Two identical wires X and Y, each of length L carry the same current I. Wire X is bent to form a square of side 'a' and wire Y is bent into a circle of radius R. If $B_X$ and $B_Y$ are the values of the magnetic field at the centres of the square and the circle respectively, the ratio $B_X:B_Y$ is

$\frac{8\sqrt{2}}{π^2}$

Side of length of square is $\frac{L}{4}$

Magnetic field due to square loop is $ B_X = 4\times \frac{\mu_0 I}{4\pi \frac{L}{8}} (sin 45^o + sin45^o) = 8\sqrt{2} \frac{\mu_0 I}{\pi L}$

Radius of the circle is $\frac{L}{2\pi}$

$B_Y = \frac{\mu_0 I}{2R} = \frac{\pi \mu_0I}{L}$

Ratio is $\frac{B_X}{B_Y} = \frac{8\sqrt{2}}{\pi^2}$

The correct answer is Option (3) → $\frac{8\sqrt{2}}{π^2}$