Question

A monoenergetic electron beam of initial energy 18 ke V moving horizontally is subjected to a horizontal magnetic field of 0.4 G normal to its initial direction. calculate the vertical deflection of the beam over a distance of 30 cm.

Answer 1

Given

$E=18Kev$

$=18\times {10}^3\times 1.6\times {10}^{-19}$

$=18\times 1.6\times {10}^{-16}J$

$B=0.4G=0.4\times {10}^{-4}T$

Distance$=30cm=0.3m$

$m_e=9.1\times {10}^{-19}$

Energy of the electron $E=\frac{1}{2}m{v}^2$

$18\times 1.6\times {10}^{-16}=\frac{1}{2}\times 9.1\times {10}^{-31}{v}^2$

$v^2=\frac{18\times 1.6\times {10}^{-16}\times 2}{9.1\times {10}^{-31}}$

Now,

$eVB=\frac{m{v}^2}{r}$

$r=\frac{mv}{Be}$

$=\frac{9.1\times {10}^{-31}\times 0.795\times {10}^8}{0.4\times {10}^{-4}\times 1.6\times {10}^{-19}}$

$=11.3m$

$\sin \theta =\frac{d}{r}$

$=\frac{0.3}{11.3}$

$\theta ={1.521}^o$

Deflection at the end of the path

$x=r-r\cos \theta$

$=r(1-\cos \theta )$

$=11.3(1-\cos (1.521\left)\right)$

$=0.0039m$

$=3.9mm$

$x\approx 4mm$

Thus the up or down deflection of the beam is approximately 4 mm.