Q 12-04-029NEETJEE MainMedium
A thin ring of radius $R$ carries a charge $q$ spread uniformly over it. It rotates about its axis with angular speed $\omega$. Its magnetic moment is
Answer: (B) $\dfrac{q\omega R^2}{2}$
The charge passes any point $\dfrac{\omega}{2\pi}$ times per second, so the ring is a current $I = \dfrac{q\omega}{2\pi}$.
$M = I \times \pi R^2 = \dfrac{q\omega}{2\pi} \times \pi R^2 = \dfrac{q\omega R^2}{2}$.
Solution by Sreeraj P, M.Sc Physics