Ampere’s Law Updated Hertz’s Discoveries Plane Electromagnetic Waves Energy in an EM Wave

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Ampere’s Law Updated
Hertz’s Discoveries
Plane Electromagnetic Waves
Energy in an EM Wave
Momentum and Radiation Pressure
The Electromagnetic Spectrum
Ampère’s Law
 
 B.ds  0 I
But there is a problem if I varies with time!
Consider a parallel plate capacitor
 
For S1,  B.d s  0 I
 
For S2,
 B.ds  0 I  0
(No current in the gap)
Call I as the conduction current.
Specify a new current that exists when a change in the
electric field occurs. Call it the displacement current.
d E
Id  0
dt
Electrical flux through S2:
 Q 
Q
 A 
 E  EA  
0
 0 A 
Id  0
d E dQ

I
dt
dt
Now the generalized form of Ampère’s Law, or Ampère-Maxwell Law becomes:
 
d E
 B.ds  0 I  I d   0 I  0 0 dt
Magnetic fields are produced by
both conduction currents and by
time-varying electrical fields.
  Q
 E.dA 
0
Gauss’ Law
 
 B.dA  0
Gauss’ Law for Magnetism –
no magnetic monopoles
 
d B
 E.ds   dt
 
d E
 B.ds  0 I  0 0 dt


 
F  qE  qv  B
Faraday’s Law
Ampère-Maxwell Law
Lorentz Force Law
By supplying short voltage bursts from the coil
to the transmitter electrode we can ionize the
air between the electrodes.
In effect, the circuit can be modeled as a LC circuit,
with the coil as the inductor and the electrodes as
the capacitor.
A receiver loop placed nearby is able to receive
these oscillations and creates sparks as well.
Circular (or Spherical in 3D) Waves
Plane Waves
 
d B
E
.
d
s



dt
 
d E
B
.
d
s


I



0
0 0

dt
E
B

x
t
B
E
   0 0
x
t
Free Space: I = 0, Q = 0
2E
  B 
  B 

E 









 
 
 0 0

x 2
x  t 
t  x 
t 
t 
2E
2E
 0 0 2
2
x
t
 
d E
B
.
d
s



0 0

dt
2B
2B
 0 0 2
x 2
t
E  Emax coskx  t 
B  Bmax cos kx  t 
c
E  Emax coskx  t 
B  Bmax cos kx  t 
1
0 0
  2f

k

k
c
f
2

c
Speed of Light
Angular Frequency
Wavelength
Wavenumber
E
B

x
t
B
E
   0 0
x
t
E  Emax coskx  t 
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B  Bmax cos kx  t 
kEmax  Bmax
E Emax 

 c
B Bmax k
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The solutions to Maxwell’s
equations in free space are
wavelike
Electromagnetic waves travel
through free space at the speed of
light.
The electric and magnetic fields of
a plane wave are perpendicular to
each other and the direction of
propagation (they are transverse).
The ratio of the magnitudes of the
electric and magnetic fields is c.
EM waves obey the superposition
principle.
a) =?, T=?
c
3 108
 
 7.5m
f 40 106
1
1
8
T 

2
.
5

10
s
6
f 40 10
b) If Emax=750N/C, then Bmax=?
Bmax 
f  40MHz
c) E(t)=? and B(t)=?
Emax
750

 2.5 106 T
8
c
3 10
Directed towards the z-direction
2
 0.838rad / m
 7.5
  2f  2 40 106  2.51108 rad / s
k
2

E  Emax coskx  t   750 N / C cos(0.838 x  2.51  108 t )
B  Bmax coskx  t   2.5  106 T cos(0.838 x  2.51  108 t )
 1  
S
EB
0
Poynting Vector (W/m2)
S is equal to the rate of EM energy flow per unit area
(power per unit area)
E2
c 2
For a plane EM Wave: S 


B
0 0c 0
EB
The average value of S
is called the intensity:
2
Emax Bmax Emax
c
2
I  S av 


Bmax
20
20 c 20
uE 
uB
0E
2
2
B2
uB 
20
2

0 0 2 1
E c


E  
20
20
2
0
E
For an EM wave, the instantaneous
electric and magnetic energies are
equal.
2
2
1
B
uE  uB   0 E 2 
2
20
u  uE  uB   0 E 
2
 
B2
Total Energy Density of an EM Wave
0
2
Bmax
1
2
2
uav   0 E av   0 Emax 
2
20
I  Sav  cuav
The intensity is c times the
total average energy density
Which gives the largest average energy density
at the distance specified and thus, at least
qualitatively, the best illumination
1. a 50-W source at a distance R.
2. a 100-W source at a distance 2R.
3. a 200-W source at a distance 4R.
P lamp= 150 W, 3% efficiency
A = 15 m2
Pav  Plamp 0.03  4.5W
2
Pav Emax
I

A 20c
Emax 
Bmax 
20cPav
 18.42V / m
A
Emax
 6.14 108 T
c
TER
p
c
A
Momentum for complete absorption
P
F 1 dp

A A dt
P
1 d  TER

A dt  c
Pressure on surface
 1 dTER dt 

A
 c
S
P
c
For complete reflection:
2TER
p
c
2S
P
c
P laser= 3 mW, 70% reflection, d = 2 mm
P
0.003
2
S 

955
W
/
m
A  0.0012


S
S
S
P   0.7  1.7
c
c
c
955
6
2
P  1.7

5
.
4

10
N
/
m
3 108
The fundamental mechanism for electromagnetic radiation is
an accelerating charge

Reading Assignment
 Chapter 35: Nature of Light and Laws of
Geometric Optics
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WebAssign: Assignment 12
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