Electromagnetic Fields
Review: Time–Dependent Maxwell’s Equations
∂B ( t )
∇ × E(t) = −
∂t
∂D ( t ) ∇ × H (t) =
+J
∂t
∇ ⋅ D( t ) = ρ
∇ ⋅ B(t ) = 0
D( t ) = ε E ( t )
B(t ) = µ H (t )
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1
Electromagnetic Fields
Electromagnetic quantities:
Vector
quantities
in space
E
H
D
B
J
∂D
∂t
ρ
ε
µ
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Electric Field
Magnetic Field
Electric Flux (Displacement) Density
Magnetic Flux (Induction) Density
Current Density
Displacement Current
Charge Density
Dielectric Permittivity
Magnetic Permeability
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Electromagnetic Fields
In free space:
ε = ε 0 = 8.854 × 10 −12 [As/Vm] or [F/m]
µ = µ 0 = 4 π × 10 −7 [Vs/Am] or [Henry/m]
In a material medium:
ε = ε r ε0
µ = µ r µ0
;
ε r = relative permittivity (dielectric constant)
µ r = relative permeability
If the medium is anisotropic, the relative quantities are tensors:
 ε xx

ε r =  ε yx

 ε zx
ε xy ε xz 

ε yy ε yz 

ε zy ε zz 
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;
 µ xx

µ r =  µ yx

 µ zx
µ xy µ xz 

µ yy µ yz 

µ zy µ zz 
3
Electromagnetic Fields
Electromagnetic fields are completely described by Maxwell’s
equations. The formulation is quite general and is valid also in the
relativistic limit (by contrast, Newton’s equations of motion of
classical mechanics must be corrected when the relativistic limit is
approached).
The complete physical picture is obtained by adding an equation
that relates the fields to the motion of charged particles.
The electromagnetic fields exert a force F on a charge q, according
to the law (Lorentz force):
F(t) = q E(t) +
Electric Force
q v ( t ) × B ( t ) = q  E ( t ) + v ( t ) × B ( t )
Magnetic Force
where v(t) is the velocity of the moving charge.
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Electromagnetic Fields
Review: Electrostatics and Magnetostatics
In the static regime, electromagnetic quantities do not vary as a
function of time. We have two main cases:
ELECTROSTATICS – The electric charges do not change
postion in time. Therefore, ρ, E and D are constant and there is no
magnetic field H, since there is no current density J.
MAGNETOSTATICS – The charge crossing a given crosssection (current) does not vary in time. Therefore, J, H and B are
constant. Although charges are moving, the steady current
maintains a constant charge density
field E is static.
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ρ in space and the electric
5
Electromagnetic Fields
The equations of electrostatics are obtained directly from Maxwell’s
equations, by assuming that ∂/∂t , J, H and B are all zero:
∇× E = 0
∇⋅ D= 0
D=ε E
The electrostatic force is simply
F=qE
We also define the electrostatic potential φ by the relationship
E = −∇φ
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Electromagnetic Fields
The electrostatic potential φ is a scalar function of space. From
vector calculus we know that
∇ × ∇φ = 0
This potential is very convenient for practical applications because
it is a scalar quantity.
The potential automatically satisfies
Maxwell’s curl equation for the electric field, since
∇ × E = −∇ × ∇φ = 0
From a physical point of view, the electrostatic potential provides
an immediate way to express the work W performed by moving a
charge from location a to location b:
b W = − F⋅d l
a
∫
Here, l is the coordinate along the path. The negative sign indicates
that the work is done against the electrical force.
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Electromagnetic Fields
By introducing the electrostatic potential, we obtain
b b
W = − q E ⋅ dl = q ∇φ ⋅ dl = q φ ( b ) − φ ( a ) = q δφ
a
a
∫
∫
[
]
NOTE: The line integral of a gradient does not depend on the path
of integration but only on the potential at the end points of the path.
The electrostatic potential is measured with respect to an arbitrary
reference value. We can assume for most problems that a
convenient reference is a zero potential at an infinite distance. In
the result above for the electrostatic work, we could set a zero
potential at the initial point of the path, so that φ(a)=0. Either
choice of potential reference would give the same potential
difference δφ.
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Electromagnetic Fields
It is quite convenient to express also the divergence equation in
terms of the electrostatic potential. If we assume a uniform material
medium:
∇ ⋅ D = ε ∇ ⋅ E = −ε ∇ ⋅ ∇φ = −ε ∇ 2 φ = ρ
This result yields the well known Poisson equation
ρ
∇ φ=−
ε
2
In the case of ρ = 0, we have the classic Laplace equation
∇ 2φ = 0
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Electromagnetic Fields
If the problem involves a non-uniform medium with varying
dielectric permittivity, a more general form of Poisson equation
must be used
∇ ⋅ ( ε ∇φ ) = −ρ
Another important equation is obtained by integrating the
divergence over a certain volume V
∫V
∇ ⋅ D dV = ∫ ρ dV
V
Gauss theorem allows us to transform the volume integral of the
divergence into a surface integral of the flux
∫V
∇ ⋅ D dV = ∫ D ⋅ dS
S
Component normal
to the surface
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Electromagnetic Fields
The volume integral of the charge density is simply the total charge
Q contained inside the volume
∫V ρ dV = Q
The final result is the integral form of Poisson equation, known as
Gauss law:
∫S
D ⋅ dS = Q
Most electrostatic problems can be solved by direct application of
Poisson equation or of Gauss law.
Analytical solutions are usually possible only for simplified
geometries and charge distributions, and numerical solutions are
necessary for most general problems.
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Electromagnetic Fields
The Gauss law provides a direct way to determine the force
between charges. Let's consider a sphere with radius r surrounding a charge Q1 located at the center. The displacement vector will
be uniform and radially directed, anywhere on the sphere surface,
so that
2 ∫S D ⋅ dS = 4π r D = Q1
Assuming a uniform isotropic medium, we have a radial electric
field with strength
D
Q1
=
E(r) =
ε 4 π ε r2
If a second charge Q2 is placed at distance r from Q1 the mutual
force has strength
Q1 Q2
F = Q2 E ( r ) =
4 π ε r2
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Electromagnetic Fields
The electrostatic potential due to a charge Q can be obtained using
the previous result for the electrostatic work:
b b
Q
φ ( b ) = φ ( a ) − E dr = φ ( a ) −
dr
a
a 4 π ε r2
∫
Q
= φ ( a) +
4πε
∫
1 1
r − r 
 a b
where r indicates the distance of the observation point b from the
charge location, and a is a reference point. If we chose the
reference point a → ∞ , with a reference potential φ(a) = 0, we can
express the potential at distance r from the charge Q as
Q
φ(r) =
4πε r
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Electromagnetic Fields
The potential φ indicates then the work necessary to move an
infinitesimal positive probe charge from distance r (point b) to
infinity (point a) for negative Q, or conversely to move the probe
from infinity to distance r for positive Q (remember that the work is
done against the field). The probe charge should be infinitesimal,
not to perturb the potential established by the charge Q.
The work per unit charge done by the fields to move a probe charge
between two points, is usually called Electromotive Force ( emf ).
Dimensionally, the emf really represents work rather than an actual
force.
The work per unit charge done against the fields represents the
voltage Vba between the two points, so that:
b
emf = E ⋅ d l = −Vba
a
∫
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Electromagnetic Fields
In electrostatics, there is no difference between voltage and
potential. To summarize once again, using formulas, we have
Potential at point “ a ”
Potential at point “ b ”
a φa = − E ⋅ d l
∞
b φb = − E ⋅ d l
∞
∫
∫
Potential difference or Voltage between “ a ” and “ b
b
(φ b − φ a ) = Vba = − a E ⋅ d l = − e.m. f .
∫
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Electromagnetic Fields
In the case of more than one point charge, the separate potentials
due to each charge can be added to obtain the total potential
1
Qi
φ ( x, y, z ) =
∑
4 π ε i ri
If the charge is distributed in space with a density
needs to integrate over the volume as
ρ(x,y,z), one
ρ ( x ', y', z ' )
1
φ ( x, y, z ) =
dV
∫
4πε V
r
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Electromagnetic Fields
In the magnetostatic regime, there are steady currents in the
system under consideration, which generate magnetic fields (we
ignore at this point the case of ferromagnetic media). The full set of
Maxwell's equations is considered (setting ∂/∂t = 0 )
∇× E = 0
∇× H = J
∇⋅ D=ρ
∇⋅B = 0
D=ε E
B=µ H
with the complete Lorentz force
F = q ( E + v × B)
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Electromagnetic Fields
It is desirable to find also for the magnetic field a potential function.
However, note that such a potential cannot be a scalar, as we found
for the electrostatic field, since
H J
Current density is a vector ≠ 0
We define a magnetic vector potential A through the relation
A B H
This definition automatically satisfies the condition of zero
divergence for the induction field, since
∇ ⋅ B = ∇ ⋅ (∇ × A) = 0
The divergence of a curl is always = 0
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Electromagnetic Fields
The vector potential can be introduced in the curl equation for the
magnetic field
1
1
1
2 ∇ × H = ∇ × B = ∇ × (∇ × A ) = ∇ (∇ ⋅ A) − ∇ A = J
µ
µ
µ
However, in order to completely specify the magnetic vector
potential, we need to specify also its divergence. First, we observe
that the definition of the vector potential is not unique since:
∇ × A' = ∇ × ( A + ∇ψ ) = ∇ × A + ∇ × (∇ψ ) = ∇ × A
ψ = Scalar function
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Always = 0
19
Electromagnetic Fields
In the magnetostatic case it is sufficient to specify (in physics
terminology: to choose the gauge)
∇⋅ A= 0
so that
∇ ⋅ A = ∇ ⋅ ( A + ∇ψ ) = ∇ ⋅ A + ∇ ⋅ ∇ψ = ∇ ⋅ A + ∇ 2ψ = 0
We simply need to make sure that the arbitrary function ψ satisfies
2 0
We can then simplify the previous result for the curl equation to
2
A J
the magnetic equivalent of the electrostatic Poisson equation.
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Electromagnetic Fields
The general solution of this vector Laplacian equation is given by
J ( x ', y', z ' )
µ
A ( x, y, z ) =
dV
∫
V
4π
r
which is similar to the formal solution obtained before for the
electrostatic potential for a distributed charge. If the current is
confined to a wire with cross-sectional area S and described by a
curvilinear coordinate l, we can write
I = I = J⋅S
with a final result
dV = S ⋅ dl
µ I dl
A ( x, y, z ) =
4 π ∫l r
(note that the total current I is constant at any wire location).
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Electromagnetic Fields
The solution for the magnetic field obtained from the vector
potential leads to the famous Biot-Savart law:
B 1
 I dl 
I
dl
=
∇×
H = = ∇× A= ∇×
∫
∫

