Electromagnetism
INEL 4152
Ch 10
Sandra Cruz-Pol, Ph. D.
ECE UPRM
Mayagüez, PR
Outline
 Plane

Wave types
general, lossy, lossless….
 Power
in a Wave
 Applications and Concepts
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10.1
Plane waves
A wave!

Start taking the curl of Faraday’s law
    Es   j  H s

Then apply the vectorial identity
    A  (  A)   2 A

And you’re left with
Ñ(Ñ × Es ) - Ñ Es = - ( jwm ) (s + jwe )Es
2
= -g 2 Es
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A Wave
 E  E  0
2
2
Let’s look at a special case for simplicity
without loosing generality:
•The electric field has only an x-component
•The field travels in z direction
Then we have
E ( z, t )
whose general solution is
E(z)  Eo e z  Eo' e z
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To change back to time domain
g 2 = ( jwm ) (s + jwe )

From phasor
g=
E xs ( z )  Eo e

 z
( jwm ) (s + jwe ) = a + jb
 Eo e
 z (  j )
…to time domain
E ( z , t )  Eo e
 z

cos(t  z ) x
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10.3-10.6
Plane wave Propagation in
Several Media
Several Cases of Media
1.
2.
3.
4.
Free space (s = 0, e = eo , m = mo )
(s = 0, e = ereo, m = mrmo or s << we )
Lossless dielectric
Lossy dielectric (general) (s ¹ 0, e = ereo, m = mr mo )
Good Conductor
(s @ ¥, e = e , m = m m or s >> we )
o
r
o
Recall: Permittivity
eo=8.854 x 10-12 [F/m]
Permeability
o= 4p x 10-7 [H/m]
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1. Free space
There are no losses, e.g.
E(z, t) = Acos(wt - b z)x
Let’s define
 The phase of the wave
 The angular frequency
 Phase constant
 The phase velocity of the wave
 The period and wavelength
 In what direction does it moves?
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Where does it moves?
E(z, t) = Acos(wt - b z)x
Pick 3 times:
 Plot
t = 0 (w t = 0)
= cos(-b z) = cos b z
t = T / 4 (w t = 90 ) = cos(90 - b z) = sin b z
o = cos(180 - b z) = -cos b z
t = T / 2 (w t = 180 )
o
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3. General Case
(Lossy Dielectrics)

In general, we had

E ( z , t )  Eo e z cos(t  z ) x
g 2 = ( jwm ) (s + jwe ) and     j
 Re  2   2   2   2 e
2 

 2   2    2   2e 2
From these we obtain
2


e
 

 1 
 
  1
2 

 e 



 
e 

 

 1 
  1
2 
e 




2
So , for a known material and frequency, we can find j
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Summary
Any medium


h
up
l
 re r 
c
2

 

 1 
  1
2 
e 





e 
2

 

 1 
  1
2 
e 




Lossless
medium
(=0)

0
w me
= c / m re r

e
/
up
=
b
f
**In free space;

e
2
j
  je
2p
Low-loss medium
(e”/e’<1/100)
 e

e
1
1
e
e
up
up
f
f
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eo =8.85
10-12 F/m
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Good conductor
(e”/e’>100)
Units
p f ms
[Np/m]
p f ms
[rad/m]

(1  j )

4p f
ms
up
f
o=4p x 10-7 H/m
[W]
[m/s]
[m]
Example: Water
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Permittivity of water vs. freq.
http://www.random-science-tools.com/electronics/water_dielectric.htm
er = 80
only here!
Example Water @ 60Hz, (Low freq.)
Find  ,  and h
 Attenuation
a=
const.
é
2
ù
æs ö
1+ ç ÷ -1ú
è we ø
úû
2 êë
w mrer ê
c
 Phase
constant
2
é
ù
æ
ö
w mrer ê
s
b=
1+ ç ÷ +1ú
è we ø
úû
c
2 êë
 For distilled water
s
mr = 1
= 370
we
er = 81
a
= 1.38 ´10 -3
s = 0.0001
b = 1.38 ´10 -3
p f ms
h=
 For
mr = 1
e r = 80
s =4
sea water
s
= 14, 986, 341
we
a = 0.03
b = 0.03
h=
Example Water @ 10GHz, (microwaves)
Find  ,  and h
 For
 Attenuation
a=
const.
é
2
ù
æs ö
1+ ç ÷ -1ú
è we ø
úû
2 êë
w mrer ê
c
mr = 1
er = 65
s = 0.0001
p f ms
 Phase
constant
2
é
ù
æ
ö
w mrer ê
s
b=
1+ ç ÷ +1ú
è we ø
úû
c
2 êë
distilled water
 For
mr = 1
e r = 62
s =4
s
= 7.4 ´10 -6
we
 
a = 1.8 ´10-3 2 e
b = 565  e
h=
sea water
s
= 0.3
we
a = 83
b = 568
h=
Intrinsic Impedance, h

