A particle of mass m and charge q is moving in a region where uniform, constant electric and mangetic fields `vec E and vec B` are present. `vec E and vec B` are parallel to each other. At time `t=0,` the velocity `vec v_0` of the particle is perpendicular to `vec E` (Assume that its speed is always `lt lt c`, the speed of light in vacuum). Find the velocity `vec v` of the particle at time `t`. You must express your answer in terms of `t, q, m,` the vector `vec v_0, vec E` and `vec B` and their magnitudes `vec v_0, vec E` and `vec B`.

Explanation

`hatj=(vecE)/E` or `(vecB)/B, hati=(vecv_0)/(v_0),hatk=(vecv_0xxvecB_0)/(v_0B)`
Force due to electric field will be along y-axis. Magnetic force will not affect the motion of changed particle in the direction of electric field (or y-axis) so.
`a_y=(F_e)/m=(qE)/m,v_y=a_yt`
`=(qE)/mT`....(i)

The changed particle under the action of magnetic field describes a circle in x-z plane (perpendicular to `vecB`) with `T=(2pim)/(Bq)` or `omega=(2pi)/T=(qB)/m`
Initially `(t=0)` velocity was along x-axis. Therefore, magnetic force `(vecF_m)` will be along positive z-axis `[vecF_m=q(vec v_0xxvecB)]`. Let it this force makes an angle `theta` with x-axis at time t, then
`theta=omegat`
`:. v_x=v_0cos omegat=v_0cos((qB)/mt)`...(ii)
`v_z=v_0sin omegat=v_0sin((qB)/mt)`...(iii)
From Eqs. (i), (ii) and (iii)
`vecv=v_xhati+v_yhatj+v_zhatk`
`:. vecv= v_0cos((qB)/mt)((vecv_0)/(v_0)) +(qE)/mt((vecE)/E)+v_0sin((qB)/mt)((vecv_0xxvecB)/(v_0B))`
Or `vecv=cos((qB)/mt)(vecv_0)+(q/mt)(vecE)+sin((qB)/mt)((vecv_0xxvecB)/B)`
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