A short bar magnet placed with its axis at `30^@` to a uniform magnetic field of 0.2 T experiences a torque of 0.060 Nm.
i. Calculate magnetic moment of the magnet
ii. Find out what orientation of the magnet corresponds to its stable equilibrium in the magnetic field.

Explanation

Here `tau =0.060 Nm, theta = 30^@ , B = 0.2 T`
As `tau =m B sin theta " " 0.060 = mxx 0.2 xx sin 30^@`
`therefore m = (0.060)/(0.2 xx 0.5) = 0.06JT^(-1)`
ii PE of magnet in field `vec B` is given by
U=-m cos `theta`
When magnet is paced parallel to `vec B , theta = 0^@` and
`U = mB cos 0^(@)`=- mB
i.e , the PE of the dipole is minimum. This is the state of stable equilibrium.