Question

A bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it, then the angle by which it is to be rotated is:

• A. $30^{\circ}$
• B. $45^{\circ}$
• C. $60^{\circ}$
• D. $90^{\circ}$

Answer 1

Answer: (C)

$\tau= MB \sin \theta$
$\tau \propto \sin \theta$
$\frac{\tau_{1}}{\tau_{2}}=\frac{\sin \theta_{1}}{\sin \theta_{2}}$
(couple considered as torque)
$\frac{\tau_{1}}{\tau_{2}}=\frac{\sin 90^{\circ}}{\sin \theta_{2}}$
$\sin \theta_{2}=\frac{\tau_{2}}{\tau_{1}}$
$\sin \theta_{2}=\frac{1}{2} $
$\Rightarrow \theta_{2}=30^{\circ}$
angle of rotation $=90^{\circ}-30^{\circ}=60^{\circ}$