An electron is released from the origin at a place where a uniform electric field E and a uniform magnetic field B exist along the negative y-axis and the nagative z-axis respectively. Find the displacement of the electron along the y-axis when its velocitybecomes perpendicular to the electric feld for the first time.

Explanation

Let us take axes as shown in
According to the right handed system, the z-exis is
upward in the figure and hence the magnetic field
is shown downwards. At any time, the velocity of the
electron may be written as
`vec mu =u_x veci+u_y vec j.`
The electric and magnetic fields may be written as
`vecE =- Evecj`
and `vecB=-B veck`,
respectively . The force on the electron is
`vecF=- e(vecE+vecu XvecB)`
`= eEvecJ+eB(u_y veci - u_x vecJ)`.
Thus, `F_x =eu_yB`,
and `F_y= e(E-u_xB)`.
The components of the acceleration are
`a_x =(du_x)/(dt) = (eB)/(m) u_y`
and `a_y =(du_y)/(dt) = (e)/(m)(E-u_xB)`.
we have,
`(d^2u_y)/(dt^2)=-(eB)/(m)(du_x)(dt)`
`= (eB)/(m). (eB)/(m)u_y`
=-omega^2u_y`
where `omega= (eB)/(m) (iii)`
This equation is similat to that for a simple harmonic
motion. thus,
`u_y = A sin)omegat+delta) (iv)`
and hence,
`(du_y)(dt) = A omega cos(omegat+s). (v)`
`At t = 0, u_y = 0` and `(du_y)/(dt) = (F_y)/ (m)= (eE)/(m)`.
Putting in (iv) and (v),
`delta = 0 and A = (eE)/(momega) = (E)/(B)`.
Thus`u_y = (E)/(B) sin omegat`.
The path of the electron will be perpendicular to the
y-axis when u_y=0. This will be the case for the first
time at t where.
`sin omegat = 0`
or, `omegat= pi`
or,`t=(pi)/(omega)=(pim)/(eB)`.
Also, `u_y = (dy)/(dt)=(E)/(B) sin omegat`.
or, `int_0^y dy= (E)/(B) int_0^t sin omegat dt`.
or,`y=(E)/(B omega) (1-cos omegat)`.
At `t =(pi)/(omega),`
`y=-(E)/(B omega)(1 - cos pi) = (2E)(B omega).`
Thus, the displacement along the y-axis is
`(2E)/B omega) = (2Em)/(BeB) =(2Em)/(eB^2).`