Classical Electrodynamics - National Chiao Tung Universityocw.nctu.edu.tw/course/classical_electrodynamics/electrodynamics-ch5.pdf · With the Gauss’s law: 0 ' 2 2 R l E r l In

Classical Electrodynamics

Chapter 5Magnetostatics, Faraday’s Law,

Quasi-Static Fields

A First Look at Quantum Physics

2011 Classical Electrodynamics Prof. Y. F. Chen

A First Look at Quantum Physics


Contents§5.1 The relationship between electric field and magnetic field

§5.2 Biot and Savart law and vetor potential

§5.3 Differential equations of magnetostatics and Ampere’s law

§5.4 Vector potential and magnetic induction for a circular current loop

§5.5 Analogy between electric dipole and magnetic dipole

§5.6 Magnetic scalar potential

§5.7 Magnetic moment

§5.8 Macroscopic equations, boundary conditions on and

§5.10 Uniformly magnetized sphere

§5.11 Final remark

§5.9 Methods of solving boundary-value problems in magnetostatics

B

H

(1) The force on a charge acted by a nearby conduction wire is:

In the inertial coordinate, the charge experiences a magnetic force: BvqF

In the relative coordinate, where the observation is performed in thecoordinate of charge, it experiences a electric force: EqF

0J R

B

+ + +

- - -0net

qv


§5. 1 The relationship between electric field and magnetic field

(2)

(x, y, z, t) (x’, y’, z’, t’)

With the transformation of coordinates in the special relativity:

222

22

2

22

1' ,

1' ,' ,' ,

1'

cvj

cv

cv

xcvt

tzzyy

cvvtxx x

With the Gauss’s law:0

2'2 lRlrE

In the relative coordinate:

22

0

22

0

2

1

121

12

cv

vrIq

cvc

vr

jRqqEF x



But when we consider the situation in the inertial coordinate:

BvqqvBvrIqF

2

0

This means that by proper transformation of the coordinate, we can justdeal with the electric force to solve the electromagnetic problems.Otherwise, the concept of the magnetic force must be introduced. Ingeneral, the force can be expressed as:

BvEqF



(1) The magnetic field generated by a short segment of wire carrying current isgiven by:

''

')'(4

)(

'''

4

33

0

30

xdxx

xxxJxB

xxxxlIdBd

(2)

AxdxJxx

xdxx

xJ

xdxx

xJB

AfAfAf

')'('

14

'')'(

4

''

1)'(4

3030

30

Where the vector potential is: '')'(

430 xd

xxxJA


§5.2 Biot and Savart law and vetor potential

§5. 3 Differential equations of magnetostatics and Ampere’s law

(1) 0 ABAB

(2)

''

1)'(4

')'('

1')'('

14

'')'('

')'(

4

'')'(

4

320

330

3230

2

30

xdxx

xJ

xdxJxx

xdxJxx

AfAfAf

xdxx

xJxdxx

xJ

ccc

xdxx

xJB


)('')'('

4

as ,0'')'('

')'('

)('')'('

4'

')'('

4

'''

)(')'('

1'4

'4'

1

030

3

03030

030

2

xJxdxxxJ

Sdaxx

xJxdxx

xJ

xJxdxxxJxd

xxxJ

AfAfAf

xJxdxJxx

xxxx



For static electromagnetics, we get the Ampere’s law:

JBJ

0 0

Generally, the Ampere’s law should be modified as the Maxwell-Ampere’s law:

tEJ

xdxx

xt

J

Jxdxxtx

B

tJ

000

3

0000

030

'')'(

41

''

)'(

4



(1)

''

4)( 0

xxldIxA

2

022

0

2

022

0

22

222

)'cos(sin2''cos

4)(

)'cos(sin2''sin

4)(

)'cos(sin2

0'sin'cos'

ˆ''cosˆ''sinˆ''

raardaIxA

raardaIxA

raar

zayaxxx

adaadaaadld

y

x

yx

aaaaaaaaaaa

rz

ry

rx

ˆsinˆcosˆ

ˆcosˆsincosˆsinsinˆ

ˆsinˆcoscosˆcossinˆ



0)'cos(sin2

')'cos(4

0)'cos(sin21sincos

4

)'cos(sin2')'sin(cos

4

0)'cos(sin214

)'cos(sin2')'sin(sin

4

2

022

0

2

0

220

2

022

0

2

0

220

2

022

0

raardIaA

raarra

Iaraar

dIaA

raarra

Iaraar

dIaAr



Expand the denominator of A with binomial expansion:

