Question

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in an equilibrium state. The energy required to rotate it by $60°$ is $W$. Now the torque required to keep the magnet in this new position is

• A. $\frac{W}{\sqrt{3}}$
• B. $\frac{\sqrt{3}}{2}W$
• C. $\sqrt{3}W$
• D. $\frac{2W}{\sqrt{3}}$

Answer 1

The correct option is C

$\sqrt{3}W$

Step 1: Given Data

Since the bar magnet is first in equilibrium, the initial angle is ${\theta }_1=0°$

Given, the final angle of rotation ${\theta }_2=60°$

Work done for this rotation $=W$

Let the magnetic moment be $m$ and the magnetic field be $B$

Step 2: Calculate the Work Done

We know that work done on a magnetic field is given as,

$W=mB\left(\cos {\theta }_1-\cos {\theta }_2\right)$

$=mB\left(\cos 0°-\cos 60°\right)$

$=mB\left(1-\frac{1}{2}\right)$

$=\frac{mB}{2}$

Step 3: Calculate the Torque

We know that torque is given as,

$\tau =mB\sin {\theta }_1$

$=mB\sin 60°$

$=mB\frac{\sqrt{3}}{2}$

$=\sqrt{3}W$

Hence, the correct answer is option (C).