A straight wire is first bent into a circle of radius ${}^{\text{'}}r^{\text{'}} \mathrm{and} \mathrm{then} \mathrm{into} \mathrm{a} \mathrm{square} \mathrm{of} \mathrm{side} {}^{\text{'}}X^{\text{'}}$ each of one turn. If currents flowing through them are in the ration 4 : 5, the ratio of their effective magnetic moment is

Given data: ratio 4: 5, radius r and side of square x

Concept used: moving charges and magnetism 

Detailed solution:

For circle:

$$\begin{gathered} M_1=I_1\times A \\ =I_1\times \pi r^2 \\ =I_1\times \pi {\left(\frac{2X}{\pi }\right)}^2 \\ =I_1\times \frac{\pi \times 4X^2}{{\pi }^2} \end{gathered}$$

For square:

 $$\begin{gathered} M_2=I_2\times A \\ =I_2\times X^2 \end{gathered}$$

Now, taking ratio of M₁ and M₂ we get: 

 $\frac{M_1}{M_2}=\frac{4}{5}\times \frac{4}{\pi }=\frac{16}{5\pi }$