Workshop: Using Visualization in Teaching Introductory E&M
AAPT National Summer Meeting, Edmonton, Alberta, Canada.
Organizers: John Belcher, Peter Dourmashkin, Carolann Koleci, Sahana Murthy
Electromagnetic Waves Supplementary Materials
Visualizations
You can find more visualizations at Light Visualizations.
ConcepTests
The following ConcepTests illustrate various aspects of traveling electromagnetic waves.
They can be used both in class and on exams.
Wavelength and Wave Number The graph shows a plot of the function y  cos(kx) .
(m)
The value of k is
1. 1/ 2 m -1
2. 1/ 4 m -1
3.  m -1
4.  / 2 m-1
5. I don’t know
Answer: 4. k   / 2 m-1 . The wavelength (from the figure) is   4 m l . Therefore
k  2 /    / 2 m-1 .So the function described by y  cos(( / 2 m-1 ) x) is equal to y  1 at
x  ..., 4 m, 0, 4 m, ...
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Propagating Plane E&M Wave The
figure shows the E (yellow) and B (blue)
fields of a plane wave. This wave is
propagating in the
1. +x direction
2. –x direction
3. +z direction
4. –z direction
5. I don’t know
Answer: 4. The wave is moving in the –z direction. The propagation direction is given by the
direction of E x B (Yellow x Blue)
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Traveling Plane Electromagnetic Wave The B field of a plane EM wave is
B( z, t )  kˆB0 sin(ky  t ) . The electric field of this wave is given by
1. E( z, t )  ˆjE0 sin(ky  t )
2. E( z, t )  ˆjE sin(ky  t )
0
3. E( z, t )  ˆiE0 sin(ky  t )
4. E( z, t )  ˆiE0 sin(ky  t )
5. I don’t know.
Answer: 4. E( z, t )  ˆiE0 sin(ky  t ) . From the argument of the sin( ky  t ) , we know the
ˆ B
ˆ  ?  kˆ  ˆj  E
ˆ  ˆi .
wave propagates in the +y direction. So we have E
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Plane Waves: Consider a plane wave whose magnetic field is given by
B  B0kˆ cos(kx  t ) .
Which is the corresponding electric field?
1. E  cB0kˆ cos(kx  t ) .
2. E  cB0kˆ cos(kx  t ) .
3. E  cB0ˆj cos(kx  t ) .
4. E  cB0ˆj cos(kx  t ) .
5. E  cB0ˆi cos(kx  t ) .
6. E  cB0ˆi cos(kx  t ) .
Answer: 4.
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Plane waves:
The magnetic vector of a plane electromagnetic wave is described as follows:
B = 1  T  cos  3 m-1 x  6·108 s -1 t  ˆj
Fill in the blanks:

 

This wave is traveling in the ___-x_______ direction with speed _____ 2·108 m s ____
The associated E field will have amplitude ____ 2·102 V m _____ along the ___z_____ axis
The way to do this is to pattern match to B = B0 cos  kx  t  ˆj . The fact that it is kx tells you
it is moving along the x-axis and the + tells you it is moving in the negative direction (as t
increases, x decreases to keep the argument constant). The speed is given by v   k . The Efield has magnitude E0 = B0  v and points along an axis such that the direction of propagation
ˆ B
ˆ  kˆ  ˆj  ˆi .
is given by E

Page 5

2/6/2016
Generating Plane Waves
A plane E&M wave is generated by shaking a sheet of positive charge up and down. At the
time shown
1. The E field is oscillating into and out of the page and the sheet is moving up
2. The E field is oscillating into and out of the page and the sheet is moving down
3. The B field is oscillating into and out of the page and the sheet is moving up
4. The B field is oscillating into and out of the page and the sheet is moving down
5. I don’t know.
Answer: 4.
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Presentation Material Problems:
Deriving the Wave Equation
In this question you are going to derive the wave equation – that is, prove that electromagnetic
radiation as you have studied it in class is a natural outcome of Maxwell’s equations.
Consider a wave traveling along the x-axis, where the magnetic field is polarized along the zaxis and the electric field along the y-axis.
(a) Use the Ampere-Maxwell law (with the
rectangular loop pictured at right) to calculate a
relationship between the spatial derivative of the
magnetic field and the time derivative of the electric field.
d
Ampere's Law:  B  d s  0 0  E  dA
dt
C
LHS :
 B  d s  B ( x, t )l  B ( x  dx, t )l
z
z
C
E y
d

