The torque acting on a magnet of magnetic length 8 cm and pole strength 20 A m placed in a magnetic field of $3 × 10^{-5} T$ at an angle of π/6 rad will be

$2.4 × 10^{-5} N m$

The correct answer is Option (2) → $2.4 × 10^{-5} N m$

Given:

Magnetic length $l = 8\ \text{cm} = 0.08\ \text{m}$

Pole strength $m = 20\ \text{A·m}$

Magnetic field $B = 3\times10^{-5}\ \text{T}$

Angle $ \theta = \frac{\pi}{6} $

Formula:

$ \tau = M B \sin\theta $

where $M = m \times l$ is the magnetic dipole moment.

Substitute:

$ M = 20 \times 0.08 = 1.6\ \text{A·m}^2 $

$ \tau = 1.6 \times 3\times10^{-5} \times \sin\frac{\pi}{6} $

$ \tau = 1.6 \times 3\times10^{-5} \times \frac{1}{2} $

$ \tau = 2.4\times10^{-5}\ \text{N·m}$

Therefore, the torque is $ \tau = 2.4\times10^{-5}\ \text{N·m} $.