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 as `B=25muT`. Earth's magnetic field is from south to north.
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Gaint in kinetic energy=loss in potential energy `Thus " " KE=U_(i)-U_(f)` `therefore " " KE=-MB "cos" (pi)/(2)=(-MB cos theta^(@))=MB` Substituting the values, we have, ` " " KE=(2.0)(25xx10^(-6))=50muJ`
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