µ µ
r
4π l r  4π l
I  1 I dl × iˆr
=−
∇  × dl =

∫
4 π l r
4 π ∫l r 2
dl
r
I
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ir
(x,y,z)
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Electromagnetic Fields
The magnetic field can also be determined by direct integration of
the curl equation over a surface
∫S
∇ × H ⋅ dS = ∫ J ⋅ dS = I
S
Stoke's theorem can be used to transform the left hand side of the
equation, to obtain the integral form of Ampere's law
∫l
H ⋅ dl = I
In many applications it is useful to determine the magnetic flux
through a given surface. The vector potential can be used to
modify a surface integral into a surface integral, using again
Stoke's theorem
Magnetic Flux= ∫
S
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B ⋅ dS = ∫ ∇ × A ⋅ dS = ∫ A ⋅ dl
S
l
23
Electromagnetic Fields
Review: Time – Varying Fields
In the dynamics case, we can distinguish between two regimes:
Low Frequency (Slowly-Varying Fields) – The displacement
current is negligible in the Maxwell’s equations, since
∂D ( t )
<< J ( t )
∂t
High Frequency (Fast-Varying Fields) – The general set of
Maxwell’s equations must be considered, with no approximations.
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Electromagnetic Fields
In the low frequency regime we use the complete set of Maxwell’s
equations, but the displacement current is omitted
B
E dt
H J
D B 0
D E
B H
F = q ( E + v × B)
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Electromagnetic Fields
The concept of low frequency and slowly-varying phenomena is
relative to the situation at hand. Any disturbance (time-variation) of
the electromagnetic field propagates at the speed of light. If a
length L is the maximum dimension of the system under study, the
maximum propagation time for a disturbance is
Maximum Propagation Time
L
td vp
Maximum Length
Phase velocity of light
We can assume slow-varying fields if the currents are practically
constant during this time period.
For sinusoidal currents, with a period of oscillation T , we have
Wavelength
Period
Frequency
1 t d
T f vp
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and
L 26
Electromagnetic Fields
The electric potential is now by itself insufficient to completely
describe the time-varying electric field, because there is also a
direct dependence on the magnetic field variations. By recalling the
definition of magnetic vector potential, we can derive a relationship
between electric field and electric potential
Time-Varying Fields
∂ B(t )
∂
∇ × E(t) = −
= − ∇ × A( t )
∂t
∂t
∂ A( t ) 

