Question

# 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

$\frac{W}{\sqrt{3}}$

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B

$\frac{\sqrt{3}}{2}W$

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C

$\sqrt{3}W$

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D

$\frac{2W}{\sqrt{3}}$

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Solution

## The correct option is C $\sqrt{3}W$Step 1: Given DataSince the bar magnet is first in equilibrium, the initial angle is ${\theta }_{1}=0°$Given, the final angle of rotation ${\theta }_{2}=60°$Work done for this rotation $=W$Let the magnetic moment be $m$ and the magnetic field be $B$Step 2: Calculate the Work DoneWe know that work done on a magnetic field is given as,$W=mB\left(\mathrm{cos}{\theta }_{1}-\mathrm{cos}{\theta }_{2}\right)$ $=mB\left(\mathrm{cos}0°-\mathrm{cos}60°\right)$ $=mB\left(1-\frac{1}{2}\right)$ $=\frac{mB}{2}$Step 3: Calculate the TorqueWe know that torque is given as,$\tau =mB\mathrm{sin}{\theta }_{1}$ $=mB\mathrm{sin}60°$ $=mB\frac{\sqrt{3}}{2}$ $=\sqrt{3}W$Hence, the correct answer is option (C).

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