Unifrom electric and magnetic fields with strength `E` and induction `B` respectively are directed along the `y` axis (Fig). A particle with specific charge `q//m` leaves the origin `O` in the direction of the `z` axis with an initial non-realtive velocity `v_(0)` find:
(a) the coordinate `y_(n)` of the particle when it crossses then `y` axis for the nth time,
(b) the angle `alpha` between the particle's velocity vector and the `y` axis at that moment.

Explanation

The equaction of motion are,
`m (dv_(x))/(dt) = -q Bv_(z), m (dy_(y))/(dt) = q E` and `m(dv_(z))/(dt) = q v_(x) B`
These equactions can be solved easily.
First, `v_(y) = (qE)/(m) t, y = (qE)/(2m) t'^(2)`
Then, `v_(x)^(2) + v_(z)^(2)` = constant `= v_(0)^(2)` as before.
Intergating again and using `x = z = 0`, at `t = 0`
`x = (v_(0))/(omega) sin omega t, z = (v_(0))/(omega) (1 - cos omega t)`
Thus, `x = z = 0` for `t = t_(n) = n (2pi)/(omega)`
At what instant, `y_(n) = (qE)/(2m) xx (2pi)/(qB//m) xx n^(2) xx (2pi)/(qB//m) = (2pi^(2) m E n^(2))/(qB^(2))`
Also, `tan alpha_(n) = (v_(x))/(v_(y))`, (`v_(z) = 0` at this moment)
`= (mv_(0))/(q E t_(n)) = (mv_(0))/(qE) xx (qB)/(m) xx (1)/(2pi n) = (B v_(0))/(2pi E n)`