Question

Question 5.4: 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?

Class 12 - Physics - Magnetism And Matter Magnetism And Matter Page-200

Answer 1

(a) An equilibrium state is stable state , if on disturbing the magnet , it comes back to same initial state . so bar magnet is in stable equilibrium at $\theta=0^{\circ}$
potential energy , $U=-\vec{M}.\vec{B}=-MBcos0^{\circ}$
here, M is the magnetic moment.e.g., M = 0.32J/T
B is the magnetic field.e.g., B=0.15T
so, U = - 0.32 × 0.15 = -0.048J

(b)A bar magnet is unstable equilibrium, if on disturbing from its position, it further gets disturb and do not come back to previous position of equilibrium.
at $\theta=180^{\circ}$ the equilibrium is unstable.
so, potential energy, U = -MBcos180°
U = MB = 0.32 × 0.15 = 0.048 J