Question

Consider a solid sphere of radius $r$ and mass $m$ which has a charge $q$ distributed uniformly over its volume. The sphere is rotated about a diameter with an angular speed $\omega$. Show that the magnetic moment $\mu$ and the angular momentum $l$ of the sphere are related as $\mu =\frac{q}{2m}l$

Answer 1

The magnetic moment acts along the axis of rotation. Consider a volume element dV.

It contains a charge, $q^{\prime }=\frac{Q}{\frac{4}{3}\pi R^3}dV$

So, the current constituted by the element can be given as:

$I=\frac{3Q}{4\pi R^3}\frac{\omega }{2\pi }dV$

and magnetic moment can be given as:

$d\mu =I\mu r^{2}si{n}^2\theta =\frac{3Q}{4\pi R^3}\frac{\omega }{2\pi }dV\pi r^{2}si{n}^2\theta$

$\Rightarrow \mu ={\int }_0^{\pi }{\int }_0^R\frac{3Q}{4\pi R^3}\left(2\pi r^{2}sin\theta d\theta dr\right)\left(\frac{\omega }{r}{r}^{2}si{n}^2\theta \right)$

$\Rightarrow \frac{3Q}{2R^3}\times \frac{\omega }{2}\times \frac{R^5}{5}\times \frac{4}{3}=\frac{Q{R}^2\omega }{5}$

$\because L=\frac{2}{5}m{R}^2\omega$

$\therefore R^2\omega =\frac{L}{2m}$

Hence, $\mu =\frac{QL}{2m}$