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adapted from http://www.labinitio.com (nz302.jpg)
Today’s lecture is brought to you by the letter P.
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Announcements
 There is lots of nice math in chapter 10! This lecture calls
your attention to those parts of the chapter that you need to
know for exams. Keep this lecture in mind when you study
chapter 10.
 Exam 3 is the two weeks from yesterday. I will need to know
by the end of next Wednesday’s lecture of any students who
have special needs different than for exam 2.
 Exam 3 will cover material through the end of today’s lecture.
Material presented in lecture next week will be covered on the
final exam.
Note!
Note: All of these mean “average:” <S> Saverage Sav Savg
Today’s agenda:
Electromagnetic Waves.
Energy Carried by Electromagnetic Waves.
Momentum and Radiation Pressure of an Electromagnetic
Wave.
rarely in the course of human events have so many starting equations been given in so little time
We began this course by studying fields that didn’t vary with
time—the electric field due to static charges, and the magnetic
field due to a constant current.
In case you didn’t notice—about a half dozen lectures ago
things started moving!
We found that changing magnetic
field gives rise to an electric field.
Also a changing electric field gives
rise to a magnetic field.
These time-varying electric and magnetic fields can propagate
through space.
Electromagnetic Waves
Maxwell’s Equations
 E  dA 
q enclosed
o
 E ds  
d B
dt
 B  dA  0
 B  d s= μ
I
0 encl
+μ 0 ε 0
dΦ E
dt
These four equations provide a complete description of
electromagnetism.
 E 
 ×E =-

0
dB
dt
 B  0
  B=
1 dE
c
2
dt
+μ 0 J
Production of Electromagnetic Waves
Apply a sinusoidal voltage to an antenna.
Charged particles in the antenna oscillate sinusoidally.
The accelerated charges produce sinusoidally varying
electric and magnetic fields, which extend throughout
space.
The fields do not instantaneously permeate all space, but
propagate at the speed of light.
y
x
z
direction of
propagation
y
x
direction of
propagation
z
This static image doesn’t show how the wave propagates.
Here are animations, available on-line:
http://phet.colorado.edu/en/simulation/radio-waves
(shows electric field only)
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=35
Here is a movie.
Electromagnetic waves are transverse waves, but are not
mechanical waves (they need no medium to vibrate in).
Therefore, electromagnetic waves can propagate in free
space.
At any point, the magnitudes of E and B (of the wave
shown) depend only upon x and t, and not on y or z. A
collection of such waves is called a plane wave.
y
x
z
direction of
propagation
Manipulation of Maxwell’s equations leads to the following
Equations on this slide are for waves
plane wave equations for E and B:
propagating along x-direction.
2
 Ey
x
2
2
= 00
 E y (x, t)
t
2
 Bz
2
x
2
2
= 00
 B z (x, t)
t
2
These equations have solutions:
E y = E m a x sin  k x -  t 
Emax and Bmax are the
electric and magnetic
field amplitudes
B z = B m a x sin  k x -  t 
where
k=
2

,
 = 2 f ,
an d
f =

k
= c.
You can verify this by direct substitution.
Emax and Bmax in these notes are sometimes written by others as E0 and B0.
You can also show that
E y
x
=-
B z
t
  E m a x sin  k x -  t  
  B m a x sin  k x -  t  
=x
t
E m ax k c o s  k x -  t  = B m ax  c o s  k x -  t 
E m ax
B m ax
=
E
B
=

