CH 14

14.
Electromagnetic Waves
14.1. Introduction
Some wave solutions to Maxwell's equations have already been encountered in the Solved Problems ofChapter 13. The present
chapter will extend the treatment of electromagnetic waves. Since most regions of interest are free of charge, it will be
assumed that charge density ρ = 0. Moreover, linear isotropic materials will be assumed, with D = ϵE, B = μH, and J = σE.
14.2. Wave Equations

With the above assumptions and with time dependence ejωt for both E and H, Maxwell's equations (Table 13-1) become
∇ × H = (σ + jωϵ)E
(1)
∇ × E = −jωμH
(2)
∇⋅E = 0
(3)
∇⋅H = 0
(4)
Taking the curl of (1) and (2),
∇ × (∇ × H) = (σ + jωϵ) (∇ × E)
∇ × (∇ × E) = −jωμ (∇ × H)
Now, in Cartesian coordinates only, the Laplacian of a vector
∇2A ≡ (∇2Ax)ax + (∇2Ay)ay + (∇2Az)az
satisfies the identity
∇ × (∇ × A) = ∇(∇ ⋅ A) − ∇2A
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Substitution for the "curl curls" and use of (3) and (4) yields the vector wave equations
∇2H = jωμ(σ + jωϵ)H ≡ γ 2H

∇2E = jωμ(σ + jωϵ)E ≡ γ 2E
The propagation constant γ is that square root of γ2 whose real and imaginary parts are positive:
γ = α + jβ
with

 μϵ
α = ω  (√1 + ( ) − 1)
σ 2
⎷ 2 ωϵ
(5)

 μϵ
β = ω  (√1 + ( ) + 1)
σ 2
⎷ 2 ωϵ
(6)
14.3. Solutions in Cartesian Coordinates

The familiar scalar wave equation in one dimension,
∂2F 1 ∂2F
=
∂z2 u2 ∂t2
has solutions of the form F = f(z − ut) and F = g(z + ut), where f and g are arbitrary functions. These represent waves traveling
with speed u in the +z and −z directions, respectively. In Fig. 14-1 the first solution is shown at t = 0 and t = t1 ; the wave has
advanced in the +z direction a distance of ut1 in the time interval t1 . For the particular choices
f(x) = Ce−j ωx/u and g(x) = De+j ωx/u
Figure 14-1
harmonic waves of angular frequency ω are obtained:
F = Cej(ωt−βz) and F = Dej(ωt+βz)
in which β ≡ ω/u. Of course, the real and imaginary parts are also solutions to the wave equation. One of these solutions,F = C
sin(ωt − βz), is shown in Fig. 14-2 at t = 0 and t = π/2ω. In this interval the wave has advanced in thepositive z direction a
distance d = u(π/2ω) = π/2β. At any fixed t, the waveform repeats itself when x changes by 2π/β; the distance
2π
λ≡
β
is called the wavelength. The wavelength and the frequency f ≡ ω/2π enjoy the relation
λf = u or λ = Tu
where T ≡ 1/f = 2π/ω is the period of the harmonic wave.
The vector wave equations of Section 14.2 have solutions similar to those just discussed. Because the unit vectors ax, ay , and
az in Cartesian coordinates have fixed directions, the wave equation forH can be rewritten in the form
∂2H ∂2H ∂2H

+ + = γ 2H
∂x2 ∂y2 ∂z2
Figure 14-2
14.4. Plane Waves
Of particular interest are solutions (plane waves) that depend on only one spatial coordinate, say z. Then the equation becomes
d2H
= γ 2H
dz2
which, for an assumed time dependence ejωt , is the vector analog of the one-dimensional scalar wave equation. Solutions are
as above, in terms of the propagation constant γ.
H(z, t) = H0e ± γzej ωtaH
The corresponding solutions for the electric field are
E(z, t) = E0e ± γzej ωtaE
The fixed unit vectors aH and aE are orthogonal and neither field has a component in the direction of propagation. This being the
case, one can rotate the axes to put one of the fields, say E, along the x axis. Then from Maxwell's equation (2) it follows that H
will lie along the ±y axis for propagation in the ±z direction.
Example
EXAMPLE 1.
Given the field E = E0 e−γzaE (time dependence suppressed), show that E can have no component in the propagation
direction, +az.
The Cartesian components of aE are found by projection:
E = E0e−γz[(aE ⋅ ax)ax + (aE ⋅ ay)ay + (aE ⋅ az)az]
From ∇ · E = 0,
∂
E0e−γz(aE ⋅ az) = 0
∂z
which can hold only if aE · az = 0. Consequently, E has no component in az.
The plane wave solutions obtained above depend on the properties μ, ϵ, and σ of the medium, because these properties are
involved in the propagation constant γ.
14.5. Solutions for Partially Conducting Media

For a region in which there is some but not much conductivity (e.g., moist earth, seawater), the solution to the wave equation in
E is taken to be
E = E0e−γzax
Then, from (2) of Section 14.2,
H=√
σ + jωϵ
E0e−γzay
jωμ
The ratio E/H is characteristic of the medium (it is also frequency-dependent). More specifically for waves E = Exax, H = Hy ay
which propagate in the +z direction, the intrinsic impedance, η, of the medium is defined by
Ex
η=
Hy
Thus,
η=√
jω μ
σ + jωϵ
where the correct square root may be written in polar form,

|η| θ , with
√μ/ϵ σ
|η| = , tan 2θ = and 0∘ < θ < 45∘
√1 + ( ωσϵ )2
4 ωϵ
(If the wave propagates in the −z direction, Ex/Hy = −η. In effect, γ is replaced by −γ and the other square root used.)
Inserting the time factor ejωt and writing γ = α + jβ results in the following equations for the fields in a partially conducting
region:
E(z, t) = E0e− αzej(ωt−βz)ax

