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)

$ \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} $