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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
E0 encl 0 0
dΦB ds=μ I +μ ε
dt
enclosed
o
qE dA
These four equations provide a complete description of electromagnetism.
0
E
02
1 dEB= +μ J
c dt
B dA 0
B 0
BdE ds
dt
dB×E=-
dt
Maxwell’s Equations
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.
direction ofpropagation
y
z
x
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=35
This static image doesn’t show how the wave propagates.
Here are animations, available on-line:
direction ofpropagation
y
z
x
Here is a movie.
http://phet.colorado.edu/en/simulation/radio-waves(shows electric field only)
Electromagnetic waves are transverse waves, but are not mechanical waves (they need no medium to vibrate in).
direction ofpropagation
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
z
x
Manipulation of Maxwell’s equations leads to the following plane wave equations for E and B:
These equations have solutions:
You can verify this by direct substitution.
2 2y y
0 02 2
E E (x,t)=
x t
2 2z z
0 02 2
B B (x,t)=
x t
y maxE =E sin kx- t
z maxB =B sin kx- t
where , ,
2
k= =2 f and f = =c.k
Emax and Bmax in these notes are sometimes written by others as E0 and B0.
Emax and Bmax are the electric and magnetic field
amplitudes
Equations on this slide are for waves propagating along x-
direction.
You can also show that
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 zE B
=-x t
max maxE k cos kx- t =B cos kx- t
.
max
max 0 0
E E 1= = =c=
B B k
max maxE sin kx- t B sin kx- t=-
x t
direction ofpropagation
y
z
x
Emax (amplitude)
E(x,t)
Today’s agenda:
Electromagnetic Waves.
Energy Carried by Electromagnetic Waves.
Momentum and Radiation Pressure of an Electromagnetic Wave.
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).
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.*
0
1S= E B
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.
z
x
y
c
SE
BBecause B = E/c we can write
These equations for S apply at any instant of time and represent the instantaneous rate at which energy is passing through a unit area.
0
1S= E B
For an EM wave
so
E B =EB
.0
EBS=
.
2 2
0 0
E cBS= =
c
The time average of sin2(kx - t) is ½, so
2 2
0 0 0
EB E cBS= = =
c
EM waves are sinusoidal.
The average of S over one or more cycles is called the wave intensity I.
2 2max max max max
average0 0 0
E B E cBI =S = S = = =
2 2 c 2
This equation is the same as 10-29 in your text, using c = 1/(00)½.
y maxE =E sin kx- t
z maxB =B sin kx- t
Notice the 2’s in this
equation.
EM wave propagating along
x-direction
Thus,
The magnitude of S is the rate at which energy is transported by a wave across a unit area at any instant:
instantaneousinstantaneous
energypowertimeS= =
area area
averageaverage
energypowertimeI = S = =
area area
Note: Saverage and <S> mean the same thing!
The energy densities (energy per unit volume) associated with electric and magnetic fields are:
Using B = E/c and c = 1/(00)½ we can write
Energy Density
2E 0
1u = E
2
2
B0
1Bu =
2
222
20 0B 0
0 0 0
EE1B 1 1 1cu = = = = E
2 2 2 2
22
B E 00
1 1Bu =u = E =
2 2remember: E and B are sinusoidal functions of
time
For an electromagnetic wave, the instantaneous energy density associated with the magnetic field equals the instantaneous energy density associated with the electric field.
22
B E 00
1 1Bu =u = E =
2 2
22
B E 00
Bu=u +u = E =
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:
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).
so we see thatRecall
The intensity of an electromagnetic wave equals the average energy density multiplied by the speed of light.
22
B E 00
Bu=u +u = E =
22 max
0 max0
B1 1u = E =
2 2
2 2max max
average0 0
E cB1 1S = S = =
2 c 2.S =c u
, 2E 0 max
1u = E
4,
2max
B0
B1u =
4and
instantaneous energy densities(E and B vary with time)
“These factors of ¼, ½, and 1 are making my brain hurt!”
Help!
It’s really not that bad.
22
B E 00
1 1Bu =u = E =
2 2
It’s really not that bad. These are the energy densities associated with E(t) and B(t) at some time t:
Add uB and uE to get the total energy density u(t) at time t:
22
B E 00
Bu=u +u = E =
Help!
22
B E 00
1 1Bu =u = E =
2 2
Again, these are the energy densities associated with E(t) and B(t) at some time t:
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).
The Emax and Bmax come from writing E = Emax sin(kx-t) and B = Bmax sin(kx-t), and canceling the sine factors.
