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

$\frac{1}{3} \pi qfL^2$

The correct answer is Option (2) → $\frac{1}{3} \pi qfL^2$

Change on element, $dq=\frac{q}{L}dx$

and,

Rod sweeps a circle of radius, $R=\frac{l}{2}$

Current, $di$ (due to frequency f) = $dq×f$

$=\frac{q}{L}fdx$

$M=\int\limits_0^1d$

$M=\int\limits_0^1\frac{q}{L}fdx×\pi x^2$

$=\frac{1}{3} \pi qfL^2$