22.3MB1 Electromagnetism
Dr Andy Harvey
e-mail a.r.harvey@hw.ac.uk
lecture notes are at:
http://www.cee.hw.ac.uk/~arharvey/22.3MB1electromagnetism.zip
Revision of Electrostatics (week 1)
Vector Calculus (week 2-4)
Maxwell’s equations in integral and differential form (week 5)
Electromagnetic wave behaviour from Maxwell’s equations (week 6-7.5)
–Heuristic description of the generation of EM waves
–Mathematical derivation of EM waves in free space
–Relationship between E and H
–EM waves in materials
–Boundary conditions for EM waves
Fourier description of EM waves (week 7.5-8)
Reflection and transmission of EM waves from materials (week 9-10)
http://www.cee.hw.ac.uk/~arharve
1
Maxwell’s equations
in differential form
t
.J  

dt
E  
B
dt
H  J 
D
.B  0
.D  
D  E
B  H
Equationof continuity
Faraday' s law
Am pere' s law
Gauss' law for m agnetostatics
Gauss' law for electrostatics
• Varying E and H fields are
coupled
NB Equations brought from elsewhere, or to be carried on to next page, highlighted in this colour
NB Important results highlighted in this colour
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2
Fields at large distances from charges and current sources
H
E
  E  
 H  J  
dt
dt
I
H
• For a straight conductor the magnetic field is
given by Ampere’s law
• At large distances or high frequencies H(t,d)
lags I(t,d=0) due to propagation time
– Transmission of field is not instantaneous
– Actually H(t,d) is due to I(t-d/c,d=0)
•
•
•
•
•
Modulation of I(t) produces a dH/dt term
dH/dt produces E
dE/dt term produces H
etc.
How do the mixed-up E and H fields spread
out from a modulated current ?
–
eg current loop, antenna etc
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3
A moving point charge:
• A static charge produces radial
field lines
• Constant velocity, acceleration and
finite propagation speed distorts
the field line
• Propagation of kinks in E field
lines which produces kinks of
2
E t and  2 E t
• Changes in E couple into H & v.v
– Fields due to E t are short range
2
2
– Fields due to  E t propagate
• Accelerating charges produce
travelling waves consisting
coupled modulations of E and H
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4
Depiction of fields propagating
from an accelerating point charge
E in plane of page
H in and out of page
• Fields propagate at c=1/3 x 108 m/sec
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5
Examples of EM waves due to
accelerating charges
•
•
•
•
Bremstrahlung - “breaking radiation”
Synchrotron emission
Magnetron
Modulated current in antennas
– sinusoidal v.important
• Blackbody radiation
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6
2.2 Electromagnetic waves in lossless media Maxwell’s equations
Constitutive relations
Maxwell
D  E   r  o E
D
 H  J 
dt
B  H   r  o H
J  sE
B
E  
dt
.D  
B
. 0
Equation of continuity

.J  
t
•
•
•
•
J
D
H
B
• E
• 
• 
• s
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SI Units
Amp/ metre2
Coulomb/metre2
Amps/metre
Tesla
Weber/metre2
Volt-Second/metre2
Volt/metre
Farad/metre
Henry/metre
Siemen/metre
7
2.6 Wave equations in free space
• In free space
– s=0 J=0
D D
– Hence:
 H  J 

dt
B
E  
dt
dt
– Taking curl of both sides of latter equation:
B


    E   
    B   o   H
t
t
t
  D 
  o 

t  t 
    E    o
 2E
t 2
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8
Wave equations in free space cont.
    E    o
 2E
t 2
• It has been shown (last week) that for any vector A
    A  .A   2 A
where
Thus:
 
2
2
x 2

2
y 2

2
z 2
is the Laplacian operator
.E   2 E    o 
 2E
t 2
• There are no free charges in free space so .E==0
and we get
2
 E   o
2
 E
t 2
A three dimensional wave equation
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9
Wave equations in free space cont.
• Both E and H obey second order partial differential wave
equations:
2

