A bar magnet of magnetic moment `2.0 A m^2` is free to rotate about a vertical axis through its centre. The magnet is released from rest from the east-west position. Find the kinetic energy of the magnet as it takes the north-south position. The horizontal component of the earth's magnetic field is `B = 25 muT`.

Explanation

The magnetic potential energy of the dipole in a uniform magnetic field is given by `U = - MB costheta`. As the earth's magnetic field is from south to north, the initial value of `theta` is `pi/2` and final value of `theta` is `0`. Thus, the decrease in magnetic potential energy during the rotation is `U_i - U_f = - MB cos (pi/2) + MB cos 0`
`= 2.0 A m^2 xx 25 muT = 50 muJ`.
Thus, the kinetic energy in the north-south position is `50 muJ`.