A magnetic needle suspended parallel to a mag | vedclass

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(D) The work done $W$ in rotating a magnetic needle in a magnetic field is given by $W = MB(\cos 	heta_{1} - \cos 	heta_{2})$.Given $	heta_{1} = 0^

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$3 	ext{ J}$

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(D) The work done $W$ in rotating a magnetic needle in a magnetic field is given by $W = MB(\cos 	heta_{1} - \cos 	heta_{2})$.Given $	heta_{1} = 0^{\circ}$ and $	heta_{2} = 60^{\circ}$,the work done is $\sqrt{3} 	ext{ J}$.Substituting the values: $\sqrt{3} = MB(\cos 0^{\circ} - \cos 60^{\circ})$.$\sqrt{3} = MB(1 - 0.5) = MB(0.5)$.Therefore,$MB = 2\sqrt{3} 	ext{ J}$.The torque $	au$ required to maintain the needle at an angle $	heta = 60^{\circ}$ is given by $	au = MB \sin 	heta$.$	au = (2\sqrt{3}) 	imes \sin 60^{\circ} = 2\sqrt{3} 	imes \frac{\sqrt{3}}{2} = 3 	ext{ N}\,	ext{m}$ (or $3 	ext{ J}$ as per the options provided).

$A$ magnetic needle suspended parallel to a magnetic field requires $\sqrt{3} \text{ J}$ of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be

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