A bar magnet of moment 2 A-m2 is free to rotate about a Vertical axis passing through its centre. The magnet is released form rest from east west direction. The K.E of the magnet as it takes north-south direction is (BH = 25 × 10–6 T)
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a
25 × 10–6 J
b
50 × 10–6 J
c
75 × 10–6 J
d
100 × 10–6 J
answer is B.
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Detailed Solution
∆KE=∆PE=MB(1-cosθ) =2 x 25 x 10-6(1-0)=50 x 10-6 J