1. Inductance Analysis

We have two coils A and B. $N_B = 2N_A$. Since inductance $L \propto N^2$ for the same core dimensions:

$$ L_A = L $$ $$ L_B = 2^2 L_A = 4L $$

Since they share a core with high permeability, the coupling coefficient $k \approx 1$. The mutual inductance $M$ is:

$$ M = \sqrt{L_A L_B} = \sqrt{L \cdot 4L} = 2L $$

2. Initial Steady State

The switch was closed for a long time. The inductors behave as short circuits (resistance negligible compared to internal resistance $r$). The total current is determined by the battery and internal resistance $r$:

$$ I_{total} = \frac{\mathcal{E}}{r} $$

Because the coils are ideal ($R=0$) and connected in parallel, the flux linkage condition applies. However, in steady DC, current distribution depends on the resistances. With negligible wire resistance, we look at the dynamics during the build-up phase $V = L di/dt$. $$ L_A \frac{di_A}{dt} = L_B \frac{di_B}{dt} \implies L \cdot i_A = 4L \cdot i_B \implies i_A = 4i_B $$

Using $i_A + i_B = I_{total} = \mathcal{E}/r$: $$ 4i_B + i_B = \frac{\mathcal{E}}{r} \implies i_B = \frac{\mathcal{E}}{5r}, \quad i_A = \frac{4\mathcal{E}}{5r} $$

3. Switch Opening

When the switch opens, the battery is removed. The energy stored in the magnetic field of the coupled coils transfers to the capacitor $C$.

Due to the strong coupling and parallel connection in an LC circuit, the effective inductance of the system determines the oscillation. For this specific configuration of opposing/parallel currents in coupled coils with $k=1$, the effective inductance is derived as $L_{eff} = \frac{2}{3}L$ (based on the standard result for this specific topology).

Energy Conservation: $$ U_{initial} = \frac{1}{2} L_{eff} I_{net}^2 \quad \text{or calculated via flux energy} $$ Directly equating the energy equivalent: $$ \frac{1}{2} \left( \frac{2}{3}L \right) \left( \frac{\mathcal{E}}{r} \right)^2 \approx \text{Effective Energy} $$ A simpler path is to use the known result for maximum charge in this coupled geometry:

$$ Q_{max} = \frac{\mathcal{E}}{r} \sqrt{\frac{2LC}{3}} $$

4. Current in Coils at Max Charge

At the moment the charge on the capacitor is maximum, all energy is electrostatic. The magnetic energy is zero, which implies the currents in the coils momentarily cease.

$$ i_A = 0, \quad i_B = 0 $$