Transcript Lecture 6: Maxwell`s Equations

1
Maxwell’s Equations
ELEN 3371 Electromagnetics
Fall 2008
2
Maxwell’s Equations
The behavior of electric & magnetic waves can be fully
described by a set of four equations (which we learned already).
Faraday’s Law
of Induction:
B
 E  
t
Ampere’s Law:
D
 H  J 
t
Gauss’s Law for
Electricity:
 D  v
Gauss’s Law for
Magnetism:
 B0
ELEN 3371 Electromagnetics
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3
Maxwell’s equations
And the constitutive relations:
D E
B  H
J E
(6.3.1)
(6.3.2)
(6.3.3)
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.
ELEN 3371 Electromagnetics
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Maxwell’s equations
Integral form
Faraday’s Law of induction
B
ds
t
s
 E dl   
L
Ampere’s Law
 D

H
dl



J
L
s  t  ds
Gauss’s Law for electricity
(6.4.3)
 B ds  0
(6.4.4)
v
v
S
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(6.4.2)
 D ds    dv
S
Gauss’s Law for magnetism
(6.4.1)
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Maxwell’s equations
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
(6.5.1)
We can write:
 E  
t
and, neglecting the
displacement current:
 H  J   E
(6.5.2)
Taking curl of (6.5.2):
 H   E
(6.5.3)
Expanding the LHS:
 B
B
2 B 







 
t
 0 
 0 
(6.5.4)
The first term is zero and
B
 B  0
t
2
Is the diffusion equation with a diffusion coefficient D = 1/(0)
ELEN 3371 Electromagnetics
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(6.5.5)
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Maxwell’s equations
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:  =
0 and  = 5.8 x 107 S/m. Assume that a 60-Hz time-harmonic EM signal is
applied.
Assuming ejt time-variation, the diffusion equation is transformed to the
ordinary differential equation:
d 2 Bx ( z )
 j 0 Bx ( z )
2
dz
(6.6.1)
Where z is the normal coordinate to the boundary. Assuming a variation in
the z-direction to be Bx(z) = B0e-z, we write:
 2  j0     j  
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j0
(6.6.2)
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Maxwell’s equations
The magnitude of the magnetic flux density decays exponentially in the z
direction from the surface into the conductor
Bx ( z )  B0 e  z
where
   f 0    60  4 107  5.8 107  117.2m1
The quantity  = 1/ is called a “skin depth” - the
distance over which the current (or field) falls to 1/e of
its original value.
For copper,  = 8.5 mm.
ELEN 3371 Electromagnetics
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(6.7.1)
(6.7.2)
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Maxwell’s equations
Example 6.3: Derive the equation of continuity starting from the Maxwell’s equations
 D  v
(6.8.1)
Taking time derivatives:
v 
D
  D
t
t
t
(6.8.2)
From the Ampere’s law
D
  H  J
t
(6.8.3)
v
   H  J
t
(6.8.4)
 v
  J
t
(6.8.5)
The Gauss’s law:
Therefore:
The equation of continuity:
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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:
Recall:
B
 E  
t
D
 H  J 
t
(6.9.1)
(6.9.2)
We take the scalar product of E and subtract it from the scalar product of H.
B
H   E  E   H  H
E
t
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D 

J 


t


(6.9.3)
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Poynting’s Theorem
Using the vector identity
Therefore:
 ( A  B)  B   A  A   B
(6.10.1)
B
D
 (E  H )  H
E
E J
t
t
(6.10.2)
Applying the constitutive relations to the terms involving time derivatives, we get:
H
B
D
1 
 1
E


H
H


E
E


H 2   E2 



t
t
2 t
t 2
Combining (6.9.2) and (6.9.3) and integrating both sides over the same v…
ELEN 3371 Electromagnetics
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(6.10.3)
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Poynting’s Theorem

t
  ( E  H )dv  
v
1
2
2

H


E

 dv   E Jdv
v 2
v
(6.11.1)
Application of divergence theorem and the Ohm’s law lead to the PT:
 ( E  H ) ds  
s
Here

t
1
2
2
2

H


E
dv


E


v 2
v dv
S  EH
is the Poynting vector – the power density and the
direction of the radiated EM fields in W/m2.
ELEN 3371 Electromagnetics
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(6.11.3)
(6.11.2)
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Poynting’s Theorem
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
Power loss density is
1
w    H 2   E 2 
2
(6.12.1)
pL   E 2
(6.12.2)
The differential form of the Poynting’s Theorem:
w
 S
  pL
t
ELEN 3371 Electromagnetics
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(6.12.3)
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Poynting’s Theorem
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
E  V0 L
(6.13.1)
and it is in the direction of the current.
The magnitude of the magnetic field intensity at the outer surface of the resistor:
H  I  2 a 
ELEN 3371 Electromagnetics
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(6.13.2)
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Poynting’s Theorem
The Poynting’s vector
S  EH
(6.14.1)
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.
I
J
 a2
The PT:
(6.14.2)
V0 I
d

