EELE 3332 Electromagnetic II Chapter 9
Transcript of EELE 3332 Electromagnetic II Chapter 9
![Page 1: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/1.jpg)
EELE 3332 – Electromagnetic II
Chapter 9
Maxwell’s Equations Islamic University of Gaza
Electrical Engineering Department
Dr. Talal Skaik
2013 1
![Page 2: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/2.jpg)
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
![Page 3: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/3.jpg)
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
![Page 4: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/4.jpg)
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
![Page 5: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/5.jpg)
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
![Page 6: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/6.jpg)
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
![Page 7: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/7.jpg)
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
![Page 8: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/8.jpg)
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
![Page 9: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/9.jpg)
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
![Page 10: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/10.jpg)
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
![Page 11: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/11.jpg)
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
![Page 12: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/12.jpg)
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
![Page 13: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/13.jpg)
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
![Page 14: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/14.jpg)
14
Maxwell’s Equations – phasor form
![Page 15: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/15.jpg)
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
![Page 16: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/16.jpg)
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
![Page 17: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/17.jpg)
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
![Page 18: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/18.jpg)
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
![Page 19: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/19.jpg)
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
![Page 20: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/20.jpg)
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
![Page 21: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/21.jpg)
21
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
![Page 22: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/22.jpg)
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
![Page 23: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/23.jpg)
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
![Page 24: EELE 3332 Electromagnetic II Chapter 9](https://reader031.fdocuments.us/reader031/viewer/2022011905/61d6e1ddd99cb74fac7f2205/html5/thumbnails/24.jpg)
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