Magnetic force on a charged particle is given by `vec F_(m) = q(vec(v) xx vec(B))` and electrostatic force `vec F_(e) = q vec (E)`. A particle having charge q = 1C and mass 1 kg is released from rest at origin. There are electric and magnetic field given by `vec(E) = (10 hat(i)) N//C for x = 1.8 m` and `vec(B) = -(5 hat(k)) T` for `1.8 m le x le 2.4 m`
A screen is placed parallel to y-z plane at `x = 3 m`. Neglect gravity forces.
The speed with which the particle will collide the screen is

Options

A. `3 m s^(-1)`
B. `6 m// s^(-1)`
C. `9 m//s^(-1)
D. `12 m// s^(-1)`

Answer

B

Explanation

Velocity at `A v=sqrt(2 as)`
`=sqrt(2xx((qE)/(m))s) = sqrt((2 xx 1 xx 10 xx 1.8)/(1))=6 m//s`
In magnetic field, speed does not change. Hence particle will collide with speed `6m//s`.