Chapter 5

Electromagnetic Fields and Waves

The main aim of this chapter is to provide overall concepts of EM waves and their characteristics. They include:

• applications of EM waves
• wave equations and solutions
• propagation characteristics
• waves in conductors and dielectrics
• polarisation
• normal and oblique incidence of EM waves
• reflection and transmission coefficients
• Brewster angle and total internal reflection
• Surface impedance and Poynting theorem
• solved problems, points/formulae to remember, objective and multiple choice questions and exercise problems.

5.1 INTRODUCTION

A wave means a recurring function of time at a point.

Definition of wave    It is defined as a physical phenomenon. In its reoccurrence, there is a time delay which is proportional to the space separation between two adjacent locations.

In general, wave is a carrier of energy or information and is a function of time as well as space.

As far as we are concerned, a wave means Electromagnetic wave (or simply EM wave). Maxwell predicted the existence of EM waves and established it through his well-known Maxwell’s equations. The same EM waves were investigated by Heinrich Hertz. Hertz conducted several experiments which could generate and detect EM waves. These radio waves are called Hertzian waves.

Examples radio, radar beams and TV signals.

5.2 APPLICATIONS OF EM WAVES

They have a wide range of applications in all types of communications like police radio, television, satellite, ionospheric, tropospheric, wireless, cellular, mobile communications and so on and in all types of radars like Doppler radar, MTI radar, speed trap radar, airport surveillance radar, weather forecasting radar, remote sensing radar, ground mapping radar, IFF radar, astronomy radar, fire control radar and so on. EM waves are also used in radiation therapy, microwave ovens and others.

Advantage in using EM waves for communication purposes is that the medium between the transmitter and receiver requires no maintenance. This is because free space is the best medium for EM wave propagation.

5.3 WAVE EQUATIONS IN FREE SPACE

Wave equations in free space are given by

 

where
and

Proof    Free space is characterised by ∈_r = 1, μ_r = 1 or ∈ = ∈_O, μ = μ₀, and σ = 0, ρ_v 0 and J = 0. Due to these characteristics of free space, Maxwell’s second equation becomes,

 

                    ∇ × E = −Ḃ

                            = −μ₀ Ḣ

 

[as B = μ₀ H]

Taking curl on both sides, we get

 

Using standard vector identity, LHS is given by

 

 

and for the first Maxwell’s equation, RHS is

 

                  −μ₀ ∇× Ḣ = −μ₀ = −μ₀ ∇₀ Ë

            ∇∇.E − ∇² E = −μ₀ ∈₀ Ë

But         ∇.D = ∇.∈₀ E = ∈₀ ∇.E = 0

 

Hence proved.

Now consider the first Maxwell’s equation

 

Taking curl on both sides

 

    ∇ × ∇ × H = ∈₀ ∇ × Ė

∇∇.H − ∇² H = ∈₀ (−) = −μ₀ ∈₀ Ḧ

               ∇² H = μ₀ ∈₀ Ḧ

 

[as ∇.H = 0]

Hence proved.

5.4 WAVE EQUATIONS FOR A CONDUCTING MEDIUM

Wave equations for a conducting medium (ρ_v = 0, σ ≠ 0 J ≠ 0) are given by

 

	∇2 E = μ ∈ Ë + μσĖ
and	∇2 H = μ ∈ Ḧ + μσḢ

Proof    Consider the second Maxwell’s equation

 

Take curl on both sides

 

 

[as ∇.E = 0]

Hence proved.

Similarly, consider the first Maxwell’s equation

 

Take curl on both sides

 

 

[as ∇.H = 0]

Hence proved.

5.5 UNIFORM PLANE WAVE EQUATION

It is defined as a wave whose electric and magnetic fields have constant amplitude over the equiphase surfaces. These waves exist only in free space at an infinite distance from the source.

A uniform plane wave propagating in x-direction has no x-components of E and H, that is, E_x = 0, H_x = 0.

The electric and magnetic fields of an EM wave are always perpendicular to each other. A typical wave is shown in Fig. 5.1.

E_x = 0 and H_x = 0 for a uniform plane wave

Proof    The plane wave equation in free space is given by

 

thata is,

As per the definition of uniform plane wave,

 

	E ≠ f (y)
and	E ≠ f (z)

Hence the wave equation becomes

that is,

Equating the respective components on both sides, we get

Also we have

 

	∇.D = 0	[as ρv = 0]
or	∇.∈0 E = 0
that is,	∇.E = 0

So,         

As

Substituting Equation (5.2) in (5.1), we get

This means that E_x should have one of the following solutions.

