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Q.

The bar magnet of magnetic moment 2.0Am2 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=25μT. Earth’s magnetic field is from south to north__________ μJ

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answer is 50.

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Detailed Solution

Gain in kinetic energy=Loss in potential energy
Thus, KE=UiUfAs,U=MB cosθ Initially, θ1=π2( for east  west direction )
And finally, θf=0( for north  south direction )
KE=MBcosπ2MBcos0=0+MB=MB Substituting the values, we have KE=2(2.0)25×106J=50μJ

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