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

(N/A) When a rod of length $d$ moves with velocity $v$ perpendicular to a magnetic field,an electromotive force (emf) is induced between the two ends of the rod.
The induced emf is $\varepsilon = Bvd$.
When the switch $S$ is closed at $t = 0$,the current in the inductor starts to increase.
Let the current in the circuit be $I$ at time $t$. According to Kirchhoff's voltage law:
$\varepsilon = V_R + V_L$
$Bvd = IR + L \frac{dI}{dt}$
Rearranging the terms,we get the linear differential equation:
$\frac{dI}{dt} + \frac{R}{L}I = \frac{Bvd}{L}$
The solution to this linear differential equation is of the form:
$I(t) = \frac{Bvd}{R} + A e^{-\frac{R}{L}t}$
Using the initial condition that at $t = 0$,$I = 0$:
$0 = \frac{Bvd}{R} + A \implies A = -\frac{Bvd}{R}$
Substituting $A$ back into the equation:
$I(t) = \frac{Bvd}{R} \left(1 - e^{-\frac{R}{L}t}\right)$
This is the required expression for the current in the rod as a function of time.