Jackson_5_13_Homework_Solution (1)

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Jackson 5.13 Homework Problem Solution
Dr. Christopher S. Baird
University of Massachusetts Lowell
PROBLEM:
A sphere of radius a carries a uniform surface-charge distribution σ. The sphere is rotated about a 
diameter with constant angular velocity ω. Find the vector potential and magnetic-flux density both 
inside and outside the sphere.
SOLUTION:
The current density in spherical coordinates is:
J= a sin r−a  
A=
0
4∫
J x '
∣x−x '∣
d x '
A=
 0
4∫
 a sin  ' r '−a   '
∣x−x '∣
d x '
Expand the denominator in spherical harmonics.
A=
0
4 ∫0
2
∫
0

∫
0
∞
 r '−a  ' 4∑
l=0
∞
∑
m=−l
l 1
2 l1
r
l
r
l1 Y lm
*  ' , 'Y lm , r '
2sin2  'dr 'd  'd  '
We must be careful and realize that  '≠  because the primed variables are being integrated over. The 
best way to handle this is to use the expansion:  '=−sin  ' i cos ' j
A=0 a∫
0
2
∫
0

∫
0
∞
 r '−a −sin ' icos  ' j ∑
l=0
∞
∑
m=−l
l 1
2l1
r l
r 'l−1
Y l m
*  ' , 'Y lm , sin
2  'dr 'd  ' d '
for r < r' (inside the sphere).
A=0 a∫
0
2
∫
0

∫
0
∞
 r '−a −sin ' icos  ' j ∑
l=0
∞
∑
m=−l
l 1
2l1
r 'l2
r l1
Y l m
*  ' , 'Y lm ,sin
2 ' dr 'd  'd  '
for r > r' (outside the sphere).
Now evaluate the delta functions and rearrange:
A=0 ∑
l=0
∞
∑
m=−l
l 1
2 l1
r l
a l−2
Y lm , I l , m (inside) 
A= 0 ∑
l=0
∞
∑
m=−l
l 1
2 l1
a l3
r l1
Y lm , I l ,m (outside)
where I l ,m=∫
0
2
∫
0

−sin ' i cos ' j Y lm
*  ' , ' sin2  'd  'd  '
Expand the definition of the spherical harmonics to solve the integrals:
I l ,m=2l14 l−m!lm!∫0
2
−sin ' i cos ' j e−im 'd  '∫
0

P l
mcos  'sin2 ' d  '
Due to the orthogonality in the first integral, all terms vanish except m = 1 and m = -1. The first integral 
can then be easily calculated to yield:
I l ,±1=2 l14  l∓1 ! l±1 ! ±i ij ∫0

P l
±1cos ' sin2  'd  '
Make the substitution x=cos , dx=−sin d  and recognize
 sin=1− x2=P11x =−2P1−1 x
I l ,1=2 l14 l−1 !l1 ! i ij ∫−1
1
P l
1x P1
1x dx
I l ,−1=2 l14  l1 ! l−1 ! −i i j −2∫−1
1
P l
−1x P1
−1x dx
Due to orthogonality, only the l = 1 term is nonzero, leading to:
I 1,±1=23 i i ±j 
The solution is now just the sum of the l = 1, m = -1 term and the l = 1, m = 1 term.
A=
0
3
a r [Y 1,1 , I 1,1Y 1,−1 , I 1,−1] (inside) 
A=
0
3
a4
r2
[Y 1,1 , I 1,1Y 1,−1 , I 1,−1] (outside)
We note that Y 1,−1=−Y 1,1
* and I 1,−1=−I 1,1
* so that:
A=
0
3
a r [Y 1,1 , I 1,1Y 1,1
*  , I 1,1
* ] (inside) 
A=
0
3
a4
r2
[Y 1,1 , I 1,1Y 1,1
*  , I 1,1
* ] (outside)
For complex numbers in general zz*=2ℜ z  so that our final solutions become
A=
0 
3
a r sin  (inside) 
A=
0 
3
a4
r2
sin  (outside)
The magnetic flux density then becomes:
B=∇×A
Expand in spherical coordinates, keeping only terms that are non-zero for this case:
B=r 1
r sin
∂
∂
sin A− 
1
r
∂
∂ r
r A
Inside:
B=2
3
a 0 [cos r−sin  ]
B=2
3
a  0 z
Outside:
B=
0 
3
a4
r 3
[2cos rsin  ]
B=
0  
3
a4
r 3
[3cos  r−z ]