E
H(z, t) = 0 e− αzej(ωt−βz− θ)ay
|η |
The factor e−αz attenuates the magnitudes of both E and H as they propagate in the +z direction. The expression for α, (5) of
Section 14.2, shows that there will be some attenuation unless the conductivity σ is zero, which would be the case only for
perfect dielectrics or free space. Likewise, the phase difference θ between E(z, t) and H(z, t) vanishes only when σ is zero.
The velocity of propagation and the wavelength are given by
ω 1
u= =
β
√ (√ 1 + ( ωϵ
σ 2
μϵ
2
) + 1)
2π 2π
λ= =
β
ω√ (√ 1 + ( ωϵ
σ 2
μϵ
2
) + 1)
If the propagation velocity is known, λf = u may be used to determine the wavelength λ. The term (σ/ωϵ) 2 has the effect of
reducing both the velocity and the wavelength from what they would be in either free space or perfect dielectrics, where σ = 0.
Observe that the medium is dispersive: waves with different frequencies ω have different velocities u.
14.6. Solutions for Perfect Dielectrics

For a perfect dielectric, σ = 0, and so
μ
α=0 β = ω√μϵ η=√ 0°
ϵ
Since α = 0, there is no attenuation of the E and H waves. The zero angle on η results in H being in time phase with E at each
fixed location. Assuming E in ax and propagation in az, the field equations may be obtained as the limits of those in Section 14.5
E(z, t) = E0ej(ωt−βz)ax
E
H(z, t) = 0 ej(ωt−βz)ay
η
The velocity and the wavelength are
ω 1 2π 2π
u= = λ= =
β √μϵ β ω√μϵ
14.6.1. Solutions in Free Space.

Free space is nothing more than the perfect dielectric for which
−7 − 12 10−9
μ = μ0 = 4π × 10 H/m ϵ = ϵ0 = 8.854 × 10 F/m ≈ F/m
36π
For free space, η = η0 ≈ 120π Ω and u = c ≈ 3 × 108 m/s.
14.7. Solutions for Good Conductors; Skin Depth
Materials are ordinarily classified as good conductors if σ ≫ ωϵ in the range of practical frequencies. Therefore, the
propagation constant and the intrinsic impedance are
α=β=√
ωμσ ωμ
γ = α + jβ = √πfμσ = η = √ 45°
2 σ
It is seen that for all conductors the E and H waves are attenuated. Numerical examples will show that this is a very rapid
attenuation. α will always be equal to β. At each fixed location H is out of time phase with E by 45° or π/4 rad. Once again
assuming E in ax and propagation in az, the field equations are, from Section 14.5,
E0 − αz j(ωt−βz− π/4)
E(z, t) = E0e− αzej(ωt−βz) ax H(z, t) = e e ay
|η |
Moreover,
2ω 2π 2π
=√
ω
u= = ωδ λ= = = 2πδ
β μσ β √πfμσ
The velocity and wavelength in a conducting medium are written here in terms of theskin depth or depth of penetration,
1
δ≡
√πfμσ
Example
EXAMPLE 2.
Assume a field E = 1.0e−αzej(ωt − βz)ax (V/m), with f = ω/2π = 100 MHz, at the surface of a copper conductor,σ = 58 MS/m,
located at z > 0, as shown in Fig. 14-3. Examine the attenuation as the wave propagates into the conductor.
Figure 14-3
At depth z the magnitude of the field is
|E| = 1.0e− αz = 1.0e− z/δ
where
1
δ= = 6.61 μm
√πfμσ
Thus, after just 6.61 micrometers the field is attenuated to e−1 = 36.8% of its initial value. At 5δ or 33 micrometers, the
magnitude is 0.67% of its initial value—practically zero.
14.8. Interface Conditions at Normal Incidence

When a traveling wave reaches an interface between two different regions, it is partly reflected and partly transmitted, with the
magnitudes of the two parts determined by the constants of the two regions. In Fig. 14-4, a traveling E wave approaches the
interface z = 0 from region 1, z < 0. Ei and Er are at z = −0, while Et is at z = +0 (in region 2). Here, i signifies "incident," r "reflected"
and t "transmitted." Normal incidence is assumed. The equations for E and H can be written as
Ei(z, t) = E0i e−γ1zej ωtax

Er(z, t) = E0r e−γ1zej ωtax
Et(z, t) = E0t e−γ2zej ωtax
Hi(z, t) = H0i e−γ1zej ωtay
Hr(z, t) = H0r eγ1zej ωtay
Ht(z, t) = H0t e−γ2zej ωtay
Figure 14-4
One of the six constants—it is almost always
E0i—may be taken as real. Under the interface conditions about to be derived, one or more of the remaining five may turn out to
be complex.
With normal incidence, E and H are entirely tangential to the interface, and thus are continuous across it. Atz = 0 this implies
E0i + E0r = E0t H0i + H0r = H0t
Furthermore, the intrinsic impedance in either region is equal to ±Ex/Hy (see Section 14.5).
Ei0 Er0 Et0

