Physics - Magnetism Question with Solution | TestHub

Let the length of the wire be $L$ and the current be $I$.

For the circular loop with two turns, the total length is $2\times 2\pi r=L$, which gives $r=\frac{L}{4\pi }$.

The magnetic moment of the circular loop is $M_1=N_{1}I{A}_1=2I(\pi r^2)=2I\pi {\left(\frac{L}{4\pi }\right)}^2=\frac{I{L}^2}{8\pi }$.

For the square loop with one turn, the perimeter is $4a=L$, which gives $a=\frac{L}{4}$.

The magnetic moment of the square loop is $M_2=N_{2}I{A}_2=1I{a}^2=I{\left(\frac{L}{4}\right)}^2=\frac{I{L}^2}{16}$.

The ratio of their magnetic moments is $\frac{M_1}{M_2}=\frac{\frac{I{L}^2}{8\pi }}{\frac{I{L}^2}{16}}=\frac{16}{8\pi }=\frac{2}{\pi }$.

Answer: $2:\pi$