Current in an RC circuit inside in a magnetic field Does my question make any sense?

Context: The capacitor is charged with charge $Q_0$. There is a magnetic field $B=B_0 \hat{z}$ perpendicular to the monitor's screen. The length of the (AB) wire is $\ell$. There is no friction.

Goal: I am trying to calculate the velocity of the (AB) wire. From the 2nd law of Newton, we have:
$$m \frac{du}{dt}=F_\text{Laplace} = B\ell I(t)$$
In my lecture notes I see that $I(t)$ is replaced by the current $I_c(t)$ which corresponds to the current due to the discharge of the capacitor, as if there was no magnetic field, i.e. as if $\mathcal{E}=0$. So it goes like:
\begin{align}
V_c(t)&=V_0 \exp(-t/RC) \Rightarrow\\dV_c/dt &= -V_0\left(\frac{1}{RC}\right) \exp(-t/RC)=-I_c(t)/C \Rightarrow\\I_c(t) &= -C dV_c/dt
\end{align}
And then substitutes $-C dV_c/dt$ in the place of $I(t)$ at the Newton's law:
\begin{align}
m \frac{du}{dt} &= B \ell \left( -C dV_c / dt \right) \Rightarrow\\
du &= -\frac{B C\ell}{m} dV_c \Rightarrow\\
u(t) - u(0) &= -\frac{BC\ell}{m} (V_c(t) - V_0)\Rightarrow\\
u(t) &= \frac{BC\ell}{m} \left(V_0 - V_c(t)\right)
\end{align}
Questions: Why use $I_c(t)$ instead of $I(t)$? We do have an induced electromotive force so it should be: $$I(t) = \frac{V_c(t)-\mathcal{E}(t)}{R}$$
What am I missing here?

After going through all the calculations he ends with the following differential equation:
$$
I(t) = \frac{V_c(t)-\mathcal{E}}{R} = \frac{V_c(t) - u(t)B\ell}{R} \Rightarrow\\
\frac{dV_c}{dt} + \underbrace{\left(\frac{\ell^2 B^2}{m R} + \frac{1}{RC}\right)}_
{k_1}V_c(t) = \underbrace{\frac{\ell^2 B^2}{m R}}_{k_2}V_0
$$
With the following solution:
$$
V_c(t) = V_0 \left(1 - \frac{k_2}{k_1}\right)e^{-k_1 t} + \frac{k_2}{k_1}V_0
$$
And after some calculations the velocity $u(t)$ is given by:
$$
u(t) = \frac{BC\ell}{m}V_0 \left(1 - \frac{k_2}{k_1}\right)(1-e^{-k_1 t})
$$
with $k_1,k_2$ being constants depending on $B,C,m,R$, etc.
This equation looks weird because there is a combination that leads to zero velocity ? ($k_1=k_2$)
I'm at a loss here! 
 A: Current flows in a closed loop, so
$$I_c = I_r\tag1$$
There is a voltage across the capacitor due to the integrated charge, across the resistor due to the instantaneous current, and across the whole loop due to the motion of the resistor. The net voltage around the loop is zero, so
$$V_c - I_r R + \mathcal{E}=0\tag2$$
The force on the wire is given by the product of field, length and current:
$$F = B\ell I\tag3$$
Where we can use either $I_c$ or $I_r$ since they are the same...
Now I disagree with the expression for the evolution of the voltage of the capacitor as a function of time - since the way it is written in your derivation 
$$V(t)=V_0 e^{-t/RC}\ \ \ \ \ \ \text{NOT VALID}\tag4$$
assumes there is no back e.m.f. from the motion of the resistor - but there is. We do know, though, that the voltage on the capacitor is a function of the current, so
$$Q = CV\\
\frac{dQ}{dt}=C\frac{dV}{dt}\\
I_c(t)=-C\frac{dV}{dt}\tag5$$
is correct - but didn't need the involved "derivation" in your work (note the sign is there because when current flows DOWN, the TOP plate will become positively charged). And since the current is flowing in a closed loop (see my equation (1) above) we can replace $I$ in equation (3) with $I_c$ (although you have to be careful of your signs).
At this point, you can make an equation relating velocity to the (remaining) voltage on the capacitor as you show in your derivation, but I don't think that you can then substitute the expression (4) for the voltage - because the rate at which the capacitor discharges depends on the voltage developed across the resistor PLUS the back EMF.
So the derivation makes sense; however, after you do this:
$$dU=-\frac{BC\ell}{m}dV_c\\
U(t) - U(0) = -\frac{BC\ell}{m}(V_c(t)-V_0)$$
you need to take into account that the time evolution of the voltage on the capacitor is not the simple function given in (4). I have not seen the rest of your work, so I can't comment on whether the final result is correct.
As for the second part of your question - how can the solution predict a velocity of $0$ when $k_1=k_2$, you figured out yourself (in comments) that this cannot happen - or at least, that it can only happen when $\frac{1}{RC}=0$, which implies infinite resistance and therefore no current flowing. At which point, indeed, the resistor will not move as you effectively have an open circuit.
I hope this helps - leave comments if you need more (or give the rest of the derivation if you still have trouble with it).