1. E_x = 0
2. E_x = a constant with time
3. E_x increases uniformly with time, that is, E_x = K_t where K is constant.

If E_x = a constant and E_x = Kt, it will not be a part of wave motion. Therefore,

 

Similarly,

 

This means that the components of electric and magnetic fields of a uniform plane wave in the direction of propagation are zero.

5.6 GENERAL SOLUTION OF UNIFORM PLANE WAVE EQUATION

The wave equation in free space is

 

Applying the conditions of uniform plane wave equation, the above equation becomes

Equating the respective components on either side and as E_x = 0, we have

Equation (5.3) has a general solution given by

 

where f₁ and f₂ are functions of (x – v₀t) and (x + v₀t) respectively.

x is the direction of propagation of the wave.

f₁(x – v₀ t) represents a forward wave and

f₂(x + v₀ t) represents a reflected wave.

This reflected wave is present when there is a conductor which acts as a reflector. Otherwise, it is absent. As we are considering free space propagation, E will be f₁(x – v₀ t) only, that is,

 

This is the solution of uniform plane wave equation in free space.

The behaviour is represented typically in Fig. 5.2.

5.7 RELATION BETWEEN E AND H IN UNIFORM PLANE WAVE

The relation between E and H is

Proof    First Maxwell’s equation is

But

Equation (5.4) becomes

Similarly,

and

Equating the respective components, we get

Writing E_y in the form of

that is,

where

From Equations (5.5) and (5.7), we have

As the constant, A cannot be a part of wave motion, we can put A = 0.

So,                

As

or,            

or,            

Similarly, if we take

From Equations (5.6) and (5.8), we have

or,

Hence proved.

 

5.8 PROOF OF E AND H OF EM WAVE BEING PERPENDICULAR TO EACH OTHER

Consider

 

As

We get

The dot product of two vectors E and H is zero only when the two vectors are perpendicular to each other. Hence proved.

5.9 WAVE EQUATIONS IN PHASOR FORM

Wave equations in free space are

                    ∇² H = μ₀ ∈₀ Ḧ and

                    ∇² E = μ₀ ∈₀ Ë

If

The wave equations become

 

These are wave equations in phasor form.

Note that a single time derivative of the field gives a factor of jω and a double time derivative of the field gives a factor of – ω².

Similarly, the wave equations in conductive medium are

 

These can be represented in the following form

where = γ propagation constant

5.10 WAVE PROPAGATION IN LOSSLESS MEDIUM

The wave equation is

 

that is,
or,
where

The y-component of E may be written as

 

where A and B are arbitrary complex constants. Then

 

If A and B are real, it becomes

 

This is the sum of two waves. They travel in opposite directions. If A = B, the waves combine together and form a standing wave. Such waves do not progress.

The wave velocity (v)    It is defined as the velocity of propagation of the wave. It is also defined as

where	ω = 2πf = angular frequency
	β = phase shift constant, radians/m

 

5.11 PROPAGATION CHARACTERISTICS OF EM WAVES IN FREE SPACE

The wave equation in free space is

 

The propagation constant, γ(m^–1)

The phase constant, β(rad/m)

The propagation characteristics of EM wave in free space are:

 

Problem 5.1    If a wave with a frequency of 100 MHz propagates in free space, find the propagation constant.

Solution    Propagation constant, γ in free space is

Problem 5.2    If H field is given by H(z, t) = 48cos(10⁸t + 40z)a_y, A/m, identify the amplitude, frequency and phase constant. Find the wavelength.

Solution    Amplitude of the magnetic field

Now wavelength,

Problem 5.3    When the amplitude of the magnetic field in a plane wave is 2 A/m, (a) determine the magnitude of the electric field for the plane wave in free space (b) determine the magnitude of the electric field when the wave propagates in a medium which is characterised by σ = 0, μ = μ₀ and ∈ = 4∈₀.

Solution    We have

1.    

   But

   
2. σ = 0, ∈_r = 4, μ_r = 1
   

Problem 5.4    If ∈_r = 9, μ = μ₀, for the medium in which a wave with a frequency, f = 0.3 GHz is propagating, determine the propagation constant and intrinsic impedance of the medium when σ = o.

Solution    The expression for propagation constant, γ is

As

Intrinsic impedance,

Problem 5.5    The wavelength of an x-directed plane wave in a lossless medium is 0.25 m and the velocity of propagation is 1.5 × 10¹⁰ cm/s. The wave has z-directed electric field with an amplitude equal to 10 V/m. Find the frequency and permittivity of the medium. The medium has μ = μ₀.

