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Announcements
 Exam 2 is one week from tomorrow.
Contact me by the end of the Wednesday’s lecture if you
have special circumstances different than for exam 1.
Today’s agenda:
Magnetic Fields.
You must understand the similarities and differences between electric fields and field lines,
and magnetic fields and field lines.
Magnetic Force on Moving Charged Particles.
You must be able to calculate the magnetic force on moving charged particles.
Motion of a Charged Particle in a Uniform Magnetic Field.
You must be able to calculate the trajectory and energy of a charged particle moving in a
uniform magnetic field.
Magnetic Forces on Currents and Current-Carrying Wires.
You must be able to calculate the magnetic force on currents.
Velocity Selector and Mass Spectrometer.
You must be able to work these detailed examples.
Magnetism
Recall how there are two kinds of electrical charge (+ and -),
and likes repel, opposites attract.
Similarly, there are two kinds of magnetic poles (north and
south), and like poles repel, opposites attract.
S
S
N
S
N
Repel
S
N
N
S
N
Repel
S
Attract
N
S
N
N
S
Attract
There is an important difference between magnetism and
electricity: it is possible to have isolated + or – electric charges,
but isolated N and S poles have “never” been observed.*
-
+
N
S
I.E., every magnet has BOTH a N and a S pole, no matter how
many times you “chop it up.”
SS N
SN
N
*But see here, here, here, and here.
Magnetic Fields
The earth has associated with it a magnetic field, with poles
near the geographic poles.
The pole of a magnet attracted to
the earth’s north geographic pole is
the magnet’s north pole.
The pole of a magnet attracted to
the earth’s south geographic pole
is the magnet’s south pole.
N
S
http://hyperphysics.phyastr.gsu.edu/hbase/magnetic/
magearth.html
Just as we used the electric field to help us “explain” and
visualize electric forces in space, we use the magnetic field to
help us “explain” and visualize magnetic forces in space.
Magnetic field lines point in the same direction that the north
pole of a compass would point. Later I’ll give a better definition for magnetic field direction.
Magnetic field lines are tangent to the magnetic field.
The more magnetic field lines in a region in space, the stronger
the magnetic field.
Outside the magnet, magnetic field lines point away
from N poles (*why?).
Huh?
*The N pole of a compass would “want to get to” the S pole of the magnet.
For those of you who aren’t going to pay attention until you
have been told the secret behind the naming of the earth’s
magnetic poles…
The north pole of a compass needle is defined as the end that points
towards the geographic “Santa Claus” north pole, which experts in the
field of geomagnetism call “the geomagnetic north” or “the Earth’s North
Magnetic Pole.”
If you think about it, the people to whom compasses meant the most—
sailors—defined magnetic north as the direction the north poles of their
compass needles pointed. Their lives depended on knowing where they
were, so I guess it is appropriate that we acknowledge their precedence.
Thus, by convention, the thing we call Earth’s North Magnetic Pole is
actually the south magnetic pole of Earth’s magnetic field.
For those of you who are normal humans, just ignore the
above.
Here’s a “picture” of the magnetic
field of a bar magnet, using iron
filings to map out the field.
The magnetic field ought to “remind”
you of the earth’s field.
We use the symbol B for magnetic field.
Remember: magnetic field lines point away from north poles,
and towards south poles.
S
N
The SI unit* for magnetic field is the Tesla.
1 kg
1 T=
Cs
These units come from the
magnetic force equation, which
appears three slides from now.
In a bit, we’ll see how the units are related to other quantities
we know about, and later in the course we’ll see an “official”
definition of the units for the magnetic field.
*Old unit, still sometimes used: 1 Gauss = 10-4 Tesla.
The earth’s magnetic field has a
magnitude of roughly 0.5 G, or
0.00005 T. A powerful permanent magnet, like the kind you
might find in headphones, might
produce a magnetic field of 1000
G, or 0.1 T.
http://liftoff.msfc.nasa.gov/academy/space/mag_field.html
The electromagnet in the basement
of Physics that my students (used
to) use in experiments can produce
a field of 26000 G = 26 kG = 2.6 T.
Superconducting magnets can produce a field of
over 10 T. Never get near an operating superconducting magnet while wearing a watch or belt
buckle with iron in it!
Today’s agenda:
Magnetic Fields.
You must understand the similarities and differences between electric fields and field lines,
and magnetic fields and field lines.
Magnetic Force on Moving Charged Particles.
You must be able to calculate the magnetic force on moving charged particles.
Motion of a Charged Particle in a Uniform Magnetic Field.
You must be able to calculate the trajectory and energy of a charged particle moving in a
uniform magnetic field.
Magnetic Forces on Currents and Current-Carrying Wires.
You must be able to calculate the magnetic force on currents.
Velocity Selector and Mass Spectrometer.
You must be able to work these detailed examples.
Magnetic Fields and Moving Charges
A charged particle moving in a magnetic field experiences a
force.
The following equation (part of the Lorentz Force Law)
predicts the effect of a magnetic field on a moving charged
particle:
F = qv  B
force on
particle
velocity of
charged particle
Oh nooo! The
little voices
are back.
magnetic field
vector
What is the magnetic force if the charged particle is at rest?
What is the magnetic force if v is (anti-)parallel to B?
Vector notation conventions:

 is a vector pointing out of the paper/board/screen (looks
like an arrow coming straight for your eye).

 is a vector pointing into the paper/board/screen (looks
like the feathers of an arrow going away from eye).
Direction of magnetic force--Use right hand rule:
fingers out in direction of v, thumb perpendicular to them
rotate your hand until your palm points in the direction of B
(or bend fingers through smallest angle from v to B)
thumb points in direction of F on + charge
Your text presents two alternative variations (curl your fingers,
imagine turning a right-handed screw). There is one other
variation on the right hand rule. I’ll demonstrate all variations in
lecture sooner or later.
still more variations: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html
Complication: your Physics
23 text uses the right
hand rule shown in the
figure, so we’ll use the
same rule in Physics 2135.
There are a number of
variations of this rule.
Unfortunately, most of the Youtube videos I find say to use your
palm for A , your thumb for B , and your outstretched fingers for
A  B . This includes the MIT Open Courseware site.
I’ll show you later that both ways are correct. I am going to use
our textbook’s technique. You can use whatever works for you!
F = q vB sinθ
Direction of magnetic force:
F
+

Fingers of right hand out in direction of velocity, thumb
perpendicular to them. Bend your fingers until they point
in the direction of magnetic field. Thumb points in
direction of magnetic force on + charge.
B

-
v
F
B
v
Force is “down”
because charge is -.
v
F?
-

B
I often do this: fingers out in direction of velocity, thumb perpendicular to them. Rotate your hand until your palm points in the
direction of magnetic field. Thumb points in direction of magnetic force on + charge.
“Foolproof” technique for calculating both magnitude and
direction of magnetic force.
F = qv  B

 ˆi


F = q det  v x

B

 x
ˆj
vy
By
kˆ  

v z 
B z  

http://www.youtube.com/watch?v=21LWuY8i6Hw
All of the right-hand rules are just techniques for determining the direction of vectors
in the cross product without having to do any actual math.
^
Example: a proton is moving with a velocity v = v0j in a region
^
of uniform magnetic field. The resulting force is F = F0i. What is
the magnetic field (magnitude and direction)?
To be worked at the blackboard.
Blue is +, red is -. Image from http://www.mathsisfun.com/algebra/matrix-determinant.htm
^
Example: a proton is moving with a velocity v = v0j in a region
^
of uniform magnetic field. The resulting force is F = F0i. What is
the magnetic field (magnitude and direction)?
v = v 0 ˆj
F = F0 ˆi
F = qv  B

