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# Give answer! 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 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

• Option 1)

• Option 2)

• Option 3)

• Option 4)

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As we learnt in

Potential Energy -

$U= -MB\cos \Theta =-\vec{M}\cdot \vec{B}$

- wherein

$\Theta$ - angle made by the dipole with the field

and

Torque -

$\vec{\tau }= \vec{M}\times \vec{B}$

$\tau = MB\sin \Theta$

- wherein

At Equilibrium

U = -MBH

Final Potential energy of dipole

U=-MBCos 60° =$\frac{-MB_{H}}{2}$

$W= U_{f}-U_{i}=\frac{-MB_{H}}{2}-(MB_{H})=\frac{MB_{H}}{2}$

Required torque $T= MB_{H}\sin 60^{\circ} = 2W \times \frac{\sqrt{3}}{2}$

$T=\sqrt{3}W$

Option 1)

Incorrect option

Option 2)

Correct option

Option 3)

Incorrect option

Option 4)

Incorrect option

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