Fields
~E and ~B are in phase, reaching their maximum and minimum values at the
same time. The electric field oscillates in the x-z plane and the magnetic field oscillates
in the y-z plane
ElectromagneticWaves :
In the absence of any source of charge or current, Maxwell’s equations in free
space are as follows :
~
∇
・
~E = 0 (1)
~
∇
・
~B = 0 (2)
~
∇
×
~E = −
∂~B
∂t
(3)
~
∇
×
~B = ǫ0μ0
∂~E
∂t
(4)
The last two equations couple the electric and the magnetic fields. If
~B is time
dependent,
~∇ × ~E is non-zero. This implies that ~E is a function of position. Further,
if
∂~B /∂t itself changes with time, so does ~∇ × ~E . In such a case ~E also
varies with time since the
~∇ operator cannot cause time variation. Thus, in general,
a time varying magnetic field gives rise to an electric field which varies both
in space and time. It will be seen that these coupled fields propagate in space.
We will first examine whether the equations lead to transverse waves. For simplicity,
assume that the electric field has only x-component and the magnetic field
only y-component. Note that we are only making an assumption regarding their
directions – the fields could still depend on all the space coordinates
x, y, z, in
addition to time
t.
Gauss’s law gives
~
∇
・
~E =
∂E
x
∂x
+
∂E
y
∂y
+
∂E
z
∂z
= 0
Since only
Ex 6= 0, this implies
∂E
x
∂x
= 0
Thus
Ex is independent of x coordinate and can be written as Ex(y, z, t). A similar
analysis shows that
By is independent of y coordinate and can be written explicitly
as
By(x, z, t).
Consider now the time dependent equations eqns. (3) and (4). The curl equation
for
~B gives, taking z-component
∂B
y
∂x
−
∂B
x
∂y
=
∂E
z
∂t
= 0
1
Since
Bx = 0, this gives
∂B
y
∂x
= 0
showing that
By is independent of x and hence depends only on z and t. In a
similar manner we can show that
Ex also depends only on z and t. Thus the fields
~E
and
~B do not vary in the plane containing them. Their only variation takes
place along the z-axis which is perpendicular to both
~E and ~B. The direction of
propagation is thus
z−direction.
Z
X
Y
E
B B
E
To see that propagation is really a wave disturbance, take y-component of Eqn.
(3) and x-component of Eqn. (4)
∂E
x
∂z
=
−
∂B
y
∂t
(5)
−
∂B
y
∂z
=
μ0ǫ0
∂E
x
∂t
(6)
To get the wave equation for
Ex, take the derivative of eqn. (5) with respect to z
and substitute in eqn. (6) and interchange the space and time derivatives,
∂
2Ex
∂z
2 = μ0ǫ0−∂2By
∂z∂t
=
−
∂
∂t
∂B
y
∂z
!
=
μ0ǫ0
∂
2Ex
∂t
2
Similarly, we can show,We get
∂
2By
∂z
2 = μ0ǫ0
∂
2By
∂t
2
2
Each of the above equations represents a wave disturbance propagating in the zdirection
with a speed
c
=
1
√
μ0ǫ0
On substituting numerical values, the speed of electromagnetic waves in vacuum
is
3 × 108 m/sec.
Consider plane harmonic waves of angular frequency
ω and wavlength λ = 2π/k.
We can express the waves as
E
x = E0 sin(kz − ωt)
B
y = B0 sin(kz − ωt)
The amplitudes
E0 an B0 are not independent as they must satisfy eqns. (5) and
(6) :
∂E
x
∂z
=
E0k cos(kz − ωt)
∂B
y
∂t
=
−B0ω cos(kz − ωt)
Using Eqn. (5) we get
E
0k = B0ω
The ratio of the electric field amplitude to the magnetic field amplitude is given
by
E
0
B
0
=
ω
k
=
c
Fields
~E and ~B are in phase, reaching their maximum and minimum values at the
same time. The electric field oscillates in the x-z plane and the magnetic field oscillates
in the y-z plane. This corresponds to a
polarized wave. Conventionally,
the plane in which the electric field oscillates is defined as the plane of polarization.
In this case it is x-z plane. The figure shows a harmonic plane wave
propagating in the z-direction. Note that
~E , ~B and the direction of propagation ˆk
form a right handed triad.
3
z
B
E
y
x
ct
Example :
The electric field of a plane electromagnetic wave in vacuum is
Ey = 0.5 cos[2π×
10
8(t − x/c)] V/m, Ex = Ez = 0. Determine the state of polarization and the
direction of propagation of the wave. Determine the magnetic field.