Solution    v = 1.5 × 10¹⁰cm/sec = 1.5 × 10⁸m/s

Frequency of the wave,

We have

Problem 5.6    Identify frequency, phase constant when the electric field of an EM wave is given by E = 5.0 sin(10⁸t – 4.0x) a_z. Also find λ.

Solution

5.12 PROPAGATION CHARACTERISTICS OF EM WAVES IN CONDUCTING MEDIUM

The wave equation in a conducting medium in phasor form is

 

where the propagation constant, γ is

or,                            γ = α + jβ

where α is called the attenuation constant, dB/m

β is called phase constant, rad/m.

Phase constant, β    is also called wave number and it is the imaginary part of propagation constant.

One solution of Equation (5.9) is

 

where x is the direction of propagation and in time varying form,

This is the equation of EM wave propagating in x-direction and attenuated by a factor e^–αx

Attenuation constant, α (dB/m)    It is defined as a constant which indicates the rate at which the wave amplitude reduces as it propagates from one point to another. It is the real part of propagation constant.

Expressions for α and β in a conducting medium

Proof    From the wave equation, propagation constant, γ

Squaring both sides, we get

 

Equating real and imaginary parts,

 

or,                   αβ = (ωμσ)/2

Thus,

From Equations (5.10) and (5.11), we get

or,

 

or,

 

Dividing this by 4,

Adding and subtracting

Taking square root on either side, we get

or,

But α cannot be (–)ve. Hence

Hence proved.

Now substituting Equation (5.12) in Equation (5.10), we get

or,

5.13 SUMMARY OF PROPAGATION CHARACTERISTICS OF EM WAVES IN A CONDUCTING MEDIUM

Problem 5.7    Earth has a conductivity of σ = 10^–2 mho/m, ∈_r = 10, μ_r = 2. What are the conducting characteristics of the earth at

1. f = 50 Hz
2. f = 1kHz
3. f = 1MHz
4. f = 100MHz
5. f = 10GHz

Solution    The parameters of earth are σ = 10^–2 mho/m, ∈_r = 10, μ_r = 2.

1. At f = 50 Hz
   

   So this is >>1. Hence it behaves like a good conductor.

2. At f = 1 kHz,
   

   This is >>1. Hence it behaves like a good conductor.

3. At f = 1 MHz,
   

   It behaves like a moderate conductor.

4. At f = 100 MHz,
   

   Earth behaves like a quasi-dielectric.

5. At f = 10 GHz,
   

   that is,   

   Earth behaves like a good dielectric.

Problem 5.8    A medium like copper conductor which is characterised by the parameters σ = 5.8×10⁷ mho/m, ∈_r = 1, μ_r = 1 supports a uniform plane wave of frequency 60 Hz. Find the attenuation constant, propagation constant, intrinsic impedance, wavelength and phase velocity of the wave.

Solution    Let us obtain the ratio

This is >>1. Therefore, it is a very good conductor.

Attenuation constant

Phase constant

Propagation constant

Wavelength

Intrinsic impedance, η

Phase velocity of wave,

 

	v = λ f = 0.0536×60=3.216m/s
α = 117.2m−1	β = 117.2m−1
γ = 117.2 + j117.2m−1	λ = 5.36cm
η = (2.022 + j2.022)μΩ	v = 3.216m/s

5.14 CONDUCTORS AND DIELECTRICS

As some media behave like good conductors at one frequency range and like good dielectrics at some other frequency range, the conventional definitions of conductors and dielectrics are not satisfactory in communication through EM waves.

The displacement current density,

 

and the conduction current density,

Definition of a good conductor

If , the medium is a good conductor.

Definition of a good dielectric

If , the medium is a good dielectric.

Dissipation factor of a dielectric material (D_f)is defined as

When D_f is small for a specific type of dielectric material, the dissipation factor is practically the same as its power factor which is equal to sin ϕ. Here, ϕ = tan^–1 D_f.

5.15 WAVE PROPAGATION CHARACTERISTICS IN GOOD DIELECTRICS

Attenuation constant, α is given by

For dielectrics,

Expanding by Binomial series, higher order terms can be neglected.

