A small coil of N turns has an effective area | vedclass

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(B) The magnetic moment of the coil is $M = NIA$.Initially,the plane of the coil is perpendicular to the magnetic field $\overrightarrow{B}$,which mea

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$2NAIB$

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(B) The magnetic moment of the coil is $M = NIA$.Initially,the plane of the coil is perpendicular to the magnetic field $\overrightarrow{B}$,which means the area vector $\overrightarrow{A}$ is parallel to $\overrightarrow{B}$. Thus,the initial angle $	heta_1 = 0^\circ$.When the coil is rotated by $180^\circ$ about the vertical axis,the area vector $\overrightarrow{A}$ becomes anti-parallel to $\overrightarrow{B}$,so the final angle $	heta_2 = 180^\circ$.The work done in rotating a magnetic dipole in a magnetic field is given by $W = MB(\cos 	heta_1 - \cos 	heta_2)$.Substituting the values: $W = (NIA)B(\cos 0^\circ - \cos 180^\circ)$.Since $\cos 0^\circ = 1$ and $\cos 180^\circ = -1$,we get $W = (NIA)B(1 - (-1)) = (NIA)B(2) = 2NAIB$.

$A$ small coil of $N$ turns has an effective area $A$ and carries a current $I$. It is suspended in a horizontal magnetic field $\overrightarrow{B}$ such that its plane is perpendicular to $\overrightarrow{B}$. The work done in rotating it by $180^\circ$ about the vertical axis is

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