= η1 = −η1 = η2
H0i H0r H0t
The five equations above can be combined to produce the following ratios in terms of the intrinsic impedances:
E0r η2 − η1 H0r η1 − η2
= =
E0i η1 + η2 H0i η1 + η2
E0t 2η2 H0t 2η1
= =
E0i η1 + η2 H0i η1 + η2
The intrinsic impedances for various materials were examined earlier. They are repeated here for reference.
partially conducting medium: η = √

jωμ
σ + jωϵ
η=√
ωμ
conducting medium: 45°
σ
η=√
μ
perfect dielectric:
ϵ
μ0
free space: η0 = √ ≈ 120π Ω
ϵ0
Example
EXAMPLE 3.
Traveling E and H waves in free space (region 1) are normally incident on the interface with a perfect dielectric (region 2) for
which ϵr = 3.0. Compare the magnitudes of the incident, reflected, and transmittedE and H waves at the interface.
120π
η2 = √
μ
η1 = η0 = 120π Ω = = 217.7 Ω
ϵ √ϵr
E0r η2 − η1 H0r η1 − η2
= = −0.268 = = 0.268
E0i η1 + η2 H0i η1 + η2
E0t 2η2 H0t 2η1
= = 0.732 = = 1.268
E0i η1 + η2 H0i η1 + η2
14.9. Oblique Incidence and Snell's Laws

An incident wave that approaches a plane interface between two different media generally will result in a transmitted wave in
the second medium and a reflected wave in the first. The plane of incidence is the plane containing the incident wave normal
and the local normal to the interface; in Fig. 14-5 this is the xz plane. The normals to the reflected and transmitted waves also
lie in the plane of incidence. The angle of incidence θi, the angle of reflection θr, and the angle of transmission θt—all defined as
in Fig. 14-5—obey Snell's law of reflection,
θi = θr
and Snell's law of refraction,
sin θ i μϵ
=√ 2 2
sin θ t μ1ϵ1
Figure 14-5
Example
EXAMPLE 4.
A wave is incident at an angle of 30° from air to Teflon, ϵr = 2.1. Calculate the angle of transmission, and repeat with an
interchange of the regions.
Since μ1 = μ2 ,
sin θ i sin 30∘ ϵ

= = √ r2 = √2.1 or θ t = 20.18∘
sin θ t sin θ t ϵr1
From Teflon to air,
sin 30∘ 1
= or θ t = 46.43∘
sin θ t √2.1
Supposing both media of the same permeability, propagation from the optically denser medium (ϵ1 > ϵ2 ) results in θt > θi.
As θi increases, an angle of incidence will be reached that results inθt = 90°. At this critical angle of incidence, instead of a
wave being transmitted into the second medium there will be a wave that propagates along the surface. The critical angle is
given by
ϵr2
θ c = sin− 1 √
ϵr1
Example
EXAMPLE 5.
The critical angle for a wave propagating from Teflon into free space is
1
θ c = sin− 1 = 43.64∘
√2.1
14.10. Perpendicular Polarization

The orientation of the electric field E with respect to the plane of incidence determines the polarization of a wave at the interface
between two different regions. In perpendicular polarization, E is perpendicular to the plane of incidence (the xz plane in Fig. 14-
6) and is thus parallel to the (planar) interface. At the interface,
E0r η2 cos θ i − η1 cos θ t

=
E0i η2 cos θ i + η1 cos θ t
and
E0t 2η2 cos θ i
=
Figure 14-6
Note that for normal incidence, θi = θt = 0° and the expressions reduce to those found in Section 14.9.
It is not difficult to show that if μ1 = μ2 ,
η2 cos θ i − η1 cos θ t ≠ 0 for any θ i
Hence, a perpendicularly polarized incident wave suffers either partial or total reflection.
14.11. Parallel Polarization

For parallel polarization, the electric field vector E lies entirely within the plane of incidence, the xz plane as shown in Fig. 14-7.
(Thus E assumes the role played by H in perpendicular polarization.) At the interface,
E0r η2 cos θ t − η1 cos θ i

=
and
E0t 2η2 cos θ i

=
E0i η1 cos θ i − η2 cos θ t
Figure 14-7
In contrast to perpendicular polarizations, if μ1 = μ2 there will be a particular angle of incidence for which there is no reflected
wave. This Brewster angle is given by
ϵ2
θ B = tan− 1√
ϵ1
Example
EXAMPLE 6.
The Brewster angle for a parallel-polarized wave traveling from air into glass for which ϵr = 5.0 is
θ B = tan− 1√5.0 = 65.91°
14.12. Standing Waves

When waves traveling in a perfect dielectric (σ1 = α1 = 0) are normally incident on the interface with a perfect conductor (σ2 = ∞,
η2 = 0), the reflected wave in combination with the incident wave produces astanding wave. In such a wave, which is readily
demonstrated on a clamped taut string, the oscillations at all points of a half-wavelength interval are in time phase. The
combination of incident and reflected waves may be written
j(ωt−βz) j(ωt−βz) −jβz jβz

