Question

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 $60^{\circ}$ 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. $\frac{\sqrt{3} W}{2}$
• D. $\frac{2W}{\sqrt{3}}$

Answer 1

Answer: (B)

$\sqrt{3} W$

Answer 2

$W = MB \left(\cos 0^{\circ} - \cos60^{\circ}\right)$
$W = MB \left(1- \frac{1}{2}\right) = \frac{MB}{2}$
Required torque for this position
$\tau = MB \sin\theta$
$= MB \sin60^{\circ}$
$= \frac{\sqrt{3}}{2} MB = \sqrt{3} W$