The ratio of the magnetic field at the cpentre of a current carrying circular wire and the magnetic field at the centre of a square coil made from the same length of wire will be
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$B_{C_{1}}=\frac{\mu_{0} i}{2 R}$
square is made from the same length of wire
$2 \pi \mathrm{R}=4 \mathrm{a}$
$\mathrm{a}=\frac{\pi \mathrm{R}}{2}$
$\mathrm{B}_{\mathrm{C}_{2}}=4 \frac{\mu_{0} \mathrm{i}}{4 \pi(\mathrm{a} / 2)}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]$
$\mathrm{B}_{\mathrm{C}_{2}}=2 \sqrt{2} \frac{\mu_{0} \mathrm{i}}{\pi \mathrm{a}}$
$\frac{\mathrm{B}_{\mathrm{C}_{1}}}{\mathrm{B}_{\mathrm{C}_{2}}}=\frac{\frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}}}{2 \sqrt{2} \frac{\mu_{0} \mathrm{i}}{\pi \mathrm{a}}} \quad\left(\mathrm{a}=\frac{\pi \mathrm{R}}{2}\right)$
$\frac{B_{C_{1}}}{B_{C_{2}}}=\frac{\pi^{2}}{8 \sqrt{2}}$
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