Question

A bar magnet of length 10 cm and having the pole strength equal to ${10}^{-3}$ weber is kept in a magnetic field having magnetic induction(B) equal to $4\pi \times {10}^{-4}$ tesla. It makes an angle of ${30}^0$ with the directon of magnetic induction.The value of the torque acting on the magnet is. $({\mu }_0=4\pi \times {10}^{-7}weber/amp\times m)$

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

Answer 1

The correct option is A $2\pi \times {10}^{-7}N\times m$

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