 ˆi


ˆ
F = F0 i = q det  v x

B

 x
ˆj
vy
By
kˆ  

v z   =  e 
B z  


 ˆi


det  0

B

 x
ˆj
v0
By
kˆ  

0 
B z  

F0ˆi =  e  ˆi  v 0  Bz  B y  0   ˆj 0  B z  B x  0   kˆ 0  B y  B x  v 0 


^
Example: a proton is moving with a velocity v = v0j in a region
^
of uniform magnetic field. The resulting force is F = F0i. What is
the magnetic field (magnitude and direction)?
F0 ˆi = ev 0B z ˆi  0jˆ  eB x v 0kˆ =  ev 0B z  ˆi +  0  ˆj+  eB x v 0  kˆ
F0 = ev 0B z
F0
Bz =
ev 0
and
and
0 = eB x v 0
Bx = 0
What is By???? It could be anything. Not enough information is
provided to find By. We can’t find the magnitude and direction
of the magnetic field using only the information given!
^
Example: a proton is moving with a velocity v = v0j in a region
^
of uniform magnetic field. The resulting force is F = F0i. What is
the minimum magnitude magnetic field that can be present?
Bz =
F0
ev 0
By = 0
B min =
and
Bx = 0
for a minimum magnitude magnetic field, so...
F0
ev 0
Homework Hint
F = q vB sinθ
A charged particle moving along or opposite to the direction of
a magnetic field will experience no magnetic force.
Conversely, the fact that there is no magnetic force along some
direction does not mean there is no magnetic field along or
opposite to that direction.
It’s OK to use F = q vB if you know that v and B are
perpendicular, or you are calculating a minimum field to
produce a given force (understand why).
Alternative* view of magnetic field units.
F = q vB sinθ
F
B=
q v sinθ
N
N
=
B  = T =
C  m/s A  m
*“Official” definition of units coming later.
Remember, units of field are
force per “something.”
Today’s agenda:
Magnetic Fields.
You must understand the similarities and differences between electric fields and field lines,
and magnetic fields and field lines.
Magnetic Force on Moving Charged Particles.
You must be able to calculate the magnetic force on moving charged particles.
Motion of a Charged Particle in a Uniform Magnetic
Field.
You must be able to calculate the trajectory and energy of a charged particle moving in a
uniform magnetic field.
Magnetic Forces on Currents and Current-Carrying Wires.
You must be able to calculate the magnetic force on currents.
Velocity Selector and Mass Spectrometer.
You must be able to work these detailed examples.
Motion of a charged particle
in a uniform magnetic field
Example: an electron travels at 2x107 m/s in a plane
perpendicular to a 0.01 T magnetic field. Describe its path.
Example: an electron travels at 2x107 m/s in a plane
perpendicular to a 0.01 T magnetic field. Describe its path.
The force on the electron
(remember, its charge is -) is
always perpendicular to the
velocity. If v and B are
constant, then F remains
constant (in magnitude).
The above paragraph is a
description of uniform
circular motion.
B
       




 
 
v
 
 





F















F
       
       - v
       
The electron will move in a circular path with a constant speed
and acceleration = v2/r, where r is the radius of the circle.
Motion of a proton in a uniform magnetic field
v
  
Bout
 FB r   
v
   + 
+
FB
FB
     
v
     
   +




      
The force is always in the
radial direction and has a
magnitude qvB. For circular
motion, a = v2/r so
mv 2
F = q vB =
r
q rB
mv
v=
r=
m
qB
Thanks to Dr. Waddill for the use of the picture and following examples.
The period T is
2πr 2πm
T=
=
v
qB
The rotational frequency f is
called the cyclotron frequency
qB
1
f= =
T 2πm
Remember: you can do
the directions “by hand”
and calculate using
magnitudes only.
Helical motion in a uniform magnetic field
If v and B are perpendicular, a
charged particle travels in a circular
path. v remains constant but the
direction of v constantly changes.
If v has a component parallel to B,
then v remains constant, and the
charged particle moves in a helical
path.
There won’t be any test problems on
helical motion.
B
v
+
v
Apply B-field
perpendicular
Electrons
confined
to move to
in aplane
plane
Thanks to Dr. Yew San Hor for this and the next slide.
Apply B-field perpendicular to plane









