An insulating rod of length l carries a charge q distributed uniformly on it. The rod is pivoted at an end and is rotated at a frequency f about a fixed perpendicular axis. The magnetic moment of the system is

At a distance ‘x’ consider a small element of width ‘dx’ .

As the element rotates, it can be considered as a current carrying loop.

Magnetic moment of the small element is 

$\mathrm{dM}=\mathrm{IA}=\frac{\mathrm{dq}}{\mathrm{T}}{\mathrm{\pi x}}^2=\left(\frac{\mathrm{q}}{\mathrm{l}}\mathrm{dx}\right){\mathrm{f\pi x}}^2$

$\Rightarrow \mathrm{M}=\frac{\mathrm{qf\pi }}{\mathrm{l}}{\int }_0^{\mathrm{l}} {\mathrm{x}}^2\mathrm{dx}=\frac{{\mathrm{q\pi fl}}^2}{3}$