Question

A short bar magnet is placed in a uniform magnetic field. To maintain equilibrium when it makes angle ${60}^{\circ }$with external field, we need to apply a torque of $40\sqrt{3}Nm$. Find the potential energy of the magnet.

• A. $40\text{J}$
• B. $-40\text{J}$
• C. $80\text{J}$
• D. $120\text{J}$

Answer 1

The correct option is B $-40\text{J}$

Torque on magnet $\tau =MB\sin 60=\frac{MB\sqrt{3}}{2}$ ...(i)
Potential energy of magnet
$U=-MB\cos 60=\frac{-MB}{2}$ .....(ii)
From (i) and (ii)
$U=-\frac{cos60}{sin60}\tau =-\frac{1}{2}\left(\frac{2\tau }{\sqrt{3}}\right)$
$=-\frac{1}{\text{/}2}\times \frac{\text{/}2}{\text{/}\sqrt{3}}40\text{/}\sqrt{3}=-40\text{J}$