=0
⇒ ∇ ×  E(t ) +

∂t 

∂A ( t )
E(t) +
= −∇φ ( t )
∂t
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Statics
E0
E 27
Electromagnetic Fields
We can also obtain an integral relation between electric field and
magnetic flux, by integrating the curl of the electric field over a
surface
∂B ( t )
∂
∫ ∫S ∇ × E ( t ) ⋅ dS = ∫ ∫S − ∂ t dS = − ∂ t ∫ ∫S B (t ) ⋅ dS
Stoke’s Theorem
Magnetic Flux
Φ(t)
∂ Φ (t)
∫ E ⋅ d l = − ∂ t
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Electromagnetic Fields
In the electrostatic case, we do not need to distinguish between
voltage and potential difference. The voltage between two points is
always defined as
b
Vba = − E ⋅ d l = − e.m. f .
a
∫
but in terms of potential φ we have
Time-Varying Fields
b
∂A Vba ( t ) = ∫  ∇φ +  ⋅ d l
a
∂t
∂ b
= φ b − φ a + ∫ A( t ) ⋅ d l
∂t a
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Statics
b
Vba = − E ⋅ d l
a
∫
= φb − φa
29
Electromagnetic Fields
Note that for time-varying fields the line integral of the magnetic
vector potential between two given points depends on the actual
path of integration. In general:
b
A ( t ) ⋅ d l ≠ A( b, t ) − A ( a, t )
a
∫
Consider now the integral of the electric field along a closed path:
Time-varying fields
∫ E ( t ) ⋅ d l ≠ 0
Statics
∫ E ( t ) ⋅ d l = 0
The closed path could be a metallic wire which confines the current
due to moving electric charge.
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Electromagnetic Fields
The line integral of the electric field gives the work necessary to
move a unit charge along the path of integration, under the
influence of time-varying electric and magnetic fields.
For a closed wire loop at rest, the work necessary to move a unit
charge once around the loop is
W = ∫
Force
Charge
⋅ d l = ∫ E ( t ) ⋅ d l = ∫ ∇ × ( E ( t )) ⋅ d S
∂ B(t ) ∂
=∫ −
⋅ d S = − ∫ B (t )⋅ d S
S
∂t
∂t S
∂
=−
Φ (t)
∂t
Magnetic Flux
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Electromagnetic Fields
As a more general case, consider a wire loop in motion. The
complete Lorentz force must be considered:
W = e. m. f . = ∫
Force
Charge
⋅ d l = ∫ ( E ( t ) + v ( t ) × B ( t )) ⋅ d l
= ∫ (∇ × ( E ( t ) + v ( t ) × B ( t ))) ⋅ d S
 ∂ B(t )
 d = ∫−
+ ∇ × ( v ( t ) × B ( t ))  ⋅ d S = − ∫ B ( t ) ⋅ d S
∂t
dt


dB ( t )
dt
Flux
Φ(t)
If the velocity of motion is constant, note that
∇ × ( v × B ( t )) =
v∇ ⋅ B − B∇ ⋅ v + ( B ⋅ ∇ ) v − ( v ⋅ ∇ ) B = − ( v ⋅ ∇ ) B
0
0
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0
32
Electromagnetic Fields
Electromagnetic Waves
For fast-varying phenomena, the displacement current cannot be
neglected, and the full set of Maxwell’s equations must be used
∂B ( t )
∇× E = −
dt
∂D ( t )
∇× H = J +
∂t
∇⋅ D=ρ
∇⋅B = 0
D=ε E
B=µ H
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F = q ( E + v × B)
33
Electromagnetic Fields
The two curl equations are analogous to the coupled (first order)
equations for voltage and current used in transmission lines. The
solutions of this system of equations are waves. In order to obtain
uncoupled (second order) equations we can operate with the curl
once more. Under the assumption of uniform isotropic medium:
∂ (∇ × B ( t ))
∂
∇ × ∇ × E ( t) = −
= −µ ∇ × H ( t )
∂t
∂t
2
∂ J ( t)
∂ E ( t)
= −µ
− µε
∂t
∂ t2
∂ (∇ × D ( t ))
∂
∇ × ∇ × H ( t) = ∇ × J +
= ∇ × J + ε ∇ × E ( t)
∂t
∂t
2
∂ H ( t)
= ∇ × J − εµ
∂ t2
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Electromagnetic Fields
From vector calculus, we also have
2
∇ × ∇ × E ( t ) = ∇∇ ⋅ E ( t ) − ∇ E ( t )
2
2
∇ × ∇ × H ( t ) = ∇∇ ⋅ H ( t ) − ∇ H ( t ) = −∇ H ( t )
1
∇ ⋅ B( t ) = 0
µ
Finally, we obtain the general wave equations
2
∂ E (t )
∂ J (t )
2
=µ
∇ E ( t ) − ∇∇ ⋅ E ( t ) − µ ε
2
∂t
∂t
2
∂ H (t )
2
= −∇ × J ( t )
∇ H (t ) − µ ε
∂t
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35
Electromagnetic Fields
In a region where the wave solution propagates away from charges
and flowing currents, the wave equations can be simplified
considerably. In such conditions, we have
0 E t / 0
J t 0
and the wave equations assume the familiar form
2
∇ E(t) − µ ε
2
∂ E(t)
=0
∂ t2
2
∂ H (t)
2
=0
∇ H (t ) − µ ε
∂t
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36
Electromagnetic Fields
When currents and charges are involved, the wave equations are
difficult to solve, because of the terms
∇ (∇ ⋅ E ( t ))
and
∇ × J (t )
It is more practical to have equations for the electric potential and
for the magnetic vector potential, which contain linear source
terms dependent on charge and current, as shown below.
We saw earlier that the divergence of the magnetic vector potential
must be specified. The simple choice made in magnetostatics of
zero divergence is not suitable for time-varying fields. Among the
possible choices, it is convenient to adopt the Lorenz gauge
Time-varying fields – Lorenz gauge
∂φ
∇ ⋅ A ( t ) = −µ ε
∂t
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Magnetostatics (d.c.)
∇⋅ A= 0
37
Electromagnetic Fields
∂ A( t )
E (t ) = −
− ∇φ
∂t
Starting from the definitions
B(t ) = ∇ × A
we obtain again the wave equation by applying the curl operation
2
µ∇ × H ( t ) = ∇ × ∇ × A ( t ) = ∇∇ ⋅ A ( t ) − ∇ A ( t ) =
2
∂ A( t )
∂φ
= µ J − εµ
− εµ∇
2
∂t
∂t
= ∇∇ ⋅ A ( t )
With the application of Lorenz gauge
2
∇ A( t ) − ε µ
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2 ∂ A( t )
∂ t2
= −µ J ( t )
38
Electromagnetic Fields
For the electric potential we have
∇ ⋅ D(t ) = ρ
ρ
⇒
∇ ⋅ E(t) =
ε
 ∂ A( t )
 ρ
∇ ⋅ E (t ) = ∇ ⋅  −
− ∇φ  =
∂t

