Question

Chapter 28, Problem 021 An electron of kinetic energy 1.19 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 26.3 cm. Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.

Answer 1

Solution:

a) using the formula of kinetic energy

=> 0.5 mv^2 = 1.19 * 10^3 * 1.6 * 10^-19 j

=> 0.5 * 9.1 * 10^-31 * V^2 = 1.19 * 10^3 * 1.6 * 10^-19 J

=> v^2 = 41.84 * 10^10

=> v = 6.47 * 10^5 m/s

b) Using the formula

=> r = mv/Bq

=> B = mv/rq

=> B = 9.1 * 10^-31 * 6.47 * 10^5 / 0.263 * 1.6 * 10^-19

=> B = 1.4 * 10^-7 T

c) frequency = v / (2*pi*r)

=> f = 6.47 * 10^5 / (2 * 3.14 * 0.263 )

=> f =3.92 * 10^5 Hz

d) T = 1/f

=> T = 1/ (3.92 * 10^5)

=> T = 2.55 * 10^-6 seconds

thanks