Chapter 32
Maxwell’s Equations and
Electromagnetic Waves
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 2
Maxwell’s Equations and EM Waves
• Maxwell’s Displacement Current
• Maxwell’s Equations
• The EM Wave Equation
• Electromagnetic Radiation
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 3
µ µ∫ ∫� �� �i i� o o C
C S
B dl = J dA = I
Something is Missing From Ampere’s Law
The surface S in the integral above can be
any surface whose boundary is C.
If the surface S2 is chosen for
use in the above integral the
result will be that the magnetic
field around C is zero. But there
is current flowing through the
wire so we know there is a
magnetic field present.
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 4
µ µ∫ ∫� �� �i i� o o C
C S
B dl = J dA = I
Something is Missing From Ampere’s Law
The surface S2 has the same boundary as S1 but there is no current
passing through S2. The charge is accumulating on the capacitor .
Maxwell noticed this deficiency in
Ampere’s law and fixed it by
defining the Displacement Current Id.
He began by taking surfaces S1 and
S2, putting them together and treating
them like one closed surface S
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 5
µ µ∫ ∫� �� �i i� o o C
C S
B dl = J dA = I
The Displacement Current
Charge is building up on the disk within the closed surface S.
Therefore there are electric field lines, E, that are crossing the
surface S. We can use Gauss’s Law here.
∂ ∂
∂ ∂
∂
∂
∫��i�
enclosed
e
S o
e
d
o o
e
d o
Qφ = E dA =
ε
φ 1 Q 1 = = I
t ε t ε
φI =ε
t
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 6
∫ ∫� �� �i i� o o C
C S
B dl = µ J dA = µ I
The Displacement Current
Maxwell fixed the problem with Ampere’s law by adding
another current to the right hand side of the equation below
∂
∂e
d o
φI =ε
t
∂
∂∫ ∫� �� �i i�
e
o o C o d o C o 0
C S
φB dl = µ J dA = µ I + µ I = µ I + µ ε
t
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 7
Displacement Current Example
In calculating the displacement
current we will be making the
approximation that the electric
field is everywhere uniform.
This requires that the plate
separation be much smaller than
R, the radius of the plate.
The surface S must not extend
past the edge of the capacitor
plates. So r must be less than R.
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 8
Displacement Current Example
In calculating the displacement
current we will need to compute
the electric flux across the
surface S.e
d o
dφI =ε
dt
ˆ∫�ie
S
φ = E ndA = EA
o o o
Qσ QAE = = =ε ε ε A
( )
d o o o
o
d EA dE d Q dQI = ε = ε A = ε A =
dt dt dt ε A dt
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 9
B-Field from the Displacement Current
In calculating the B-Field from the
displacement current we will be
making the same approximations
that were made in the last example:
the electric field is everywhere
uniform.
∂
∂∫��i�
e
o C o 0
C
φB dl = µ I + µ ε
t
( )∫��i�
C
B dl = B 2πr
There is no current through S so IC is zero
( )∫��i�
eo o
C
dφB dl = B 2πr = 0 + µ ε
dt
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 10
B-Field from the Displacement Current
( )∫��i�
eo o
C
dφB dl = B 2πr = 0 + µ ε
dt
2 2
e
o
σφ = AE = πr E = πr
ε2
2 2
e 2 2
o o o
σ Q Qrφ = πr = πr =
ε ε πR ε R
The size of S will vary so φe will depend on r
( )
=
2 2
o o o2 2
o
o o
2 2
d Qr r dQB 2πr = µ ε = µ
dt dt ε R R
µ µr dQ rB = I
2π dt 2πR R
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 11
Maxwell’s Equations
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 12
Maxwell’s Equations
( )∂
∂∫��i�
e
o C d o C o 0
C
φB dl = µ I + I = µ I + µ ε
tAmpere’s Law
Faraday’s Law
∫� insiden0S
QE dA =
εGauss’s Law
∫� nS
B dA = 0 No name - there are no magnetic monopoles
[ ] [ ]∂ ∂
∂ ∂∫ ∫��i�
m n
C
φ B= E dl = 0 - = 0 - dA
t tε
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 13
Maxwell’s Equations
( )∂
∂∫ ∫��i�
n
o C d o C o 0
C S
EB dl = µ I + I = µ I + µ ε dA
tAmpere’s Law
Faraday’s Law
∫ ∫� � insidenS V0 0
Q1E dA = ρdV =
ε εGauss’s Law
∫� nS
B dA = 0 No name - there are no magnetic monopoles
∂ ∂
∂ ∂∫ ∫��i�
m n
C
φ B= E dl = - = - dA
t tε
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 14
EM Wave Equation
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 15
Conservative Forces and Potentials
from Vector Analysis
Work around a closed loop = 0
Stokes Theorem
Therefore a potential function V exists for a conservative force.
