A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (m_e = 9.11 × 10^–31 kg). [Note: Data in this exercise are so chosen that the answer will give you an 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.]

Energy of the electron beam(E) = 18keV = 18000eV = 18000 × 1.6 × 10^-19 = 2.88 × 10^-15J

Mass of electron (m_e) = 9.11 × 10^–31 kg

Horizontal magnetic field (B) = 0.04 G = 0.04 × 10^-4T

Distance of deflection from the beam (r) = 30 cm = 0.3m

Let velocity of electrons be v.

∴ Energy of the beam = Kinetic energy of electrons

⇒ E = 1/2 m_ev²

⇒

⇒

∴ v = 7.95 × 10⁷m/s

∵ The electron beam deflects along a circular path in a magnetic field.

Let the radius of the circular path be r m.

The force due to magnetic field on the electron = qvBsinθ = e × v × B × sin 90°

Here, q = charge on the electron,

θ = angle between velocity and magnetic field.

∵ The force due to magnetic field on electron = centripetal force on electron

⇒

⇒

⇒

⇒ r = 11.3 m

Let the deflection of electron beam in upward and downward directions be x = r(1-cosΦ),

where Φ = angle of deflection.

∵

⇒ = 0.026

⇒ Φ = sin^-1(0.026) = 1.521°

Thus, Deflection(x) = r(1- cosΦ) = 11.3 × (1- cos(1.521)°)

= 11.3 × (1- 0.90)

= 11.3 × 0.01

= 1.1310^-3m