A magnetic needle lying parallel to a magneti | vedclass

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(A) The work done $W$ in rotating a magnetic needle from an angle $	heta_1$ to $	heta_2$ in a magnetic field $B$ is given by $W = MB(\cos 	heta_1 -

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$\sqrt{3} W$

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

$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be

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