Now Equation (5.13) becomes

The phase shift constant, β is given by

that is,

The velocity of propagation is

Intrinsic or characteristic impedance of general medium, η

5.16 SUMMARY OF THE PROPAGATION CHARACTERISTICS OF EM WAVES IN GOOD DIELECTRICS

1. Attenuation constant, dB/m
2. Phase shift constant,
3. Velocity of the wave,
4. Intrinsic impedance,

5.17 WAVE PROPAGATION CHARACTERISTICS IN GOOD CONDUCTORS

The propagation constant, γ is given by

Thus,         

or,               

The velocity of the wave in a conductor is

and the intrinsic impedance of the conductor is

5.18 SUMMARY OF CHARACTERISTICS OF WAVE PROPAGATION IN GOOD CONDUCTORS

5.19 DEPTH OF PENETRATION, δ (M)

The conductivity of the medium attenuates the wave during propagation. At radio frequencies, the rate of attenuation is more in good conductors.

Proof of   

Let the wave attenuation be represented by

 

z being the direction of propagation.

At                      z = τ

As per the definition of δ, we have

 

From Equations (5.14) and (5.15), we have

or,

Hence proved.

Problem 5.9    Find the depth of penetration, δ of an EM wave in copper at f = 60 Hz and f = 100 MHz. For copper, σ = 5.8 × 10⁷ mho/m, μ_r = 1, ∈_r = 1.

Solution    For copper, at f = 60 Hz,

Therefore, at f = 60 Hz, copper is a very good conductor.

The depth of penetration,

At f = 100 MHz,

Copper is a very good conductor at f = 100 MHz.

The depth of penetration,

5.20 POLARISATION OF A WAVE

 

The polarisation of a composite wave is the direction of the electric field.

Types of Polarisations

These are of three types, namely

1. Linear polarisation
2. Circular polarisation
3. Elliptical polarisation

Linear Polarisation

A wave is said to be linearly polarised if the electric field remains along a straight line as a function of time at some point in the medium. Linear polarisation of a wave is again of three types, namely

1. Horizontal polarisation
2. Vertical polarisation
3. Theta polarisation

When a wave travels in z-direction with and fields lying in xy-plane, if = 0 and is present, it is said to be x-polarised or horizontally polarised.

If is only present and = 0 the wave is said to be vertically (y-polarised) polarised. On the other hand, if and are present and are in phase then the wave is said to be θ-polarised. This is given by

Circular Polarisation

 

Let E of a uniform plane wave travelling in the z-direction be represented by

 

and in time varying form

 

As the wave moves in z-direction, and lie in x-y plane.

Here, E_c is a complex vector, that is, E_c can be written as

 

where E₁ and E₂ are real vectors.

At some point in space (say z = 0) becomes

 

As per the definition of circular polarisation, the electric vector at z = 0 is expressed as

 

that is,

 

So,

This represents a circle.

Elliptical Polarisation

 

Here E_c can be written as

 

 

or,

This is the equation of an ellipse and hence the wave is said to be elliptically polarised.

5.21 SOURCES OF DIFFERENT POLARISED EM WAVES

Horizontal dipole produces horizontally polarised waves.

Vertical dipole produces vertically polarised waves.

Inclined dipole produces θ-polarised waves.

Circular slots produce circularly polarised waves.

Elliptical slots produce elliptically polarised waves.

5.22 DIRECTION COSINES OF A VECTOR FIELD

Consider a vector E which is arbitrarily oriented with respect to the Cartesian coordinate axes. Assume E makes angles θ_x, θ_y and θ_z with x, y and z-axes (Fig. 5.3). Then the component of a vector in a given direction is the projection of the vector, E on a line in that direction, that is,

cos θ_x, cos θ_y and cos θ_z are known as direction cosines of the vector along the coordinate axes.

Problem 5.10    A vessel under sea water requires a minimum signal level of 20μ V/m. What is the depth in the sea that can be reached by a 4.0 MHz plane wave from an aeroplane? The wave has an electric field intensity of 100 V/m. The propagation is vertical into the sea. For sea water, σ = 4 mho/m, m μ_r = 1, ∈_r = 81.

Solution    At f = 4.0 MHz = 4.0 × 10⁶ Hz

For sea water

Hence sea water is a good conductor.

Intrinsic impedance of sea water,

η₁ (free space) = 377Ω

α₁ (free space) = 0

β₁ (free space)

Transmission coefficient

Transmission coefficient

The transmitted electric field

 

Propagation takes place in the form of . Therefore, the distance at which the signal becomes 20μ V/m is found out from

or,

But

that is, the signal will reach the vessel when it is at a depth of 1.977 m.

