Two metal bars are fixed vertically and are connected on the top by a capacitor `C`. A sliding conductor of length land mass m slides with its ends in contact with the bars. The arrangement is placed in a uniform horizontal magnetic field directed normal to the plane of the figure. The conductor is released from rest. Find the displacement `x(t)` of the conductor as a function of time t.

Explanation

Let at anytime `t` , velocity of wire is `v`
Induced emf `e=Bvl`
Change on capacitor `q=Ce+CBlv`
Induced current `I=(dq)/(dt)=CBl(dv)/(dt)=Cbla` , anticlockwise
Magnetic force `F_(m)=Bil=CB^(2)l^(2)a` lt upward
`mg-F_(m)=ma`
`mg-CB^(2)l^(2)a=ma`
`a=(mg)/((m+CB^(2)l^(2))` (a : constant acceleration)
`x=-(1)/(2)at^(2)`
`=(mgt^(2))/(2(m+Cb^(2)l^(2)))`