Q. A magnetic needle lying parallel to a magnetic field required $W$ units of work to turn it through $60{^\circ}$. The torque required to maintain the needle in this position will be
A) $\sqrt{3}$ $W$
B) $W$
C) $\sqrt{3}\frac{W}{2}$
D) 2 $W
$
Explanation
(a) $W = BM\ \cos 60{^\circ} = \frac{MB}{2};MB = 2W$
$\tau = MB\ sin\theta = (2W)sin60{^\circ} = 2W\frac{\sqrt{3}}{2} = W\sqrt{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.