Question

A bar magnet has a magnetic moment of $200$ A $m^2$. The magnet is suspended in a magnetic field of $0.30$N $A^{-1}{m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of ${30}^o$, will be:

• A. $30$ N m
• B. $30\sqrt{3}$ N m
• C. $60$ N m
• D. $60\sqrt{3}$ N m

Answer 1

Answer: (A)

$30$ N m

Torque experienced by a magnet suspended in a uniform magnetic field B is given by
$\tau =MB\sin \theta$
Here, $M=200A{m}^2,B=0.30N{A}^{-1}{m}^{-1}$ and $\theta ={30}^o$
$\therefore \tau =200\times 0.30\times \sin {30}^o$
$\tau =30$N m