E  dA   E y  l dx   l dx

dt
t
t
Equating the left and right hand sides and dividing through by l dx :
RHS :

E y
Bz ( x  dx, t )  Bz ( x, t )
B
  z   0 0
dx
x
t
b) Use Faraday’s Law (with the rectangular loop
pictured at left) to calculate a relationship between
the spatial derivative of the electric field and the
time derivative of the magnetic field.
Faraday's Law:
d
 E  ds   dt  B  dA
C
LHS :
 E  d s  E ( x  dx, t )l  E ( x, t )l
y
y
C
B
d
B  dA  ldx z

dt
t
Equating the left and right hand sides and dividing through by l dx : (c) Now you have two
partial differential
E y ( x  dx, t )  E y ( x, t ) E y
Bz
equations, relating


spatial derivatives and
dx
x
t
time derivatives of E
RHS :
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and B. Unfortunately they are coupled (that is, E depends on B and B depends on E). To
solve these types of equations we need to decouple them – to get an equation that only has E
in is, and another that only has B in it.
To do this, take an x-derivative of one of your equations and a time derivative of the other.
Since the derivatives commute (that is, taking the space derivative followed by the time
derivative is the same as taking the time derivative followed by the space derivative) you will
find that you can substitute one equation into the other and get an equation that only has E in
it or only has B in it. If you change which equation you do which derivative to you will get
the other.
These two new equations are Wave Equations! They are second order differential equations
(two spatial derivatives on one side, two time derivatives on the other).
Start with the two equations that we derived in (a) and (b):
E y
E y
B
B
 z   0 0
 z
x
t
x
t
2
2 Ey
  E y   E y   Bz    Bz 










0 0




x  x  x 2
x  t  t  x 
t 2
2 Ey
2 Ey
  0 0
x 2
t 2
If we instead take the spatial derivative of the first equation we get our wave eqn for B:
  Bz   2 Bz   E y 
  E y 
1  2 Bz
 
 



t  t  t 2
t  x 
x  t  0 0 x 2
 2 Bz
 2 Bz



0 0
x 2
t 2
Note that these two equations are essentially “the same” – they only differ in which
components of the electric/magnetic fields they show propagating.
(d) Finally, let’s prove that the pictured wave is actually a solution to the equations you just
derived. We’ll just do the electric field. Write an equation (looking at the picture) for what E
is as a function of time and position. Remember that it is a traveling wave (you should be
able to tell from the picture in what direction it is traveling). You don’t know the amplitude,
period or wavelength of the wave, just use variables for these, but make sure that your wave
equation is satisfied (it should force a well-known relationship between your period and
wavelength).
Looking at the electric field, it is a sinusoidal traveling wave polarized along the y-axis and
traveling along the +x direction (we know it is +x because the cross product of E and B give
you the direction of travel):
2 
 2
E  ˆjf  x  vt   ˆjE0 sin  kx  t   ˆjE0 sin 
x
t
T 
 
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You could use any of these forms (or cosines). The first (with f(x-vt)) is completely generic
while the next two assume a sinusoidal variation. Since no indication of the time of the
picture was given, the phase (whether it is sin, cos or something in between) is undetermined.
Since the question mentioned wavelength and period I’ll use the last form, substituting it into
the wave equation for E we derived in (c):
2 
 2
E  ˆjf  x  vt   ˆjE0 sin  kx  t   ˆjE0 sin 
x
t
T 
 
LHS :
2 Ey
x 2

2
2
 2
E0 sin 
x
2
x
T
 
2

 2 
 2
t   
x
 E0 sin 
T

  
 