( )
( )
( )
⋅
⋅ = ∇ ⋅
∇ ⋅ ⇒ ∇
∇ ∇ ∇
∫
∫ ∫
∫
��
�� � � �
� � � ��
� � � �
�
�
C
C S
S
W = F dl = 0
F dl × F da
× F da = 0 × F = 0
F = - V since × V = 0
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 16
Vector Analysis
ψ��
φ and are scalar functions
F and G are vector functions
∇
∇ ⋅
∇
�
� � � �
� � � �
φ= grad φ= gradient of φ
F = div F = divergence of F
× F = curl F = curl of F
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 17
Vector Analysis
Gradient
Divergence
Curl
ˆˆ ˆφ φ φ∂ ∂ ∂
∇∂ ∂ ∂
�φ= i + j + k
x y z
∂∂ ∂∇ ⋅
∂ ∂ ∂
� �yx z
FF FF = + +
x y z
ˆˆ ˆ
∂ ∂ ∂∇ ×
∂ ∂ ∂
� �
x y z
i j k
F =x y z
F F F
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 18
Vector Identities
( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
∇ ∇ ∇
∇ ∇ ∇ ∇ ∇
∇ ∇ ∇
� � �
� � � � �� � � � � � � � � �i i i
� � �� � �i i i
fg = f g + g f
A B = B A+ A B + B× × A + A× × B
fA = f A+ f A
( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( ) ( ) ( )
( )
∇ ∇ ∇
∇ ∇ ∇
∇ ∇ ∇ ∇ ∇
∇ ∇ ∇ ∇ ∇
� � �� � � � � �i i i
� � �� � �
� � � � �� � � � � � � � � �i i i i
� � �� � � �i
2
A× B = B × A - A × B
× fA = f × A+ f × A
× A× B = B A - A B + B A - A B
× × A = A - A
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 19
Vector Identities
( ) ˆ
ˆ
ˆ
∂ ∂ ∂∇ ∂ ∂ ∂
∂ ∂ ∂ + ∂ ∂ ∂
∂ ∂ ∂+ ∂ ∂ ∂
� � �i
x x xx y z
y y y
x y z
z z zx y z
B B BA B = A + A + A i
x y z
B B BA + A + A j
x y z
B B BA + A + A k
x y z
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 20
Maxwell’s Equations:
Integral Form to Differential Form
Stokes Theorem
Divergence Theorem
ˆ∇∫ ∫�� � �i i�
C S
E dl = × E ndA
ˆ ∇∫ ∫� � �i i�
S V
F ndA = FdV
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 21
Maxwell’s Equations
∂
∂∫ ∫��i�
n
o C o 0
C S
EB dl = µ I + µ ε dA
tAmpere’s Law
Faraday’s Law
∫ ∫� �nS V0
1E dA = ρdV
εGauss’s Law
∫� nS
B dA = 0 No name - there are no magnetic monopoles
∂
∂∫ ∫��i�
n
C
BE dl = - dA
t
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 22
Maxwell’s Equations
ˆ =∫ ∫ ∫�i� � �n
S S V0
1E dA = E ndA ρdV
εGauss’s Law
Use the Divergence Theorem to recast the surface
integral into a volume integral
ˆ ρ∇ =∫ ∫ ∫� � �i i�
oS V V
1E ndA = EdV dV
ε
ρ
ρ
∇ =
∇ =
∫� �i
� �i
oV
o
E - dV 0ε
E - 0ε
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 23
Maxwell’s Equations
∂
∂∫ ∫��i�
n
o C o 0
C S
EB dl = µ I + µ ε dA
tAmpere’s Law
Faraday’s Law
∫ ∫� �nS V0
1E dA = ρdV
εGauss’s Law
∫� nS
B dA = 0
∂
∂∫ ∫��i�
n
C
BE dl = - dA
t
ρ∇� �i
o
E =ε
∇ =� �iB 0
∂∇ ×
∂
�� � B
E + = 0t
0 0µ ε
∂∇ ×
∂
�� � �
o m
1 EB - = µ J
t
Integral form Differential form
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 24
Wave Eqn from Maxwell’s Eqn
The differential form of Maxwell’s equations brings out
the symmetry and non-symmetry of the E and B fields
We will use the following vector identity with the E-field
( )∇ ∇ ∇ ∇ ∇� � �� � � �
i2
× × A = A - A
( )∇ ∇ ∇ ∇ ∇� � � � � � �
i2
× × E = E - E ∂∇ ×
∂
�� � B
E + = 0t
( ) ∂∇ ∇ ∇ ∇∂
� � � � � �i
2E - E = × B
t
( ) ∂ ∂
∇ ∇ ∇ + ∂ ∂
�� � � � �
i2
f
EE - E = µJ εµ
t t
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 25
Wave Eqn from Maxwell’s Eqn
( ) ∂ ∂
∇ ∇ ∇ + ∂ ∂
�� � � � �
i2
f
EE - E = µJ εµ
t t
These are the source
terms
ρ
ε
∂∂ ∇ + ∇
∂ ∂
��� �2
f2
2
JEE -εµ = µ
t t
∂∇
∂
∂∇
∂
��
��
22
o o 2
22
2 2
o o
EE - ε µ = 0
t
1 E 1E - = 0; where c =
c t ε µ
In free space there are no sources
This is the form of a wave equation
is the speed of light
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 26
Solutions of the Wave Equation
In free space the solutions of the wave equations show
that E and B are in phase.