Lorentz Force Law


If both electric and magnetic fields are present, F = q E + v  B .
Applications
See your textbook for numerical calculations related the next
two slides.
If I have time, I will show the mass spectrometer today.
The energy calculation in the mass spectrometer example is
often useful in homework.
Velocity Selector
       
Bout
----------------
  v     
+ q
E
+
FE = qE
       

FB = qv  B
       
Thanks to Dr. Waddill for the use of the picture.
When the electric and magnetic forces balance then the charge
will pass straight through. This occurs when FE = FB or
qE = qvB
or
E
v=
B
Mass Spectrometers
Mass spectrometers separate charges of different mass.
When ions of fixed energy enter a region of constant magnetic
field, they follow a circular path.
V
The radius of the path
depends on the
mass/charge ratio and
speed of the ion, and
the magnitude of the
magnetic field.
x = 2r
and
mv
r=
qB
Worked example with numbers at the end of this lecture.
     B
S
+q
x
    
    
r
    
    
    
    
Thanks to Dr. Waddill for the use of the picture.
Today’s agenda:
Magnetic Fields.
You must understand the similarities and differences between electric fields and field lines,
and magnetic fields and field lines.
Magnetic Force on Moving Charged Particles.
You must be able to calculate the magnetic force on moving charged particles.
Motion of a Charged Particle in a Uniform Magnetic Field.
You must be able to calculate the trajectory and energy of a charged particle moving in a
uniform magnetic field.
Magnetic Forces on Currents and Current-Carrying
Wires.
You must be able to calculate the magnetic force on currents.
Velocity Selector and Mass Spectrometer.
You must be able to work these detailed examples.
Magnetic Forces on Currents
So far, I’ve lectured about magnetic forces on moving charged
particles.
F = qv  B
Actually, magnetic forces were observed on current-carrying
wires long before we discovered what the fundamental charged
particles are.
Experiment, followed by theoretical understanding, gives
F = IL  B.
For reading clarity, I’ll use L
instead of the l your text uses.
If you know about charged particles, you can derive this from the equation for the
force on a moving charged particle. It is valid for a straight wire in a uniform
magnetic field.
Here is a picture to help you visualize. It came from
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html.
Homework Hint
F = ILB sin F max = ILB
For fixed L, B: this is the biggest force
you can get from a given current I.
For bigger F, you need bigger I.
F
       
       
B
I L
       
       
F = IL  B
Valid for straight wire, length L inside region of magnetic field,
constant magnetic field, constant current I, direction of L is
direction of conventional current I.
You could apply this equation to a beam of charged particles moving through
space, even if the charged particles are not confined to a wire.
What if the wire is not straight?
B


I
 





















 
ds
 





dF
 






dF = I ds  B
F =  dF

F = I ds  B

Integrate over the part
of the wire that is in the
magnetic field region.
Homework Hint: if you have a tiny piece of a wire, just calculate dF; no need to integrate.
Note: I generally use ds for an infinitesimal piece of wire,
instead of dl.
Font choice may make “l” look like “1.”
It’s a pain to search the fonts for a script lowercase l: l.
Example: a wire carrying current I consists of a semicircle of
radius R and two horizontal straight portions each of length L.
It is in a region of constant magnetic field as shown. What is
the net magnetic force on the wire?
I
       
B
       
R
       
 L     L 
y
x
       
There is no magnetic force on the portions of the wire outside
the magnetic field region.
First look at the two
straight sections.
F = IL  B
L  B, so
F1 =F2 = ILB
I
y
     
F1
     
R
     
 L    
 
F2
B
 
 
L 
       
x
Next look at the
semicircular section.
Calculate the
infinitesimal force dF
on an infinitesimal ds
of current-carrying
wire.
I
y
    
