Q. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is
A) $\frac{W}{\sqrt{3}}$
B) $\sqrt{3}W$
C) $\frac{\sqrt{3}}{2}W$
D) $\frac{2W}{\sqrt{3}}$
Explanation
[Solution] $W = MB(1 - \cos 60^\circ) = MB/2$; $\tau = MB\sin 60^\circ = \sqrt{3}W$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.