A wire of length L m carrying a current of I | vedclass

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(B) Let the radius of the circle be $r$. The circumference of the circle is equal to the length of the wire,so $2\pi r = L$. Thus,the radius is $r = \

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$\frac{IL^2}{4\pi}$

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(B) Let the radius of the circle be $r$. The circumference of the circle is equal to the length of the wire,so $2\pi r = L$. Thus,the radius is $r = \frac{L}{2\pi}$. The area $A$ of the circle is given by $A = \pi r^2 = \pi \left(\frac{L}{2\pi}\right)^2 = \frac{\pi L^2}{4\pi^2} = \frac{L^2}{4\pi}$. The magnetic moment $M$ is defined as $M = IA$. Substituting the value of $A$,we get $M = I \left(\frac{L^2}{4\pi}\right) = \frac{IL^2}{4\pi}$.

$A$ wire of length $L$ $m$ carrying a current of $I$ $A$ is bent in the form of a circle. Its magnitude of magnetic moment will be

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