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Q.

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 600 is W. Now the torque required to keep the magnet in this new position is

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a

W3

b

3W

c

3W2

d

2W3

answer is B.

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Detailed Solution

At equilibrium, potential energy of dipole,Ui=MBHFinal potential energy of dipole,Uf=-MBHcos60°=-MBH2W=Uf-Ui=-MBH2--MBH=MBH2………..(2) Required torque, τ=MBHsin60°τ=2W×32  Using eqn. (2) =3W
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