If we divide E by H, we get units of ohms and
the definition of the intrinsic impedance of a
medium at a given frequency.
given E = Eoe-g z x
Find
|E|
j
h

H (use Maxwell)
|H |
  je
E(z, t) = Eoe-a z cos(w t - b z)x
Eo - a z
H (z, t) = e cos(w t - b z - qh ) ŷ
h
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 h h
[]
*Not in-phase
for a lossy
medium
Applet
See Applet by Daniel Roth at
http://fipsgold.physik.uni-kl.de/software/java/oszillator/index.html
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Note…
in time domain :
E(z, t) = Eoe cos(w t - b z)x
E
H (z, t) = o e-a z cos(w t - b z - qh ) ŷ
-a z
h
phasor :
E(z) = Eo e-a z e- j ( b z ) x
Eo -a z - j ( b z-qh )
H (z) =
e e
ŷ

h
E and H are perpendicular to one another
 Travel is perpendicular to the direction of
propagation
 The amplitude is related by the impedance
 And so is the phase
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Loss Tangent

If we divide the conduction current by the
displacement current
J cs
J ds
 Es



 tan   loss tangent
je Es e
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Relation between tan and ec
 

  H  E  je E  je 1  j
E

e 

 je c E
The complex permittivi ty is
 

e c  e 1  j   e ' je ' '
e 

e" 
The loss tangent can be defined also as tan   
e ' e
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2. Lossless dielectric
(  0, e  e r e o ,   r o or   e )
 Substituting
in the general equations:
  0,    e

1
2p
u 
l


e
 o
h
0
e
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Review: 1. Free Space
(  0, e  e o ,   o )

Substituting in the general equations:
  0,    e   / c

1
2p
u 
c
l


 oe o
h
o o
0  120p   377 
eo

E ( z , t )  Eo cos(t  z ) x V / m
H ( z, t ) 
Eo
ho
cos(t  z ) yˆ
A/ m
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4. Good Conductors
(  , e  e o ,   r o )

Substituting in the general equations:
  

2

2
u 


l
2p
Is water a good
conductor???


h
45o


E ( z , t )  Eo e z cos(t   z ) x [V / m]
H ( z, t ) 
Eo
e z cos(t  z  45o ) yˆ [ A / m]

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Summary
Any medium


h
uc
l
 re r 
c
2

 

 1 
  1
2 
e 





e 
2

 

 1 
  1
2 
e 




Lossless
medium
(=0)

0
w me
= c / m re r

e
/
up
=
b
f
**In free space;

e
2
j
  je
2p
Low-loss medium
(e”/e’<1/100)
 e

e
1
1
e
e
up
up
f
f
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eo =8.85
10-12 F/m
UPRM
Good conductor
(e”/e’>100)
Units
p f ms
[Np/m]
p f ms
[rad/m]

(1  j )

4p f
ms
up
f
o=4p x 10-7 H/m
[W]
[m/s]
[m]
Ejemplo
 In
a free space,
H =10cos(10 t - b x)ẑ A / m
8
 Find Jd
E=
and E
10 b
we
cos(10 t - b x) ŷ V / m
8
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¶D e¶E
Jd =
=
¶t
¶t
Skin depth, d
E ( z, t )  Eo e
z

cos(t  z ) x [V / m]
We know that a wave attenuates in a lossy medium
until it vanishes, but how deep does it go?

Is defined as the
depth at which the
electric amplitude is
decreased to 37%
1
e  0.37  (37%)
e
z
e
1
at z  1 /   d
d  1 /  [m]
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Microwave Oven
Most food are lossy media at
microwave frequencies,
therefore EM power is lost
in the food as heat.

