Question
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?
Solution
(a) For stable equilibrium: Magnetic moment is parallel to .
Then,
Potential energy,
= – 4.8 x 10–2 J
(b) For unstable equilibrium : Magnetic moment is antiparallel to .
Then, θ = 180°
Potential energy ,U’ = - MB cos
=- 0.32 x 0.15 (-1)
= 4.8 x 10–2 J.
Then,
Potential energy,
= – 4.8 x 10–2 J
(b) For unstable equilibrium : Magnetic moment is antiparallel to .
Then, θ = 180°
Potential energy ,U’ = - MB cos
=- 0.32 x 0.15 (-1)
= 4.8 x 10–2 J.
