1
Today’s agendum:
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.
Magnetic Flux and Gauss’ “Law” for Magnetism.
You must be able to calculate magnetic flux and recognize the consequences of Gauss’
“Law” for Magnetism.
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.
2
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
3
There is one 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
4
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.
5
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?).
*The N pole of a compass would “want to get to” the S pole of the magnet.
6
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.
7
We use the symbol B for magnetic field.
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 two 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.
8
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
An electromagnet like this one 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
9
buckle with iron in it!
Today’s agendum:
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.
Magnetic Flux and Gauss’ “Law” for Magnetism.
You must be able to calculate magnetic flux and recognize the consequences of Gauss’
“Law” for Magnetism.
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.
10
Magnetic Fields and Moving Charges
A charged particle moving in a magnetic field experiences a
force.
The magnetic force equation predicts the effect of a
magnetic field on a moving charged particle.
F =qv B
force on
particle
velocity of
charged particle
magnetic field
vector
What is the force if the charged particle is at rest?
11
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).
12
Cross product as presented in Physics 103.
F = q vB sinθ = q v B = q vB
Magnitude of magnetic force (and the meaning of v and B):
+

v
F = q v B
B
v
+

v
B
B
F = q vB
13
Direction of magnetic force--Use right hand rule:
fingers out, thumb perpendicular to them
rotate your hand until your palm points in the direction of B
(curl fingers through smallest angle from v to B)
thumb points in direction of F
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
14
F = q vB sinθ = q v B = q vB
Direction of magnetic force:
F
+

B

-
v
B
v
F
v
F?
-

B
15
“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  

16
Alternative* view of magnetic field units.
F = q vB sinθ
F
B=
q v sinθ
N
N
=
B  = T =
C  m/s A  m
Remember, units of field are
force per “something.”
*“Official” definition of units coming later.
17
Today’s agendum:
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.
Magnetic Flux and Gauss’ “Law” for Magnetism.
You must be able to calculate magnetic flux and recognize the consequences of Gauss’
“Law” for Magnetism.
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.
18
Magnetic Flux and Gauss’ “Law” for Magnetism
Magnetic Flux
We have used magnetic field lines to visualize magnetic fields
and indicate their strength.
We are now going to count the
number of magnetic field lines passing
through a surface, and use this count
to determine the magnetic field.
B
19
The magnetic flux passing through a surface is the number of
magnetic field lines that pass through it.
Because magnetic field lines are
drawn arbitrarily, we quantify
magnetic flux like this: M=BA.
If the surface is tilted, fewer lines cut
the surface.
A
B
B

If these slides look familiar, refer back to lecture 4!
20
We define A to be a vector having a
magnitude equal to the area of the
surface, in a direction normal to the
surface.
A

B

The “amount of surface” perpendicular
to the magnetic field is A cos .
Because A is perpendicular to the surface, the amount of A
parallel to the electric field is A cos .
A = A cos  so M = BA = BA cos .
Remember the dot product from Physics 103? M  B  A
21
If the magnetic field is not uniform, or the surface is not flat…
divide the surface into
infinitesimal surface
elements and add the
flux through each…
dA
B
M  lim
Ai 0
 B  A
i
i
i
 M   B  dA
your starting equation
sheet has…
 B   B  dA
22
If the surface is closed (completely encloses a volume)…
…we count lines going out
as positive and lines going
in as negative…
B
dA
 
 M   B  dA
a surface integral, therefore a
double integral 
But there are no magnetic monopoles in nature (experimental
fact). If there were more flux lines going out of than into the
volume, there would be a magnetic monopole inside.
23
Therefore
B
dA
 
 M   B  dA  0
Gauss’ “Law” for Magnetism!
Gauss’ “Law” for magnetism is not very useful in this course.
The concept of magnetic flux is extremely useful, and will be
used later!
24
You have now learned Gauss’s “Law” for both electricity and
magnetism.
  q enclosed
 E   E  dA 
o
 
 M   B  dA  0
These equations can also be written in differential form:

E 
0
B  0
25
Today’s agendum:
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.
Magnetic Flux and Gauss’ “Law” for Magnetism.
You must be able to calculate magnetic flux and recognize the consequences of Gauss’
“Law” for Magnetism.
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.
26
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.
27
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 an acceleration
equal to v2/r, where r is the radius of the circle.
28
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
v=
q rB
m
mv
r=
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
29
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.
B
v
+
v
There won’t be any test problems on
helical motion.
30
Lorentz Force “Law”


If both electric and magnetic fields are present, F = q E+ v  B .
Applications
The applications in the remaining slides are in your textbook
sections for the next lecture.
If I have time, I will show them today.
The energy calculation in the mass spectrometer example is
often useful in homework.
31
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
32
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.
     B
S
+q
x
    
    
r
    
    
    
    
33
Thanks to Dr. Waddill for the use of the picture.
Example: Ions from source S enter a region of constant magnetic
field B that is perpendicular to the ions 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
Radius of ion path:
x = 2r
and
mv
r=
qB
Unknowns are m and v.
S
     B
    
    
+q
r
x
    
    
    
    
34
Conservation of energy gives speed of ion. The ions leave the
source with approximately zero kinetic energy
Ki +Ui =K f +Uf
0
K f =Ki + Ui -Uf 
Caution! V is potential,
v (lowercase) is speed.
K f =- Uf -Ui  =-q Vf - Vi  =-qV
V
     B
    
    
K f = -qV
S
1
mv 2 = -qV
2
-2qV
v=
m
+q
r
A proton accelerates when it
goes from a more positive V
to a less positive V; i.e., when
V is negative. That’s what
this minus sign means.
x
    
    
    
    
35
-2qV
v=
m
2mv
x = 2r =
qB
2m -2qV
x=
qB
m
2 -2mV
x=
B
q
V
S
     B
    
    
+q
r
x
B2 x 2q
m=8V
    
    
    
    
36
B2 x 2q
m=8V
 0.08 T  1.7558 m 1.6022 10-19 C 
2
m=-
2
8  -1000 V 
V
-25
m=3.9515 10 kg
1 atomic mass unit
(amu) equals
1.66x10-27 kg, so
m = 238.04 u
S
uranium-238!
     B
    
    
+q
r
x
    
    
    
    
37
Today’s agendum:
Review and some interesting consequences of
F=qvxB.
You must understand the similarities and differences between electric forces and magnetic
forces on charged particles.
Magnetic forces on currents and current-carrying wires.
You must be able to calculate the magnetic force on currents.
Magnetic forces and torques on current loops.
You must be able to calculate the torque and magnetic moment for a current-carrying wire
in a uniform magnetic field.
Applications: galvanometers, electric motors, rail guns.
You must be able to use your understanding of magnetic forces and magnetic fields to
describe how electromagnetic devices operate.
38
Reminder: signs
F = qv  B
Include the sign on q, properly account for the directions of any
two of the vectors, and the direction of the third vector is
calculated “automatically.”
F = q vB sinθ = q v B = q vB
If you determine the direction “by hand,” use the magnitude of
the charge.
Everything in this equation is a magnitude.
The sign of r had better be +!
mv
r=
qB
39
Magnetic and Electric Forces
The electric force acts in the direction of the electric field.
FE =qE
The electric force is nonzero even if v=0.
The magnetic force acts perpendicular to the magnetic field.
FB =qv B
The magnetic force is zero if v=0.
FB  v = 0  = q  0   B = 0
40
Magnetic and Electric Forces
The electric force does work in displacing a charged particle.
FE =qE
E
+
F
D
WF =F  D=FD=qED
The magnetic force does no work in displacing a charged
particle!
B v
FB =qv B
WF =F  ds =0
+ds
Amazing!
F
41
Today’s agendum:
Review and some interesting consequences of F=qvxB.
You must understand the similarities and differences between electric forces and magnetic
forces on charged particles.
Magnetic forces on currents and current-carrying
wires.
You must be able to calculate the magnetic force on currents.
Magnetic forces and torques on current loops.
You must be able to calculate the torque and magnetic moment for a current-carrying wire
in a uniform magnetic field.
Applications: galvanometers, electric motors, rail guns.
You must be able to use your understanding of magnetic forces and magnetic fields to
describe how electromagnetic devices operate.
42
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.
43
Here is a picture to help you visualize. It came from
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html.
44
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.
45
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.
46
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.
47
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
48
Next look at the
semicircular section.
Calculate the
incremental force dF
on an incremental ds
of current-carrying
wire.
I
y
    
