Lecture 06 maxwell eqn

ELEN 3371 Electromagnetics Fall 2008
1
Maxwell’s Equations

2
Maxwell’s equations
The behavior of electric and magnetic waves can be fully described by a set
of four equations (which we have learned already).
B
E
t
∂
∇× = −
∂
D
H J
t
∂
∇× = +
∂
vD ρ∇ =g
0B∇ =gGauss’s Law for magnetism
Gauss’s Law for electricity
Ampere’s Law
Faraday’s Law of induction (6.2.1)
(6.2.2)
(6.2.3)
(6.2.4)

3
And the constitutive relations:
D Eε=
B Hµ=
J Eσ=
They relate the electromagnetic field to the properties of the material, in
which the field exists. Together with the Maxwell’s equations, the
constitutive relations completely describe the electromagnetic field.
Even the EM fields in a nonlinear media can be described through a
nonlinearity existing in the constitutive relations.
(6.3.1)
(6.3.2)
(6.3.3)

4
L s
B
E dl ds
t∆ ∆
∂
= −
∂∫ ∫g gÑ
Gauss’s Law for magnetism
Gauss’s Law for electricity
Ampere’s Law
Faraday’s Law of induction (6.4.1)
(6.4.2)
(6.4.3)
(6.4.4)
Integral form
L s
D
H dl J ds
t∆ ∆
∂ 
= − + ÷
∂ 
∫ ∫g gÑ
v
S v
D ds dvρ
∆ ∆
=∫ ∫gÑ
0
S
B ds
∆
=∫ gÑ

5
Example 6.1: In a conductive material we may assume that the conductive
current density is much greater than the displacement current density. Show that
the Maxwell’s equations can be put in a form of a Diffusion equation in this
material. B
E
t
∂
∇× = −
∂
H J Eσ∇× ≈ =
We can write:
and, neglecting the
displacement current:
(6.5.1)
(6.5.2)
Taking curl of (6.5.2): H Eσ∇×∇× = ∇×
Expanding the LHS: 2
0 0
B B B
t
σ
µ µ
    ∂
∇ ∇ −∇ = − ÷  ÷
∂   
g
(6.5.3)
(6.5.4)
The first term is zero and 2
0
B
B
t
µ σ
∂
∇ =
∂
(6.5.5)
Is the diffusion equation with a diffusion coefficient D = 1/(σµ0)

6
Example 6.2: Solve the diffusion equation for the case of the magnetic flux
density Bx(z,t) near a planar vacuum-copper interface, assuming for copper: µ =
µ0 and σ = 5.8 x 107
S/m. Assume that a 60-Hz time-harmonic EM signal is
applied.
Assuming ejωt
time-variation, the diffusion equation is transformed to the
ordinary differential equation:
2
02
( )
( )x
x
d B z
j B z
dz
µ σω=
2
0 0j j jγ ωµ σ γ α β ωµ σ= ⇒ = + =
Where z is the normal coordinate to the boundary. Assuming a variation in
the z-direction to be Bx(z) = B0e-γz
, we write:
(6.6.1)
(6.6.2)

7
The magnitude of the magnetic flux density decays exponentially in the z
direction from the surface into the conductor
0( ) z
xB z B e α−
=
where
7 7 1
0 60 4 10 5.8 10 117.2f mα π µ σ π π − −
= = × × × × × =
The quantity δ = 1/α is called a “skin depth” - the
distance over which the current (or field) falls to 1/e of
its original value.
For copper, δ = 8.5 mm.
(6.7.1)
(6.7.2)

8
Example 6.3: Derive the equation of continuity starting from the Maxwell’s equations
The Gauss’s law: vD ρ∇ =g
Taking time derivatives:
v D
D
t t t
ρ∂ ∂ ∂
= ∇ = ∇
∂ ∂ ∂
g g
From the Ampere’s law
D
H J
t
∂
= ∇× −
∂
Therefore:
v
H J
t
ρ∂
= ∇ ∇× −∇
∂
g g
The equation of continuity:
v
J
t
ρ∂
= −∇
∂
g
(6.8.1)
(6.8.2)
(6.8.3)
(6.8.4)
(6.8.5)

