Chapter 5Magnetostatics

Lecturer: Lee, Sang Young (Professor at Konkuk Univ., Seoul, Korea)

Download site for lecture notes: http://konkuk.ac.kr/~sylee

Textbook: Introduction to Electrodynamics(3rd edition)by D.J. Griffiths

1. The Lorentz Force Law

5.1(1) Magnetic Fields

Inside materials Spin (S) & orbital angular momentum (L)

B

of electrons

-g J ; J L S

e

2m

J

B

; spin magnetic moment of free electron

g ; g- factor

B

r

IB

2

5.1.2 Magnetic Forces

,

,

[ ( )]

B

E

The forceonachargeQ moving withvelocity v ina magnetic field B;

F Q v B Lorentz forcelaw

In presenceof both Eand B thenet forceon Qis expressed by

F Q E v B

F

;

( ) ( ) 0 ( . 5.3)

B

B

B B

F

Note F does not do work

F dl F vdt Q v B vdt Ex

1. The Lorentz Force Law

5.1.3 Currents and the magnetic force on a segment of current-carrying wire

Unit of currents; 1A 1

1)

~

( )

( )

B

CS

Caseof alinechargetravelling downa wire

linechargedensity

dq dI vdt v

dt dt

dF dl v B

I

(Note. / / )

( )

( )B

dl B v dl

Forceontheinfinitesimal chargedq dl indl

F I dl B I dl B

Therefore,

1. The Lorentz Force Law

2)

:

:

Case for charge flowing over a surface

surfacechargedensity

v velocity of c

( )

~

;

B

harge

Q wv t a

dQI vw Kw thecurrent flowing throughthe width w

dt

K v Surface current density thecurrent per unit width

dF dQv B

( )

( ) ( )B

dav B K B da

F v B da K B da

w

vΔt

1. The Lorentz Force Law

3)

:

Case for charge flowing throughathree diimensional region

volumechargedensity

:

/

;

v velocity of charrge

Q vA t

dQ dt vA JA

Charge per unit time flowing thro

B

S

B

ugh A

Current definition of J ; J ρv

I J da

dF dQv B d v B J Bd F J Bd

A

vΔt

S

J

1. The Lorentz Force Law

4 *

Ω

)Continuity equation

Charge flowing outof per unit time

J da

0

s

s v

d J da ρdτ

dt

dρ dρJ da Jdτ ( )dτ ( case for steady currents)

dt dt

dρJ Continuity equation ~ "Contribution of Maxwell"

dt

1 2Volume

1. The Lorentz Force Law

2. Biot – Savart Law ~ Applicable for magnetostatics

Themagneticfieldof a steadycurrent can beobtained byusing theBiot-Savart law :

Condition for a steadycurrent : 0 0

(cf. A moving point chargedoesnot producea staticfield.)

(1) Biot - Sabar

ρ J (or )

t

t law

(

B'

0 0

2 2

'

'

'

)4 4

: ~ An element of length along the wire

~ the vector from the source of current (r ) to the point r,

. ., r r

μ μId Id rr )

π π

d

i e

r rr r

r

r7

20N : permaeability of vacuum ( 4 10 )

A

4

[ ] (A m) ; 1 Tesla 1N (A m) (SI units)

10 gauss (cgs units)

(Note, [ ] [ ] [ ][ ] [ ] [ ][ ])

B

B N

F Qv B B F Q V F I L

Themagnitudeof themagneticfield

that exertsa forceof 1Non a1m-long wire

1Tesla

(Ex. 5.5)

sin ( , ) sin sin sin(d d d d d x d

r r )

2

2

2

cos

tan seccos

d

d zz z

d

z

y

xI

z

P

r

ld

2. Biot – Savart Law ~ Applicable for magnetostatics

2

2

2 2

0

2

0 0

2 2 2

2

0 0 02

2

2

From cos ,

1 cos

Therefore, from ,4

cos coscos

4 4 cos

cossin

4 4 2

Direction : along the x a

π

π

π

π

θ z

θ

z

μ IdlB

π

μ μ IIdl θ z θB dθ θ

π π θ z

μ I μ I μ Iθdθ θ

π z πz πz

r

r

r

r

r

2

2

0

0

xis (Right hand rule!)

Note: Expression of Bdue to thesurfacecurrent ( )or the volume current clensity (J)

4

4

This willbeusedin deriving theAmpere la

K

μ KB da , K K(r )

π

μ Jor B dτ , J J(r )

π

r

rr

rw from the Biot-Savart law.

