First slide
Magnetism
Question

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

Moderate
Solution

    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 = UiUf

KE = [-MBcos90o] - [-MBcos0o]

KE = MB = 2×25 = 50μJ

                                                

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