A short bar magnet is placed in a uniform magnetic field. The torque on it is \(0.15\ \mathrm{N\,m}\) when its axis makes an angle of \(37^\circ\) with the field. What is the torque on it when its axis makes an angle of \(53^\circ\) with the field? (Take \(\sin37^\circ=0.6\) and \(\sin53^\circ=0.8\).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(0.20\ \mathrm{N\,m}\)
A magnet at an angle \( \theta \) to a uniform field feels a torque \( \tau=mB\sin\theta \). Here \( \theta \) is the angle between the magnetic moment and the field. The magnet and the field stay the same, so \( mB \) does not change. We first find \( mB \) from the first position.
Given:
\( \tau_1=0.15\ \mathrm{N\,m} \)
\( \theta_1=37^\circ \)
\( \theta_2=53^\circ \)
\( \begin{aligned} mB &= \frac{\tau_1}{\sin\theta_1} \\ &= \frac{0.15}{0.6} \\ &= 0.25\ \mathrm{N\,m} \end{aligned} \)
\( \begin{aligned} \tau_2 &= mB\sin\theta_2 \\ &= 0.25\times0.8 \\ &= 0.20\ \mathrm{N\,m} \end{aligned} \)
NCERT (2026-27):
The dipole in a uniform magnetic field – “The torque on the needle is …”
\( \boldsymbol{\tau}=\mathbf{m}\times\mathbf{B} \qquad (5.2) \)
“In magnitude τ = mB sinθ”
“Here τ is restoring torque and θ is the angle between m and B.”