A rod had a total charge $Q$ uniformly distributed along its length $L.$ If the rod rotates with angular velocity $\omega$ about its end, compute its magnetic moment.

  

We can visualize the rod to consist of differential elements $\mathrm{dQ}$, which constitute a series of concentric current loops. The charge per unit length of the rod $\lambda$,

$\lambda =\frac{Q}{L}$

So the charge on a differential element of length $\mathrm{dl}$

$\mathrm{dq}=\mathrm{\lambda dl}$

The current $\mathrm{dI}$ due to rotation of this charge is given by

$dI=\frac{dq}{(2\pi /\omega )}=\frac{\omega }{2\pi }dq=\frac{\omega }{2\pi }\lambda \mathrm{d}\ell$

The magnetic moment of this differential current loop,

$d\mu =dI\left(\pi {\ell }^2\right)=\left(\frac{\omega }{2\pi }\lambda d\ell \right)\pi {\ell }^2=\frac{\omega \lambda }{2}{\ell }^2 \mathrm{d}\ell$

to find total magnetic moment, we integrate

$\mu =\frac{\omega \lambda }{2}{\int }_0^L{\ell }^2 \mathrm{d}\ell =\frac{\omega \lambda {\mathrm{L}}^3}{6}$

Substituting for $\lambda$, we obtain

$\mu =\frac{\mathrm{Q}\omega {\mathrm{L}}^2}{6}$