Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

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Electrical Engineering (EE) : Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The document Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev is a part of the Electrical Engineering (EE) Course Electromagnetic Theory.
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Propagation of Electromagnetic Waves in a Conducting Medium 

We will consider a plane electromagnetic wave travelling in a linear dielectric medium such as air along the z direction and being incident at a conducting interface. The medium will be taken to be a linear medium. So that one can describe the electrodynamics using only the E and H vectors. We wish to investigate the propagation of the wave in the conducting medium.

As the medium is linear and the propagation takes place in the infinite medium, the vectors Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev  and  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev are still mutually perpendicular. We take the electric field along the x direction, the magnetic field along the y- dirrection and the propagation to take place in the z direction. Further, we will take the conductivity to be finite and the conductor to obey Ohm’s law,  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev Consider the pair of curl equations of Maxwell. 

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Let us take Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev to be respectively in x, y and z direction. We the nhave,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

i.e.,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

and

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

i.e.

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

We take the time variation to be harmonic  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev so that the time derivative is equivalent to a multiplication by  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev. The pair of equations (1) and (2) can then be written as

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

We can solve this pair of coupled equations by taking a derivative of either of the equations with respect to z and substituting the other into it,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Define, a complex constan γ through

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

in terms of which we have,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

In an identical fashion, we get

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Solutions of (3) and (4) are well known and are expressed in terms of hyperbolic functions,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

where A, B, C and D are constants to be determined. If the values of the electric field at z=0 is E0 and that of the magnetic field at z=0 is H0 we have A E0 and C =H0

In order to determine the constants B and D, let us return back to the original first order equations (1) and (2)

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Substituting the solutions for E and H

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

This equation must remain valid for all values of z, which is possible if the coefficients of sinh and cosh terms are separately equated to zero,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The former gives,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

where

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Likewise, we get,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Substituting these, our solutions for the E and H become,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The wave is propagating in the z direction. Let us evaluate the fields when the wave has reached  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

If ℓ is large, we can approximate

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

we then have,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The ratio of the magnitudes of the electric field to magnetic field is defined as the “characteristic impedance” of the wave

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Suppose we have lossless medium, σ=0, i.e. for a perfect conductor, the characteristic impedance is

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

If the medium is vacuum,  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev gives η≈377Ω. The characteristic impedance, as the name suggests, has the dimension of resistance.

In this case,  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Let us look at the full three dimensional version of the propagation in a conductor. Once again, we start with the two curl equations,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Take a curl of both sides of the first equation,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

As there are no charges or currents, we ignore the divergence term and substitute for the curl of H from the second equation,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

We take the propagating solutions to be

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

so that the above equation becomes,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

so that we have, the complex propagation constant to be given by

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

so that

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

k is complex and its real and imaginary parts can be separated by standard algebra,

we have

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Thus the propagation vector β and the attenuation factor α are given by

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The ration  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev determines whether a material is a good conductor or otherwise. Consider a good conductor for which σ>> ω∈.For this case, we have,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The speed of electromagnetic wave is given by

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The electric field amplitude diminishes with distance as  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev The distance to which the field penetrates before its amplitude diminishes be a factor e-1 is known as the “skin depth” , which is given by

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The wave does not penetrate much inside a conductor. Consider electromagnetic wave of frequency 1 MHz for copper which has a conductivity of approximately  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev Substituting these values, one gets the skin depth in Cu to be about 0.067 mm. For comparison, the skin depth in sea water which is conducting because of salinity, is about 25 cm while that for fresh water is nearly 7m. Because of small skin depth in conductors, any current that arises in the metal because of the electromagnetic wave is confined within a thin layer of the surface.

Reflection and Transmission from interface of a conductor 

Consider an electromagnetic wave to be incident normally at the interface between a dielectric and a conductor. As before, we take the media to be linear and assume no charge or current densities to exist anywhere. We then have a continuity of the electric and the magnetic fields at the interface so that 

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The relationships between the magnetic field and the electric field are given by

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

the minus sign in the second relation comes because of the propagation direction having been reversed on reflection.

Solving these, we get,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The magnetic field expressions are given by interchanging η1 and  η2 in the above expressions.

