JEE Main 2026 Jan 24 Shift 1, Physics Q21: Torque and Work Done in a Magnetic Field

Convert: $B=800$ G $=0.08$ T.

Find magnetic moment from torque: $\tau =\mu B\sin \theta$.

$0.016=\mu \times 0.08\times \sin 30^{\circ }=\mu \times 0.04\Rightarrow \mu =0.4$ J/T.

Most stable to most unstable means rotating from $\theta =0$ (aligned) to $\theta =180^{\circ }$ (anti-aligned). Using $U=-\mu B\cos \theta$:

$W=U_f-U_i=(+\mu B)-(-\mu B)=2\mu B=2\times 0.4\times 0.08=0.064\text{J}=64\times 10^{-3}$ J.

$\therefore \alpha =64$.