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Maxwell’s Equations and Electromagnetic Waves
- Chapter
- 13.1 The Displacement Current 13-
- 13.2 Gauss’s Law for Magnetism 13-
- 13.3 Maxwell’s Equations 13-
- 13.4 Plane Electromagnetic Waves 13-
- 13.4.1 One-Dimensional Wave Equation 13-
- 13.5 Standing Electromagnetic Waves 13-
- 13.6 Poynting Vector 13-
- Example 13.1: Solar Constant............................................................................. 13-
- Example 13.2: Intensity of a Standing Wave...................................................... 13-
- 13.6.1 Energy Transport 13-
- 13.7 Momentum and Radiation Pressure................................................................ 13-
- 13.8 Production of Electromagnetic Waves 13-
- Animation 13.1: Electric Dipole Radiation 1.................................................... 13-
- Animation 13.2: Electric Dipole Radiation 2.................................................... 13-
- Animation 13.3: Radiation From a Quarter-Wave Antenna 13-
- 13.8.1 Plane Waves............................................................................................. 13-
- 13.8.2 Sinusoidal Electromagnetic Wave 13-
- 13.9 Summary......................................................................................................... 13-
- 13.10 Appendix: Reflection of Electromagnetic Waves at Conducting Surfaces 13-
- 13.11 Problem-Solving Strategy: Traveling Electromagnetic Waves 13-
- 13.12 Solved Problems 13-
- 13.12.1 Plane Electromagnetic Wave 13-
- 13.12.2 One-Dimensional Wave Equation 13-
- 13.12.3 Poynting Vector of a Charging Capacitor.............................................. 13-
- 13.12.4 Poynting Vector of a Conductor 13-
- 13.13 Conceptual Questions 13-
- 13.14 Additional Problems 13-
- 13.14.1 Solar Sailing........................................................................................... 13-
- 13.14.2 Reflections of True Love 13-
- 13.14.3 Coaxial Cable and Power Flow.............................................................. 13-
- 13.14.4 Superposition of Electromagnetic Waves.............................................. 13-
- 13.14.5 Sinusoidal Electromagnetic Wave 13-
- 13.14.6 Radiation Pressure of Electromagnetic Wave........................................ 13-
- 13.14.7 Energy of Electromagnetic Waves......................................................... 13-
- 13.14.8 Wave Equation....................................................................................... 13-
- 13.14.9 Electromagnetic Plane Wave 13-
- 13.14.10 Sinusoidal Electromagnetic Wave 13-
If the surface bounded by the path is the flat surface , then the enclosed current
is
S 1
I (^) enc= I. On the other hand, if we choose to be the surface bounded by the curve,
then since no current passes through. Thus, we see that there exists an
ambiguity in choosing the appropriate surface bounded by the curve C.
S 2
I (^) enc = 0 S 2
Maxwell showed that the ambiguity can be resolved by adding to the right-hand side of the Ampere’s law an extra term
0 E d
d I dt
which he called the “ displacement current .” The term involves a change in electric flux. The generalized Ampere’s (or the Ampere-Maxwell) law now reads
E d
d d I I I dt
∫ B^^ ⋅^ s =^ +^ =^ +
G G
v (13.1.4)
The origin of the displacement current can be understood as follows:
Figure 13.1.2 Displacement through S 2
In Figure 13.1.2, the electric flux which passes through S 2 is given by
0
E S
Q
d EA
Φ = ∫∫ E ⋅ A = =
G G
w (13.1.5)
where A is the area of the capacitor plates. From Eq. (13.1.3), we readily see that the displacement current Id is related to the rate of increase of charge on the plate by
0
E d
d dQ I dt dt
However, the right-hand-side of the expression, , is simply equal to the conduction
current,
dQ / dt I. Thus, we conclude that the conduction current that passes through S 1 is
precisely equal to the displacement current that passes through S 2 , namely I = Id. With
the Ampere-Maxwell law, the ambiguity in choosing the surface bound by the Amperian loop is removed.
