$A$ short bar magnet of magnetic moment $M=0.32 \; J \, T^{-1}$ 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?

(N/A) Given:
Magnetic moment,$M = 0.32 \; J \, T^{-1}$
Magnetic field,$B = 0.15 \; T$
$(a)$ Stable equilibrium occurs when the magnetic moment is aligned with the magnetic field $(\theta = 0^{\circ})$.
Potential energy $U = -M B \cos \theta = -0.32 \times 0.15 \times \cos(0^{\circ}) = -0.048 \; J = -4.8 \times 10^{-2} \; J$.
$(b)$ Unstable equilibrium occurs when the magnetic moment is aligned opposite to the magnetic field $(\theta = 180^{\circ})$.
Potential energy $U = -M B \cos \theta = -0.32 \times 0.15 \times \cos(180^{\circ}) = -0.048 \times (-1) = 0.048 \; J = 4.8 \times 10^{-2} \; J$.