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Lecture 20: July 10th 2009
Physics for Scientists and Engineers II
Physics for Scientists and Engineers II , Summer Semester 2009
1
Transmission line economics
RT
RL
Consumer runs vacuum cleaner (1KW power)
Pconsumer  I ΔVC  I 2 RL  I 2 
Plost  I 2 RT 
Pconsumer
RL
Pconsumer
RT
RL
Pgenerator  I 2 RT  RL  
Pconsumer
RT  RL   Pconsumer RT  Pconsumer  Plost  Pconsumer
RL
RL
Physics for Scientists and Engineers II , Summer Semester 2009
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Transmission line economics – using transformers
IG 
N2
N N
IT  2 1 I  I
N1
N1 N 2
N1
IT 
RT
N2
N2
Pconsumer  I 2 RL  I 2 
2
I
N1
RL
Step - down trans former N2 , N1
Step - up transform er N1  N2
N
2
Plost  I T RT   1
 N2
N1
I
N2
Pconsumer
RL
2
2

N 
N  P
I  RT   1  I 2 RT   1  consumer RT
RL

 N2 
 N2 
2
2
2








Pconsumer  N1
N1
N2
2

 RT  
 
 RL 
Pgenerator  I Req 

RL   N 2 
 N 2   N1 


2
2



Pconsumer   N1 
N
R

 RT  RL   Pconsumer  1  T  Pconsumer  Plost  Pconsumer


RL   N 2 
 N 2  RL


Physics for Scientists and Engineers II , Summer Semester 2009
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Example: Problem 43 in Chapter 33
Transmissi on line has resistance per unit length : 4.50x10 -4

m
Length of transmiss ion line : 400 miles (6.44x10 5 m)
Need to transmit 5.00MW of power
Vgenerator  4.50kV transforme d up to 500KV
What is the loss of power in the transmiss ion line?

 6.44 105 m  2.90 10 2 
m
5.00 MW

 1.11103 A
4.50kV
RT  4.50 10  4
IG 
Ptransmitted
VG
IT 
N1
V
4.50kV
IG  1 IG 
1.11103 A  10.0 A
N2
V2
500kV
Plost  I T2 RT  10.0 A  2.90 10 2   29.0kW
2
Plost
Ptransmitted
29.0 103

 0.00580  0.580%
5.00 106
Physics for Scientists and Engineers II , Summer Semester 2009
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Example: Problem 43 in Chapter 33
Without tr ansformer :

 6.44 105 m  2.90  10 2 
m
5.00 MW

 1.11103 A
4.50kV
RT  4.50 10  4
IG 
Ptransmitted
VG
I T  I G  1.11 103 A


2
Plost  I T2 RT  1.11 103 A  2.90  10 2   358MW  PGenerator
 This cannot be. How much can be transmitt ed?
Physics for Scientists and Engineers II , Summer Semester 2009
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Example: Problem 43 in Chapter 33
I max 
V
RL  RT
 V
2
Pload  I max RL  
 RL  RT
Find power maximum :
2

RL
 RL  V 2 2
2
RL  2 RT RL  RT

dPload
R  2 RT RL  RT  RL 2 RL  2 RT 
 V 2 L
0
2
2 2
dRL
R  2R R  R
2
2

L
T
L
T

 RL  2 RT RL  RT  RL 2 RL  2 RT   0
2
2
 RT  RL  0  RL  RT
2
2
2
 4.50kV 
 4.50kV 
2
 RL  
Pload max  I max RL  
 290  17.5kW
 580 
 RL  290  
(and this is ONLY when R L  RT . In all other cases even less power can be
2
transmitte d). BTW, same principle as with internal resistance of a battery, where
we found that the maximum power is transferr ed to load when R L  Rint ernal .
Physics for Scientists and Engineers II , Summer Semester 2009
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The 500000 Volt Tesla Coil – featuring Adam Beehler succesfully trying to not electrocute himself
Physics for Scientists and Engineers II , Summer Semester 2009
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Chapter 34: Electromagnetic Waves
Remember Ampere' s law?
 B  d s  0 I , where I 
dq
is a conduction current (created by charge carriers)
dt
I
Path P
I
Capacitor plates
S1
S1
Problem : I only passes through S1 but both S1 and S 2 are bound by the same path.
However, while a conduction current passes through S1 none passes through S 2 .
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General Form of Ampere’s Law
 B ds   I
0
, is clearly inadequate for this situation.
James Clark Maxell : Generalize d Ampere' s law to include
time - varying electric fields. He defined a displaceme nt current :
d E
Id   0
, where  0  permittivi ty of free space.
dt
and
E   E  d A
is the electric flux.
General form of Ampere' s Law
(Ampere - Maxwell Law) :
 B  d s   0  I  I d    0 I   0 0
Physics for Scientists and Engineers II , Summer Semester 2009
d E
dt
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Calculating displacement current: Example
I
I
S1
S2
Surface S1 :
 B  d s   I  0    I
Surface S2 :
 B  d s   0 0  I d    0 0
0
0
d E
d
 0 0 EA
dt
dt
d q
dq
  0 0
 0
 0 I
dt  0
dt
Physics for Scientists and Engineers II , Summer Semester 2009
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Major Point
General form of Ampere' s Law
(Ampere - Maxwell Law) :
 B  d s   0  I  I d    0 I   0 0
d E
dt
Magnetic fields are produced both by conduction currents and
by time-varying electric fields.
Physics for Scientists and Engineers II , Summer Semester 2009
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Example: Chapter 34, Problem 1
A 0.100 - A current is charging a capacitor that has square plates 5.00 cm on each side. The
plate separation is 4.00mm. Find a) the time rate of change of electric flux between th e plates
and b) the displaceme nt current between th e plates.
Plate Capacitor : E 
q
q
  E  EA 
0 A
0
d E d q
1 dq 1