E(z, t) = [E0ie + E0re ]ax = ej ωt(E0ie + E0re )ax
Since η2 = 0,
E0r/E0i = −1 and
−jβz jβz
E(z, t) = ej ωt(E0ie − E0ie )ax = −2jE0i sin βzej ωtax
Taking the real part,
E(z,t) = 2E0i sin βz sin ωt ax
The standing wave is shown in Fig. 14-8 at time intervals of T/8, where T = 2π/ω is the period. At t = 0, E = 0 everywhere; at t =
1(T/8), the endpoints of the E vectors lie on sine curve 1; at t = 2(T/8), they lie on sine curve 2; and so forth. Sine curves 2 and 6
form an envelope for the oscillations; the amplitude of this envelope is twice the amplitude of the incident wave. Note that
adjacent half-wavelength segments are 180° out of phase with each other.
Figure 14-8
14.13. Power and the Poynting Vector

Maxwell's first equation for a region with conductivity σ is written and then E is dotted with each term
∂E
∇ × H = σE + ϵ
∂t
∂E
E ⋅ (∇ × H) = σE 2 + E ⋅ ϵ
∂t
where, as usual, E2 = E · E. The vector identity ∇ · (A × B) = B · (∇ × A) − A · (∇ × B) is employed to change the left side of the
equation.
∂E
H ⋅ (∇ × E) − ∇ ⋅ (E × H) = σE 2 + E ⋅ ϵ
∂t
By Maxwell's second equation,
∂H μ ∂H2
H ⋅ (∇ × E) = H ⋅ (−μ )=−
∂t 2 ∂t
Similarly,
∂E ϵ ∂E 2
E⋅ϵ =
∂t 2 ∂t
Substituting and rearranging terms,
ϵ ∂E2 μ ∂H 2
σE 2 = − − − ∇ ⋅ (E × H)
2 ∂t 2 ∂t
Integration of this equation throughout an arbitrary volume v gives
∫ σE 2 dv = − ∫ ( ) dv − ∮ (E × H) ⋅ dS
ϵ ∂E2 μ ∂H 2
+
v v 2 ∂t 2 ∂t S
where the last term has been converted to an integral over the surface ofv by use of the divergence theorem.
The integral on the left has the units of watts and is the usual ohmic term representing energy dissipated per unit time as heat.
This dissipated energy has its source in the integrals on the right. Because ϵE2 /2 and μH2 /2 are the densities of energy stored
in the electric and magnetic fields, respectively, the volume integral (including the minus sign) gives the decrease in this stored
energy. Consequently, the surface integral (including the minus sign) must be the rate of energy entering the volume from
outside. A change of sign then produces the instantaneous rate of energy leaving the volume:
P(t) = ∮ (E × H) ⋅ dS = ∮ P ⋅ dS
S S
where = E × H is the Poynting vector, the instantaneous rate of energy flow per unit area at a point.
In the cross product that defines the Poynting vector, the fields are supposed to be in real form. If, instead, E and H are
expressed in complex form and have the common time-dependence ejωt , then the time-average of is given by
1
Pavg = Re (E × H*)
2
where H* is the complex conjugate of H. This follows the complex power of circuit analysis, S = ½ VI*, of which the power is the
real part, P = ½ Re VI*.
For plane waves, the direction of energy flow is the direction of propagation. Thus, the Poynting vector offers a useful,
coordinate-free way of specifying the direction of propagation, or of determining the directions of the fields if the direction of
propagation is known. This can be particularly valuable where incident, transmitted, and reflected waves are being examined.
14.14. SOLVED PROBLEMS

14.1. A traveling wave is described by y = 10 sin (βz − ωt). Sketch the wave at t = 0 and at t = t1 , when it has advanced λ/8,
if the velocity is 3 × 108 m/s and the angular frequency ω = 106 rad/s. Repeat for ω = 2 × 106 rad/s and the same t1 .
The wave advances λ in one period, T = 2π/ω. Hence,
T π
t1 = =
8 4ω
λ π
= ct1 = (3 × 108) = 236 m
8 4(106)
The wave is shown at t = 0 and t = t1 in Fig. 14-9(a). At twice the frequency, the wavelength λ is one-half, and the phase shift
constant β is twice the former value. See Fig. 14-9(b). At t1 the wave has also advanced 236 m, but this distance is nowλ/4.
Figure 14-9
14.2. In free space, E(z, t) = 10 3 sin(ωt − βz)ay (V/m). Obtain H(z, t).
Examination of the phase, ωt − βz, shows that the direction of propagation is +z. Since E × H must also be in the +z
direction, H must have the direction −ax. Consequently,
Ey 103
= η0 = 120π Ω or Hx = − sin (ωt − βz) (A/m)
−Hx 120π
and
103
H(z,t) = − sin (ωt − βz)ax (A/m)
120π
14.3. For the wave of Problem 14.2 determine the propagation constant γ, given that the frequency is f = 95.5 MHz.
In general, γ = √jωμ (σ + jωϵ).In free space, σ = 0 so that

