W13D2:
Maxwell’s Equations and
Electromagnetic Waves
Today’s Reading Course Notes: Sections 13.5-13.7
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Announcements
No Math Review next week
PS 10 due Week 14 Tuesday May 7 at 9 pm in boxes outside 32-082 or
26-152
Next Reading Assignment W13D3 Course Notes: Sections 13.9, 13.11,
13.12
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Outline
Maxwell’s Equations and the Wave Equation
Understanding Traveling Waves
Electromagnetic Waves
Plane Waves
Energy Flow and the Poynting Vector
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Maxwell’s Equations in Vacua
1. òò E × d A =
S
Qin0
e0
2. òò B × d A = 0
(Gauss's Law)
(Magnetic Gauss's Law)
S
dF B
3. ò E × d s = dt
C
0
dF E
4. ò B × d s = m0 I enc + m0e 0
dt
C
(Faraday's Law)
(Ampere - Maxwell Law)
No charges or currents
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Wave Equations: Summary
Electric & magnetic fields travel like waves satisfying:
 Ey
2
x 2
1  Ey
 2
c t 2
 Bz
2
with speed
c
2
x 2
1  Bz
 2
c t 2
2
1
0 0
But there are strict relations between them:
¶E y
¶Bz
=¶t
¶x
¶E y
¶Bz
= - m0 e 0
¶x
¶t
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Understanding Traveling Wave
Solutions to Wave Equation
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Example: Traveling Wave
Consider
(xvt )2 /a2
y(x,t)  y0 e
The variables x and t
appear together as x - vt
(xvt )2 /a2
y(x  vt)  y0e
At t = 0:
(x)2 /a2
y(x  vt)  y0 e
At vt = 2 m:
(x(2 m))2 /a2
y(x  vt)  y0 e
At vt = 4 m:
(x(4 m))2 /a2
y(x  vt)  y0 e
y(x  vt)
is traveling in the positive xdirection
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Direction of Traveling Waves
Consider
(xvt )2 /a2
y(x,t)  y0 e
The variables x and t
appear together as x + vt
(xvt )2 /a2
y(x  vt)  y0e
At t = 0:
(x)2 /a2
y(x  vt)  y0 e
At vt = 2 m:
(x(2 m))2 /a2
y(x  vt)  y0 e
At vt = 4 m:
(x(4 m))2 /a2
y(x  vt)  y0 e
y(x  vt)
is traveling in the negative xdirection
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General Sol. to One-Dim’l Wave Eq.
Consider any function of a single variable, for example
y(u )  y0e
u 2 / a2
u
u
 1 and
 v
Change variables. Let u  x  vt then
x
t
and
y (u )  y ( x, t )  y0e
 x  vt  / a 2
2
Now take partial derivatives using the chain rule
y y u y


 f
x u x u
and
 2 y f f u f  2 y



 2
2
x
x u x u u
and
2 y
f
 f u
2 y
2  f
 v
 v
v
 2
2
t
u t
u u
t
Similarly
 y  y u
y

 v
 vf
t u t
u
Therefore
2 y 1 2 y
 2 2
2
x
v t
y(x,t) satisfies the wave equation!
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Generalization
Take any function of a single variable y (u ) , where
Then y(x  vt) or y(x  vt) (or a linear
combination) is a solution of the one-dimensional
wave equation
u = x ± vt
1 2 y(x,t) 2 y(x,t)

2
2
v
t
x 2
y(x  vt) corresponds to a wave traveling in the
positive x-direction with speed v and
y(x  vt) corresponds to a wave traveling in the
negative x-direction with speed v
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Group Problem: Traveling Sine Wave
Let
y(u)  y0 sin(ku) ,
where
u  x  vt
.
Show that
y(x,t)  y(x  vt)  y0 sin(k(x  vt))
satisfies
1 2 y(x,t) 2 y(x,t) .

