Q 12-05-048JEE MainJEE Main 2022 (26 Jun, Shift 2)Easy
A bar magnet having a magnetic moment of $2.0\times10^5\ \text{J T}^{-1}$ is placed along the direction of a uniform magnetic field of magnitude $B = 14\times10^{-5}\ \text{T}$. The work done in rotating the magnet slowly through $60^\circ$ from the direction of field is
Answer: (A) $14\ \text{J}$
$$W = mB(\cos0^\circ - \cos60^\circ) = 2\times10^5\times14\times10^{-5}\times\frac12 = 14\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics