A bar magnet of magnetic moment  2Am2 is free to rotate about a vertical axis passing through  its centre. The magnet is released from rest from east-west position. Then the K.E of  the magnet as it takes North-South position is ----------- Take :  BH=25μT

The amount of work done by magnetic field is converted into rotational kinetic energy of bar magnet.                                                  Initial angle, θ1=90°Final Angle , θ2=0°  Kinetic energy = Work done by Magnetic field ⇒KE = Ui−Uf⇒KE = [-MBcos90o] - [-MBcos0o]⇒KE = MB = 2×25 = 50μJ