PHY 184
Week 6 Spring 2007
Lecture 22
Title: The Lorentz Force = q v x B
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Announcements
 Homework Set 5 is due next Tuesday at 8:00 am.
 The correction set to Midterm 1 is due next
Tuesday at 6:00pm.
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Review
 The force that a magnetic field exerts on a charge moving
with velocity v is given by

 
FB  qv  B
 The direction is sideways (RH rule).
 The magnitude of the force
FB  qvBsin 
 If the charge moves perpendicular to the magnetic field
then
F  qvB
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Review (2)
 The unit of magnetic field strength the tesla (T)
1T=1
Ns
N
1
Cm
Am
 Another unit of magnetic field strength that is
often used but is not an SI unit is the gauss (G)
1 G = 10-4 T
10 kG = 1 T
 Typically the Earth’s magnetic field is about 0.5 G
at the surface.
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Example: Cathode Ray Tube




Consider the cathode-ray tube used for lecture demonstrations.
In this tube electrons form an electron beam when accelerated horizontally by a
voltage of 136 V in an electron gun.
The mass of an electron is 9.109410-31 kg while the elementary charge is
1.602210-19 C.
(a) Calculate the velocity of the electrons in the beam after leaving the electron
gun.
K  U  qV
1 mv 2
2
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 eV
implies
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v  6.92  10 6 m/s
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Cathode Ray Tube (2)
 (b) If the tube is placed in a uniform magnetic field, in what direction is
the electron beam deflected?
downward
 (c) Calculate the magnitude of acceleration of an electron if the field
strength is 3.65×10-4T.
F  ma  qvB




19
6
4
qvB 1.6022 10 C 6.92 10 m/s 3.65 10 T
14
2
a


4.44
10
m/s
m
9.1094 10 31 kg
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Particle Orbits in a Uniform B
 Tie a string to a rock and twirl it at constant speed in a
circle over your head.
 The tension of the string provides the centripetal force
that keeps the rock moving in a circle.
 The tension on the string always points to the center of
the circle and creates a centripetal acceleration.
 A particle with charge q and mass m moves with velocity v
perpendicular to a uniform magnetic field B.
 The particle will move in a circle with a constant speed v
and the magnetic force F = qvB will keep the particle
moving in a circle.
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Particle Orbits in Uniform B (2)
 Recall centripetal acceleration
 Newton;’s second law
a = v2/r
F=ma
 So, for charged particle q in circular motion in magnetic field B
v2
m  qv B
r
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Example - Moving Electrons
 In this photo, an electron beam is initially accelerated by an electric field.
 Then the electrons move in a circle perpendicular to the constant
magnetic field created by a pair of Helmholtz coils.
Is the magnetic field into
the page or out of the
page?
F
(Remember that the
magnetic force on an
electron is opposite that
on a proton.)
v
The magnetic field is out
of the page.
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Clicker Question
 The figure shows the circular
paths of two particles that travel
at the same speed in a uniform
magnetic field (directed into the
page). One particle is a proton and
the other is an electron. Which
particle follows the smaller circle?
A) the electron
B) the proton
C) Both, proton and electron travel
along on the same circle
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Clicker Question
 The figure shows the circular paths of two particles that
travel at the same speed in a uniform magnetic field
(directed into the page). One particle is a proton and the
other is an electron. Which particle follows the smaller
circle?
A) the electron
mv
r
qB
… for same speeds, r is proportional to m
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Example: Mass Spectrometer (1)
 Suppose that B=80mT,
V=1000V. A charged ion
(1.6022 10-19C) enters the
chamber and strikes the
detector at a distance
x=1.6254m. What is the mass
of the ion?
 Key Idea: The uniform
magnetic field forces the ion
on a circular path and the ion’s
mass can be related to the
radius of the circular
trajectory.
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Example: Mass Spectrometer (2)
 Suppose that B=80mT,
V=1000V. A charged ion
(1.6022 10-19C) enters
the chamber and strikes
the detector at a
distance x=1.6254m.
What is the mass of
the ion?
 From the figure: r=0.5x
 Also need the velocity v
after the ion is
accelerated by the
potential difference V
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mv 2
 evB
r
2
1
mv
2
 eV
Now solve for m
3.4x10-25 kg
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Example: The Time Projection Chamber
 In high-energy nuclear physics, new forms of matter
are studied by colliding gold nuclei at very high energies
 In particle physics, new elementary particles are
created and studied by colliding protons and antiprotons at the highest energies
 In these collisions, many particles are created that
stream away from the interaction point at high speeds
 A simple particle detector is not sufficient to measure
and identify these particles
 A device that can help physicists study these collisions
is the time projection chamber (TPC)
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Example: The Time Projection Chamber (2)
 The STAR TPC consists of a large cylinder filled with a
carefully chosen gas (90% argon, 10% methane) that allows
free electrons to drift without recombining
 As created charged particles pass through the gas, the
particles ionize the atoms of the gas, releasing free
electrons
 An electric field is applied between the center of the TPC
and the caps of the cylinder that exerts an electric force
on these freed electrons, making them drift to the end-caps
of the TPC, where they are recorded electronically
 Using the drift time and the recording positions, the
computer software reconstructs the trajectories that the
produced particles took through the TPC
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Example: The Time Projection Chamber (3)
 The TPC sits inside a giant
solenoid magnet shown to
the right with the magnetic
field pointing along the
beam direction
 The produced charged
particles have a component
of their velocity that is
perpendicular to the
magnetic field and thus
have circular trajectories
when viewed end-on
 From the radius of
curvature one can extract
the particle momentum
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Events in the TPC
 Here are two events in the STAR TPC at Brookhaven
National Lab
 Magnetic field is directed into the screen
Let’s analyze this track
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Example - Momentum of a Track
 Calculate the momentum of this track
r = 2.3 m
mv / r  evB
2
mv  erB  1.8  10
19
kg m/s
19
v mv
1.8  10