Problem 5.11    The electric field of a plane wave propagating in a medium is given by E = 4.0e^–αx cos(10⁹πt) – βx)a_z V/m. The medium is characterised by ∈_r = 49, μ_r = 4 and σ = 4 mho/m. Find the magnetic field of the wave.

Solution    Here ω = π×10⁹

The ratio      

This is neither << 1 nor >>

Problem 5.12    An elliptical polarised wave has an electric field of

 

Find the power per unit area conveyed by the wave in free space.

 

Solution	Ex = sin(ωt − βz), V/m
	Ey = 2sin(ωt − βz + 75°), V/m

According to Poynting theorem, we have

5.23 WAVES ON A PERFECT CONDUCTOR—NORMAL INCIDENCE

1. When a wave in air is incident on a perfect conductor normally, it is entirely reflected.
2. As neither E nor H can exist in a perfect conductor, none of the energy is transmitted through it.
3. As there are no losses within a perfect conductor, no energy is absorbed in it.
4. When an EM wave travelling in one medium is incident upon a second medium, it is partially reflected and partially transmitted.

Total fields of a wave at any point after reflection with normal incidence on a perfect conductor

Resultant electric field,

 

E_i is the amplitude of the electric field of the incident wave, z = direction of propagation.

Resultant magnetic field,

 

H_i is the amplitude of the magnetic field of the incident wave.

Let the electric field of the incident wave be

Then the electric field of the reflected wave is

 

The boundary condition is

 

This requires that

 

At z = 0

 

This means, the amplitudes of incident and reflected electric field strengths are equal but with a phase reversal on reflection.

Now

 

In time varying form

 

This obviously represents a standing wave. The variation of E_R is shown in Fig. 5.4.

Conclusions

1. The magnitude of electric field varies sinusoidally with distance from the reflecting plane.
2. E_R = 0 at the surface of the conductor (z = 0) and also at
3. (E_R)_max = 2E_i.
4. E_R is maximum at

Resultant magnetic field, H_R

Let us write H_R as H_R(z) = H_ie^–jβz + H_re^jβz

At the surface of a perfect conductor,

 

As J_s is not specified, we cannot use this boundary condition. Suppose H_i = –H_r. It leads to identical directions of incident and reflected powers. This cannot be true. Therefore, H_i and H_r should be the same at z = 0,

 

In time varying form,

 

This also represents a standing wave. The variation of the H_R is shown in Fig. 5.5.

Conclusions

1. Magnitude of H_R varies cosinusoidally with distance.
2. H_R = 0 at
3. (H_R)_max = 2H_i
4. H_R is maximum at from the surface.

5.24 WAVES ON DIELECTRIC—NORMAL INCIDENCE

When an EM wave is incident normally on the surface of a dielectric, reflection and transmission take place.

For a perfect dielectric, σ = 0. Hence, there is no loss or no absorption of energy in it.

 

That is, the reflection coefficient

Reflection coefficient for

Reflection coefficient for

where

 

 

Transmission coefficient

Transmission coefficient for E is

and transmission coefficient for H is

Expressions for reflection and transmission coefficients are:

where η₁ and η₂ are intrinsic impedances of medium 1 and medium 2 respectively.

Proof    Let ∈₁, μ₁, η₁ be the permittivity, permeability and intrinsic impedance of medium 1. ∈₂, μ₂, η₂ are the values for medium 2. We know that

 

At the boundary of a dielectric, the tangential components of E and H are continuous, that is,

 

From the above equations, we have

So,

Now consider,

So,

From the above equations

Hence proved.

Similarly, consider

5.25 OBLIQUE INCIDENCE OF A PLANE WAVE ON A BOUNDARY PLANE

Reflection and transmission of a wave depend on

1. the type of polarisation of a wave
2. the medium of the boundary.

   General polarisations, namely, parallel and perpendicular, are considered.

Parallel Polarisation

 

Perpendicular Polarisation

 

Plane of Incidence

 

It is described in Fig. 5.6 in which x-y is the plane of incidence.

5.26 OBLIQUE INCIDENCE OF WAVE ON PERFECT CONDUCTOR

When a wave is incident on a perfect conductor, it is reflected back into the same medium. The resultant fields depend on the type of polarisation.

Parallel Polarisation

The incident and reflected electric fields are shown in Fig. 5.7.

The incident and reflected magnetic fields are shown in Fig. 5.8.

