Q 12-05-067JEE MainJEE Main 2020 (4 Sep, Shift 1)Medium
A small bar magnet placed with its axis at $30^\circ$ with an external magnetic field of $0.06\ \text{T}$ experiences a torque of $0.018\ \text{N m}$. The minimum work required to rotate it from its stable to unstable equilibrium position is:
Answer: (C) $7.2\times10^{-2}\ \text{J}$
$\tau = mB\sin30^\circ \Rightarrow m = \dfrac{0.018}{0.06\times0.5} = 0.6\ \text{A m}^{2}$.
From $\theta = 0$ (stable) to $\theta = 180^\circ$ (unstable): $W = mB(\cos0 - \cos180^\circ) = 2mB = 2\times0.6\times0.06 = 7.2\times10^{-2}$ J.
Solution by Sreeraj P, M.Sc Physics