Answer –

Energy of the electron beam is given by E = 18 keV

Or, E = 18 x 10³ eV = 18 x 10³ x 1.6 x 10^-19 J

Magnetic field is given by B = 0.04 G

Mass of the electron is given by m_e = 9.11 × 10^–31 kg

Distance to which the beam travels is represented by d = 30 cm = 0.3 m

Then, the kinetic energy of the electron beam can be written as –

E = (1/2) mv²

Where v is given by –

\[v=\sqrt{\frac{2\times 18\times {{10}^{3}}\times 1.6\times {{10}^{-19}}}{9.11\times {{10}^{-31}}}}=0.795\times {{10}^{8}}m{{s}^{-1}}\]

It can be concluded that the electron beam deflects along the circular path of radius r.

In such a case, the centripetal force is balanced by the force due to the applied magnetic field

Therefore, we can write –

Bev = mv²/r

Or, r = mv/Be

= (9.11 x 10^-31 x 0.795 x 10⁸)/(0.04 x 10^-4 x 1.6 x 10^-19)

= (7.24 x 10^-23)/(0.064 x 10^-23)

r = 113.125

Suppose that the up and down deflection of the beam is given by

x = r (1 – cosθ)

Where, θ represents the angle of deflection

Therefore, sin θ = d/r

= 0.3/113.12 = 0.0026

θ = sin ^-1 (0.0026) = 0.1489⁰

x = r (1 – cos θ) = 113.12 (1 – cos 0.1489⁰)

= 113.12 (1- 0.999)

= 113.12 x 0.01

x = 1.13 mm, which is the required answer.