13.2 Gauss’s Law for Magnetism
We have seen that Gauss’s law for electrostatics states that the electric flux through a closed surface is proportional to the charge enclosed (Figure 13.2.1a). The electric field lines originate from the positive charge (source) and terminate at the negative charge (sink). One would then be tempted to write down the magnetic equivalent as
0
m B S
Q
d
Φ = ∫∫ B ⋅ A =
G^ G
w (13.2.1)
where is the magnetic charge (monopole) enclosed by the Gaussian surface. However,
despite intense search effort, no isolated magnetic monopole has ever been observed. Hence, and Gauss’s law for magnetism becomes
Q m
Qm = 0
B^0
S
Φ = ∫∫ B ⋅ d A =
G G
w (13.2.2)
Figure 13.2.1 Gauss’s law for (a) electrostatics, and (b) magnetism.
This implies that the number of magnetic field lines entering a closed surface is equal to the number of field lines leaving the surface. That is, there is no source or sink. In addition, the lines must be continuous with no starting or end points. In fact, as shown in Figure 13.2.1(b) for a bar magnet, the field lines that emanate from the north pole to the south pole outside the magnet return within the magnet and form a closed loop.
13.3 Maxwell’s Equations
We now have four equations which form the foundation of electromagnetic phenomena:
13.4 Plane Electromagnetic Waves
To examine the properties of the electromagnetic waves, let’s consider for simplicity an
electromagnetic wave propagating in the + x -direction, with the electric field E
G
pointing
in the + y- direction and the magnetic field B
G
in the + z -direction, as shown in Figure 13.4. below.
Figure 13.4.1 A plane electromagnetic wave
What we have here is an example of a plane wave since at any instant both E and B
G G
are uniform over any plane perpendicular to the direction of propagation. In addition, the wave is transverse because both fields are perpendicular to the direction of propagation,
which points in the direction of the cross product E × B
G G
Using Maxwell’s equations, we may obtain the relationship between the magnitudes of the fields. To see this, consider a rectangular loop which lies in the xy plane, with the left side of the loop at x^ and the right at x^ +^ ∆ x^. The bottom side of the loop is located at^ ,
and the top side of the loop is located at
y y + ∆ y , as shown in Figure 13.4.2. Let the unit
vector normal to the loop be in the positive z -direction, n ˆ = k ˆ.
Figure 13.4.2 Spatial variation of the electric field E
G
Using Faraday’s law
d d dt
∫ E^ ⋅^ s^ = −^ ∫∫ B^ ⋅ d A
G G G G
v (13.4.1)
the left-hand-side can be written as
( ) ( ) [ ( ) ( )] (^) ( y y y y y
E
d E x x y E x y E x x E x y x y x
∆ ∆ ∆ ∆ ∆ ∆ ∆ (^) )
E s
G G
v (13.4.2)
where we have made the expansion
y^ (^ )^ y ( )^ y
E
E x x E x x x
On the other hand, the rate of change of magnetic flux on the right-hand-side is given by
z ( d B d dt t
∆ x ∆ y )
B A
G G
Equating the two expressions and dividing through by the area ∆ x ∆ y yields
E (^) y Bz x t
The second condition on the relationship between the electric and magnetic fields may be deduced by using the Ampere-Maxwell equation:
0 0
d d dt
∫ B^ ⋅^ s^ =^ μ ε^ ∫∫ E^ ⋅^ d A
G G G G
v (13.4.6)
Consider a rectangular loop in the xz plane depicted in Figure 13.4.3, with a unit normal
n ˆ^ =^ ˆ j.