 ε
∂
ρ
2
−∇ φ − ∇ ⋅ A ( t ) =
∂t
ε
After applying the Lorenz gauge once more, we arrive at the
potential wave equation
∂ 2φ
ρ
∇ φ − εµ
=−
2
ε
∂t
2
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39
Electromagnetic Fields
In engineering it is very important to consider time-harmonic fields
with a sinusoidal time-variation. If we assume a steady-state
situation (after all transients have died out) most physical situations
may be investigated by considering one single frequency at a time.
This assumption leads to great simplifications in the algebra. It is
also realistic, because in practical electromagnetics applications
we often have a dominant frequency (carrier) to consider.
The time-harmonic fields have the form
E ( t ) = E0 cos ( ω t + ϕ E )
H ( t ) = H0 cos ( ω t + ϕ H )
We can use the complex phasor representation
{
jϕ E jω t
E ( t ) = Re E0 e
e
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}
{
j ϕ H j ωt
H ( t ) = Re H0 e
e
}
40
Electromagnetic Fields
We define
jϕ E
= phasor of E ( t )
E = E0 e
jϕ H
= phasor of H ( t )
H = H0 e
Maxwell’s equations can be rewritten for phasors, with the timederivatives transformed into linear terms
jω E = phasor of
2
− ω E = phasor of
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∂ E(t)
∂t
2 ∂ E(t)
∂ t2
41
Electromagnetic Fields
In phasor form, Maxwell’s equations become
∇ × E = − jωµ H
∇ × H = J + jωε E
∇⋅D = ρ
∇⋅B = 0
D=εE
B=µH
F = q (E + v × B)
where all electromagnetic quantities are phasors and functions only
of space coordinates.
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42
Electromagnetic Fields
Let’s consider first vacuum as a medium. The wave equations for
phasors become Helmholtz equations
2
∇ E + ω µ0 ε0 E = 0
2
2
∇ H + ω µ0 ε0 H = 0
2
The general solutions for these differential equations are waves
moving in 3-D space. Note, once again, that the two equations are
uncoupled.
This means that each equation contains all the necessary
information for the total electromagnetic field and one only needs to
solve the equation for one field to completely specify the problem.
The other field is obtained with a curl operation by invoking one of
the original Maxwell equations.
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43
Electromagnetic Fields
At this stage we assume that a wave exists, and we do not yet
concern ourselves with the way the wave is generated. So, for the
sake of understanding wave behavior, we can restrict the Helmhlotz
equations to a simple case:
• We assume that the wave solution has an electric field which is
uniform on the { x , y }-plane and has a reference positive
orientation along the x-direction. Then, we verify that this is a
reasonable choice corresponding to an actual solution of the
Helmholtz wave equations. We recall that the Laplacian of a
scalar is a scalar
∇2 f =
∂2 f
∂x
2
+
∂2 f
2
∂y
+
∂2 f
∂ z2
and that the Laplacian of a vector is a vector
2
∇ E = iˆx∇ 2 E x + iˆy∇ 2 E y + iˆz∇ 2 E z
© Amanogawa, 2001 – Digital Maestro Series
44
Electromagnetic Fields
The Helmholtz equation becomes:
2
∂
Ex ˆ
2
2
∇ E + ω µ0 ε0 E =
ix + ω2µ 0 ε 0 ( E x iˆx ) = 0
∂ z2
Only the x-component of the electric field exists (due to the chosen
orientation) and only the z-derivative exists, because the field is
uniform on the { x , y }-plane.
We have now a one-dimensional wave propagation problem
described by the scalar differential equation
∂2 Ex
∂ z2
© Amanogawa, 2001 – Digital Maestro Series
+ ω2 µ 0 ε 0 E x = 0
45
Electromagnetic Fields
This equation has a well known general solution
A exp ( − jβ z ) + B exp ( jβ z )
where the propagation constant is
ω
β = ω µ0 ε0 =
c
The wave that we have assumed is a plane wave and we have
verified that it is a solution of Helmholtz equation. The general
solution above has two possible components
A exp ( − j β z )
B exp ( j β z )
Forward wave, moving along positive
z
Backward wave, moving along negative
z
For the simple wave orientation chosen here, the problem is
mathematically identical to the one solved earlier for voltage
propagation in a homogeneous transmission line.
© Amanogawa, 2001 – Digital Maestro Series
46
Electromagnetic Fields
If a specific electromagnetic wave is established in an infinite
homogeneous medium, moving for instance along the positive
direction, only the forward wave should be considered.
A reflected wave exists when a discontinuity takes place along the
path of the forward wave (that is, the material medium changes
properties, either abruprtly or gradually).
We can also assume that the amplitude of the forward plane wave
solution is given and that it is in general a complex constant fixed
by the conditions that generated the wave
A = E0 e jϕ
We can write at last the phasor electric field describing a simple
forward plane wave solution of Helmholtz equation as:
E x ( z) = E0 e jϕ e− jβ z iˆx
© Amanogawa, 2001 – Digital Maestro Series
47
Electromagnetic Fields
The corresponding time-dependent field is obtained by applying the
inverse phasor transformation
{
}
{
Ex ( z , t ) = Re E x ( z ) e jω t iˆx = Re E0 e jϕ e− jβ z e jω t iˆx
}
= E0 cos ( ω t − β z + ϕ ) iˆx
The phasor magnetic field is obtained directly from the Maxwell
equation for the electric field curl
(
)
jϕ − jβ z ˆ
∇ × E = ∇ × E0 e e
ix = − jωµ 0 H
H=−
(
∇ × E0 e jϕ e− jβ z iˆx
© Amanogawa, 2001 – Digital Maestro Series
jωµ 0
)
48
Electromagnetic Fields
We then develop the curl as
 iˆx
iˆy iˆz 


