One end of a horizontal track of gauge l and negligible resistance, is connected to a capacitor of capacitance C charged to voltage Vo. The inductance of the assembly is negligible. The system is placed in a homogeneous, vertical magnetic field of induction B, as shown in the figure.

A frictionless conducting rod of mass m and resistance R is placed perpendicularly onto the track. The polarity of the capacitor is such that the rod is repelled from the capacitor when the switch is turned over. What is the maximum velocity of the rod?

At the instant when the capacitor is connected, a current I=V₀/R starts flowing in the rod, which experiences a force F=BlI and an initial acceleration a=BlV₀/mR. In accordance with Lenz’s law, the voltage induced in the moving rod causes the current flowing in the rod to decrease. The charge Q on the capacitor decreases and consequently so does the voltage across it. Meanwhile the voltage induced in the rod increases, until the two voltages cancel out each other. The rod then continues with its maximum velocity given by

$Bl{v}_{\text{max}}=\frac{Q_{\text{min}}}{C}$ ………..$\left(1\right)$

The equation of motion of the rod is 

$m\frac{dv}{dt}=ma=BlI=-\frac{dQ}{dt}$………….$\left(2\right)$

where the acceleration and the current have been expressed as the rates of change in velocity and charge, respectively. The proportionally between the two rates of change holds throughout. The speed of the rod increases from zero to $v_{max}$, whilst the charge on capacitor decreases from $Q_0=C{V}_0$ to $Q_{min}$. Equation $\left(2\right)$ can therefore be written as 

$m{v}_{\max }=Bl(Q_0-Q_{\min })$

The final velocity and the residual charge on the capacitor can be calculated using equations $\left(1\right),\left(2\right)$

$v_{max}=\frac{BlC{V}_0}{m+B^2{l}^{2}C}$ and $Q_{min}=\frac{B^2{l}^2{C}^2{V}_0}{m+B^2{l}^{2}C}$