2πf 2π(95.5 × 106)
γ = jω √μ0ϵ0 = j ( )=j 8
= j(2.0)m− 1
c 3 × 10
Note that this result shows that the attenuation factor is α = 0 and the phase-shift constant is β = 2.0 rad/m.
14.4. Examine the field
E(z,t) = 10 sin (ωt + βz)ax + 10 cos(ωt + βz)ay
in the z = 0 plane, for ωt = 0, π/4, π/2, 3π/4, and π.
The computations are presented in Table 14-1.
Table 14-1
ωt E x = 10 sin ωt E y = cos ωt E = E xa x + E ya y
0 0 10 10a y
π 10 10 ax + ay
4 10 ( )
√2 √2 √2
π 10 0 10a x
2
3π 10 −10 ax − ay
10 ( )
4 √2 √2 √2
π 0 −10 10(−a y)
As shown in Fig. 14-10, E(x, t) is circularly polarized. In addition, the wave travels in the −az direction.
Figure 14-10
14.5. An H field travels in the −az direction in free space with a phase-shift constant of 30.0 rad/m and an amplitude of
(1/3π) A/m. If the field has the direction −ay when t = 0 and z = 0, write suitable expressions for E and H. Determine the
frequency and wavelength.
In a medium of conductivity σ, the intrinsic impedance η, which relates E and H, would be complex, and so the phase of E and
H would have to be written in complex form. In free space this restriction is unnecessary. Using cosines,
1
H(z, t) = − cos (ωt + βz) ay
3π
For propagation in the −z direction,
Ex
= −η0 = −120π Ω or Ex = +40 cos (ωt + βz) (V/m)
Hy
Thus,
E( , ) = 40 cos ( + )a (V/m)
E(z,t) = 40 cos (ωt + βz)ax (V/m)
Since β = 30 rad/m,
2π π c 3 × 108 45
λ= = m or f= = = × 108 Hz
β 15 λ π/15 π
14.6. Determine the propagation constant γ for a material having μr = 1, ϵr = 8, and σ = 0.25 pS/m, if the wave frequency is
1.6 MHz.
In this case,
σ 0.25 × 10− 12
= ≈ 10−9 ≈ 0
ωϵ 2π(1.6 × 106)(8)(10−9 /36π)
so that
√μrϵr
α ≈ 0 β ≈ ω√μϵ = 2πf = 9.48 × 10− 2 rad/m
c
and γ = α + jβ ≈ j9.48 × 10−2 m−1. The material behaves like a perfect dielectric at the given frequency. Conductivity of the
order of 1 pS/m indicates that the material is more like an insulator than a conductor.
14.7. Determine the conversion factor between the neper and the decibel.
Consider a plane wave traveling in the +z direction whose amplitude decays according to
E = E0e− αz
From Section 14.13, the power carried by the wave is proportional to E2 , so that
P = P0e− 2αz
Then, by definition of the decibel, the power drop over the distancez is 10 log10(P0 /P) dB. But
P0 10 P 20
10 log10 = ln 0 = (αz) = 8.686 (αz)
P 2.3026 P 2.3026
Thus, αz nepers is equivalent to 8.686(αz) decibels; i.e.,
1 Np = 8.686 dB
14.8. At what frequencies may the earth be considered a perfect dielectric, if σ = 5 × 10−3 S/m, μr = 1, and ϵr = 8? Can α be
assumed zero at these frequencies?
Assume arbitrarily that
σ 1
≤
ωϵ 100
marks the cutoff. Then
ω 100σ
f= ≥ = 1.13 GHz
2π 2πϵ
For small σ/ωϵ,

 μϵ
α = ω (√ 1 + ( ) − 1)
σ 2
⎷ 2 ωϵ
μϵ 1 σ 2
[ ( ) ] = √ = √ r (120π) = 0.333 Np/m
σ μ σ μ
≈ ω√
2 2 ωϵ 2 ϵ 2 ϵr
Thus, no matter how high the frequency, α will be about 0.333 Np/m, or almost 3 db/m (see Problem 14.7); α cannot be
assumed zero.
14.9. Find the skin depth δ at a frequency of 1.6 MHz in aluminum, whereσ = 38.2 MS/m and μr = 1. Also find γ and the
wave velocity u.
1
δ= = 6.44 × 10−5 m = 64.4 μm
√πfμσ
Because α = β = δ−1,
γ = 1.55 × 104 + j1.55 × 104 = 2.20 × 104∠45° (m− 1)
and
ω
u= = ωδ = 647 (m/s)
β
14.10. A perpendicularly polarized wave propagates from region 1 (ϵr1 = 8.5, μr1 = 1, σ1 = 0) to region 2, free space, with an
angle of incidence of 15°. Given
E0i = 1.0 μV/m, find:
E0r,
E0t,
H0i,
H0r, and
H0t.
The intrinsic impedances are
η0 120
η1 = = = 129 Ω and η2 = η0 = 120π Ω
√ϵr1 √8.5
and the angle of transmission is given by
sin 15° ϵ
=√ or θ t = 48.99°
sin θ t 8.5ϵ0
Then
E0r η2 cos θ i − η1 cos θ t

