Download Note 06 Electromagnetic Waves - Physics and Engineering Physics

EP225 Note No. 6
Electromagnetic Waves
6.1 Basic Requirements of Media to Accommodate Electromagnetic Waves
For mechanical waves, a wave medium should have both mass density (inertia) and
elasticity which, respectively, store kinetic and potential energies. In electromagnetic waves,
electric and magnetic energies are carried by waves and for a medium to accommodate
electromagnetic waves, it should be able to store electric and magnetic energies. In other
words, a medium should have capacitance and inductance. Vacuum is characterized by the
electric permittivity "0 = 8:85 10 12 F/m (capacitance per unit length) and magnetic
permeability 0 = 4
10 7 H/m (inductance per unit length) and thus satis…es the basic
requirements for wave medium. Electromagnetic waves in vacuum propagate at the speed
c= p
1
"0
108 m s
= 3:0
1
0
and characterized by the impedance,
Z0 =
r
0
"0
= 377
In insulators having a permittivity "; these basic quantities are accordingly modi…ed,
c= p
Z=
1
"
r
0
0
"
6.2 LC Transmission Line (analogue of mass-spring transmission line)
A transmission line having an inductance per unit length L=l and capacitance per unit
length C=l can model various aspects of electromagnetic waves. We consider a segment dx
of transmission line. The inductance of the segment is
L
dx
l
and the capacitance is
C
dx
l
If voltage and current waves in the form
V (x; t); i(x; t)
are assumed, the Kirchho¤’s voltage theorem applied to the segment reads
V (x:t) = V (x + dx; t) + VL = V (x + dx; t) +
1
L
@
dx i(x; t)
l
@t
Expanding V (x + dx; t) as
V (x + dx; t) ' V (x; t) +
we …nd
@V (x; t)
=
@x
@V
dx
@x
L@
i(x; t)
l @t
(1)
Fig. 6.1. LC transmission line.
Similarly, Kirchho¤’s current theorem applied to the junction reads
i(x; t) = i(x + dx; t) + iC = i(x + dx; t) +
@V
C
dx
l
@t
where
C @V
dx
l
@t
is the (displacement) current ‡owing through the capacitor. Eq. (8) yields
iC =
@i
=
@x
C @V
l @t
(2)
Eliminationof the current between Eqs. (1) and (2) leads to
@2V
L C @2V
=
;
@x2
l l @t2
(3)
and elimination of the voltage leads to
@2i
L C @2i
=
:
@x2
l l @t2
2
(4)
Both the voltage and current waves thus propagate at the velocity
1
c= q
(5)
LC
l l
If applied to vacuum, L=l =
0;
C=l = "0 ; we obtain
c= p
1
"0
0
The ratio of the voltage wave and current wave is constant,
V (x; t) = Z i(x; t)
where
Z=
s
L=l
C=l
(6)
is the characteristic impedance of the transmission line.
The electric energy density associated with the wave is identical to the magnetic energy
density,
1C 2 1C 2 2 1L 2
V =
Z i =
i
2 l
2 l
2l
and the total energy density is
C 2
V
(J m 1 )
l
This energy equipartition is similar to that in mechanical waves in which potential and
kinetic energy densities are identical.
Example 1 A dc voltage source of 6 V is suddenly connected to a 50 coaxial cable …lled
with te‡on (" = 2"0 ): The capacitance per unit length of the cable is C=l = 94.4 pF/m.
Find the velocity of the voltage and current waves and the magnitude of the current wave.
Calculate the electric and magnetic energies per unit length of the cable. The cable is
terminated with a 50 resistor and there is no need to consider re‡ection.
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Solution Both waves propagate at the velocity
p
1
"
0
c
= p = 2:12
2
108 m s
1
The amplitude of the current pulse is i = V =Z = 6 V=50
density is
= 0:12 A. The electric energy
1C 2
F
1
V =
94:4 10 12
2 l
2
m
= 1:70 10 9 J m 1
36 V2
The inductance per unit length of the cable is
C
L
= Z 2 = (50 )2 94:9
l
l
= 2:36 10 7 H m 1
10
12
Fm
1
The magnetic energy per unit length is
1
1L 2
i =
2:36 10 7 H/m
2l
2
= 1:70 10 9 J m 1
(0:12 A)2
which is identical to the electric energy density as expected.
In the above Example, the power supply provides energy at a rate
P = V i = 0:72 W
as the wave propagates along the cable. The energy appears as magnetic and electric energy
stored in the cable which increases with time. Note that there is no dissipation of energy.
The power indicates the rate of energy transfer from the power supply to the cable and
eventually to the load resistor.
The capacitance and inductance per unit length of a coaxial cable …lled with an insulator
having permittivity " are
C
2 "
L
=
;
= 0 ln(a=b)
l
ln(a=b)
l
2
where a is outer radius and b is the inner radius of the cable. Those of a parallel wire
transmission line are
C
"0
'
;
l
ln(d=a)
L
'
l
0
ln(d=a);
d
a
where d is the distance between the wires and a is the wire radius. Derivation of these
formulae is given in Chapter 9 of the textbook.
