Chapter 35
Electromagnetic Fields and
Waves
Galilean Relativity Why do E and B depend on the observer?
Maxwell’s displacement current (it isn’t a real current)
Electromagnetic Waves
Chapter 35. Electromagnetic
Fields and Waves
Topics:
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•
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•
•
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E or B? It Depends on Your Perspective
The Field Laws Thus Far
The Displacement Current
Maxwell’s Equations
Electromagnetic Waves
Properties of Electromagnetic Waves
Polarization
Preview of what is coming
New!
Gives rise to
Electromagnetic
waves.
Moving observers do not agree on the
values of the Electric and magnetic fields.
Observer S says:
r
r r r
F = q E + v¥ B
Observer S’ says:
r
r
F = qE
How can both be right?
(
)
q makes E and B
v=0 for him, q makes E
Option A: There is a preferred reference frame (for example S).
The laws only apply in the preferred frame. But, which frame?
Option B: No frame is preferred. The Laws apply in all frames.
The electric and magnetic fields have different values for different
observers.
Qu ickTime™ and a
TIFF (Uncompressed) dec ompressor
are nee ded to see this picture.
Extended Option B: No frame is preferred. The Laws apply in all
frames. All observers agree that light travels with speed c. Einstein’s
postulates Special Relativity
Which diagram
shows the fields in
frame S´?
Qu ickTime™ and a
TIFF (Uncompressed) dec ompressor
are nee ded to see this picture.
What we know about fields - so far
Integrals over closed surfaces
Gauss’ Law:
r Qin
—
 E  dA 
0
Gauss’ Law:
r
—
 B  dA  0
Integrals around closed loops
Faraday’s Law:
r r
d
E

d
S


 m through
—
 loop
dt
Ampere’s Law:
r
r
—
 B(r)  ds =0 Ithrough
Integrals around closed loops
Faraday’s Law:
r r
d
E

d
S


 m through
—
 loop
dt
Ampere’s Law:
r
r
—
 B(r)  ds =0 Ithrough
Faraday’s Law for Stationary Loops
r r
EMF  —
 loop E  dS  
 r r
B  dA

t
Area
related by right hand rule
r
dA
r
dS
Which area should I
pick to evaluate the
flux?
A. The minimal area
as shown
B. It doesn’t matter
Faraday’s Law, no problem
r r
d
—
 loop E  dS   dt m through
QuickTime™ and a
decompressor
are needed to see this picture.
 mthrough 

Surface
r
Because for a closed surface —
 B  dA  0
 mthrough 

Surface A
r r
B dA =

Surface B
r r
B dA
r r
B dA
QuickTime™ and a
decompressor
are needed to see this picture.
r
r
B(
r)

d
s =0 I through
—

As long as net current leaving surface is zero, again no problem
QuickTime™ and a
decompressor
are needed to see this picture.
Current through surface B is zero
QuickTime™ and a
decompressor
are needed to see this picture.
James Clerk Maxwell
(1831–1879)
en.wikipedia.org/
wiki/James_Clerk_M
axwell
Thus, for any closed surface
r r
0 —
 E  dA  Qin
d r r
I through   0 —
E  dA  Qin

dt
Maxwell said to add this term
r r
F e = ÚE³dA
S
The new term is called the displacement current.
It is an unfortunate name, because it is not a current.
Maxwell is known for all of the following except:
A. Proposing the displacement current and unifying
electromagnetism and light.
B. Developing the statistical theory of gases.
C. Having a silver hammer with which he would bludgeon
people.
D. Studying color perception and proposing the basis for
color photography.
r r
—B³dS =Bq (r)2pr
Ú
r r
F e = ÚE ³dA = pr 2 Ez
S
r r
dF e
B³d
S
=
m
(I
+
e
)
0
through
0
—
Ú
dt
Bq (r) =
Recall from Faraday:
Eq (r) = -
r ŽBz
2 Žt
m0 e0 r ŽEz
2 Žt
Faraday: time varying B makes an E
r r
d r r
—
 loop E  dS   dt S B  dA
example
Eq (r) = -
r ŽBz
2 Žt
Ampere-Maxwell: time varying E makes a B
r r
d
B
³
d
S
=
m
e
0 0
—
Ú
dt
r r
ÚE ³ dA
S
example
m e r ŽEz
Bq (r) = 0 0
2 Žt
Put together, fields can sustain themselves - Electromagnetic Waves
r r
d
B
³
d
S
=
m
e
0 0
—
Ú
dt
r r
ÚE ³ dA
S
Based on the arrows, the
electric field coming out
of the page is
A. increasing
B. decreasing
C. not changing
D. undetermined
Properties of electromagnetic waves:
Waves propagate through vacuum (no medium is required like
sound waves)
All frequencies have the same propagation speed, c.
Electric and magnetic fields are oriented transverse to the direction
of propagation. (transverse waves)
Waves carry both energy and momentum.
mynasadata.larc.nasa.gov/ ElectroMag.html
35.5 - Electromagnetic Waves
1. Assume that no currents or charges exist nearby, Q=0 I=0
2. Try a solution where:
r r
E(r,t) = Ey (x,t)j
r r
B(r, t) = Bz (x, t)k
Only y-component, depends only on x,t
Only z-component, depends only on x,t
3. Show that this satisfies Maxwell Equations provided the
following is true:
ŽBz (x, t) ŽEy (x, t)
Faraday
=
Žt
Žx
ŽEy (x, t)
ŽBz (x, t)
= m0e0
Ampere-Maxwell
Žx
Žt
4. Show that the solution of these are waves with speed c.
Maxwell’s Equations in Vacuum
Qin =0, Ithrough = 0
Integrals over closed surfaces
Gauss’ Law:
r
—
 E  dA  0
Gauss’ Law:
r
—
 B  dA  0
Integrals around closed loops
Faraday’s Law:
Ampere’s Law:
r r
d
E

d
S


 m through
—
 loop
dt
d
r
r
—
 B(r)  ds =0 0 dt ethrough
Do our proposed solutions satisfy the two Gauss’ Law Maxwell
Equations? r
r
r
E(r,t) = Ey (x,t)j
r
B(r, t) = Bz (x, t)k
r
—
 E  dA  0
Ans: Yes, net flux entering any volume is zero
r
—
 B  dA  0
What about the two loop integral equations?
Faraday’s Law
r r
r r
d
—
 loop E  dS   dt Surface
 B  dA
Apply Faraday’s law to loop
r r
Ey (x)
—
 loop E  dS  h  Ey (x  x)  Ey (x) ; hx x