= = 0.623 or E0r = 0.623 μV/m
E0t 2η2 cos θ i
= = 1.623 or E0t = 1.623 μV/m
Finally,
H0i = E0i/η1 = 7.75 n A/m,
H0r, = 4.83 nA/m, and
H0t = 4.31 nA/m.
14.11. Calculate the intrinsic impedance η, the propagation constant γ, and the wave velocity u at a frequency f = 100 MHz
for a conducting medium in which σ = 58 MS/m and μr = 1
γ = √ωμσ 45° = 2.14 × 105 45° m− 1
η=√
ωμ
45° = 3.69 × 10−3 45° Ω − 1
σ
1
α = β = 1.51 × 105 δ = = 6.61 μm u = ωδ = 4.15 × 103 m/s
α
14.12. A plane wave traveling in the +z direction in free space (z < 0) is normally incident at z = 0 on a conductor (z > 0) for
which σ = 61.7 MS/m, μr = 1. The free-space E wave has a frequency f = 1.5 MHz and an amplitude of 1.0 V/m; at the
interface it is given by
E(0,t) = 1.0 sin 2πftay (V/m)
Find H(z, t) for z > 0.
For z > 0, in complex form,
E (z, t) = 1.0e− αz e j(2π ft−βz)ay (V/m)
where the imaginary part will ultimately be taken. In the conductor,
α = β = √πfσ = √π(1.5 × 106)(4π × 10−7)(61.7 × 106) = 1.91 × 104
η=√
ωμ
45° = 4.38 × 10− 4ej π/4
σ
Then, since Ey /(−Hx) = η,
H(z, t) = −2.28 × 103e− αz ej(2πft−βz− π/4)ax (A/m)
or, taking the imaginary part,
H(z,t) = −2.28 × 103e− az sin(2πft − βz − π/4)ax (A/m)
where f, α, and β are as given above.
14.13. In free space, E(z, t) = 50 cos (ωt − βz)ax (V/m). Find the average power crossing a circular area of radius 2.5 m in
the plane z = const.
In complex form,
E = 50e j(ωt−βz) ax (V/m)
and since η = 120π Ω and propagation is in the +z direction,
5 j(ωt − βz)
H= e ay (A/m)
12π
Then,
1 1 2
Pavg = Re (E × H*) = (50) ( ) az W/m2
2 2 12π
The flow is normal to the area, and so
1 5
Pavg = (50) ( ) π (2.5)2 = 65.1 W
2 12π
14.14 A voltage source, v, is connected to a pure resistor R by a length of coaxial cable, as shown in Fig. 14-11(a). Show
that use of the Poynting vector in the dielectric leads to the same instantaneous power in the resistor as methods of
circuit analysis.
Figure 14-11
From Problem 8.9 and Ampère's law,
v i
E= ar and H= aϕ
r ln (b/a) 2πr
where a and b are the radii of the inner and outer conductors, as shown inFig. 14-11 (b). Then
This is the instantaneous power density. The total instantaneous power over the cross section of the dielectric is
2π b vi
P (t) = ∫ ∫ az ⋅ r dr dϕ az = vi
0 a 2πr2 ln (b/a)
which is also the circuit-theory result for the instantaneous power loss in the resistor.
14.15. Determine the amplitudes of the reflected and transmitted E and H fields at the interface shown in Fig. 14-12, if
E0i = 1.5 × 10−3 V/m in region 1, in which ϵr1 = 8.5, μr1 = 1, and σ1 = 0. Region 2 is free space. Assume normal incidence.
μ0μr1
η1 = √ = 129 Ω η2 = 120π Ω = 377 Ω
ϵ0ϵr1
η2 − η1 i
E0r = E = 7.35 × 10− 4 V/m
η2 + η1 0
2η2
E0t = E i = 2.24 × 10−3 V/m
η2 + η1 0
E0i
H0i = = 1.16 × 10−5 A/m
η1
η1 − η2 i
H0r = H = −5.69 × 10−6 A/m
η1 + η2 0
2η1
H0t = H i = 5.91 × 10−6 A/m
η1 + η2 0
Figure 14-12
14.16. The amplitude of Ei in free space (region 1) at the interface with region 2 is 1.0 V/m. If
H0r = −1.41 × 10−3 A/m, ϵr2 = 18.5, and σ2 = 0, find μr2.
From
E0r E0r η2 − 377

= −120π Ω = −377 Ω and =
H0r E0i 377 + η2
E0i 1.0 −377 (377 + η2)
= = or η2 = 1234 Ω
H0r −1.41 × 10 −3 η2 − 377
Then
μ0μr2
1234 = √ or μr2 = 198.4
ϵ0(18.5)
14.17. A normally incident E field has amplitude

E0i = 1.0 V/m in free space just outside of seawater in which ϵr = 80, μr = 1, and σ = 2.5 S/m. For a frequency of 30 MHz, at
what depth will the amplitude of E be 1.0 m V/m?
Let the free space be region 1 and the seawater be region 2.
η1 = 377 Ω η2 = 9.73 43.5° Ω
Then the amplitude of E just inside the seawater is

E0t.
E0t 2η2
= or E0t = 5.07 × 10− 2 V/m
E0i η1 + η2
From
γ = √jωμ(σ + jωϵ)
= 24.36 46.53° m− 1
a = 24.36 cos 46.53° = 16.76 Np/m
Then, from
10−3 = (5.07 × 10− 2)e−16.76z
z = 0.234 m.
14.18. A traveling E field in free space, of amplitude 100 V/m, strikes a sheet of silver of thickness 5 μm, as shown in Fig.
14-13. Assuming σ = 61.7 MS/m and a frequency f = 200 MHz, find the amplitudes |E2 |, |E3 |, and |E4 |.
Figure 14-13
For the silver sheet at 200 MHz, η = 5.06 × 10−3

45° Ω.
E2 2(5.06 × 10−3 45° )

