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(D) Let the length of the wire be $L$. For the circular coil,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 coil,the circumference is $L = 2\pi r$,so the radius is $r = \frac{L}{2\pi}$. The magnetic moment $M$ is given by $M = I 	imes A = I 	imes \pi r^2$. Substituting $r$,we get $M = I 	imes \pi \left( \frac{L}{2\pi} ight)^2 = \frac{I L^2}{4\pi}$ ..... $(i)$. For the square coil,the perimeter is $L = 4a$,where $a$ is the side length,so $a = \frac{L}{4}$. The magnetic moment $M'$ is given by $M' = I 	imes A' = I 	imes a^2$. Substituting $a$,we get $M' = I 	imes \left( \frac{L}{4} ight)^2 = \frac{I L^2}{16}$ ..... $(ii)$. Dividing equation $(ii)$ by equation $(i)$,we get $\frac{M'}{M} = \frac{I L^2 / 16}{I L^2 / 4\pi} = \frac{4\pi}{16} = \frac{\pi}{4}$. Therefore,$M' = \frac{\pi}{4} M$.

$A$ 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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