A square frame of side l carries a current i. | vedclass

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(A) For a square frame of side $l$,the distance from the centre to any side is $d = \frac{l}{2}$.The magnetic field due to one side at the centre is $

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$\frac{8 \sqrt{2}}{\pi^2}$

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(A) For a square frame of side $l$,the distance from the centre to any side is $d = \frac{l}{2}$.The magnetic field due to one side at the centre is $B_1 = \frac{\mu_0 i}{4 \pi d} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_0 i}{4 \pi (l/2)} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{2 \pi l} \sqrt{2} = \frac{\sqrt{2} \mu_0 i}{2 \pi l}$.Since there are $4$ sides,the total magnetic field at the centre is $B = 4 	imes B_1 = 4 	imes \frac{\sqrt{2} \mu_0 i}{2 \pi l} = \frac{2 \sqrt{2} \mu_0 i}{\pi l}$.For the circular coil,the perimeter is equal to the perimeter of the square,so $2 \pi r = 4l$,which gives $r = \frac{2l}{\pi}$.The magnetic field at the centre of the circular coil is $B^{\prime} = \frac{\mu_0 i}{2r} = \frac{\mu_0 i}{2(2l/\pi)} = \frac{\mu_0 i \pi}{4l}$.Now,the ratio is $\frac{B}{B^{\prime}} = \frac{2 \sqrt{2} \mu_0 i / \pi l}{\mu_0 i \pi / 4l} = \frac{2 \sqrt{2}}{\pi} 	imes \frac{4}{\pi} = \frac{8 \sqrt{2}}{\pi^2}$.

$A$ square frame of side $l$ carries a current $i$. The magnetic field at its centre is $B$. The same current is passed through a circular coil having the same perimeter as the square. The field at the centre of the circular coil is $B^{\prime}$. The ratio of $\frac{B}{B^{\prime}}$ is

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