Question

A short bar magnet of magnetic moment m = 0.32 JT–1 is placed in auniform 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 themagnet in each case?

Answer 1

Given: The magnetic moment of the bar magnet is 0.32  JT −1 and uniform magnetic field is 0.15 T.

The potential energy of the bar magnet is given as,

U=−mBcosθ(1)

Where, U is the potential energy of the bar magnet, m is the magnetic moment, B is the magnetic field and θ is the angle between magnetic field and magnetic moment of the bar magnet.

a)

A bar magnet is most stable when the uniform magnetic field is parallel to the magnetic moment of the bar magnet.

So, the angle between magnetic field and magnetic moment in this case will be 0°.

By substituting the given values in the expression (1), we get

U=−0.32×0.15( cos0° ) =−0.048 J

Thus, the potential energy of bar magnet in stable orientation is −0.048 J.

b)

A bar magnet is in unstable equilibrium when the uniform magnetic field of the bar magnet is antiparallel to the magnetic moment of the bar magnet.

So, the angle between magnetic field and magnetic moment in this case will be 180°.

By substituting the given values in the expression (1), we get

U=−0.32×0.15( cos180° ) =−0.048( −1 ) =0.048 J

Thus, the potential energy of bar magnet in unstable orientation is 0.048 J.