Find depth of penetration for
potato which at 2.45 GHz
has the complex permittivity
given.
e c  e o (30  j1)
g=
( jwmo ) ( jwec )
2p f
( j - 30)
c
= 202 - j195 [/m]
=
The power reaches the inside
as soon as the oven in
turned on!
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d =1/ a = 5mm
Short Cut!

You can use Maxwell’s or use
 1

H  kˆ  E
h


E  h kˆ  H
where k is the direction of propagation of the wave,
i.e., the direction in which the EM wave is
traveling (a unitary vector).
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Waves - Summary

Static charges > static electric field, E
 Steady current > static magnetic field, H
 Static magnet > static magnetic field, H

Time-varying current > time varying E(t) & H(t) that are
interdependent > electromagnetic wave

Time-varying magnet > time varying E(t) & H(t) that are
interdependent > electromagnetic wave
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EM waves don’t need a medium
to propagate

Sound waves need a
medium like air or water
to propagate
 EM waves don’t. They
can travel in free space in
the complete absence of
matter.
Look at a “wind wave”; the
energy moves, the plants
stay at the same place.
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Exercises: Wave Propagation in
Lossless materials

A wave in a nonmagnetic material is given by

H  zˆ50 cos(109 t  5 y ) [mA/m]
Find:
(a) direction of wave propagation,
(b) wavelength in the material
(c) phase velocity
(d) Relative permittivity of material
(e) Electric field phasor
Answer: +y,
l=1.26m,
up= 2x108 m/s,
2.25,

 j5 y
E   xˆ12.57e
[V/m]
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10.7-10.8
Power and Poynting Vector
Power in a wave

A wave carries power and transmits it
wherever it goes
The power density per
area carried by a wave
is given by the
Poynting vector.
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Poynting Vector Derivation

Start with E dot Ampere’s
E 

E    H  E  e
t 

E    H   E  E  E  e

Apply vector identity
   A  B   B    A  A    B  or in this case :
  H  E   E    H   H    E 

And end up with:
1 E
H    E     H  E   E  e
2 t
2
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2
E
t
Poynting Vector Derivation…

Substitute Faraday in 1rst term
H 
1 E 2

2
H  
    H  E   E  e

t
2 t


H    H  H 

As in derivative of square function : H    

t  2
t

and if invert the order, it' s (-)
  H  E     E  H 

 H 2
2 t
   E  H   E 2 
e E 2
2 t
Rearrange
 e E 2  H 2 
  E 2
  E  H   

t
2Electromagnetics
t
 2 Cruz-Pol,
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Poynting Vector Derivation…

Taking the integral wrt volume
¶ æe 2 m 2ö
2
Ñ
×
E
´
H
dv
=
E
+
H
dv
s
E
ç
÷
ò
ò è 2 2 ø ò dv
¶t
v
v
v
(

)
Applying Theorem of Divergence
ò(
S
¶ æe 2 m 2ö
E ´ H × dS = - ò ç E + H ÷dv - ò s E 2 dv
ø
¶t v è 2
2
v
)
Total power across
surface of volume

Rate of change of
stored energy in E or H
Ohmic losses due to
conduction current
Which means that the total power coming out of a
volume is either due to the electric or magnetic field
energy variations or is lost in ohmic losses.
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Power: Poynting Vector

Waves carry energy and information
 Poynting says that the net power flowing out of a
given volume is = to the decrease in time in
energy stored minus the conduction losses.
  
P  EH
2
[W/m ]
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Represents the
instantaneous
power density
vector associated
to the
electromagnetic
wave.
Bluetooth antenna in car
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Cellphone power radiation
Source: AnSys= Fred German
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Time Average Power density

The Poynting vector averaged in time is



T
T 

 *


1
1
1
Pave   P dt   E  H dt  Re Es  H s
T0
T0
2


For the general case wave:
Es = Eoe-a z e- jb z x̂ [V / m]
Hs =
Eo
h
e-a z e- jb z ŷ [A / m]
Eo2 -2a z
Pave =
e cosqh ẑ
2h
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[W/m 2 ]
Total Power in W
The total power through a surface S is
Pave = ò Pave × dS [W ]
S