E
 2 E   o 2
 2 H   o
t
 2H
t 2
• What does this mean
– dimensional analysis ?
Volts/metre
metre
2
  o
Volts/metre
seconds2
– o has units of velocity-2
– Why is this a wave with velocity 1/ o ?
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10
The wave equation
 A   o
2
2A
v
t 2
• Why is this a travelling wave ?
zˆ
• A 1D travelling wave has a solution of the form:
A  Ao f ( z  vt)
Constant for a
travelling wave
2A
2A
2





A
f
z

vt

A
v
f z  vt 
o
o
2
2
z
t
Substitute back into the above EM 3D wave equation
Ao f z  vt   Ao v 2 of z  vt 
This is a travelling wave if v  1 o
• In free space v  1 o o  2.9979108 m/sec  c
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11
Wave equations in free space cont.
 2E
2
2
 E   o 2
Ex  sin z  vt 
t

• Substitute this 1D expression into 3D ‘wave equation’ (Ey=Ez=0):


2


cos

z

vt



sinz  vt 
2
z
z


2 2




sin

z

vt



v
cos

z

vt



v sinz  vt 
2
t
t
sin  z  vt   
  2 sinx  vt    o 2 v 2 sinx  vt 
v
1
 o
• Sinusoidal variation in E or H is a solution ton the wave equation
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12
Summary of wave equations in free space cont.
• Maxwells equations lead to the three-dimensional wave equation to
describe the propagation of E and H fields
• Plane sinusoidal waves are a solution to the 3D wave equation
• Wave equations are linear
– All temporal field variations can be decomposed into Fourier components
• wave equation describes the propagation of an arbitrary disturbance
– All waves can be written as a superposition of plane waves (again by Fourier
analysis)
• wave equation describes the propagation of arbitrary wave fronts in free space.
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13
Summary of the generation of
travelling waves
• We see that travelling waves are set up when  2 E t 2  0
– accelerating charges
– but there is also a field due to Coulomb’s law:
q
E
4 o r 2
– For a spherical travelling wave, the power carried by the travelling
wave obeys an inverse square law (conservation of energy)
• P a E2 a 1/r2
– to be discussed later in the course
• Ea1/r
– Coulomb field decays more rapidly than travelling field
– At large distances the field due to the travelling wave is much
larger than the ‘near-field’
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14
Heuristic summary of the generation of travelling waves
E t due to charge
constant velocity (1/ r2)
E due to stationary
charge (1/r2)
 2E t 2
 2 H t 2
kinks due to charge
acceleration (1/r)
due to E (1/r)
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15
2.8 Uniform plane waves - transverse relation of E and H
• Consider a uniform plane wave, propagating in the z
direction. E is independent of x and y
E
0
x
E
0
y
In a source free region, .D= =0 (Gauss’ law) :
E x E y E z
.E 


0
x
y
z
E is independent of x and y, so
E y
E x
 0,
0
x
y

E z
0
z
 Ez  0
( E z  constis nota wave)
• So for a plane wave, E has no component in the direction of
propagation. Similarly for H.
• Plane waves have only transverse E and H components.
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16
Orthogonal relationship between E and H:
• For a plane z-directed wave there are no variations along x and y:
  H  a x
 A Ay 

  A  a x  z 


y

z


 A A 
ay x  z  
x 
 z
H y
H x
ay
z
z
D

t
E y
 Ex
E z 

   a x
 ay
 ay
t
t
t 

• Equating terms:
H y
E x


z
t
E y
H x

z
t
•
 Ay Ax 

a z 

y 
 x
 H  J 
D
dt
• and likewise for   E   o H t :
E y
 o
H x
t
H y
z
E x
 o
z
t
Spatial rate of change of H is proportionate to the temporal rate of change
of the orthogonal component of E & v.v. at the same point in space
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17
Orthogonal and phase relationship between E and H:
• Consider a linearly polarised wave that has a transverse
component in (say) the y direction only:
E y  Eo f z  vt 


