Question

A bar magnet of magnetic dipole moment `10 Am^(2)` is in stable equilibrium . If it is rotated through `30^(@)` , find its potential energy in new position [Given , B = 0.2 T]

Answer 1

`M = 10 A m^(2)`
B = 0.2 T
U = `- MB cos theta`
= `-10 xx 0.2 xx (sqrt3)/(2)`
`= - sqrt3 J`