Question 2

The self-inductance of the larger coil is $L = \mu_0 N^2 S d$.

For the smaller coil, the length is $l_2 = d/2$, the cross-sectional area is $S_2 = S/2$, and the turns per unit length is $n_2 = 2N$.

Its self-inductance is $L_2 = \mu_0 n_2^2 S_2 l_2 = \mu_0 (2N)^2 (S/2) (d/2) = \mu_0 (4N^2) \frac{Sd}{4} = \mu_0 N^2 S d = L$.

The mutual inductance $M$ between the coils is given by $M = \mu_0 n_1 n_2 S_2 l_2 = \mu_0 N (2N) (S/2) (d/2) = \frac{1}{2} \mu_0 N^2 S d = \frac{L}{2}$.

Let the current in the larger coil be $i_1$ and in the smaller shorted coil be $i_2$. Since the smaller coil is shorted and has zero resistance, the net EMF in it must be zero:

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

The voltage $V$ across the larger coil is given by:

$V = L_1 \frac{di_1}{dt} + M \frac{di_2}{dt} = L \frac{di_1}{dt} + \frac{L}{2} \left( -\frac{1}{2} \frac{di_1}{dt} \right) = \left( L - \frac{L}{4} \right) \frac{di_1}{dt} = \frac{3L}{4} \frac{di_1}{dt}$

Thus, the effective inductance across the larger coil or for the circuit containing the capacitor and larger coil is $L_{eq} = \frac{3L}{4}$.

The resonant frequency of the circuit is $\omega = \frac{1}{\sqrt{L_{eq} C}} = \frac{1}{\sqrt{\frac{3LC}{4}}} = \frac{2}{\sqrt{3LC}}$.