E y
t
 vEo f z  vt 

H x

z
H x  vEo  f z  vt dz  const  vEo f  z  vt 
H y

E x
t
E y
z
H x

z
t
 vEy
Hx  

Ey
o
• Similarly
E y

Hy 
Ex
o
 o
H x
t
H y
z
E x
 o
z
t
• H and E are in phase and orthogonal
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18

Hy 
Ex
o

Hx  
Ey
o
• The ratio of the magnetic to electric fields strengths is:
E x2  E y2
H x2  H y2

E

H
o


which has units of impedance
Note:
E
E


B o H
1
 o o
c
Volts/ metre

amps/ metre
• and the impedance of free space is:
o

o
4 107
 120  377
1
109
36
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19
Orientation of E and H
• For any medium the intrinsic impedance is denoted by 

Ey
Hx
Ex
Hy

and taking the scalar product
E.H  E x H x  E y H y
 H y H x  H x H y  0
so E and H are mutually orthogonal
• Taking the cross product of E and H we get the direction
of wave propagation

E  H  a z Ex H y  E y H x

 a z H y2
E  H  a zH
 H x2


2
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

A  B  a x Ay B z  Az B y 
a y  Az B x  Ax B z  

a z Ax B y  Ay B x

20
A ‘horizontally’ polarised wave
• Sinusoidal variation of E and H
• E and H in phase and orthogonal
Hy

Hy 
Ex
o
Ex
E H
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21
A block of space containing an EM plane wave
• Every point in 3D space is characterised by
– Ex, Ey, Ez
– Which determine
• Hx, Hy, Hz
• and vice versa
– 3 degrees of freedom

Ex
Hy 
E H
Hy
http://www.cee.hw.ac.uk/~arharve
Hx  

Ex
o

Ey
o
22
An example
application of
Maxwell’s
equations:
The Magnetron
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23
The magnetron
• Locate features:
Lorentz force F=qvB
Poynting vector S
Displacement current, D
Current, J
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24
Power flow of EM radiation
• Intuitively power flows in the direction of
propagation
– in the direction of EH

E  H  a z Ex H y  E y H x

 a zH 2
 az
E2

• What are the units of EH?
– H2=.(Amps/metre)2 =Watts/metre2 (cf I2R)
– E2/=(Volts/metre)2 / =Watts/metre2 (cf I2R)
• Is this really power flow ?
http://www.cee.hw.ac.uk/~arharve
25
Power flow of EM radiation cont.
• Energy stored in the EM field in the thin box is:
uE 
dU  dU E  dU H  u E  u H Adx
 E 2  o H 2 
 Adx
dU  

 2

2


uH 
 E 2 Adx
Hy 
• Power transmitted through the box is dU/dt=dU/(dx/c)....

E 2
2
o H 2
2

Ex
o
dx
Ex
E H
Hy
Area A
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26
Power flow of EM radiation cont.
dU  E 2 Adx
dU
E 2
S

Adx 
Adt Adx c 

E2

o

W/m2
• This is the instantaneous power flow
– Half is contained in the electric component
– Half is contained in the magnetic component
• E varies sinusoidal, so the average value of S is obtained
2
as:
z  vt 
E  E sin
o

S
Eo2 sin2 z  vt 
S
Eo2



RMS
Eo2

Eo2
sin z  vt  
2
2
• S is the Poynting vector and indicates the direction and magnitude of
power flow in the EM field.
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27
Example problem
• The door of a microwave oven is left open
– estimate the peak E and H strengths in the aperture of the door.
– Which plane contains both E and H vectors ?
– What parameters and
equations are required?
•
•
•
•
Power-750 W
Area of aperture - 0.3 x 0.2 m
impedance of free space - 377 
Poynting vector:
S
E2

