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Question

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 60 is W. Now the torque required to keep the magnet in this new position is

A
2W3
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B
3W2
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C
3W
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D
W3
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Solution

The correct option is C 3W
Work done by the earth's magnetic field to rotate a bar magnet is,

W=MBH(cosθ1cosθ2)

W=MBH(1cos60) [ θ1=0 ; θ2=60]

W=MBH(112)

W=MBH2

MBH=2W

The required torque to keep the magnet in equilibrium is,

τ=MBHsin600

=2W×32

=3W

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (C) is the correct answer.

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