 0.36
 27
8
c mc 1.67  10  3.0  10
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Orbits in a Constant Magnetic Field
 If a particle performs a complete circular orbit inside a
constant magnetic field, then the period of revolution of
the particle is just the circumference of the circle divided
by the speed
2 r 2 m
T

v
qB
 From the period we can get the frequency and angular
frequency
1
qB
f  
T 2 m
qB
  2 f 
m
 The frequency of the rotation is independent of the speed
of the particle.
Isochronous orbits.
Basis for the cyclotron.
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Cyclotrons
 A cyclotron is a particle accelerator
 The D-shaped pieces (descriptively
called “dees”) have alternating
electric potentials applied to them
such that a positively charged particle
always sees a negatively charged dee
ahead when it emerges from under
the previous dee, which is now
positively charged
 The resulting electric field
accelerates the particle
 Because the cyclotron sits in a strong
magnetic field, the trajectory is
curved
 The radius of the trajectory is
proportional to the momentum, so the
accelerated particle spirals outward
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Example: Deuteron in Cyclotron
 Suppose a cyclotron is operated at frequency f=12
MHz and has a dee radius of R=53cm. What is the
magnitude of the magnetic field needed for deuterons
to be accelerated in the cyclotron (m=3.34 10-27kg)?
 Key Idea: For a given frequency f, the magnetic field
strength, B, required to accelerate the particle
depends on the ratio m/q (or mass to charge):
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Special Clicker
 Suppose a cyclotron is operated at frequency f=12 MHz and
has a dee radius of R=53cm. What is the kinetic energy of
the deuterons in this cyclotron when they travel on a
circular trajectory with radius R (m=3.34 10-27kg, B=1.57 T)?
A) 0.9 10-14 J
B) 8.47 10-13 J
C) 2.7 10-12 J
D) 3.74 10-13 J
mv
r
qB
implies
RqB
v
 3.99  107 m/s
m
K  12 mv 2  2.7  10 12 J
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K500 Superconducting Cyclotron
 Movie from Nova program “The Nucleus Factory”
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