Resonant Frequency of Circuit with Concentric Coupled Coils

To find the resonant frequency of the circuit, we first determine the self-inductance of both coils and the mutual inductance between them.

1. Self-Inductance of the Larger Coil ($L_1$)

For the larger coil (coil 1):

• Number of turns per unit length: $n_1 = N$
• Total length: $d_1 = d$
• Cross-sectional area: $A_1 = S$

The self-inductance $L_1$ of coil 1 is given by: $L_1 = \mu_0 n_1^2 A_1 d_1 = \mu_0 N^2 S d = L$

2. Self-Inductance of the Smaller Coil ($L_2$)

For the smaller coil (coil 2):

• Number of turns per unit length: $n_2 = 2N$
• Total length: $d_2 = \frac{d}{2}$
• Cross-sectional area: $A_2 = \frac{S}{2}$

The self-inductance $L_2$ of coil 2 is given by: $L_2 = \mu_0 n_2^2 A_2 d_2 = \mu_0 (2N)^2 \left(\frac{S}{2}\right) \left(\frac{d}{2}\right) = \mu_0 N^2 S d = L$

3. Mutual Inductance ($M$)

Since the smaller coil is completely inside the larger coil, the magnetic field produced by current $i_1$ flowing in coil 1 is uniform inside it: $B_1 = \mu_0 n_1 i_1 = \mu_0 N i_1$

The total magnetic flux $\Phi_{21}$ linked with coil 2 due to coil 1 is: $\Phi_{21} = N_2 B_1 A_2 = \left(2N \cdot \frac{d}{2}\right) (\mu_0 N i_1) \left(\frac{S}{2}\right) = \frac{1}{2} \mu_0 N^2 S d \, i_1 = \frac{1}{2} L i_1$

Thus, the mutual inductance $M$ between the two coils is: $M = \frac{\Phi_{21}}{i_1} = \frac{L}{2}$

4. Effective Inductance of the System ($L_{\text{eff}}$)

The ends of the smaller coil are short-circuited (neglecting resistance), so the induced potential difference across coil 2 is zero: $L_2 \frac{di_2}{dt} + M \frac{di_1}{dt} = 0$

Substituting $L_2 = L$ and $M = \frac{L}{2}$: $L \frac{di_2}{dt} + \frac{L}{2} \frac{di_1}{dt} = 0 \implies \frac{di_2}{dt} = -\frac{1}{2} \frac{di_1}{dt}$

The potential drop across coil 1 (the main circuit loop) is: $v_1 = L_1 \frac{di_1}{dt} + M \frac{di_2}{dt}$

Substituting $L_1 = L$, $M = \frac{L}{2}$, and $\frac{di_2}{dt} = -\frac{1}{2} \frac{di_1}{dt}$: $v_1 = L \frac{di_1}{dt} + \left(\frac{L}{2}\right) \left(-\frac{1}{2} \frac{di_1}{dt}\right) = \left(L - \frac{L}{4}\right) \frac{di_1}{dt} = \frac{3}{4} L \frac{di_1}{dt}$

Therefore, the effective inductance of the larger coil connected in the circuit is: $L_{\text{eff}} = \frac{3}{4} L$

5. Resonant Frequency

The resonant frequency $\omega$ of the $L_{\text{eff}}C$ circuit is given by: $\omega = \frac{1}{\sqrt{L_{\text{eff}} C}} = \frac{1}{\sqrt{\frac{3}{4} L C}} = \frac{2}{\sqrt{3 LC}}$

Thus, the correct option is C.