2 Ey
2

 2 
t   
 Ey

  
2
 2 
Similarly, RHS : 0 0 2   0 0 
 Ey
t
 T 
For the equation to be satisfied then we have:
2

 2 
 2 

 E y    0 0 
 Ey  
T
  
 T 
2
2
1
 0 0
c
The ratio of wavelength to period is the speed of the wave which is the speed of light.
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Electromagnetic Plane Wave
An electromagnetic plane wave is propagating in vacuum has a magnetic field given by
B  B0 f (ax  bt )ˆj
1
f (u )  
0
0  u 1
otherwise
where a and b are positive quantities. The kˆ direction is out of the paper.
(a) What condition between a and b must be met in order for this wave to satisfy Maxwell’s
equations?
Differentiating By with respect to x gives
B y
x
 2 By
x 2
 B0
 f (u )
(ax  bt )  f (u )
 aB0
x
u
u
 2 f (u )
(ax  bt )  2 f (u )
2

a
B
0
x
u 2
u 2
 aB0
Similarly, differentiating By with respect to t yields
B y
t
 By
 B0
2
t 2
 bB0
 f (u )
 (ax  bt )  f (u )
 bB0
t
u
u
 2 f (u )
(ax  bt )  2 f (u )
2

b
B
0
t
u 2
u 2
Thus,
 2 By
x 2
2
 2 f (u )
1  By
b2
2
 2
 (a  2 ) B0
0
c t 2
c
u 2
or
b
c
a
(b) What is the magnitude and direction of the E field of this wave?
E0
 c  E0  cB0
B0
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
The electromagnetic wave propagates in the x-direction with the B field pointing in the ydirection. Since the direction of propagation is along the direction of the cross product E  B ,

the electric field E must point in the +z direction (so that (kˆ )  ( ˆj)  ˆi ).
(c) What is the magnitude and direction of the energy flux (power per unit area) carried by
the incoming wave, in terms of B0 and universal quantities?
 1  


Since S 
E  B with E and B in phase, the magnitude of the Poynting vector is
0
S0 
E 0 B0
0

cB0
2
0
S points in the direction of propagation (x-direction).
(d) This wave hits a perfectly conducting sheet and is reflected. What is the pressure (force
per unit area) that this wave exerts on the sheet while it is impinging on it?
With total reflection, the pressure exerted by the wave on a perfectly conducting sheet is
P
Page 11
2
2 I 2S 0 2  cB0

 
c
c
c  0
 2 B0 2


0

2/6/2016
Wave Equation in Differential Form
Consider a plane electromagnetic wave with the electric and magnetic fields given by
E( x, t )  Ez ( x, t )kˆ , B( x, t )  By ( x, t )ˆj
Applying arguments similar to that presented in Section13.4 of the Course Notes, show that
the fields satisfy the following relationships:
Ez By

,
x
t
By
 0 0
x
Ez
t
Consider a rectangular loop in the xz plane depicted in the figure below, with a unit normal
nˆ  ˆj .
y
By(x)
Ez(x)
z
x
x
z
Ez(x+x)
Using Faraday’s law
d
 E  d s   dt  B  dA
the left-hand-side can be written as
 E  d s  E ( x)z  E ( x  x)z    E ( x  x)  E ( x) z  
z
z
z
z
Ez
xz
x
where we have made the expansion
E z ( x  x)  E z ( x) 
E z
x  ...
x
On the other hand, the rate of change of magnetic flux on the right-hand-side is given by

Page 12
 By
d
B  dA   

dt
 t

 xz

2/6/2016
Equating the two expressions and dividing through by the area xz yields
E z B y

x
t
The second condition on the relationship between the electric and magnetic fields may be
deduced by using the Ampere-Maxwell equation:
 Bd s   
0 0
d
E  dA
dt 
Consider a rectangular loop in the xy plane depicted in the figure below, with a unit normal
nˆ  kˆ .
y
By(x)
x
By(x+x)
y
Ez(x)
z
x
The line integral of the magnetic field is
 By
x
 E  d s  B ( x  x)y  B ( x)y   B ( x  x)  B ( x) y  
y
y
y
y

 xy

On the other hand, the time derivative of the electric flux is
 0 0
d
 E
E  dA  0 0  z

dt
 t

 xy

Equating the two equations and dividing by xy , we have
B y
x
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  0 0
E z
t
2/6/2016
Traditional Analytic Problems:
Electromagnetic Waves The magnetic vector of a plane electromagnetic wave is described
as follows:
B = 100 mT  sin  3 m-1  x   9·108 s -1  t  ˆj
(a) What is the wavelength  of the wave?

