\(O D=\frac{l}{2}\)

\(B=4\left[\frac{\mu_0 i}{4 \pi\left(\frac{l}{2}\right)}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)\right]\)

\(B=2 \sqrt{2} \frac{\mu_0 i}{\pi !}\)

For circle

\(2 \pi r=4l\)

\(r=\frac{2l}{\pi}\)

So \(B^{\prime}=\frac{\mu_0 i}{2 r}=\frac{\mu_0 i}{2\left(\frac{2 l}{\pi}\right)}\)

\(\Rightarrow B^{\prime}=\frac{\mu_0 i \pi}{4 I}\)

\(\Rightarrow\) So \(\frac{B}{B^{\prime}}=\frac{8 \sqrt{2}}{\pi^2}\)