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Q.

The work done in rotating the magnet from the direction of the uniform field by 600 is W. The potential energy of the magnet at the position where it experiences half the maximum couple position is

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a

−3W

b

32W

c

W/2

answer is D.

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Detailed Solution

detailed_solution_thumbnail

W=MB(1−cos60)=MB2
τ=MBsinθ
MB2=MBsinθ1⇒θ1=30
And U = -M.B = -MB cos θ1
⇒U=−3W

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