2. Biot – Savart Law ~ Applicable for magnetostatics

Homework No.1:

Problems 4, 5, 6, 11, 12.(Due date: Sep. 15, 2009)

3. The divergence and Curl of B

5.3.1 Straight-LineCurrents 0a

00 0

0b

0

at r =2

2 2

at r =2

a

aa

b

b

IB R

R

IB d l B dl R I

R

IB R

R

B d l I

insideof r=Ra

binsideofr=R

dl

aR

a

b

IbR

0Generally, : ~ the totalcurrent enclosed by theintegration path B dl I I dl

0

and

( ) , we get

~ Ampere law (for thecaseof straight-linecurrents)

Note: Ampere law is applicable tosteadycurrents.

From I J da

B d l B da

B J

I

1I

nI

n

iiIldB

10

B

0

:

Does the relation ( ) 0 always hold?

Question

B J

The divergence and Curl of B5.3.2

According to theBiot-Savart law, due to the volume current density isexpressed by,B J

'0 0

2 2

( ')( ) '

4 4

J J rB r d d

r rr r

'r

'd

r

( ')r r r ( ') ( ') ( ')yx x y y z zx z r

'

' ' ', cf .y yx z x z

x y z x y z

BofdivergencetheTakei )0

2

0

2 2

( ( ') ) '4

{ ( ) ( )} '4

B J r d

J J d

r

r r

r

r r

2

'( '), ( ') 0. : ( ') 0.

( ) 0 ( . ( ) )

Since J J r J r Note J r

cf fA f A A f

rr (Problem 1.62)

0 B

)()()( BAABBA

BofcurltheTakeii)

0

2

0

2 2 2 2

0

2 2

( ) '4

{( ) ( ) ( ) ( )} '4

{ ( ) ( ) } '4

B J d

J J J J d

J J d

r

r r r r

r r

r

r r r r

r r

( )

( ) ( )

( ) ( )

C D

D C C D

C D D C

2

3

2

' 3 '00

. ( ) ' 0 1

4 ( ) 2 ( , .1.100)

( )4 ( ) ' ( )4

Note J d

See Eq

B J r r r d J r

r

r

r

r

r

2(1), i.e., ( ) ' 0 Verification of J d

rr

2 2

3

( ) ' ( ') '..................(3)

( ')

J d J d

J d

r r

rr

r r

3 3 3 3

3 3 3

' ( ') '( ') ( ') ( ') ( ')

' ( ') ( ')( ') ' [ ] ( ' )

x x y y z zJ J y J J

x x x x x xJ J J

x z

r r r r

r r r

r

3 3

3 3

( ') 0. Therefore, ' ( ') 0

' ( ') ( ) ' [ ]

( ') ( '), ' [ ] ' '

rJ r

t

x x x xJ J

x x x xJ d J da

For steady currents,

Also

r r

r r

Forsufficientlylarge τ', J can be0at thesurface (If J 0, takelarger τ'untilJ= 0.)

J

5.3.3 Applications of Ampere's Law

0

0

0.

( )

: The total current passing through the s

t

B J

B da B d l J da

J da

Forsteady currents,

urface enclosed by d l

(Ex. 5.7)

r

B

I0

0 0

, 0 from the symmetry

and 0 from the Biot-Savart Law.

, 2 2 2

ˆ ˆ with r z r

z

ITherefore B d l dl rB I rB I B

r

B rB B zB B

B

B

(Ex.5.9)

o

' ' '

ˆ ˆ 1

Rotate the coil by 180 : ( ' . '

ˆ ˆ ' 2

ˆ

r z

r z

r

B rB B zB

J J J B B B

B rB B zB

B rB B

' '

Outside of the coil, 0 0 (

ˆ

Meanwhile, since , ( ) . Therefore, 0

0 Inside the coil, 0 0 ( 0)

in

z

r r r r r r

in

B B bd I

zB

B B B B B B

B B B ad I

0)

( ) ( ) 0

( ) ( )

This is also true for going to infinity(i.e., ) ( ) ( ) 0

( ( ) 0)

B d l B a L B b L

B a B b

b b B a B b

B

0

0 0 0

0 0

0

ˆ Also, inside the coil,

~ number of turns/length

ˆ ; inside

0 ; outside

tot

B zB

B dl B L μ I μ nLI ; n

B μ nIz

B

Homework No.2:Ex. 5.8, Problems 13(a), 13(b), 14, 15, 16(a)-(b), 16(c)(Due date: Sep. 29, 2009)

(1) Vector potential

In Electrostatics, ~Scalar potential

( )

5.4. Magnetic Vector Potential

E V ; V

electric Potential

0

(Or, 0 )

E

E E

0

For the magnetic field , 0 always holds true.

