Wed. 5.4.3 Multipole Expansion of the Vector Potential...

Wed. Thurs. Fri.

5.4.3 Multipole Expansion of the Vector Potential 6.1 Magnetization

HW8

Mon. Wed.

Review Exam 2 (Ch 3 & 5)

Relating Current, Potential, and Field

JA o

2

d

JA o

r

4

AB

24

1ˆ

r

r

dBA

24

1ˆ

r

r

dJB o

JB o

ldBadJIo

1

ldAadBB

J

B A

Finding A from J

Find the vector potential for a current I along the z axis from z1 to z2.

ldI

dJ

A oo

rr

44

22 sz r

I z

dzdl

z1 z2

r r

z’

s

zszzI

Az

zˆln

4

2

1

220

2

1

ˆ4 22

0

z

z

zsz

IdzA

.ˆln4 22

11

22

220 zszz

szzIA

Finding J from Vector Potential What current density would produce the vector potential

(where k is a constant) in cylindrical coordinates?

A k ˆ

JA o

2

So, convert to Cartesian 0,cos,sin kA

zAyAxAA zyxˆˆˆ 2222

where

2

2

22

2 sinsin1sin1

z

kk

ss

ks

ssAx

2

sin...

sk

0,cos

,sin

22

2

sk

skA

0,cos,sin 2

2 s

kA

2s

k Jo

ˆ

2

1

s

kJ

o

so

One component at a time

2

2

22

2 coscos1cos1

z

kk

ss

ks

ssAy

2

cos...

sk

and then



A k ˆ

JA o

2

AB

BJo

1

Alternatively,

zsAss

ˆ1

zks

ssˆ

1

ˆ1

0

s

Bz

ˆ1

0

s

k

s

zs

kˆ

ˆ2

0s

k

Ex. 5.11: What’s the magnetic potential of a sphere with surface charge density constant s rotating at w.

r

adKrA o

4

𝑥

𝑦

𝑧

R Rdq’

Rsin(q’)d’ r q’

’

r

vK

s

w qq ddRad sin2

What is ? v

If rotating about z, it would simply be . wq ˆsin R

zww

vRr

wqw ˆsinIf this would have been

Generally, vr

w

0,cos,0,sin ww

qqq cos,sinsin,cossinRr

qqqww cos,sinsin,cossin0,cos,0,sin Rrv

qqqqw sinsinsin,cossincossincos,coscossinRv

q cos222 RrrRr

For the four terms to v, there will be four integrals. All but one has a factor of

0cos

2

0

d or 0sin

2

0

d

yRrrR

ddRRrA o ˆ

cos2

sincossin2

0 022

2

4

q

q

qqqswleaving


𝑥

𝑦

𝑧

R Rdq’


’

r

w

y

RrrR

ddRRrA o ˆ

cos2

sincossin2

0 022

2

4

q

q

qqqsw

yRrrR

dRrA o ˆ

cos2

sincossin2

022

3

4

q

q

qqqsw

yRrrR

dRrA o ˆ

cos2

coscossin

0cos

122

3

2

q

q

qqsw

yRrrR

dRrA o ˆ

2sin

1

122

3

2

sw

yRrrRrR

RrrRRrA o ˆ2

3sin

1

1

22

22

223

2

sw

yRrrRRrrRRr

rA o ˆ2sin6

1

1

2222

2

ws

yrRRrrRrRRrrRRr

rA o ˆsin6

222222

2

w

s


𝑥

𝑦

𝑧

R Rdq’


’

r

w

If R>r, then rRrR

yrRRrrRrRRrrRRr

rA o ˆsin6

2222

2 w

s

32222 2rrRRrrRrRRrrR

yrRrA o ˆsin3

ws

If R<r, then RrrR

32222 2RRrRrrRrRRrrR

yRr

rA o ˆsin3

4

2w

s

Recognizing that these can be written generally yrr ˆsinww

RrrRr

RrrR

rAo

o

3

3

4

3

ws

ws

Motivating Electric Potential, Physically

b

adFW

2121

Object 2 is the “system”, 1 is “external.” Work done by

object 1 when exerting force on object 2 which moves

from a to b

Objects 1 and 2 are the “system”. Change in their

potential as they interact while separating from a to b

b

adFEP

2121,..

