\(L = 4a\)
\(B = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi i}}{r}\)\( = \frac{{{\mu _0}}}{{4\pi }}.\frac{{4{\pi ^2}i}}{r}\)
\(B = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \,i}}{a}\)\(B = \frac{{{\mu _0}}}{{4\pi }}.\frac{{8\sqrt 2 \,i}}{a}\)
\(\frac{{{B_{circular}}}}{{{B_{square}}}} = \frac{{{\pi ^2}}}{{8\sqrt 2 }}\)