There are two current-carrying planar coils made from identical wires of length $L$. $C_1$ is circular (radius $R$) and $C_2$ is square (side $a$). They are constructed such that they have the same frequency of oscillation when placed in the same uniform magnetic field $B$ and carry the same current $I$. Find $a$ in terms of $R$.

(A) The frequency of oscillation of a magnetic dipole in a uniform magnetic field $B$ is given by $f = \frac{1}{2\pi} \sqrt{\frac{mB}{I_0}}$,where $m$ is the magnetic moment and $I_0$ is the moment of inertia.
Since the frequency $f$ and magnetic field $B$ are the same for both coils,we must have $\frac{m_1}{I_1} = \frac{m_2}{I_2}$.
For the circular coil $C_1$: Number of turns $N_1 = \frac{L}{2\pi R}$. Magnetic moment $m_1 = N_1 I A_1 = \left(\frac{L}{2\pi R}\right) I (\pi R^2) = \frac{LIR}{2}$. Moment of inertia $I_1 = \frac{M R^2}{2}$.
For the square coil $C_2$: Number of turns $N_2 = \frac{L}{4a}$. Magnetic moment $m_2 = N_2 I A_2 = \left(\frac{L}{4a}\right) I (a^2) = \frac{LIa}{4}$. Moment of inertia $I_2 = \frac{M a^2}{6}$ (about the axis passing through the center and parallel to a side).
Equating the ratios: $\frac{m_1}{I_1} = \frac{m_2}{I_2} \Rightarrow \frac{LIR/2}{MR^2/2} = \frac{LIa/4}{Ma^2/6} \Rightarrow \frac{LI}{MR} = \frac{3LI}{2Ma} \Rightarrow \frac{1}{R} = \frac{3}{2a} \Rightarrow a = 1.5R$.