Q.

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 25×103Nm. 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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a

zero 

b

0.01 J

c

1.50 J

d

0.25 J

answer is D.

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Detailed Solution

τ=MBsin60oMB=2τ3

Now Ui=MBcos60o=τ3andUf=MB(90o60o)=τ

W=UfUi==τ(τ3)=(313)τ=0.01J

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