Surface
r
B  dA  hxBz (x,t)
ŽEy (x, t)
Žx
ŽBz (x, t)
=Žt
Ampere-Maxwell Law:
r r
d
r r
—
 B(r)  ds =00 dt Surface
 E  dA
Apply Ampere-Maxwell law to loop:
Bz (x)
r
r
—
 B(r)  ds = h Bz (x  x)  Bz (x) ; hx x

Surface
r
E  dA ; hxEy (x,t)
ŽEy (x, t)
ŽBz (x, t)
= m0e0
Žx
Žt
Combine to get the Wave Equation:
ŽEy (x, t)
ŽBz (x, t)
= m0e0
#1 Žx
Žt
#2
ŽEy (x, t)
Žx
ŽBz (x, t)
=Žt
Differentiate #1 w.r.t. time
2
Ž
Ey (x,t)
ŽB
(x,t)
Ž
z
= m0e0
Žx Žt
Žt 2
Use #2 to eliminate Bz
Ž2 Ey (x,t)
Žx
2
= m0e0
Ž2 Ey (x,t)
Žt
2
The Wave Equation
Solution of the Wave equation
Ž2 Ey (x,t)
Žx
2
= m0e0
Ž2 Ey (x,t)
Žt 2
Ey (x, t) = f+ (x - vemt) + f- (x + vemt)
Where f+,- are any two functions you like,
and
v = 1/ m e
em
0 0
vem is a property of space.
vem = 2.9979¥ 108 m / s
f+ ,- Represent forward and backward propagating wave
(pulses). They depend on how the waves were launched
Shape of pulse determined by source of wave.
Ey (x, t) = f+ (x - vem t)
Speed of pulse determined by medium
vem = 1 / m0e0
What is the magnetic field of the wave?
Ey (x, t) = f+ (x - vemt) + f- (x + vemt)
Magnetic field can be found from either equation #1 or #2
#2
ŽEy (x, t)
Žx
ŽBz (x, t)
=Žt
This gives:
Bz (x,t) =
1
( f+ (x - vemt)- f- (x + vemt))
vem
Notice minus sign
E and B fields in waves and Right Hand Rule:
f+ solution
f- solution
Wave propagates in E x B direction
Do the pictures below depict possible electromagnetic waves?
A. Yes
B. No
A. Yes
B. No
Special Case Sinusoidal Waves
Ey (x, t) = f+ (x - vem t) = E0 cos[k(x - vem t)]
Ey (x, t)
E0
Wavenumber and
wavelength
k = 2p / l
l = 2p / k
- E0
These two contain
the same
information
Special Case Sinusoidal Waves
Ey (x, t) = f+ (x - vem t) = E0 cos[k(x - vem t)]
Ey (x, t)
2p = kvemT
E0
Introduce
w = 2p / T
f = 1/T
- E0
Different ways of saying the same thing:
w / k = vem
fl = vem
Energy Density and Intensity of EM Waves
Energy density associated with electric and magnetic fields
uE 
For a wave:
r 2
0 E
2
uB 
r 2
B
2 0
r
1 r
B
E   0 0 E
vem
Thus:
uE  u B
Units: J/m3
Energy density in electric and magnetic fields are equal for a wave in
vacuum.
Wave Intensity - Power/area
Energy density inside cube
r 2
r 2
0 E
B
uE 
 uB 
2
2 0
vem
In time t=L/vem an
amount of energy
r 2
U  V(uE  uB )  AL 0 E
comes through the area A.
Volume
V=AL
Area - A
Length - L
I=Power/Area
U
I

tA
0 r 2
E
0
Poynting Vector
The power per unit area flowing in a given direction
1 r r
S
EB
0
0 r 2
S I
E
0
What are the units of
I-
W/m2,
E - V/m
0
0
Ans: Ohms
0
 377
0
Antennas
Simple dipole
AM radio transmitter
Power
radiated in all
horizontal
directions
Earth is a
good
conductor
Image
charges
The amplitude of the oscillating electric field
at your cell phone is 4.0 µV/m when you are
10 km east of the broadcast antenna. What is
the electric field amplitude when you are 20
km east of the antenna?
A. 4.0 µV/m
B. 2.0 µV/m
C. 1.0 µV/m
D. There’s not enough information to tell.
Polarizations
We picked this combination
of fields: Ey - Bz
Could have picked this
combination of fields: Ez - By
These are called plane polarized. Fields lie in plane
Integral versus Differential Versions of Maxwell’s Equations
Integral
Differential
r Qin
—
 E  dA 
r 
E 
r
—
 B  dA  0
r
B  0
0
r r
r r
d
—
 loop E  dS   dt Surface
 B  dA
r r
d
r r
—
 B(r)  ds =0 (Ithrough  0 dt surface
 E  dA)
Charge density
0
r
r
B
E 
t
r
r
r
E
  B  0 ( J   0
)
t
Current density