= whence |E2| ≈ 2.68 × 10−3V/m
E1 377 + 5.06 × 10−3 45° )
Within the conductor,
α = β = √πfμσ = 2.21 × 105
Thus, in addition to attenuation there is phase shift as the wave travels through the conductor. Since |E3 | and |E4 | represent
maximum values of the sinusoidally varying wave, this phase shift is not involved.
5 −6)
|E3| = |E2|e− αz = (2.68 × 10−3)e−(2.21× 10 )(5× 10 = 8.88 × 10− 4 V/m
and
E4 2(377)
= whence |E4| ≈ 1.78 × 10−3 V/m
E3 377 + 5.06 × 10−3 45°
14.15. SUPPLEMENTARY PROBLEMS

14.19. Given
E(z,t) = 103 sin(6 × 108t − βz) ay (V/m)
in free space, sketch the wave at t = 0 and at time t1 when it has traveled λ/4 along the z axis. Find t1 , β, and λ.
14.20. In free space,
8
H(z, t) = 1.0ej(1.5× 10 t+βz)ax (A/m)
Obtain an expression for E (z, t) and determine the propagation direction.
Schaum's Electromagnetics Supplementary Problem 14-20: Electromagnetic

Waves
This video illustrates different factors in electric and magnetic waves

representations.
Sudipa Mitra-Kirtley, Ph.D, Professor, Physics and Optical Engineering, Rose-Hulman

Institute of Technology
2013
14.21. In free space,
H(z,t) = 1.33 × 10− 1 cos (4 × 107t − βz) ax (A/m)
Obtain an expression for E(z, t). Find β and λ.
14.22. A traveling wave has a velocity of 106 m/s and is described by
y = 10 cos (2.5z + ωt)
Sketch the wave as a function of z at t = 0 and t = t1 = 0.838 μs. What fraction of a wavelength is traveled between these two
times?
14.23. Find the magnitude and direction of
E(z,t) = 10 sin (ωt − βz) ax − 15 sin(ωt − βz) ay (V/m)
at t = 0, z = 3λ/4.
Schaum's Electromagnetics Supplementary Problem 14-23: Electromagnetic

Waves
This video illustrates time evolution of electric field representation.
Sudipa Mitra-Kirtley, Ph.D, Professor, Physics and Optical Engineering, Rose-Hulman

Institute of Technology
2013
14.24. Determine γ at 500 kHz for a medium in whichμr = 1, ϵr = 15, σ = 0. At what velocity will an electromagnetic wave
travel in this medium?
14.25. An electromagnetic wave in free space has a wavelength of 0.20 m. When this same wave enters a perfect
dielectric, the wavelength changes to 0.09 m. Assuming that μr = 1, determine ϵr and the wave velocity in the dielectric.
14.26. An electromagnetic wave in free space has a phase-shift constant of 0.524 rad/m. The same wave has a phase-
shift constant of 1.81 rad/m upon entering a perfect dielectric. Assuming that μr = 1, find ϵr and the velocity of propagation.
14.27. Find the propagation constant at 400 MHz for a medium in whichϵr = 16, μr = 4.5, and σ = 0.6 S/m. Find the ratio of
the velocity u to the free-space velocity c.
14.28. In a partially conducting medium, ϵr = 18.5, μr = 800, and σ = 1 S/m. Find α, β, η, and the velocity u, for a frequency
of 109 Hz. Determine H(z, t), given
E(z,t) = 50.0e− αz cos (ωt − βz) ay (V/m)
14.29. For silver, σ = 3.0 MS/m. At what frequency will the depth of penetrationδ be 1 mm?
14.30. At a certain frequency in copper (σ = 58.0 MS/m) the phase-shift constant is 3.71 × 105 rad/m. Determine the
frequency.
14.31. The amplitude of E just inside a liquid is 10.0 V/m and the constants areμr = 1, ϵr = 20, and σ = 0.50 S/m. Determine
the amplitude of E at a distance of 10 cm inside the medium for frequencies of (a) 5 MHz, (b) 50 MHz, and (c) 500 MHz.
14.32. In free space, E(z, t) = 1.0 sin (ωt − βz)ax (V/m). Show that the average power crossing a circular disk of radius 15.5
m in a z = const. plane is 1 W.
14.33. In spherical coordinates, the spherical wave
100 0.265
E= sin θ cos (ωt − βr)a0 (V/m) H= sin θ cos (ωt − βr) aϕ (A/m)
r r
represents the electromagnetic field at large distances r from a certain dipole antenna in free space. Find the average power
crossing the hemispherical shell r = 1 km, 0 ≤ θ ≤ π/2.
14.34. In free space, E(z, t) = 150 sin (ωt − βz)ax (V/m). Find the total power passing through a rectangular area, of sides
30 mm and 15 mm, in the z = 0 plane.
14.35. A free space-silver interface has

E0i = 100 V/m on the free-space side. The frequency is 15 MHz and the silver constants areϵr = μr = 1 and σ = 61.7 MS/m.
Determine
E0r and
E0t at the interface.
14.36. A free space-conductor interface has
H0i = 1.0 A/m on the free-space side. The frequency is 31.8 MHz and the conductor constants areϵr = μr = 1 and σ = 1.26
MS/m. Determine
H0r and
H0t and the depth of penetration of Ht.
14.37. A traveling H field in free space, of amplitude 1.0 A/m and frequency 200 MHz, strikes a sheet of silver of thickness
5 μm with σ = 61.7 MS/m, as shown in Fig. 14-14. Find
H0t just beyond the sheet.
Figure 14-14
14.38. A traveling E field in free space, of amplitude 100 V/m, strikes a perfect dielectric, as shown in Fig. 14-15.
Determine
E0t.
Figure 14-15
14.39. A traveling E field in free space strikes a partially conducting medium, as shown in Fig. 14-16. Given a frequency of
500 MHz and
E0i = 100 V/m, determine
E0t and
H0t.
Figure 14-16
14.40. A wave propagates from a dielectric medium to the interface with free space. If the angle of incidence is the critical
angle of 20°, find the relative permittivity.
14.41. Compute the ratios

