Let I be the length of metal wire.
When wire is bent into a circular coil of radius \(r\), then
\(r=\frac{l}{2 \pi}\)
\(\therefore\) Area, \(\mathrm{A}=\pi\left(\frac{l}{2 \pi}\right)^{2}=\pi \frac{l^{2}}{4 \pi^{2}}\)
\(\therefore\) Magnetic dipole moment associated with circular coil,
\(\mu_{\mathrm{c}}-\mathrm{iA}=\mathrm{i} \pi\left(\frac{l^{2}}{4 \pi^{2}}\right)\)
\(\mu_{\mathrm{c}}=\frac{\mathrm{i \textrm {i } ^ { 2 }}}{4 \pi} \quad \ldots . .(\mathrm{i})\)
When metal wire Is bent into a square coil then side of square
\(\mathrm{a}=\frac{l}{4}\)
\(\therefore\) Area, \(\mathrm{A}=\mathrm{a}^{2}=\frac{l^{2}}{16}\)
\(\therefore\) Magnetic dipole moment associated with square coil,
\(\mu_{\mathrm{s}}=\mathrm{i} \mathrm{A}=\mathrm{i} \frac{l^{2}}{16}\)
\(\begin{aligned}
&\therefore \frac{\mu_{\mathrm{c}}}{\mu_{\mathrm{s}}}=\frac{\frac{\mathrm{i} l^{2}}{4 \pi}}{\frac{\mathrm{i} l^{2}}{16}} \\
&=\frac{16}{4 \pi}=\frac{4}{\pi}
\end{aligned}\)
Hence, the ratio of magnetic dipole moment of circular coil and square coil is \(4: \pi\).