Q 12-04-181JEE MainJEE Main 2020 (9 Jan, Shift 2)Medium
A small circular loop of conducting wire has radius $a$ and carries current $I$. It is placed in a uniform magnetic field $B$ perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period $T$. The mass of the loop is $m$ then:
Answer: (C) $T = \sqrt{\dfrac{2\pi m}{IB}}$
Restoring torque for a small twist $\theta$: $\tau = -\mu B\theta$ with $\mu = I\pi a^2$.
Moment of inertia of a ring about a diameter: $I_d = \dfrac{ma^2}{2}$.
$$T = 2\pi\sqrt{\frac{I_d}{\mu B}} = 2\pi\sqrt{\frac{ma^2/2}{I\pi a^2B}} = 2\pi\sqrt{\frac{m}{2\pi IB}} = \sqrt{\frac{2\pi m}{IB}}$$
Solution by Sreeraj P, M.Sc Physics