k
=c=
1
 00
.
At every instant, the ratio of the magnitude of the electric field
to the magnitude of the magnetic field in an electromagnetic
wave equals the speed of light.
y
direction of
propagation
x
z
Emax (amplitude)
E(x,t)
Today’s agenda:
Electromagnetic Waves.
Energy Carried by Electromagnetic Waves.
Momentum and Radiation Pressure of an Electromagnetic
Wave.
Energy Carried by Electromagnetic Waves
Electromagnetic waves carry energy, and as they propagate
through space they can transfer energy to objects in their path.
The rate of flow of energy in an electromagnetic wave is
described by a vector S, called the Poynting vector.*
S=
1
0
EB
The magnitude S represents the rate at which energy flows
through a unit surface area perpendicular to the direction of
wave propagation (energy per time per area).
Thus, S represents power per unit area. The direction of S is
along the direction of wave propagation. The units of S are
J/(s·m2) =W/m2.
*J. H. Poynting, 1884.
For an EM wave E  B = EB
y
E
B
z
S=
1
0
so S =
EB
EB
0
.
S
c
x
Because B = E/c we can write
S=
E
2
0c
=
cB
2
0
.
These equations for S apply at any instant of time and
represent the instantaneous rate at which energy is passing
through a unit area.
S=
EB
0
=
E
2
0c
=
cB
2
0
EM waves are sinusoidal. E y = E m a x sin  k x -  t 
B z = B m a x sin  k x -  t 
EM wave propagating
along x-direction
The average of S over one or more cycles is called the wave
intensity I.
The time average of sin2(kx - t) is ½, so
I = S average = S =
E m ax B m ax
2 0
2
=
E m ax
2 0 c
2
=
cB m ax
2 0
Notice the 2’s in
this equation.
This equation is the same as 10-29 in your text, using c = 1/(00)½.
The magnitude of S is the rate at which energy is
transported by a wave across a unit area at any instant:
 e n e rgy
tim e
S=