 H 2
W/m2
http://www.cee.hw.ac.uk/~arharve
28
Solution
Power  SA 
E2

A
 H 2 A
Watts
Power
750
E 
 377
 2,171V/m
A
0.3.0.2
H
E


2170
 5.75A/m
377
B   o H  4  107  5.75  7.2μTesla
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29
• Suppose the microwave source is omnidirectional and displaced
horizontally at a displacement of 100 km. Neglecting the effect of
the ground:
• Is the E-field
a) vertical
b) horizontal
c) radially outwards
d) radially inwards
e) either a) or b)
• Does the Poynting vector point
a) radially outwards
b) radially inwards
c) at right angles to a vector from the observer to the source
• To calculate the strength of the E-field should one
a) Apply the inverse square law to the power generated
b) Apply a 1/r law to the E field generated
c) Employ Coulomb’s
1/r2 law
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30
Field due to a 1 kW omnidirectional generator (cont.)
Power  S 
E 
P
4r 2

103
 
5 2
4 10
 5.97 nW/m2
Power
Power
750
 

377
A
4r 2
4 105
 
http://www.cee.hw.ac.uk/~arharve
2
 1.5 mV/m
31
4.1 Polarisation
• In treating Maxwells equations we referred to
components of E and H along the x,y,z directions
– Ex, Ey, Ez and Hx, Hy, Hz
• For a plane (single frequency) EM wave
– Ez=Hz =0
– the wave can be fully described by either its
E components or its H components
– It is more usual to describe a wave in terms of
its E components
• It is more easily measured
– A wave that has the E-vector in the x-direction
only is said to be polarised in the x direction
Hy 
Hx  

Ex
o

Ey
o
• similarly for the y direction
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32
Polarisation cont..
• Normally the cardinal axes are Earth-referenced
– Refer to horizontally or vertically polarised
– The field oscillates in one plane only and is referred to as linear polarisation
• Generated by simple antennas, some lasers, reflections off dielectrics
– Polarised receivers must be correctly aligned to receive a specific polarisation
A horizontal polarised wave generated by a horizontal dipole and
incident upon horizontal and vertical dipoles
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33
Horizontal and vertical linear polarisation
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34
Linear polarisation
• If both Ex and Ey are present and in phase then components
add linearly to form a wave that is linearly polarised signal
at angle
  tan
1
Ey
Ex

Horizontal
polarisation
Vertical
polarisation
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Co-phased vertical+horizontal
=slant Polarisation
35
Slant linear polarisation
Slant polarised waves generated by co-phased horizontal and vertical
dipoles and incident upon horizontal and vertical dipoles
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36
Circular polarisation
LHC polarisation
NB viewed as approaching wave
Eh  Eoh sin  z  vt 


Ev  Eov sin   z  vt  
2

 Eov cos  z  vt 
RHC polarisation
Eh  Eoh sin  z  vt 


Ev  Eov sin   z  vt  
2

  Eov cos  z  vt 
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37
Circular polarisation
LHC
RHC
LHC & RHC polarised waves generated by quadrature -phased horizontal
and vertical dipoles and incident upon horizontal and vertical dipoles
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38
Elliptical polarisation - an example


Ev  1.5 sin   z  vt  
3

Eh  1.0 sin  z  vt 
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39
Constitutive relations
• permittivity of free space 0=8.85 x 10-12 F/m
• permeability of free space o=4x10-7 H/m
• Normally r (dielectric constant) and r
D  E   r  o E
B  H   r  o H
J  sE
– vary with material
– are frequency dependant
• For non-magnetic materials r ~1 and for Fe is ~200,000
• r is normally a few ~2.25 for glass at optical frequencies
– are normally simple scalars (i.e. for isotropic materials) so
that D and E are parallel and B and H are parallel
• For ferroelectrics and ferromagnetics r and r depend on the
relative orientation of the material and the applied field:
 B x    xx
  
