Question

A bar magnet has a magnetic moment of $200A{m}^2$. The magnet is suspended in a magnetic field of $0.30N{A}^{-1}{m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of ${30}^0$, will be then

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

Answer 1

Answer: (A)

$30$Nm

Torque experienced by a magnet suspended in a uniform magnetic field B is given by
$T=MBsin\theta$
Here, $M=200A{m}^2,B=0.30N{A}^{-1}{m}^{-1}and\theta ={30}^0$
$\therefore T=200\times 0.30\times sin{30}^0$
$T=30Nm$