Solution :
Comparing with the standard form for a harmonic wave
ω
= 2π × 108 rad/s
k
= 2π × 108/c
so that
λ = c/108 = 3 m. the direction of propagation is x-direction. Since
the electric field oscillates in x-y plane, this is the plane of polarization. Since
~B
must be perpendicular to both the electric field direction and the direction of
propagation,
~B has only z-component with an amplitude B0 = E0/c ≃ 1.66 ×
10
−9 T. Thus
B
z = 1.66 × 10−9 cos[2π × 108(t − x/c)] T
Exercise :
The magnetic field of a plane electromagnetic wave is given by
B
y = Bz = 10−8 sin[
2
π
3
x
− 2π × 108t] T
Determine the electric field and the plane of polarization. (Ans. Strength of
electric field is
3√2 V/m)
Plane, Circular and Elliptic Polarization :
4
We have shown that
~E and ~B lies in a plane perpendicular to the direction of propagation,
viz., the plane of polarization. This does not imply that the direction of
these fields are constant in time. If it so happens that the successive directions of
~E
(and hence
~B) remains parallel, the electric vectors at different points in space
along the direction of propagation at a given time will lie in a plane. (Equivalently,
the directions of electric vectors at a given point in space at different times will be
parallel). Such a situation is called a
plane polarized wave.
Z
X
y
X
y
Z
E− parallel to x−direction
E parallel to y−direction
X
y
Z
q
E inclined at an angle
q to x−axis
A plane polarized wave propagating in z-direction can be described by an electric
field given by
~E
=
0 sin ωtˆn = E0 sin ωt cos θˆi + E0 sin ωt sin θˆj
We know that because of linearity of the wave equation, any linear combination
of two solutions of the wave equation is also a solution of the wave equation.
This allows us to construc new states of polarization of the electromagnetic wave
5
from the plane polarized wave. Suppose we have two plane polarized waves of
equal amplitude one with the electric vector parallel to the x-direction and the
other parallel to y-direction, the two waves having a phase difference of
π/2. The
resultant, which is also a solution of the wave equation, has
E
x = E0 sin ωt
E
y = E0 cos ωt
The tip of the resultant electric vector
~E describes a circle of radius E0, which
is independent of
t. The state of polarization is known as circular polarization.
Looking along the direction of propagation if the radius vector is moving clockwise,
the polarization is called
right circularly polarized and if anticlokwise it is
called left circularly polarized.
x
y
x
y
w
t=0
w
t=p /2
w
t
w
t
w
t=0
w
t=p /2
Right Circularly polarized Left Circularly polarized
In the same way if the two plane polarized solutions have a phase difference of
π/2
but have different amplitudes
a and b, they produce what is known as elliptically
polarized light. In this case, we have
E
x = a sin ωt
E
y = b cos ωt
so that the equation to the trajectory is given by
E
2
x
a
2 +
E
2
y
b
2 = 1
6
x
y
x
y
w
t=0
w
t=p /2
w
t
w
t
w
t=0
w
t=p /2
Right Elliptically polarized Left Elliptically polarized
b
a
a
b
Wave Equation in Three Dimensions :
We can obtain the wave equation in three dimensions by using eqns. (1) to (4).
On taking the curl of both sides of eqn. (3), we get
~
∇
×
(~∇ × ~E ) = −
∂
∂t
(
~∇ × ~B)
Using the operator identity
~
∇
×
(~∇ × ~E ) = ~∇(~∇ ・ ~E ) − ~∇2~E = −~∇2~E
wherein we have used
~∇ ・ ~E = 0, and substituting eqn. (4) we get
~
∇
2
~E = μ0ǫ0
∂
2~E
∂t
2
A three dimensional harmonic wave has the form
sin(~k ・ ~r − ωt) or cos(~k ・ ~r − ωt) Instead of using the trigonometric form, it is convenient to use the complex
exponential form
f
(~r, t) = exp(i~k ・ ~r − ωt)
and later take the real or imaginary part of the function as the case may be. The
derivative of
f(~r, t) is given as follows :
∂
∂x
f
(~r, t) =
∂
∂x
exp(
ikxx + ikyy + ikzz − iωt) = ikxf(~r, t)
7
Since
∇
= ˆı
∂
∂x
+ ˆ
∂
∂y
+ ˆ
k
∂
∂z
we have,
∇
f(~r, t) = i~kf(~r, t)
In a similar way,
∂
∂t
f
(~r, t) = −iωf(~r, t)
Thus for our purpose, the differential operators
∇ and ∂/∂t may be equivalently
replaced by
∂
∂t
→ −iω
~
∇
→
i~k
Using these, the Maxwell’s equations in free space become
~k
・ ~E = 0 (7)
~k
・ ~B = 0 (8)
~k
× ~B = −μ0ǫ0~E (9)
~k
× ~E = B (10)
We can see that
~E , ~B and ~k form a mutually orthogonal triad. The electric field
and the magnetic field are perpendicular to each other and they are both perpendicular
to the direction of propagation.
