A magnetic needle lying parallel to a magneti | vedclass

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(A) The work done $W$ in rotating a magnetic dipole of magnetic moment $M$ in a magnetic field $B$ from an angle $	heta_1 = 0^{\circ}$ to $	heta_2 =

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$\sqrt{3} W$

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(A) The work done $W$ in rotating a magnetic dipole of magnetic moment $M$ in a magnetic field $B$ from an angle $	heta_1 = 0^{\circ}$ to $	heta_2 = 60^{\circ}$ is given by:$W = MB(1 - \cos 60^{\circ}) = MB(1 - 0.5) = \frac{MB}{2}$.From this, we find $MB = 2W$.The torque $	au$ required to maintain the needle at $	heta = 60^{\circ}$ is given by:$	au = MB \sin 60^{\circ}$.Substituting $MB = 2W$ and $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$:$	au = (2W) 	imes \frac{\sqrt{3}}{2} = \sqrt{3} W$.

$A$ magnetic needle lying parallel to a magnetic field is turned through $60^{\circ}$. The work done on it is $W$. The torque required to maintain the magnetic needle in the position mentioned above is

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