Q 12-04-233JEE MainJEE Main 2018 (16 Apr, Shift 1)Easy
A charge $q$ is spread uniformly over an insulated loop of radius $r$. If it is rotated with an angular velocity $\omega$ with respect to normal axis then magnetic moment of the loop is:
Answer: (B) $\dfrac12q\omega r^2$
The charge passes any point once per revolution, so the equivalent current is
$$I = \frac{q}{T} = \frac{q\omega}{2\pi}$$
$$m = I\cdot\pi r^2 = \frac{q\omega}{2\pi}\pi r^2 = \frac12q\omega r^2$$
Solution by Sreeraj P, M.Sc Physics