Generation of Electromagnetic Waves :
We have looked for solutions to Maxwell’s equations in free space which does
not have any charge or current source. In the presence of sources, the solutions
become complicated. If
ρ = constant, i.e. if ~ J = 0, we only have a steady electric
field. If
ρ varies uniformly with time, we have steady currents which gives
us both a steady electric field as well as a magnetic field. Clearly, time varying
electric and magnetic fields may be generated if the current varies with time, i.e.,
if the charges accelerate. Hertz confirmed the existence of electromagnetic waves
in 1888 using these principles. A schematic diagram of Hertz’s set up is shown in
the figure.
8
Source
Primary
Plate
Plate
Secondary
Gap
The radiation will be appreciable only if the amplitude of oscillation of charge is
comparable to the wavelength of radiation that it emits. This rules out mechanical
vibration, for assuming a vibrational frequency of 1000 cycles per second,
the wavelength work out to be 300 km. Hertz, therefore, made the oscillating
charges vibrate with a very high frequency. The apparatus consists of two brass
plates connected to the terminals of a secondary of a transformer. The primary
consists of an LC oscillator circuit, which establishes charge oscillations at a frequency
of
ω = 1/√LC. As the primary circuit oscillates, oscillations are set up
in the secondary circuit. As a result, rapidly varying alternating potential difference
is developed across the gap and electromagnetic waves are generated. Hertz
was able to produce waves having wavelength of 6m. It was soon realized that
irrespective of their wavelength, all electromagnetic waves travel through empty
space with the same speed, viz., the speed of light.
9
Depending on their wavelength range, electromagnetic waves are given different
names. The figure shows the electromagnetic spectrum. What is known as visible
light is the narrow band of wavelength from 400 nm (blue) to 700 nm (red). To
its either side are the infrared from 700 nm to 0.3 mm and the ultraviolet from 30
nm to 400 nm. Microwaves have longer wavelength than the infrared (0.3 mm to
300 mm) and the radio waves have wavelengths longer than 300 mm. The television
broadcast takes place in a small range at the end of the microwave spectrum.
Those with wavelengths shorter than ultraviolet are generally called
rays. Prominent
among them are x-rays with wavelengths 0.03 nm t0 30 nm and
γ-rays with
wavelengths shorter than 0.03 nm.
Poynting Vector :
Electromagnetic waves, like any other wave, can transport energy. The power
through a unit area in a direction normal to the area is given by
Poynting vector,
given by
~S
=
1
μ
0
~E
×
~B
As
~E , ~B and ~k form a right handed triad, the direction of ~S is along the direction
of propagation. In SI units
~S ismeasured in watt/m2.
The magnitude of
~Sfor the electromagnetic wave travelling in vacuum is given
10
by
S
=
EB
μ
0
=
E
2
cμ
0
where we have used the relationship between
E and B in free space. For harmonic
waves, we have
S
=
E
2
0
cμ
0
sin
2(kx − ωt)
The average power transmitted per unit area, defined as the
intensity is given by
substituting the value 1/2 for the average of the square of sine or cosine function
I
=
E
2
0
2
cμ0
Example :
Earth receives 1300 watts per squar meter of solar energy. assuming the energy
to be in the form of plane electromagnetic waves, compute the magnitude of the
electric and magnetic vectors in the sunlight.
Solution :
From the expression for the average Poynting vector
E
2
0
2
cμ0
= 1300
which gives
E0 = 989 V/m. The corresponding rms value is obtained by dividing
by
√2, Erms = 700 V/m. The magnetic field strength is Brms = Erms/c =
2
.33 × 10−6 T.
Exercise :
A 40 watt lamp radiates all its energy isotropically. Compute the electric field at
a distance of 2m from the lamp. (Ans. 30 V peak)
Radiation Pressure :
We have seen that electric field, as well as magnetic field, store energy. The
energy density for the electric field was seen to be
(1/2)ǫ0E2 and that for the
magnetic field was found to be
(1/2)B2/2μ0. For the electromagnetic waves,
where
E/B = c, the total energy density is
u
=
1
2
ǫ
0E2 +
B
2
2
μ0
=
ǫ0E2
where we have used
c2 = 1/μ0ǫ0.
In addition to carrying energy, electromagnetic waves carry momentum as well.
11
The relationship between energy (
U) and momentum (p) is given by relatistic
relation for a massless photons as
p = U/c. Since the energy density of the
electromagnetic waves is given by
ǫ0E2, the momentum density, i.e. momentum
per unit volume is
|
p |=
ǫ
0E2
c
=
ǫ0 | ~E × ~B |
Since the direction of momentum must be along the direction of propagation of
the wave, the above can be converted to a vector equation
~p
= ǫ0~E × ~B
If an electromagnetic wave strikes a surface, it will thus exert a pressure. Consider
the case of a beam falling normally on a surface of area
A which absorbs the wave.
