Physics - Magnetism Question with Solution | TestHub

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, ${\mu }_{\mathrm{c}}-\mathrm{iA}=\mathrm{i}\pi \left(\frac{l^2}{4{\pi }^2}\right)$ ${\mu }_{\mathrm{c}}=\frac{\mathrm{i}{\text{i }}^2}{4\pi }\dots ..(\mathrm{i})$ 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, ${\mu }_{\mathrm{s}}=\mathrm{iA}=\mathrm{i}\frac{l^2}{16}$ $\begin{array}{rl}& \therefore \frac{{\mu }_{\mathrm{c}}}{{\mu }_{\mathrm{s}}}=\frac{\frac{\mathrm{i}{l}^2}{4\pi }}{\frac{\mathrm{i}{l}^2}{16}}\\ & =\frac{16}{4\pi }=\frac{4}{\pi }\end{array}$ Hence, the ratio of magnetic dipole moment of circular coil and square coil is $4:\pi$.