9
Poynting’s Theorem
It is frequently needed to determine the direction the power is flowing. The
Poynting’s Theorem is the tool for such tasks.
We consider an arbitrary
shaped volume:
B
E
t
∂
∇× = −
∂
D
H J
t
∂
∇× = +
∂
Recall:
We take the scalar product of E and subtract it from the scalar product of H.
B D
H E E H H E J
t t
∂ ∂ 
∇× − ∇× = − − + ÷
∂ ∂ 
g g g g
(6.9.1)
(6.9.2)
(6.9.3)

10
Using the vector identity
( )A B B A A B∇ × = ∇× − ∇×g g g
Therefore:
( )
B D
E H H E E J
t t
∂ ∂
∇ × = − − −
∂ ∂
g g g g
Applying the constitutive relations to the terms involving time derivatives, we get:
( ) ( )2 21 1
2 2
B D
H E H H E E H E
t t t t
µ ε µ ε
∂ ∂ ∂ ∂
− − = − + = − +
∂ ∂ ∂ ∂
g g g g
(6.10.1)
(6.10.2)
(6.10.3)
Combining (6.9.2) and (6.9.3) and integrating both sides over the same ∆v…

11
( )2 21
( )
2v v v
E H dv H E dv E Jdv
t
µ ε
∂
∇ × = − + −
∂∫ ∫ ∫V V V
g g
Application of divergence theorem and the Ohm’s law lead to the PT:
( )2 2 21
( )
2s v v
E H ds H E dv E dv
t
µ ε σ
∂
× = − + −
∂∫ ∫ ∫V V V
gÑ
Here S E H= ×
is the Poynting vector – the power density and the
direction of the radiated EM fields in W/m2
.
(6.11.1)
(6.11.2)
(6.11.3)

12
The Poynting’s Theorem states that the power that leaves a region is
equal to the temporal decay in the energy that is stored within the
volume minus the power that is dissipated as heat within it – energy
conservation.
EM energy density is 2 21
2
w H Eµ ε = + 
Power loss density is
2
Lp Eσ=
The differential form of the Poynting’s Theorem:
L
w
S p
t
∂
∇ + = −
∂
g
(6.12.1)
(6.12.2)
(6.12.3)

13
Example 6.4: Using the Poynting’s Theorem,
calculate the power that is dissipated in the
resistor as heat. Neglect the magnetic field that
is confined within the resistor and calculate its
value only at the surface. Assume that the
conducting surfaces at the top and the bottom of
the resistor are equipotential and the resistor’s
radius is much less than its length.
The magnitude of the electric field is
0E V L=
and it is in the direction of the current.
(6.13.1)
The magnitude of the magnetic field intensity at the outer surface of the resistor:
( )2H I aπ= (6.13.2)

14
S E H= × (6.14.1)The Poynting’s vector
is into the resistor. There is NO energy stored in the
resistor. The magnitude of the current density is in the
direction of a current and, therefore, the electric field.
2
I
J
aπ
= (6.14.2)
The PT:
20 0
2
0 0
2 (0 0)
2 v
V VI d I
aL dv a L
L a dt a L
V I V I
π π
π π
− = − + −
⇒ − = −
∫V
(6.14.3)
(6.14.4)
The electromagnetic energy of a battery is completely absorbed with
the resistor in form of heat.

15
Example 6.5: Using Poynting’s Theorem,
calculate the power that is flowing through
the surface area at the radial edge of a
capacitor. Neglect the ohmic losses in the
wires, assume that the radius of the
plates is much greater than the
separation between them: a >> b.
Assuming the electric field E is uniform and confined between the plates, the total
electric energy stored in the capacitor is:
2
2
2
E
W a b
ε
π=
The total magnetic energy stored in the capacitor is zero.
(6.15.1)

16
The time derivative of the electric energy is
2dW dE
a bE
dt dt
επ− = − (6.16.1)
This is the only nonzero term on the RHS of PT since an ideal capacitor does not
dissipate energy.
We express next the time-varying magnetic field intensity in terms of the
displacement current. Since no conduction current exists in an ideal capacitor:
s
E
H dl ds
t
ε
∂
=
∂∫ ∫V
g gÑ (6.16.2)
2
2
2
dE a dE
aH a H
dt dt
ε
π ε π= ⇒ =
Therefore:
(6.16.3)