∂
∂
∂
jϕ − jβ z ˆ

∇ × E0 e e
=
ix = det
 ∂x
∂ y ∂ z


Ex ( z)
0 
 E x ( z) 0
=
(
)
(
) iˆy − (
∂ E0 e jϕ e− jβ z
∂z
= − jβ E0 e jϕ e− jβ z iˆy
© Amanogawa, 2001 – Digital Maestro Series
∂ E0 e jϕ e− jβ z
∂y
) iˆz =
=0
49
Electromagnetic Fields
The final result for the phasor magnetic field is
− j β E0 e jϕ e− jβ z ˆ
H y ( z) = −
iy =
jωµ
ω µ 0ε 0
=
E0 e jϕ e− jβ z iˆy =
ω µ0
ε0
ε0
jϕ − jβ z ˆ
E x ( z ) iˆy
E0 e e
iy =
=
µ0
µ0
We define
µ0
= η0 ≈ 377 Ω =
ε0
© Amanogawa, 2001 – Digital Maestro Series
Intrinsic impedance of vacuum
50
Electromagnetic Fields
We have found that the fields of the electromagnetic wave are
perpendicular to each other, and that they are also perpendicular
(or transverse) to the direction of propagation.
x
E
z
H
y
© Amanogawa, 2001 – Digital Maestro Series
51
Electromagnetic Fields
Electromagnetic power flows with the wave along the direction of
propagation and it is also constant on the phase-planes. The
power density is described by the time-dependent Poynting vector
P ( t ) = E ( t )× H ( t )
The Poynting vector is perpendicular to both field components, and
is parallel to the direction of wave propagation.
When the wave propagates on a general direction, which does not
coincide with one of the cartesian axes, the propagation constant
must be considered to be a vector with amplitude
| | and direction parallel to the Poynting vector.
© Amanogawa, 2001 – Digital Maestro Series
52
Electromagnetic Fields
The condition of mutual orthogonality between the field
components and the Poynting vector is general and it applies to
any plane wave with arbitrary direction of propagation. The mutual
orientation chosen for the reference directions of the fields follows
the right hand rule.
E( x, y, z)
x
y
H ( x, y, z)
z
, P
© Amanogawa, 2001 – Digital Maestro Series
53
Electromagnetic Fields
So far, we have just verified that electromagnetic plane waves are
possible solutions of the Maxwell equations for time-varying fields.
One may wonder at this point if plane waves have practical physical
relevance.
First of all, we should notice that plane waves are mathematically
analogous to the exponential basis functions used in Fourier
analysis. This means that a general wave, with more than one
frequency component, can always be decomposed in terms of
plane waves.
• For periodic signals, we have a discrete set of waves which are
harmonics of the fundamental frequency (analogy with Fourier
series).
• For general signals, we must consider a continuum of
frequencies in order to decompose in terms of elementary plane
waves (analogy with Fourier transform).
© Amanogawa, 2001 – Digital Maestro Series
54
Electromagnetic Fields
From a physical point of view, however, the properties of a plane
wave may be somewhat puzzling.
Assume that a steady-state plane wave is established in an ideal
infinite homogeneous medium. On any plane perpendicular to the
direction of propagation (phase-planes), the electric and magnetic
fields have uniform magnitude and phase.
The electromagnetic power, flowing with a phase-plane of the wave,
is obtained by integrating the Poynting vector, which is also
uniform on each phase-plane. For a plane where the Poynting
vector is non-zero, the total power carried by the wave is infinite
plane
P t plane
E t H t In many practical cases, we approximate an actual wave with a
plane wave on a limited region of space, thus considering an
appropriate finite power.
© Amanogawa, 2001 – Digital Maestro Series
55
Electromagnetic Fields
Review of Boundary Conditions
Consider an electromagnetic field at the boundary between two
materials with different properties. The tangent and the normal
component of the fields must me examined separately, in order to
understand the effects of the boundary.
Medium 1
ε 1 ; µ1
Hn1
Medium 2
ε2; µ2
Hn 2
H1
boundary
© Amanogawa, 2001 – Digital Maestro Series
H2
H t1
Ht 2
56
Electromagnetic Fields
Tangential Magnetic Field
Medium 1
ε 1 ; µ1
boundary
H n3
Medium 2
ε2; µ2
H t1
H n4
Ht2
b
a
y
.
z
x
Ampère’s law for the boundary region in the figure can be written as
H y H x
H
J z j E z
x
y
© Amanogawa, 2001 – Digital Maestro Series
57
Electromagnetic Fields
In terms of finite differences approximation for the derivatives
H n4 Hn 3 Ht 1 Ht 2
J z j E z
b
a
If one lets the boundary region shrink, with
than b,
a
going to zero faster
H n 3 Hn 4
H t 2 H t1 lim ( J z a j E z a a
)
b
a 0
for materials with finite conductivity
H t 2 H t1 0
Tangential components are conserved
for perfect conductors
H t 2 H t1 lim ( J z a) Js
a 0
© Amanogawa, 2001 – Digital Maestro Series
(surface current)
58
Electromagnetic Fields
For a general boundary geometry
nˆ (H t1 H t 2 ) J s
nˆ unit vector normal to the surface
In the case of a perfect conductor, the electromagnetic fields go
immediately to zero inside the material, because the conductivity is
infinite and attenuates instantly the fields. The surface current is
confined to an infinitesimally thin “skin”, and it accounts for the
discontinuity of the tangential magnetic field, which becomes
immediately zero inside the perfect conductor.
For a real medium, with finite conductivity, the fields can penetrate
over a certain distance, and there is a current distributed on a thin,
but not infinitesimal, skin layer. The tangential field components on
the two sides of the interface are the same. Nonetheless, the
perfect conductor is often a good approximation for a real metal.
© Amanogawa, 2001 – Digital Maestro Series
59
Electromagnetic Fields
Tangential Electric Field
Medium 1
ε 1 ; µ1
boundary
E n3
Medium 2
ε2; µ2
E t1
Et2
b
E n4
a
y
.
z
x
Faraday’s law for the same boundary region can be written as
E y E x
E
j H z
x
y
© Amanogawa, 2001 – Digital Maestro Series
60
Electromagnetic Fields
In terms of finite differences approximation for the derivatives
E n4 En 3 Et 1 Et 2
j H z
b
a
If one lets the boundary region shrink, with
than b,
a
going to zero faster
E n 3 En 4
E t 2 E t1 lim ( j H z a a
)
b
a 0
E t 2 E t1 0
Tangential components are conserved
For a general boundary geometry
nˆ (E t1 E t 2 ) 0
© Amanogawa, 2001 – Digital Maestro Series
61
Electromagnetic Fields
Normal components
D n1
Medium 1
