(D) At equilibrium,the initial potential energy of the dipole is $U_{i} = -MB_{H} \cos 0^{\circ} = -MB_{H}$.The final potential energy of the dipole a

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(D) At equilibrium,the initial potential energy of the dipole is $U_{i} = -MB_{H} \cos 0^{\circ} = -MB_{H}$.The final potential energy of the dipole after rotating by $60^{\circ}$ is $U_{f} = -MB_{H} \cos 60^{\circ} = -\frac{MB_{H}}{2}$.The work done $W$ is the change in potential energy: $W = U_{f} - U_{i} = -\frac{MB_{H}}{2} - (-MB_{H}) = \frac{MB_{H}}{2}$.From this,we get $MB_{H} = 2W$.The torque required to keep the magnet in this new position is $\tau = MB_{H} \sin 60^{\circ}$.Substituting $MB_{H} = 2W$,we get $\tau = (2W) \times \frac{\sqrt{3}}{2} = \sqrt{3} W$.

Class 12 Physics Magnetism and Matter Magnetic field due to magnetic dipole and Dipole in Magnetic Field and Poential Energy and Work Done

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$A$ 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 $60^{\circ}$ is $W$. Now,the torque required to keep the magnet in this new position is

NEET 2016 All NEET

A
$\frac{\sqrt{3} W}{2}$
B
$\frac{2W}{\sqrt{3}}$
C
$\frac{W}{\sqrt{3}}$
D
$\sqrt{3} W$

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Magnetic field due to magnetic dipole and Dipole in Magnetic Field and Poential Energy and Work Done

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$A$ 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 $60^{\circ}$ is $W$. Now,the torque required to keep the magnet in this new position is

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