Generally

Electrically

21221 rEqF

b

a

b

adrEqdrEqEP

21221221,..

Combining:

b

adrE

q

EPV

21

2

21

1

,..

thus

Akin to Potential Energy

Physical Meaning of Vector Potential Akin to potential momentum

From __future__ import time-varying electric

AvVqAqpdt

d

Consider your “system” a particle interacting with electric and magnetic fields (really interacting with other charges via their electric and magnetic fields)

netFpdt

d EqBvq

t

AVE

t

AVqAvq

Avt

AA

dt

d

for any vector

By Product rule (4)

AvA

dt

dVqAvq

AvvAvAAvAv

Derivative with respect to potential not source velocity

VqAdt

dAvqp

dt

d

Consider your “system” a particle and the fields. The force is negative gradient the potential energy

""U"" p

if 0 AvVq

then constAqpAqp ffii

‘potential momentum’

Finding Vector Potential

Charged particle outside a disappearing solenoid

adBdA

.ˆ2

,ˆ22

0

0

RssnIR

RsnIsAinitially

q

0finallyA

AvVqAqvmdt

d

initially

0

ffii AqvmAqvm

fvmsnIRq

ˆ22

0

fvsnIRm

q ˆ22

0

x

Memory Lane: Multi-pole Expansion of Scalar Potential

1

r

1

r

r

r

n 0

n

Pn cos q

q cosnPObservation location

nth Legendre polynomial

P0 1

P1 u u

2

13 2

2

uuP

r

dqrV

o41)(

Continuous charge distribution

014

1 cos1

)(n

n

n

ndrPr

rrV

oq

q

drPr

r

rrV

n

n

n

o

0

41 cos

1)(

Re-ordering sums

𝑥

𝑦

𝑧

r

r

r

q

...2

1cos3cos

3

22

2 r

drr

r

drr

r

dr qq

...ˆ

)(

24

14

1

r

rp

r

QrV

oo

drrp

Electric Dipole Moment

Multi-pole Expansion of Vector Potential

1

r

1

r

r

r

n 0

n

Pn cos q

q cosnPObservation location

nth Legendre polynomial

P0 1

P1 u u

2

13 2

2

uuP

r

vdqrA o

4)(

Continuous current distribution

014

cos1

)(n

n

n

ndrJPr

rrA o q

q

drJP

r

r

rrA

n

n

n

o

0

4cos

1)(

Re-ordering sums

𝑥

𝑦

𝑧

r

r

r

q

...2

1cos3cos

3

22

2 r

drJr

r

drJr

r

drJ qq

Monopole Dipole term term


Observation location

r

vdqrA o

4)(


𝑥

𝑦

𝑧

r

r

r

q

...cos

24r

drJr

r

drJo

q

Monopole Dipole term term

Monopole’s integral

ldrI ld

dt

dq

dt

lddq

vdq

If we were to divide by Q, we’d have the charge-averaged velocity.

If the source is stationary, the current is steady, then the average velocity is just 0.

drJ

0

zzyyxxzz

yy

xx

zyx ˆˆˆˆˆˆˆˆˆ



r

vdqrA o

4)(


𝑥

𝑦

𝑧

r

r

r

q

...cos

24r

drJro

q

Dipole term Dipole’s integral

q drJr

cos drJrr

ˆ lIdrr

ˆ

Steady current

ldrrI

ˆ

mathland

adrrIldrrI r

ˆˆ

Stokes’ to Pr. 1.61e using Rule’s 7 and then 1:

rrrrrrrrrr rrrrrˆˆˆˆˆ

curlless

Derivative w/ respect to source not observation location

zzyyxxzyx

ˆˆˆ

zyx ˆˆˆ r

adrIldrrI

ˆˆ

adrIldrrI

ˆˆ arI ˆ

Product rule (4)

raI ˆ



r

vdqrA o

4)(


𝑥

𝑦

𝑧

r

m

...