F1 dF ds d
    
R
    
 L   
  
F2
B
  
  
 L 
       
x
ds subtends the angle from  to +d.
Why did I call that
angle  instead of ?
The infinitesimal force is dF =I ds  B.
Because we usually
use  for the angle
in the cross product.
ds  B, so dF = I ds B.
Arc length ds =R d.
Finally, dF =I R d B.
Calculate the ycomponent of F.
dFy = I R d B sin
I

Fy =  dFy
y
0

Fy =  I R d B sin
0
dFy 
 
d
F1 dF
ds
 
R

 
 L   
  
 L 
       
x

Fy = I R B  sind
0

Fy =  -I R B cos  0
Fy = 2 I R B
  
F2
B
  
Interesting—just the force on a
straight horizontal wire of length 2R.
Does symmetry give
you Fx immediately?
Or, you can calculate
the x component of F.
I
y
dFx = -I R d B cos

Fx = - I R d B cos
 

F1 dF ds d
 
dFx
R

 
 L   
  
F2
B
  
  
 L 
       
x
0

Fx = -I R B  cosd
0

Fx = -  I R B sin 0
Fx = 0
Sometimes-Useful Homework Hint
Symmetry is your friend.
Total force:
F = F1 + F2 + Fy
F = ILB + ILB + 2IRB
I
y
Fy
     
F1 dF ds
     
R
     
 L    
 
F2
B
 
 
L 
       
F = 2IB L + R 
F = 2IB L + R  ˆj
x
We probably should write
the force in vector form.
Possible homework hint: how would the result differ if the magnetic field were directed along
the +x direction? If you have difficulty visualizing the direction of the force using the right
hand rule, pick a ds along each different segment of the wire, express it in unit vector
notation, and calculate the cross product.
Example: a semicircular closed loop of radius R carries current
I. It is in a region of constant magnetic field as shown. What
is the net magnetic force on the loop of wire?
FC
       
B
       
R
       
y
x
   I    
       
We calculated the force on the semicircular part in the previous
example (current is flowing in the same direction there as
before).
F =2 I R B
C
Next look at the
straight section.

FS = IL  B

L  B, and L=2R so

y
FS = 2IRB

FC
      
B
      
R
      
  I    
       
FS
x
Fs is directed in the –y direction (right hand rule).
Fnet =FS +Fc = -2IRB ˆj+2IRB ˆj = 0
The net force on the closed loop is zero!
This is true in general for closed loops
in a uniform magnetic field.
Today’s agenda:
Magnetic Fields.
You must understand the similarities and differences between electric fields and field lines,
and magnetic fields and field lines.
Magnetic Force on Moving Charged Particles.
You must be able to calculate the magnetic force on moving charged particles.
Motion of a Charged Particle in a Uniform Magnetic Field.
You must be able to calculate the trajectory and energy of a charged particle moving in a
uniform magnetic field.
Magnetic Forces on Currents and Current-Carrying Wires.
You must be able to calculate the magnetic force on currents.
Velocity Selector and Mass Spectrometer.
You must be able to work these detailed examples.
Velocity Selector
       
Bout
----------------
  v     
+
E
       

       
The magnetic field is uniform inside the dashed-line region.
What speed will allow the proton to pass undeflected through
region of uniform electric field?
Velocity Selector
       
Bout
FE = qE
----------------
  v     
+
E
+
       
v
FB = qv  B

       
When the electric and magnetic forces balance then the charge
will pass straight through.
FE = FB
q E = q vB
or
E
v=
B
Velocity Selector
       