Note that the units now are in Watts
Note that power nomenclature, P is not cursive.
Note that the dot product indicates that the surface
area needs to be perpendicular to the Poynting
vector so that all the power will go thru. (give example
of receiver antenna)
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PE 10.7
In free space, solar TEM wave at some latitude &
season can be measured to be H=0.2 cos (t-x) z
A/m. Find the total power passing through a solar
panel of side 10cm on plane x + ẑ =1
Answer;
 square
(.10)2
Ptot = 2.4p
2
=53mW
plate at ẑ = 3
Answer; Ptot = 0mW!
x
Hz
Ey
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Solar Panels
Optimum inclination of solar panels
for maximum power transmission
depends on geographical location.
http://www.altenergymag.com/e
magazine.php?issue_number=0
5.08.01&article=gonzalez
Exercises: Power
2. A 5GHz wave traveling In a nonmagnetic medium with

er=9 is characterized by E  yˆ 3 cos(t  x)  zˆ 2 cos(t  x)[V/m]
Determine the direction of wave travel and the average
power density carried by the wave
 Answer:
Eo2 -2a z
Pave =
e cosqh âk = - x̂0.05 [W/m 2 ]
2 h1
TEM wave
Transverse ElectroMagnetic = plane wave
 There
are no fields parallel to the direction of
propagation,
 Only perpendicular (=transverse).

If have an electric field Ex(z)
…then must have a corresponding magnetic field Hy(z)
 The
direction of propagation is âE ´ âH = âk
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Polarization:
Why do we care??
 Antenna applications –

Antenna can only TX or RX a polarization it is designed to support.
•
Straight wires, square waveguides, and similar rectangular systems support
linear waves (polarized in one direction)
• Round waveguides, helical or flat spiral antennas produce circular or elliptical
waves.

Remote Sensing and Radar Applications –



Many targets will reflect or absorb EM waves differently for different
polarizations.
Using multiple polarizations can give more information and improve
results.
Absorption applications –

Human body, for instance, will absorb waves with E oriented from
head to toe better than side-to-side, esp. in grounded cases. Also,
the frequency at which maximum absorption occurs is different for
these two polarizations. This has ramifications in safety guidelines
and studies.
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Outline
 Origin
 Maxwell
Equations explain waves
 Phasors and Time Harmonic fields

Maxwell eqs for time-harmonic fields
 Cases

general, lossy, lossless….
 Wave
Power
 Applications and Concepts
Cell phone & brain

Computer model for Cell phone
Radiation inside the Human Brain
 SAR=Specific Absorption Rate [W/Kg]
FCC limit 1.6W/kg, ~.2mW/cm2 for
30mins
http://www.ewg.org/cellphoneradiation/Get-a-SaferPhone/Samsung/Impression+%28SGH-a877%29/
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Human absorption
 30-300
MHz is
where the human
body absorbs RF
energy most
efficiently




* The FCC limit in the US for
public exposure from
cellular telephones at the ear
level is a SAR level of 1.6
watts per kilogram (1.6 W/kg)
as averaged over one gram
of tissue.
http://handheld-safety.com/SAR.aspx
http://www.fcc.gov/Bureaus/Engineering_Technology/Docume
nts/bulletins/oet56/oet56e4.pdf
http://www.rfcafe.com/references/electrical/fcc-maximumpermissible-exposure.htm
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ICNIRP= Intl Commission of Non-ionizing radiation protection
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Exercise: Cellphone
Tower Safe Power Level
At microwave frequencies, the power density
considered safe for humans is 1
mW/cm2.(SAR varies by frequency and
country) A cellphone tower radiates a wave
with an electric field amplitude E that decays
with distance as |E(R)|=3000/R [V/m], where R
is the distance in meters. What is the radius
of the unsafe region?
 Answer: 34.54 m
Radiación en
Teléfonos
móviles
En 2011, la
Organización Mundial
de la Salud de la ONU
clasificó a los teléfonos
celulares en Categoría
de Peligro de Cáncer, a
pesar de que producen
radiación no-ionizante
6 Tips para Protegerte:
Mantén las llamadas cortas.
2. Mantén distancia de 1“ (2.5 cm)
del cuerpo)
3. Usa el cable auricular
4. Usa bocina altoparlante,
5. Apaga el bluetooth, GPS y WiFi
cuando no lo estés usando
6. Envía texto en vez
7. *Coteja el nivel de radiación en
tu modelo. Depende de la región.
Los mismos modelos en Europa
emiten menos radiación debido a
exigencias de sus leyes. Busca
en sarvalues.com SAR levels
1.
Radar bands
Band Name
Nominal Freq
Range
HF, VHF, UHF
3-30 MHz0, 30-300 MHz, 3001000MHz
Specific Bands
138-144 MHz
216-225, 420-450 MHz
890-942
Application
TV, Radio,
Clear air, soil moist
L
1-2 GHz (15-30 cm)
1.215-1.4 GHz
S
2-4 GHz (8-15 cm)
2.3-2.5 GHz
2.7-3.7>
C
4-8 GHz (4-8 cm)
5.25-5.925 GHz
TV stations, short range
Weather
X
8-12 GHz (2.5–4 cm)
8.5-10.68 GHz
Cloud, light rain, airplane
weather. Police radar.
Ku
12-18 GHz
13.4-14.0 GHz, 15.7-17.7
K
18-27 GHz
24.05-24.25 GHz
Ka
27-40 GHz
33.4-36.0 GHz
V
40-75 GHz
59-64 GHz
W
75-110 GHz
millimeter
110-300 GHz
76-81 GH, 92-100 GHz
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Weather observations
Cellular phones
Weather studies
Water vapor content
Cloud, rain
Intra-building comm.
Rain, tornadoes
Tornado chasers
Electromagnetic Spectrum
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Radares UPRM
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Decibel Scale

