Two long horizontal frictionless rails 0.50 m apart are connected at one end by an ideal capacitor of capacitance 0.80 F. A uniform vertical magnetic…

At speed $v$ the induced emf is $Blv$, and with no resistance it equals the capacitor voltage, so the charge is $q=CBlv$. The current is $i=dq/dt=CBl\,a$. The magnetic force on the rod is $iBl=CB^2l^2a$, opposing the pull. Newton's second law: $F-CB^2l^2a=ma$, so $a=\dfrac{F}{m+CB^2l^2}$. Here $CB^2l^2=0.8\times4\times0.25=0.80\ \mathrm{kg}$, so $a=2.0/(0.20+0.80)=2.0\ \mathrm{m\,s^{-2}}$, constant. Then $i=CBla=0.8\times2.0\times0.5\times2.0=1.6\ \mathrm{A}$. The value $2.0\ \mathrm{A}$ is the naive $F/Bl$, which would hold only if the rod were not accelerating, and ignores the effective mass added by the capacitor. Energy check: the work done by the force goes partly into kinetic energy and partly into the capacitor, and the two stores, $\tfrac12mv^2$ and $\tfrac12CB^2l^2v^2$, grow in the ratio $m:CB^2l^2=0.2:0.8$, so only one fifth of the work goes into kinetic energy here.