2
2
2


m
-1
k
3
3m


(b) What is the velocity v of the wave?
v

k
9·10 s   3 10
3 m 
8

-1
8
-1
m
s
(c) Write an expression for the time & position dependent electric field E of the wave.
E = -  3 107 V m  sin 3 m-1  x  9·108 s -1  t  kˆ
(d) Show that for any point in this wave, the density of the energy in the electric field equals
the density of the energy in the magnetic field.
It is easy to show this generically, recalling the relationship between the magnitude of the
electric field and the magnetic field, and the value of the speed of light:
uE  12  0 E 2 = 12  0 E02 sin 2  kx  t   12  0  cB0  sin 2  kx  t 
2
uB  12 01 B 2 =
1
2
01 B02 sin 2  kx  t   12 01 01 0 B02 sin 2  kx  t   12  0c 2 B02 sin 2  kx  t 
They are equal!
(e) What is the time-averaged total (electric plus magnetic) energy density u in this wave?
u  2uE   0 E02 sin 2  kx  t  
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u  12  0 E02
2/6/2016
(f) This wave is totally reflected by a thin conducting sheet lying in the y-z plane. What is the
resulting time-averaged radiation pressure on the sheet?
 0.1 T 
2S E0 B0 B02
N




 104 2
7
-1
c
0c
0  410 T m A 
m
2
Preflect
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Standing Waves The electric field of an electromagnetic wave is given by the superposition
of two waves
E  E0 cos(kz  t ) ˆi  E0 cos(kz  t ) ˆi .
(1.1)
You may find the following identities useful
cos(kz  t )  cos(kz ) cos(t )  sin(kz ) sin(t )
(1.2)
sin(kz  t )  sin(kz ) cos(t )  cos(kz ) sin(t ) .
(1.3)
a) What is the associated magnetic field B( x, y, z , t ) ?
Solution: The electric field in Eq. (1.1) is the superposition of two traveling waves, one in the
positive z -direction and the other in the negative z -direction. The amplitude of the
associated magnetic field is diminished by a factor of 1/ c . By the right hand rule, the
magnetic field associated with the wave traveling in the positive z -direction points in the  ˆj
(when the cos factor is positive), and the magnetic field associated with the wave traveling in
the negative z -direction points in the  ˆj (when the cos factor is positive). Hence the
magnetic field is given by
B
E0
(cos(kz  t )  cos(kz  t )) ˆj
c
(1.4)
Using the identity in Eq. (1.2), Eq. (1.4) can be rewritten as
B
E0
(cos(kz  t )  cos(kz  t )) ˆj
c
E0
[cos(kz ) cos(t )  sin(kz) sin(t )]  [cos(kz) cos(t )  sin( kz) sin(t )] (ˆj)
c
2E
 0 sin(kz ) sin(t )ˆj
c

(1.5)
In a similar fashion, using Eq. (1.2), the electric field in Eq. (1.1) can be rewritten as
E  E0 (cos(kz  t )  cos(kz  t )) ˆi
 E0 [cos(kz ) cos(t )  sin(kz ) sin(t )]  [cos(kz ) cos(t )  sin( kz ) sin(t )] (ˆi )
(1.6)
 2 E0 cos(kz ) cos(t )ˆi
Equivalently, Faraday’s Law in differential form for the plane waves traveling in the  z direction is given by
Page 16
2/6/2016
B
Ex
 y
z
t
(1.7)
Eq. (1.7) can be integrated with respect to time to find the electric field
By   
Ex
dt .
z
(1.8)
The partial derivative in the integrand can be calculated using Eq. (1.6),
Ex
 2kE0 sin(kz ) cos(t ) .
z
(1.9)
Substituting Eq. (1.9)into the integrand in Eq. (1.8)and integrated yields
By  2kE0 sin(kz )  cos(t )dt 
2kE0

