A loop in the form of four connected semi-cir | vedclass

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(B) The loop consists of four semi-circles of radius $r = a/2$. The total area $A$ enclosed by the loop can be calculated by considering the four semi

Vedclass · Accepted Answer

$\left( \frac{\pi}{2} + 1 \right) a^2 I \hat{k}$

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(B) The loop consists of four semi-circles of radius $r = a/2$. The total area $A$ enclosed by the loop can be calculated by considering the four semi-circles and the central square of side $a$.The area of four semi-circles of radius $r = a/2$ is $4 	imes \frac{1}{2} \pi r^2 = 2 \pi (a/2)^2 = 2 \pi (a^2/4) = \frac{\pi a^2}{2}$.The central square has an area of $a^2$.Total area $A = \frac{\pi a^2}{2} + a^2 = \left( \frac{\pi}{2} + 1 ight) a^2$.Using the right-hand rule for the current direction shown in the figure (counter-clockwise),the magnetic moment vector $\vec{M} = I \vec{A}$ points in the positive $z$-direction.Therefore,$\vec{M} = \left( \frac{\pi}{2} + 1 ight) I a^2 \hat{k}$.

$A$ loop in the form of four connected semi-circular wires carrying current $I$ lies in the $x-y$ plane as shown in the figure. The unit vector $\hat{k}$ is coming out of the plane of the paper. The magnetic moment of the current loop is

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