A bar magnet is suspended by a string in a uniform magnetic field making an angle of 60o with the field. Torque required to hold the magnet in that position is 25×10−3 N−m. Then the work done by an external agent to rotate the magnet clockwise through an angle of 90o is (Take 3=1.7)

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1.50 J

zero

0.25 J

0.01 J

answer is D.

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Detailed Solution

τ=MBsin⁡60o⇒MB=2τ3Now Ui=−MBcos⁡60o=−τ3 and Uf=−MB(90o−60o)=−τ∴ W=Uf−Ui==−τ−(−τ3)=−(3−13)τ=0.01 J

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