A bar magnet having a magnetic moment of 2 × 10^4 JT^ -1 is free to...

Key Idea The work done in rotating a magnetic dipole against the torque acting on it, when placed in magnetic field is stored inside it in the form of potential energy. When magnetic dipole is rotated from initial position $\theta=\theta_1$ to final position $\theta=\theta_2$, then work done =$\mathrm{MB}$\left(\cos \theta_1-\cos\theta_2\right) $$\begin{aligned} & =\ operatorname{MB}\l \left(1-\frac{1}{2}\right)\\ & =\frac{2 \times 10^4 \times 6 \times 10^{-4}}{2}= 6 \mathrm{ J} \end{aligned}$$