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 $=90^{\circ }-30^{\circ }=60^{\circ }$