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A bar magnet of magnetic moment 2 Am² is free to rotate about a vertical axis passing through its center. The magnet is released from rest from east - west position. Then the KE of the magnet as it takes N-S position is $(B_{H} = 25 μT)$

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12.5 $μJ$

50 $μJ$

100 $μJ$

25 $μJ$

answer is B.

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Detailed Solution

$K . E = W = MB (\cos θ_{1} - \cos θ_{2})$

$K . E = 2 \times 25 \times 10^{- 6} = 50 μ J$

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