Question

Question 5.22: 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 (me= 9.11 × 10 −19 C). [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.]

Class 12 - Physics - Magnetism And Matter Magnetism And Matter Page-203

Answer 1

kinetic energy of electron is 18keV
e.g., K.E = 1/2mv² = 18keV = 18000 eV
=> 1/2 × (9.1 × 10^-31) ×v² = 18000 × 1.6 × 10^-19
=> 9.1 × 10^-31 × v²= 36000 × 1.6 × 10^-19
=> v² = 36000 × 1.6 × 10^-19/(9.1 × 10^-31)
=> v = 8 × 10^7 m/s

we know, velocity in x direction remains constant.
hence, time taken to cross 30cm distance inx direction is t = x/v = 30 × 10^-2/(8 × 10^7)
= 3/8 × 10^-8 sec

A magnetic force F = Bve is acting in vertical direction, which provides acceleration in vertical direction,
so, a = Bve/m = (0.4 × 10^-4 × 8 × 10^7 × 1.6 × 10^-19)/(9.1 × 10^-31)
a = 5.63 × 10¹⁴ m/s²
now, vertical deflection,
$Y=U_yt+\frac{1}{2}a_yt^2$
= 0 + 1/2 × 5.63 × 10¹⁴ × (3/8 × 10^-8)²
≈ 4mm

hence, deflection in vertical direction is 4mm