Download CH 8: Magnetic Fields

CH 8: Magnetic Fields
8.4 – Motion of Charged Particles
in Magnetic Fields
Agenda

Warm Up / Homework Check

Clicker Questions from yesterday

Lesson on Section 8.4

Textbook Questions
Learning Goals

Analyse and solve problems involving
the motion of charged particles in
magnetic fields
Warm Up Question
A 155mm part of a wire has a mass of
0.30kg and carries an electric current of
3.5 A. The conventional current passes
through a uniform magnetic field of 2.5 T.
The direction of the wire and the magnetic
field are shown in Figure 7.
A. What is the magnitude of the magnetic
force on the wire?
B. Use the right-hand rule to determine
the direction of the magnetic force.
Warm Up Question
Earth’s magnetic field exerts a force of
1.4 x 10-5 N on a 5 cm segment of wire in
a truck motor. The motor wire is positioned
at an 23° angle to Earth’s magnetic field,
which has a magnitude of 4.9 x 10-5 T at
the truck’s location. Calculate the current
in the wire.
4. Clicker Question
5. Clicker Question
7. Clicker Question
8. Clicker Question
9. Clicker Question
Mass Spectrometer
http://www.youtube.com/watch?v=_L4U6I
mYSj0
Motion of a Charged Particle in
a Magnetic Field

















v


v

v
F




F
F
v












Magnetic force provides the
centripetal acceleration.
Motion of a Charged Particle in
a Magnetic Field
If m, q and B are held constant, higher speed  larger radius.










2r




r

106 m/s
F1


2x106 m/s
F2












Example
A Bainbridge mass spectromoter is used to measure the mass of zinc ions
(Zn+). The ions enter a magnetic field, of strength B=1.8T with a speed
of 4.2 x 104 ms-1 and they move in a circular path of radius r = 1.7cm.
What is the mass of the zinc ions?
An electron in a magnetic Field
An electron starts from rest. A horizontally directed
electric field accelerates the electron through a
potential difference of 40V. The electron then leaves
the electric field and moves into a magnetic field.
The magnetic field strength is 0.T, directed into the
page (Figure 3), and the mass of the electron is 9.11
x 10-31 kg.
A. Determine the speed of the electron at
the moment it enters the magnetic field.
B. Determine the magnitude and direction
of the magnetic force on the electron.
C. Determine the radius of the electron’s
circular path.
Mass Spectrometer: Separating Isotopes
Cyclotron Particle Accelerator
http://www.youtube.c
om/watch?v=cNnNM
2ZqIsc
 TRIUMF National
laboratory in
Vancouver has the
largest cyclotron at
18m

Use of Magnetic Fields in
Cyclotrons
Particles moving in a cyclotron are subjected to a uniform magnetic field
which causes the particles to move in a circular path.
Each time the charged particle enters a Dee it follows a semi-circular path
and each time it traverses to the other Dee its speed is greater than before
as it is subjected to an Electric field. Its energy has increased by W=qV.
Use of Magnetic Fields in
Cyclotrons
The period of the charged
particle is independent of its
speed, this can be shown by
the following,
Circumfere nce of a circle,
C  2r (length of one semicircle  r )
 time to traverse one semicircle
r
t
v
mv
but r 
qB
rqB
v 
m
rm m
thus t 

rqB qB
 time taken to complete one full revolution
2m
t
qB
Use of Magnetic Fields in
Cyclotrons
The Kinetic Energy can be found by,
Ek 
1
2
mv
2
 rqB 
 2 m

 m 
q 2 B 2r 2

2m
1
2
mv
as r 
qB
Use of Magnetic Fields in
Cyclotrons
Example,
A cyclotron is used to accelerate dueterons (q = 1.6 x 10-19C) of mass
3.34 x 10-27 kg. The radius of the cyclotron is 30cm and the
magnetic field applied to it has strength 1.6T. Find
(a) The period of the particles in their circular path.
(b) The energy (in eV) of the particles leaving the cyclotron.
Use of Magnetic Fields in
Cyclotrons
(a) The period of the particles in their circular path.
2m
T
qB
2    3.34 10  27

1.6 10 19 1.6
 8.2 10 8 s
Use of Magnetic Fields in
Cyclotrons
(b) The energy (in eV) of the particles leaving the cyclotron.
q 2 B 2r 2
K
2m
(1.6 10 19 ) 2  (1.6) 2  (0.3) 2

2  3.34 10  27
 8.83 10 13 J
change into eV 
8.83 10-13

1.6 10 19
 5.52 106 eV
 5.52 MeV
Homework
pg. 404 #2-6
Quiz on Chapter 8 Tomorrow?