 B y     yx
 B   
 z   zx
 xy
 yy
 zy
 xz  H x 


 yz  H y 

 zz  H z 
At

microwave    j

frequencies: ij 
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 0
 j

0
0 

0 
 o 
40
Constitutive relations cont...
• What is the relationship between  and refractive index for non
magnetic materials ?
– v=c/n is the speed of light in a material of refractive index n
v
1
 o o r

c
n
n  r
– For glass and many plastics at optical frequencies
• n~1.5
• r~2.25
• Impedance is lower within a dielectric

o r
 o r
What happens at the boundary between materials of different
n,r,r ?
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41
Why are boundary conditions important ?
• When a free-space electromagnetic wave is incident
upon a medium secondary waves are
– transmitted wave
– reflected wave
• The transmitted wave is due to the E and H fields at
the boundary as seen from the incident side
• The reflected wave is due to the E and H fields at the
boundary as seen from the transmitted side
• To calculate the transmitted and reflected fields we
need to know the fields at the boundary
– These are determined by the boundary conditions
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42
Boundary Conditions cont.
1,1,s1
2,2,s2
• At a boundary between two media, r,rs are different on either
side.
• An abrupt change in these values changes the characteristic
impedance experienced by propagating waves
• Discontinuities results in partial reflection and transmission of
EM waves
• The characteristics of the reflected and transmitted waves can be
determined from a solution of Maxwells equations along the
boundary
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43
2.3 Boundary conditions
• The tangential component of E is continuous
at a surface of discontinuity
– E1t,= E2t
E1t, H1t
1,1,s1
E2t, H2t
2,2,s2
D1n, B1n
1,1,s1
D2n, B2n
2,2,s2
• Except for a perfect conductor, the
tangential component of H is continuous at a
surface of discontinuity
– H1t,= H2t
• The normal component of D is continuous at
the surface of a discontinuity if there is no
surface charge density. If there is surface
charge density D is discontinuous by an
amount equal to the surface charge density.
–
D1n,= D2n+s
• The normal component of B is continuous at
the surface of discontinuity
– B1n,= B2n
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44
Proof of boundary conditions - continuity of Et
x
y
E x1
E y1
E y3
E y2
E y4
1,1,s1
2,2,s2
E x2
• Integral form of Faraday’s law:
B
.dA
A t
 E.ds  
0
0
0
0
B
y
y
y
y
E y2
 E y1
 Ex1x  E y3
 E y4
 Ex 2 x   z xy
2
2
2
2
t
0
As y  0, Bz t xy  0
That is, the tangential
component of E is
E x1x  E x 2 x  0
continuous

E E


x1
x2
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45
Proof of boundary conditions - continuity of Ht
x
y
H x1
H y1
H y4
1,1,s1
2,2,s2
0
0
H y3
H y2
H x2
• Ampere’s law
0
H y2
0

H.ds 
 D

 J .dA

A  t


y
y
y
y
 Dz

 H y1
 H x1x  H y 3
 H y4
 H x 2 x  
 J z xy
2
2
2
2
 t

0
As y  0, Dz t  J z xy  0 That is, the tangential
H x1x  H x 2 x  0

H x1  H x 2
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component of H is
continuous
46
Proof of boundary conditions - Dn
Dn1
y
x
1,1,s1
2,2,s2
z
Dn 2
• The integral form of Gauss’ law for electrostatics is:
. A   dV
 Dd
V
applied to the box gives
Dn1xy  Dn2 xy  edge   s xy
As dz  0, edge  0 hence
Dn1  Dn 2   s
The change in the normal component of D at a
boundary is equal to the surface charge density
http://www.cee.hw.ac.uk/~arharve
47
Proof of boundary conditions - Dn cont.
Dn1  Dn 2   s
• For an insulator with no static electric charge s=0
Dn1  Dn 2
• For a conductor all charge flows to the surface and for an
infinite, plane surface is uniformly distributed with area
charge density s
In a good conductor, s is large, D=E0 hence if
medium 2 is a good conductor
Dn1   s
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48
Proof of boundary conditions - Bn
• Proof follows same argument as for Dn on page 47,
• The integral form of Gauss’ law for magnetostatics is
 B.dA  0
– there are no isolated magnetic poles
Bn1xy  Bn 2 xy  edge  0