 I
dt
dt  0  0 dt  0
(b) I d   0
0.100 A
8.854 10
12
Nm 2
 1.13 10
Cs
10
C2
Nm 2
d E
1
  0 I  0.100 A
dt
0
Physics for Scientists and Engineers II , Summer Semester 2009
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A Complete Description of All Classical Electromagnetic Interactions in a
Vacuum
Maxwell' s Equations in free space
Gauss' s law 
q
 E  dA  
0
Gauss' s law in magnetism 
 B  dA  0
 E  ds  
d B
dt
 B  d s   0 I  I d    0 I   0 0
F  qE  q v  B
Physics for Scientists and Engineers II , Summer Semester 2009
Faraday' s law 
d E
dt
Ampere - Maxwell law 
Lorentz force law 
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Maxwell’s Prediction
Maxwell' s Equations in free space when q  0 and I  0
Gauss' s law 
 E  dA  0
 B  dA  0
 E  ds  
Gauss' s law in magnetism 
d B
dt
 B  d s   0 0
Faraday' s law 
d E
dt
Ampere - Maxwell law 
Can be combined  Result: Wave equations for electric and magnetic fields
Solution of wave equations shows that electric and magnetic field waves travel
(“electromagnetic radiation”) with the speed of light, leading Maxwell to predict
that light is a form of electromagnetic radiation.
Physics for Scientists and Engineers II , Summer Semester 2009
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Hertz’s Experiment
Hertz showed with this apparatus that electromagnetic waves exist.
Physics for Scientists and Engineers II , Summer Semester 2009
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Hertz’s Experiment
Details of the apparatus:
How it works: Battery charges the capacitor in the primary circuit on the left. The interrupter
periodically creates a short circuit across that capacitor and alternating charging and
discharging of capacitor occurs. A large periodic change of flux occurs in the primary winding.
The secondary winding picks up this large flux change and transforms the primary voltage
fluctuation into a much higher voltage fluctuation in the secondary circuit. The secondary
circuit is tuned (LC) creating a very large voltage across the spark balls.  High electric fields
occur between the spark balls, causing a sudden charge transfer (spark).  The electric field
breaks down but is then built up again. This fluctuating electric field induces a fluctuating
magnetic field (Ampere-Maxwell law), which in turn creates a fluctuating electric field
(Faraday’s law)….. electromagnetic radiation, which is then picked up by the detection ring
causing it to spark across it’s spark gaps (detection ring is also tuned to the same frequency).
Physics for Scientists and Engineers II , Summer Semester 2009
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Hertz’s Experiment
Other experiments by Hertz:
em wave generator
Standing
em wave
metal plate reflects
em wave.
Nodes can be detected  wavelength can be determined
Since frequency is know, the speed of the em wave can be calculated
using v   f . The result was v  c, the speed of visible light.
Physics for Scientists and Engineers II , Summer Semester 2009
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Plane versus Spherical Waves
Let’s first look at mechanical waves:
Imagine looking down on a pond. Someone throws rocks into the middle of
The pond, creating spherical waves in two dimensions.
Direction of wave propagation
If instead, the person throws very long metal bars into the pond,
plane waves are created.
Metal
bar
All parts of the wave front have the same
phase.
Physics for Scientists and Engineers II , Summer Semester 2009
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Plane Electromagnetic Waves
Plane linearly polarized electromagnetic waves with propagation in x-direction.
Properties:
1)Wherever in the yz plane em wave comes from, it propagates in the x-direction.
(all “rays” of this wave are parallel).
2) All em waves coming from different parts of the yz plane are all in phase.
3) Electric and magnetic waves oscillate perpendicular to each other (take this by faith for now).
4) Electric field is in the same direction (here in y-direction)  “linearly polarized”.
5) Magnetic field is in the same direction (here in z-direction)  “linearly polarized”.
6) And B are only function of x and t and do not depend on y and z  E(x,t), B(x,t).
y
E
E
x
z
c
c
B
B
Physics for Scientists and Engineers II , Summer Semester 2009
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