Find the current in the sliding rod $AB$ (resistance $= R$) for the arrangement shown in the figure. The magnetic field $\vec{B}$ is constant and directed out of the paper. The parallel wires have no resistance,and the velocity $v$ is constant. The switch $S$ is closed at time $t = 0$.

(N/A) rod of length $d$ moves with velocity $v$ perpendicular to a magnetic field $B$,inducing an electromotive force (emf) $\varepsilon = Bvd$ between its ends.
When the switch $S$ is closed at $t = 0$,the capacitor begins to charge. Let $Q(t)$ be the charge on the capacitor at time $t$.
Applying Kirchhoff's voltage law to the circuit:
$\varepsilon = V_R + V_C$
$Bvd = IR + \frac{Q}{C}$
Since $I = \frac{dQ}{dt}$,we have:
$R \frac{dQ}{dt} + \frac{Q}{C} = Bvd$
$\frac{dQ}{dt} + \frac{Q}{RC} = \frac{Bvd}{R}$
This is a first-order linear differential equation. The general solution for $Q(t)$ is:
$Q(t) = BvdC + A e^{-t/RC}$
At $t = 0$,the charge $Q(0) = 0$,so $0 = BvdC + A$,which gives $A = -BvdC$.
Thus,$Q(t) = BvdC(1 - e^{-t/RC})$.
The current $I(t)$ is the derivative of charge with respect to time:
$I(t) = \frac{dQ}{dt} = \frac{d}{dt} [BvdC(1 - e^{-t/RC})]$
$I(t) = BvdC \cdot \left( \frac{1}{RC} \right) e^{-t/RC}$
$I(t) = \frac{Bvd}{R} e^{-t/RC}$