The potential energy of a magnetic dipole in its most stable position in a uniform magnetic field is \(-0.12\ \mathrm{J}\). The magnitude of the torque on the dipole when its magnetic moment is perpendicular to the field is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(0.12\ \mathrm{N\,m}\)
In a uniform field, a magnetic dipole has potential energy \( U=-mB\cos\theta \). It is least at \( \theta=0^\circ \), the most stable position, where \( U=-mB \). The torque on the dipole is \( \tau=mB\sin\theta \).
\( \begin{aligned} -mB &= -0.12\ \mathrm{J} \\ mB &= 0.12\ \mathrm{J} \end{aligned} \)
When the moment is perpendicular to the field, \( \theta=90^\circ \):
\( \begin{aligned} \tau &= mB\sin90^\circ \\ &= 0.12\times1 \\ &= 0.12\ \mathrm{N\,m} \end{aligned} \)
NCERT (2026-27):
The dipole in a uniform magnetic field – “… potential energy is minimum (= –mB) at θ = 0° (most stable position) …”
The dipole in a uniform magnetic field – “… In magnitude τ = mB sinθ Here τ is restoring torque and θ is the angle between m and B.”