Download Chapter 26 Problem 23 † Given f = 2.4 GHz = 2.4 × 10 me = 9.11

Chapter 26
Problem 23
†
Given
f = 2.4 GHz = 2.4 × 109 Hz
me = 9.11 × 10−31 kg
qe = −1.60 × 10−19 C
Solution
a) Find the magnetic field strength.
The effect that the magnetic field has on the moving electrons is that it will change the direction of the
electrons without changing their speed. Therefore, the electrons will move in a circle at a constant speed.
The magnetic force acting on the electrons will equal the centripetal acceleration times the mass of the
electron.
Fm = qvB = m
v2
r
Solving for r gives
r=
mv
qB
(1)
The time for an electron to make one circle is
T =
2πr
v
Therefore, the frequency at which the electron travels this circle is
f=
1
v
=
T
2πr
Substituting in the radius of the circle gives
f=
2π
v
mv
qB
=
qB
2πm
The strength of the magnetic field can now be found since the frequency is known.
B=
2π(9.11 × 10−31 kg)(2.4 × 109 Hz)
2πmf
=
q
(1.60 × 10−19 C)
B = 8.6 × 10−2 T = 86 mT
b) Find the energy of the electrons if the diameter of the orbit is 2.5mm.
From equation 1 solve for the velocity of the electron.
v=
rqB
m
v=
(1.25 × 10−3 m)(1.60 × 10−19 C)(8.6 × 10−2 T )
(9.11 × 10−31 kg)
v = 1.89 × 107 m/s
†
Problem from Essential University Physics, Wolfson
The maximum kinetic energy of each electron is then
K = 12 mv 2 = 12 (9.11 × 10−31 kg)(1.89 × 107 m/s)2
K = 1.62 × 10−16 J
Converting this answer to electron volts gives
1eV
K=
1.62 × 10−16 J = 1.01 × 103 eV = 1.01 keV
1.60 × 10−19 J