Upload
others
View
0
Download
0
Embed Size (px)
Citation preview
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
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
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
Ex. 5.11: What’s the magnetic potential of a sphere with surface charge density constant s rotating at w.
𝑥
𝑦
𝑧
R Rdq’
Rsin(q’)d’ r q’
’
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
Ex. 5.11: What’s the magnetic potential of a sphere with surface charge density constant s rotating at w.
𝑥
𝑦
𝑧
R Rdq’
Rsin(q’)d’ r q’
’
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
Multi-pole Expansion of Vector Potential
Observation location
r
vdqrA o
4)(
Continuous current distribution
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
Multi-pole Expansion of Vector Potential
Observation location
r
vdqrA o
4)(
Continuous current distribution
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
Multi-pole Expansion of Vector Potential
Observation location
r
vdqrA o
4)(
Continuous current distribution
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)
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
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