Solution

Suppose the steady-state speed of the tape is $v$. Then, in a short time interval $\Delta t$ the charge $-qv\Delta t/l$ is removed from the region between the plates upward along the tape. Equivalently, conventional current $I=qv/l$ flows upward through the resistor, which generates heat at the rate of $I^2R=q^2v^2R/l^2$. Since this must equal equal the power input $Fv$ of the system, we have

$$\frac{q^2v^2R}{l^2}=Fv\qquad\Rightarrow\qquad v=\frac{Fl^2}{q^2R},$$

and thus $I=Fl/(qR)$.