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

<p>To determine the potential energy of a magnetic dipole in a uniform magnetic field, we start with the torque equation: </p> <p>$ \tau = M \times B = MB \sin 60^\circ $</p> <p>Given that $\tau = 80 \sqrt{3} \, \text{N m}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we can equate and solve for $MB$:</p> <p>$ MB \frac{\sqrt{3}}{2} = 80 \sqrt{3} $</p> <p>Solving for $MB$, we find:</p> <p>$ MB = 160 $</p> <p>The potential energy $U$ of the dipole in the magnetic field is given by:</p> <p>$ U = -M \cdot B = -MB \cos 60^\circ $</p> <p>Substituting in $\cos 60^\circ = \frac{1}{2}$, we calculate:</p> <p>$ U = -160 \times \frac{1}{2} = -80 \, \text{J} $</p> <p>Thus, the potential energy of the dipole is $-80 \, \text{J}$.</p>