F1 dF ds d
    
R
    
 L   
  
F2
B
  
  
 L 
       
x
ds subtends the angle from  to +d.
The incremental force is dF = I ds  B.
ds  B, so dF = I ds B.
Arc length ds =R d.
Finally, dF =I R d B.
49
Calculate the ycomponent of F.
dFy =I R d B sin

Fy =  dFy
I
y
0

Fy =  I R d B sin
0
dFy 
 
d
F1 dF
ds
 
R

 
 L   
  
F2
B
  
  
 L 
       
x

Fy = I R B  sind
0

Fy =  -I R B cos 0
Fy =2 I R B
Interesting—just the force on a
straight horizontal wire of length 2R.
50
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
51
Total force:
F =F1 + F2 + Fy
F = ILB + ILB + 2IRB
F =2IB L + R 
I
y
Fy
     
F1 dF ds
     
R
     
 L    
 
F2
B
 
 
L 
       
x
52
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).
53
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.
54
Today’s agendum:
Review and some interesting consequences of F=qvxB.
You must understand the similarities and differences between electric forces and magnetic
forces on charged particles.
Magnetic forces on currents and current-carrying wires.
You must be able to calculate the magnetic force on currents.
Magnetic forces and torques on current loops.
You must be able to calculate the torque and magnetic moment for a current-carrying wire
in a uniform magnetic field.
Applications: galvanometers, electric motors, rail guns.
You must be able to use your understanding of magnetic forces and magnetic fields to
describe how electromagnetic devices operate.
55
Magnetic Forces and Torques on Current Loops
We showed (not in general, but illustrated the technique) that
the net force on a current loop in a uniform magnetic field is
zero.
motion.
No net force means no motion.NOT.
Example: a rectangular current loop of area A is placed in a
uniform magnetic field. Calculate the torque on the loop.
L
I
B
W
Let the loop carry a
counterclockwise
current I and have
length L and width W.
The drawing is not meant to imply
that the top and bottom parts are
outside the magnetic field region.
56
There is no force on the “horizontal”
segments because the current and
magnetic field are in the same
direction.
The vertical segment on the left
“feels” a force “out of the page.”
L
FL
FR B
I
W
The vertical segment on the right
“feels” a force “into the page.”
The two forces have the same magnitude: FL = FR = I L B.
Because FL and FR are in opposite directions, there is no net
force on the current loop, but there is a net torque.
57
Top view of current loop, looking
“down,” at the instant the magnetic
field is parallel to the plane of the loop.
In general, torque is  = r  F .
W
1
R = FR  WILB
2
2
L =
FR
IL
FL W
2
B
 IR
W
2
W
1
FL  WILB
2
2
net = R  L = WILB =IAB
area of loop = WL
58
When the magnetic field is not parallel
to the plane of the loop…
W
1
R = FR sin   WILB sin 
2
2
L =
W
1
FL sin   WILB sin 
2
2
FR
W
sin 
2

IL
FL
 IR
B

W
2
A
net = R  L = WILB sin  =IAB sin 
Define A to be a vector whose magnitude is the area of the
loop and whose direction is given by the right hand rule (cross
A into B to get ). Then
 = IA  B .
59
Magnetic Moment of a Current Loop
W
sin 
2
 = IA  B
Alternative way to get direction of A:
curl your fingers (right hand) in
direction of current; thumb points in
direction of A.
IA is defined to be the magnetic
moment of the current loop.
FR

