Question

A short bar magnet placed with its axis at ${30}^o$ with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to $4.5\times {10}^{-2}$ J. What is the magnitude of magnetic moment of the magnet?

Answer 1

Given data

angle between magnetic moment and magnetic field $\theta ={30}^0$

External magetic field $B=0.25T$

Torque $\tau =4.5\times {10}^{-2}J$

The torque experienced by a bar magnet placed at an angle $\theta$ with the external magnetic field $B$ is given by

$\tau =MB\sin \theta$

$⟹4.5\times {10}^{-2}=M\times 0.25sin{30}^{\circ }$

$⟹M=0.36J{T}^{-1}$