Question

A bar magnet of length $0.1m$ and with a magnetic moment of $5{Am}^2$ is placed in a uniform magnetic field of intensity $0.4T$, with its axis making an angle of ${60}^o$ with the field. What is the torque on the magnet?

Answer 1

Magnetic moment, $M=5A{m}^2$

Magnetic field of intensity, $B=0.4T$

Angle between $M$ and $B$, $\theta ={60}^o$

Torque acting on the magnet, $\tau =MB\sin \theta$ $(\because \sin {60}^o=0.866)$

$\therefore$ $\tau =5\times 0.4\times 0.866=1.732$ $N-m$