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EELE 3332 – Electromagnetic II Chapter 9 Maxwell’s Equations Islamic University of Gaza Electrical Engineering Department Dr. Talal Skaik 2013 1
EELE 3332 – Electromagnetic II
Chapter 9
Maxwell’s Equations Islamic University of Gaza
Electrical Engineering Department
Dr. Talal Skaik
2013 1
2
9.6 Time Varying Potentials
For Electrostatic fields, we defined electric scalar potential (V).
For Magnetostatic fields, we defined magnetic vector potential (A).
For Electromagnetic Fields: We need to define (A and V) for time
varying fields……
2
0B=0 B= A, A= J
JA=
4 Rv
dv
2E=0 E= , V=
V=4 R
V
V
v
V
dv
3
Time Varying Potentials
Since B=0 holds for time varying field, then the relation
B= A
also holds for time varying field.
Using Faraday's law, E=
E= A =0
B
t
Aor E
t t
Since the curl of the gradient of a scalar is zero, then
E (1)
AE V
t
Aor V
t
So, we can determine the vector fields B and E from the potentials
A and V:
However, we still need to find some expressions for A and V
suitable for time-varying fields.
4
EA
Vt
Time Varying Potentials
B= A
5
2
2
We know that D= is valid for time-varying conditions.
E=
using eq (1) E
E= A
E= V A
V+ A (2)
V
V
V
V
V
AV
t
Vt
t
or
t
Time Varying Potentials
6
2
2
Now take the curl of B= A B= A= H
Dsince H J , and D= E, and E
DA= J J
A= J+ = J
Reca
AV
t t
AV
t t t
A V AV
t t t t
2 2
22
2
22
2
ll that Laplacian of a vector is given by:
A= A A A= A A
A A= J
A+ A= J (3)
V A
t t
V A
t t
Time Varying Potentials
We have now reduced the set of four Maxwell eqs. to two eqs. But they are still coupled. The uncoupling can be achieved by using Lorenz condition for potentials.
7
2
22
2
V+ A (2)
A+ A= J (3)
V
t
V A
t t
Time Varying Potentials
Lorenz condition for potentials (relates V and A):
A= (4)V
t
8
22
2
Using equ (4) A=
Substitute (4) in equations (2) and (3),
= (5)
V
V
t
VV
t
2
2
2 A = J (6)
which are the wave equations to be discussed in the next chapter...
A
t
Time Varying Potentials
2 2
Recall that E and B can be obtained by:
E , B= A
Notice that for static conditions (special cases of time varying potentials):
E , B= A, and = , AV
AV
t
V V
= J
9
9.7 Time Harmonic Fields A time-harmonic field is one that varies periodically or sinusoidally with time.
Sinosoids are easily expressed in phasors, which are more
convenient to work with.
A phasor is a complex number that contains the amplitude and the
phase of a sinusoidal oscillation.
2 2
1
cos sin
1 is , is the real part of , is the imaginary part of
is the magnitude of given by
and is the phase of z, given by: tan
jz x jy r re r j
where
j x z y z
r z r z x y
y
x
10
Time Harmonic Fields
•Addition and subtraction are better performed in rectangular form.
•Multiplication and division are better done in polar form.
Rectangular Form
Polar Formj
z x jy
z r re
1 1 1 1 1 2 2 2 2 2
1 2 1 1 1 2
1 2 1 1 1 2
1 2 1 2
, = , =
( ) ( )
( ) ( )
(
Given z x jy r z x jy r z x jy r
Addition z z x x j y y
Subtraction z z x x j y y
Multiplication z z r r
1 2
1 11 2
2 2
)
( )
( / 2)
( ) j
z rDivision
z r
Square root z r
Complex Conjugate z x jy r re
11
Time Harmonic Fields
( )
To introduce the time element, we let:
where may be a function of time or space coordinates or a constant.
The real and imginary part of cos( ) sin( )
are given by:
j j t
t
re re r t jr t
0 0
0 0
Re( ) cos( )
Im( ) sin( )
: current ( ) cos( ) equals real part of
: current ( ) sin( ) equals imaginary part of
j
j
j j t
j
re r t
re r t
Example I t I t I e e
Example I t I t I e
90
0 also equals real part of because sin =cos( 90)
j t
j j t j
e
I e e e
12
Time Harmonic Fields
0 0
0 0
The Phasor is defined by dropping the time factor
For example, ( ) cos( ) Re( )
The phasor current is :
and hence,
( ) Re(
j t
j j t
j
s
e
I t I t I e e
I I e I
I t I
j t
s
)
If a vector ( , , , ) is a time-harmonic field, the phasor form of A
is ( , , ). The two quantities are related by
A=Re(A e )
j t
s
s
e
A x y z t
A x y z
13
Time Harmonic Fields j t
s
j t j t
s s
A=Re(A e )
ANotice that Re(A e ) Re( A e )
, taking the derivative of the instataneous quantity is equivalent
to multiplying its phasor form by
jt t
Hence
j
s
s
s s
j t
s
A A
:
instantaneous A( , , , ) time dependent
phasor A ( , , ) time invariant
It is easier to work with A and obtain A from A whenever necessary
using A=Re(A e )
jt
Notes
x y z t
x y z
14
Maxwell’s Equations – phasor form
15
Evaluate the complex numbers:
Example 9.5
*
1 2
1/2
2
(3 4)( ) =
( 1 6)(2 )
1( ) =
4 8
j ja z
j j
jb z
j
*
1 2
1 2 2
1
( ) Method 1 (Working in rectangular form)
(3 4) (3 4) 4 3 4 3 =
( 1 6)(2 ) ( 1 6)(4 4 1) ( 1 6)(3 4 ) 27 14
( 4 3)( 27 14) 150 25=
( 27 14)( 27 14) 27 14
=0.1622 0.027 0.1644 9
a
j j j j j jz
j j j j j j j
j j jz
j j
z j
.46
16
Example 9.5 - Continued
1
1 21 1
2
1 2
1 1
1
Method 2 (Working in polar form)
1 90 5 tan ( 4 / 3) =
1 36 tan ( 6) 4 1 tan ( 1/ 2)
1 90 5 53.13 = 5 26.265 5 53.13
37 99.46 5 26.265
1 5 1 = (90 53.13 99.46 53.13) = ( 9.46)
3737 (5)
1= (cos 9.46
37
z
z
z z
z j
1
sin 9.46)
=0.1622 0.027z j
17
1/2
2
1/2 1/21
2 1
2
2
1( ) =
4 8
2 tan (1) 2(45 ( 63.43)
8080 tan ( 2)
0.15811 (108.4 / 2)
0.3976 (54.2)
jb z
j
z
z
z
Example 9.5 - Continued
18
8 2 /3
s
Given that 10cos(10 -10 60 ) and B (20 / ) 10
Express A in phasor form and B in instantaneous form.
