Subject Name: Electromagnetic Field Theory

Subject Code: EC-5001

Semester: 5th
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ELECTROMAGNETIC THEORY
Unit 4
Circular and elliptic polarization, resolution in terms of linear polarized waves and vice- versa. Plane
waves in lossy medium, low loss dielectric, good conducting and ionized media, complex permittivity,
loss tangent, skin depth, transmission line analogy, boundary conditions at perfect conductor surface,
surface current density Interference of two plane waves traveling at oblique directions.

Introduction
Maxwell s Equations and EM Waves
Disclaimer: These lecture notes are not meant to replace the course textbook. The content may be
incomplete. Some topics may be unclear. These notes are only meant to be a study aid and a
supplement to your own notes. Please report any inaccuracies to the professor.

Maxwell s Equations
Let s summarize all the electromagnetic equations we have learned so far, both integral and differential
forms:
1. Gauss Law for electric field:
 E    E  dA  enc E 
q ρ
S ε0 or ε0
2. Gauss Law for magnetic field:

 B    B  dA  0
S or B  0 (no magnetic monopole charges)

3. Faraday s Law of Induction:

 B
 C
E  ds  
t S
B  dA
or
E  
t

4. Ampere s Law and Maxwell s Law of Induction:

E
 B  ds  µ0ε 0
dt S
E  dA  µ0ienc   B  µ0ε 0  µ0 j
d
C
or t

Derivation of Electromagnetic Wave Equation

Now let s see how we can combine the differential forms of Maxwell s equations to derive a set of
differential equations (wave equations) for the electric and magnetic fields. Let s assume we solve these
equations in a region without any electric charges present (=0) or any currents (j=0).
Start with Maxwell s Law:
E
  B  µ0ε 0
t
Now take the curl of this equation:

     B   µ0ε 0    E 
t
Now it can be shown as a proof in vector calculus that:
     B       B    2B
where   B  0 by Gauss Law for magnetic fields. And since Faraday s Law tells us:
B
E  
t
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then we get:
 2B
 B  µ 0ε 0 2
t
2

which is a second order differential equation for each of the 3 components of the magnetic field. (It is a
wave equation it turns out).
Now we can follow a similar derivation for the electric field starting with Faraday s Law:
B
E  
t
Now take the curl of this equation:

    E     B 
t
Now using the same vector calculus proof:
     E       E    2E
where   E  0 by Gauss Law for electric fields in vacuum. And since Maxwell s Law tells us:
E
  B  µ0ε 0
t
then we get:
 2E
 E  µ 0ε 0 2
t
2

Interestingly this is the exact same differential wave equation as for magnetic fields!

Solution to Wave Equation

The general form of the wave equation is:

 2 2 2  1 2
 2   2 

 x y z  v 2 t 2
2
F F

v  c  3 108 m/s
1
µ 0ε 0
where we have expanded the Laplacian (  ) operator and defined .
2

Solutions to this partial differential equation have the general form:

F  x, t   Fm sin  k  x  ωt  φ 
But let s use complex notation to see how the solution works:
F  x, t   Fm exp i  k  x  ωt  
where we just take the imaginary or real part to get the physical solution.
Plugging into the wave equation yields:
i 2  k x 2  k y 2  k z 2  F  x, t   2 
iω  F  x, t 
1 2

 k2 
ω2
v2

v
ω
k
Thus, the solution to the wave equation that is a consequence of Maxwell s equations in vacuum is a
sinusoidally varying function for both the electric and magnetic fields. It is a traveling wave solution,
which becomes more apparent if we write the solution in this form:

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E  x, t   E m exp ik   x  vt   (or E m sin k   x  vt   )

B  x, t   B m exp ik   x  vt   (or B m sin k   x  vt   )

Here:
Em , Β m  amplitude of electric and magnetic fields
ω angular frequency (rad/s) = π / f
f  cyclic frequency (s -1 , Hz )
T period (s)
k wavenumber (m -1 )  2π / λ
λ wavelength (m)

v
ω 1
= =c phase velocity of wave (m/s)
k ε 0 µ0
Note that Maxwell s Equations predict a unique velocity for the electromagnetic waves, which is just c,
the speed of light. Thus, for Maxwell s equations to be correct in all reference frames we are led to
Einstein s theory of Special Relativity!

The wavenumber k is actually a vector, as is the velocity v. They both point in the direction of the
traveling wave. Notice that as time increases, k  x must increase proportionately to maintain the same
phase.