From Ampere law, B μ ,

2 B ( ) ( ) - ~ vector

B B

B A

J

A A A

0

identity

. A cannot be uniquely defined from B, because

there can be many different 's that satisfy

a

μ J

Note

A B A

A A - λ

llows the same in that

.

B

A A - λ A

So, we can choose the kind of 's that satisfy 0.

What if ' does not satisfy ' 0,while 'satisfies '.

Then, use the following procedure

This is called the Coulomb gauge.

A A

A A A B A

2

2

for finding

that satisfies the Coulomb gauge 0.

i) Solve for .

ii) Then, .

0.

A A

A

A A

A A

2

0

0

2

0

Note. If 0, then .

B μ B ( )

( ) - = .

A A J

J A

A A μ J

2

0

0i

Therefore, from [ ] [ ],

( ) 0( )

4 .

x y z x y z

ii

A yA A J yJ J

If J rJ rA d

at rr r

x z x z

0

'0

0

( )

4

( ),

4

' , : ' ~ direction of the current I

4

(

J rA d

r r

K ror da

r r

Idlor dl

r r

2

0 0

0

if 01cf. . )

at4

1 ( )

4

(Here, the condition of ( ) = 0 sh

Poisson s eqn V V dr

rd

r r

V

r

0 0

ould be satisfied. See p.87, 88)

. * Lorenz gauge : V

cf At

'r

'd

r

( ')r r r

0

(Ex. 5.12)

Find the vector potential of an infinite solenoid

with n turns per unit lenght, radius R, and current I.

i)

A

B d a Φ ( A ) d a A d

cf B d ( B ) d a μ J

0

0 0

ii)

(ii) (i) (ii) (i)

The direction of is the same as the current direction.

d a μ I

B A , μ I Φ (or μ J B )

* A

0

2

0

0

2

0

2

0

i) For ,

Therefore, 2

ˆ 2

(Note: The direction of

ii) For ,

2

ˆ

2

(Note: T

is the same as that of the current .)

r R B μ nI

A d B d a μ nI (πr ) πr A

μ nIA r

A

r R

A d μ nI (πR ) πr A

μ nI RA

r

he direction of is the same as that of the current.)A

I

R

I

2 12 1

2 1

- Boundary condition for the normal components

(i) 0

( )

( )

(h 0 is usen n

B

B d

B d s B S B S

contributions from the side wall

SB SB

2 1

d.)

0

(This condition always holds true.)n nB B

(2)Magnetostatic Boundary Conditions

.

0

0

1 2

Boundary condition for the transverse components

ii)

( )

( )

ˆ ˆ ( ) ( )

B J

B da J da

B da B d

B Rx B Rx

1 2 0

1 21 2

0( )

0 ; ( )

Here, as h 0, 0 and ( ) 0

if there is no surface current density .

x x

t tx x

K

B h

h B da B R B R J da

J da B da

B BB B

I

1 2 0

20 0 0

2

01 2

(Correct Eq. 5.73 in the t

ˆ If surface current density 0 ( . .,if ( ) ( ) ),

( ) ( )

ˆ ˆ ( ) ( )

ˆ ˆ ( )

x x

R

R

x x

K i e J K y K y z

B da B B R J da

K y z z dxdy KR

z KzB B

2 2 01 1

2 2 21 1 1

21 2 1 2 1

2 1 0 2 1

ext !!!)

ˆ ˆ ( ) ( )

ˆ ˆ ˆ ˆ ˆ ˆ ( ( )) ( ( )) ( )( )

ˆ ˆ ˆ ˆ ˆ( ) ( ) ( )( )

ˆ( )

n n t t

y n KB B B B

n n n n n nB B B B B B

n n n t n nB B B B B B

B B n K B B

0

1 2

ˆ

ˆ ˆ Here, and 0 are used along with .n n n t

K n

n t n tB B B B B

I

1 2 1 2

1 2

iii) 0 ( ) -(a)

iv) From ,

( )

A

Here, 0 as 0.

Therefor

Boundary conditions (continued)

n n

B

B

x x

A or

A B

A da B da

d

h

R RA A

A A A A

1 2 21

1 2

, )1 2

e, which means that ---(b)

From (a) and (b), we get the condition of

at the boundany.

A( Ax x n nR R orA A A A

A A

Homework No.3:Problems 20, 22, 25(a), 25(b), 26, 31(a)(Due date: Oct. 6, 2009)

I

2 1

2 2 22 2

1 1 11 1

2 1

0 0 0

0

Boundary conditions (continued)

v) n y x

( )

( )

x -

No

z y z

z y z

K K z K

x xA A AB Ay y y

x xA A AB Ay y y

A A K Kn n

B B

2 1

1t 2t

te. Boundary conditions in electrostatics.