Figure 13.4.3 Spatial variation of the magnetic field B
G
The line integral of the magnetic field is
Recall that the general form of a one-dimensional wave equation is given by
2 2 2 2 2
( , ) x t 0 x v t
⎜ −^ ⎟ =
⎝ ∂^ ∂ ⎠
where v is the speed of propagation and ψ ( , ) x t is the wave function, we see clearly that
both E (^) y and Bz satisfy the wave equation and propagate with the speed of light:
8 7 12 2 2 0 0
2.997 10 m/s (4 10 T m/A)(8.85 10 C /N m )
v c
μ ε π −^ −
= = = ×
× ⋅ × ⋅
Thus, we conclude that light is an electromagnetic wave. The spectrum of electromagnetic waves is shown in Figure 13.4.4.
Figure 13.4.4 Electromagnetic spectrum
13.4.1 One-Dimensional Wave Equation
It is straightforward to verify that any function of the form ψ ( x ± vt )satisfies the one-
dimensional wave equation shown in Eq. (13.4.14). The proof proceeds as follows:
Let x ′ = x ± vt which yields ∂ x ′/ ∂ x = 1 and ∂ x ′/∂ = ± t v. Using chain rule, the first two partial derivatives with respect to x are
( x ) x x x x x
∂ ψ ′^ ∂ ψ ∂ ′ ∂ ψ
∂ ∂ ′^ ∂ ∂ ′
2 2 2 2
x^2 x x x x x x^2
Similarly, the partial derivatives in t are given by
x v t x t x
2 2 2 2 2
x v v v t t x x t x
2 2
∂ ∂ ⎝ ∂ ′^ ⎠ ∂ ′ ∂ ∂′^
Comparing Eq. (13.4.17) with Eq. (13.4.19), we have
2 2 2 2 2 2
x ' x v t^2
which shows that ψ ( x ± vt )satisfies the one-dimensional wave equation. The wave
equation is an example of a linear differential equation, which means that if ψ 1 ( , ) x t and
ψ 2 ( , ) x t are solutions to the wave equation, then ψ 1 ( , ) x t ± ψ 2 ( , ) x t is also a solution. The
implication is that electromagnetic waves obey the superposition principle.
One possible solution to the wave equations is
0 0
0 0
( , )ˆ cos ( )ˆ cos( )
( , ) ˆ^ cos ( ) ˆ cos( )
y
z
E x t E k x vt E kx t ˆ
B x t B k x vt B kx t ˆ
E j j j
B k k
G
G
k
where the fields are sinusoidal, with amplitudes E 0 and B 0. The angular wave number k is
related to the wavelength λ by
k
and the angular frequency ω is
v
ω kv π π f
where f is the linear frequency. In empty space the wave propagates at the speed of light,
. The characteristic behavior of the sinusoidal electromagnetic wave is illustrated in Figure 13.4.5.
v = c
- The ratio of the magnitudes and the amplitudes of the fields is
0 0
E E
c B B k
4. The speed of propagation in vacuum is equal to the speed of light, c = 1/ μ 0 ε 0.
- Electromagnetic waves obey the superposition principle.
