A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me `= 9.11 × 10^(–31) kg)`. [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]

Explanation

Energy of an electron beam, E=18KeV=`18xx10^(3)eV`
Charge on an electron, e`=1. 6xx10^(-19)C`
`E=18xx10^(3)xx1.6xx10^(-19)J`
Mgnetic field, B=0.04G
Mass of an electron, `m_(e)=9.11xx10^(-19)kg`
Distance up to which the electron beam travesls, d=30cm=0.3m
We can write the kinetic energy of the electron beam as:
`E=(1)/(2)mv^(2)`
`v=sqrt((2E)/(m))`
`=sqrt(2xx18xx10^(3)xx1.6xx10^(-19)xx10^(-15))/(9.11xx10^(-31))=0.795xx10^(8)m//s`
The electron beam deflects along a circular path of radius r
The force due to the magnetic field balances the centripetal for
`BeV=(mv^(2))/(r)`
`thereforer=(mv)/(Be) `
`=(9.11xx10^(-34)xx0.795xx10^(8))/(0.4xx10^(-4)xx1.6xx10^(-19))=11.3m`
Let the up and down deflection of the electron beam be `x=r(1-cos tehta)` ltbr. where,
`theta=` Angle of declination
`sintehta=(d)/(r)`
`=(0.3)/(11.3)`
`theta=sin^(-1)(0.3)/(11.3)=1.521^(@)`
=0.0039m=3.9mm
Therefore, the up and down deflectionof the beam is 3.9mm.