Two identical wires A and B, each of length '$l$', carry the same current $I$. Wire A is bent into a circle of radius $R$ and wire B is bent to form a square of side '$a$'. If $B_A$ and $B_B$ are the values of magnetic field at the centres of the circle and square respectively, then the ratio $\frac{B_A}{B_B}$ is

The correct option is D $\frac{{\pi }^2}{8\sqrt{2}}$
For wire A: $2\pi R=l$
$\Rightarrow$ R = $\frac{l}{2\pi }$
Magnetic field at the centre of the ring is given by
$B_A=\frac{{\mu }_{0}I}{2R}$
For wire B: $4a=l$
$a=l/4$

Magnetic field due to wire of length $L$ at a distance $d$ is given by
$B_B=\frac{{\mu }_{0}I}{4\pi d}(cos{\theta }_1+cos{\theta }_2)$
$\Rightarrow B_B=\frac{2\sqrt{2}{\mu }_{0}I}{\pi l}$ $(\because d=\frac{l}{8})$

$B_B$ net at the centre of the square is $\frac{8\sqrt{2}{\mu }_{0}I}{\pi l}$ (Due to all four wires)

Now, $\frac{B_A}{B_B}=\frac{{\pi }^2}{8\sqrt{2}}$