Q 12-05-070JEE MainJEE Main 2019 (10 Jan, Shift 1)Medium
A magnet of total magnetic moment $10^{-2}\,\hat i\ \text{A m}^2$ is placed in a time varying magnetic field, $B\,\hat i\,(\cos\omega t)$ where $B = 1\ \text{T}$ and $\omega = 0.125\ \text{rad s}^{-1}$. The work done for reversing the direction of the magnetic moment at $t = 1$ second, is:
Answer: (B) $0.02\ \text{J}$
At $t = 1\ \text{s}$ the field is $B\cos(0.125) = 0.992\ \text{T}$ along $\hat i$, parallel to the moment.
Work to turn the moment from $0^\circ$ to $180^\circ$:
$$W = U_f - U_i = mB - (-mB) = 2mB = 2\times10^{-2}\times0.992 \approx 0.02\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics