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Q.

A bar magnet of magnetic moment 5 Am2 is suspended in a uniform magnetic field of strength 2T making an angle 600 with the direction of the field. The potential energy of the magnet in that position is

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a

5 J

b

–5 J

c

10 J

d

–10 J

answer is B.

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Detailed Solution

U=−MBcos⁡θ=−5×2×12=−5J

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