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(D) Let the length of the wire be $L$. For the circular loop,the circumference is $L = 2\pi r$,so the radius is $r = \frac{L}{2\pi}$. The magnetic mom

Vedclass · Accepted Answer

$\frac{\pi}{4} M$

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(D) Let the length of the wire be $L$. For the circular loop,the circumference is $L = 2\pi r$,so the radius is $r = \frac{L}{2\pi}$. The magnetic moment of the circular loop is $M = I A = I(\pi r^2) = I \pi \left(\frac{L}{2\pi}\right)^2 = \frac{IL^2}{4\pi}$. When the wire is reshaped into a square of side $a$,the perimeter is $4a = L$,so $a = \frac{L}{4}$. The area of the square is $A' = a^2 = \left(\frac{L}{4}\right)^2 = \frac{L^2}{16}$. The new magnetic moment is $M' = I A' = I \frac{L^2}{16}$. From the expression for $M$,we have $IL^2 = 4\pi M$. Substituting this into the expression for $M'$,we get $M' = \frac{4\pi M}{16} = \frac{\pi}{4} M$.

$A$ thin circular wire carrying a current $I$ has a magnetic moment $M$. The shape of the wire is changed to a square and it carries the same current. It will have a magnetic moment

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