A monoenergetic electron of initial energy `18 keV` moving horizontally is subjected to a horizontal magnetic field of `0*4G` normal to its initial direction. Calculate the vertical deflection of the beam over a distance of `30cm`.
Mass of electron`=9*1xx10^(-31)kg`,
charge of electron `=1*6xx10^(-19)C`.

Answer

`4mm`

Explanation

Here, K.E. of electron, `K=18keV`
`=18xx10^3xx1*6xx10^(-19)J`
`=18xx1*6xx10^(-16)J`
`B=0*4G=0*4xx10^-4T`,
`m=9*1xx10^(-31)kg`, `e=1*6xx10^(-19)C`.
`K=1/2mv^2`
or `mv=sqrt(2mK)`
Radius of the circular path of electron in magnetic field is
`r=(mv)/(eB)=(sqrt(2mK))/(eB)`
`=(sqrt(2xx9*1xx10^(-31)xx(18xx1*6xx10^(-16))))/((1*6xx10^(-19))xx(0*4xx10^4))`
`=11*32m`
It means the electron will move in a circle of radius `11*32m` in a magnetic field as shown in figure. As the electron covers a distance `CD(=30cm)`, it moves down through a vertical distance `=CA`. Let `theta` be the angle subtended by arc CD at the centre O, then

`CA=OC-OA=r-rcostheta=r(1-costheta)`
As, angle `=(arc)/(radius)`,
so `theta=(30xx10^-2)/(11*32)=0*0265rad`
`=(0*0265xx180^@)/(pi)=1*52^@`
`cos theta=cos 1*52^@=0*99965`
`:. CA=11*32(1-0*99965)=3*9744xx10^-3m`
`~~4xx10^-3m=4mm`