Magnetism and Capacitor Story

Nice problem.

Conceptually, the following is what is happening. At $t=0$ when the switch closes, the capacitor begins to discharge. Since a resistor is present, the discharge will not be instantaneous. So the discharging current will begin to flow in a clockwise manner. As the current flows through the rod of length $D$ , a voltage will be induced in the rod as per Faraday's law. Let us break this down into mathematical steps as follows.

At an arbitrary time $t$ , let the charge on the capacitor be $Q$ and the current through the circuit be $I$ . SInce it is a discharging current flowing clockwise, we have:

$\dot{Q} = -I \ \dots \ (1)$

Now applying Kirchoff's voltage law to the circuit gives:

$V_{induced} -IR + \frac{Q}{C}=0 \ \dots \ (2)$

Now, looking at the dynamics of the rod, since the current flows clockwise, it flows down along the rod. The rod, as a result, experiences a magnetic force acting to the right, with magnitude:

$F = IDB$

Let the velocity of the rod be $v$ . Applying Newton's second law:

$m\dot{v} = IDB \ \dots \ (3)$

Say the rod has moved a distance $x$ from the left end at time $t$ . The magnetic flux through the rectangular circuit is:

$\phi = BDx$

Now, applying Faraday's law gives:

$V_{induced} = -\dot{\phi} = -BDv \ \dots \ (4)$

Now, looking at equation $3$ , we see that:

$m\dot{v} = IDB \implies m\dot{v} = -\dot{Q}DB$

at $t=0$ , I assume that the rod is at rest and the capacitor voltage is $V$ . By integrating the above equation and applying initial conditions, one gets:

$mv = CVBD - QBD \ \dots \ (5)$

Looking at equation $2$ , it can be re-written as:

$V_{induced} -IR + \frac{Q}{C}=0 \implies -BDv-IR + \frac{Q}{C}=0$

Replacing equation $3$ above gives:

$\implies -BDv-\frac{Rm\dot{v}}{BD} + \frac{Q}{C}=0$

Replacing $Q$ with equation $5$ above gives:

$-BDv-\frac{Rm\dot{v}}{BD} + \frac{CVBD-mv}{BDC}=0$

Now, at steady state, the acceleration $\dot{v}$ becomes zero and the velocity reaches its final value. Performing this step in the above equation, and solving, gives the final velocity which is:

$\boxed{v_{final} = \frac{CVBD}{m+CB^2D^2}}$