IL
FL
 IR
B

W
2
A
 = IA
 =  B
Your starting equation sheet has:
=N I A
(N=1 for a single loop)
60
Energy of a magnetic dipole in a magnetic field
You don’t realize it yet, but we have been talking about
magnetic dipoles for the last 5 slides.
A current loop, or any other body that experiences a magnetic
torque as given above, is called a magnetic dipole.
Energy of a magnetic dipole?
You already know this:
Today:
Electric Dipole
Magnetic Dipole
 = p E
 =  B
U= - p E
U= -  B
61
Today’s agendum:
Review and some interesting consequences of F=qvxB.
You must understand the similarities and differences between electric forces and magnetic
forces on charged particles.
Magnetic forces on currents and current-carrying wires.
You must be able to calculate the magnetic force on currents.
Magnetic forces and torques on current loops.
You must be able to calculate the torque and magnetic moment for a current-carrying wire
in a uniform magnetic field.
Applications: galvanometers, electric motors, rail
guns.
You must be able to use your understanding of magnetic forces and magnetic fields to
describe how electromagnetic devices operate.
62
The Galvanometer
Now you can understand how a galvanometer works…
When a current is passed through a coil connected to a needle,
the coil experiences a torque and deflects. See the link below
for more details.
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/galvan.html#c1
63
Electric Motors
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/mothow.html#c1
64
Hyperphysics has nice interactive graphics showing how dc and
ac motors work.
65
We’ve been working with the effects of magnetic fields
without considering where they come from. Today we
learn about sources of magnetic fields.
Today’s agendum:
Magnetic Fields Due To A Current.
You must be able to calculate the magnetic field due to a moving charged particle.
Biot-Savart “Law.”
You must be able to use the Biot-Savart “Law” to calculate the magnetic field of a currentcarrying conductor (for example: a long straight wire).
66
But first, a note on the right hand rule.
I personally find the three-fingered axis system
often (but not always) to be the most useful.


“In F = IL  B and F = qv  B does it matter
which finger I use for what?”
F = IL  B
F = qv  B
 
 


No, as long as
you keep the
right order. All
three of these
will work:

 










67


This works:


This doesn’t:

Switching only two is wrong!

“The right-hand rule is unfair! Physics is discriminating against
left-handers!”
No, you can get the same results with left-hand axes and left
hand rules. See this web page.
68
Today’s agendum:
Magnetic Fields Due To A Current.
You must be able to calculate the magnetic field due to a moving charged particle.
Biot-Savart “Law.”
You must be able to use the Biot-Savart “Law” to calculate the magnetic field of a currentcarrying conductor (for example, a long straight wire).
69
Biot-Savart “Law”: magnetic field of a current element
Let’s start with the magnetic field of a moving charged particle.
r
B
rˆ
+
v
It is experimentally
observed that a moving
point charge q gives rise
to a magnetic field
μ 0 qv  rˆ
B=
.
2
4π r
0 is a constant, and its value is
0=4x10-7 T·m/A
Remember: the direction of r is always from the source point (the thing that
causes the field) to the field point (the location where the field is being measured.
70
Example: proton 1 has a speed v0 (v0<<c) and is moving along
the x-axis in the +x direction. Proton 2 has the same speed
and is moving parallel to the x-axis in the –x direction, at a
distance r directly above the x-axis. Determine the electric and
magnetic forces on proton 2 at the instant the protons pass
closest to each other.
y
This is example 28.1 in your text.
FE
The electric force is
v0
1 q1q2 ˆ
FE =
r
2
4  r
2
E
r
1
2
1 e ˆ
FE =
j
2
4 r
z
rˆ
v0
x
71
At the position of proton 2 there is a magnetic field due to
proton 1.
 q1 v1  rˆ
B1 =
4 r2
 ev 0 ˆi  ˆj
B1 =
4 r 2
y
FE
v0
  ev 0 ˆ
B1 =
k
2
4 r
2
B1
r
1
z
rˆ
v0
x
72
Proton 2 “feels” a magnetic force due to the magnetic field of
proton 1.
FB = q2 v2 B1
 
