A short bar magnet of magnetic moment \( m=0.45\,\mathrm{J/T} \) is placed in a uniform magnetic field of 0.22 T. If the bar is free to rotate in the plane of the field, what is the potential energy of the magnet in stable and unstable equilibrium respectively?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 9.9\times10^{-2}\,\mathrm{J} \) (Unstable); \( -9.9\times10^{-2}\,\mathrm{J} \) (Stable)
The potential energy of a magnetic dipole in a uniform field is \( U = -mB\cos\theta \), where \( \theta \) is the angle between the magnetic moment and the field.
Stable equilibrium is the position of minimum energy, \( \theta = 0^\circ \) (moment along the field).
\( U_{\mathrm{stable}} = -mB = -(0.45)(0.22)\,\mathrm{J} = -9.9\times10^{-2}\,\mathrm{J} \)
Unstable equilibrium is the position of maximum energy, \( \theta = 180^\circ \) (moment opposite to the field).
\( U_{\mathrm{unstable}} = +mB = +9.9\times10^{-2}\,\mathrm{J} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
