Q. A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through $60{^\circ}.$ The torque required to keep the needle in this position will be
A) $2W$
B) W
C) $\frac{W}{\sqrt{2}}$
D) $\sqrt{3}$W
Explanation
(d) $W = mB\cos\theta$
$\ \ \ \ \ \ = mB\cos{60{^\circ}}$
$\ \ \ \ \ \ = mB \times \frac{1}{2}$
$\ \ \tau = mB\sin\theta$
$\ \ \ \ \ = mB\sin{60{^\circ}}$
$\ \ \ \ \ = \sqrt{3}$ W $\lbrack\because mB = 2W\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.