A loop carrying current I lies in the x-y pla | vedclass

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(C) The magnetic moment $\vec{\mu}$ of a current loop is given by $\vec{\mu} = I \vec{A}$,where $\vec{A}$ is the area vector.From the figure,the loop

Vedclass · Accepted Answer

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

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(C) The magnetic moment $\vec{\mu}$ of a current loop is given by $\vec{\mu} = I \vec{A}$,where $\vec{A}$ is the area vector.From the figure,the loop consists of a central square of side $a$ and four semicircles of diameter $a$ (radius $r = a/2$) attached to its sides.The total area $A$ is the sum of the area of the square and the four semicircles:$A = a^2 + 4 	imes \left( \frac{1}{2} \pi r^2 ight) = a^2 + 2 \pi \left( \frac{a}{2} ight)^2 = a^2 + 2 \pi \left( \frac{a^2}{4} ight) = a^2 + \frac{\pi a^2}{2} = a^2 \left( 1 + \frac{\pi}{2} ight)$.Since the current $I$ flows in a clockwise direction in the $x$-$y$ plane,by the right-hand rule,the area vector $\vec{A}$ points into the plane,i.e.,in the $-\hat{k}$ direction.Therefore,the magnetic moment is $\vec{\mu} = I \vec{A} = -I a^2 \left( 1 + \frac{\pi}{2} ight) \hat{k} = -\left( 1 + \frac{\pi}{2} ight) a^2 I \hat{k}$.

$A$ loop 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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