  ev 0 ˆ 
ˆ
FB = ev 0  i  
k
2
 4 r

y
FE
v0
 e2 v 02 ˆ
FB =
j
2
4 r
2
FB
B1
What would proton 1 “feel?”
r
1
Caution! Relativity overrules Newtonian mechanics!
However, in this case, the force is “equal & opposite.”
z
rˆ
v0
x
73
Both forces are in the +y direction. The ratio of their
magnitudes is
   e2 v 02 

2 
4

r
FB 

=
FE  1 e2 

2 
4

r



y
FE
FB
=  v 20
FE
v0
2
FB
B1
Later we will find that
r
1
1
 = 2
c
z
rˆ
v0
x
74
FB v 20
Thus
= 2
FE c
If v0=106 m/s, then
10 

FB
-5
=

1.11

10
FE  3  10 8 2
6 2
y
FE
Don’t you feel sorry for the poor,
weak magnetic force?
v0
2
FB
B1
r
1
z
rˆ
v0
x
75
Today’s agendum:
Magnetic Fields Due To A Current.
You must be able to calculate the magnetic field due to a moving charged particle.
Biot-Savart “Law.”
You must be able to use the Biot-Savart “Law” to calculate the magnetic field of a currentcarrying conductor (for example, a long straight wire).
76
From the equation for the magnetic field of a moving charged
particle, it is “easy” to show that a current I in a little length dl
of wire gives rise to a little bit of magnetic field.
r
dB
rˆ
dl
I
μ 0 I d  rˆ
dB =
4π r 2
The Biot-Savart “Law”
If you like to be more precise in your language, substitute
“an infinitesimal” for “a little length” and “a little bit of.”
I often use ds instead of dl because the script l does not
display very well.
You may see the equation written using r = r rˆ .
77
Applying the Biot-Savart “Law”
dB 
r

rˆ
ds
I
μ 0 I ds  rˆ
r
ˆ
dB =
where
r
=
4π r 2
r
dB =
μ 0 I ds sin θ
4π
r2
B =  dB
78
Example: calculate the magnetic field at point P due to a thin
straight wire of length L carrying a current I. (P is on the
perpendicular bisector of the wire at distance a.)
y
dB
P
r
rˆ 
ds
x
μ 0 I ds  rˆ
dB =
4π r 2
a

z
ds  rˆ = ds sinθ kˆ
x
I
dB =
μ 0 I ds sinθ
4π
r2
L
ds is an infinitesimal quantity in the direction of dx, so
μ 0 I dx sinθ
dB =
4π
r2
79
sinθ =
a
r
2
r
rˆ 
ds
x
L
μ 0 I dx sinθ
4π
r2
μ 0 I dx a μ 0 I dx a
dB =
=
3
4π r
4π  x 2 + a2 3/2
a

z
dB =
r = x +a
y
dB
P
2
x
I
μ0
I dx a
-L/2 4π
2
2 3/2
x +a 
B=
L/2
μ 0I a L/2
dx
B=
4π -L/2  x 2 + a2 3/2
80
μ 0I a L/2
dx
B=
4π -L/2  x 2 + a2 3/2
y
dB
P
r
rˆ 
ds
x
look integral up in tables, use the
web,or use trig substitutions
a
x

z

I
L
dx
x
2

2 3/2
+a
=
x
a x +a
2
2

2 1/2
L/2
μ 0I a
x
B=
4π a2  x 2 +a2 1/2
-L/2

μ 0I a 
L/2
=
4π  a2 L/2 2 + a2



1/2

-L/2


1/2 
2
2
2
a  -L/2  + a



81
y
dB
P
r
rˆ 
ds
x


μ 0I a 
2L/2

B=
1/2
4π  a2 L2 /4 + a2  


a

z
x
I
B=
μ 0I L
1
4πa L2 /4 + a2 1/2
μ 0I L
1
B=
2πa L2 + 4a2
μ 0I
B=
2πa
1
4a2
1+ 2
L
82
y
dB
P
r
rˆ 
ds
a