Now both the electric and magnetic fields have the same form of the wave solution. These solutions
have field directions Em and Bm, and the solution itself fills all space (a little unrealistic!) But the
magnetic and electric fields are also related.
Let s explore some properties of the derived electromagnetic wave:

Consider Gauss Law:

  E  0     Em exp  ik  x  ωt    0
 k  Em  0

Which implies that the electric field direction is perpendicular to the velocity direction,or wave
direction. We can derive the same thing for magnetic field direction:

  B  0     B m exp  ik  x  ωt    0
 k  Bm  0

Now consider Faraday s Law:

B
E  
t
 ik  Em exp  ik  x  ωt   iω B m exp  ik  x  ωt 
 k  Em  ωB m

Note that the last step can only be satisfied if the electric and magnetic waves have exactly the same
time and space form (same phase, velocity, wavenumber).

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Now both Em and Bm are perpendicular to k, and by this equation Bm is perpendicular to Em as well!
 k  Em  B m

This is an important feature of electromagnetic waves. They are transversely polarized, and the electric
and magnetic components are perpendicular to each other.

Moreover, the magnitudes are related:

 c
Em ω
Bm k

Thus, there is only one independent wave solution. If you know the form of the electric field, the
magnetic field is completely specified by the above relations.

Note that there is another way to represent the above equation if we substitute in for the angular
frequency and the wavenumber:


ω 2π f
k  2π / λ 
 fλ c (or v more generally)
Now the wide range of electromagnetic wave phenomena are described by just different choices of the
frequency (or alternatively wavelength, as it is not independent). Lowest in frequency (largest
wavelength) are radio waves (kHz-MHz), increasing to microwaves (GHz), infrared radiation (10 14 Hz),
visible light (~1015Hz), ultraviolet light, x-rays, and finally gamma rays!

Polarisation of plane wave:

The polarisation of a plane wave can be defined as the orientation of the electric field vector as a
function of time at a fixed point in space. For an electromagnetic wave, the specification of the
orientation of the electric field is sufficent as the magnetic field components are related to electric field
vector by the Maxwell's equations.
Let us consider a plane wave travelling in the +z direction. The wave has both E x and Ey components.

..........................................(4.1)
The corresponding magnetic fields are given by,

Depending upon the values of Eox and Eoy we can have several possibilities:
1. If Eoy = 0, then the wave is linearly polarised in the x-direction.
2. If Eoy = 0, then the wave is linearly polarised in the y-direction.
3. If Eox and Eoy are both real (or complex with equal phase), once again we get a linearly polarised wave

with the axis of polarisation inclined at an angle , with respect to the x-axis. This is shown in fig
4.4.

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Fig 4.1 : Linear Polarisation

4. If Eox and Eoy are complex with different phase angles, will not point to a single spatial direction.
This is explained as follows:

Let

Then,

and ....................................(4.2)

To keep the things simple, let us consider a =0 and . Further, let us study the nature of the
electric field on the z =0 plain.

From equation (4.2) we find that,

.....................................(4.3)

and the electric field vector at z = 0 can be written as

.............................................(4.4)

Assuming , the plot of for various values of t is hown in figure 4.5.

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Figure 4.5 : Plot of E(o,t)

From equation (4.47) and figure (4.5) we observe that the tip of the arrow representing electric field
vector traces qn ellipse and the field is said to be elliptically polarised.

Figure 4.6: Polarisation ellipse

The polarisation ellipse shown in figure 4.6 is defined by its axial ratio(M/N, the ratio of semimajor to
semiminor axis), tilt angle (orientation with respect to xaxis) and sense of rotation(i.e., CW or CCW).
Linear polarisation can be treated as a special case of elliptical polarisation, for which the axial ratio is
infinite.

In our example, if , from equation (4.47), the tip of the arrow representing electric field
vector traces out a circle. Such a case is referred to as Circular Polarisation. For circular polarisation the
axial ratio is unity.

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Figure 4.7: Circular Polarisation (RHCP)

Further, the circular polarisation is aside to be right handed circular polarisation (RHCP) if the electric
field vector rotates in the direction of the fingers of the right hand when the thumb points in the
direction of propagation-(same and CCW). If the electric field vector rotates in the opposite direction,
the polarisation is asid to be left hand circular polarisation (LHCP) (same as CW).
In AM radio broadcast, the radiated electromagnetic wave is linearly polarised with the field vertical
to the ground( vertical polarisation) where as TV signals are horizontally polarised waves. FM broadcast
is usually carried out using circularly polarised waves.
In radio communication, different information signals can be transmitted at the same frequency at
orthogonal polarisation ( one signal as vertically polarised other horizontally polarised or one as RHCP
while the other as LHCP) to increase capacity. Otherwise, same signal can be transmitted at orthogonal
polarisation to obtain diversity gain to improve reliability of transmission.

Behaviour of Plane waves at the inteface of two media:

We have considered the propagation of uniform plane waves in an unbounded homogeneous medium.
In practice, the wave will propagate in bounded regions where several values of will be present.
When plane wave travelling in one medium meets a different medium, it is partly reflected and partly
transmitted. In this section, we consider wave reflection and transmission at planar boundary between
two media.

