Question

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in the equilibrium state. The energy required to rotate it by ${60}^{\circ }$ is$\text{W}$. Now, the torque required to keep the magnet in this new position is

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

Answer 1

The correct option is B

$\sqrt{3}W$
Energy required to rotate the magnet by $\theta$ from equilibrium state is given by,

$W=MB(1-\cos \theta )$

And, when the magnet is at an orientation $\theta$ with the direction of field, torque applied on it by the field

$\overrightarrow{\tau }=|\overrightarrow{M}\times \overrightarrow{B}|=MB\sin \theta$

So, we need to apply an equal but opposite torque $MB\sin \theta$ to hold it in the new state.

Thus, the required torque is given by,

${\tau }_{req}=MB\sin \theta =MB\sin {60}^{\circ }$

${\tau }_{req}=\frac{\sqrt{3}MB}{2}$

Given that, $W=MB(1-\cos {60}^{\circ })$

$W=\left(\frac{MB}{2}\right)$

So, ${\tau }_{req}=\sqrt{3}\left(\frac{MB}{2}\right)=\sqrt{3}W$

Hence, option $\left(B\right)$ is the correct answer.