Problem 6 Solution
1. Initial State (Switch at 1):
The capacitor is connected to the battery of EMF $\mathcal{E}$. In steady state, the charge on the capacitor is: $$ Q_0 = C\mathcal{E} $$
The capacitor is connected to the battery of EMF $\mathcal{E}$. In steady state, the charge on the capacitor is: $$ Q_0 = C\mathcal{E} $$
2. Switch thrown to 2 (Impulse):
The charged capacitor discharges through the conducting rod. Since resistance is negligible, this discharge is extremely rapid. The current $i(t)$ flowing through the rod in the presence of the magnetic field $B$ creates an impulsive magnetic force $F_m = i l B$.
The total impulse $J$ imparted to the rod is: $$ J = \int F_m dt = \int (i l B) dt = l B \int i dt = l B Q_0 $$ Substituting $Q_0$: $$ J = l B C \mathcal{E} $$
The charged capacitor discharges through the conducting rod. Since resistance is negligible, this discharge is extremely rapid. The current $i(t)$ flowing through the rod in the presence of the magnetic field $B$ creates an impulsive magnetic force $F_m = i l B$.
The total impulse $J$ imparted to the rod is: $$ J = \int F_m dt = \int (i l B) dt = l B \int i dt = l B Q_0 $$ Substituting $Q_0$: $$ J = l B C \mathcal{E} $$
3. Initial Velocity for SHM:
This impulse provides an initial velocity $v_0$ to the rod of mass $m$: $$ J = m v_0 \implies v_0 = \frac{l B C \mathcal{E}}{m} $$ Immediately after the discharge, the capacitor is empty, and the rod acts as a mass attached to two springs (parallel configuration), effectively $k_{eq} = k + k = 2k$.
This impulse provides an initial velocity $v_0$ to the rod of mass $m$: $$ J = m v_0 \implies v_0 = \frac{l B C \mathcal{E}}{m} $$ Immediately after the discharge, the capacitor is empty, and the rod acts as a mass attached to two springs (parallel configuration), effectively $k_{eq} = k + k = 2k$.
4. Oscillation Dynamics:
The rod performs Simple Harmonic Motion (SHM) with angular frequency: $$ \omega = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{2k}{m}} $$ Using conservation of energy to find the amplitude $A$: $$ \frac{1}{2} m v_0^2 = \frac{1}{2} (2k) A^2 $$ $$ A = v_0 \sqrt{\frac{m}{2k}} = \left( \frac{l B C \mathcal{E}}{m} \right) \sqrt{\frac{m}{2k}} = \frac{B l C \mathcal{E}}{\sqrt{2km}} $$
The rod performs Simple Harmonic Motion (SHM) with angular frequency: $$ \omega = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{2k}{m}} $$ Using conservation of energy to find the amplitude $A$: $$ \frac{1}{2} m v_0^2 = \frac{1}{2} (2k) A^2 $$ $$ A = v_0 \sqrt{\frac{m}{2k}} = \left( \frac{l B C \mathcal{E}}{m} \right) \sqrt{\frac{m}{2k}} = \frac{B l C \mathcal{E}}{\sqrt{2km}} $$
Final Answer:
The motion is Simple Harmonic Oscillation given by: $$ y = A \sin(\omega t) = \frac{BlC\mathcal{E}}{\sqrt{2km}} \sin\left(\sqrt{\frac{2k}{m}}t\right) $$
The motion is Simple Harmonic Oscillation given by: $$ y = A \sin(\omega t) = \frac{BlC\mathcal{E}}{\sqrt{2km}} \sin\left(\sqrt{\frac{2k}{m}}t\right) $$