Page 190 - Electromagnetics

we have
" # 2 π
−A x a r < 3 3
= µ 0 M 0 sin θ sin θ dθ ·
2
A y 3 r 8π
> 0
π sin φ π sin φ

· e − jφ e jφ dφ + e jφ e − jφ dφ .
−π cos φ −π cos φ
Carrying out the integrals we find that
2
" # " #
−A x a r < sin φ
= µ 0 M 0 sin θ
A y 3 r 2 cos φ
>
or
2
a r < ˆ
A = µ 0 M 0 sin θφ.
3 r 2
>
Finally, B =∇ × A gives

2µ 0 M 0 ˆ z, r < a,
3
B = 3 (3.199)
µ 0 M 0 a ˆ r 2 cos θ + θ sin θ , r > a.
ˆ
3r 3
Hence B within the sphere is uniform and in the same direction as M, while B outside
the sphere has the form of the magnetic dipole field with moment

4 3
m = πa M 0 .
3
We can also compute B by first finding the scalar potential through direct computation
of the integral (3.126). Substituting for ρ Ms from (3.197), we have

1 ρ Ms (r ) 1 π π M 0 cos θ

m (r) = dS = sin θ dθ dφ .
4π S |r − r | 4π −π 0 |r − r |

This integral has the form of (3.100)with f (θ) = M 0 cos θ. Thus, from (3.102),
a 2 r <
m (r) = M 0 cos θ . (3.200)
3 r 2
>
The magnetic field H is then

M 0
− ˆ z, r < a,
H =−∇ m = 3 .
3
ˆ
M 0 a ˆ r 2 cos θ + θ sin θ , r > a.
3r 3
Inside the sphere B is given by B = µ 0 (H + M), while outside the sphere it is merely
B = µ 0 H. These observations lead us again to (3.199).
Since the scalar potential obeys Laplace’s equation both inside and outside the sphere,
as a last approach to the problem we shall write m in terms of the separation of variables
solution discussed in § A.4. We can repeat our earlier arguments for the dielectric sphere
in an impressed electric field (§ 3.2.10). Copying equations (3.109) and (3.110), we can
write for r ≤ a
∞

n
m1 (r,θ) = A n r P n (cos θ), (3.201)
n=0

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