Q. A magnet of length 0.1 m and pole strength $10^{- 4}$ A-m is kept in a magnetic field of 30 Wb $m^{- 2}$ at an angle $30{^\circ}.$ The couple acting on it is ……..$\times 10^{- 4}$ Nm.
A) 7.5
B) 3.0
C) 4.5
D) 1.5
Explanation
(d) Torque, $\tau = MB\sin\theta$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = m \times (2l) \times B\sin\theta$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10^{- 4} \times 0.1 \times 30\sin{30{^\circ} = 1.5 \times 10^{- 4}\ N - m}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.