The magnetic field at the centre of coil of $n$ $turns$, bent in the form of a square of side $2l$ , carrying current $i$, is
Diffcult
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(a) Magnetic field due to one side of the square at centre $O$
${B_1} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2i\sin {{45}^o}}}{{a/2}}$
${B_1} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2i\sin {{45}^o}}}{{a/2}}$
$==>$ ${B_1} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \,i}}{a}$
Hence magnetic field at centre due to all side
$B = 4{B_1} = \frac{{{\mu _0}(2\sqrt 2 i)}}{{\pi a}}$
Magnetic field due to $n$ $turns$
${B_{net}} = nB = \frac{{{\mu _0}2\sqrt 2 ni}}{{\pi a}} = \frac{{{\mu _0}2\sqrt 2 ni}}{{\pi (2l)}}$$ = \frac{{\sqrt 2 {\mu _0}ni}}{{\pi l}}$
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