6.3 Electric and Magnetic Fields, Poynting Vector
The voltage and current wave employed are convenient for analyzing wave propagation
along transmission lines. However, there are cases in which the concept of voltage and current
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becomes vague. For example, it is di¢ cult to assign voltage and current for electromagnetic
waves in free space. There are more fundamental physical quantities behind voltage and
current. They are the electric and magnetic …elds, E (V m 1 ) and H (A m 1 ) or B (Wb
m 2 ):
The electric …eld in the coaxial cable is radially outward, and its magnitude can be found
from Gauss’law,
q=l
Er =
(V m 1 )
(7)
2 "r
where q=l is the linear charge density. The magnetic …eld is azimuthal, and its magnitude
can be found from Ampere’s law,
B =
0i
2 r
(Tesla = Weber m 2 ); or H =
B
0
=
i
2 r
(A m 1 )
(8)
Note that the …elds are perpendicular to each other, and both are normal to the cable axis
which is the direction of wave propagation. Since q=l = i=c; we …nd
r
Er
1
0
=
=
= constant
(9)
H
c"
"
This simple relationship between the two …elds holds in general for electromagnetic waves.
For plane waves in vacuum propagating in the z direction, the electric …eld is in the x
direction and magnetic …eld is in the y direction;
r
Ex
0
= Z0 =
= 377
(10)
Hy
"0
In the …gure, the z axis is horizontal, x axis vertical, and y axis out of page.
[sin(x
t); x; 0]
1.0
0.5
0.0
0 0.0
0.5
1.0
-0.5
-0.5
-1.0
1
2
z
3
-1.0
Fig. 6.2. Plane sinusoidal electromagentic wave propagating in z direction. Electric …eld is
in the x direction and magnetic …eld in y direction.
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Since the energy density associated with an electric …eld is
1 2
"E
2
(11)
and that associated with a magnetic …eld is
1
2
0H
2
(12)
an equal amount of energy density is shared by the electric and magnetic …elds,
1 2 1
"E =
2
2
0H
2
(13)
The total energy density associated with an electromagnetic wave is therefore,
2
1 2
"E = "E 2 (J m 3 )
2
(14)
Multiplying this by the wave propagation velocity, we can de…ne the intensity of electromagnetic wave,
2
I = c"E
r
" 2
E
=
0
2
E
(W m 2 )
Z
=
(15)
However, since
E = ZH
the intensity can be rewritten as
I = EH
The vector de…ned by
S=E
H (W m 2 )
(16)
is called Poynting vector. Its magnitude is the intensity (power density, W m 2 ) carried by
electromagnetic …elds and it direction indicates the direction of energy ‡ow.
Example 2 Find the electric and magnetic …elds associated with a 1 kW (rms) laser beam
1 cm in diameter.
Solution The rms (root-mean-square) intensity of the laser beam is
I=
103 W
= 12:7
(5 10 3 m)2
From
I=
E2
= 12:7
Z
6
106 W m
106 W m
2
2
we …nd
p
ZI
p
377
12:7 106 W m 2
=
= 6:9 104 V m 1 .
p
The peak electric …eld is E0 = 2 Erms = 9:8 104 V/m. The magnetic …eld is
Erms =
Erms
Z
= 183 A m 1 ,
Hrms =
and the magnetic ‡ux density is
Brms = 0 Hrms (= Erms =c)
= 2:3 10 4 Tesla = (Weber m 2 )
6.4 Wave Re‡ection at Impedance Discontinuity
As mechanical waves su¤er re‡ection at an impedance discontinuity, re‡ection of electromagnetic waves also occurs wherever an impedance discontinuity exists. If a transmission
line having an impedance Z1 is connected to another with an impedance Z2 ; the re‡ected
voltage wave is given by
Z 2 Z1
Vr =
Vi
(17)
Z 1 + Z2
and transmitted wave in line 2 is
2Z2
Vt =
Vi
(18)
Z1 + Z2
These readily follow from the energy (power) conservation,
Vi2
V2 V2
= r + t
Z1
Z1
Z2
(19)
and the continuity of the voltage at the connection,
Vi + Vr = Vt
(20)
Almost identical formulae emerge for re‡ection/transmission of the electric …eld for normal
incidence,
Z 2 Z1
Er =
Ei
(21)
Z 1 + Z2
2Z2
Et =
Ei
(22)
Z1 + Z2
at a boundary between two dielectric media with permittivities "1 and "2 ; respectively, and
corresponding impedances, Z1 and Z2 :
The re‡ected current wave is
Z 1 Z2
ir =
ii
Z 1 + Z2
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and re‡ected magnetic …eld wave is
Hr =
Z 1 Z2
Hi
Z 1 + Z2
because the power and power density in terms of current wave and magnetic …eld wave are
Zi2 (W) and ZH 2 (W m 2 )
respectively.