The incident magnetic field is given by

where

          a_n = unit vector normal to the plane

            r = (x, y, z) is a radius vector on the plane

        a_n.r = x cos θ_x + y cos θ_y + z cos θ_z

where

θ_x, θ_y and θ_z are the angles made by a unit vector normal to the plane with x, y and z-axes (Fig. 5.9).

            a_n . r = x sin θ_i − y cos θ_i

                H_I = H_i e^{−γ(xsinθ − ycosθ)}

Similarly,            

Here,

are the angles made by the unit vector normal to the plane with x, y and z-axes (Fig. 5.10).

        H_R = H_r e^{−γ (xsin θ + y cos θ)}

        H_T = H_I + H_R

             = H_i e^{−γ (xsin θ − y cos θ)} + H_r e^{−γ (xsin θ + y cos θ)}

At the surfaces of the conductor, E_i = –E_r. In order to satisfy the direction of power flow, H_i must be equal to H_r that is, H_i = H_r . Now

 

Here, γ = propagation constant = α + jβ

But in free space, α = 0.

Hence             γ = jβ

The expression for H_T is given by

where,             β_y = β cosθ, β_x = β sinθ

Conclusions

1. The maximum value of H_T is 2H_i.
2. The maximum occurs at y = 0, at and at its even multiples.
3. The minimum is zero and it occurs at and at its odd multiples.

In order to find the resultant electric field, we can make use of the relations between the magnetic and electric fields.

            E_I = ηH_I

            E_x = η cosθ H_I, for the incident wave

                = −η cosq H_R, for the reflected wave

Similarly,

            E_y = η sin θ H_I, for the incident wave

                = η cos θ H_R, for the reflected wave

The resultant E_x is

Similarly,

Perpendicular Polarisation

For the incident wave,

For the reflected wave,

But E_i = E_r

            E_T = E_i [e^{−jβ (x sin β − y cos β)} + e_{−jβ (x sin θ + y cos θ)}]

                    = 2j E_i sin (βy cos θ) e^{−jβx sin θ}

or,            

where             β_y = β cosθ, β_x = β sinθ

5.27 OBLIQUE INCIDENCE OF A PLANE WAVE ON DIELECTRIC

When a wave is incident on a dielectric, a part of it is reflected and a part of it is transmitted through the dielectric. If θ_i, θ_r and θ_t are the angles of the incident, reflected and transmitted rays, θ_i = θ_r. The angles θ_i and θ_t are related by Snell’s law, that is,

Parallel Polarisation

Consider Fig. 5.12.

The boundary condition on E is

Dividing both sides by E_i, we get

By the law of conservation of energy, incident energy is equal to the sum the of reflected and transmitted energies, that is,

From Equations (5.16) and (5.17)

Simplifying this expression, we get

But

For most of the dielectrics,

Hence the reflection coefficient is

as

or,

Perpendicular Polarisation

Consider Fig. 5.13 in which E is z-directed and x-y is the plane of incidence.

From the boundary condition, we have

From the law of conservation of energy, we have

On simplification of the above expression, we get

5.28 BREWSTER ANGLE

Brewster angle for parallel polarisation

For parallel polarisation, we have

At Brewster angle,

Simplifying this, we get

or,

Brewster angle,         

Brewster angle for perpendicular polarisation

We have

At Brewster angle,

or,

Hence, the condition for no reflection in perpendicular polarisation is ∈_r1 = ∈_r2.

5.29 TOTAL INTERNAL REFLECTION

Definition Total internal reflection is said to exist if

1. the angle of incidence is very high
2. medium 1 is denser than medium 2 and
3. 

For total internal reflection,

The concept of total internal reflection is often used in binocular optics. For these applications, glass prisms are used to shorten the instrument.

5.30 SURFACE IMPEDANCE

For a flat thick conducting sheet, the current density (Fig. 5.14) is given by

As the conductor considered is thick, the depth of penetration is much smaller compared to the thickness of the conductor.

The surface current density is

The current density at the surface, J₁ is given by

 

From the above expressions, we have

For a perfect conductor, σ is very high.

Problem 5.13    A perpendicularly polarised wave is incident at an angle of θ_i = 15°. It is propagating from medium 1 to medium 2. Medium 1 is defined by ∈_r1 = 8.5, μ_r1 = 1, σ₁ = 0 and medium 2 is free space. If E_i = 1.0 mV/m, determine E_r, H_i, H_r.

Solution    The intrinsic impedance of medium 1 is given by

As medium 2 is free space,

 

By Snell’s law

The reflection coefficient for electric field is

Now            

or,

and similarly,            

or,                

5.31 POYNTING VECTOR AND FLOW OF POWER

When EM waves travel from one point to another, there will be energy flow across the surface involved.

