Transcript Energy Flux - Purdue Physics

Please Fill the online Course
evaluation.
So far we have inputs from only 41 students
http://www.purdue.edu/eval/
This is the official survey requested by
university
1
Official Exam
• Saturday, May 10, 1:00-3:00 pm
• room: EE129
Final Exam
Makeup Exam (include crew team member):
• Saturday, May 03, 2:00-4:00 pm
• room: physics building, Room238.
• send me request with valid reasons and proofs by April 28. After that, no
requests will be approved.
Exam for ADA students:
• Saturday, May 03, 1:00-4:00 pm
• room: physics building, Room238.
AOB
• multiple choice problem.
• Prepare your own scratch paper, pens, pencils, erasers, etc.
• Use only pencil for the answer sheet
• Bring your own calculators
• No cell phones, no text messaging which is considered cheating.
• No crib sheet of any kind is allowed. Equation sheet will be provided.
2
Last Time
Maxwell Equations – complete!
Wave solutions
3
Today
Accelerated Charges
Energy and Poynting Vector
Momentum and Poynting Vector
Re-Radiation (scattering)
Polarized Light
Why the sky is blue
4
Accelerated Charges
Electromagnetic pulse can propagate in space
How can we initiate such a pulse?
1. Transverse pulse
propagates at speed of
light
2. Since E(t) there must
be B
3. Direction of v is given
 
by: E  B
5
Magnitude of the Transverse Electric Field
We got the direction.
Magnitude can be derived
from Gauss’s law*
Field ~ -qa

Eradiative 

1  qa
40 c 2 r
1. The direction of the transverse
field is opposite to qa
2. The electric field falls off at a rate 1/r
* This was first shown by Edward Purcell, BSEE Purdue,
6 Nobeal Laureate!
iClicker Question
A proton is briefly accelerated as shown below. What
is the direction of the radiative electric field that will
be detected at location A?
B
A
A
+
D
C
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Sinusoidal Electromagnetic Radiation
Acceleration:
d2y
a  2   ymax  2 sin t 
dt


1  qa
Eradiative 
40 c 2 r

f 
2
T  1/ f

1 qymax  2
Eradiative 
sin t  ĵ
2
40 c r
Sinusoidal E/M field
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Sinusoidal E/M Radiation: Wavelength
Instead of period can
use wavelength:
  cT  c
f

f 
2
T  1/ f
Example of sinusoidal E/M
radiation:
atoms
radio stations
E/M noise from AC wires
Freeze picture in time:
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Energy of E/M Radiation
A particle will experience electric
force during a short time d/c:
Felec  qE
 d
p  p  0  Felec t  qE  
 c
What will happen to the ball?
It will oscillate
Energy was transferred from E/M field to the ball
2
p 2  qEd   1 
K  K  0 

 

2m  c   2m 
Amount of energy in
the pulse is ~ E2
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Energy Flux
There is E/M energy stored in the pulse:
Energy
  0 E 2 (J/m 3 )
Volume
Pulse moves in space:
there is energy flux
Units: J/(m2s) = W/m2
During t:


Energy   0 E 2 A  ct 
Energy
flux 
 c 0 E 2
 A  t 
flux 
1
0
EB
used: E=cB, 00=1/c2
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Energy Flux: The Poynting Vector
flux 
1
0
EB
 
The direction of the E/M radiation was given by E  B
Energy flux, the “Poynting vector”:
 1  
S
EB
0
John Henry
Poynting
(1852-1914)
(in W/m 2 )
• S is the rate of energy flux in E/M radiation
• It points in the direction of the E/M radiation
Note energy flux = [Energy/(time * Area)]
Energy flux = Intensity = [Power/Area]
Same Thing!
12
Example
In the vicinity of the Earth, the energy intensity of radiation
emitted by the sun is ~1400 W/m2. What is the approximate
magnitude of the electric field in the sunlight?
Solution:
flux 
E
1
0
EB  c 0 E 2
flux
 725 N/C
c 0
Note: this is an average (rms) value
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Momentum of E/M Radiation
• E field starts motion
• Moving charge in magnetic field:

 
Fmag  qv  B
Fmag
What if there is negative charge?

 
Fmag   q v  B
‘Radiation pressure’:
What is its magnitude?
Average speed: v/2
v
vE
Fmag  q B  q
2
2 c
Fmag
vE
v
q
/( qE ) 
 1
Felec
2c
2c
Fmag
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Momentum Flux
Net momentum:
in transverse direction: 0
in longitudinal direction: >0
Relativistic energy:
 
E   pc  mc
2
2
2 2
Quantum view: light consists of photons with zero mass: E 2   pc 2
Classical (Maxwell): it is also valid, i.e. momentum = energy/speed
 1  
S
EB
0
Momentum flux:

S
1  

E  B (in N/m 2 )
c 0 c
Units of Pressure
15
Exercise: Solar Sail
What is the force due to sun light on a sail with the area 1 km2
near the Earth orbit (1400 W/m2)?
Solution:
E