sin(kz ) sin(t ) .
(1.10)
Finally,
1 k
 .
c 
(1.11)
Substituting Eq. (1.11)into Eq. (1.10)yields
By 
2 E0
sin(kz ) sin(t ) ,
c
(1.12)
in agreement with Eq. (1.5) for the y -component of the magnetic field.
b) What is the energy per unit area per unit time (the Poynting vector S ) transported by
this wave?
Solution: The Poynting vector is given by
S
1
0
EB .
(1.13)
Substituting Eq. (1.5) and Eq. (1.6) into Eq. (1.13) yields
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S
2E
2 E0 cos(kz ) cos(t )ˆi  0 sin(kz )sin(t )ˆj
0
c
1
Eo2

(2 cos(kz )sin(kz ))(2 cos(t )sin(t )) kˆ
c 0
.
(1.14)
Recall the identity
2 cos(a) sin(a))  sin(2a) .
(1.15)
Use Eq. (1.15) twice in Eq. (1.14) yields
Eo2
S
sin(2kz )sin(2t ) kˆ .
c 0
(1.16)
c) What is the time average of the Poynting S vector? Explain your answer, (note: you
may be surprised by your answer, but try to explain it). Recall that the time average of
the Poynting vector is given by
T
S 
1
Sdt .
T 0
(1.17)
Solution: Substitute Eq. (1.16) into Eq. (1.17) yielding
T  2 / 
E2
S  o sin(2kz )
c0T

sin(2t )dt kˆ .
(1.18)
0
Integrate Eq. (1.18) finding
S 
Eo2
T  2 / 
sin(2kz ) cos(2t ) 0
kˆ
c0 2T
Eo2

sin(2kz )(cos(4 )  cos(0)) kˆ .
c0 2T
(1.19)
0
Since the wave described by Eq. (1.5) and Eq. (1.6) is a plane standing wave, the wave is not
propagating, therefore there is no time averaged energy transport.
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Concept Rich Problems
Solar Observatory. Your project goal is to design a solar observatory that does not have to
orbit but rather can just sit still relative to the sun. You decide that you want the observatory
to be a solar sail ship such that the the gravitation force on the ship will balance a force that
arises from radiation pressure on a perfectly reflective sail of area Asail . You need to estimate
min
the mass mship of the observatory to determine the minimum area Asail
for the sail in order to
exactly balance the attractive gravitational force from the sun.
Some possibly useful numbers: G  6.67 1011 N m2 kg -2 , the speed of light is
c  3 108 m s . Sail materials have to be very lightweight. Current materials have surface
mass densities of about 1 g/m2, but proposed materials are projected to have densities as low
as 0.05 g/m2.The sun has mass msun  2 1030 kg and radius rsun  7 108 m . The time
averaged radiative power of the sun is Psun  3.9 1026 W and has a rotation period of about
30 days.
Answer: We shall divide this problem into several questions that will lead us to finding the
minimum area for the sail.
a) How is the radiation pressure related to the time averaged Poynting vector
T
1
S   S dt associated with the radiation from the sun?
T 0
b) How is the radiation pressure related to the radiative force acting on the sail if we
assume it is perfectly reflecting?
c) What is the minimum area Asail for the sail in order to exactly balance the gravitational
attraction from the sun?
d) We will then estimate the mass of the ship and determine the area of the sail.
.
The time averaged Poynting vector at the outermost part of Earth’s atmosphere is equal to the
time averaged power of the Sun Psun divided by the area of a spherical shell with radius r
equal to the distance from the Sun.
Psun
 S .
4 r 2
The radiation pressure for a perfectly reflecting sail is given by
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rad
Preflecting
 2 S /c 
Psun
2 r 2c
rad
Since pressure is equal to force Freflecting
on the sail divided by the area of the sail Asail , the
force on the sail due to solar radiation is given by
rad
rad
. Freflecting
 Preflecting
Asail 
Psun Asail
2 r 2c
The gravitation force on the ship is
Fgravity 
Gmsun mship
r2
Assume that the radiation force balances the gravitational force,
rad
Freflecting
 Fgravity .
then
min
Gmsun mship
Psun Asail

