The torque acting on a magnet of magnetic length 8 cm and pole strength 20 A m placed in a magnetic field of \( 3 \times 10^{-5}\,\mathrm{T} \) at an angle of \( \pi/6 \) rad will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2.4 \times 10^{-5}\,\mathrm{N\,m} \)
The magnetic moment of a bar magnet is pole strength times magnetic length, and the torque in a uniform field is \( \tau = mB\sin\theta \).
\( \displaystyle m = q_m \times 2l = 20\,\mathrm{A\,m} \times 0.08\,\mathrm{m} = 1.6\,\mathrm{A\,m^2} \)
\( B = 3\times10^{-5}\,\mathrm{T} \)
\( \theta = \pi/6 = 30^\circ,\ \sin 30^\circ = 0.5 \)
\( \displaystyle \tau = 1.6 \times 3\times10^{-5} \times 0.5 = 2.4\times10^{-5}\,\mathrm{N\,m} \)