2E 0 max
1u = E
4,
2max
B0
B1u =
4
Help!
These are the average energy densities associated with E(t) and B(t) over one or more complete cycles.
Add uE and uB to get the total average energy density over one or more cycles:
2E 0 max
1u = E
4,
2max
B0
B1u =
4
2 22 2max max
E B 0 max 0 max0 0
B B1 1 1 1u = u + u = E + = E =
4 4 2 2
Help!
Summary:
22
B E 00
1 1B (t)u (t)=u (t)= E (t)=
2 2At time t:
2E 0 max
1u = E
4,
2max
B0
B1u =
4Average:
22
00
B (t)u(t)= E (t)=At time t:
22 max
0 max0
B1 1u = E =
2 2Average:
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.
Station
Satellite
*In problems like this you need to ask whether the power is radiated into all space or into just part of space.
Strategy: we want Emax, Bmax. We are given average power. From average power we can calculate intensity, and from intensity we can calculate Emax and Bmax.
From the average power we can calculate intensity, and from intensity we can calculate Emax and Bmax.
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.
R
Station
SatelliteAll the radiated power passes through the hemispherical surface* so the average power per unit area (the intensity) is
2
average
power PI = =
area 2 RToday’s lecture is
brought to you by the letter P.
All the radiated power passes through the hemispherical surface* so the average power per unit area (the intensity) is
Area=4R2/2
4
-7 222 5
average
5.00 10 Wpower PI = = = =7.96 10 W m
area 2 R 2 1.00 10 m
*In problems like this you need to ask whether the power is radiated into all space or into just part of space.
R
Station
Satellite
2max
0
E1I = S =
2 c
max 0E = 2 cI
-7 8 -7= 2 4 10 3 10 7.96 10
-2 V=2.45 10 m
-2
-11maxmax 8
V2.45 10E mB = = =8.17 10 Tc 3 10 m s
You could get Bmax from I = c Bmax2/20, but that’s a lot more work
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.
2E 0 max
1u = E
4
2 -12 -2
E
1u = 8.85 10 2.45 10
4
-15E 3
Ju =1.33 10
m
2max
B0
B1u =
4
2
-11
B -7
8.17 101u =
4 4 10
-15B 3
Ju =1.33 10
m
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
Today’s lecture is brought to you by the
letter P.
2
Sdp=
dV c
where dp is the momentum carried in the volume dV.
This equation is not on your equation sheet, but you have
permission to use it for tomorrow’s homework (if
needed) (10.28)
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:
The direction of the momentum change of the object is in the direction of the incident radiation.
incident
Up = (total absorption)
c
When the momentum carried by an electromagnetic wave is absorbed at a surface, pressure is exerted on that surface.
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.
The direction of the momentum change of the object is again in the direction of the incident radiation.
2 Up = (total reflection along incident path)
c
incident
reflected
The radiation pressure on the object is the force per unit area:
From Newton’s 2nd Law (F = dp/dt) we have:
For total absorption,
So
Radiation Pressure
FP=
A
F 1 dpP= =
A A dt
Up=
c
dU1 dp 1 d U 1 SdtP= = = =A dt A dt c c A c
incident
(Equations on this slide involve magnitudes of vector quantities.)
This is the instantaneous radiation pressure in the case of total absorption:
S(t)P(t)=
c
For the average radiation pressure, replace S by <S>=Savg=I:
averagerad
S IP = =
c 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.
rad
IP = (total absorption)
c
rad
2IP = (total reflection)
c
incident
incident
reflected
absorbed
Using the arguments above it can also be shown that:
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 2 32
WPower=IA= 1.4 10 4.0 m =5.6 10 W=5.6 kWm
Assuming total absorption of the radiation:
32
average -6rad 8
W1.4 10S I mP = = = =4.7 10 Pamc c 3 10 s
-6 2 -52rad
NF=P A= 4.7 10 4.0 m =1.9 10 Nm
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.
I know you are mostly engineers, and think applications are important…
So I found you a revolutionarynew application that uses electro- magnetic waves.
Revolutionary Application of 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.”
*2014 price
New starting equations from this lecture:
0
1S= E B
2 2max max
average0 0
E cB1 1S = =
2 c 2
max
max 0 0
E E 1= =c=
B B
22
B E 00
1 1Bu =u = E =
2 2
22 max
0 max0
B1 1u = E =
2 2
U 2 Up = or
c c rad
I 2IP = or
c c
, ,
2
k= =2 f f = =ck
There are even more on your starting equation sheet; they are derived from the above!
