A bar magnet of magnetic moment M is freely suspended in a uniform magnetic field of strength B. The work done in rotating the magnet through an angle $\mathrm{\theta }$ is

Torque acting on the bar magnet is given by $\tau =MB\cos \theta$.

$\mathrm{dW}=\mathrm{MBsin}\mathrm{\theta d\theta }$

Now, the work done in rotating the magnet through an angle θ from initial position (θ₁=0) is given by $\mathrm{W}=\mathrm{MB}\left(\cos {\mathrm{\theta }}_1-\cos \mathrm{\theta }\right)$

$\mathrm{W}=\mathrm{MB}\left(\cos 0^{\circ }-\cos \mathrm{\theta }\right)\Rightarrow \mathrm{W}=\mathrm{MB}\left(1-\cos \mathrm{\theta }\right)$