, we find the corresponding field byintegrating over z and replacing I 0 with NI=l

H D OzNIa2

4�l

Z l=2

�l=2

Z 2�

0

1 � r=a cos��.r2 C a2/ � 2ra cos� C z2

�3=2 d� dzC Or

NIa2

4�l

Z l=2

�l=2

Z 2�

0

z.1 � r=a cos�/�.r2 C a2/ � 2ra cos� C z2

�3=2 d� dz� To evaluate a numerical integration over both � and z is re-

quired, but this is easily done in Python/Mathematica/Matlab

� Of primary interest here is the Oz term, as the flux normal to thesolenoid cross section contributes to flux linkages and is usedin the inductance calculation

� A comparison of the exact Hz field versus r is made with theexact field on the z axis and the l � a approximation

� For plotting purposes the approximate fields are held constanton the radial 0 � r � a and then set to zero

� Note here a D 1

5-47


Figure 5.31: Python code using scipy.integrate.dblquad() forthe numerical integration.

� Plots of the normalized Hz are now shown for l D 2, 5, and10 in the next three figures.

� It is observed that the exact field becomes more constant versusr and l increases; note the solenoid extends abouve and belowthe x � y plane by l=2

5-48

5.9. APPENDIX: SOLENOID EXACT MAGNETIC FIELD OFF AXIS

0 1 2 3 4 5r

0.1

0.0

0.1

0.2

0.3

0.4

0.5

Norm

aliz

ed M

agneti

c Fi

eld

Hz

Exact Field in x− y Plane

Exact Field on z Axis

Approx: lÀ a on z axis

Figure 5.32: Normalized Hz comparison plots for l D 2 and a D 1.

0 1 2 3 4 5r

0.05

0.00

0.05

0.10

0.15

0.20

Norm

aliz

ed M

agneti

c Fi

eld

Hz





5-49


0 1 2 3 4 5r

0.02

0.00

0.02

0.04

0.06

0.08

0.10

Norm

aliz

ed M

agneti

c Fi

eld

Hz





5-50

Magnetostatics - ece.uccs.edu

Documents

Transcript of Magnetostatics - ece.uccs.edu