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(B) For the square frame of side $a$,the magnetic field at the centre $O$ is the sum of the fields produced by its four sides.Each side acts as a fini

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$\frac{16 \sqrt{2}}{\pi^2}$

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(B) For the square frame of side $a$,the magnetic field at the centre $O$ is the sum of the fields produced by its four sides.Each side acts as a finite wire of length $a$ at a perpendicular distance $r = a/2$ from the centre.The magnetic field due to one side is $B_1 = \frac{\mu_0 I}{4 \pi r} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_0 I}{4 \pi (a/2)} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 I}{2 \pi a} (\frac{2}{\sqrt{2}}) = \frac{\sqrt{2} \mu_0 I}{\pi a}$.Since there are $4$ sides,the total magnetic field is $B = 4 	imes B_1 = \frac{4 \sqrt{2} \mu_0 I}{\pi a}$.For the circular coil,the perimeter is equal to the perimeter of the square,so $2 \pi R = 4 a$,which gives $R = \frac{2 a}{\pi}$.The magnetic field at the centre of the circular coil is $B^{\prime} = \frac{\mu_0 I}{2 R} = \frac{\mu_0 I}{2 (2 a / \pi)} = \frac{\mu_0 \pi I}{4 a}$.Taking the ratio,$\frac{B}{B^{\prime}} = \frac{4 \sqrt{2} \mu_0 I / \pi a}{\mu_0 \pi I / 4 a} = \frac{16 \sqrt{2}}{\pi^2}$.

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