o j x
z s x yA t x a j a e a
Example 9.6
( 10 60 ) 8
(60 10 )
(60 10 )
s
2 /3 2 /3
/2 2 /3
( /2)
A Re 10 , where 10 .
Hence, A Re 10
or A 10
B (20 / ) 10 20 10
B 20 10
B Re[B ] Re[20 1
o
o
o
j t x
z
j x j t
z
j x
z
j x j x
s x y x y
j j x
s x y
j t j t
s x
e a
e a e
e a
j a e a j a e a
e a e a
e e a
( 2 /3)0 ]
20cos( / 2) 10cos( 2 / 3)
20sin 10cos( 2 / 3)
j t x
y
x y
x y
e a
B t a t x a
B t a t x a
19
6 60
0
The electric field and msgnetic field in free space are given by
50E= cos(10 ) V/m , H= cos(10 ) A/m
Express these in phasor form and determine the constants
and such that the fi
Ht z a t z a
H
elds satisfy Maxwell's equations
Example 9.7
j t j t 6
s s
j z j z0s s
0 0
The instantaneous forms of E and H are written as
E=Re(E e ), H=Re(H e ) , where 10
50 HE e , H e (1)
In free space, 0, =0, = and = , v
a a
0 s
0 s
0 s 0 s
So Maxwell's equations become
D= E=0 E =0 (2)
B= H=0 H =0 (3)
EH= E+ H =j E (4)
t
0 s 0 s
HE= E =- H (5)j
t
s
s
j z j z0 0s
Substitution eq (1) into (2) and (3), it is verified two Maxwell's equation are verified.
1E = E 0
1H = 0
H e e (6)
Substituting equations (1)
s
sH
H jHa a
j z j z0s 0 s 0
0 0
s 0 s
and (4) into equation (6):
50H e j E j e
50 (7)
Similarly, substituting equation (1) into equation (5):
E = H
jHa a
H
j
j z j z00
00 0
0
50 He e
50 50 H (8)
j a j a
H
20
Example 9.7 continued
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Example 9.7 continued
00 0
0
2 2 00 0 0 0
0
2 2
0 0
63
0 0 8
5050 (7) , (8)
Multiplying equations (7) and (8):
50(50) 50 / 0.1326
120
Dividing equation (7) by (8):
=
10= 3.33 10
3 10
HH
H H
c
Method 1 (time domain)--(the harder way!)
It is evident that gauss law is satisfied: E= 0
H 1From Faraday's law, E= E
E=
0 0
E=20 cos(1
y
x y z
y y
x z
y
E
y
H dtt
a a a
E EBut a a
x y z z x
E
8
8 8
8
0 ) 0
20 20 cos(10 ) sin(10 ) (1)
10
x
x x
t z a
Hence H t z dt a t z a
22
0 0
8
In a medium characterised by =0, = , =4 and
E=20sin(10 ) V/m
Calculate and H.
yt z a
Example 9.8
23
0
28
8
2 28 8
8 16
It is verified that H= 0
E 1H= E+ E H ( =0)
20 H= cos(10 ) 0
10
0 0
20 20E cos(10 ) sin(10 )
10 10
the
x
x y z
x xy z y
x
y y
H
x
dtt
a a a
H HBut a a t z a
x y z z y
H
t z dt a t z a
Since
8
2 88 8
0 016
8
7 8
8
given is E=20sin(10 )
20 10 (2) 220 10 10 4
10 3
20(2 / 3) 2From equ (1): H= sin(10 )
4 .10 (10 ) 3
1 2H= sin(10 ) A/m
3 3
y
x
x
t z a
c
zt a
zt a
Example 9.8 continued
24
j t 8
s s
s
ss s s
2
ss s s 2
Method 2 : (Using Phasors)
E=Im(E e ) E 20 where =10
Again E 0
E 1 20E = H H
H 1 20H =j E E
j j
Comp
j z
y
ys
ys j z
x x
j z
xsy y
e a
E
dy
Ej a e a
j j z
H ea a
dz
s
2 2
2 2
s 8 7
8
s
aring this with E , we have:
20 20 220 20
3
20(2 / 3) 1H
10 (4 10 ) 3
1H Im(H ) sin(10 ) A/m
3
j zj z
y y
j zj z
x x
j t
x
ea e a
ea e a
e t z a
Example 9.8 continued
