Question

An electron of kinetic energy 1.73 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 21.2 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

kinetic energy KE = (1/2)*m*v^2

given KE = 1.73 KeV = 1.73*10^3*1.6*10^-19 J

m = mass of electron = 9.11*10^-31 Kg

v = speed of electron

therefore

(1/2)*9.11*10^-31*v^2 = 1.73*10^3*1.6*10^-19

v = 2.465*10^7 m/s

====================

part(b)

In the magnetic field

magnetic force Fb = centripetal force

Fb = Fc

q*v*B = m*v^2/r

radius r = m*v/(q*B)

magnetic field B = m*v/(r*q)

B = (9.11*10^-31*2.465*10^7)/(21.2*10^-2*1.6*10^-19)

B = 6.62*10^-4 T

========================

part(c)

frequency f = qB/(2*pi*m)

frequency f = 1.6*10^-19*6.62*10^-4/(2*pi*9.11*10^-31) = 18.5 MHz

===================================

part(d)

time period T = 1/f = 5.405*10^-8 s