.
2
2 r c
r2
So we can solve for the minimum sail area
min
Asail

2 cGmsun mship
Psun
.
Note that the ratio
mship
min
sail
A

Psun
3.9 1026 Watts

2 Gmsun c 2  6.67 1011 N m2 kg -2  2 1030 kg
3 10
8
1.5
m s
g
m2
Since this is more dense than the sail material itself, this can work. Let’s break the mass of
the observatory into the part the equipment is in (which I’ll estimate has a mass of about 100
tons) and the sail itself, for which I’ll use a pretty realistic mass density of 0.5 g/m2.
mship
min
sail
A

Page 20
mequipment
min
sail
A


105 kg
mequipment
mequipment
msail
g
g
min
 min  1.5 2 
 1 2  Asail 
 3
 108 m 2
min
2
2
Asail
m
Asail
m
1g m
10 kg m
2/6/2016
This corresponds to a sail that is 104 m (10 km) on a side. This is large, but not totally
unreasonable. And certainly if I made the payload lighter and used a lighter sail material this
number would come down (1 metric ton and 0.05 g/m2 leads to a sail less than a km on a
side.)
Extra Note: We can easily get power from solar power. We might need 100 kW, but even as
far away as the Earth we have 1.4 kW/m2, meaning our sail will catch order 1011 Watts! Even
if we only absorb a small fraction of this we will still have plenty!
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Solve the Energy Crisis Energy is probably the biggest issue facing the world right now (if
you are looking for soething important to do, think about this – the very existence of human
life may hinge on our solving this problem). In a nutshell, we currently use fossil fuel to
generate energy. This creates greenhouse gasses, which lead to global warming.
a) What is the average power the average American uses (averaged over the year)? Include
all energy use, not just electricity. What does that work out to for all of America?
There are several ways to think about this, here is one. I figure that personally I use 1-2 kW
of electrical energy continuously (this is the equivalent of 10-20 100 Watt light bulbs, which
when you add up the computer, lights, dishwasher, etc. seems reasonable). I also figure that I
spend about the same amount per month in 4 other categories: gas (rather, my wife does since
I take the train, but this is an average), heating my house, food and other expenses (clothing,
etc). Clearly these expenses aren’t totally energy expenses and cost does not equal amount,
but different types of energy have very similar costs and so I figure I must use around 5-10
kW of energy in total.
According to recent DOE numbers the average per capita energy use is 11 kW, about what I
estimated. Since the US population is about 300 million, this corresponds to energy usage of
about 3 TW (terawatts).
b) Let’s say that we want to get that all from nuclear power. Assuming that an average
nuclear power plant generates 1 GW of power, how many do we need to power the US?
We would need about 3000 nuclear power plants to supply all of our energy. There are
currently about 440 plants in the world, about 100 in the united states. So aside from safety
and environmental issues, this would be a huge task to build up this many nuclear power
plants.
c) If we were to get it from solar energy, how large an area would we need to cover with
solar cells? The solar power at the earth is about 1.4 kW/m2.
The most efficient solar cells today convert about 25% of incoming power to usable energy.
Let’s be optimistic and say that doubles to 50% (this might actually be realistic), but then
don’t forget that we only have sunlight about half a day. So 4x3 TW/1.4 kW/m2 means we
need 1010 m2 which is about two million acres or about 4000 square miles or about three times
the area of Rhode Islands (I think Rhode Island prides itself on being a useful unit of
measure). This might actually be doable – there are vast tracts of open land in the Midwest
that could probably be used pretty effectively.
d) What happens to these numbers if we also want power for China and India to use energy at
the same per capita rate?
This is where things get dicey. The current population of China is 1.3 billion and of India is
1.1 billion, meaning that we will need to multiply the above numbers by 9
((1.3+1.1+0.3)/0.3).
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Of course, the current energy usage of these countries is currently much less than the US, with
