In a television set, electrons are accelerated from rest through a potential difference of 20 kV. The electrons then pass through a 0.41-T magnetic field that deflects them to the appropriate spot on the screen. Find the magnitude of the maximum magnetic force that an electron can experience.

Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) = 20kv = 20,000v and strength of magnetic field (B) = 0.41 T.

The maximum force experienced by a charge of magnitude (q) is given as

Fmax = qvB

Where v = velocity of electron.

The velocity of the electron can be gotten by using the work energy theorem.

The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.

mv²/2 = qV.

Where m = mass of an electronic charge = 9.11*10^-31 kg, q = magnitude of an electronic charge = 1.609*10^-19 c, v = velocity of electron, V = potential difference = 20,000v.

By substituting the parameters, we have that

(9.11*10^-31 * v²) / 2 = 1.609*10^-19 * 20000

(9.11*10^-31 * v²) = 1.609*10^-19 * 20000 * 2

v² = (1.609*10^-19 * 20000 * 2) / 9.11*10^-31

v² = 64.36*10^ (-16) / 9.11*10^-31

v² = 7.0647*10^15

v = √7.0647*10^15

v = 8.40*10^7 m/s

Fmax = 1.609*10^-19 * 8.40*10^7 * 0.41

Fmax = 5.54*10^-12 N