Bout
FE = qE
----------------
  v     
+
E
       

       
+
v
FB = qv  B
This only works if the electric and magnetic fields are
perpendicular and oriented so that qv  B is antiparallel to E.
Also, we simplified the calculation by making v perpendicular
to B.
Mass Spectrometers
Mass spectrometers separate charges of different mass.
When ions of fixed energy enter a region of constant magnetic
field, they follow a circular path.
The radius of the path
depends on the
mass/charge ratio and
speed of the ion, and
the magnitude of the
magnetic field.
x = 2r
and
mv
r=
qB
Worked example with numbers to follow.
V
S
+q
x
    
B
    
    
R
    
    
    
    
Thanks to Dr. Waddill for the use of the picture.
Example: ions from source S enter a region of uniform magnetic field
B that is perpendicular to the ion path. The ions follow a semicircle
and strike the detector plate at x = 1.7558 m from the point where
they entered the field. If the ions have a charge of 1.6022 x 10-19 C,
the magnetic field has a magnitude of B = 80.0 mT, and the
accelerating potential is V = 1000.0 V, what is the mass of the ion?
V
S
    
B
    
+q
x
 
R
 
 
 












    
Step 1: find the final speed of the ions if they have a charge of
1.6022 x 10-19 C and the accelerating potential is V = 1000.0 V?
V
Let’s focus on the part of
the diagram where the
ions from the source are
being accelerated.
S
    
B
    
+q
If the ions are “boiling” off a source
filament, their initial velocity directions
will be almost random, so their average
speed will be almost zero. Let’s assume
the ions start at rest (corresponding to
zero average speed).
x
    
R
    
    
    
    
Step 1: find the final speed of the ions if they have a charge of
1.6022 x 10-19 C and the accelerating potential is V = 1000.0 V?
V
Ef  Ei   Wother if
0
0
Kf  Uf   Ki  Ui    Wother if
S
+q
vi = 0
+q v = ?
Kf  Uf +Ui    Uf  Ui   U
1
mv 2  U  qV
2
2qV
v
m
Notes: we don’t know m, so let’s keep this symbolic for now. Besides,
we want to solve for m later. Also, V must be negative, which makes
sense (a + charge would be accelerated away from a higher potential).
Don’t get your v’s and V’s mixed up! Use a script v for speed!
Step 2: find the radius of the circular path followed by the ions in the
region of uniform magnetic field of magnitude B.
V
S
    
B
    
+q
x
    
R
    
    
    
    
Step 2: find the radius of the circular path followed by the ions in the
region of uniform magnetic field of magnitude B.
Let’s focus on the
magnetic field region of
the diagram.
  
+q   
v
  
R
  
x
  
 
B
 
 
 
 
    
    
Step 2: find the radius of the circular path followed by the ions in the
region of uniform magnetic field of magnitude B.
For motion in a circular path...
    
B
+q
    
v2
FB  ma  m
R
FB  qv  B  qvBsin 90
2
v
q vB  m
R
x
    
v
FB R
    
    
    
    
mv
R
qB
On an exam, we’ll probably let you get away without using the absolute value signs,
even though it makes our brains ache when we do that.
Step 3: solve for m, using q=1.6022x10-19 C, V=1000.0 V,
B = 80.0 mT, and R=x/2=1.7558/2=0.8779 m.
V
Now we can put it all
together.
S
2qV
v
m
mv
R
qB
    
B
    
+q
We know q, R, B, and V. We have two
unknowns, v and m, and two equations,
so we can solve for m.
x
    
R
    
    
    
    
 2qV 
m 
  2qV  m  2V  m
2 2
m
v
m


R2  2 2 

2
2
2
2
2
q
B
q B
q B
q B
2
Step 3: solve for m, using q=1.6022x10-19 C, V=-1000.0 V,
B = 80.0 mT, and R=x/2=1.7558/2=0.8779 m.
R
2
V
2V  m


q B2
S
    
B
    
+q
q B2 R 2
m
 2V 
x
1.6022 10  80 10   0.8779 

m
19
3 2
2
    
R
    
    
    
    
  2  1000 
m  3.9515 1025 kg
uranium-238!
1 atomic mass unit (u) equals 1.66x10-27 kg, so m=238.04 u.