(B) The work done in rotating a magnetic dipole in a uniform magnetic field is given by $W = MB(1 - \cos \theta)$.Given $\theta = 60^{\circ}$, we have

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(B) The work done in rotating a magnetic dipole in a uniform magnetic field is given by $W = MB(1 - \cos \theta)$.Given $\theta = 60^{\circ}$, we have:$W = MB(1 - \cos 60^{\circ}) = MB(1 - 0.5) = \frac{MB}{2}$.Therefore, $MB = 2W$.The torque $\tau$ on the magnetic needle is given by $\tau = MB \sin \theta$.Substituting the values, $\tau = (2W) \sin 60^{\circ} = 2W \times \frac{\sqrt{3}}{2} = \sqrt{3} W$.

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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$A$ magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of $60^{\circ}$ and the work done is $W$. The torque on the magnetic needle at this position is

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A
$2 \sqrt{3} W$
B
$\sqrt{3} W$
C
$\frac{\sqrt{3}}{2} W$
D
$\frac{\sqrt{3}}{4} W$

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Magnetism and Matter

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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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$A$ magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of $60^{\circ}$ and the work done is $W$. The torque on the magnetic needle at this position is

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