Electromagnetic Induction

Concept:The resonant frequency is found by modeling the coupled coils as an effective LC circuit with mutual inductance.Explanation:The outer solenoid has self-inductance $L = \mu_0 N^2 S d$ (given).The inner solenoid has $2N$ turns per unit length, length $d/2$, area $S/2$, so its self-inductance $L_2 = \mu_0 (2N)^2 (S/2)(d/2) = \mu_0 N^2 S d = L$.When current $I_1$ flows in the outer coil, the magnetic field inside is $B_1 = \mu_0 N I_1$.Flux through the inner coil: $\Phi_{21} = (Nd) (\mu_0 N I_1)(S/2) = \frac{1}{2}\mu_0 N^2 S d I_1$.Thus mutual inductance $M = \Phi_{21}/I_1 = L/2$.For the inner shorted loop (no resistance), KVL: $-L \frac{dI_2}{dt} - M \frac{dI_1}{dt} = 0 \implies \frac{dI_2}{dt} = -\frac{1}{2}\frac{dI_1}{dt}$. With zero initial currents, $I_2 = -\frac{1}{2}I_1$.For the outer loop containing capacitor C: $\frac{q}{C} - L\frac{dI_1}{dt} - M\frac{dI_2}{dt} = 0$.Substitute $M$ and $\frac{dI_2}{dt}$: $\frac{q}{C} - L\frac{dI_1}{dt} - \frac{L}{2}\left(-\frac{1}{2}\frac{dI_1}{dt}\right) = \frac{q}{C} - \frac{3}{4}L\frac{dI_1}{dt} = 0$.Using $I_1 = -\frac{dq}{dt}$, we get $\frac{3}{4}L\frac{d^2q}{dt^2} + \frac{q}{C} = 0$, or $\frac{d^2q}{dt^2} + \frac{4}{3LC}q = 0$.Comparing with SHM, angular resonant frequency $\omega_0 = \sqrt{\frac{4}{3LC}} = \frac{2}{\sqrt{3LC}}$.Answer:$\frac{2}{\sqrt{3LC}}$ (Option C).