z
x
When L, B =
x
μ 0I
B=
2πa
1
4a2
1+ 2
L
I
μ 0I
.
2πa
μ 0I
or B =
2 πr
83
Magnetic Field of a Long Straight Wire
We’ve just derived the equation for the magnetic
field around a long, straight wire*
μ0 I
B=
2 πr
with a direction given by a “new” right-hand
rule.
I
B
r
*Don’t use this equation unless you have a long, straight wire!
84
Looking “down” along the wire:
I
B
The magnetic field is not
constant.
At a fixed distance r from the wire, the magnitude of the
magnetic field is constant.
The magnetic field direction is always tangent to the
imaginary circles drawn around the wire, and perpendicular
to the radius “connecting” the wire and the point where the
field is being calculated.
85
Today’s agendum:
Force Between Current-Carrying Conductors.
You must be able to begin with starting equations and calculate forces between currentcarrying conductors.
Magnetic Fields Due To A Current Loop.
You must be able to apply the Biot-Savart “Law” to calculate the magnetic field of a current
loop.
86
Magnetic Field of a Current-Carrying Wire
It is experimentally observed that parallel wires exert forces on
each other when current flows.
I1
I2
F12

F21

I1
I2
F12

F21

“Knowledge advances when Theory does battle with Real Data.”—Ian Redmount
87
We showed that a long straight wire carrying a current
I gives rise to a magnetic field B at a distance r from
the wire given by
μ0 I
B=
2 πr
The magnetic field of one wire exerts a force on a
nearby current-carrying wire.
d
I1
L
I2
F12

F21

I
B
r
The magnitude of the force depends on
the two currents, the length of the wires,
and the distance between them.
μ I I L
F= 0 1 2
2πd
The wires are electrically neutral,
so this is not a Coulomb force.
Remember, the direction of the field is given by another (different) right-hand rule: grasp the wire and point the thumb of your
88
right hand in the direction of I; your fingers curl around the wire and show the direction of the magnetic field.
Example: use the expression for B due to a current-carrying
wire to calculate the force between two current-carrying wires.
d
F12 =I1L1 B2
I1
L
μ I
Bˆ 2 = 0 2 kˆ
2 πd
μ I
F12 = I1Ljˆ  0 2 kˆ
2 πd
μ 0 I1I2L ˆ
F12 =
i
2 πd
I2
F12
B2
y
L1

L2

x
F12 μ 0 I1I2 ˆ
The force per unit length of wire is
=
i.
L
2 πd
89
d
F21 =I2L2 B1
I1
L
μ I
B1 = - 0 1 kˆ
2πd
 μ0 I1 ˆ 
ˆ
F21 = I2Lj   
k
 2πd 
μ 0 I1I2L ˆ
F21 = i
2 πd
I2
F12
F21
B1
y
L1

L2

x
μ 0 I1I2 ˆ
F21
The force per unit length of wire is
=i.
L
2πd
90
If the currents in the wires are in the opposite direction, the
force is repulsive.
d
L
I1
I2
F12
F21
y
L1

L2

91
F12 = F21 =
μ 0 I1I2L
2 πd
d
I1
4 π 10-7 I1I2L
L
-7
F12 = F21 =
= 2 10 I1I2
2πd
d
The official definition of the Ampere: 1 A is
the current that produces a force of 2x10-7
N force per meter of length between two
long parallel wires placed 1 meter apart in
empty space.
I2
F12
F21
L1