Fig 4.8 : Normal Incidence at a plane boundary

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Case1: Let z = 0 plane represent the interface between two media. Medium 1 is characterised by
and medium 2 is characterized by .
Let the subscripts 'i' denotes incident, 'r' denotes reflected and 't' denotes transmitted field components
respectively.

The incident wave is assumed to be a plane wave polarized along x and travelling in medium 1 along
direction. From equation (4.5) we can write

..................(4.6.a)

......................(4.6.b)

where and .

Because of the presence of the second medium at z =0, the incident wave will undergo partial reflection
and partial transmission.

The reflected wave will travel along in medium 1.

The reflected field components are:

...............................................(4.7a)

.........(4.7b)

The transmitted wave will travel in medium 2 along for which the field components are

............................................(4.8a)

............................................(4.8b)

where and

In medium 1,
and
and in medium 2,
and
Applying boundary conditions at the interface z = 0, i.e., continuity of tangential field components and
noting that incident, reflected and transmitted field components are tangential at the boundary, we can
write

&
From equation 4.49 to 4.51 we get,
................................................................(4.9a)

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..............................................................(4.9b)
Eliminating Eto ,

or,

or,

...............(4.10)
is called the reflection coefficient.
From equation (4.10), we can write

or,

........................................(4.11)
is called the transmission coefficient.
We observe that,

........................................(4.12)
The following may be noted
(i) both and T are dimensionless and may be complex
(ii)
Let us now consider specific cases:
Case I: Normal incidence on a plane conducting boundary
The medium 1 is perfect dielectric and medium 2 is perfectly conducting .

From (4.11) and (4.12)

= -1
and T =0
Hence the wave is not transmitted to medium 2, it gets reflected entirely from the interface to the
medium 1.

& .................................(4.13)
Proceeding in the same manner for the magnetic field in region 1, we can show that,

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...................................................................................(4.14)
The wave in medium 1 thus becomes a standing wave due to the super position of a forward travelling
wave and a backward travelling wave. For a given ' t', both and vary sinusoidally with distance
measured from z = 0. This is shown in figure 4.9.

Figure 4.9: Generation of standing wave

Zeroes of E1(z,t) and Maxima of H1(z,t).

Maxima of E1(z,t) and zeroes of H1(z,t).

.......(4.15)

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Case2: Normal incidence on a plane dielectric boundary

If the medium 2 is not a perfect conductor (i.e. ) partial reflection will result. There will be a
reflected wave in the medium 1 and a transmitted wave in the medium 2.Because of the reflected
wave, standing wave is formed in medium 1.
From equation (4.16(a)) and equation (4.17) we can write

..................(4.18)
Let us consider the scenario when both the media are dissipation less i.e. perfect dielectrics (
)

..................(4.19)
In this case both and become real numbers.

..................(4.20)
From (4.20), we can see that, in medium 1 we have a traveling wave component with amplitude TE io
and a standing wave component with amplitude 2JEio.
The location of the maximum and the minimum of the electric and magnetic field components in the
medium 1from the interface can be found as follows.
The electric field in medium 1 can be written as

..................(4.21)
If i.e. >0
The maximum value of the electric field is

..................(4.22)
and this occurs when

or , n = 0, 1, 2, 3.......................(4.23)

The minimum value of is

.................(4.24)
And this occurs when

or , n = 0, 1, 2, 3.............................(4.25)
For i.e. <0

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The maximum value of is which occurs at the zmin locations and the minimum value of

is which occurs at zmax locations as given by the equations (4.24) and (4.25).

From our discussions so far we observe that can be written as

.................(4.26)
The quantity S is called as the standing wave ratio.
As the range of S is given by
From (4.62), we can write the expression for the magnetic field in medium 1 as

.................(4.27)

From (4.68) we find that will be maximum at locations where is minimum and vice versa. In
medium 2, the transmitted wave propagates in the + z direction.
Oblique Incidence of EM wave at an interface. So far we have discuss the case of normal incidence
where electromagnetic wave traveling in a lossless medium impinges normally at the interface of a
second medium. In this section we shall consider the case of oblique incidence. As before, we consider
two cases
i. When the second medium is a perfect conductor.
ii. When the second medium is a perfect dielectric.
A plane incidence is defined as the plane containing the vector indicating the direction of propagation of
the incident wave and normal to the interface. We study two specific cases when the incident electric
field is perpendicular to the plane of incidence (perpendicular polarization) and is parallel to the
plane of incidence (parallel polarization). For a general case, the incident wave may have arbitrary
polarization but the same can be expressed as a linear combination of these two individual cases.
Oblique Incidence at a plane conducting boundary
i. Perpendicular Polarization
The situation is depicted in figure 4.10.

