Exercise 4.1.18: Spinning charged spherical shell

A point of the shell at the polar angle $\theta$ moves on a circle of radius $R\sin \theta$ with speed $\omega R\sin \theta$:

The ring between $\theta$ and $\theta +d\theta$ has radius ${\rho }^{\prime }=R\sin \theta$, width $Rd\theta$, current $dI=\sigma \omega R\sin \theta \cdot Rd\theta$, and lies at the height $h=R\cos \theta$ above the centre. The on-axis loop formula with ${\rho }^{\prime 2}+h^2=R^2$ gives

and ${\int }_0^{\pi }{\sin }^3\theta d\theta =4/3$:

Each ring has the moment $dm=\pi {\rho }^{\prime 2}dI=\pi \sigma \omega R^4{\sin }^3\theta d\theta$, so $m=\frac{4}{3}\pi \sigma \omega R^4$, and with $Q=4\pi R^2\sigma$, $m=Q{R}^2\omega /3$.

$\omega =1047$ rad/s: $B(0)=\frac{2}{3}\times 4\pi \times 10^{-7}\times 27\times 10^{-6}\times 1047\times 0.1=2.4$ nT and $m=1.2\times 10^{-5}$ A m${}^2$.

Units: ${\mu }_0\sigma \omega R$ is (H/m)(C/m${}^2$)(1/s)(m) $=$ T. Inside a spinning shell the field turns out to be uniform and equal to the centre value, $\frac{2}{3}{\mu }_0\sigma \omega R$, while outside it is exactly a dipole field of moment $m$, the classical model of a spinning charged particle. The nanotesla numbers show again how weak convection currents are: $27$ μC/m${}^2$ at $10000$ rpm is only $J_S=2.8$ mA/m at the equator.