Bn1  Bn 2
The normal component of B at a boundary is
always continuous at a boundary
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49
2.6 Conditions at a perfect conductor
• In a perfect conductor s is infinite
• Practical conductors (copper, aluminium silver) have
very large s and field solutions assuming infinite s can
be accurate enough for many applications
– Finite values of conductivity are important in calculating
Ohmic loss
• For a conducting medium
– J=sE
• infinite s infinite J
• More practically, s is very large, E is very small (0) and J is finite
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50
2.6 Conditions at a perfect conductor
• It will be shown that at high frequencies J is confined to a surface
layer with a depth known as the skin depth
• With increasing frequency and conductivity the skin depth, dx
becomes thinner
Current sheet
dx
dx
Lower frequencies,
smaller s
Higher frequencies,
larger s
• It becomes more appropriate to consider the current density in
terms of current per unit with:
limJdx  J s A/m
dx  0
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51
Conditions at a perfect conductor cont. (page 47 revisited)
x
y
H x1
H y1
H y3
H y2
H y4
1,1,s1
2,2,s2
H x2
• Ampere’s law:
0
H y2
0

H.ds 
 D

 J .dA

A  t
 0

Jszx
y
y
y
y
 D

 H y1
 H x1x  H y 3
 H y4
 H x 2 x   z  J z xy
2
2
2
2
 t

As y  0,
0
Dz t xy  0,
H x1  H x 2  J sz
J z xy  xJ sz
That is, the tangential component of H is
discontinuous by an amount equal to the
surface current density
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52
Conditions at a perfect conductor cont. (page 47 revisited)
cont.
• From Maxwell’s equations:
– If in a conductor E=0 then dE/dT=0
H
– Since
  E  
dt
Hx2=0 (it has no time-varying component and also
cannot be established from zero)
H x1  J sz
The current per unit width, Js, along the surface of a
perfect conductor is equal to the magnetic field just
outside the surface:
• H and J and the surface normal, n, are mutually
perpendicular: J  n  H
s
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53
Summary of Boundary conditions
At a boundary between non-conducting media
n  E1  E 2   0
E t1  E t 2
H t1  H t 2
Dn1  Dn 2

n  H1  H 2   0
n.D1  D 2   0
n.B1  B 2   0
Bn1  Bn 2
At a metallic boundary (large s)
n  E1  E 2   0
n  H1  H 2   0
n.D1  D 2    s
n.B1  B 2   0
At a perfectly conducting boundary
n  E1  0
n  H1  J s
n.D1   s
n.B1  0
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54
2.6.1 The wave equation for a conducting medium
• Revisit page 8 derivation of the wave equation with J0
D
 H  J 
dt
B
E  
dt
As on page 8:
B


    E   
    B   o   H
t
t
t

D 
    E   o  J 

t 
t 
J
 2D
  o
 o 2
t
t
E
 2E
   os
  o 2
t
t
http://www.cee.hw.ac.uk/~arharve
J  sE
D  E
55
2.6.1 The wave equation for a conducting medium
cont.
E
 2E
    E    os
  o 2
t
t
E
 2E
.E   E    os
  o 2
t
t
2
In the absence of sources
.E    0
hence:
E
 2E
 E   os
  o 2
t
t
2
• This is the wave equation for a decaying wave
– to be continued...
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56
Reflection and refraction of plane waves
• At a discontinuity the change in ,  and s results
in partial reflection and transmission of a wave
• For example, consider normal incidence:
Incidentwave  Ei e j t  z 
Reflected wave  Er e j t  z 
• Where Er is a complex number determined by the
boundary conditions
http://www.cee.hw.ac.uk/~arharve
57
Reflection at a perfect conductor
• Tangential E is continuous across the
boundary
– see page 45
• For a perfect conductor E just inside the
surface is zero
– E just outside the conductor must be zero
Ei  E r  0
 Ei   E r
• Amplitude of reflected wave is equal to
amplitude of incident wave, but reversed in
phase
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58
Standing waves
• Resultant wave at a distance -z from the
interface is the sum of the incident and
reflected waves
ET z, t   incidentwave reflected wave
 Ei e j t  z   Er e j t  z 