These equations describe plane waves that are uniform
through out any plane perpendicular to the x-axis.
( )
x xo
y yo
E E= sin kx -ωt
B B
2π 2πk = ; ω= = 2πf
λ T
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 27
Plane Polarized Waves
( )
x xo
y yo
E E= sin kx -ωt
B B
2π 2πk = ; ω= = 2πf
λ T
Examining the E and B components
show that this represents a plane
polarized wave.
The E vector is oriented in the x
direction and the B vector is
oriented in the y direction.
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 28
Relationships Between E and B Vectors
oo o o
Ek 1B = E = E =
ω c c
E = cB
The Poynting Vector describes the propagation of the
electromagnetic energy � ��
o
E× BS =
µ
With E in the x-direction and B in the y-direction the
energy flows in the z-direction.
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 29
Relationships Between E and B Vectors
The Poynting Vector describes the propagation of the
electromagnetic energy � ��
o
E× BS =
µ
( ) ( )
( )
ˆ ˆ
ˆ
� �
� �
o o
2
o o
E× B = E sin kx -ωt i× B sin kx -ωt j
E × B = E B sin kx -ωt k
The energy is proportional to E and B and is flowing
in the z-direction, perpendicular to E and B.
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 30
The Principle of Invariance
The laws of Physics should be the same for all
non-accelerated observers.
Einstein’s fundamental postulate of relativity can be stated:
“It is physically impossible to detect the uniform motion of a
frame of reference from observations made entirely within
that frame.”
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 31
The Principle of Invariance
The laws of Physics should be the same for all
non-accelerated observers.
If two observers watch the motion of an object from
two different inertial reference systems (no
acceleration), moving at a relative velocity v, they
should find the same laws of Physics
F1 = m1a1 and F2=m2a2
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 32
Galilean and Lorentz Transformations
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 33
Galilean and Lorentz Transformations
The inertial reference frames are related by a Galilean transformation.
Newton’s laws are invariant under these transformations but not Maxwell’s
Equations
Prior to Einstein’s Theory of Special Relativity it was determined that a
Lorentz transformation kept Maxwell’s equation invariant. However, no one
knew exactly what they meant.
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 34
Galilean and Lorentz Transformations
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 35
Electromagnetic Radiation
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 36
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 37
These are all different forms of
electromagnetic radiation.
Anytime you accelerate or
decelerate a charged particle it
gives off electromagnetic radiation.
Electrons circulating about their
nuclei don't give off radiation
unless they change energy levels.
Thermal motion gives off continuous EM radiation. Example –
Infrared radiation which peaks below the visible spectrum.
Electromagnetic Radiation
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 38
Electric Dipole Radiation
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 39
Electric Dipole Radiation
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 40
Dipole Antenna - Radiation Distribution
Note the different orientation
of the angle measurement
2
2
sinI(θ)
r
θ∝
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 41
Dipole Antenna - Radiation Distribution
2
o 2
sin θI(θ) = I
r
In these problems you will need to determine the value of
Io or else take a ratio so that the Io factor will cancel out.
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 42
Dipole Antenna - Radiation Distribution
2
o 2
sin θI(r,θ) = I
r
(a.) Find I1 at r1 = 10m and θ = 90o
(b.) Find I2 at r2 = 30m and θ = 90o
(c.) Ratio of I2 / I1
=2
o o1 o 2
I1I = I(r = 10,θ = 90 ) = I
10 100
=2
o o2 o 2
I1I = I(r = 30,θ = 90 ) = I
30 900
(a.)
(b.)
