A short bar magnet of magnetic moment m = 0.32 JT –1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Which orientation would correspond to its stable equilibrium?
Stable equilibrium occurs when the magnetic moment m is parallel to the magnetic field B (θ = 0°). The potential energy is U = –mB = – = –⁻² J.
Which orientation would correspond to its unstable equilibrium?
Unstable equilibrium occurs when the magnetic moment m is anti-parallel to the magnetic field B (θ = 180°). The potential energy is U = +mB = + = +⁻² J.
Explanation
According to the textbook context, the potential energy of a bar magnet in a uniform magnetic field is given by U = –m·B = –mB cos θ. Equation (5.3) explicitly states that potential energy is minimum (= –mB) at θ = 0° (most stable position) and maximum (= +mB) at θ = 180° (most unstable position). At θ = 0°, the magnetic moment aligns parallel to the field, resulting in minimum potential energy and stable equilibrium. At θ = 180°, the magnetic moment is anti-parallel to the field, resulting in maximum potential energy and unstable equilibrium.