A particle having mass m and charge `q` is released from the origin in a region in which ele field and magnetic field are given by `B=-B_0hatj`and `E=E_0hatk`.
Find the y- component of the velocity and the speed of the particle as a function of it z-coordinate.

Explanation

In Fig shown `B_0` is outward. As soon as we release the
particle at origin. Electric field will apply force `F_e=qE_0` on the
particle. Due to this force, the charge particle will gain accelera-
tion and hence velocity along z-axis. And as it gains velocity
along z-axis, then magnetic field will apply the force `F_b` on charge particle. Now both the forces are always in x-z plane, so
path of charge will be in x-z plane as shown. Let at any time its
velocity is `vecv=v_xhati+v_zhatk`. Let `veca=a_xhati+a_zhatk` is the acceleration
at any time.

Then `vecF_(net)=mveca`
`implies qE_0hatk-q(v_xhati+v_zhatk)xxB_0hatj=m(a_xhati+a_zhatk)`
`implies qE_0hatk-qv_xB_0hatk+qv_zB_0hati=m(a_xhati+a_zhatk)`
`implies ma_x=qv_zB_0 and ma_z=q(E_0-v_xB_0)...(i)`
`implies m(dv_x)/(dt)=qv_zB_0 and m(dv_z)/(dt)=q(E_0-v_xB_0)...(ii)`
From Eq. (ii), `(dv_z)/(dt)=q/m(E_0-v_xB_0)implies (d^2v_z)/(dt^2)=(-qB_0)/m (dv_x)/(dt)`
`implies (d^2v_z)/(dt^2)=(-qB_0)/m (qB_0)/mv_z=-((qB_0)/m)^2v_z`
Solution of the above equation is : `v_z=v_(z0) sin (omegat+phi) .....(iii)`
(where `omega=(qB_0)/m`)
Puting at `t=0, v_z=0, we get phi=0`
Differentiating Eq. (iii),
`(dv_z)/(dt)=v_(z0)omegacos(omegat+phi) implies a_z=omegav_(z0) cos omegat`
at `t=0, a_z=(F_e)/m=(qE_0)/m`
`implies (qE_0)/m=(qB_0)/m v_(z0) cos 0^@ implies v_(z0)=(E_0)/(B_0)`
So Eq. (iii) can be written as : `v_z=(E_0)/(B_0) sin(omegat)....(iv)`
Puting the value of `v_z` in Eq. (ii), `m(dv_x)/(dt)=(qE_0)/(B_0) (sin omegat)B_0`
`implies int_0^vdv_x=int_0^t(qE_0)/m sin omegatdt implies v_x=(qE_0)/(momega)[1-cos omegat]`
`implies v_x=(qE_0)/(m(qB_(0))//m)[1-cos omegat]implies v_x=(E_0)/(B_0)[1-cosomegat]....(v)`
Equations (iv) and (v) give velocity as a function of time.
From Eq.(v), `(dx)/(dt)=(E_0)/(B_0)(1-cos omegat)`
`implies int_0^xds=(E_0)/(B_0)int_0^t(1-cosomegat)dt`
`implies x=(E_0)/(omegaB_0)[omegat-sinomegat]....(vi)`
From Eq.(iv)
`(dz)/(dt)=(E_0)/(B_0) sin omegat implies int_0^zdz=(E_0)/(B_0)int_0^tsinomegatdt`
`implies z=(E_0)/(omegaB_0)[1-cosomegat]....(vii)`
Equations (vi) and (vii) give the path of the verticle.
We see that at `t=2pi`
`x=(E_0)/(B_0)2pi, z=0`

This path is known as cycloided path. This is the same kind of path as followed by a point on the rim of a purely rolling wheel, which is expalined as below.
Let a wheel is rolling purely with angular velocity `omega` as shown.
Let at `t=0`, a point P on the rim is at the bottom-most position of
the wheel. After time t, the wheel turns by an angle `theta` as shown
and point P comes at point P'. Let us find the coordinates of P'.

`x=R theta-R sin theta=Romegat-Rsin omegat=R[omegat-sinomegat]...(viii)`
`z=R-Rcos theta=R[1-cosomegat]....(ix)`
Equations (viii) and (ix) resemble Eqs (vi) and (vii).
Hence, the resulting path followed by charge is cycloidal.
(b) To find velocity as function of z-coordinate.
From equations (iv) and (v)
`v=sqrt(v_x^2+v_z^2)=(E_0)/(B_0)sqrt((1-cosomegat)^2+(sinomegat)^2)`
`=(E_0)/(B_0)sqrt((1-cos omega t)^2+(1-cos omega t)(1+cos omega t))`
`=(E_0)/(B_0) sqrt(1-cos omega t)^2=(E_0)/(B_0)sqrt((z omega B_0)/(E_0))2 [From Eq. (vii)]`
`=sqrt((E_0^2)/(B_0^2) (zqB_0)/m (B_0)/(E_0) 2)=sqrt((2qE_0Z)/m)`
Alternate Method to find velocity: Since the magnetic field does
not perform any work, therefore, whatever has been the gain in
kinetic energy it is only because of the work done by electric
field. Applying work-energy theoram, `W_E=DeltaK`
`implies qE_0z=1/2mv^2-0 or v=sqrt((2qE_0Z)/m)`