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.

Answer

B, C

Explanation

Let `v` be the velocity at some instant. Then motional emf `V=Bvl`
Charge stored in capacitor `q=CV=(CBl)v`
Current in the wire `=(dq)/(dt)=(CBl)(dv)/(dt)`
Magnetic force `F_m=ilB=(CB^2l^2)(dv)/(dt)`(upwards)
`:.` Net force `F_("net") = mg - F_(m)`
or `m=(dv)/(dt) = mg -(CB^(2)l^(2)) (dv)/(dt)`
`:. (dv)/(dt) =`acceleration, `a=(mg)/(m+CB^2l^2)`
Since, `a =`constant
`:. x=1/2at^2=(mg t^2)/(2(m+CB^2l^2))`