35. Max Current in Coil
Step 1: Initial State (Switch Open)
The batteries are connected in a single loop but with opposing polarities. The $2\mathcal{E}$ battery dominates.
- Net EMF = $2\mathcal{E} – \mathcal{E} = \mathcal{E}$.
- Equivalent Capacitance (Series) = $C/2$.
- Initial Charge Magnitude: $Q_i = C_{eq} V = (C/2)\mathcal{E} = 0.5 C\mathcal{E}$.
Polarity Check: The dominant battery ($2\mathcal{E}$) pushes current clockwise (up left, down right).
Right Cap: Top Plate is Positive ($+0.5C\mathcal{E}$).
Left Cap: Top Plate is Negative ($-0.5C\mathcal{E}$).
Right Cap: Top Plate is Positive ($+0.5C\mathcal{E}$).
Left Cap: Top Plate is Negative ($-0.5C\mathcal{E}$).
Initial Energy: $U_i = \frac{Q_i^2}{2C_{eq}} = \frac{(0.5C\mathcal{E})^2}{C} = 0.25 C\mathcal{E}^2$.
Step 2: Max Current State (Switch Closed)
The switch closes, connecting the center points via the inductor. At maximum current, the EMF across the inductor is zero ($V_L = L di/dt = 0$). The top and bottom wires become equipotential.
The circuit splits into two independent loops:
- Left Loop: Cap $C$ connects to Battery $\mathcal{E}$. It charges to $Q_{L} = 1.0 C\mathcal{E}$. (Top Plate Positive).
- Right Loop: Cap $C$ connects to Battery $2\mathcal{E}$. It charges to $Q_{R} = 2.0 C\mathcal{E}$. (Top Plate Positive).
Final Potential Energy: $U_f = \frac{1}{2}C(\mathcal{E})^2 + \frac{1}{2}C(2\mathcal{E})^2 = 2.5 C\mathcal{E}^2$.
Step 3: Work Done by Batteries
We calculate work as $W = \Delta Q \cdot \mathcal{E}$.
- Left Battery ($\mathcal{E}$): The top plate charge changes from $-0.5 C\mathcal{E}$ to $+1.0 C\mathcal{E}$.
$\Delta Q_L = +1.5 C\mathcal{E}$.
$W_L = (1.5 C\mathcal{E}) \times \mathcal{E} = 1.5 C\mathcal{E}^2$. - Right Battery ($2\mathcal{E}$): The top plate charge changes from $+0.5 C\mathcal{E}$ to $+2.0 C\mathcal{E}$.
$\Delta Q_R = +1.5 C\mathcal{E}$.
$W_R = (1.5 C\mathcal{E}) \times 2\mathcal{E} = 3.0 C\mathcal{E}^2$.
Total Work Input: $W = 4.5 C\mathcal{E}^2$.
Step 4: Energy Conservation
$$ U_i + W = U_f + E_{inductor} $$ $$ 0.25 C\mathcal{E}^2 + 4.5 C\mathcal{E}^2 = 2.5 C\mathcal{E}^2 + \frac{1}{2} L I_{max}^2 $$ $$ 4.75 C\mathcal{E}^2 – 2.5 C\mathcal{E}^2 = \frac{1}{2} L I_{max}^2 $$ $$ 2.25 C\mathcal{E}^2 = \frac{1}{2} L I_{max}^2 $$ $$ \frac{9}{4} C\mathcal{E}^2 = \frac{1}{2} L I_{max}^2 \implies I^2 = \frac{9 C \mathcal{E}^2}{2 L} $$
$$ I_{max} = 3 \mathcal{E} \sqrt{\frac{C}{2L}} $$