Question

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

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

Answer 1

The correct option is A 30 Nm

$\text{Given:-}$ M=200 A $m^2$, B= 0.30 N $A^{-1}{m}^{-1}$ and $\theta ={30}^0$

$\text{To find:-}$torque

$\text{Solution:-}$

Torque experienced by a magnet suspended in a uniform magnetic field B is given by $\tau =MBsin\theta$

$\therefore \tau =200\times 0.30\times sin{30}^0$

$\tau =30Nm$

$\text{Hence the correct option is A}$