Q 12-05-031JEE MainJEE Main 2024 (4 Apr, Shift 2)Easy
The magnetic moment of a bar magnet is $0.5\ \text{A m}^2$. It is suspended in a uniform magnetic field of $8\times10^{-2}\ \text{T}$. The work done in rotating it from its most stable to its most unstable position is
Answer: (A) $8\times10^{-2}\ \text{J}$
From $\theta = 0^\circ$ to $\theta = 180^\circ$:
$$W = mB(\cos0^\circ - \cos180^\circ) = 2mB = 2\times0.5\times8\times10^{-2} = 8\times10^{-2}\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics