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A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in an 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

W3

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B

32W

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C

3W

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D

2W3

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Solution

The correct option is C

3W


Step 1: Given Data

Torque on a bar magnet placed in a uniform magnetic field

Since the bar magnet is first in equilibrium, the initial angle is θ1=0°

Given, the final angle of rotation θ2=60°

Work done for this rotation =W

Let the magnetic moment be m and the magnetic field be B

Step 2: Calculate the Work Done

We know that work done on a magnetic field is given as,

W=mBcosθ1-cosθ2

=mBcos0°-cos60°

=mB1-12

=mB2

Step 3: Calculate the Torque

We know that torque is given as,

τ=mBsinθ1

=mBsin60°

=mB32

=3W

Hence, the correct answer is option (C).


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