From Faraday’s law, we can write:
\[\dot{q}R + \frac{q}{C} = vBl\tag1\]
Where \(v\) is the velocity of the rod at any general instant.
And from the Lorentz force and Newton’s law,
\[F = -iBl \Longleftrightarrow \dot{v} = \frac{-\dot{q}Bl}{m}\tag2\]
Differentiating eq (1), and substituting \(\dot{v}\) from eq (2) into it gives us
\[\ddot{q}R + \dot{q} \left( \frac{1}{C} + \frac{B^2l^2}{m} \right) = 0\tag3\]
For simplicity let’s define:
\[\kappa \overset{\text{def}}{=}\frac{1}{C} + \frac{B^2l^2}{m}\]
Solving the differential eq (3),
\[q = C_1 e^{\alpha_1 t} + C_2 e^{\alpha_2 t}\]
Where \(\alpha_1\) and \(\alpha_2\) are the roots of the equation \(D^2R+D\kappa = 0\). Substituting boundary condition, i.e. at \(t=0\), charge on capacitor \(q=0\), and initial current \(i_0 = \frac{v_0 Bl}{R}\), we get:
\[q = \frac{v_0 Bl}{\kappa}\left( 1 - e^{-\frac{\kappa}{R} t} \right)\tag4\]
Here, we see that \[\tau \overset{\text{def}}{=} R / \kappa\] plays the role of time constant.
Note that the final charge on the capacitor, at \(t=\infty\):
\[q\big|_{t=\infty} = \frac{v_0 B l}{\kappa}\]
Further, current as a function of time:
\[i = \dot{q} = \frac{v_0 B l R }{\kappa^2} e^{-t/\tau}\tag5\]
Substituting this is eq (2), and integrating with appropriate limits, we get velocity of the rod as a function of time,
\[v = \frac{v_0 B^2 l^2 R^2}{\kappa^3 m} \left( 1 - e^{-t/\tau}\right) + v_0\tag6\]
Further, we can say that:
\[P = i^2 R\]
\[\implies \int dH = \frac{v_0^2B^2l^2R^3}{\kappa^4} \int_0^\infty e^{-2t/\tau}dt\]
\[\implies H = \frac{v_0^2B^2l^2R^4}{2\kappa^5}\tag7\]
This is the total heat lost due to current in the resistor(rod).
We can also find the final kinetic energy of the rod.
\[K_{f,rod} = \frac{1}{2} m v_0^2 \left( \frac{B^2l^2R^2}{\kappa^3m} + 1\right) ^2\tag8\]
Also, The final energy stored in the capacitor,
\[E_{capacitor} = \frac{q^2}{2C}\]
\[E = \frac{v_0^2B^2l^2}{2C\kappa^2}\tag9\]