Question

A bar magnet is held at a right angle to a uniform magnetic field. The couple acting on the magnet is to be halved by rotating it from this position. The angle of rotation is

• A. $60^\circ $
• B. $45^\circ $
• C. $30^\circ $
• D. $75^\circ $

Answer 1

Answer: (A)

Torque, $\tau=MBsin \theta $
Where $\theta =90^\circ $
$\therefore \tau=MB$
Given, $\tau_{2}=\frac{1}{2 \, }\tau_{1}$
$\therefore MB \, sin \theta =\frac{1}{2} \, MB$
$\therefore sin \theta = \frac{1}{2} \, $
$\Rightarrow \theta =30^\circ $
Angle of rotation $=90^\circ -30^\circ =60^\circ $