(a)
E =\left\{\begin{array}{ll} 0 , & (r<R) \\\frac{1}{4 \pi \epsilon_{0}} \frac{Q}{r^{2}} \hat{ r }, & (r>R)\end{array}\right\} ; B =\left\{\begin{array}{ll}\frac{2}{3} \mu_{0} M \hat{ z }, & (r<R) \ \\frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}}[2 \cos \theta \hat{ r }+\sin\theta \hat{ \theta }], & (r>R)\end{array}\right\} (Ex. 6.1)
\text { (where } \left.m=\frac{4}{3} \pi R^{3} M\right) ; g =\epsilon_{0}( E \times B )=\frac{\mu_{0}}{(4 \pi)^{2}} \frac{Q m}{r^{5}}(\hat{ r } \times \hat{ \theta }) \sin \theta, \text { and }(\hat{ r } \times \hat{ \theta })=\hat{\phi} , so
\ell = r \times g =\frac{\mu_{0}}{(4 \pi)^{2}} \frac{m Q}{r^{4}} \sin \theta(\hat{ r } \times \hat{ \phi }).
\text { But } \left.(\hat{ r } \times \hat{ \phi })=-\hat{ \theta }, \text { and only the } z \text { component will survive integration, so (since }(\hat{ \theta })_{z}=-\sin \theta\right) :
L =\frac{\mu_{0} m Q}{(4 \pi)^{2}} \hat{ z } \int \frac{\sin ^{2} \theta}{r^{4}}\left(r^{2} \sin \theta d r d \theta d \phi\right) . \quad \int_{0}^{2 \pi} d \phi=2 \pi ; \quad \int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{4}{3} ; \quad \int_{R}^{\infty} \frac{1}{r^{2}} d r=\left.\left(-\frac{1}{r}\right)\right|_{R} ^{\infty}=\frac{1}{R} .
L =\frac{\mu_{0} m Q}{(4 \pi)^{2}} \hat{ z }(2 \pi)\left(\frac{4}{3}\right)\left(\frac{1}{R}\right)=\frac{2}{9} \mu_{0} M Q R^{2} \hat{ z } .
(b) Apply Faraday’s law to the ring shown:
\oint E \cdot d l =E(2 \pi r \sin \theta)=-\frac{d \Phi}{d t}=-\pi(r \sin \theta)^{2}\left(\frac{2}{3} \mu_{0} \frac{d M}{d t}\right)
\Rightarrow E =-\frac{\mu_{0}}{3} \frac{d M}{d t}(r \sin \theta) \hat{\phi} .
The force on a patch of surface (da) is (d a) \text { is } d F =\sigma E d a=-\frac{\mu_{0} \sigma}{3} \frac{d M}{d t}(r \sin \theta) d a \hat{ \phi }\left(\sigma=\frac{Q}{4 \pi R^{2}}\right) .
The torque on the patch is d N = r \times d F =-\frac{\mu_{0} \sigma}{3} \frac{d M}{d t}\left(r^{2} \sin \theta\right) d a(\hat{ r } \times \hat{ \phi }) . \text { But }(\hat{ r } \times \hat{ \phi })=-\hat{ \theta } , and we want only the z component \left(\hat{ \theta }_{z}=-\sin \theta\right) :
N =-\frac{\mu_{0} \sigma}{3} \frac{d M}{d t} \hat{ z } \int r^{2} \sin ^{2} \theta\left(r^{2} \sin \theta d \theta d \phi\right) .
\text { Here } r=R ; \int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{4}{3} ; \int_{0}^{2 \pi} d \phi=2 \pi, \text { so } N =-\frac{\mu_{0} \sigma}{3} \frac{d M}{d t} \hat{ z } R^{4}\left(\frac{4}{3}\right)(2 \pi)=-\frac{2 \mu_{0}}{9} Q R^{2} \frac{d M}{d t} \hat{ z } .