 Ei e  jz  e jz e jt
 2 jEi sin z e jt
e j  e j
sin 
2j
and if Ei is chosen to be real
ET z , t   Re 2 jEi sin z cost  j sint 
 2 Ei sin z sint
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59
Standing waves cont...
ET z, t   2Ei sin z sint
• Incident and reflected wave combine to produce
a standing wave whose amplitude varies as a
function (sin z) of displacement from the
interface
• Maximum amplitude is twice that of incident
fields
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60
Reflection from a perfect conductor
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61
Reflection from a perfect conductor
• Direction of propagation is given by EH
If the incident wave is polarised along the y axis:
Ei  a y E yi
 H i  a x H xi
then


From page 18
E  H   a y  a x E yi H xi
 a z E yi H xi
That is, a z-directed wave.
For the reflected wave Ε  H  a z E yi H xi and Er  a y E yi
So H r  a x H xi  H i and the magnetic field is
reflected without change in phase
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62
Reflection from a perfect conductor
• Given that cos 
e
j
e
2
 j
derive (using a similar method that used
for ET(z,t) on p59) the form for HT(z,t)
H T z, t   H i e j t  z   H r e j t  z 


 H i e jz  e  jz e jt
 2 H i cos z e jt
As for Ei, Hi is real (they are in phase), therefore
H T z, t   Re2H i cos z cost  j sint   2 H i cos z cost
http://www.cee.hw.ac.uk/~arharve
63
Reflection from a perfect conductor
H T z, t   2H i cos z cost
• Resultant magnetic field strength also has a
standing-wave distribution
• In contrast to E, H has a maximum at the surface
and zeros at (2n+1)/4 from the surface:
resultant wave
free space
resultant wave
E [V/m]
H [A/m]
z [m]
z [m]
silver
z = 0
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free space
silver
z = 0
64
Reflection from a perfect conductor
ET z, t   2 Ei sin z sint
H T z, t   2H i cos z cost
• ET and HT are /2 out of phase( sint  cost   / 2)
• No net power flow as expected
– power flow in +z direction is equal to power flow in - z direction
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65
Reflection by a perfect dielectric
• Reflection by a perfect dielectric (J=sE=0)
– no loss
• Wave is incident normally
– E and H parallel to surface
• There are incident, reflected (in medium 1)and
transmitted waves (in medium 2):
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66
Reflection from a lossless dielectric
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67
Reflection by a lossless dielectric
Ei  1 H i
j


s  j o r
Er  1 H r
Et   2 H t


• Continuity of E and H at boundary requires:
Ei  E r  Et
Hi  Hr  Ht
Which can be combined to give
1
1
H i  H r  Ei  E r   H t 
Et 
1
1
Ei  E r  
1
2
Ei  E r 
1
2
  2 Ei  Er   1 Ei  Er 
 Ei  2  1   Er  2  1 

1
2
E i  E r 
E r  2  1
E 

Ei  2  1
The reflection coefficient
http://www.cee.hw.ac.uk/~arharve
68
Reflection by a lossless dielectric
Ei  E r  Et
Hi  Hr  Ht
• Similarly
E
Et E r  Ei E r
 2  1  2  1
2 2



1 


Ei
Ei
Ei
 2  1  2  1  2  1
E
2 2

 2  1
The transmission coefficient
http://www.cee.hw.ac.uk/~arharve
69
Reflection by a lossless dielectric
• Furthermore:
Hr
E
  r  H
Hi
Ei
H t 1 Et 1 2 2
21



H
H i  2 Ei  2  2  1  2  1
And because =o for all low-loss dielectrics
1   2 n1  n2
Er
E 


 H
Ei
n

n
1   2
1
2
E 
H 
2 1
Er
2n1


Ei
1   2 n1  n2
2 2
1   2

2n2
n1  n2
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70
The End
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71