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 (m_(e) = 9.11 xx 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

Consider uniform magnetic field `overset(to) (B) (.)` with a width `x lt R` where R is the radius of curvature of circular path followed by charge q (here charge e of electron), entering the magnetic field `overset(to) (B)` perpendicularly as shown in the figure. Here, width of magnetic field is `x= DH= CG and x lt R`.
Now, we have the formula,
`R = (mv sin 90^@)/( "Be") = (mv ) /( "Be") = (p)/( "Be") = ( sqrt(2mK) )/( "Be")`
`therefore R= ((sqrt(2 xx 9.1 xx 10^(-31) xx 18 xx 10^(3) xx 1.6 xx 10^(-19) ))/( 0.4 xx 10^(-4) xx 1.6 xx 10^(-19) ))= 11.3` m
Here width of region of magnetic field is `x= 0.3 m rArr x lt R`.
In right angled `Delta ACG`, we have `sin alpha = (x)/( R) = (0.3)/( 11.3) = 0.0265`
As shown in the figure, D is the point of entry of electron in the magnetic field and G is the point of exit of electron from the magnetic field. Here, perpendicular distance between two horizontal lines passing through points D and G is `d = DC = HG`, which is called linear deflection of electron (here vertically upward), which is to be found out.
From the diagram,
`d- DC = AD = R - R cos alpha`
`therefore d= R(1- cos alpha ) = R (1- sqrt(1 - sin^(2) alpha))`
`therefore d= 11.3 { 1 - sqrt(1- (0.0265)^(2) ) }`
`= 0.003956 m= 3.956` mm