11.1.20∗. Referring to Problem 11.1.19, find the dependence of the conductor speed on time with zero initial speed in the case when the upper ends of the rails are closed: a) on the resistance $R$; b) on the capacitance $C$.
Solution
For problem $11.1.20$
In addition to gravity, the conductor experiences upward magnetic force $F=IlB$, where $I$ is the current through the conductor. The current flows in the circuit, because there is an induced EMF $\mathcal{E}=Blv$, where $v$ is the downward speed of the conductor, as a result of increasing area of the circuit loop and hence magnetic flux.
In the case when the circuit is closed with the resistor, $I=\mathcal{E}/R=Blv/R$, and the equation of motion of the conductor becomes
$$m\dot v=mg-B^2l^2v/R.$$
Subject to the initial condition $v=0$ at $t=0$, we have the solution
$$v=\frac{mgR}{B^2l^2}(1-e^{-B^2l^2t/(mR)}).$$
In the case when the circuit is closed with the capacitor, $I=C(d\mathcal{E}/dt)=CBl\dot v$, and the equation of motion of the conductor becomes