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(C) The magnetic moment $\vec \mu$ of a current-carrying loop is given by $\vec \mu = i \vec A$,where $i$ is the current and $\vec A$ is the area vect

Vedclass · Accepted Answer

$iab(\sin 30^\circ \hat i + \cos 30^\circ \hat k)$

Vedclass · Answer

(C) The magnetic moment $\vec \mu$ of a current-carrying loop is given by $\vec \mu = i \vec A$,where $i$ is the current and $\vec A$ is the area vector.The area vector $\vec A$ has a magnitude equal to the area of the loop $(A = a 	imes b)$ and its direction is perpendicular to the plane of the loop.The loop lies in a plane that makes an angle of $30^\circ$ with the $xy$-plane,or more specifically,the normal to the loop makes an angle with the axes.Looking at the geometry,the side of length $b$ is along the $y$-axis. The side of length $a$ makes an angle of $30^\circ$ with the $x$-axis in the $xz$-plane.The normal vector $\vec n$ to the loop can be found by taking the cross product of the two vectors representing the sides of the rectangle.Vector along side $b$ is $\vec b = b \hat j$.Vector along side $a$ is $\vec a = a \cos 30^\circ \hat i + a \sin 30^\circ \hat k$.The area vector $\vec A = \vec a 	imes \vec b = (a \cos 30^\circ \hat i + a \sin 30^\circ \hat k) 	imes (b \hat j)$.Using the cross product rules ($\hat i 	imes \hat j = \hat k$ and $\hat k 	imes \hat j = -\hat i$):$\vec A = ab \cos 30^\circ \hat k - ab \sin 30^\circ \hat i$.Wait,checking the orientation: The magnetic moment direction is determined by the right-hand rule. For the given current direction,the normal vector points in the direction of $(\sin 30^\circ \hat i + \cos 30^\circ \hat k)$.Thus,$\vec \mu = iab(\sin 30^\circ \hat i + \cos 30^\circ \hat k)$.

Current $i$ is flowing in the rectangular loop placed in the $xyz$ plane as shown in the figure. Find the magnetic moment of the loop.

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