ˆ24

r

raIo

Magnetic Dipole Moment

aIm

...

ˆ)(

24r

rmrA o

Dipole term for a loop Observation location

𝑥

𝑦

𝑧

r

q

Magnetic Dipole Moment

aIm

...

ˆ)(

24r

rmrA o

r

m

I

zRIm ˆ2

...

ˆˆ)(

2

2

4r

rzRIrA o

...ˆsin

)(2

2

4

q

r

RIrA o

Same direction as current

Dipole term for a cylindrical shell spinning at w with surface charge density sObservation

location

𝑥

𝑦

𝑧

r

...

ˆ)(

24r

rmrA o

zRdImd loopˆ2

w

L

R

dz

Imagine as a stack of differentially-thin current loops

loopshell rAdrA )()(

...

ˆ)(

24r

rmdrAd

loop

loopo

v=wR

KdzdI loop vdzs

dzRdI loop ws

zRdzRmd loopˆ2ws

...

ˆˆ)(

2

3

4r

rzdzRrAd o

loop

sw

...ˆsin

)(2

3

4

qsw

r

dzRrAd o

loop

qsw ˆsin

)(2

3

4r

LRrA o

shell

Dipole term for a solid cylinder spinning at w with volume charge density Observation

location

𝑥

𝑦

𝑧

r

...

ˆ)(

24r

rmrA o

w

L

R

dz

Imagine as a collection of concentric cylinders

qsw ˆsin

)(2

3

4r

LrrAd o

shell

shellcylinder rAdrA )()(

qw ˆsin

)(2

3

4r

LrrdrAd o

shell

qw ˆsin

)(0

2

3

4

R

cylinderr

LrrdrA o

qw ˆsin

)(0

3

24

R

cylinder rdrr

LrA o

qw ˆ4

sin2

4

4r

LRo

Dipole term for a spherical shell spinning at w with surface charge density sObservation

location

𝑦

𝑧

r

...

ˆ)(

24r

rmrA o

w

R

Imagine as a collection of coaxial loops

loopsphere rAdrA )()(

𝑥

...

ˆ)(

24r

rmdrAd

loop

loopo

q’

Rsinq’ q Rddl

KdldIloop

zadImd looplooploopˆ

q Rddl

2sinq Raloop

wqss sinRvK

...

ˆˆsinsin)(

2

2

4r

rzRRdRrAd o

loop

qqwqs

Dipole term for a spherical shell spinning at w with surface charge density sObservation

location

𝑦

𝑧

r

...

ˆ)(

24r

rmrA o

w

R

Imagine as a collection of coaxial loops

loopsphere rAdrA )()(

𝑥

q’

Rsinq’ q Rddl

...

ˆˆsin)(

2

34

4r

rzdRrAd o

loop

qqsw

qqqsw

sinˆsin)(0

3

2

4

4 dr

RrA o

sphere

qqqq0

2

0

3 cossinsin dd

1

1

2 coscos1 qq d

383

313

31 1111

qsw ˆsin

2)(

2

4

3r

RrA o

sphere

Which is actually the exact solution!

Wed. Thurs. Fri.

5.4.3 Multipole Expansion of the Vector Potential 6.1 Magnetization

HW8

Mon. Wed.

Review Exam 2 (Ch 3 & 5)



A k ˆ

JA o

2

So, convert to Cartesian 0,cos,sin kA

0,,

2/1222/122 yx

x

yx

yk

zAyAxAA zyxˆˆˆ 2222

where

222

2

222

22

yx

y

yyx

y

xkAx

2/322

2

2/1222/322

2 1

yx

y

yxyyx

xy

xkAx

2/322...

yx

yk

2/322

2

yx

xkAy

0,,

2/3222/322

2

yx

x

yx

ykA

0,,

2/1222/1222

2

yx

x

yx

y

s

kA

2s

k Jo

ˆ

2

1

s

kJ

o

similarly

so

One component at a time

Wed. 5.4.3 Multipole Expansion of the Vector Potential...

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