Physics 712
Chapter 5 Problems
4. [5] Consider a localized static current distribution. By considering the expression
3
3
3
    xi x j J  x  d x , show  xi J j  x  d x   x j J i  x  d x .
We first use the divergence theorem to turn it into a surface integral, which then vanishes
because the currents at infinity vanish:
    x x J  x  d x  
3
i
j
S
xi x j J  x   nˆ da  0 .
We now distribute the gradient using the product rule, using the fact that a static current
distribution has   J  0 , and then rewriting the dot products as sums to figure out what is going
on.
0      xi x j J  x   d 3 x    x j J  x  xi  xi J  x  x j  xi x j   J  x   d 3 x
x j  3

x
3
    x j J k  x  i  xi J k  x 
 d x     x j J k  x   ik  xi J k  x   jk  d x
xk
xk 
k
k

   x j J i  x   xi J j  x   d 3 x   x j J i  x  d 3 x   xi J j  x  d 3 x .
The desired identity is now a trivial step away.
5. [5] For each of the four loops in problem 2, find the magnetic dipole moment m and the
resulting magnetic flux density far from the loop.
The magnetic dipole moment is just m  IAzˆ , because ẑ is the normal to the plane of the
loop. The area is most easily computed by finding the area of one triangle and then multiplying
them by the number of triangles. The area of one triangle is
1
1C
C

Aone  bd 
2
2 N 2 N tan  N 
Multiplying by the number of triangles N, our answer is
IC 2 zˆ
m
4 N tan  N 
Again, we can just substitute in our four values of N, using again N   for the circle:
m3 
IC 2 zˆ
IC 2 zˆ
IC 2 zˆ
IC 2 zˆ
, m4 
, m6 
, m 
.
16
4
12 3
8 3
The magnetic flux density is just
B
0 3  rˆ  m  rˆ  m
.
4
r3