13.5 Standing Electromagnetic Waves
Let us examine the situation where there are two sinusoidal plane electromagnetic waves, one traveling in the + x -direction, with
E 1 (^) y ( , ) x t = E 10 (^) cos( k x 1 − ω 1 t ), B 1 (^) z ( , ) x t = B 10 (^) cos( k x 1 − ω 1 t ) (13.5.1)
and the other traveling in the − x -direction, with
E 2 (^) y ( , ) x t = − E 20 (^) cos( k x 2 + ω 2 t ), B 2 (^) z ( , ) x t = B 20 (^) cos( k x 2 + ω 2 t ) (13.5.2)
For simplicity, we assume that these electromagnetic waves have the same amplitudes ( E 10 (^) = E 20 (^) = E 0 , B 10 (^) = B 20 (^) = B 0 ) and wavelengths ( k 1 (^) = k 2 (^) = k , ω 1 = ω 2 = ω). Using the
superposition principle, the electric field and the magnetic fields can be written as
E (^) y ( , ) x t = E 1 (^) y ( , ) x t + E 2 (^) y ( , ) x t = E 0 (^) [ cos( kx − ω t ) − cos( kx + ω t )] (13.5.3)
and
Bz ( , ) x t = B 1 (^) z ( , ) x t + B 2 (^) z ( , ) x t = B 0 (^) [ cos( kx − ω t ) + cos( kx + ω t )] (13.5.4)
Using the identities
cos( α ± β ) = cos α cos β ∓ sin α sinβ (13.5.5)
The above expressions may be rewritten as
0 [^ ] 0
( , ) cos cos sin sin cos cos sin sin 2 sin sin
E y x t E kx t kx t kx t kx t E kx t
and
0 [^ ] 0
( , ) cos cos sin sin cos cos sin sin 2 cos cos
B z x t B kx t kx t kx t kx t B kx t
One may verify that the total fields E (^) y ( , ) x t and Bz ( , ) x t still satisfy the wave equation
stated in Eq. (13.4.13), even though they no longer have the form of functions of kx ± ω t.
The waves described by Eqs. (13.5.6) and (13.5.7) are standing waves , which do not propagate but simply oscillate in space and time.
Let’s first examine the spatial dependence of the fields. Eq. (13.5.6) shows that the total electric field remains zero at all times if sin kx = 0 , or
, 0,1, 2, (nodal planes of ) 2 / 2
n n n x n k
π π λ π λ
= = = = E
G
The planes that contain these points are called the nodal planes of the electric field. On the other hand, when sin kx = ± 1 , or
, 0,1, 2, (anti-nodal planes of ) 2 2 2 / 2 4
n x n n n k
π π λ π λ
E
G
the amplitude of the field is at its maximum. The planes that contain these points are
the anti-nodal planes of the electric field. Note that in between two nodal planes, there is an anti-nodal plane, and vice versa.
For the magnetic field, the nodal planes must contain points which meets the condition cos kx = 0. This yields
, 0,1, 2, (nodal planes of ) 2 2 4
n x n n k
π = ⎛⎜^ + ⎞⎟^ = ⎛⎜^ + ⎞⎟λ = ⎝ ⎠ ⎝ ⎠
B
G
Similarly, the anti-nodal planes for B
G
contain points that satisfy cos kx = ± 1 , or
, 0,1, 2, (anti-nodal planes of ) 2 / 2
n n n x n k
π π λ π λ
= = = = B
G
Thus, we see that a nodal plane of E
G
corresponds to an anti-nodal plane of B , and vice versa.
G
For the time dependence, Eq. (13.5.6) shows that the electric field is zero everywhere
when sin ω t = 0 , or
2 2 0 0
E B 2
B
dU uA dx u u A dx ε E A dx
where
2 2 0 0
E B
B
u ε E u μ
are the energy densities associated with the electric and magnetic fields. Since the electromagnetic wave propagates with the speed of light c , the amount of time it takes for the wave to move through the volume element is dt = dx / c. Thus, one may obtain the rate of change of energy per unit area, denoted with the symbol S , as
2 2 0 (^20)
dU c B S E A dt
The SI unit of S is W/m^2. Noting that E = cB and c = 1/ μ 0 ε 0 , the above expression
may be rewritten as
2 2 2 0 (^20 )
c B cB
S ε E c ε 0 E^2
0
EB
In general, the rate of the energy flow per unit area may be described by the Poynting
vector S
G
(after the British physicist John Poynting), which is defined as
0
μ
S = E × B
G G G
with S pointing in the direction of propagation. Since the fields and
G
E
G
B
G
are
perpendicular, we may readily verify that the magnitude of S
G
is
0 0
EB
S
×
E B
S
G G
G
As an example, suppose the electric component of the plane electromagnetic wave is
E = E 0 cos( kx − ω t )ˆ j
G
. The corresponding magnetic component is B = B 0 cos( kx − ω t ) k ˆ
G
and the direction of propagation is + x. The Poynting vector can be obtained as
0 0 2 0 0 0 0
cos( ) cos( ) cos ( )
E B
E kx ω t B kx ω t kx ω t
S = − j × − k = − i
G
Figure 13.6.2 Poynting vector for a plane wave
As expected, S points in the direction of wave propagation (see Figure 13.6.2).