ε 1 ; µ1
ρs +
boundary
Medium 2
ε2; µ2
+ +
+
Bn1
w
+ +
D n2
Bn 2
Area
y
.
z
x
Consider a small box that encloses a certain area of the interface
with
s interface charge density
© Amanogawa, 2001 – Digital Maestro Series
62
Electromagnetic Fields
Integrate the divergence of the fields over the volume of the box:
D dr Volume
dr
Volume
theorem
D nˆ ds Flux of D out of the box
B dr 0
Divergence
Surface
Volume
theorem
B nˆ ds Flux of B out of the box
Divergence
Surface
© Amanogawa, 2001 – Digital Maestro Series
63
Electromagnetic Fields
If the thickness of the box tends to zero and the charge density is
assumed to be uniform over the area, we have the following fluxes
D-Flux out of box = Area (D1 n D2n ) = Total interface charge = Area s
B-Flux out of box = Area (B1 n B2 n ) 0
The resulting boundary conditions are
D1n D 2n s
B1n B 2n 0
The discontinuity in the normal component of the displacement
field D is equal to the density of surface charge.
The normal components of the magnetic induction field B are
continuous across the interface.
© Amanogawa, 2001 – Digital Maestro Series
64
Electromagnetic Fields
For isotropic and uniform values of ε and µ in the two media
D n1 Dn 2 1En1 2 En 2 s
Bn1 Bn 2 1Hn1 2 Hn 2 0
Even when the interface charge is zero, the normal components of
the electric field are discontinuous at the interface, if there is a
change of dielectric constant .
The normal components of the magnetic field have a similar
discontinuity at the interface due to the change in the magnetic
permeability. In many practical situations, the two media may have
the same permeability as vacuum, µ0, and in such cases the normal
component of the magnetic field is conserved across the interface.
© Amanogawa, 2001 – Digital Maestro Series
65
Electromagnetic Fields
SUMMARY
If medium 2 is
perfect conductor
H t1
Ht2
E t1
Et2
H n1
H n2
E n1
E n2
ε1, µ1
ε2, µ2
ε1, µ1
ε2, µ2
ε1, µ1
ε2, µ2
ε1, µ1
ε2, µ2
© Amanogawa, 2001 – Digital Maestro Series
H t1 H t 2
E t1 E t 2
1H n1 2 Hn 2
1E n1 2 En 2 + s
nˆ H t1 J s
Ht2 0
E t1 0
Et2 0
H n1 0
H n2 0
E n1 s 1
E n2 0
66
Electromagnetic Fields
Examples:
An infinite current sheet generates a plane wave (free space on
both sides)
x
Js
-z
+z
y
H
Js ( t ) Jso cos( t ) iˆx
Phasor J s Jso iˆx
The E.M. field is transmitted on both sides of the infinitesimally thin
sheet of current.
© Amanogawa, 2001 – Digital Maestro Series
67
Electromagnetic Fields
BOUNDARY CONDITIONS
nˆ (H t1 H t 2 ) J s
H t1 H t 2 Jso iˆx
E t1 E t 2
E t1 0 H t 1
Symmetry H t1 H t 2
Jso
Jso
H1 H2 2
2
© Amanogawa, 2001 – Digital Maestro Series
68
Electromagnetic Fields
A semi-infinite perfect conductor medium in contact with free space
has uniform surface current and generates a plane wave
x
Free Space
Perfect
Conductor
Js
-z
+z
y
H
J s J so cos( t ) iˆx
The E.M. field is zero inside the perfect conductor. The wave is only
transmitted into free space.
© Amanogawa, 2001 – Digital Maestro Series
69
Electromagnetic Fields
BOUNDARY CONDITIONS
nˆ (H t1 H t 2 ) J s
H t1 H t 2 H t1 0 Jso iˆx
Et2 0
Asymmetry H t1 H t 2
H t1 Jso
Ht2 0
© Amanogawa, 2001 – Digital Maestro Series
70
Electromagnetic Fields
Doppler Effect
E
zo
H
z
Observer
z zo vo t
An observer moves along the direction of propagation of the
electromagnetic wave, with constant velocity vo .
Because of its movement, the observer will detect phase planes of
the wave at a different rate than in stationary position, as
obs t ( zo vo t ) t vo t zo
© Amanogawa, 2001 – Digital Maestro Series
71
Electromagnetic Fields
The observer detects an angular frequency
vo
obs vo vo (1 )
vp
vp
vo
fobs f (1 )
vp
The frequency deviation (Doppler shift) is
v0
f f
vp
If the observer moves on a different direction, one has to take into
account the projection of the velocity along the wave propagation
direction. The Doppler shift becomes
vo i p
f f vp
© Amanogawa, 2001 – Digital Maestro Series
72
Electromagnetic Fields
zo
Stationary observer
z
E
Plane wave
source
H
z zo vs t
If the observer is stationary and the electromagnetic wave moves
with uniform velocity, the true and the measured frequencies are
related through the phase relationship
obs
vs t
t obs t obs z obs t vs t obs ( t )
vp
vp
Frame of reference of
moving E.M. source
Frame of reference of
fixed observer
© Amanogawa, 2001 – Digital Maestro Series
73
Electromagnetic Fields
vs
obs (1 ) obs vs
vp
1
vp
The frequency recorded by the observer is
fobs f
vs
1
vp
In most practical cases, the velocity of the source is much smaller
than the phase velocity of the wave
vs v p © Amanogawa, 2001 – Digital Maestro Series
vs
fobs f (1 )
vp
74
Electromagnetic Fields
Electromagnetic Waves in Material Media
In a material medium free charges may be present, which generate
a current under the influence of the wave electric field. The current
Jc is related to the electric field E through the conductivity σ as
Jc E
The material may also have specific relative values of dielectric
permittivity and magnetic permeability
ε = εr εo
© Amanogawa, 2001 – Digital Maestro Series
µ = µr µo
75
Electromagnetic Fields
Maxwell’s equations become
∇ × E = − jωµ H
σ ∇ × H = σ E + jωε E = jω(ε − j )E
ω
In phasor notation, it is as if the material conductivity introduces an
imaginary part for the dielectric constant ε. The wave equation for
the phasor electric field is given by
2
∇ × ∇ × E = ∇∇ ⋅ E − ∇ E = − jωµ ∇ × H
= − jωµ(J c + jωε E)
2
⇒ ∇ E = jωµ(σ + jωε)E
We have assumed that the net charge density is zero, even if a
conductivity is present, so that the electric field divergence is zero.
© Amanogawa, 2001 – Digital Maestro Series
76
Electromagnetic Fields
In 1-D the wave equation is simply
2
∂ Ex
2
= jωµ(σ + jωε )E x = γ E x
∂ z2
with general solution
E x ( z) = A exp(−γ z) + B exp( γ z)
σ + jωε
1 ∂E x
=
H y ( z) = −
( A exp(−γ z) − B exp( γ z) )
jωµ ∂ z
jωµ
1
= ( A exp(−γ z) − B exp( γ z) )
η
These resemble the voltage and current solutions in lossy
transmission lines.
© Amanogawa, 2001 – Digital Maestro Series
77
Electromagnetic Fields
The intrinsic impedance of the medium is defined as
η = η e jτ =
jωµ
σ + jωε
For the propagation constant, one can obtain the real and imaginary
parts as
γ=
jωµ(σ + jωε ) = α + jβ
2