In many applications need comparison of two
powers, a power ratio, e.g. reflected power,
attenuated power, gain,…
 The decibel (dB) scale is logarithmic
G
P1
P2
 V12 /R 
 P1 
V 
G[dB]  10 log    10 log  2   20 log  1 
 P2 
 V2 
 V2 /R 

Note that for voltages, fields, and electric
currents, the log is multiplied by 20 instead of 10.
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Attenuation rate, A

Represents the rate of decrease of the magnitude
of Pave(z) as a function of propagation distance`
æ Pave (z) ö
-2 a z
A = 10 log ç
÷ = 10 log ( e )
è Pave (0) ø
= -20a z log e = -8.68a z = -a dB z [dB]
where
a dB[dB/m] = 8.68a[Np/m]
Eo2 -2a z
Pave =
e cosqh
2 h1
Submarine antenna
A submarine at a depth of 200m uses an
antenna array to receive signal
transmissions at 1kHz.
 Determine the power density incident upon
the submarine antenna due to the EM wave
with |Eo|= 10V/m.

[At 1kHz, sea water has er=81, =4].
Eo2 -2a z
Pave =
e cosqh ẑ
2h

[W/m 2 ]
At what depth the amplitude of E has
decreased to 1% its initial value at z=0 (sea
surface)?
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a = f pms = .126
h = (1+ j)
a
s
= 0.044Ð45o W
Summary
Any medium


h
uc
l
 re r 
c
2

 

 1 
  1
2 
e 





e 
2

 

 1 
  1
2 
e 




Lossless
medium
(=0)

0
w me
= c / m re r

e
/
up
=
b
f
**In free space;

e
2
j
  je
2p
Low-loss medium
(e”/e’<1/100)
 e

e
1
1
e
e
up
up
f
f
Cruz-Pol,xElectromagnetics
eo =8.85
10-12 F/m
UPRM
Good conductor
(e”/e’>100)
Units
p f ms
[Np/m]
p f ms
[rad/m]

(1  j )

4p f
ms
up
f
o=4p x 10-7 H/m
[W]
[m/s]
[m]
Exercise: Lossy media
propagation
For each of the following determine if the material is low-loss dielectric,
good conductor, etc.
(a) Glass with r=1, er=5 and =10-12 S/m at 10 GHZ
(b) Animal tissue with r=1, er=12 and =0.3 S/m at 100 MHZ
(c) Wood with r=1, er=3 and =10-4 S/m at 1 kHZ
Answer:
(a)
(b)
(c)
low-loss,  8.4x1011 Np/m,  468 r/m, l 1.34 cm, up1.34x108, hc168 
general,  9.75, 12, l52 cm, up0.5x108 m/s, hc39.5j31.7 
Good conductor,  6.3x104,  6.3x104, l 10km, up0.1x108, hc6.281j 
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Who was NikolaTesla?
 Find
out what inventions he made
 His relation to Thomas Edison
 Why is he not well known?
http://www.youtube.com/watch?
v=i-j-1j2gD28&feature=related
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Pure Water is blue
 Is
turquoise blue due to its
electromagnetic spectrum emission
 http://ece.uprm.edu/~pol/OceanBlue
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