G
The intensity of the wave, I , defined as the time average of S , is given by
2 2 0 0 2 0 0 0 0 0 0
cos ( ) 2 2 2
E B E B E cB I S kx t c
where we have used
cos (^2 ) 1 2
kx − ω t = (13.6.9)
To relate intensity to the energy density, we first note the equality between the electric and the magnetic energy densities:
2 2 2 2 0 2 0 0 0
B
B E c E E u c
= = = = = uE (13.6.10)
The average total energy density then becomes
2 0 2 0 0 2 (^2 ) 0 0
u u (^) E u (^) B E E
B B
Thus, the intensity is related to the average energy density by
I = S = c u (13.6.12)
Example 13.1: Solar Constant
At the upper surface of the Earth’s atmosphere, the time-averaged magnitude of the
Poynting vector, S = 1.35 × 10 W m^32 , is referred to as the solar constant.
Figure 13.6.3 (a) a spherical wave, and (b) plane wave.
Example 13.2: Intensity of a Standing Wave
Compute the intensity of the standing electromagnetic wave given by
E y ( , ) x t = 2 E 0 cos kx cos ω t , Bz ( , ) x t = 2 B 0 sin kx sin ω t
Solution:
The Poynting vector for the standing wave is
0 0 0 0 0 0 0 0 0 0
(2 cos cos ) (2 sin sin )
(^4) ˆ (sin cos sin cos )
(sin 2 sin 2 )ˆ
E kx t B kx t
E B kx kx t t
E B kx t
×
= = ×
E B
S j k
i
= i
G G
G
The time average of S is
0 0 0
sin 2 sin 2 0
E B
S kx ω
= t = (13.6.15)
The result is to be expected since the standing wave does not propagate. Alternatively, we may say that the energy carried by the two waves traveling in the opposite directions to form the standing wave exactly cancel each other, with no net energy transfer.
13.6.1 Energy Transport
Since the Poynting vector S represents the rate of the energy flow per unit area, the rate of change of energy in a system can be written as
G
dU d dt
= − ∫∫ S ⋅ A
G G
w (13.6.16)
where , where is a unit vector in the outward normal direction. The above
expression allows us to interpret S
d A = dA^ ˆ
G
n n ˆ G as the energy flux density, in analogy to the current
density J in
G
dQ I d dt
= = ∫∫ J ⋅ A
G G
If energy flows out of the system, then S = S n ˆ
G
and dU / dt < 0 , showing an overall
decrease of energy in the system. On the other hand, if energy flows into the system, then
S = S ( −ˆ)and , indicating an overall increase of energy.
G
n 0
dU / dt >
As an example to elucidate the physical meaning of the above equation, let’s consider an inductor made up of a section of a very long air-core solenoid of length l , radius r and n turns per unit length. Suppose at some instant the current is changing at a rate. Using Ampere’s law, the magnetic field in the solenoid is
dI / dt > 0
C
∫ B^ ⋅^ d^ s =^ Bl^ =^ μ NI
G G
v
or
B = μ 0 nI k ˆ
G
Thus, the rate of increase of the magnetic field is
0
dB dI n dt dt
According to Faraday’s law:
B C
d d dt
= ∫ E ⋅ s = −
G G
v (13.6.20)
changing magnetic flux results in an induced electric field., which is given by
( 2 ) 0 2
dI E r n r dt
π = −μ ⎛⎜^ ⎞⎟π
or
nr dI dt
E φ
G