 

If E and H are instantaneous, P is also instantaneous.

Proof    First Maxwell’s equation is

 

or,             J = ∇ × H − ∈ Ė

that is,          E.J = E.∇ × H − ∈ E.Ë

But

   ∇.(E × H) = H.∇ × E − E.∇ × H

               E.J = H.∇ × E − ∇.E × H − ∈ E.Ė

As

Here

Taking volume integral,

By divergence theorem,

The left hand side represents energy dissipated in the volume.

The second term on the right hand side represents the rate at which the stored energy in magnetic and electric fields is changing. (–)ve sign indicates decrease. Therefore, by the law of conservation of energy, the rate of energy dissipation in the volume is equal to the rate at which the stored energy in static electric and magnetic fields in the volume is decreasing plus the rate at which the energy is entering the volume from outside. Therefore,

represents inward power flow, or

represents outward power flow.

E × H represents power flow per unit area, or, E × H is in watts/m².

 

Problem 5.14    The magnetic field, H of a plane wave has a magnitude of 5 mA/m in a medium defined by ∈_r = 4, μ_r = 1. Determine (a) the average power flow (b) the maximum energy density in the plane wave.

Solution    (a) We have

(b) The maximum energy density of the wave is

Problem 5.15    A plane wave travelling in a medium of ∈_r = 1, μ_r = 1 has an electric field intensity of . Determine the energy density in the magnetic field and also the total energy density.

Solution    The electric energy density is given by

As the electric energy density is equal to that of the magnetic field for a plane travelling wave,

So the total energy density,

Problem 5.16    The conductivity of sea water, σ = 5 mho/m, ∈_r = 80. What is the distance, an EM wave can be transmitted at 25 kHz and 25 MHz when the range corresponds to 90% of attenuation?

Solution    If the wave is moving in x-direction, we have

 

that is,

Hence

or                        

Problem 5.17    A plane wave with a frequency of 2 MHz is incident upon a copper conductor normally. The wave has an electric field amplitude of E = 2 mV/m. Copper has μ_r = 1, ∈_r = 1 and σ = 5.8 × 10⁷ mho/m. Find the average power density absorbed by copper.

Solution    Copper is a good conductor

5.32 COMPLEX POYNTING VECTOR

Complex Poynting vector is useful to find average power flow. We have,

If

 

1. Characteristic impedance of a medium is	(Yes/No)
2. Velocity of propagation of uniform plane wave and its phase velocity are identical.	(Yes/No)
3. Brewster angle is the angle of reflection.	(Yes/No)
4. P × H gives average power.	(Yes/No)
5. E = ej2x ay means that E = sin (ωt + 2x) ay.	(Yes/No)
6. E and H in good conductors are in time phase.	(Yes/No)
7. Power density is represented by Poynting vector.	(Yes/No)
8. Complex Poynting vector is E × H*.	(Yes/No)
9. The units of Poynting vector are watts.	(Yes/No)
10. Depth of penetration is nothing but α.	(Yes/No)
11. β has the unit of radian.	(Yes/No)
12. Unit of α and β are the same.	(Yes/No)
13. Unit of the propagation constant is m–1.	(Yes/No)
14. Brewster angle is the same as critical angle.	(Yes/No)
15. In circular polarisation, Ex and Ey components have the same magnitude.	(Yes/No)
16. In elliptical polarisation, Ex and Ey components have the same magnitude.	(Yes/No)
17. Horizontal polarisation is said to be linear polarisation.	(Yes/No)
18. When an EM wave is incident on a perfect conductor normally, standing waves are produced.	(Yes/No)
19. According to Snell’s law, the angle of incidence and the angle of reflection are the same.	(Yes/No)
20. Polarisation and the direction of propagation of an EM wave are one and the same.	(Yes/No)
21. In perpendicular polarisation with oblique incidence on a dielectric, there exists Brewster angle.	(Yes/No)
22. If E = cos (6 × 107 t–βz) ax, β is _______.
23. If the attenuation of a plane wave in a medium is 22.5 × 103 m–1, the depth of penetration is _______.
24. If the depth of penetration of a plane wave in a medium is 2 mm, the attenuation constant is _______.
25. Brewster angle is given by _______.