ω µε 
σ
 
α=
1 +   − 1

2 
 ωε 


2


ω µε 
σ
 
1 +   + 1
β=

2 
 ωε 


© Amanogawa, 2001 – Digital Maestro Series
1/ 2
1/ 2
78
Electromagnetic Fields
Phase velocity and wavelength are now functions of frequency
vp =
λ=
ω
β
2π
β


µε 

2
=
=


µε 

2
f
1+
( )
1+

+ 1

2
σ
ωε
( )
σ
ωε
2
−1 / 2

+ 1

−1 / 2
The intrinsic impedance of the medium is complex as long as the
conductivity is not zero.
The phase angle of the intrinsic
impedance indicates that electric field and magnetic field are out of
phase. Considering only the forward wave solutions
E x ( z) = A exp( −γ z) = A exp( −α z) exp( − jβ z)
H y ( z) =
1
η
A exp( −γ z − jτ ) =
© Amanogawa, 2001 – Digital Maestro Series
1
η
A exp( −α z) exp( − jβ z − jτ )
79
Electromagnetic Fields
In time-dependent form
Ex ( z, t ) = Re { A exp( jθ ) exp( −α z) exp( − jβ z) exp( jω t)}
= A exp( −α z) cos(ω t − β z + θ)
H y ( z, t ) =
=
1
η
1
η
{
}
Re A exp( jθ ) exp( −α z) exp( − jβ z − jτ ) exp( jω t)
A exp( −α z) cos(ω t − β z + θ − τ )
where the integration constant has been assumed to be in general a
complex quantity as
A = A exp( jθ)
© Amanogawa, 2001 – Digital Maestro Series
80
Electromagnetic Fields
Classification of materials
Perfect dielectrics - For these materials σ = 0
Propagation constant
β = ω ε rε o µ rµ o
α=0
Medium Impedance
µ rµ o
jωµ
η=
=
ε rε o
jωε
© Amanogawa, 2001 – Digital Maestro Series
Phase velocity
ω
1
vp = =
β
µ r µ oε r ε o
Wavelength
2π v p
1
λ=
=
=
β
f
f µ r µ oε r ε o
81
Electromagnetic Fields
Imperfect dielectrics – For these materials σ ≠ 0 but (σ/ωε)<<1
γ=
σ
jωµ(σ + jωε) = jω µε 1 − j
ωε
σ µ
≈
+ jω µε + 2 ε
σ µ
α≈
2 ε
1
ω
vp = ≈
β
µε
jωµ
η=
=
σ + jωε
© Amanogawa, 2001 – Digital Maestro Series
β ≈ ω µε
2π
1
λ=
≈
β
f µε
σ 
jωµ 
1 − j 
ωε 
jωε 
−1
2
µ
≈
ε
82
Electromagnetic Fields
If (σ/ωε)<<1, the errors made in the approximations for α, β, vp
and λ are very small, since only terms of order (σ/ωε) or higher
appear in the expansions. The error is slightly higher fo the
2
medium impedance η since the expansion contains a term of order
(σ/ωε).
The simple rule of thumb is that approximations for imperfect
dielectric can be applied when
σ
≤ 0.1
ωε
When the condition above is verified, the imperfect dielectric
behaves in all respects like a perfect dielectric, except for an
attenuation term in the fields.
The quantity σ/ωε is called Loss Tangent.
© Amanogawa, 2001 – Digital Maestro Series
83
Electromagnetic Fields
Good conductors – For these materials σ ≠ 0 but (σ/ωε)>>1
γ=
jωµ (σ + jωε ) ≈
jωµσ = ωµσ j
π
1 
 1
= ωµσ exp( j ) = ωµσ
+ j
= πf µσ (1 + j )
 2
4
2
α ≈ πf µσ
β ≈ πf µσ
ω
4 πf
vp = ≈
β
µσ
jωµ
η=
≈
σ + jωε
2π
λ=
≈
β
4π
f µσ
ωµ
π
jωµ
=
exp( j )
σ
σ
4
ωµ  1
π fµ
1 
=
+ j
=
(1 + j )
σ  2
σ
2
© Amanogawa, 2001 – Digital Maestro Series
84
Electromagnetic Fields
The simple rule of thumb is that approximations for good conductor
can be applied when
σ
≥ 10
ωε
Note that for a good conductor the attenuation constant α and the
propagation constant β are approximately equal.
The medium impedance η has nearly equal real and imaginary
parts, therefore its phase angle is approximately 45°.
This means that in a good conductor the electric and magnetic
fields have always a phase difference τ = 45° = π /4.
© Amanogawa, 2001 – Digital Maestro Series
85
Electromagnetic Fields
Also, in a good conductor the fields attenuate very rapidly. The
distance over which fields are attenuated by a factor exp(−1.0) is
1
1
=δ=
= Skin depth
α
π f µσ
A typical good conductor is copper, which has the following
parameters:
σ = 5.80 × 10 7 [S/m]
ε ≈ εo
µ ≈ µo
© Amanogawa, 2001 – Digital Maestro Series
86
Electromagnetic Fields
Copper remains a good conductor at extremely high frequencies.
Another good conductor example is sea water at relatively low
frequencies
σ ≈ 4.0 [S/m]
ε ≈ 80ε o
µ ≈ µo
At a frequency of 25 kHz
σ
≈ 36, 000
ωε
© Amanogawa, 2001 – Digital Maestro Series
87
Electromagnetic Fields
Perfect conductor - For this ideal material σ →
∞
For this material, the attenuation is also infinite and the skin depth
goes to zero. This means that the electromagnetic field must go to
zero below the perfect conductor surface.
General medium - When a material is not covered by one of the limit
cases, the complete formulation must be used. We can classify a
material for which the conditions (σ/ωε)<<1 or (σ/ωε)>>10
invalid as a general medium.
are
The simple rule of thumb for general medium is
σ
> 0.1
10 >
ωε
© Amanogawa, 2001 – Digital Maestro Series
88
Electromagnetic Fields
Power Flow in Electromagnetic Waves
The time-dependent power flow density of an electromagnetic wave
is given by the instantaneous Poynting vector
P ( t ) = E ( t )× H ( t )
For time-varying fields it is important to consider the time-average
power flow density
1 T
1 T
P( t ) = ∫ P( t ) dt = ∫ E( t ) × H ( t ) dt
T 0
T 0
where T is the period of observation.
© Amanogawa, 2001 – Digital Maestro Series
89
Electromagnetic Fields
Consider time-harmonic fields represented in terms of their phasors
E( t ) Re E exp( j t ) Re{E} cos t Im{E} sin t
H ( t ) Re H exp( j t ) Re{H} cos t Im{H} sin t
The time-dependent Poynting vector can be expressed as the sum
of the cross-products of the components
E( t ) × H ( t ) = Re{E} × Re{H} cos2 ωt
+ Im{E} × Im{H} sin 2 ωt
− ( Re{E} × Im{H} + Im{E} × Re{H} ) cos ωt sin ωt
(Note that:
cos ωt sin ωt = 1 sin 2ωt )
2
© Amanogawa, 2001 – Digital Maestro Series
90
Electromagnetic Fields
The time-average power flow density can be obtained by integrating
the previous result over a period of oscillation T . The pre-factors
containing field phasors do not depend on time, therefore we have
to solve for the following integrals:
T
1 T 2
1  t sin 2ωt 
1
=
cos ωt dt =  +
∫

