Two vertical conducting rails separted by distance `1.0m` are placed parallel to `z`-axis as shown in figure. At `z=0`, a capacitor of `0.15 F` is connected between the rails and a metal rod of mass `100g` placed across the rails slides down along the rails. if a constant magnetic fields of `2.0` `T` exists perpendicular to the plane of the rails, what is the acceleration of the rod?

Options

A. `2.5m//s^(2)`
B. `zero`
C. `9.8m//s^(2))`
D. `1.4m//s^(2))`

Answer

D

Explanation

Motional emf induced across capacitor, `e=Blv`
:. Charge trored in capacitor, `Q=Ce=C(Blv)`
current in the rod `I=(dQ)/(dt)=CBl(dv)/(dt)=Cbla`
force oppoosing the downward motion, `F_(m)=Bll`
`F_(m)=B(Cbla)l=B^(2)l^(2)Ca`
From `FBD` of rod
`ma=mg-B^(2)l^(2)Ca`
or `a=(mg)/((m+B^(2)l^(2)C))`
so, `a=(0.1xx9.8)/((0.1+2^(2)xx1^(2)xx0.15))=1.4m//s^(2)`