A magnetic needle lying parallel to a magneti | vedclass

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

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$\sqrt{3} W$

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(A) The work done in rotating a magnetic dipole in a magnetic field is given by $W = MB(1 - \cos 	heta)$.Given $	heta = 60^{\circ}$,we have $W = MB(1 - \cos 60^{\circ}) = MB(1 - 0.5) = \frac{MB}{2}$.From this,we find $MB = 2W$.The torque required to maintain the needle at an angle $	heta$ is $	au = MB \sin 	heta$.Substituting $	heta = 60^{\circ}$ and $MB = 2W$,we get $	au = (2W) \sin 60^{\circ}$.$	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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