17
(6.17.1)
The power flow would be:
( )S
s
P E H ds= ×∫V
gÑ
In our situation: 2 rds ab uπ=
and 1rS u = −g
2
2S abEH
dE
P a bE
dt
εππ= −− =Therefore:
We observe that S
dW
P
dt
= −
The energy is conserved in the circuit.
(6.17.2)
(6.17.3)
(6.17.4)
(6.17.5)

18
Time-harmonic EM fields
Frequently, a temporal variation of EM fields is harmonic; therefore,
we may use a phasor representation:
( , , , ) Re ( , , )
( , , , ) Re ( , , )
j t
j t
E x y z t E x y z e
H x y z t H x y z e
ω
ω
 =  
 =  
It may be a phase angle between the electric and the magnetic fields
incorporated into E(x,y,z) and H(x,y,z).
(6.18.1)
(6.18.2)
Maxwell’s Eqn in
phasor form:
( ) ( )E r j H rωµ∇× = −
( ) ( ) ( )H r j E r J rωε∇× = +
( ) ( )vE r rρ ε∇ =g
( ) 0B r∇ =g
(6.18.3)
(6.18.4)
(6.18.5)
(6.18.6)

19
Power is a real quantity and, keeping in mind that:
Re ( ) Re ( ) Re ( ) ( )j t j t j t j t
E r e H r e E r e H r eω ω ω ω
     × ≠ ×     
Since [ ]
*
Re
2
A A
A
+
=
complex conjugate
Therefore:
[ ] [ ]
* *
* * * *
( ) ( ) ( ) ( )
Re ( ) Re ( )
2 2
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
4
E r E r H r H r
E r H r
E r H r E r H r E r H r E r H r
   + +
× = × ÷  ÷
   
× + × + × + ×
=
Taking the time average, we obtain the average power as:
*1
( ) Re ( ) ( )
2
avS r E r H r = × 
(6.19.1)
(6.19.2)
(6.19.3)
(6.19.4)

20
Therefore, the Poynting’s theorem in phasors is:
( ) ( )* 2 2 2
( ) ( )
s v v
E r H r ds j H E dv E dvω µ ε σ× = − − −∫ ∫ ∫V V V
gÑ (6.20.1)
Total power radiated
from the volume
The power dissipated
within the volume
The energy stored
within the volume
Indicates that the power (energy) is reactive

21
Example 6.6: Compute the frequency at which the conduction current equals the
displacement current in copper.
Using the Ampere’s law in the phasor form, we write:
( ) ( ) ( )H r J r j E rωε∇× = +
Since J Eσ=
σ ωε=
and ( ) ( ) ( ) ( )dJ r J r E r j E rσ ωε= ⇒ =
Therefore:
Finally:
7
18
90
5.8 10
1.04 10
12 2 2 10
36
f Hz
ω σ
π πε π
π
−
×
= = = ≈ ×
× ×
At much higher frequencies, cooper (a good conductor) acts like a dielectric.
(6.21.1)
(6.21.2)
(6.21.3)
(6.21.4)
(6.21.5)

22
Example 6.7: The fields in a free space are:
4
10cos ;
3 120
z
x
u Ez
E t u H
π
ω
π
× 
= + = ÷
 
Determine the Poynting vector if the frequency is 500 MHz.
In a phasor notation: 4 4
3 3
10
( ) 10 ( )
120
z z
j j
x yE r e u H r e u
π π
π
= =
And the Poynting vector is:
2
*1 10
( ) Re ( ) ( ) 0.133
2 2 120
av z zS r E r H r u u
π
 = × = =  ×
HW 5 is ready 
(6.22.3)
(6.22.2)
(6.22.1)

23
What is diffusion equation?
The diffusion equation is a partial differential equation which
describes density fluctuations in a material undergoing diffusion.
Diffusion is the movement of
particles of a substance from an
area of high concentration to an
area of low concentration, resulting
in the uniform distribution of the
substance.
Similarly, a flow of free charges in a material, where a charge difference
between two locations exists, can be described by the diffusion equation.
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Lecture 06 maxwell eqn

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