Figure 6.10: Perpendicular Polarization

As the EM field inside the perfect conductor is zero, the interface reflects the incident plane wave.
and respectively represent the unit vector in the direction of propagation of the incident and
reflected waves, is the angle of incidence and is the angle of reflection.
We find that

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............................(4.28)
Since the incident wave is considered to be perpendicular to the plane of incidence, which for the
present case happens to be xz plane, the electric field has only y-component.
Therefore,

The corresponding magnetic field is given by

...........................(4.29)
Similarly, we can write the reflected waves as

...................................................(4.30)
Since at the interface z=o, the tangential electric field is zero.

............................................(4.31)
Consider in equation (4.72) is satisfied if we have

..................................(4.32)
The condition is Snell's law of reflection.

..................................(4.33)

..................................(4.34)
The total electric field is given by

..................................(4.35)
Similarly, total magnetic field is given by

.............................(4.36)
From eqns (4.37) and (6.38) we observe that
1. Along z direction i.e. normal to the boundary
y component of and x component of maintain standing wave patterns according to and
where . No average power propagates along z as y component of and x
component of are out of phase.

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2. Along x i.e. parallel to the interface

y component of and z component of are in phase (both time and space) and propagate with phase
velocity

.............................(4.39)

The wave propagating along the x direction has its amplitude varying with z and hence constitutes a non
uniform plane wave. Further, only electric field is perpendicular to the direction of propagation (i.e.
x), the magnetic field has component along the direction of propagation. Such waves are called
transverse electric or TE waves.
ii. Parallel Polarization:
In this case also and are given by equations (4.69). Here and have only y component.

Figure 4.11: Parallel Polarization

With reference to fig (4.11), the field components can be written as:

Incident field components:

............................(4.40)
Reflected field components:

............................(4.41)
Since the total tangential electric field component at the interface is zero.

Which leads to and as before.

Substituting these quantities in (4.40) and adding the incident and reflected electric and magnetic field
components the total electric and magnetic fields can be written as

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...........................(4.42)

Once again, we find a standing wave pattern along z for the x and y components of and , while a

non uniform plane wave propagates along x with a phase velocity given by where .
Since, for this propagating wave, magnetic field is in transverse direction, such waves are called
transverse magnetic or TM waves.

Oblique incidence at a plane dielectric interface. We continue our discussion on the behavior of plane
waves at an interface; this time we consider a plane dielectric interface. As earlier, we consider the two
specific cases, namely parallel and perpendicular polarization.

Fig 4.12: Oblique incidence at a plane dielectric interface

For the case of a plane dielectric interface, an incident wave will be reflected partially and transmitted
partially.
In Fig(4.12), corresponds respectively to the angle of incidence, reflection and transmission.
1. Parallel Polarization
As discussed previously, the incident and reflected field components can be written as

..........................(4.44)

..........................(4.45)
In terms of the reflection coefficient

..........................(4.46)
The transmitted filed can be written in terms of the transmission coefficient T

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..........................(4.47)
We can now enforce the continuity of tangential field components at the boundary i.e. z=0

..........................(4.48)
If both and are to be continuous at z=0 for all x , then form the phase matching we have

We find that

..........................(4.49)
Further, from equations (4.49) and (4.48) we have

..........................(4.50)

or ..........................(4.51)

..........................(4.52)
From equation (4.90) we find that there exists specific angle for which = 0 such that

or .........................(4.53)

Further, .........................(4.54)
For non magnetic material
Using this condition

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.........................(4.55)
From equation (4.93), solving for we get

This angle of incidence for which = 0 is called Brewster angle. Since we are dealing with parallel
polarization we represent this angle by so that

2. Perpendicular Polarization

For this case

.........................(4.56)

.........................(4.57)

.........................(4.58)

Using continuity of field components at z=0

.........................(4.59)
As in the previous case

.........................(4.60)
Using these conditions we can write

.........................(4.61)
From equation (4.61) the reflection and transmission coefficients for the perpendicular polarization can
be computed as

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.........................(4.62)
We observe that if = 0 for an angle of incidence

Again

or

or

or .........................(4.63)
We observe if i.e. in this case of non magnetic material Brewster angle does not exist as the
denominator or equation (4.63) becomes zero. Thus for perpendicular polarization in dielectric media,
there is Brewster angle so that can be made equal to zero.
From our previous discussion we observe that for both polarizations

If

For ;

The incidence angle for which i.e. is called the critical angle of incidence. If
the angle of incidence is larger than total internal reflection occurs. For such case an evanescent
wave exists along the interface in the x direction (w.r.t. fig (4.12)) that attenuates exponentially in the
normal i.e. z direction. Such waves are tightly bound to the interface and are called surface waves.
-----------------******-----------

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