S
1  

EB
c 0 c
flux
 725 N/C
c 0

S
1
E
E2
6
2

E 

4
.
65

10
N/m
c 0 c c 0 c 2
Note: What if we have a reflective surface?
 9.3  106 N/m 2
Total force on the sail: F  9.3 N
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Effect of Radiation on a Neutral Atom
Main effect:
brief electric kick sideways
Neutral atom: polarizes
Electron is much lighter than nucleus:
can model atom as outer electron
connected to the rest of the atom by a
spring:
F=eE
17
Radiation and Neutral Atom: Resonance
E y  E0 sin t 
Fy  eE y  F0 sin t 
Amplitude of oscillation will depend
on how close we are to the natural
free-oscillation frequency of the ballspring system
Resonance
This is why you can tune a radio
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Importance of Resonance
E/M radiation waves with frequency ~106 Hz has big effect on
mobile electrons in the metal of radio antenna:
can tune radio to a single frequency
E/M radiation with frequency ~ 1015 Hz has big effect on organic
molecules:
retina in your eye responds to visible light but not radio waves
Very high frequency (X-rays) has little effect on atoms and can
pass through matter (your body):
X-ray imaging
19
Cardboard
Why there is no light going through a cardboard?
Electric fields are not blocked by matter
Electrons and nucleus in cardboard reradiate light
Behind the cardboard reradiated E/M field cancels original field
20
Effect of E/M Radiation on Matter
1. Radiative pressure – too small to be observed in most cases
2. E/M fields can affect charged particles: nucleus and electrons
http://www.youtube.com/watch?v=dL9_Tldmrhs
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Electromagnetic Spectrum
22
Color Vision
Three types of receptors (cones) in retina
which incorporate three different organic
molecules which are in resonance with red,
green and blue light frequencies (RGBvision):
Response spectra for three
types of receptors
Max response wavelengths:
S – 440 nm (6.81.1014 Hz)
M – 540 nm (5.56.1014 Hz)
L – 560 nm (5.36.1014 Hz)
Refers
23 to length of cone
Polarizer: polarization by absorption
An electric field component parallel to the
transmission axis is passed by a polarizer; a
component perpendicular to it is absorbed.
transmission
axis
dichroism (tourmaline, polaroid,…)
So if linearly polarized beam with E is
incident on a polarizer as shown,
E y  E cos 
I  I 0 cos 2 
Zero if =/2, I0 if =0
If unpolarized beam is incident instead,
The intensity will reduce by a factor of two.
I  I 0 cos 2 
 I0 / 2
The light will become polarized along the transmission
axis
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iClicker Questions
•
A beam of un-polarized lights with
intensity I is sent through two polarizers
with transmission axis perpendicular to
each other. What’s the outgoing light
intensity?
a)
b)
c)
d)
½I
2I
0
1.5 I
25
Example: two polarizers
This set of two linear
polarizers produces LP
(linearly polarized)
light. What is the final
intensity?
– P1 transmits 1/2 of
the unpolarized
light:
I1 = 1/2 I0
– P2 projects out the
E-field component
parallel to x’ axis:
E2  E1 cos 
I  E2
1
I 2  I1 cos   I 0 cos 2 
2
2
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= 0 if  = /2
(i.e., crossed)
7B-22 Polarizer Effects
27
Polarized Light
Polaroid sunglasses and camera filters:
reflected light is highly polarized: can block it
Considered: using polarized car lights and polarizers-windshields
28
Polarized E/M Radiation
AC voltage
(~300 MHz)
What will happen if distance is increased twice?
E/M radiation can be polarized
along one axis…
no
light
…and it can be unpolarized:
29
UHF Transmitter and Dipole
Receiver (6D-17)
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iClicker question
In which of these situations will the
bulb light?
A)
B)
C)
D)
E)
A
B
C
None
B and C
31
Why the Sky is Blue
Why there is light coming from the sky?
Why is it polarized?
Why is it blue?
x  x0 sin  t 
d2x
E ~ a  2  y0 2 sin  t 
dt
Energy flux: ~ E 2 ~  4
Ratio of blue/red frequency is ~2  scattering intensity ratio is 16
Why is sun red at sunset? Is its light polarized?
Why are distant mountains blue?
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Why is the sunset red?
The shorter wavelengths of blue light
are scattered by gas molecules in the
atmosphere more than longer
wavelengths such as red light.
When the sun is low on the horizon, the
light must pass through more
atmosphere than when the sun is
directly above.
By the time the sun’s light reaches our
eyes, the shorter wavelengths such as
blue and yellow have been removed by
scattering, leaving only orange and red
light coming straight from the sun.
Today
Accelerated Charges
Energy and Poynting Vector
Momentum and Poynting Vector
Re-Radiation (scattering)
Polarized Light
Why the sky is blue
34
Please Fill the online Course
evaluation.
So far we have inputs from only 41 students
http://www.purdue.edu/eval/
This is the official survey requested by
university
35
Official Exam
• Saturday, May 10, 1:00-3:00 pm
• room: EE129
Final Exam
Makeup Exam (include crew team member):
• Saturday, May 03, 2:00-4:00 pm
• room: physics building, Room238.
• send me request with valid reasons and proofs by April 28. After that, no
requests will be approved.
Exam for ADA students:
• Saturday, May 03, 1:00-4:00 pm
• room: physics building, Room238.
AOB
• multiple choice problem.
• Prepare your own scratch paper, pens, pencils, erasers, etc.
• Use only pencil for the answer sheet
• Bring your own calculators
• No cell phones, no text messaging which is considered cheating.
• No crib sheet of any kind is allowed. Equation sheet will be provided.
36