Download PH0101 Unit 2 Lecture 3

PH0101 Unit 2 Lecture 3
Maxwell’s equations in free space
Plane electromagnetic wave equation
Characteristic impedance
Poynting vector
Physical significance
PH0101 UNIT 2 LECTURE 3
1
Representation of EM Waves in free space:
In free space, the volume charge density (ρ) = 0 and
conduction current density (J1) = 0 (since  = 0 )
PH0101 UNIT 2 LECTURE 3
2
Maxwell’s equations in free space
In free space the Maxwell’s Equations becomes,
= 0
(1)
0=
 D
 B
(2)
 E
= 
 H
=
B


=
t
t
D
t
=
0
(μ0 H)
(3)
E
t
(4)
PH0101 UNIT 2 LECTURE 3
3
From eqn (4)
 H
=
D
t
=  0 E
t
[Since D = ε0 E]
Differentiating above eqn with respect to time,

2D
2E
(   H )
0
t 2
t 2
t
PH0101 UNIT 2 LECTURE 3
(5)
4

H
2E
(   H ) 
 0
t
t 2
From eqn (3)

E

=
 0 H
Taking curl on both sides of above eqn

 (   E )    0 (   H )
PH0101 UNIT 2 LECTURE 3
(6)
5
 (   E )   (   E )   2 E
=   2E [since   E= 0]
(7)
Using equation (7) in

 (   E )    0 (   H )
 2 E =   0

2

( H )

E  0 (   H )  0  0
PH0101 UNIT 2 LECTURE 3
2E
t 2
(8)
6
The above equation is free space electromagnetic equation.
2
In one dimension,
2 y
y
2 
t
C
2
x
2
(9)
Comparing (9) with standard mechanical wave equation,
 E
1  E
 0  0
(or)

.
x 2
t 2
t 2  0  0 x 2
2
 E
2
 E
2
PH0101 UNIT 2 LECTURE 3
2
(10)
7
C2 
1
0  0
( or ) C 
1
0  0
 3 10 8
m/s
(11)
The velocity of electromagnetic wave in free space.
Similarly, the wave equation in terms of H can be written as,
2H =
0  0
2H
t
PH0101 UNIT 2 LECTURE 3
2
(12)
8
In a medium of magnetic permeability  and electric permittivity ,
the wave equation becomes,
2H=
 E =
2

2H
t 2


2
t
E
(13)
(14)
2
The velocity of electromagnetic wave in any medium is,
1
C=

PH0101 UNIT 2 LECTURE 3
(15)
9
Worked Example 2.1:
An electromagnetic wave of frequency f = 3.0 MHz
passes from vacuum into a non – magnetic medium
with relative permittivity 4. Calculate the increment
in its wavelength. Assume that for a non – magnetic
medium μr=1.
Solution
Frequency of the em wave = f = 3.0 MHz = 3  10 6 Hz
Relative permittivity of the non – magnetic medium = εr = 4
Relative permeability of the non – magnetic medium = μr = 1
Velocity of em wave in vacuum = C =
PH0101 UNIT 2 LECTURE 3
1
0  0
10
 Wavelength of the EM wave in
vacuum = λ =
Velocity of em wave in
non- magnetic medium =
C 
C
1

.
f
f
1


1
 0 0
1
 0  r  0 r
Wavelength of the em wave in non-magnetic medium =
C
1

 

.
f
f
1
 0  r  0 r
PH0101 UNIT 2 LECTURE 3
11
 Therefore the change in wavelength =
1

  
.
f
=
1
 0 0
=




1
r r

 1



3  10 8  1
 1  50 m
6 
3  10  4 
i.e. the wavelength decreased by 50 m.
PH0101 UNIT 2 LECTURE 3
12
Worked Example 2.2
 Prove that the current density is irrotational.
We know that , J = σ E
curl J = curl (σ E) = σ (curl E) [ since σ is a constant]
= σ curl ( - grad V )
[E = - grad V]
= - σ curl ( grad V)
= 0 [ as curl ( grad V) = 0]
i.e. the current density is irrotational.
PH0101 UNIT 2 LECTURE 3
13
Characteristic Impedance
The solution of the equation for the electric component in the
electromagnetic wave is,
2
Ey = Eo sin
(ct – x)

(1) For magnetic component,
Hz = HO sin
2

(ct – x)
(2)
Differentiating equation (1) with respect to time,
E y
2
 2c 
 E0 
(ct  x)
 cos
t

