Question

A wire of mass $m$ and length $l$ can slide freely on a pair of smooth, vertical rails (Figure). A magnetic field $B$ exists in the region in the direction perpendicular to the plane of the rails. The rails are connected at the top end by a capacitor of capacitance $C$. Find the acceleration of the wire neglecting any electric resistance.

Answer 1

Let the rod has a velocity $v$ at any instant
then at the point
$e=Blv$
now $q=c\times potential=cd=CBlv$
current $I=\frac{dq}{dt}=\frac{d}{dt}CBlv=CBl\frac{dv}{dt}=CBla$ where $a$ is acceleration
from figure, force due to magnetic field and gravity are opposite to each other
so $mg-IlB=ma\Rightarrow mg-CBla\times lB=ma\Rightarrow ma+C{B}^2{l}^{2}a=mg\Rightarrow a(m+C{B}^2{l}^2)=mg\Rightarrow a=\frac{mg}{m+C{B}^2/2}$