L2

92
Today’s agendum:
Force Between Current-Carrying Conductors.
You must be able to begin with starting equations and calculate forces between currentcarrying conductors.
Magnetic Fields Due To A Current Loop.
You must be able to apply the Biot-Savart “Law” to calculate the magnetic field of a current
loop.
93
Magnetic Field of a Current Loop
A circular ring of radius a carries a current I as shown.
Calculate the magnetic field at a point P along the axis of the
ring at a distance x from its center.
Complicated diagram!
You are supposed to
visualize the ring
lying in the yz plane.
dl is in the yz plane. rˆr
is in the xy plane and
is perpendicular to dl.
Thus d  rˆ = d .
y
I
dl ˆ
r
dB

dBy
r
a
90-
x
z

P dBx
Also, dB must lie in the xy plane (perpendicular to dl) and is
94
also perpendicular to r.
x
μ 0 I d  rˆ
dB =
4π r 2
dB =
y
I
μ0 I d
4π r 2
μ0 I d
dB =
4π  x 2  a2 
dl ˆ
r
dB

dBy
r
a
90-
x
z

P dBx
x
μ0 I d
μ0 I d
a
dB x =
cos =
2
2
4π  x  a 
4π  x 2  a2   x 2  a2 1/2
μ0 I d
μ0 I d
x
dB y =
sin =
2
2
4π  x  a 
4π  x 2  a2   x 2  a2 1/2
By symmetry, By will be 0. Do you see why?
95
y
I dl
rˆ
r
dBy
dBz
x
z
x
P dBx
When dl is not centered at z=0, there will be a z-component
to the magnetic field, but by symmetry Bz will still be zero.
96
y
μ0
Iad
dB x =
4π  x 2  a2 3/2
I
dl ˆ
r
dB

r
a
Bx =

dB x
dBy
90-
x
ring
z

P dBx
x
I, x, and a are constant as you integrate around the ring!
μ0
Ia
Bx =
4π  x 2  a2 3/2
Bx =
μ 0 I a2
2x  a
2

μ0
Ia
ring d = 4π x 2  a2 3/2 2a


2 3/2
97
y
At the center of the
ring, x=0.
I
B x,center =
2
μ0 I a
2 a
dl ˆ
r
dB

r
a

dBy
2 3/2
90-
x
z

P dBx
x
μ 0 I a2
μ0 I
B x,center =
=
2a3
2a
For N tightly packed concentric rings (a coil)…
B x,center =
μ0 N I
2a
98
Magnetic Field at the center of a Current Loop
A circular ring of radius a lies in the xy plane and carries a
current I as shown. Calculate the magnetic field at the center
of the loop.
One-page derivation, so you
won’t have to go through all the
above steps (to be worked at the
blackboard)!
The direction of the magnetic
field will be different if the plane
of the loop is not in the xy plane.
x
I
dl
a
z
y
I’ll explain the “funny” orientation
of the axes in lecture.
99
Today’s agendum:
Ampere’s “Law.”
You must be able to use Ampere’s “Law” to calculate the magnetic field for high-symmetry
current configurations.
Solenoids.
You must be able to use Ampere’s “Law” to calculate the magnetic field of solenoids and
toroids. You must be able to use the magnetic field equations derived with Ampere’s “Law”
to make numerical magnetic field calculations for solenoids and toroids.
100
Ampere’s “Law”
Just for kicks, let’s evaluate the line integral
along the direction of B over a closed circular
path around a current-carrying wire.
 
 B  d s  B ds  B2r 

B

ds
I
B
ds
r
   o I 
 B  ds   2r 2r   o I
The above calculation is only for the special case of a long
straight wire, but you can show that the result is valid in
general.
101
 
 B  ds  o I
Ampere’s “Law”
I is the total current that passes through a surface bounded by
the closed (and not necessarily circular) path of integration.
Ampere’s “Law” is useful for calculating the magnetic field due
to current configurations that have high symmetry.
The current I passing through a loop is positive if the direction
of integration is the same as the direction of B from the right
hand rule.
I
positive I
B
r
I
ds
negative I
B
r
ds
102
Your text writes
 
 B  d s   o Iencl
because the current that you use is the current “enclosed” by
the closed path over which you integrate.
 
d E
General form of Ampere’s “Law”:  B  d s   o I encl  o
dt
The reason for the 2nd term on the right will become apparent later. Ignore it for now.
If your path includes more than one source of current, add all
the currents (with correct sign).
 
