Question

A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of $0.04G$ normal to the initial direction . estimate the up or down deflection of the beam over a distance of $30cm(m_e=9.11\times {10}^{-31}kg)$ [ note : data in this exercise are so chosen that the answer will give you ah 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 ].

Answer 1

$M=9.11\times {10}^{-31}kg$
$C=1.6\times {10}^{-19}C$
$B=0.40G=0.4\times {10}^{-4}T$
$\frac{1}{2}m{V}^2=18KeV=18\times 1.6\times {10}^{-16}J$
Suppose that the electron moves along a circular path of radius $r$ under the effect of magnetic field. Then
$\frac{m{V}^2}{r}=BeV$
or $r=\frac{mV}{Be}$
$=\frac{\sqrt{2m\times \frac{1}{2}m{V}^2}}{Be}$
$=\frac{\sqrt{2\times 9.11\times {10}^{-31}\times 18\times 1.6\times {10}^{-16}}}{0.40\times {10}^{-4}\times 1.6\times {10}^{-19}}$
=$11.32m.$
Figure shows an electron covering a circular path $PQC=30cm=0.3m$ and of radius $OP=OQ=11.32m$
It follows the electron will undergo a deflection $PA$ in downward direction, while it covers a distance $PQ$. If $\theta$ is angle subtended by $PQ$
at the point $O$, then
$PA=OP-OA=r-rcos\theta$
=$r(1-cos\theta )$
Now,$\theta =\frac{PQ}{OP}=\frac{0.30}{11.32}=0.265rad={152}^o$
$\therefore =11.32(1-0.9996)$
=$4.53\times {10}^{-3}m$
=$4.53mm$.