A charged particle is projected with a certain velocity between the plates with opposites charges. The magnetic field is also applied between the plates so that the particle moves undeflected and hits the plate `A`. The force experienced by the plate is `F_(1)`. Now if the particle is projected with double of the initial velocity with smaller conditions, the force experienced by the plate is `F_(1)//F_(2)` is :

Options

A. `2`
B. `(1)/(2)`
C. `(1)/(4)`
D. `1.0`

Answer

D

Explanation

`qE=qvBv=E//B`
`F=(mv)/(Deltat)=(mvl)/(q)` (since `I=(q)/(Deltat)`)
` = (mI)/(q).(E)/(B)`
When the particle remains undeflected by adjusting `E` and `B` , force is independent of velocity.