There are two current-carrying planar coil made each from identical wires of length L. ${\mathrm{C}}_{1}$ is circular (radius R) and ${\mathrm{C}}_{2}$ is square (side a). They are so constructed, that they have the same frequency of oscillation when they are placed in the same uniform B and carry the same current. Find a in terms of R.

Hint: Use the formula of frequency of oscillation of a bar magnet.
Step 1: ${\mathrm{C}}_{1}$ = circular coil of radius R, length L,
number of turns per unit length, ${\mathrm{n}}_{1}=\frac{\mathrm{L}}{2\mathrm{\pi R}}$

${\mathrm{C}}_{2}$ = square of side a and perimeter L, number of turns per unit length, ${\mathrm{n}}_{2}=\frac{\mathrm{L}}{4\mathrm{a}}$

${\mathrm{m}}_{1}=\frac{\mathrm{L}\cdot \mathrm{I}\cdot {\mathrm{\pi R}}^{2}}{2\mathrm{\pi R}}$
${\mathrm{m}}_{2}=\frac{\mathrm{L}}{4\mathrm{a}}\cdot \mathrm{I}\cdot {\mathrm{a}}^{2}$

Step 2:

Step 3:
Plugging the values by Eqs. (i), (ii), (iii) and (iv);
$\begin{array}{l}\frac{\mathrm{LIa}\cdot 2}{4×\mathrm{LIR}}=\frac{{\mathrm{Ma}}^{2}\cdot 2}{12\cdot {\mathrm{MR}}^{2}}\\ \frac{\mathrm{a}}{2\mathrm{R}}=\frac{{\mathrm{a}}^{2}}{6{\mathrm{R}}^{2}}\\ 3\mathrm{R}=\mathrm{a}\end{array}$
Thus, the value of a is 3R.