Q. A closely wound solenoid of 2000 turns and area of cross-section $1.5 \times 10^{- 4}m^{2}$ carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{- 2}$ tesla making an angle of $30{^\circ}$ with the axis of the solenoid. The torque on the solenoid will be
A) $3 \times 10^{- 3}N.m$
B) $1.5 \times 10^{- 3}N.m$
C) $1.5 \times 10^{- 2}N.m$
D) $3 \times 10^{- 2}N.m$
Explanation
(c) $M = 2000 \times 1.5 \times 10^{- 4} \times 2 = 6 \times 10^{- 1}$
$\tau = MB\sin 30 = 0.6 \times 5 \times 10^{- 2} \times \frac{1}{2}$
$\tau = 1.5 \times 10^{- 2}Nm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.