per capita usage about a tenth of ours in China and less than a twentieth in India. If they
manage to rise to the US energy “standard of living,” which China has publicly stated as a
goal, then we are in for some serious problems.
e) What will we need 30 years from now assuming current population growth rates?
The US has the highest fertility rate of the industrial world at about 2.1 births per woman. We
have moderate immigration numbers of about a million a year. So the projected population in
2035 is about 400 million. The projected population of both India and China will be about 1.5
billion by that time. So this is an increase of another 25%. Compared to the numbers in (d)
though, the issue isn’t so much population growth as per capita energy usage.
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Amazing Communication Cell phones are pretty cool. So is the deep space network used to
communicate with the voyager spacecraft
(http://voyager.jpl.nasa.gov/news/profiles_dsn.html), which are currently about 80 AU away
(1 AU = 150 million km). Let’s think about both.
a) About how much power does a cell phone use? Think about how often you need to
charge your cell phone and how much energy could realistically be stored in it (see for
example your work on problem 2)
Typical new cell phones broadcast at about ¼ Watt. Older mobile phones (as well as boosters
for cell phones that some people get for their cars) broadcast at 3 W.
b) Assuming that most of the power from the cellphone is used in signal transmission (which
is becoming a progressively worse assumption, but use it anyway), and knowing the
average size of a cell phone cell (26 km2), what kind of signal strength (power per unit
area) is needed at the receiver in order to still “have signal?”
We’ll pretend that the cell is circular (in reality they try to make them hexagonal). That
means that you are never further than about r  A   3 km from the base station. The
signal strength at the base station could then be as weak as
2
S  P A   0.25 W  4  3 km   2.2 nW/m 2
c) Let’s compare that with a radio, to see if its in the same ballpark. FCC regulations prevent
broadcasts at powers above 100 kW. How far away from a radio transmitter can you still
hear the station? How much power density are you then receiving?
You might be able to hear a station about 200 km away. Beyond 100 miles the curvature of
the Earth begins to kill you (FM radio waves go in a straight line and will just go off into
space,as opposed to AM and CB radio waves that can bounce off the ionosphere and explain
why you can occasionally get good signals from around the globe using those technologies).
2
S  P A  100 kW  4  200 km   200 nW/m 2
So this is a little bigger than what we calculated in (b) but not outrageously out of line,
especially considering that line of sight rather than power limitations are probably largely
responsible for the loss of signal.
d) What kind of electric fields do these power densities correspond to?
From the Poynting vector:
1 mV/m for cell phones
E02
1
S
E0 B0 
 E0  2o cS  
2 o
2 o c
12 mV/m for FM radio
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e) The voyager spacecraft have 20 Watt transmitters (3 meter dishes broadcasting at 2.3
GHz). The dishes are aimed at the Earth. How wide an angular dispersion could they
have such that there is still enough power at the Earth to receive the signal?
It’s pretty amazing that you can use a 20 Watt transmitter from so far and still get a signal, but
they can aim really well. What is the possible angular spread? We don’t want a power per
unit area below about 1 nW/m2, meaning the 20 W can spread out over 2 x 1010 m2 (a circle of
radius about 80 km), meaning an angular dispersion of:
80 km
1
tan    

 7  109 rad  0.4μdeg
6
80 AU 150 10
Pretty darn impressive, especially the part about aiming to that accuracy. I would guess that
in reality they don’t do that well and instead have a much better receiving antenna, meaning
that the signal could be a few orders of magnitude weaker, but I can’t find any info about that
to confirm it. I do know, however, that the antenna for sending signals to spacecraft is big (70
m, 20 kW). Kind of sad that they don’t have the power of a good radio station, but the
antenna is MUCH more directional.
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