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 me = 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.