Question

Chapter 21, Problem 19 An a-particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference that has a value of 1.42 x 10° V and then enters a uniform magnetic field whose magnitude is 2.78 T. The a particle moves perpendicular to the magnetic field at all times. What is (a) the speed of the a-particle, (b) the magnitude of the magnetic force on it, and (c) the radius of its circular path? Parallel plate capacitor B (into screen) (a) Number (b) Number (c) Number Click if you would like to Show Work fo Unit Uni Unit m/s No units m/s 2 kg n: Open Show Work

Answer 1

a) Using work done through the potential difference is W = change in kinetic energy

q*V = 0.5*m*v^2

2*1.6*10^-19*1.42*10^6 = 0.5*6.64*10^-27*v^2

v = 1.16*10^7 m/s is the answer for a)

b) force on it is F = q*v*B = 2*1.6*10^-19*1.16*10^7*2.78 = 1.03*10^-11 N

c) radius is r = (m*v)/(q*B) = (6.64*10^-27*1.16*10^7)/(2*1.6*10^-19*2.78)

r = 0.0865 m = 8.65 cm