L =\int N d t=-\frac{2 \mu_{0}}{9} Q R^{2} \hat{ z } \int_{M}^{0} d M=\frac{2 \mu_{0}}{9} M Q R^{2} \hat{ z } (same as (a))
(c) Let the charge on the sphere at time t be q(t); the charge density is \sigma=\frac{q(t)}{4 \pi R^{2}} . The charge below (“south of”) the ring in the figure is
q_{s}=\sigma\left(2 \pi R^{2}\right) \int_{0}^{\pi} \sin \theta^{\prime} d \theta^{\prime}=\left.\frac{q}{2}\left(-\cos \theta^{\prime}\right)\right|_{\theta} ^{\pi}=\frac{q}{2}(1+\cos \theta) .
So the total current crossing the ring (flowing “north”) is I(t)=-\frac{1}{2} \frac{d q}{d t}(1+\cos \theta) , and hence
K (t)=\frac{I}{2 \pi R \sin \theta}(-\hat{ \theta })=\frac{1}{4 \pi R} \frac{d q}{d t} \frac{(1+\cos \theta)}{\sin \theta} \hat{ \theta } . The force on a patch of area da is dF = (K × B) da.
B _{ ave }=\left[\frac{2}{3} \mu_{0} M \hat{ z }+\frac{\mu_{0}}{4 \pi} \frac{\frac{4}{3} \pi R^{3} M}{R^{3}}(2 \cos \theta \hat{ r }+\sin \theta \hat{ \theta })\right] \frac{1}{2}=\frac{\mu_{0} M}{6}[2 \hat{ z }+2 \cos \theta \hat{ r }+\sin \theta \hat{ \theta }] ;
K \times B =\frac{1}{4 \pi R} \frac{d q}{d t} \frac{\mu_{0} M}{6} \frac{(1+\cos \theta)}{\sin \theta}[2(\hat{ \theta } \times \hat{ z })+2 \cos \theta \underbrace{(\hat{ \theta } \times \hat{ r })}_{-\hat{\phi}}].
d N =R \hat{ r } \times d F =\frac{\mu_{0} M}{24 \pi}\left(\frac{d q}{d t}\right) \frac{(1+\cos \theta)}{\sin \theta} 2[\underbrace{\hat{ r } \times(\hat{ \theta } \times \hat{ z })}_{\hat{ \theta }(\hat{ r } \cdot \hat{ z })-\hat{ z }(\hat{ r } \cdot \hat{ \theta })}-\cos \theta \underbrace{(\hat{ r } \times \hat{ \phi })}_{-\hat{ \theta }}] R^{2} \sin \theta d \theta d \phi
=\frac{\mu_{0} M}{12 \pi}\left(\frac{d q}{d t}\right)(1+\cos \theta) R^{2}[\cos \theta \hat{ \theta }+\cos \theta \hat{ \theta }] d \theta d \phi=\frac{\mu_{0} M R^{2}}{6 \pi}\left(\frac{d q}{d t}\right)(1+\cos \theta) \cos \theta d \theta d \phi \hat{ \theta } .
The x and y components integrate to zero; (\hat{ \theta })_{z}=-\sin \theta, \text { so }\left(\text { using } \int_{0}^{2 \pi} d \phi=2 \pi\right) :
N_{z}=-\frac{\mu_{0} M R^{2}}{6 \pi}\left(\frac{d q}{d t}\right)(2 \pi) \int_{0}^{\pi}(1+\cos \theta) \cos \theta \sin \theta d \theta=-\left.\frac{\mu_{0} M R^{2}}{3}\left(\frac{d q}{d t}\right)\left(\frac{\sin ^{2} \theta}{2}-\frac{\cos ^{3} \theta}{3}\right)\right|_{0} ^{\pi}
=-\frac{\mu_{0} M R^{2}}{3}\left(\frac{d q}{d t}\right)\left(\frac{2}{3}\right)=-\frac{2 \mu_{0}}{9} M R^{2} \frac{d q}{d t} . \quad N =-\frac{2 \mu_{0}}{9} M R^{2} \frac{d q}{d t} \hat{ z } .
Therefore
L =\int N d t=-\frac{2 \mu_{0}}{9} M R^{2} \hat{ z } \int_{Q}^{0} d q=\frac{2 \mu_{0}}{9} M R^{2} Q \hat{ z } (same as (a)).
(I used the average field at the discontinuity—which is the correct thing to do—but in this case you’d get the same answer using either the inside field or the outside field.)