 B  d s   o I1  I 2 
I1
I2
ds
103
Example: a cylindrical wire of radius R carries
a current I that is uniformly distributed over
the wire’s cross section. Calculate the
magnetic field inside and outside the wire.
I
R
Cross-section of the wire:
 direction of I
B
R
r
 
 
 
r 2
r2
 B  ds  o Iencl  o I R 2  o I R 2
104
Over the closed circular path r:
 
 B  d s  B ds  B2r 
Solve for B:
 direction of I
B
R
r
r2
2πrB = μ 0 I 2
R
μ0 I
r2
r
B = μ0 I
=
μ
I
=
r
0
2
2
2
2πrR
2πR
2πR
B is linear in r.
105
Outside the wire:
 
 B  d s  B ds  B2r    o I
μ0 I
B=
2 πr
 direction of I
R
A lot easier than using
the Biot-Savart “Law”!
(as expected).
r
B
Plot:
B
R
r
106
Calculating Electric and Magnetic Fields
Electric Field
Magnetic Field
in general: Coulomb’s “Law”
in general: Biot-Savart “Law”
for high symmetry
configurations: Gauss’ “Law”
for high symmetry
configurations: Ampere’s “Law”
This analogy is rather flawed because Ampere’s “Law” is not
really the “Gauss’ “Law” of magnetism.”
107
Today’s agendum:
Ampere’s “Law.”
You must be able to use Ampere’s “Law” to calculate the magnetic field for high-symmetry
current configurations.
Solenoids.
You must be able to use Ampere’s Law to calculate the magnetic field of solenoids and
toroids. You must be able to use the magnetic field equations derived with Ampere’s “Law”
to make numerical magnetic field calculations for solenoids and toroids.
108
Magnetic Field of a Solenoid
A solenoid is made of many loops of wire, packed closely
together. Here’s the magnetic field from a single loop of wire:
Some images in this
section are from
hyperphysics.
109
Stack many loops to make a solenoid:
This ought to remind you of the magnetic field of a bar magnet.
110

B

           I



l
You can use Ampere’s “Law” to calculate the magnetic field of a
solenoid.

 
 
 
 
 B  ds   B  ds   B  ds   B  ds   B  ds
1
2
3
4

 B  ds  B  0  0  0   o Iencl
B = μ0 N I
N is the number of loops
enclosed by our surface.
111

B

           I



l
B = μ0
N
I
B = μ0 n I
Magnetic field of a solenoid of
length l , N loops, current I.
n=N/l (number of turns per
unit length).
The magnetic field inside a long solenoid does not depend on the position
inside the solenoid (if end effects are neglected).
112
A toroid* is just a solenoid “hooked up” to itself.

 B  ds   o Iencl   o NI
 
 B  d s  B ds  B2r 
B 2πr  = μ0 N I
μ0 N I
B=
2 πr
Magnetic field
inside a toroid of N
loops, current I.
The magnetic field inside a toroid is not subject to end effects, but is not
constant inside (because it depends on r).
*Your text calls this a “toroidal solenoid.”
113
Example: a thin 10-cm long solenoid has a total of 400 turns of
wire and carries a current of 2 A. Calculate the magnetic field
inside near the center.
B = μ0
N
I

-7 T  m   400 
B =  4 π ×10
2 A 

A   0.1 m

B = 0.01 T
114
“Help! Too many similar starting equations!”
μ0 I
B=
2 πr
long straight wire
μ0 N I
B=
2a
center of N loops of radius a
B = μ0
N
use Ampere’s “Law” (or note the lack of N)
I
probably not a starting equation
solenoid, length l, N turns
field inside a solenoid is constant
B = μ0 n I
solenoid, n turns per unit length
field inside a solenoid is constant
μ0 N I
B=
2 πr
toroid, N loops
field inside a toroid depends on position (r)
“How am I going to know which is which on the next exam?”
115