The force exerted on the surface is equal to the rate of change of momentum of the
wave. The momentum change per unit time is given by the momentum contained
within a volume
cA. The pressure, obtained by dividing the force by A is thus
given by
P
= cp = cǫ0EB = ǫ0E2
which is exactly equal to the energy density
u.
If on the other hand, the surface reflects the wave, the pressure would be twice the
above value.
The above is true for waves at normal incidence. If the radiation is diffuse, i.e., if it
strikes the wall from all directions, it essentially consists of plane waves travelling
in all directions. If the radiation is isotropic, the intensity of the wave is the
same in all directions. The contribution to the pressure comes from those waves
which are travelling in a direction which has a component along the normal to the
surface. Thus on an average a third of the radiation is responsible for pressure.
The pressure for an absorbing surface is
u/3 while that for a reflecting surface is
2
u/3.
The existence of radiaton pressure can be verified experimentally. The curvature
of a comet’s tail is attributed to the radiation pressure exerted on the comet by
solar radiation.
Exercise :
Assuming that the earth absorbs all the radiation that it receives from the sun,
calculate he radiation pressure exerted on the earth by solar radiation. (Ans.
Assuming diffuse radiation
1.33 × 10−6 N/m2)
Wave Propagation in Matter :
12
Inside matter but in region where there are no sources, the relevant equations are
∇ ・
~D = 0
∇ ・
~B = 0
∇ ×
~E = −
∂B
∂t
∇ ×
~H =
∂D
∂t
which results in a travelling wave with speed
v = 1/√ǫμ where ~B = μ~H and
~D
=
ǫ~H .
Waves in a conducting Medium :
If the medium is conducting, we need to include the effect due to conduction
current. The two curl equations become
∇ ×
~E = −
∂~B
∂t
=
−μ
∂~H
∂t
∇ ×
~H =
∂~D
∂t
+
~ J = ǫ
∂~E
∂t
+
σ~E
where we have used Ohm’s law as another constitutive relation. Thus, we have,
∇ ×
(∇ × ~E) = −μ
∂
∂t
(
∇ × ~H )
Using the expansion for
∇ × (∇ × ~E) = ∇(∇ ・ ~E ) − ∇2E, we get
1
ǫ
∇(∇ ・ ~D) − ∇2~E = −μǫ∂2~E
∂t
2 − μσ
∂E
∂t
It may be noted that though
∇ ・ ~D term equals ρf , free charges, if they exist in a
conductor soon depletes. This may be seen from the equation of continuity. Using
∂ρ
∂t
=
−∇ ・ ~ J
and using Ohm’s law one can see that if at time
t = 0 there exists some free charge
ρ
0
f
, the charge drops exponentially to 1/e th of its value after a time ǫ/σ which is
very small for conductors. Thus the wave equation that we have is
∇
2~E = μǫ∂2~E
∂t
2 + μσ
∂E
∂t
13
and a similar equation for
~H , i.e.
∇
2~H = μǫ∂2~H
∂t
2 + μσ
∂t
∂t
In practice, most generators produce voltages and currents which vary sinusoidally.
Corresponding electric and magnetic fields also vary simusoidally with time. This
implies
∂E/∂t −→ −iωE and ∂2E/∂t2 −→ −ωE. The wave equation becomes
∇
2~E = −μǫω2~E + iμσωE ≡ γ2E
where
γ
2 = −μǫω2 + iμσω
is a complex constant. We may write
γ = α + iβ, squaring and equating it to the
expression for
γ2, we can see that α and β have the same sign. When we take the
square root of
γ2 and write it as γ we assume that α and β are positive. One can
explicitly show that
α
= ω
vuuut
μǫ
2
s
1 +
σ
2
ω
2ǫ2 − 1
β
= ω
vuuut
μǫ
2
s
1 +
σ
2
ω
2ǫ2 + 1
We have seen that in Maxwell’s equation,
σ~E is conduction current density while
the term
iωǫE is the displacement current density. This suggest that the value of
the ratio
σ/ωǫis a good measure to divide whether a material is a dielectric or a
metal. For good conductors this ratio is much larger than 1 over the entire radio
frequency spectrum. For instance, at a frequency as high as 30, 000 MHz, Cu has
σ/ωǫ
≃ 108 while for the same frequency this ratio is 0.0002 for mica.
For good conductors, we take
σ/ωǫ≫ 1. We have, in this limit
α
= β =
r
ωσμ
2
The velocity of the wave in the conductor is
v
=
ω
β
=
s
2
ω
μσ
14
The wave is greatly attenuated inside a conductor.
Depth of penetration or skin
depth
δ is defined as the distance at which an eletromagnetic wave would have
attenuated to 1/e of its value on the surface.
δ
=
1
α
=
s
2
ωμσ
For conductors like copper this distance is typically less than a fraction of a millimeter.
15
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