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

` B = 0.40 G = 0.40 xx 10^(-4) T, m = 9.1 xx 10^(-31) kg, e = 1.6 xx 10^(-19) C, E = 18 keV = 18 xx 1000 xx 1.6 xx 10^(-19) J = 18 xx 1.6 xx 10^(-16)J`
Under the influence of magnetic field, the electron will follow a circular path. The radius of the circular path is given by
`r = (sqrt(2mE))/(Be)`
` r = (sqrt(2 xx 9.1 xx 10^(-31) xx 18 xx 1.6 xx 10^(-16) ))/(0.40 xx 10^(-4) xx 1.6 xx 10^(-19) ) = 11.3 m `
The deflection as per Fleming.s left-hand rule will be up or down. From right-angled triangle CED,

` (CE)/(CD) = cos theta`
`CE = CD cos theta `
` CE = R cos theta `
`AE = CA -CE`
`AE = R - R cos theta = R (1- cos theta)`
`y = R (1 -cos theta)`
From `deltata CED`
`sin theta = (0.3)/(11.3)`
`therefore theta = 1.52`
` therefore cos theta = 0.9996`
` therefore y = r - r cos theta `
`= 11.3 (1 - 0.9996)`
` = 3.98 xx 10^(-3) m`
`y = 3.98mm`
= 4mm