Question

A bar magnet of length $10\, cm$ and having the pole strength equal to $10^{-3} Wb$ is kept in a magnetic field having magnetic induction $(B)$ equal to $4 \pi \times 10^{-3} T$. It makes an angle of $30^{\circ}$ with the direction of magnetic induction. The value of the torque acting on the magnet is

• A. $2 \pi \times 10^{-7} Nm$
• B. $2 \pi \times 10^{-5} Nm$
• C. $0.5\, Nm$
• D. $0.5 \times 10^{2} Nm$

Answer 1

Answer: (A)

Torque, $\tau=M B \sin \theta$
$=0.1 \times 10^{-3} \times 4 \pi \times 10^{-3} \times \sin 30^{\circ}$
$=10^{-7} \times 4 \pi \times \frac{1}{2}$
$=2 \pi \times 10^{-7} N m$