Question

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

Answer 1

Torque acting on a bar magnet kept in a magnetic field $\overrightarrow{B}$ is $\tau =\overrightarrow{m}\times \overrightarrow{B}=mBsin\theta$

Hence work done in moving from $\theta =0^{\circ }$ to $\theta ={60}^{\circ }$ is $W=mBcos{60}^{\circ }=\frac{1}{2}mB$

Hence torque acting when magnet is placed at ${60}^{\circ }=\frac{\sqrt{3}}{2}mB=\sqrt{3}W$