A bar magnet lies along a uniform magnetic field of 0.2 T. An external torque turns it slowly until its magnetic moment points opposite to the field. The work done by the torque is 1.0 J. The magnetic moment of the magnet is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2.5\ \mathrm{J\,T^{-1}}\)
The potential energy of a magnet in a uniform field is \( U=-mB\cos\theta \). When the magnet is turned slowly, the work done by the external torque is equal to the gain in this energy.
\( \begin{aligned} U_i &= -mB\cos0^\circ \\ &= -mB \end{aligned} \)
\( \begin{aligned} U_f &= -mB\cos180^\circ \\ &= +mB \end{aligned} \)
\( \begin{aligned} W &= U_f-U_i \\ &= 2mB \end{aligned} \)
\( \begin{aligned} m &= \frac{W}{2B} \\ &= \frac{1.0}{2\times0.2} \\ &= 2.5\ \mathrm{J\,T^{-1}} \end{aligned} \)
NCERT (2026-27):
The dipole in a uniform magnetic field – “The magnetic potential energy Um is given by”
\( \displaystyle U_m=\int\tau(\theta)\,d\theta=\int mB\sin\theta\,d\theta=-mB\cos\theta=-\mathbf{m}\cdot\mathbf{B} \qquad (5.3) \)
“… Equation (5.3) shows that potential energy is minimum (= –mB) at θ = 0° (most stable position) and maximum (= +mB) at θ = 180° (most unstable position).”