1. Poynting vector is given by
   1. E × H
   2. E. H
   3. H × E
   4. H. E
2. Poynting vector gives
   1. rate of energy flow
   2. direction of polarisation
   3. electric field
   4. magnetic field
3. E.H of a uniform plane wave is
   1. EH
   2. 0
   3. ηE²
   4. ηH²
4. Absolute permeability of free space is
   1. 4π × 10^–7 A/m
   2. 4π × 10^–7 H/m
   3. 4π × 10^–7 F/m
   4. 4π × 10^–7 H/m²
5. For a uniform plane wave in the x-direction
   1. E_x = 0
   2. H_x = 0
   3. E_x = 0 and H_x = 0
   4. E_z = 0
6. Depth of penetration in free space is
   1. infinity
   2. 1/α
   3. 0
   4. small
7. Complex Poynting vector, P is
   1. P = E × H*
   2. P = E × H*
   3. 
   4. E × H
8. Uniform plane wave is
   1. longitudinal in nature
   2. transverse in nature
   3. neither transverse nor longitudinal
   4. x-directed
9. The direction of propagation of EM wave is obtained from
   1. E × H
   2. E.H
   3. E
   4. H
10. The velocity of an EM wave is
    1. inversely proportional to β
    2. inversely proportional to α
    3. directly proportional to β
    4. directly proportional to α
11. Velocity of the wave in an ideal conductor is
    1. zero
    2. very large
    3. moderate
    4. small
12. If E = 2 V/m of a wave in free space, (H) is
    1. 
    2. 60π A/m
    3. 120π A/m
    4. 240π A/m
13. If wet soil has σ = 10^–2 mho/m, ∈_r = 15, μ_r = 1, f = 60 Hz, it is a
    1. good conductor
    2. good dielectric
    3. semi-conductor
    4. magnetic material
14. If wet soil has σ = 10^–2 mho/m, ∈_r = 15, μ_r = 1, at 10 GHz, it is a
    1. good conductor
    2. good dielectric
    3. semi-conductor
    4. semi-dielectric
15. The cosine of the angle between two vectors is
    1. sum of the products of the directions of the two vectors
    2. difference of the products of the directions of the two vectors
    3. product of the products of the directions of the two vectors
    4. zero
16. If E is a vector, then ∇.∇ × E is
    1. 0
    2. 1
    3. does not exist
    4. none of these
17. Velocity of an EM wave in free space is
    1. independent of f
    2. increases with increase in f
    3. decreases with increase in f
    4. zero
18. The direction of propagation of an EM wave is given by
    1. the direction of E
    2. the direction of H
    3. the direction of E × H
    4. the direction of E.H
19. For uniform plane wave propagating in z-direction
    1. E_x = 0
    2. H_x = 0
    3. E_y = 0, H_y = 0
    4. E_z = 0, H_z = 0
20. For free space,
    1. σ = ∞
    2. σ = 0
    3. J ≠ 0
    4. μ_r = 0
21. Velocity of propagation of an EM wave is
    1. 
    2. 
    3. 
    4. 
22. The intrinsic impedance of the medium whose σ = 0, ∈_r = 9, μ_r = 1 is
    1. 40πΩ
    2. 9Ω
    3. 120πΩ
    4. 60πΩ
23. The wavelength of a wave with a propagation constant = 0.1π + j0.2π is
    1. 10 m
    2. 20 m
    3. 30 m
    4. 25 m
24. Electric field just above a conductor is always
    1. normal to the surface
    2. tangential to source
    3. zero
    4. ∞

1. Find the depth of penetration for copper at 20 MHz. For copper

    

   σ = 5.8 × 10⁷ mho/m, μ_r = 1

    

2. What is the velocity of propagation of a uniform plane wave in a medium whose ∈_r = 10, μ_r = 3?
3. Conduction current in a copper wire is 4.0 amps. Find the displacement current in the wire at 1 MHz and 10 MHz. For copper ∈_r = 1, μ_r = 1, σ = 5.8 × 10⁷ mho/m.
4. An EM wave in free space is incident normally on a dielectric whose ∈_r = 5.0. Find the reflection and transmission coefficients.
5. If an electric field in free space is E = 2.0 cos (ωt–βz) a_x V/m, find the average power flowing across a square whose side is 2 m. The square is in z = a constant plane.
6. Derive the condition under which the electric field E = k cos (3 × 10⁸ t – z) a_y exists in a source free dielectric medium. Here k is a constant, β is a constant.
7. If the electric and magnetic fields in a medium (μ_r = 1 and ∈_r = 4) are given by E = 10 cos (10⁸t–βz) a_x and find out η, f and β.