; ~surface free charge densityD D

0 E E

n n f fPD

E

I

00 3

* A useful tip related with the Coulomb gauge

From the Bio-savart law,

( ) ( )Jwe get 0 and .4

If the Coulomb gauge, 0, is used, 0 and

r r rB B J dB

A A A

r r

3

.

( ) ( )1 BTherefore, can be written as . 4

B

r r rA A d

r r

(3) Multipole expansion of the vector potential

3 0 00

0

0

01

( ) ' '' . (Similar to Eq. 3.94 on p. 154)d

4 4 4

Therefore, assuming that ', due to the current loop can be written

'

4

4

'{n

n

n

J J S d r d rA r I

r r A

d rI

A

I r

r

r rr

r

n(cos )

n

'} (1)

nP dr A

c

r

r r

I

0

r r r

dr d

0

1

0 n

0 1 2

0 0

0

0

0

2

'

4

'{ (cos )} (1)

4

( ) ~ Legendre polynomial (See e.g., Table 3.1)

1( ) 1, ( ) , ( ) (3 1) etc.

2

In Eq. (1), = 04 4

'

''

n

n n

n

n

n

d rA

r

P x

x x x x xP P P

I IddA

r r

I

Idr A

rr

Pr

r

0 0

1 11 2 2

0

1

( 0), and

' 1 = (cos ) ' (cos ) .

4 4

Note. Monopole term (~ 1/r) does not exist, and the first non-trivial term is the dipole term.

: Dipole term ~ 1/

'

' '

:

d

I Ird r dA P P

r r

A

A

r

r r

2 3

2 r Quadrupole term ~ 1/r , : , etc.A

Multipole expansion of the vector potential (continued)

1

0 012 21

0 02 2

Re-expression of the dipole term :

Since cos = ',

' (cos ) ' 'cos '4 4

'( ' ) ' ( ' ) '. (2)4 4

A

r r

I Ir dr r drA

r rI I

r r r dr r r drr r

P

Calculation of :

Since a (b c) b(a c) - c(a b),

( ' ') '( ') '( ')

( ' ) ' { '( ')} { ( ' ')}

r r dr r r dr dr r r

r r dr r r dr r r dr

(3)

Multipole expansion of the vector potential (continued)

1Re-expression of the dipole term (continued)

( ' ) ' { '( ')} { ( ' ')} (3)

Calculation of :

From [( ' ) '] ( ' )

A

r r dr r r dr r r dr

d r r r d r r

' ( ' ) '

( ' ) ' d[( ' ) '] - ' ( ' )

( ' ) ' d[( ' ) '] - ' ( ' )

( 0)

- { '

r r r dr

r r dr r r r r d r r

r r dr r r r r d r r

r

,

( ' ' ') - '( ')

Therefore, '( ') ( ' ) ' ---(4)

dr r r dr r r dr

r r dr r r dr

0

1 2( ' ) '. (2)

( ' ) ' { '( ')} { ( ' ')} (3)

'( ') ( ' ) ' =-

4 r r d r

r r d r r r d r r r d r

r r d r r r d r

IA

r

0

1 2

---(4)

From (3) and (4), ( ' ) ' { ( ' ')

1( ' ) ' { ( ' ')} (5)

2

1Therefore,

2

( ' ')4

r r d r r r d r

r r d r r r d r

r r d rI

Ar

0 01 2 2

( ' ')1

If we use an expression of ,2

( ) ( )(6) due to the magnetic dipole moment

4 4

Note. i) is called the magnetic dipole moment.

r d r dI I a m

r m m rmA A

r r

m

' '

' '

1 Significance of ; ( ) , and

2

1 ( ) ( ~ vector area of the current loop).

2

ii) is independent of the choice of origin.

Proof

r d r

r d r

m d a

a a

m I a

m

'

' '

'

: If we take O' as the origin, then

" , and " '

" " ( ) ' ' '

'+0

r

r r

r

r c dr dr

r dr c dr dr c dr

dr

' = ' ; Q.E.Dr dr

c

r

r r

I

0

r r r

dr d

0

20

2 23

1 2 2 1 1

iii) Since the quadrupole term is expressed by

' (cos ) ',4

' for ' due to ~ . ~

The magnetic field of a magnetic dipole :

If we

Ir d rpA

r

rr rA A A A

r

B m

A A

0 0

2

0

2

3

place at the origin (note that is independent of the origin),

( )

4

sin

4

(2cos sin )

4

m m

m r

r

mA

r

mB A r

r

z

x

y

rm

y

z z

y

(a) Field of a pure dipole (b)Field of a physical dipole

Field of a dipole