4ω  0 2
T 0
T 2
T
1 T 2
1  t sin 2ωt 
1
=
sin ωt dt =  −
∫

T 0
T 2
4ω  0 2
2
T
1 T
1  sin ωt 
=0
cos ωt ⋅ sin ωt dt = 

∫
T 0
T  2ω 
0
© Amanogawa, 2001 – Digital Maestro Series
91
Electromagnetic Fields
The final result for the time-average power flow density is given by
1 T
P( t ) = ∫ E( t ) × H ( t ) dt
T 0
1
= (Re{E} × Re{H} + Im{E} × Im{H} )
2
Now, consider the following cross product of phasor vectors
*
E × H = Re{E} × Re{H} + Im{E} × Im{H}
+ j ( Im{E} × Re{H} − Re{E} × Im{H} )
© Amanogawa, 2001 – Digital Maestro Series
92
Electromagnetic Fields
By combining the previous results, one can obtain the following
time average rule
{
*
1 T
1
P( t ) = ∫ E( t ) × H ( t ) dt = Re E × H
T 0
2
}
We also call complex Poynting vector the quantity
1 *
P = E×H
2
NOTE: the complex Poynting vector is not the phasor of the timedependent power nor that of the time-average power density!
P( t ) = Re {P}
( don't try P( t ) = Re {P exp( jωt )} )
Phasor notation cannot be applied to the product of two timeharmonic functions (e.g., P( t )), even if they have same frequency.
© Amanogawa, 2001 – Digital Maestro Series
93
Electromagnetic Fields
Consider a 1-D electro-magnetic wave moving along the z-direction,
with a specified electric field amplitude Eo
E x ( z) = Eo exp(−αz) exp( − jβz)
Eo
H y ( z) =
exp( −αz) exp(− jβz) exp( − jτ )
η
The time-average power flow density is
*
*
1
1 
E

−αz − jβ z o −αz jβ z jτ 
P( t ) = Re E × H = Re  Eo e e
e e e 
η
2
2 

{
}
−2 α z
−2 α z
e
1
1
2e
2
= Eo
Re e jτ = Eo
cos τ
η
η
2
2
{ }
Power in a lossy medium decays as exp(-2α z)!
© Amanogawa, 2001 – Digital Maestro Series
94
Electromagnetic Fields
Consider the same wave, with a specified amplitude for the
magnetic field
H y ( z) = Ho exp( −αz) exp(− jβz)
E x ( z) = η Ho exp(−αz) exp( − jβz) exp( jτ )
The time-average power flow density is expressed as
{
1
P( t ) = Re η Ho e−αz e− jβ z Ho* e−αz e jβ z e jτ
2
1
2 −2 α z
= η Ho e
cos τ
2
}
If α is the attenuation constant for the electromagnetic fields
⇒ 2α is the attenuation constant for power flow.
© Amanogawa, 2001 – Digital Maestro Series
95
Electromagnetic Fields
If the wave is generated by an infinitesimally thin sheet of uniform
current Jso (embedded in an infinite material with conductivity σ)
we have for propagation along the positive z-direction (normal to
the plane of the current sheet):I
Jso
Jso
Ho =
Eo = η
2
2
2
Jso
η e−2αz cos τ
P( t ) =
8
For this ideal case, an identical wave exists, propagating along the
negative z-direction and carrying the same amount of power.
© Amanogawa, 2001 – Digital Maestro Series
96
Electromagnetic Fields
Poynting Theorem
Consider the divergence of the time-dependent power flow density
∇ ⋅ P ( t ) = ∇ ⋅ ( E( t ) × H ( t ) ) = H ( t ) ⋅ ∇ × E ( t ) − E( t ) ⋅ ∇ × H ( t )
The curls can be expressed by using Maxwell’s equations
∂H
∂E
− σE ( t ) ⋅ E ( t ) − ε E( t ) ⋅
∇ ⋅ P( t ) = −µ H ( t ) ⋅
∂t
∂t
∂ 1
∂ 1

2
2 
= − σE ( t ) −  ε E ( t )  −  µ H 2 ( t ) 
∂t  2
∂t  2


Density of
dissipated
power
Rate of change
of stored electric
energy density
Rate of change
of stored magnetic
energy density
This is the differential form of Poynting Theorem.
© Amanogawa, 2001 – Digital Maestro Series
97
Electromagnetic Fields
Now, integrate the divergence of the time-dependent power over a
specified volume V to obtain the integral form of Poynting theorem
∫ ∇ ⋅ P( t) dV = ∫∫ P (t ) ⋅ ds = Power Flux through S
V
S
= −∫
V
∂
σE ( t ) dV −
∂t
2
Power dissipated
in volume
© Amanogawa, 2001 – Digital Maestro Series
∫
V
1
∂
2
ε E ( t ) dV −
∂t
2
Rate of change
of electric energy
stored in volume
∫
V
1
µ H 2 (t ) dV
2
Rate of change
of magnetic energy
stored in volume
98
Electromagnetic Fields
Typical applications
Pin ( t )
α=?
Pout ( t )
1 m2
L
Pout ( t ) = Pin ( t ) exp( −2αL)
1  Pout ( t ) 
⇒α=−
ln  
2 L  Pin ( t ) 
© Amanogawa, 2001 – Digital Maestro Series
 Watts 


2
 m 
 Nepers 
 m 
99
Electromagnetic Fields
Example:
 Watts   Watts 
Pin ( t ) = 30 
 ; Pout ( t ) = 5 
 ; L = 20 m
2
2
 m 
 m 
 Nepers 
⇒ α = 0.0448 
 m 
Pay attention to the logarithms:
 Pout ( t ) 

ln   = − ln 
 Pin ( t ) 

© Amanogawa, 2001 – Digital Maestro Series
Pin ( t ) 

Pout ( t ) 
100
Electromagnetic Fields
SURFACE A
SURFACE B
Pin ( t ) Power IN
A
Pout ( t ) Power OUT
B
Power dissipated
between A and B?
L
Area = Area(A) = Area(B)
Power IN = ∫∫ P ( t ) A dS = P ( t ) A ⋅ Area
A
Power OUT =
∫∫
B
P ( t ) B dS = P ( t ) B ⋅ Area
P ( t ) B = P ( t ) A exp( −2αL )
Power dissipated = Power IN − Power OUT
© Amanogawa, 2001 – Digital Maestro Series
101
Electromagnetic Fields
Example
2
Area = 5 m ;
L = 1.0 cm;
f = 1.0 GHz; Eo = 10 V/m
ε = ε o ; µ = µ o ; σ = 0.45755 S/m
σ
⇒
= 8.2244637 General Lossy medium
ωε
η = 130.88∠0.725rad = 130.88∠ 41.534
α = 40.0 Ne/m;
Pin ( t) = 0.286 W/m2 ;
Pout ( t ) B = Pin ( t ) A exp(−2αL) = 0.12845 W/m2 ;
Power IN = Area ⋅ Pin ( t ) = 1.43 W
Power OUT = Area ⋅ P ( t ) B = 0.6423 W
Power dissipated = Power IN − Power OUT = 0.7876 W
© Amanogawa, 2001 – Digital Maestro Series
102