a re a


 po w e r 

=


a
re
a

 in sta n ta n e o u s
 in sta n ta n e o u s
Thus,
 e n e rg y
tim e
I= S = 

a re a


 pow er 

=


 a re a  a v e ra g e
 a v e ra g e
Note: Saverage and <S> mean the same thing!
Energy Density
The energy densities (energy per unit volume) associated
with electric and magnetic fields are:
uE =
1
2
 0E
2
uB =
1B
2
2 0
Using B = E/c and c = 1/(00)½ we can write
uB =
1B
2
2 0
=
1
E c 
2
0
uB = uE =
2
1
2
=
1  0  0E
2
0
2
1B
 0E =
2
=
1
2
 0E
2
2
2 0
remember: E and B are
sinusoidal functions of time
uB = uE =
1
2
2
 0E =
1B
2
2 0
For an electromagnetic wave, the instantaneous energy density
associated with the magnetic field equals the instantaneous
energy density associated with the electric field.
Hence, in a given volume the energy is equally shared by the
two fields. The total energy density is equal to the sum of the
energy densities associated with the electric and magnetic
fields:
2
u = u B + u E =  0E =
B
2
0
2
u = u B + u E =  0E =
B
2
instantaneous energy densities
(E and B vary with time)
0
When we average this instantaneous energy density over one
or more cycles of an electromagnetic wave, we again get a
factor of ½ from the time average of sin2(kx - t).
uE =
1
4
2
 0E
2
m ax
,
uB =
Recall S average = S =
1 B m ax
4 0
2
1 E m ax
2 0c
, and
u =
1
2
2
2
 0 E m ax =
1 B m ax
2 0
2
=
1 cB m ax
2
0
so we see that S = c u .
The intensity of an electromagnetic wave equals the average
energy density multiplied by the speed of light.
Help!
“These factors of ¼, ½, and 1 are making my brain hurt!”
It’s really not that bad. These are the energy densities
associated with E(t) and B(t) at some time t:
uB = uE =
1
2
2
 0E =
1B
2
2 0
Add the two to get the total energy density u(t) at time t:
2
u = u B + u E =  0E =
B
2
0
Help!
Again, these are the energy densities associated with E(t)
and B(t) at some time t:
uB = uE =
1
2
2
 0E =
1B
2
2 0
If you average uB and uE over one or more cycles, you get an
additional factor of ½ from the time average of sin2(kx-t).
uE =
1
4
2
2
 0E m ax
uB =
1 B m ax
4 0
,
The Emax and Bmax come from writing E = Emax sin(kx-t) and
B = Bmax sin(kx-t), and canceling the sine factors.
Help!
These are the average energy densities associated with E(t)
and B(t) over one or more complete cycles.
uE =
1
4
2
2
 0E m ax
uB =
1 B m ax
4 0
,
Add the two to get the total average energy density over one
or more cycles:
u = uE + uB =
1
4
2
2
 0 E m ax +
1 B m ax
4 0
=
1
2
2
2
 0 E m ax =
1 B m ax
2 0
Help!
Summary:
At time t:
Average:
u B (t) = u E (t) =
uE =
1
4
2
1
2
 0 E (t) =
2
1 B (t)
2
0
2
2
 0E m ax
uB =
1 B m ax
4 0
,
2
At time t:
Average:
2
u(t) =  0 E (t) =
u =
1
2
B (t)
0
2
2
 0 E m ax =
1 B m ax
2 0
If you use a starting equation that is not valid for the problem scenario, you will get incorrect results!
Example: a radio station on the surface of the earth radiates a
sinusoidal wave with an average total power of 50 kW.*
Assuming the wave is radiated equally in all directions above
the ground, find the amplitude of the electric and magnetic
fields detected by a satellite 100 km from the antenna.
Strategy: we want Emax, Bmax. We
are given average power. From
average power we can calculate
intensity,
frompower
intensity
From the and
average
we we
cancan
calculate
max and B
max.from
calculate E
intensity,
and
Satellite
Station
intensity we can calculate Emax
and Bmax.
*In problems like this you need to ask whether the power is radiated into all space or into just part of space.
Example: a radio station on the surface of the earth radiates a
sinusoidal wave with an average total power of 50 kW.*
Assuming the wave is radiated equally in all directions above
the ground, find the amplitude of the electric and magnetic
fields detected by a satellite 100 km from the antenna.
Area=4R2/2
All the radiated power passes
through the hemispherical
surface* so the average power
per unit area (the intensity) is
Satellite
R
Station
 pow er
I=
 area
 5.00  10 W 
4
P

-7
2
=
=
=
7.96

10
W
m

2
2
5
2 R
 average
2  1.00  10 m 
Today’s lecture is brought
to you by the letter P.
*In problems like this you need to ask whether the power is radiated into all space or into just part of space.
2
I= S =
E m ax =
=
1 E m ax
Satellite
2 0c
R
2  0 cI
Station
2  4   10
= 2.45  10
B m ax =
E m ax
c
=

-2
-7
V
 3 10
8
  7 .9 6  1 0
-7

m
2.45  10
 3  10
8
-2
V
m
m s

= 8.17  10
-11
You could get Bmax from I = c Bmax2/20, but that’s a lot more work
T
Example: for the radio station in the example on the previous
two slides, calculate the average energy densities associated
with the electric and magnetic field.
uE =
uE =
1
4
1
4
2
2
 0E m ax
uB =
 8.85  10   2.45  10 
u E = 1.33  10
-12
-15
J
m
3
-2
2
uB =
1 B m ax
4 0
1  8.17  10
4
-11

