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Q.
Work done in rotating a bar magnet from $0$ to angle $120^{\circ}$ is
Magnetism and Matter
Solution:
Work done in rotating a magnet (from angle 0 to $\theta$ ) is given by
$W=\int\limits_{0}^{\theta} \tau d \theta$
where, $\tau=$ torque and $d \theta=$ angular change.
Also, $\tau =M B \sin \theta=\int\limits_{0}^{\theta} M B \sin \theta d \theta$
$=M B \int\limits_{0}^{\theta} \sin \theta d \theta=M B(-\cos \theta)_{0}^{\theta}$
$=M B\left(-\cos 120^{\circ}+\cos 0^{\circ}\right)$
$\Rightarrow M B\left(1+\frac{1}{2}\right)=\frac{3}{2} M B$