(c.)o
2
o1
II 1900= =
II 9100
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 43
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 44
Electric - Dipole Antenna
http://www.austincc.edu/mmcgraw/physics_simulations.htm
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 45
http://www.falstad.com/mathphysics.html
Oscillating Ring Antenna
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 46
http://www.falstad.com/mathphysics.html
Oscillating Ring Pair Antenna
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 47
Electric - Dipole Antenna
Plane wave – Far from source antenna - “Far Field”
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 48
Magnetic - Loop Antenna
Plane wave – Far from source antenna - “Far Field”
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 49
Magnetic - Loop Antenna
( )
( )
( )
( )
∂
∂
∂
∂
∂
∂
∂ ∂
2m
2
rms
o
o
o
o rmsrms
rms
rms
d BAdφ B= - = - = - πr
dt dt t
B= πr
t
B = B sin kx -ωt
B= -ωB cos kx -ωt
t
ωBB= ωB -cos kx - ωt = = ωB
t 2
ε
ε
c
∂=
∂
= =
2 2
rms
rms
22 2 2rms
rms rms
rms
rms
B= πr πr ωB
t
E 2π= πr ωB πr ω r fE
c
ε
ε
Find εrms ?Find εrms ?
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 50
Magnetic - Loop Antenna
22
rmsrms
2π= r f E
cε
rms
r = 10.0cm
N = 1
E = 0.150 V/m
f : (a.) 600 kHz; (b.) 60.0 MHz
( ) ( ) =2
2 3
rms
2π= 0.10 600x10 0.150 59.2µV
cε(a.)
(b.) ( ) ( ) =2
2 6
rms
2π= 0.10 60x10 0.150 5.92mV
cε
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 51
Energy and Momentum in an
Electromagnetic Wave
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 52
Energy and Momentum in an
Electromagnetic Wave
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 53
The Poynting Vector describes the propagation of the
electromagnetic energy � ��
o
E× BS =
µ
avg avg
avg avg
avg
avg
U u LAP = = = u Ac
∆t L c
P I = = u c
A
Uavg is the total energy and uavg is the energy density.
I is the intensity, the average power per unit area.
E-M Energy and Momentum
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 54
E-M Energy and Momentum
22
e o m
o
1 Bu = ε E and u =
2 2µ
These are the electric and magnetic energy densities
Since E = cB
( )2
2 22
m o e2
o o o
E cB E 1u = = = = ε E = u
2µ 2µ 22µ c
Therefore the energy density can be expressed in different ways.
=2
2
e m o
o o
B EBu = u + u = ε E =
µ µ c
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 55
E-M Energy and Momentum
The energy density
=2
2
e m o
o o
B EBu = u + u = ε E =
µ µ c
�rms rms o o
avg avg
o o
E B E B1I = u c = = = S
µ 2 µ
The intensity I is the energy/(m2 sec) = power/m2;
� ��
o
E× BS =
µ
This is the Poynting vector, its magnitude is the intensity.
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 56
Radiation Pressure pr
=
ir avg
2 2
o o rms rms o or 2
o o o o
Momentum Ip = = u
Unit Area Unit Time c
E B E B E BIp = = = = =
c 2µ c µ c 2µ c 2µ
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 57
Radiation Pressure pr - Example
A lightbulb emits spherically symmetric electromagnetic waves in al
directions. Assume 50W of electromagnetic radiation is emitted. Find (a)
the intensity, (b) the radiation pressure, (c) the electric and magnetic field
magnitudes at 3.0m from the bulb.
The energy spreads out uniformly over a sphere of radius r.
The surface area of the sphere is 4πr2.
2 2
Power 50 WI = Intensity = = = 0.442
Area 4πr m(a.)
(b.)-9
r 8
I 0.442p = = = 1.47x10 Pa
c 3.0x10
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 58
Radiation Pressure pr - Example
A lightbulb emits spherically symmetric electromagnetic waves in al
directions. Assume 50W of electromagnetic radiation is emitted. Find (a)
the intensity, (b) the radiation pressure, (c) the electric and magnetic field
magnitudes at 3.0m from the bulb.
(c.) Remember
2 2
o or o o2
o o
E BIp = = = and E = cB
c 2µ c 2µ
( )( )( )
( )
-9 -7
o o r
-8
o
8 -8
0 o
0
B = 2µ p = 1.47x10 2 4πx10
B = 6.08x10 T
E = cB = 3.0x10 6.08x10
VE = 18.2
m
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 59
Extra Slides
MFMcGraw-PHY 2426 Chap32a-Maxwell's Eqn-Revised: 7/13/2013 60