2
 4   10 
u B = 1.33  10
-7
-15
J
m
3
If you are smart, you will write <uB> = <uE> = 1.33x10-15 J/m3 and be done with it.
Today’s agenda:
Electromagnetic Waves.
Energy Carried by Electromagnetic Waves.
Momentum and Radiation Pressure of an
Electromagnetic Wave.
Momentum and Radiation Pressure
EM waves carry linear momentum as well as energy.
The momentum density carried by an electromagnetic wave is
dp
dV
=
S
c
2
This equation is not on your
equation sheet, but you have
permission to use it for tomorrow’s
homework (if needed) (10.28)
where dp is the momentum carried in the volume dV.
Today’s lecture is brought
to you by the letter P.
When the momentum carried by an electromagnetic wave is
absorbed at a surface, pressure is exerted on that surface.
If we assume that EM radiation is incident on an object for a
time t and that the radiation is entirely absorbed by the
object, then the object gains energy U in time t.
Maxwell showed that the momentum
change of the object is then:
p =
U
c
incident
(to ta l a bso rptio n )
The direction of the momentum change of the object is in the
direction of the incident radiation.
If instead of being totally absorbed the radiation is totally
reflected by the object, and the reflection is along the incident
path, then the magnitude of the momentum change of the
object is twice that for total absorption.
incident
reflected
p =
2U
c
(to tal reflectio n alo n g in ciden t p ath )
The direction of the momentum change of the object is again
in the direction of the incident radiation.
Radiation Pressure
The radiation pressure on the object is the force per unit area:
P=
From Newton’s
2nd
F
A
Law (F = dp/dt) we have: P =
For total absorption,  p =
F
A
=
1 dp
A dt
U
c
incident
So
1 dp
1 d U
P=
=

A dt
A dt  c
 dU

S
 1
d
t

=
=
 c A  c


(Equations on this slide involve magnitudes of vector quantities.)
This is the instantaneous radiation pressure in the case of total
absorption:
P(t) =
S (t)
c
For the average radiation pressure, replace S by <S>=Savg=I:
Prad =
S average
c
=
I
c
Electromagnetic waves also carry momentum through space
with a momentum density of Saverage/c2=I/c2. This is not on your
equation sheet but you have special permission to use it in
tomorrow’s homework, if necessary.
Today’s lecture is brought
to you by the letter P.
Prad =
I
c
(to ta l a bso rptio n )
incident
absorbed
Using the arguments above it can also be shown that:
Prad =
2I
c
(to ta l re fle ctio n )
incident
reflected
If an electromagnetic wave does not strike a surface, it still
carries momentum away from its emitter, and exerts Prad=I/c
on the emitter.
Example: a satellite orbiting the earth has solar energy
collection panels with a total area of 4.0 m2. If the sun’s
radiation is incident perpendicular to the panels and is
completely absorbed find the average solar power absorbed
and the average force associated with the radiation pressure.
The intensity (I or Saverage) of sunlight prior to passing through
the earth’s atmosphere is 1.4 kW/m2.

3
Pow er = IA = 1.4  10 W
m
2

 4.0 m
2
 = 5.6  10 W = 5.6 kW
3
Assuming total absorption of the radiation:
Prad =
S average
c

I
= =
3
1.4  10 W
c

F = Prad A = 4.7  10
-6
 3  10
N
m
2

8
m
m
s
2

 4.0 m
2
 = 4.7  10
-6
Pa
 = 1.9  10 N
-5
Caution! The letter P
(or p) has been used
in this lecture for
power, pressure, and
momentum!
That’s because today’s lecture is
brought to you by the letter P.
Revolutionary Application of Electromagnetic Waves
I know you are mostly engineers, and
think applications are important…
So I found you a revolutionary
new application that uses electromagnetic waves.
The UFO Detector. Only $48.54 at
amazon.com (just search for ufo
detector).
“The UFO detector continually monitors its
surrounding area for any magnetic and
electromagnetic anomalies.”
New starting equations from this lecture:
S=
E m ax
B m ax
k=
2

=
E
B
1
0
2
S average =
EB
=c=
1
,  = 2 f , f =
p =
U
c
or
2U
c
k
2 0c
uB = uE =
0 0

1 E m ax
=c
u =
1
2
1
2
2
=
1 cB m ax
2
2
 0E =
0
1B
2 0
2
2
 0 E m ax =
Pra d =
I
c
or
2
1 B m ax
2 0
2I
c
There are even more on your starting equation sheet; they are derived from the above!