  
PH0101 UNIT 2 LECTURE 3
(3)
14
For three dimensional variation of H
ii

 
H
 H

xx
H
Hx
x
jj


yy

H
H yy
kk

zz
HH
z z
H =
 H z H y   H z H x   H y H x 
H z
i


 k

  j
 

z   x
z   x
y 
x
 y
(4)
PH0101 UNIT 2 LECTURE 3
15
Since H varies only in the Z – direction and wave traveling along
X – axis, the component of H other than
H z
x
becomes zero in equation (4)
From the fourth law of free space Maxwell’s equation,
 H 0
E y
t
(5)
From equations (4) and (5),
Hz
Ey
  0
x
t
PH0101 UNIT 2 LECTURE 3
(6)
16
Substituting
E y
2
 2c 
 E0 
(ct  x)
 cos
t

  
in Eqn (6)
Hz
2c 
2

   0  E0
(ct  x)
 cos
x
 


(7)
Integrating with respect to x,
Hz =
2
 2c  

 E0 0 
sin
(
ct

x
)







  2 


  
PH0101 UNIT 2 LECTURE 3
(8)
17
Hz =
Hz =
c0E0 sin
1
 0 0
2

0E0 sin
(ct – x)
2

Hz =
0
2
E0 sin
0

Hz =
0
0
(ct – x)
(ct – x)
Ey
(9)
PH0101 UNIT 2 LECTURE 3
18
Ey
Hz
0
 Z
0

(10)
Characteristic Impedance of the medium
For free space, Z =
For any medium, Z =
0
0


PH0101 UNIT 2 LECTURE 3
= 376.8 
ohm
19
Worked Example 2.3
 Electromagnetic radiation propagating in free space
has the values of electric and magnetic fields 86.6 V
m – 1 and 0.23 A m – 1 respectively. Calculate the
characteristic impedance.
Solution:
Electric field intensity = E =86.6 V m – 1
Magnetic field intensity = H = 0.23 A m – 1
E 86.6

Characteristic impedance = Z = H 0.23
Z = 376. 52 ohm
PH0101 UNIT 2 LECTURE 3
20
Exercise Problem
 In a plane electromagnetic wave, the electric field
oscillates sinusoidally at a frequency of 2  10 10Hz
and amplitude of 48 V m – 1. What is the wavelength of
the wave? What is the amplitude of the oscillation of
the magnetic field?
E0
C
7
2
Hint:    1.5  10 m and B0 
 1.6  10 T
f
C
PH0101 UNIT 2 LECTURE 3
21
Poynting vector ( P )
 Poynting vector represent the rate of energy
flow per unit area in a plane electromagnetic
wave.
P 
1
0
EB  E  H
The direction of (P ) gives the direction in which the energy is
transferred. Unit: W/m2
PH0101 UNIT 2 LECTURE 3
22
Representation of Poynting vector
Y
Ey
P
Hz
X
Z
PH0101 UNIT 2 LECTURE 3
23
Expression for energy density
We know
P E  H
(1)
Taking divergence on (1)
.(E  H)  H.(  E)  E.(  H)
P E  H
=
B
D
 H.
 E.
t
t
PH0101 UNIT 2 LECTURE 3
(2)
24
.
.(E  H) 
B
D 

  H.
 E.

t
t 

==
E
H 

   0 E.
 0 H.

t
t 

==
E 1
H 
1
   0 (2E ).   0 (2H ). 
t 2
t 
2
PH0101 UNIT 2 LECTURE 3
(3)
25
.(E  H)
1
( E ) 2
( H ) 2
1
   2  0 . t  2  0 . t

==




 1
1
2
2 


E


H


0
0
t  2
2

(4)
Integrating the equation (4) over the volume V, we get
 1
1
2
2
 .(E  H )   t   2  0 E  2  0 H dV

V
V
PH0101 UNIT 2 LECTURE 3
(5)
26
Applying divergence theorem to the LHS of Eqn (5), we get
 1
1
2
2
 ( E  H ).dS   t   2  0 E  2  0 H dV
 (6)
S
V
The term on the RHS within the integral of the
equation (6) represents the sum of the energies of
electric and magnetic fields.
Hence the RHS of the equation (6) represents the
rate of flow of energy over the volume V.
PH0101 UNIT 2 LECTURE 3
27
UE 
Energy associated with the electric field
and that of the magnetic field U m 
Um 
[as
E
B
0 H
2
2


 C
B

2
2
2
B2

2 0
2
C
2 0
and
0 H
0E2

1
0E2 UE
2
C 
1
 0 0
]
which shows that instantaneous energy density associated
with electric field i.e. energy is equally shared by the two
fields.
PH0101 UNIT 2 LECTURE 3
28
Significance of P
The Vector P = E X H has interpreted as representing
 Eenergy
 H passing through the unit area of
the amount of P
field
surface in unit time normally to the direction of flow of energy.
This statement is termed as Poynting’s theorem and the vector
P is called Poynting Vector.
The direction of flow of energy is perpendicular to vectors E
and H
E X H
i.e., in the direction of the vector E  H
PH0101 UNIT 2 LECTURE 3
29
PH0101 UNIT 2 LECTURE 3
30