5.22 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 × 1031 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.]

The energy of an electron beam,
Charge on an electron, $e=1.6×{10}^{-19}C,$
, Magnetic field, B =0.04 G,
Mass of an electron,
Distance up to which the electron beam travels, d= 30 cm =0.3 m
We can write the kinetic energy of the electron beam as:
$\begin{array}{l}E=\frac{1}{2}m{v}^{2}\\ v=\sqrt{\frac{2E}{m}}\\ =\sqrt{\frac{2×18×{10}^{3}×1.6×{10}^{-19}×{10}^{-15}}{9.11×{10}^{-31}}}=0.795×{10}^{8}\mathrm{m}/\mathrm{s}\end{array}$
The electron beam deflects along a circular path of radius, r.
The force due to the magnetic field balances the centripetal force of the path.

$\begin{array}{l}BeV=\frac{m{v}^{2}}{r}\\ \therefore r=\frac{mv}{Be}\\ =\frac{9.11×{10}^{-31}×0.795×{10}^{8}}{0.4×{10}^{-4}×1.6×{10}^{-19}}=11.3\mathrm{m}\end{array}$
Let the up and down deflection of the electron beam be x=r(1-cos)
Where $\mathrm{\theta }$=Angle of declination

Therefore, the up and down deflection of the beam is 3.9 mm.

Distance up to which the electron beam travels, d = 30 cm = 0.3 m
We can write the kinetic energy of the electron beam as:
$\begin{array}{l}\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^{2}\\ \mathrm{v}=\sqrt{\frac{2\mathrm{E}}{\mathrm{m}}}\\ =\sqrt{\frac{2×18×{10}^{3}×1.6×{10}^{-19}×{10}^{-15}}{9.11×{10}^{-31}}}=0.795×{10}^{8}\mathrm{m}/\mathrm{s}\end{array}$
The electron beam deflects along a circular path of radius, r.
The force due to the magnetic field balances the centripetal force of the path.

$\mathrm{BeV}=\frac{{\mathrm{mv}}^{2}}{\mathrm{r}}$
$\therefore \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Be}}$
$=\frac{9.11×{10}^{-31}×0.795×{10}^{8}}{0.4×{10}^{-4}×1.6×{10}^{-19}}=11.3\mathrm{m}$
Let the up and down deflection of the electron beam be x=r(1 - cos)$\theta$
Where, $\mathrm{\theta }$=Angle of declination
$\mathrm{sin}\theta =\frac{d}{r}$
$=\frac{0.3}{11.3}$
$\theta ={\mathrm{sin}}^{-1}\frac{0.3}{11.3}=1.521°$
And x=11.3(1-cos1.521$°$)
=0.0039 m = 3.9mm
Therefore, the up and down deflection of the beam is 3.9 mm.