There exists a horizontal magnetic field in space of value \( B = 9\times10^{-4}\,\mathrm{T} \). A magnetic bar of magnetic moment \( 9\times10^{3}\,\mathrm{J/T} \) is placed at an angle \( 60^\circ \) to the field and is free to rotate. Finally, it comes to a position parallel to the field. Work done by magnetic field on the magnetic bar will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( -4.05\,\mathrm{J} \)
A magnet of moment \( m \) in a uniform field \( B \) has potential energy \( U = -mB\cos\theta \). The textbook result for the work done in rotating a magnetic dipole from angle \( \theta_1 \) to angle \( \theta_2 \) with the field is
\( \displaystyle W = mB\,(\cos\theta_1 - \cos\theta_2) \)
\( m = 9\times10^{3}\,\mathrm{J\,T^{-1}} \)
\( B = 9\times10^{-4}\,\mathrm{T} \)
\( \theta_1 = 60^\circ,\ \theta_2 = 0^\circ \)
\( \displaystyle mB = 9\times10^{3} \times 9\times10^{-4} = 8.1\,\mathrm{J} \)
\( \displaystyle W = 8.1\,(\cos60^\circ - \cos0^\circ) = 8.1\,(0.5 - 1) = -4.05\,\mathrm{J} \)
The magnet turns to the position parallel to the field, which is its position of lowest potential energy (stable equilibrium). Its potential energy falls:
\( U_1 = -mB\cos60^\circ = -8.1 \times 0.5 = -4.05\,\mathrm{J} \)
\( U_2 = -mB\cos0^\circ = -8.1\,\mathrm{J} \)
\( \Delta U = U_2 - U_1 = -8.1 - (-4.05) = -4.05\,\mathrm{J} \)
This change of \( -4.05\,\mathrm{J} \) is the value given by \( W = mB(\cos\theta_1 - \cos\theta_2) \).