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Let us look at consequence of this. Consider a good conductor such as copper. We can see that ηis a small complex number. For instance, taking the wave frequency to be 1 MHz and substituting conductivity of Cu to be  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev we can calculate ηto be approximately  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRevElectromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev whereas the vacuum impedance η1= 377 Ω. This implies 

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

which shows that a good metal is also a good reflector. On the other hand, if we calculate the transmission coefficient we find it to be substantially reduced, being only about  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

For the transmitted magnetic field, the ratio  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev is approximately +2. Though E is reflected with a change of phase, the magnetic field is reversed in direction but does not undergo a phase change. The continuity of the magnetic field then requires that the transmitted field be twice as large.

Surface Impedance

As we have seen, the electric field is confined to a small depth at the conductor interface known as the skin depth. We define surface impedance as the ratio of the parallel component of electric field that gives rise to a current at the conductor surface,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

where kis the surface current density.

Assuming that the current flows over the skin depth, one can write, for the current density, (assuming no reflection from the back of this depth)

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Since the current density has been taken to decay exponentially, we can extend the integration to infinity and get

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The current density at the surface can be written as  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev For a good conductor, we have,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Thus

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

where the surface resistance Rs and surface reactance Xs are given by 

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The current profile at the interface is as shown.

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Tutorial Assignment

  1. A 2 GHz electromagnetic propagates in a non-magnetic medium having a relative permittivity of 20 and a conductivity of 3.85 S/m. Determine if the material is a good conductor or otherwise. Calculate the phase velocity of the wave, the propagation and attenuation constants, the skin depth and the intrinsic impedance.
  2. An electromagnetic wave with its electric field parallel to the plane of incidence is incident from vacuum onto the surface of a perfect conductor at an angle of incidence θ. Obtain an expression for the total electric and the magnetic field.

Solutions to Tutorial Assignments 

1. One can see that  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev The ratio of conductivity σ ω∈ is 1.73 which says it is neither a good metal nor a good dielectric. The propagation constant β and the attenuation constant α are given by, 

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The intrinsic impedance is

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The phase velocity is  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The skin depth is  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

2. The case of p polarization is shown.

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Let the incident plane be y-z plane. Let us look at the magnetic field. We have, since both the incident and the reflected fields are in the same medium,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Let us write the incident magnetic field as

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The reflected magnetic field is given by

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Since the tangential components of the electric field is continuous, we have,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

As  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev we have  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev and consequently,  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev Thus the total magnetic field can be written as

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The electric field has both y and z components,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The reflected electric field also has both components,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Adding these two the total electric field, has the following components,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Self Assessment Questions

  1. For electromagnetic wave propagation inside a good conductor, show that the electric and the magnetic fields are out of phase by 450 .
  2. A 2 kHz electromagnetic propagates in a non-magnetic medium having a relative permittivity of 20 and a conductivity of 3.85 S/m. Determine if the material is a good conductor or otherwise. Calculate the phase velocity of the wave, the propagation and attenuation constants, the skin depth and the intrinsic impedance.
  3. An electromagnetic wave with its electric field perpendicular to the plane of incidence is incident from vacuum onto the surface of a perfect conductor at an angle of incidence θ. Obtain an expression for the total electric and the magnetic field.

Solutions to Self Assessment Questions

1. From the text, we see that the ratio of electric field to magnetic field is given by

   Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

For a good conductor, we can approximate this by  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

2. One can see that  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev The ratio of conductivity  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev which says it is r a good metal. The propagation constant β and the attenuation constant α are given by,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The intrinsic impedance is

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The phase velocity is  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The skin depth is  Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

3. The direction of electric and magnetic field for s polarization is as shown below.

Let the incident plane be y-z plane. The incident electric field is

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

taken along the x direction and is given by

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The minus sign comes because at z=0, for any y, the tangential component of the electric field must be zero.

The total electric field is along the x direction and is given by the sum of the above,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

which is a travelling wave in the y direction but a standing wave in the z direction. Since the wave propagates in vacuum, we have,

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

The magnetic field has both y and z components.

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev
Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev
Electromagnetic Waves (Part - 2) Electrical Engineering (EE) Notes | EduRev

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