A bar magnet held perpendicular to a uniform magnetic field experiences a torque $τ_1$. By what angle should it be rotated to change the experienced torque to $0.5 τ_1$?

π/3

The correct answer is Option (3) → π/3

Torque on a magnetic dipole in a uniform magnetic field:

$\tau = m B \sin\theta$

Initially: $\tau_1 = m B \sin 90^\circ = m B$

Required torque: $\tau_2 = 0.5 \tau_1 = 0.5 m B$

So, $\tau_2 = m B \sin\theta = 0.5 m B \Rightarrow \sin\theta = 0.5$

⇒ $\theta = \pi/6$ (angle with the magnetic field)

Initially, the magnet was perpendicular to the field ($\theta_0 = \pi/2$):

Therefore, Rotation angle = $\theta_0 - \theta = \pi/2 - \pi/6 = \pi/3$

∴ The bar magnet should be rotated by π/3 to reduce the torque to 0.5τ1