E0r/E0i and
E0t/E0i for normal incidence and for oblique incidence atθi = 10°. For region 1, assume ϵr1 = 8.5, μr1 = 1, and σ1 = 0. Region
2 is free space.
14.42. A parallel-polarized wave propagates from air into a dielectric at a Brewster's angle of 75°. Find ϵr.
14.16. ANSWERS TO SUPPLEMENTARY PROBLEMS

14.19. t1 = 2.62 ns, β = 2 rad/m, λ = πm. See Fig. 14-17.
Figure 14-17
14.20. E0 = 377 V/m, −az
14.21.
E0 = 50V/m, ( 304 ) rad/m, 15π m
14.22. ⅓. See Fig. 14.18.
Figure 14-18
14.23. 18.03 V/m, 0.555ax − 0.832ay
14.24. j4.06 × 10−2 m−1, 7.74 × 107 m/s
14.25. 4.94, 1.35 × 108 m/s
14.26. 11.9, 8.69 × 107 m/s
14.27. 99.58
60.34° m−1, 0.097
14.28. 1130 Np/m, 2790 rad/m, 2100
22.1° Ω, 2.25 × 106 m/s, 2.38 × 10−2e−αz cos(ωt − 0.386 − βz) (− ax) (A/m)
14.29. 84.4 kHz
14.30. 601 MHz
14.31. (a) 7.32 V/m; (b) 3.91 V/m; (c) 1.42 V/m
14.33. 55.5 W
14.34. 13.4 mW
14.35. −100 V/m, 7.35 × 10−4

45° V/m
14.36. 1.0 A/m, 2.0 A/m, 80 μm
14.37. 1.78 × 10−5 A/m
14.38. 59.7 V/m
14.39. 19.0 V/m, 0.0504 A/m
14.40. 8.55
14.41. For normal incidence,

E0i/E0r = 0.490 and
E0t/E0i = 1.490. At 10°,
E0r/E0i = 0.539 and
E0t/E0i = 1.539.
14.42. 13.93

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14.

14.2. Wave Equations

Taking the curl of (1) and (2),

Now, in Cartesian coordinates only, the Laplacian of a vector

∇2A ≡ (∇2Ax)ax + (∇2Ay)ay + (∇2Az)az

satisfies the identity

∇2H = jωμ(σ + jωϵ)H ≡ γ 2H

14.3. Solutions in Cartesian Coordinates

f(x) = Ce−j ωx/u and g(x) = De+j ωx/u

harmonic waves of angular frequency ω are obtained:

F = Cej(ωt−βz) and F = Dej(ωt+βz)

where T ≡ 1/f = 2π/ω is the period of the harmonic wave.

∂2H ∂2H ∂2H

H(z, t) = H0e ± γzej ωtaH

The corresponding solutions for the electric field are

E(z, t) = E0e ± γzej ωtaE

The Cartesian components of aE are found by projection:

E = E0e−γz[(aE ⋅ ax)ax + (aE ⋅ ay)ay + (aE ⋅ az)az]

which can hold only if aE · az = 0. Consequently, E has no component in az.

14.5. Solutions for Partially Conducting Media

Then, from (2) of Section 14.2,

where the correct square root may be written in polar form,

E(z, t) = E0e− αzej(ωt−βz)ax

The velocity of propagation and the wavelength are given by

14.6. Solutions for Perfect Dielectrics

The velocity and the wavelength are

14.6.1. Solutions in Free Space.

For free space, η = η0 ≈ 120π Ω and u = c ≈ 3 × 108 m/s.

14.7. Solutions for Good Conductors; Skin Depth

|E| = 1.0e− αz = 1.0e− z/δ

14.8. Interface Conditions at Normal Incidence

Ei(z, t) = E0i e−γ1zej ωtax

One of the six constants—it is almost always

E0i + E0r = E0t H0i + H0r = H0t

Ei0 Er0 Et0

partially conducting medium: η = √

14.9. Oblique Incidence and Snell's Laws

and Snell's law of refraction,

sin θ i sin 30∘ ϵ

From Teflon to air,

14.10. Perpendicular Polarization

E0r η2 cos θ i − η1 cos θ t

It is not difficult to show that if μ1 = μ2 ,

η2 cos θ i − η1 cos θ t ≠ 0 for any θ i

14.11. Parallel Polarization

E0r η2 cos θ t − η1 cos θ i

E0t 2η2 cos θ i

θ B = tan− 1√5.0 = 65.91°

14.12. Standing Waves

j(ωt−βz) j(ωt−βz) −jβz jβz

E(z,t) = 2E0i sin βz sin ωt ax

14.13. Power and the Poynting Vector

By Maxwell's second equation,

Substituting and rearranging terms,

Integration of this equation throughout an arbitrary volume v gives

14.14. SOLVED PROBLEMS

The wave advances λ in one period, T = 2π/ω. Hence,