Use the result of Ex. 5.6 to calculate the magnetic field at the centerof a uniformly charged spherical shell, of radius Rand total charge Q,spinning atconstant angular velocity $\mathit{\omega }$.

There isa uniformly charged spherical shell, of radius Rand total charge Q,spinning at

constant angular velocity $\omega$.

The magnetic field at a distance $z$ on the axis of a circular coil of radius $a$ and carrying current$I$ is

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}}\frac{{\mathbf{a}}^{\mathbf{2}}}{{\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}\mathbf{)}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}$ …… (1)

Here, ${\mu }_0$ is the permeability of free space.

Consider a ring on the surface of the sphere at an angle $\theta$ from the center.

From equation (1), the magnetic field at the center from that ring is

$\begin{array}{c}dB=\frac{{\mu }_{0}dI}{2}\frac{{\left(R\sin \theta \right)}^2}{[{\left(R\sin \theta \right)}^2+{\left(R\cos \theta \right)}^2]}\\ =\frac{{\mu }_0}{2R}{\sin }^2\theta dI\mathrm{.....}\left(2\right)\end{array}$

Solve as:

$\begin{array}{c}dI=\frac{Q}{4\pi R^2}\omega R\sin \theta Rd\theta \\ =\frac{Q\omega }{4\pi }\sin \theta d\theta \end{array}$

To get the field from the full spherical surface, substitute this in equation (2) and integrate from $0$ to $\pi$,

$\begin{array}{c}B=\frac{{\mu }_0}{2R}\underset{0}{\overset{\pi }{\int }}{\sin }^2\theta \frac{Q\omega }{4\pi }\sin \theta d\theta \\ =\frac{{\mu }_{0}Q\omega }{8\pi R}\underset{0}{\overset{\pi }{\int }}{\sin }^3\theta d\theta \\ =\frac{{\mu }_{0}Q\omega }{8\pi R}\times \frac{4}{3}\\ =\frac{{\mu }_{0}Q\omega }{6\pi R}\end{array}$

Thus, the field is $\frac{{\mu }_{0}Q\omega }{6\pi R}$.