A square shaped wire loop of side L is carrying a current I. What is the magnetic field at the point of intersection of diagonals of the square wire loop?

$\frac{2\sqrt{2}μ_0I}{π L}$

The correct answer is Option (2) → $\frac{2\sqrt{2}μ_0I}{π L}$

The side length of the square loop, $AB = L$

$∴BC=\frac{L}{2}$

∴ In ΔOAD

$⇒\frac{AD}{x}=\tan 45°$

$⇒x=\frac{L}{2}$

Magnetic field at O due to AB,

${I_0}'=\frac{μ_0I}{4\pi x}[\sin θ_1+\sin θ_2]$

$=\frac{μ_0I}{\pi L\sqrt{2}}$

∴ $I_0$ due to 4 such wire,

$I_0=4{I_0}'=\frac{2\sqrt{2}μ_0I}{π L}$