1) The document contains information about a student named Devendra Keer who is in the 4th semester of his BSC in computer science at Rai Saheb Bhanwar Singh College. 2) It discusses Maxwell's equations and provides examples of applying them, including to derive the diffusion equation, solve for magnetic flux density near a metal interface, and derive the continuity equation. 3) It also covers the Poynting theorem, showing how it relates the energy, power, and fields in a volume, and provides examples of applying it to calculate power dissipated in a resistor and flowing through a capacitor.

2. NAME : DEVENDRA KEER
CLASS : BSC IV SEM (C.S.)
COLLEGE : RAI SAHEB BHANWAR
SINGH COLLEGE
NASRULLAGANJ
SUBMITTED TO: GYAN RAO DHOTE SIR
RAI SAHEB BHANWAR SINGH
COLLEGE NASRULLAGANJ 2

3. RAI SAHEB BHANWAR SINGH
COLLEGE NASRULLAGANJ 3

5. Instructor:
Dr. Gleb V.
Tcheslavski
Contact: gleb@ee.lamar.edu
Office Hours: Room 2030
Class web site:
www.ee.lamar.edu/gleb/em/
Index.htm

6. 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  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)

7. 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 dv
 
 gÑ
0
S
B ds

 gÑ

8. 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 E  
We can write:
and, neglecting the
displacement
current:
(6.5.1)
(6.5.2)
Taking curl of
(6.5.2):
H E  
Expanding the LHS: 2
0 0
B B B
t

 
    
       
   
g
(6.5.3)
(6.5.4)
The first term is zero
and
2
0
B
B
t
 

 

(6.5.5)
Is the diffusion equation with a diffusion coefficient D = 1/(0)

9. 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:
2
02
( )
( )x
x
d B z
j B z
dz
 
2
0 0j j j           
Where z is the normal coordinate to the boundary. Assuming a
variation in the z-direction to be Bx(z) = B0e-z, we write:
(6.6.1)
(6.6.2)

10. 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 

where
7 7 1
0 60 4 10 5.8 10 117.2f m       
        
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.
(6.7.1)
(6.7.2)

11. Example 6.3: Derive the equation of continuity starting from the Maxwell’s equ
The Gauss’s
law:
vD  g
Taking time
derivatives:
v D
D
t t t
  
   
  
g g
From the Ampere’s
law
D
H J
t

  

Therefore: v
H J
t

   

g g
The equation of continuity: v
J
t

 

g
(6.8.1)
(6.8.2)
(6.8.3)
(6.8.4)
(6.8.5)

12. 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)

13. 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
   
   
       
   
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…

14.  2 21
( )
2v v v
E H dv H E dv E Jdv
t
 

     
  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
  

    
  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
)

15. 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 E    
Power loss density is
2
Lp E
The differential form of the Poynting’s Theorem:
L
w
S p
t

   

g
(6.12.1
)
(6.12.2
)
(6.12.3
)

16. 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 a (6.13.2
)

17. 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
a
 (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
 
 
     
  
V
(6.14.3
)
(6.14.4
)
The electromagnetic energy of a battery is completely absorbed
with the resistor in form of heat.

18. 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


The total magnetic energy stored in the capacitor is zero.
(6.15.1
)

19. The time derivative of the electric energy is
2dW dE
a bE
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:
s
E
H dl ds
t



 V
g gÑ (6.16.2
)
2
2
2
dE a dE
aH a H
dt dt

    
Therefore:
(6.16.3
)

20. (6.17.1
)
The power flow would be:
 S
s
P E H ds V
gÑ
In our situation: 2 rds ab u 
and 1rS u  g
2
2S abEH
dE
P a bE
dt
  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
)

21. 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


   
   
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 r  
( ) ( ) ( )H r j E r J r  
( ) ( )vE r r  g
( ) 0B r g
(6.18.3
)
(6.18.4
)
(6.18.5
)
(6.18.6
)

22. 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 e   
            
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
)

23. Therefore, the Poynting’s theorem in phasors is:
   * 2 2 2
( ) ( )
s v v
E r H r ds j H E dv E dv         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

24. 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 J E
 
and ( ) ( ) ( ) ( )dJ r J r E r j E r   
Therefore:
Finally:
7
18
90
5.8 10
1.04 10
12 2 2 10
36
f Hz
 
  



     
 
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
)

25. Example 6.7: The fields in a free space are:
4
10cos ;
3 120
z
x
u Ez
E t u H



 
    
 
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
 

  
And the Poynting vector is:
2
*1 10
( ) Re ( ) ( ) 0.133
2 2 120
av z zS r E r H r u u

       
HW 5 is ready

(6.22.3
)
(6.22.2
)
(6.22.1
)

26. 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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