Question

Two parallel long smooth conducting rails seperated by a distance $l$ are connected by a movable conducting connector of mass ${}^{\prime }{m}^{\prime }$. Terminals of the rails are connected by the resistor $R$ and the capacitor $C$ as shown. A uniform magnetic field $B$ perpendicular to the plane of the rails is switched on. The connector is dragged by a constant force $F$. If the force $F$ is applied at $t=0$, then find the terminal velocity of the connector (in $m/s$. Given that $FR=4B^2{l}^2.$

Answer 1

In steady state, capacitor will act like an open switch.

Let the terminal velocity of the connector be $v$.

Current induced in the resistor $i=\frac{\text{Induced emf}}{R}=\frac{Blv}{R}$

Power required to move the connector with velocity $v$ will be equal to the power dissipated in the resistor (so that there is no increase in kinetic energy of the connector).

$Fv=i^{2}R={\left(\frac{Blv}{R}\right)}^{2}R$
$\Rightarrow Fv=\frac{B^2{l}^2{v}^2}{R}$
$\Rightarrow v=\frac{FR}{B^2{l}^2}=4$
(Given $FR=4B^2{l}^2$)