Question

A bar magnet is held at right angles 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)

$60^{\circ}$

Answer 2

$\tau = MB \sin\,\theta$
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 is = $90^\circ-30^{\circ}=60^{\circ}$