A bar magnet is held in a uniform magnetic field. In one position, the torque on it is \(0.25\ \mathrm{N\,m}\) and its potential energy is \(-0.25\ \mathrm{J}\). The zero of potential energy is taken when the magnet is perpendicular to the field. The angle between the magnetic moment and the field is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(45^\circ\)
A magnet in a uniform field feels a torque \( \tau=mB\sin\theta \). Its potential energy is \( U=-mB\cos\theta \). Here \( \theta \) is the angle between \( \mathbf{m} \) and \( \mathbf{B} \). Dividing the torque by \( -U \) removes \( mB \) and leaves only the angle.
Given:
\( \tau=0.25\ \mathrm{N\,m} \)
\( U=-0.25\ \mathrm{J} \)
\( \begin{aligned} \tan\theta &= \frac{mB\sin\theta}{mB\cos\theta} \\ &= \frac{\tau}{-U} \\ &= \frac{0.25}{0.25} \\ &= 1 \end{aligned} \)
The potential energy is negative, so \( \cos\theta \) is positive. The angle is therefore less than \( 90^\circ \).
\( \theta=45^\circ \)
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”
“The magnetic potential energy Um is given by”
\( U_m=\int \tau(\theta)\,d\theta=\int mB\sin\theta\,d\theta= \)\( -mB\cos\theta \)\( =-\mathbf{m}\cdot\mathbf{B} \qquad (5.3) \)