Question

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).

Answer 1

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, m_e = 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 = `1/2 "mv"^2`

v = `sqrt((2"E")/"m")`

= `sqrt((2 xx 18 xx 10^3 xx 1.6 xx 10^-19 xx 10^-15)/(9.11 xx 10^-31))`

= 0.795 × 10⁸ m/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.

BeV = `"mv"^2/"r"`

∴ r = `"mv"/"Be"`

= `(9.11 xx 10^-31 xx 0.795 xx 10^8)/(0.4 xx 10^-4 xx 1.6 xx 10^-19)`

= 11.3 m

Let the up and down deflection of the electron beam be x = r(1 − cos θ)

Where,

θ = Angle of declination

sin θ = `"d"/"r"`

= `0.3/11.3`

θ = `sin^-1  0.3/11.3` = 1.521°

And x = 11.3(1 − cos 1.521°)

= 0.0039 m

= 3.9 mm

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