Q. A bar magnet of length 10 cm and having 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 f the torque acting on the magnet is
A) 0.5 Nm
B) $2\pi \times 10^{- 5}$ Nm
C) $\pi \times 10^{- 5}$ Nm
D) 0.5 $\times 10^{- 5}$ Nm
Explanation
(a) In SI unit of pole strength is Amp-meter. Here, the pole strength is given in Weber, which is the unit of $(\mu_{0}m)$.
$\therefore\mu_{0}m = 10^{- 3}$Wb
$m = \frac{10^{- 3}}{\mu_{0}}$
Magnetic moment of magnet
$M = m \times 2l = \frac{10^{- 3}}{\mu_{0}}(0.1) = \frac{10^{- 4}}{\mu_{0}}$.
Torque, $\tau = MBsin\theta$
$= \frac{10^{- 4}}{4\pi \times 10^{- 7}}\left( 4\pi \times 10^{- 3} \right) \times \frac{1}{2} = 0.5$ Nm
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.