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A bar magnet of magnetic moment 2Am² 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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25µJ

50µJ

100µJ

12.5µJ

answer is B.

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Detailed Solution

$W = MB (\cos θ_{1} - \cos θ_{2}), h e r e θ_{1} = 0^{0} a n d θ_{2} = 90^{0} △ U = - W a n d △ K + △ U = 0$

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