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 axis as shown in figure.According to the right-handed system, the `Z`-axis 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
`vecu=u_(x)hveci+u_(y)vecj`
The electric and magnetic fields may be written as
`vecE=-Evecj`
and `vecB=-Bveck`
respectively.The force on the electron is
`vecF=e(vecE+vecuxxvecB)`
`=eEvecj+eB(u_(y)veci-u_(x)vecj)`
Thus, `F_(x)=eu_(y)B`
and `F_(y)=e(E-u_(x)B)`.
The components of the acceleration are
`a_(x)=(du_(x))/(dt)=(eB)/mu_(y)`
and `a_(y)=(du_(y))/(dt)=e/m(E-u_(x)B)`
We have, `(d^(2)u_(y))/(dt^(2))=-(eB)/m(du_(x))/(dt)=-(eB)/m.(eB)/m"u"_(y)`
where `omega=(eB)/m`
This equation is similar to that for a simple harmonic motion.Thus,
`u_(y)=A sin (omegat+delta)`...(iv)
and hence, `(du_(y))/(dt)=A omega cos(omegat+delta)`...(v)
At `t=0,u_(y) ` and `(du_(y))/ (dt)=F_(y)/(dt)=(eE)/m`.
Putting in (iv) and (v).
`delta=0` and `A=(eE)/(momega)E/B`.
Thus, `u_(y)=E/B sin omega t`
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 `underset(0)overset(y)(dy)=E/B sin omegat dt` or `y=E/(Bomega)(1-cos omega)`
At `t=pi/omega,y=E/(Bomega)(1-cos pi)=(2E)/(Bomega)`
Thus, the displacement along the `Y` -axis is
`(2E)/(Bomega)=(2Em)/(BeB)=(2Em)/(eB^(2))`.