Example 3 Calculate the fraction of power re‡ection for light falling normal at a glass
surface. Assume that the glass has " = 2:6"0 :
Solution The impedance of glass is
r
0
Zg =
= 234 :
2:6"0
The electric …eld of the re‡ected wave is
Zg Z0
Ei
Zg + Z0
234 377
=
Ei
234 + 377
=
0:23Ei :
Er =
Therefore, the intensity ratio, or the fraction of power re‡ection, is
( 0:23)2 = 0:053 = 5:3 %
Example 4 A 10 V dc generator with an internal impedance of 50 is suddenly connected
to a 50 coaxial cable terminated by a load resistance of RL = 20 : The coaxial cable is
30 m long and …lled with te‡on. Analyze how the voltage and current waves evolve. Is the
…nal current consistent with what is expected from dc theory?
Solution Right after connection is made, the generator sees the generator resistance and line
impedance in series, 50 + 50 = 100 . Waves have not reached the load yet and a voltage
pulse of 5 V and current pulse of 5 V=50 = 0:1 A start propagating along the cable. This
lasts until the pulses reach the load end which occurs at = L=c = 30 m/2.12 108 m/sec
= 0:14 sec. The re‡ected voltage wave at the load end is
Vr =
20 50
20 + 50
5V =
8
2:14 V
Therefore, after re‡ection the voltage in the cable is 5 2:14 = 2:86 V. The current in the
load resistor is 2.86 V/20 = 0:143 A. Since the generator impedance is matched with the
impedance of the coaxial cable, there is no further re‡ection and they are the …nal voltage
and current. From dc theory in which the cable impedance and initial transient are entirely
neglected, the current is
10 V
idc =
= 0:143 A
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which agrees with the …nal current in the wave theory.
Example 5 Repeat the preceding problem when the generator impedance is 30 . In this
case, there are multiple re‡ections. In summing a series, use
1 + x + x2 + x3
1
=
1
x
Solution Right after connection, a voltage wave of V = 6.25 V starts propagating toward
the load. After re‡ection at the load end, the voltage is
(1 +
L )V
where
20 50
= 0:428
20 + 50
When the re‡ected wave reaches the generator, a fraction of
=
L
G
=
30 50
=
30 + 50
0:25
is re‡ected. The voltage after the second re‡ection is
(1 +
L
+
L G )V
This process continues, and the …nal voltage is
V…nal = (1 + L + L G + 2L G +
1+ L
=
V
1
L G
1 0:428
=
6:25
1 0:428 0:25
= 4:0 V
2 2
L G
+
3 2
L G
+
)V
This is consistent with dc theory. The …nal current is 0.2 A.
6.5 Impedance Matching
In communication engineering (and in optics too), it is important to achieve impedance
matching between a generator and load for most e¢ cient power transfer. Suppose we are
given two transmission lines with di¤erent impedances Z1 and Z2 : If they are directly connected, wave re‡ection is bound to occur. For sinusoidal wave in steady state (i.e., long
after initial transient), there is a way to avoid re‡ection by inserting a third transmission
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p
line with an impedance Z3 = Z1 Z2 between the two lines. For an incident voltage wave
Vi ; the re‡ection at the …rst boundary is
Vr1 =
Z 3 Z1
Vi
Z 1 + Z3
Vr2 =
Z 2 Z3
Vi
Z 2 + Z3
and that at the second boundary is
Note that we are seeking a condition to avoid re‡ection and thus can assume a priori the same
incident wave amplitude at both boundaries. If the transmission line inserted is a quarter
wavelength long, the wave re‡ected at the second boundary changes its sign relative to the
wave re‡ected at the …rst boundary because it makes a round trip over a half wavelength.
(Note that k 2 = ; and thus sin( + ) = sin :) Therefore, if
Vr1 + ( Vr2 ) = 0
the two re‡ected waves cancel each other, and in e¤ect, re‡ection can be avoided. Solving
Z 3 Z1
Z2 Z3
=
Z 1 + Z3
Z2 + Z3
we …nd
Z32 = Z1 Z2
or
Z3 =
p
Z 1 Z2
(23)
The condition to eliminate wave re‡ection is that the line to be inserted have an impedance
of geometric mean of the given line impedances and its length be a quarter wavelength. Note
that the wavelength is that in the transmission to be inserted.
Fig. 6.3. If Z3 =
p
Z1 Z2 and the length of the line to be inserted is =4; wave re‡ection can
be avoided.
Example 6 Find the permittivity " and thickness of a dielectric …lm to be coated on a glass
surface in order to avoid re‡ection of light at air = 550 nm. Assume that the permittivity
of glass is "g = 2:5 "0 :
Solution Let the …lm impedance be Zf : From
p
Z f = Za Z g
10
where Za =
p
0 ="0 ; Zg
=
p
0 ="g ;
we …nd
Zf =
=
r
r
0
p
"0 "g
p
2:5"0
0
Therefore, the permittivity of the …lm should be
p
"f =
2:5"0
= 1:58"0 :
The speed of light in the …lm is
p
1
"f
0
c
1:58
= 2:39 108 m s 1 .
= p
and the wavelength in the …lm is
f
2:39
= 438 nm.
3
= 550
The required coating thickness is
t=
f
4
= 110 nm.
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