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Transcript
```Maxwell’s Equations
q
S E  dA  εo
 B  dA  0
Gauss's law  electric 
Gauss's law in magnetism
S
dB
 E  ds   dt
dE
 B  ds  μo I  εo μo dt
Ampere-Maxwell law
•The two Gauss’s laws are symmetrical, apart from the absence of the term for
magnetic monopoles in Gauss’s law for magnetism
•Faraday’s law and the Ampere-Maxwell law are symmetrical in that the line
integrals of E and B around a closed path are related to the rate of change of
the respective fluxes
•
•
•
•
•
•
•
•
Gauss’s law (electrical):
The total electric flux through any
closed surface equals the net charge
inside that surface divided by eo
This relates an electric field to the
charge distribution that creates it
Gauss’s law (magnetism):
The total magnetic flux through
any closed surface is zero
This says the number of field lines
that enter a closed volume must
equal the number that leave that
volume
This implies the magnetic field
lines cannot begin or end at any
point
Isolated magnetic monopoles have
not been observed in nature
q
S E  dA  εo
 B  dA  0
S
•
•
•
•
This describes the creation of an electric field by a
changing magnetic flux
The law states that the emf, which is the line
integral of the electric field around any closed
path, equals the rate of change of the magnetic flux
through any surface bounded by that path
One consequence is the current induced in a
conducting loop placed in a time-varying B
•
The Ampere-Maxwell law is a generalization of
Ampere’s law
•
It describes the creation of a magnetic field by an
electric field and electric currents
The line integral of the magnetic field around any
closed path is the given sum
•
dB
 E  ds   dt
dE
 B  ds  μo I  εo μo dt
Maxwell’s Equation’s in integral form
Q 1
 A E  dA  eo  eo

A
B  dA  0

V
dV
Gauss’s Law
Gauss’s Law for Magnetism
d B
d
 C E  d   dt   dt A B  dA

d E
dE 
C B  d  o Iencl  oeo dt  o A  J  eo dt   dA
Ampere’s Law
Maxwell’s Equation’s in free space
(no charge or current)


A
A
E  dA  0
Gauss’s Law
B  dA  0
Gauss’s Law for Magnetism
d B
d
 C E  d   dt   dt A B  dA
d E
d
 C B  d  oeo dt  oeo dt A E  dA
Ampere’s Law
Hertz’s Experiment
•
•
•
•
•
An induction coil is connected to a
transmitter
The transmitter consists of two spherical
electrodes separated by a narrow gap
The discharge between the electrodes
exhibits an oscillatory behavior at a very
high frequency
Sparks were induced across the gap of the
receiving electrodes when the frequency of
the transmitter
In a series of other experiments, Hertz also
showed that the radiation generated by this
equipment exhibited wave properties
–
•
Interference, diffraction, reflection, refraction
and polarization
He also measured the speed of the radiation
Implication
• A magnetic field will be produced in empty space if there
is a changing electric field. (correction to Ampere)
• This magnetic field will be changing. (originally there
was none!)
• The changing magnetic field will produce an electric field.
• This changes the electric field.
• This produces a new magnetic field.
• This is a change in the magnetic field.
An antenna
Hook up an
AC source
We have changed the magnetic
field near the antenna
An electric field results! This is
the start of a “radiation field.”
Look at the cross section
Accelerating
electric charges
give rise to
electromagnetic
waves
Called:
“Electromagnetic Waves”
E and B are perpendicular (transverse)
We say that the waves are “polarized.”
E and B are in phase (peaks and zeros align)
Angular Dependence of Intensity
• This shows the angular
intensity produced by a dipole
antenna
• The intensity and power
radiated are a maximum in a
plane that is perpendicular to
the antenna and passing
through its midpoint
• The intensity varies as
(sin2 θ / r2
Harmonic Plane Waves
At t = 0
E
l  spatial period or
wavelength
l
x
l
2 l 
v   fl 

T
T 2 k
At x = 0
E
t
T
T  temporal period
d B
 C E  d   dt

C
E  d   E  dE  y  Ey  dEy
d B dB

dxy
dt
dt
dB
dEy  
dxy
dt
dE
dB

dx
dt
d E
 C B  d  oeo dt

C
B  d  Bz   B  dB z  dBz
d E dE

dxz
dt
dt
dE
dBz   o eo
dxz
dt
dB
dE
 o eo
dx
dt
Fields are functions of both
position (x) and time (t)
dE
dB

dx
dt
Partial derivatives
are appropriate
B
E
  o e o
x
t
dB
dE
 o eo
dx
dt
E
 B

2
x
x t
2
E
B

x
t
 B
2E
 o eo 2
t x
t
2E
2E
 o eo 2
2
x
t
This is a wave
equation!
The Trial Solution
• The simplest solution to the partial differential
equations is a sinusoidal wave:
– E = Emax cos (kx – ωt)
– B = Bmax cos (kx – ωt)
• The angular wave number is k = 2π/λ
– λ is the wavelength
• The angular frequency is ω = 2πƒ
– ƒ is the wave frequency
The trial solution
E  E y  Eo sin  kx  t 
2E
2E
 o eo 2
2
x
t
2E
2
 k E o sin  kx  t 
2
x
2E
2
  E o sin  kx  t 
2
t
k 2Eo sin  kx  t   oeo2Eo sin  kx  t 
2
1

2
k
oe o
The speed of light
l
2 l 
v   fl 

T
T 2 k

1
vc 
k
o e o
The electromagnetic spectrum
l
2 l 
v   fl 

T
T 2 k
Another look
dE
dB

dx
dt
B  Bz  Bo sin  kx  t 
E  E y  Eo sin  kx  t 
d
d
E o sin  kx  t    Bo sin  kx  t 
dx
dt
Eo k cos  kx  t   Bo cos  kx  t 
Eo 
1
 c
Bo k
o e o
Energy in Waves
1
1 2
2
u  e0 E 
B
2
2 0
u  e0 E 2
Eo 
1
 c
Bo k
o e o
1 2
u B
0
e0
u
EB
0
Poynting Vector

1
S
EB
0

EB E 2 c B 2
S


μo μo c
μo
S  cu
• Poynting vector points in the direction the wave moves
• Poynting vector gives the energy passing through a unit
area in 1 sec.
• Units are Watts/m2
Intensity
• The wave intensity, I, is the time average of
S (the Poynting vector) over one or more
cycles
• When the average is taken, the time average
of cos2(kx - ωt) = ½ is involved
2
2
E max Bmax E max
c Bmax
I  S av 


 cu ave
2 μo
2 μo c
2 μo
F 1 dp
P 
A A dt
Maxwell showed:
U
p 
c
by an object)
1 dU Save
P

Ac dt
c
What if the radiation reflects off an object?
Pressure and Momentum
• For a perfectly reflecting surface,
p = 2U/c and P = 2S/c
• For a surface with a reflectivity somewhere
between a perfect reflector and a perfect absorber,
the momentum delivered to the surface will be
somewhere in between U/c and 2U/c