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(C) The magnetic field produced by a single side of the square (e.g.,side $AB$) at the centre is given by the formula for a finite wire: $B_1 = \frac{

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$\frac{2 \sqrt{2} \mu_0 i}{\pi a}$

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(C) The magnetic field produced by a single side of the square (e.g.,side $AB$) at the centre is given by the formula for a finite wire: $B_1 = \frac{\mu_0 i}{4 \pi r} (\sin \phi_1 + \sin \phi_2)$.Here, the distance from the centre to the side is $r = \frac{a}{2}$, and the angles subtended at the centre by the ends of the side are $\phi_1 = 45^{\circ}$ and $\phi_2 = 45^{\circ}$.Substituting these values: $B_1 = \frac{\mu_0 i}{4 \pi (a/2)} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_0 i}{2 \pi a} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{2 \pi a} (\frac{2}{\sqrt{2}}) = \frac{\sqrt{2} \mu_0 i}{2 \pi a} = \frac{\mu_0 i}{\sqrt{2} \pi a}$.Since there are $4$ identical sides, the total magnetic field at the centre is $B = 4 B_1 = 4 	imes \frac{\mu_0 i}{2 \sqrt{2} \pi a} = \frac{2 \sqrt{2} \mu_0 i}{\pi a}$.

$A$ wire in the form of a square of side $a$ carries a current $i$. Then, the magnetic induction at the centre of the square is (Magnetic permeability of free space $= \mu_0$)

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