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:
Answer & explanation
Correct answer: option 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) = $dq×f$
$=\frac{q}{L}fdx$
$M=\int\limits_0^1d$
$M=\int\limits_0^1\frac{q}{L}fdx×\pi x^2$
$=\frac{1}{3} \pi qfL^2$