2
2
v
t
x 2
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Wavelength and Wave Number:
Spatial Periodicity
Consider
Fix
t  0:
y(x,t)  y0 sin(k(x  vt))
y(x,0)  y0 sin(kx)
 is called the wavelength, k is called the wave number
When x  
 k   2
 k  2 / 
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Concept Question: Wave Number
The graph shows a plot of
the function
y(x,0)  cos(kx)
The value of k is
1. k = 2p / (2 m)
2. k = 2p / (1 m)
3. k = 2p / (0.5 m)
4. k = 2p / (4 m)
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Concept Q. Answer: Wave Number
Answer: 4.
k = 2p / (4 m)
Wavelength is 4 m so wave number is
k = 2p / l = 2p / (4 m)
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Period: Temporal Periodicity
Consider
Fix
x  0:
y(x,t)  y0 sin(k(x  vt))
y(0,t)  y0 sin(kvt)   y0 sin(kvt)
When t  T  kvT  2  2 vT /   2
T   / v
T is called the period
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Do Problem 1
In this Java Applet
http://web.mit.edu/8.02t/www/applets/superposition.htm
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Traveling Sinusoidal Wave: Summary
y(x,t)  y0 sin(k(x  vt))
Two periodicities:
Spatial period : Wavelength  ; Temporal period T.
Wave Number : k = 2p / l
Dispersion Relation : T = l / v
Direction of Propagation : + x - direction
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Traveling Sinusoidal Wave
Alternative form:
y(x,t)  y0 sin(k(x  vt))  y0 sin(kx   t)
Wave Number : k  2 / 
Angular Frequency :   2 / T
Dispersion Relation :   vT    kv
Frequency :
f  1/ T  v   f
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Plane Electromagnetic Waves
http://youtu.be/3IvZF_LXzcc
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Electromagnetic Waves: Plane
Sinusoidal Waves
Watch 2 Ways:
1) Sine wave
traveling to right
(+x)
2) Collection of out
of phase
oscillators (watch
one position)
Don’t confuse vectors with heights – they are magnitudes of
electric field (gold) and magnetic field (blue)
http://youtu.be/3IvZF_LXzcc
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Electromagnetic Spectrum
H
z
Wavelength and frequency are related by:
f c
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Traveling Plane Sinusoidal
Electromagnetic Waves
E = E0 sin(kx - w t) ĵ
B = B0 sin(kx - w t) k̂
are special solutions to the 1-dim wave equations
 Ey
2
x 2
1  Ey
 2
c t 2
2
 Bz
1  Bz
 2 2
2
x
c t
2
2
where
k º 2p / l ,
w º 2p / T,
c=l/T
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Group Problem: 1 Dim’l Sinusoidal EM
Waves
Show that in order for the fields
E = E0 sin(kx - w t) ĵ,
B = B0 sin(kx - w t) k̂
to satisfy either condition below
¶E y
¶Bz
=¶t
¶x
¶Bz
1 ¶E y
=- 2
¶x
c ¶t
then
B0  E0 / c
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Group Problem: Plane Waves
æ 2p
ö
E(x, y, z,t) = E y,0 sin ç (x - ct)÷ ĵ
è l
ø
æ 2p
ö
1
B(x, y, z,t) = E y,0 sin ç (x - ct)÷ k̂
c
è l
ø
1) Plot E, B at each
of the ten points
pictured for t = 0
2) Why is this a
“plane wave?”
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Electromagnetic Radiation:
Plane Waves
Magnetic field vector uniform on infinite plane.
http://youtu.be/3IvZF_LXzcc
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Direction of Propagation
E = E0 sin(kx - w t) ĵ; B = B0 sin(kx - w t)k̂ Þ dir(E ´ B) = î
Special case
generalizes
dir E
î
ĵ
k̂
ĵ
k̂
î
dir B
ĵ
k̂
î
î
ĵ
k̂
dir E ´ B
k̂
î
ĵ
- k̂
- î
- ĵ
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Concept Question: Direction of
Propagation
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
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Concept Question Answer:
Propagation
Answer: 4. The wave is moving in the –z direction
The propagation
direction is given by
the dir E ´ B
(Yellow x Blue)
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Properties of 1 Dim’l EM Waves
1. Travel (through vacuum) with
speed of light
c
1
m
 3.0  10
s
0 0
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2. At every point in the wave and any instant of time,
electric and magnetic fields are in phase with one another,
amplitudes obey
E0 / B0  c
3. Electric and magnetic fields are perpendicular to one another,
and to the direction of propagation (they are transverse):
4. Direction of propagation = Direction of E ´ B.
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Concept Question: Traveling Wave
The B field of a plane EM wave is B( y,t) = B0 sin(ky - w t)k̂
The electric field of this wave is given by
1. E( y,t) = E0 sin(ky - w t) ĵ
2. E( y,t) = E0 sin(ky - w t)(- ĵ)
3. E( y,t) = E0 sin(ky - w t)î
4. E( y,t) = E0 sin(ky - w t)(- î)
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Concept Q. Ans.: Traveling Wave
Answer: 4. E( y,t) = E0 sin(ky - w t)(- î)
From the argument of the sin(ky   t) , we know the
wave propagates in the positive y-direction.
So we have Ê ´ B̂ = ?´ k̂ = ĵ Þ Ê = - î
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Concept Question EM Wave
The electric field of a plane wave is:
E(z,t) = E0 sin(kz + w t) ĵ
The magnetic field of this wave is given by:
1. B(z,t) = B0 sin(kz + w t)î
2. B(z,t) = B0 sin(kz + w t)(- î)
3. B(z,t) = B0 sin(kz + w t)k̂
4. B(z,t) = B0 sin(kz + w t)(-k̂)
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Concept Q. Ans.: EM Wave
Answer: 1. B(z,t) = B0 sin(kz + w t)î
From the argument of the sin(kz   t) , we know
the wave propagates in the negative z-direction.
So we have Ê ´ B̂ = ĵ ´ ? = -k̂ Þ B̂ = î
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Energy in EM Waves:
The Poynting Vector
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Energy in EM Waves
Energy densities:
Consider cylinder:
1
1
2
2
uE   0 E , uB 
B
2
2 0
2

