Question

In a certain region of space, there exists a uniform and constant electric field of magnitude E along the positive y-axis of a coordinate system. A charged particle of mass m and charge -q (qgt0) is projected form the origin with speed 2v at an angle of `60^@` with the positive x-axis in x-y plane. When the x-coordinate of particle becomes `sqrt3mv^2//qE` a uniform and constant magnetic. field of strength B is also switched on along positive y-axis.
z-coordinate of the particle as a function of time after the magnetic field is switched on is
A. `(mv)/(qB)[1-cos((qB)/mt)]`
B. `-(mv)/(qB)[1+cos((qB)/mt)]`
C. `-(mv)/(qB)[1-cos((qB)/mt)]`
D. `(mv)/(qB)[1+cos((qB)/mt)]`

Answer 1

Correct Answer - c
First particle will travel along parabolic path OA. Let time
from O to A is t. `a_y=(-qE)/m`

`x=(sqrt3mv^2)/(qE)=(2v cos 60^@)t_0 implies t_0 =(sqrt3mv)/(qE)`
`v_y=u_y+a_yt_0=2v sin 60^@-(qE)/m (sqrt2mv)/(qE)=0`
Hence at point A, velocity will be purely along x-axis and it will
be `2v cos60^@=v`. Now magnetic field is switched on along y-axis.
Now its path will be helical as shown below with increasing pitch
towards negative y-axis.

`r=(mv)/(qB)`
`x=x_0+r sin theta=(2v cos 60^@)t_0+(mv)/(qB) sin omegat`
`=vsqrt3(mv)/(qB) +(mv)/(qB)sin ((qB)/mt)`
z-coordinate: `z=-(r-rcos theta)=-(mv)/(qB)[1-cos ((qB)/mt)]`