Question

In a chambe of a uniform magnetic field of `8*0G[1G=10^-4T]` is maintained. An electron with a speed of `4*0xx10^6ms^-1` enters the chamber in a direction normal to the field
(i) Describe the path of electron.
(ii) What is the frequency of revolution of electron?
(iii) What happens to the path of the electron if it progressively loses its energy due to collisions with the atoms or molecules of the environment?

Answer 1

Here, `B=8*0G=8*0xx10^-4T`,
`v=4*0xx10^6ms^-1`, `theta=90^@`.
If r is the radius of circular path, then
`Bevsin90^@=(mv^2)/(r)` or `r=(mv)/(Be)`
`:. r=((9*1xx10^(-31))xx(4*0xx10^6))/((8*0xx10^-4)xx(1*6xx10^-19))`
`=2*8xx10^-2m=2*8cm`
The sense of rotation of electron in a circular path can be predicted from the direction of centripetal force, `vecF=-e(vecvxxvecB)`. If we see along the direction of `vecB`, the electron will be revolving in clockwise direction.
(ii) The frequency of revolution of electron
`v=(eB)/(2pim)=((1*6xx10^(-19))xx(8*0xx10^-4))/(2xx3*14xx(9*1xx10^(-31))`
(iii) If the electron loses its energy in successive collisions, then electron loses its speed progressively. If after collision, the velocity of electron remains in the same plane of the initial circular orbit, the radius of the circualr orbit will decrease in proportion to the decreased speed. If it is not so then the path of the electron will be helical between two collisions.