1
B
2
dU  (uE  uB ) Adz    0 E   Acdt
2
0 
What is rate of energy flow per unit area?
2
 c
B
EB 
1 dU c 
2
  0 E 
S
   0 cEB 


2

A dt
2
c0 

0 
EB
EB
2

 0 0 c  1 
0
2 0


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Poynting Vector and Intensity
Direction of energy flow = direction of wave propagation
S=
E´B
m0
: Poynting vector
units: Joules per square meter per sec
Intensity I:
I S 
E0 B0
2 0

2
0
E
2 0 c

2
0
cB
2 0
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Group Problem: Poynting Vector
An electric field of a plane wave is given by the
expression
E( y,t) = E0 sin(ky + w t)k̂
Find the Poynting vector associated with this plane
wave.
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Appendix A
Standing Waves
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Standing Waves
What happens if two waves headed in opposite
directions are allowed to interfere?
E1 = E0 sin(kx - w t)
E2  E0 sin(kx   t)
Superposition : E  E1  E2  2E0 sin(kx)cos( t)
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Standing Waves
E1 = E0 sin(kx - w t)
E2  E0 sin(kx   t)
Superposition :
E  E1  E2
E  2E0 sin(kx)cos( t)
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Standing Waves
Most commonly seen in resonating
systems:
Musical Instruments, Microwave Ovens
E  2E0 sin(kx)cos( t)
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Standing Waves
Do Problem 2 In the Java Applet
http://web.mit.edu/8.02t/www/applets/superposition.htm
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Appendix B
Radiation Pressure
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Momentum & Radiation Pressure
EM waves transport energy: S =
They also transport momentum:
E´B
m0
U
p
c
F 1 dp 1 dU S
And exert a pressure: P 



A A dt cA dt
c
This is only for hitting an absorbing surface. For
hitting a perfectly reflecting surface the values are
doubled, as follows:
2U
2S
Momentum transfer: p 
; Radiation pressure: P 
c
c
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Problem: Catchin’ Rays
As you lie on a beach in the bright midday sun,
approximately what force does the light exert on you?
The sun:
Total power output ~ 4 x 1026 Watts Distance from
Earth 1 AU ~ 150 x 106 km
Speed of light c = 3 x 108 m/s
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