Question

An electron accelerated by a potential difference `V= 3` volt first enters into a uniform electric field of a parallel plate capacitor whose plates extend over a length `l= 6 cm` in the direction of initial velocity. The electric field is normal to the direction of initial velocity and its strength varies with time as `E= alpha t,` where `alpha= 3600 V m^-1 s^-1.` Then the electron enters into a uniform magnetic field of induction `B = pi xx 10^-9 T.` Direction of magnetic field is same as that of the electric field. Calculate pitch of helical path traced by the electron in the magnetic field. (Mass of electron, `m = 9 xx 10^-31 kg`)

Answer 1

Since, electron is accelerated thorugh a potential difference
V, therefore, its initial velcity `v_0` is given by `1/2mv_0^2=eV`
`v_0=sqrt((2eV)/m)....(i)`
Since, initial velocity is paralle to plates or normal to the
direction of electric field, therefore, component of velocity
parallel to plates remains constant as `v_0`.

Hence, time taken by the electron to cross the electric field is
`t_0=1/(upsilon_0)`
Now consider motion of electron , normal to plates.
At some instant t, its acceleration `=(eE)/m=(e alpha t)/m`
Let velocity component normal to plates to `v_y`
Then this acceleration is equal to `d/(dt)upsilon_y`
`d/(dt)upsilon_y=(ealphat)/m` or `d upsilon_y=(ealphat)/mdt`
But at initial moment`t=0, v_y=0 implies int_0^(v_y)dupsilon_y=(ealpha)/(m) int_0^(t_0) tdt`
`upsilon_y=(ealpha)/(2m)t_0^2=(alphal^2)/(4V)....(ii)`
Angular deviation `theta` of electron form its initial direction of motion is shown in Fig.
Now electron enters into magnetic field. Pitch of its helical path
`P=(2pim)/(eB)vcos(90- theta)`

`=(2pim)/(eB)vsin theta=(2pim)/(eB)v_y`
`=(pimalphaI^2)/(2eBV)=1.214 cm`.