The electric field acts positive `x`-axis. A charged particle of charge `q` and mass `m` is released from origin and moves with velocity `vec(v)=v_(0)hatj` under the action of electric field and magnetic field, `vec(B)=B_(0)hati`. The velocity of particle becomes `2v_(0)` after time `(sqrt(3)mv_(0))/(sqrt(2)qE_(0))`. find the electric field:

Options

A. `(sqrt(2))/(sqrt(3)) E_(0)hati`
B. `(sqrt(3))/(sqrt(2)) E_(0)hati`
C. `(sqrt(3)) E_(0)hati`
D. `(sqrt(2)) E_(0)hati`

Answer

D

Explanation

`vec(v)` is perpendicular to both `vec(E)` and `vec(B)`.
Path of particle is helix with increasing pitch
`v=(v_(x)^(2)+v_(y)^(2)+v_(z)^(2))^(1//2)`
Here, `v_(x)^(2)=((qE)/mt)^(2)` and `v_(y)^(2)+v_(z)^(2)=v_(0)^(2)`
Also `v=2v_(0)`
`(2v_(0))^(2)=(q^(2)E^(2)t^(2))/(m^(2))+v_(0)^(2)`
`t=(sqrt(3)mv_(0))/(qE)=(sqrt(3)mv_(0))/(sqrt(2)qE_(0))` (given)
`vec(E)=sqrt(2) E_(0)hati`