(a) From Eq. 5.70 and Prob. 5.37,
B = \nabla \times A =\frac{2 \mu_{0} R \omega \sigma}{3}(\cos \theta \hat{ r }-\sin \theta \hat{ \theta })=\frac{2}{3} \mu_{0} \sigma R \omega \hat{ z }=\frac{2}{3} \mu_{0} \sigma R \omega (5.70)
\left\{\begin{array}{l}r<R: E = 0 , B =\frac{2}{3} \mu_{0} \sigma R \omega \hat{ z }, \text { with } \sigma=\frac{e}{4 \pi R^{2}} \\r>R: E =\frac{1}{4 \pi \epsilon_{0}} \frac{e}{r^{2}} \hat{ r }, B=\frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}}(2 \cos \theta \hat{ r }+\sin \theta \hat{ \theta }), \text { with } m=\frac{4}{3} \pi \sigma \omega R^{4}\end{array}\right.
The energy stored in the electric field is (Ex. 2.9):
W_{E}=\frac{1}{8 \pi \epsilon_{0}} \frac{e^{2}}{R} .
The energy density of the internal magnetic field is:
u_{B}=\frac{1}{2 \mu_{0}} B^{2}=\frac{1}{2 \mu_{0}}\left(\frac{2}{3} \mu_{0} R \omega \frac{e}{4 \pi R^{2}}\right)^{2}=\frac{\mu_{0} \omega^{2} e^{2}}{72 \pi^{2} R^{2}}, \text { so } W_{B_{ in }}=\frac{\mu_{0} \omega^{2} e^{2}}{72 \pi^{2} R^{2}} \frac{4}{3} \pi R^{3}=\frac{\mu_{0} e^{2} \omega^{2} R}{54 \pi} .
The energy density in the external magnetic field is:
u_{B}=\frac{1}{2 \mu_{0}} \frac{\mu_{0}^{2}}{16 \pi^{2}} \frac{m^{2}}{r^{6}}\left(4 \cos ^{2} \theta+\sin ^{2} \theta\right)=\frac{e^{2} \omega^{2} R^{4} \mu_{0}}{18\left(16 \pi^{2}\right)} \frac{1}{r^{6}}\left(3 \cos ^{2} \theta+1\right) , so
W_{B_{\text {out }}}=\frac{\mu_{0} e^{2} \omega^{2} R^{4}}{(18)(16) \pi^{2}} \int_{R}^{\infty} \frac{1}{r^{6}} r^{2} d r \int_{0}^{\pi}\left(3 \cos ^{2} \theta+1\right) \sin \theta d \theta \int_{0}^{2 \pi} d \phi=\frac{\mu_{0} e^{2} \omega^{2} R^{4}}{(18)(16) \pi^{2}}\left(\frac{1}{3 R^{3}}\right)(4)(2 \pi)=\frac{\mu_{0} e^{2} \omega^{2} R}{108 \pi} .
W_{B}=W_{B_{ in }}+W_{b_{ out }}=\frac{\mu_{0} e^{2} \omega^{2} R}{108 \pi}(2+1)=\frac{\mu_{0} e^{2} \omega^{2} R}{36 \pi} ; W=W_{E}+W_{B}=\frac{1}{8 \pi \epsilon_{0}} \frac{e^{2}}{R}+\frac{\mu_{0} e^{2} \omega^{2} R}{36 \pi} .
(b) Same as Prob. 8.10(a), with Q \rightarrow e \text { and } m \rightarrow \frac{1}{3} e \omega R^{2}: L =\frac{\mu_{0} e^{2} \omega R}{18 \pi} \hat{ z } .
\text { (c) } \frac{\mu_{0} e^{2}}{18 \pi} \omega R=\frac{\hbar}{2} \Rightarrow \omega R=\frac{9 \pi \hbar}{\mu_{0} e^{2}}=\frac{(9)(\pi)\left(1.05 \times 10^{-34}\right)}{\left(4 \pi \times 10^{-7}\right)\left(1.60 \times 10^{-19}\right)^{2}}=9.23 \times 10^{10} m / s .
\frac{1}{8 \pi \epsilon_{0}} \frac{e^{2}}{R}\left[1+\frac{2}{9}\left(\frac{\omega R}{c}\right)^{2}\right]=m c^{2} ;\left[1+\frac{2}{9}\left(\frac{\omega R}{c}\right)^{2}\right]=1+\frac{2}{9}\left(\frac{9.23 \times 10^{10}}{3 \times 10^{8}}\right)^{2}=2.10 \times 10^{4};
R=\frac{\left(2.10 \times 10^{4}\right)\left(1.6 \times 10^{-19}\right)^{2}}{8 \pi\left(8.85 \times 10^{-12}\right)\left(9.11 \times 10^{-31}\right)\left(3 \times 10^{8}\right)^{2}}=2.95 \times 10^{-11} m ; \quad \omega=\frac{9.23 \times 10^{-10}}{2.95 \times 10^{-11}}=3.13 \times 10^{21} rad / s .
Since \omega R, the speed of a point on the equator, is 300 times the speed of light, this “classical” model is clearly unrealistic.