......sin

815010

4

sin

42cos2cos21

22cos1cos Note

......')'(cossin2!3

125

23

21

')'(cossin2!2

123

21

')'(cossin2!1

121')'cos(

4

')'cos(sin21)'cos(4

2

222

322

20

2

0

22

0

22

0

4

2

0

43

22

2

0

32

22

2

0

222

2

022

0

21

22

2

022

0

arra

ar

rIa

ddd

dar

ra

dar

ra

dar

radar

Ia

dar

raar

IaA



30

302

44sin assume & let

rrm

rmrAaraIm

(2)

aarmB

rm

rmr

rB

rm

rmrr

rrB

Arr

arara

rAhAhAhqqq

ahahah

hhhAB

r

r

r

ˆsinˆcos24

4cos2

4sinsin

sin1

4sin

4sin1

sin00

ˆsinˆˆ

sin1

ˆˆˆ1

30

30

30

30

30

2

332211

321

332211

321



(1) Electric dipole:

30

30

')ˆ(ˆ3

41

')'(

41

xxpnpnE

xxxxp

(2) magnetic dipole:

30

30

')ˆ(ˆ3

4

')'(

4

xxmnmnAB

xxxxmA


§5.5 Analogy between electric dipole and magnetic dipole

§5. 6 Magnetic scalar potential

(1)MBHJ

0 :space freein ,0 0 If

To illustrate this concept, we consider the problem as follows. For themagnetic induction at the point P with coordinate produced by an incrementof current at , the magnetic induction can be explicitly expressed as:

x

'lId

'x

dIxdR

RldI

xdxx

xxldIxdBxdd MM

44

')'(

4

3

30

where the solid angle is:

3

)(R

RAd

4

40

0IBI

MM


(2) As an example, find the magnetic induction at a point on the z-axis:

2322

0

2

0 2322

03

0

22

ˆ2

'ˆˆ0

4''''

'')'(

4

ˆˆ)'()'(')'(

' ,ˆˆ'

ˆ)'()'()'(

az

aaI

adaz

aaaIdzddxx

xxxJB

aaazzaIxxxJ

azxxaaazxx

azaIxJ

z

z

z

z

(i) Directly find magnetic induction:

'xx



(ii) With the concept of the magnetic scalar potential:

2322

20

0z

22

220

2

0 2322

3

2

12

4

12'

'''

'''

ˆ'ˆ ,)(

az

aIz

B

azzII

azz

z

ddz

ddzAdR

aazRR

RAd

M

M

a

z



(iii) Since this problem has the property of symmetry, we can expand themagnetic scalar potential with the help of the Legendre polynomial. Besides,the magnetic scalar potential at any point can be obtained with the knowingthe magnetic scalar potential on the z-axis:

(a) For r < a: l

ll

lM PrA )(cos

l

llMl zAzPzr )( 1)1( , 0

Expand M(z) with the binomial expansion, and note that the constant termof the M(z) can be dropped without loss:

......!3

125

23

21

!21

23

21

!11

211

2

12

)(

642

21

2

az

az

az

aIz

az

aIzzM



......)(cos165

)(cos83)(cos

2)(cos

2) ,(

...... ,325 ,

163 ,

4 ,

2

......325

163

42

77

7

55

5

33

3

1

7151321

7

7

5

5

3

3

Par

ParP

arP

arIr

aIA

aIA

aIA

aIA

aIz

aIz

aIz

aIz

M



(b) For r > a: l

llM zBAz 10

1)(

l

llMl zAzPzr )( 1)1( , 0

Expand M(z) with the binomial expansion, and note that the constant termof the M(z) can be dropped without loss:

...... ,325 ,

163 ,

4 ,

2

......325

163

42

......!3

125

23

21

!21

23

21

!11

211

2

12

)(

65

43

210

6

6

4

4

2

2

642

21

2

aIBaIBaIBIA

zIa

zIa

zIaI

za

za

zaI

zaIzM



The figure below shows the simulation of the magnetic scalar potentialviewed from radial axis with the parameters of I = 0.1 and a = 1:

......)(cos165

)(cos83)(cos

21

2) ,(

56

6

34

4

12

2

Pr

a

PraP

raIrM




For a localized current density, we can use Taylor expansion:

......'1'

13

xxx

xxx

0'')'( 3 ldIxdxJ no magnetic monopole

......'')'(4

)(

......'')'(4

'')'(

4)(

33

0

33

030

xdxxJxxxA

xdxxxJx

xdxx

xJxA

ii


O( )J x

xx

P

In the text book, it is pointed out that:

