Question

A short bar magnet experiences a torque of magnitude $0.64J$. When it is placed in a uniform magnetic field of $0.32T$, making an angle of ${30}^{\circ }$ with the direction of the field. The magnetic moment of the magnet is

• A. $2A{m}^2$
• B. $4A{m}^2$
• C. $6A{m}^2$
• D. None of these

Answer 1

Answer: (B)

$4A{m}^2$

Torque, $\tau =0.64J,B=0.32T,\theta ={30}^{\circ }$
Torque, $\tau =MB\sin \theta$
$0.64=M\times 0.32\sin {30}^{\circ }$
$0.64=M\times 0.32\times \frac{1}{2}$
$M=\frac{2\times 0.64}{0.32}=4A{m}^2$.