A magnetic dipole experiences a torque of $$80\sqrt{3}$$ N m when placed in uniform magnetic field in such a way that dipole moment makes angle of 60° with magnetic field. The potential energy of the dipole is:

A magnetic dipole with dipole moment $$M$$ in a uniform magnetic field $$B$$ experiences a torque $$\tau = MB\sin\theta$$ and has potential energy $$U = -MB\cos\theta$$.

Given $$\tau = 80\sqrt{3}$$ N m and $$\theta = 60°$$:

$$MB\sin 60° = 80\sqrt{3}$$

$$MB \cdot \dfrac{\sqrt{3}}{2} = 80\sqrt{3}$$

$$MB = 160$$

The potential energy is: $$U = -MB\cos 60° = -160 \cdot \dfrac{1}{2} = -80$$ J

Hence, the correct answer is Option D.