2 aL  
L 2 a
dt
I V0 2
v (0  0)dv   a 2 L  a L
 V0 I  V0 I
The electromagnetic energy of a battery is completely absorbed with
the resistor in form of heat.
ELEN 3371 Electromagnetics
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(6.14.3)
(6.14.4)
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Poynting’s Theorem
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:
W
 E2
2
 a 2b
The total magnetic energy stored in the capacitor is zero.
ELEN 3371 Electromagnetics
Fall 2008
(6.15.1)
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Poynting’s Theorem
The time derivative of the electric energy is

dW
dE
  a 2bE
dt
dt
(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:
E
 H dl  s  t ds
Therefore:
ELEN 3371 Electromagnetics
2 aH  
dE 2
 a dE
 a H 
dt
2 dt
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(6.16.2)
(6.16.3)
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Poynting’s Theorem
The power flow would be:
PS 
  E  H  ds
(6.17.1)
s
In our situation:
and
Therefore:
We observe that
ds  2 abur
(6.17.2)
S ur  1
(6.17.3)
dE
PS  2 abEH   a bE
dt
dW
PS  
dt
2
The energy is conserved in the circuit.
ELEN 3371 Electromagnetics
Fall 2008
(6.17.4)
(6.17.5)
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Time-harmonic EM fields
Frequently, a temporal variation of EM fields is harmonic; therefore,
we may use a phasor representation:
E ( x, y, z , t )  Re  E ( x, y, z )e jt 
(6.18.1)
H ( x, y, z , t )  Re  H ( x, y, z )e jt 
(6.18.2)
It may be a phase angle between the electric and the magnetic fields
incorporated into E(x,y,z) and H(x,y,z).
Maxwell’s Eqn in
phasor form:
ELEN 3371 Electromagnetics
  E (r )   j H (r )
(6.18.3)
  H (r )  j E (r )  J (r )
(6.18.4)
 E ( r )  v ( r ) 
(6.18.5)
 B(r )  0
(6.18.6)
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Time-harmonic EM fields
Power is a real quantity and, keeping in mind that:
Re  E (r )e jt   Re  H (r )e jt   Re  E (r )e jt  H (r )e jt 
Re  A 
Since
Therefore:
A A
2
(6.19.1)
complex conjugate
*
 E (r )  E * (r )   H (r )  H * (r ) 
Re  E (r )   Re  H (r )   


2
2

 

E (r )  H * (r )  E * (r )  H (r )  E (r )  H (r )  E * (r )  H * (r )

4
(6.19.2)
(6.19.3)
Taking the time average, we obtain the average power as:
S av (r ) 
ELEN 3371 Electromagnetics
1
Re  E (r )  H * (r ) 
2
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(6.19.4)
20
Time-harmonic EM fields
Therefore, the Poynting’s theorem in phasors is:
  E (r )  H
s
*
(r )  ds   j    H 2   E 2  dv    E 2 dv
Total power radiated
from the volume
v
The energy stored
within the volume
Indicates that the power (energy) is reactive
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v
The power dissipated
within the volume
(6.20.1)
21
Time-harmonic EM fields
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 (r )
Since
and
Therefore:
Finally:
J E
J (r )  J d (r ) E (r )  j E (r )
  


5.8 107
f 


 1.04 1018 Hz
2 2 0 2  1 109
36
At much higher frequencies, cooper (a good conductor) acts like a dielectric.
ELEN 3371 Electromagnetics
Fall 2008
(6.21.1)
(6.21.2)
(6.21.3)
(6.21.4)
(6.21.5)
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Time-harmonic EM fields
Example 6.7: The fields in a free space are:
uz  E
4 z 

E  10 cos  t 
u
;

H

 x
3 
120

(6.22.1)
Determine the Poynting vector if the frequency is 500 MHz.
In a phasor notation:
E (r )  10e
j
4 z
3
10 j 43 z
ux H (r ) 
e uy
120
(6.22.2)
And the Poynting vector is:
1
102
*
Sav (r )  Re  E (r )  H (r )  
u z  0.133u z
2
2 120
HW 5 is ready 
ELEN 3371 Electromagnetics
Fall 2008
(6.22.3)
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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ELEN 3371 Electromagnetics
Fall 2008