A monoenergetic (18 keV) electron beam initially

Solution:

Energy of an electron beam, E = 18 keV = 18 × 103 eV

Charge on an electron, e = 1.6 × 10−19 C

E = 18 × 103 × 1.6 × 10−19 J

Magnetic field, B = 0.04 G

Mass of an electron, me = 9.11 × 10−19 kg

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:

$E=\frac{1}{2} m v^{2}$

$v=\sqrt{\frac{2 E}{m}}$

$=\sqrt{\frac{2 \times 18 \times 10^{3} \times 1.6 \times 10^{-19} \times 10^{-15}}{9.11 \times 10^{-31}}}=0.795 \times 10^{8} \mathrm{~m} / \mathrm{s}$

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.

$\mathrm{BeV}=\frac{m v^{2}}{r}$

$\therefore r=\frac{m v}{B e}$

$=\frac{9.11 \times 10^{-31} \times 0.795 \times 10^{8}}{0.4 \times 10^{-4} \times 1.6 \times 10^{-19}}=11.3 \mathrm{~m}$

Let the up and down deflection of the electron beam be $x=r(1-\cos \theta)$

Where,

θ = Angle of declination

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

$=\frac{0.3}{11.3}$

$\theta=\sin ^{-1} \frac{0.3}{11.3}=1.521^{\circ}$

And $x=11.3\left(1-\cos 1.521^{\circ}\right)$

$=0.0039 \mathrm{~m}=3.9 \mathrm{~mm}$

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