Question

Electrons in old-fashioned cathode ray tubes—as in CRT monitors—are accelerated, starting from rest, by a potential difference of 20.0 kV.

a. What is their speed after being accelerated by this potential?

b. If the Earth’s magnetic field is at a right angle to the electron beam and its magnitude is 0.600 gauss, what is the corresponding magnetic force on an electron?

c. Estimate (pretending that the magnetic force is approximately constant as the electron moves through the tube) the displacement that results from this force when the beam travels 15.0 cm through the tub

Answer 1

a) given
delta_v = 20 kV = 20*10^3 V

we know,
magnitude of charge of electron, q = 1.6*10^-19 c
mass of electron, m = 9.1*10^-31 kg

workdone electron = gain in kinetic energy

q*delta_V = (1/2)*m*v^2

v^2 = 2*q*delta_V/m

v = sqrt(2*q*delta_V/m)

= sqrt(2*1.6*10^-19*20*10^3/(9.1*10^-31))

= 8.39*10^7 m/s <<<<<<<--------Answer

b) B = 0.6 Gauss = 0.6*10^-4 T

magnetic force on electron, F = q*v*B*sin(theta)

= 1.6*10^-19*8.39*10^7*0.4*10^-4*sin(90)

= 5.37*10^-16 N <<<<<<<--------Answer

c) acceleration of the electron, a = F/m

= 5.37*10^-16/(9.1*10^-31)

= 5.90*10^14 m/s^2

time taken for the elctron to travel 15 cm,

t = d/v

= 0.15/(8.39*10^7)

= 1.788*10^-9 s

displacement due to magnetic force, d = (1/2)*a*t^2

= (1/2)*5.90*10^14*(1.788*10^-9)^2

= 9.43*10^-4 m <<<<<<<<<--------------Answer