'''')'( 33 xdxJxxdxxJxj

jiji

0''' 3 xdJxJx ijji

')'('21

'''21'')'(

3

33

xdxJxx

xdJxJxxxdxxJxj

ijjiji

It is customary to define the magnetic moment: ')'('21 3xdxJxm

Consequently, the magnetic dipole vector potential is: 30

4)(

xxmxA

And the magnetic induction outside the localized source is: 30 )ˆ(ˆ3

4)(

xmnmnxB



B

H

(1) ABB

0

(2) Magnetization: i

ii mNM

(3) With the bulk magnetization and a macrosopic current density:

SxdxxxM

fAAfAf

xdxx

xMxx

xJ

xdxx

xxxMxx

xJxA

xxVxxxM

xxVxJxA

as ,0'')'('

'')('

''

1')'(')'(

4

''

')'(')'(

4)(

'')'(

4')'(

4)(

3

30

33

0

300


§ 5.8 Macroscopic equations, boundary conditions on and

interpret:

''

)'(')'(4

30 xdxx

xMxJ

MJxM

)(

JMBH

JJB M

0

0

If the material is linear:

cdiamagneti :

tic paramagne: :

0

0

HB

(4) Boundary conditions: (the same discussion as for the electrostatics)

densitycurrent surface a is where, :

: 0

12

21

KKHHJH

BBB

tt

nn


B

H

§ 5.8 Macroscopic equations, boundary conditions on and

(1) Generally applicable method of the vector potential:

equation Poisson :

0 :gauge Coulomb choose

11

if

0

2

2

JA

A

AAA

JABH

HB

ABB

(2)

equation Laplace: 0

0 if

0 entialscalar pot magnetic 0

2

2

M

M

M

BHB

HHJ


§ 5.9 Methods of solving boundary-value problems in magnetostatics

(3) Hard ferromagnetic 0 given, JM

equation Poisson :

defineand ,0

0

20

MM

M

M

MMHB

HJ

(i) In free space and no surface contribution:

''

)'('41'

')'(

41)( 33 xd

xxxMxd

xxxx M

M

(ii) With surface contribution:

'

')'(ˆ

41'

')'('

41)( 3 da

xxxMnxd

xxxMxM



Moreover:

......(*)'

')'(

41

)(

')'('

141

0 ,0'ˆ')'('

')'('

')'('

1'41'

')'('

41

'')('

''

)'('41)(

3

3

3

33

3

xdxxxM

MfMffM

xdxMxx

SdanxxxMxd

xxxM

xdxMxx

xdxxxM

MfMffM

xdxxxMxM



Note that the expression (*) is generally applicable even for the limit ofdiscontinuous contributions magnetic surface charge density:

'')'('

41'

')'('

41

')'('

1'41

')'('

141

'')'(

41)(

33

3

3

3

xdxxxMxd

xxxM

xdxMxx

xdxMxx

xdxxxMxM

Surface charge contribution



Consider a sphere of radius a, with a uniform permanent magnetization ofmagnitude M0 and parallel to the z-axis:

M

0 2and symmetry,-

) ,()' ,'(12

4)'(cos

)'(cos'

1 :remember

'''sin''cos

4'

''cos

41)(

'cosˆand ,0

ˆ

2

00,

*

01

2020

0

0

mde

YYl

P

Prr

xx

ddxx

aMdaxx

Mx

MMnM

zMM

mim

lm

l

lmlml

lll

l

M

M



cos3

)(

1 12

2)()( :ityorthogonalBy

'''sin'cos)'(cos)(cos4

)(

)(cos4

12) ,(

1

20

'

1

1-'

2

0 001

20

0

l

l

M

llll

ll

ll

l

M

ll

rraMx

ll

dxxPxP

ddPPrraMx

PlY

(1) Inside the sphere: cos3

) ,( , 0 rMrrrar M

(2) Outside the sphere: cos3

) ,( , 2

30

raMrarrr M




The magnetic scalar potential and the lines of and are shown below. The linesof are continuously closed paths, but those of terminate on the surfacebecause there is an effective surface charge density.

B

H

B

H


Find the magnetic dipole moment of a uniformly charged spherical shell of radiusa rotating with angular frequency about the z-axis:

zaaM

aVmM

adam

adaadm

adaqTT

qtqdI

aA

ˆ

34sin

sin22

sin

sin222

2sin

4

0

34

2

2

We can use this model to explain some magnetized behavior of the atomic system.

z

e


§5.11 Final remark

Documents

Classical Electrodynamics - National Chiao Tung Universityocw.nctu.edu.tw/course/classical_electrodynamics/electrodynamics-ch5.pdf · With the Gauss’s law: 0 ' 2 2 R l E r l In