A circular coil having N turns and radius r c | vedclass

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Question

(D) The magnetic moment $\vec{M}$ of a coil with $N$ turns,area $A = \pi r^2$,and current $I$ is given by $\vec{M} = NIA \hat{n}$,where $\hat{n}$ is t

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$B \pi r^2 I N$

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(D) The magnetic moment $\vec{M}$ of a coil with $N$ turns,area $A = \pi r^2$,and current $I$ is given by $\vec{M} = NIA \hat{n}$,where $\hat{n}$ is the unit vector normal to the plane of the coil.Since the coil is in the $XZ$ plane,its normal vector is along the $Y$-axis,so $\vec{M} = NI(\pi r^2) \hat{j}$.The magnetic field is $\vec{B} = B \hat{i}$.The torque $\vec{	au}$ on the coil is given by $\vec{	au} = \vec{M} 	imes \vec{B}$.Substituting the values: $\vec{	au} = (NI \pi r^2 \hat{j}) 	imes (B \hat{i})$.Using the cross product rule $\hat{j} 	imes \hat{i} = -\hat{k}$,we get $\vec{	au} = -NI \pi r^2 B \hat{k}$.The magnitude of the torque is $|\vec{	au}| = NI \pi r^2 B$.

$A$ circular coil having $N$ turns and radius $r$ carries a current $I$. It is held in the $XZ$ plane in a magnetic field $\vec{B} = B\hat{i}$. The torque on the coil due to the magnetic field is

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