A bar magnet held perpendicular to a uniform magnetic field experiences a torque \( \tau_1 \). By what angle should it be rotated to change the experienced torque to \( 0.5\,\tau_1 \)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \pi/3 \)
The torque on a bar magnet of moment \( m \) in a uniform field \( B \) is
\( \displaystyle \tau = mB\sin\theta \)
where \( \theta \) is the angle between the magnet's axis and the field.
Held perpendicular to the field, \( \theta = 90^\circ \):
\( \displaystyle \tau_1 = mB\sin90^\circ = mB \)
For the torque to become \( 0.5\,\tau_1 \):
\( \displaystyle mB\sin\theta = 0.5\,mB \;\Rightarrow\; \sin\theta = 0.5 \;\Rightarrow\; \theta = 30^\circ \)
The magnet must turn from \( 90^\circ \) to \( 30^\circ \), so the angle of rotation is
\( \displaystyle 90^\circ - 30^\circ = 60^\circ = \frac{\pi}{3} \)
(Turning the other way, to \( 150^\circ \), where \( \sin\theta \) is again 0.5, is also a rotation of \( 60^\circ \).)