(a) A phonograph record of radius R, carrying a uniform surface charge σ, is rotating at constant angular velocity ω. Find its magnetic dipole moment.
(b) Find the magnetic dipole moment of the spinning spherical shell in Ex. 5.11. Show that for points r > R the potential is that of a perfect dipole.
(a) For a ring, m=I \pi r^{2} . \text { Here } I \rightarrow \sigma v d r=\sigma \omega r d r, \text { so } m=\int_{0}^{R} \pi r^{2} \sigma \omega r d r=\pi \sigma \omega R^{4} / 4 .
(b) The total charge on the shaded ring is dq = \sigma(2 \pi R \sin \theta) R d \theta . The time for one revolution is d t=2 \pi / \omega .
So the current in the ring is I=\frac{d q}{d t}=\sigma \omega R^{2} \sin \theta d \theta . The area of the ring is \pi(R \sin \theta)^{2} , so the magnetic moment of the ring is d m=\left(\sigma \omega R^{2} \sin \theta d \theta\right) \pi R^{2} \sin ^{2} \theta , and the total dipole moment of the shell is
m=\sigma \omega \pi R^{4} \int_{0}^{\pi} \sin ^{3} \theta d \theta=(4 / 3) \sigma \omega \pi R^{4}, \text { or } m =\frac{4 \pi}{3} \sigma \omega R^{4} \hat{ z } .
The dipole term in the multipole expansion for A is therefore A _{\text {dip }}=\frac{\mu_{0}}{4 \pi} \frac{4 \pi}{3} \sigma \omega R^{4} \frac{\sin \theta}{r^{2}} \hat{\phi}=\frac{\mu_{0} \sigma \omega R^{4}}{3} \frac{\sin \theta}{r^{2}} \hat{\phi} , which is also the exact potential (Eq. 5.69); evidently a spinning sphere produces a perfect dipole field, with no higher multipole contributions.
A (r, \theta, \phi)= \begin{cases}\frac{\mu_{0} R \omega \sigma}{3} r \sin \theta \hat{ \phi }, & (r \leq R), \\ \frac{\mu_{0} R^{4} \omega \sigma}{3} \frac{\sin \theta}{r^{2}} \hat{ \phi }, & (r \geq R).\end{cases} (5.69)
