Transcript EM_Course_Module_2 - University of Illinois at Urbana

Fundamentals of Electromagnetics
for Teaching and Learning:
A Two-Week Intensive Course for Faculty in
Electrical-, Electronics-, Communication-, and
Computer- Related Engineering Departments in
Engineering Colleges in India
by
Nannapaneni Narayana Rao
Edward C. Jordan Professor Emeritus
of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign, USA
Distinguished Amrita Professor of Engineering
Amrita Vishwa Vidyapeetham, India
Program for Hyderabad Area and Andhra Pradesh Faculty
Sponsored by IEEE Hyderabad Section, IETE Hyderabad
Center, and Vasavi College of Engineering
IETE Conference Hall, Osmania University Campus
Hyderabad, Andhra Pradesh
June 3 – June 11, 2009
Workshop for Master Trainer Faculty Sponsored by
IUCEE (Indo-US Coalition for Engineering Education)
Infosys Campus, Mysore, Karnataka
June 22 – July 3, 2009
2-2
Module 2
Maxwell’s Equations
in Integral Form
2.1 The line integral
2.2 The surface integral
2.3 Faraday’s law
2.4 Ampere’s circuital law
2.5 Gauss’ Laws
2.6 The Law of Conservation of Charge
2.7 Application to static fields
2-3
Instructional Objectives
8. Evaluate line and surface integrals
9. Apply Faraday's law in integral form to find the
electromotive force induced around a closed loop, fixed or
revolving, for a given magnetic field distribution
10. Make use of the uniqueness of the magnetomotive force
around a closed path to find the displacement current
emanating from a closed surface for a given current
distribution
11. Apply Gauss’ law for the electric field in integral form to
find the displacement flux emanating from a closed
surface bounding the volume for a specified charge
distribution within the volume
12. Apply Gauss’ law for the magnetic field in integral form
to simplify the problem of finding the magnetic flux
crossing a surface
2-4
Instructional Objectives (Continued)
13. Apply Gauss' law for the electric field in integral form,
Ampere's circuital law in integral form, the law of
conservation of charge, and symmetry considerations, to
find the line integral of the magnetic field intensity
around a closed path, given an arrangement of point
charges connected by wires carrying currents
14. Apply Gauss’ law for the electric field in integral form to
find the electric fields for symmetrical charge
distributions
15. Apply Ampere’s circuital law in integral form, without
the displacement current term, to find the magnetic fields
for symmetrical current distributions
2-5
2.1 The Line Integral
(EEE, Sec. 2.1; FEME, Sec. 2.1)
2-6
The Line Integral:
Work done in carrying a charge from A to B in
an electric field:
B
E2
A
E1 a
2
a1 l2
l1
n
WAB   dWj
j1
2-7
dWj   qE j cos a j  l j 
 qE j l j cos a j
 qE j l j
n
WAB  q
E
j
• l j
j 1
VAB
WAB


q
n
E
j 1
j
• l j
(Voltage between
A and B)
2-8
In the limit n   ,
B
VAB   E • dl
A
= Line integral of E
from A to B.

E  d l = Line integral of E
C
around the closed
path C.
2-9
A
If
R
C
C
L
B
 ARBLA E

C
E dl = 0
B
then  E d l
A
is independent of
the path from A to B
(conservative field)
d l ARB E d l BLA E d l
 ARB E d l  ALB E d l  0
ARB E
d l = ALB E d l
2-10
F  yza x  zxa y  xya z ,

(1,2,3)
F d l along the straight line paths,
(0,0,0)
from (0, 0, 0) to (1, 0, 0), from (1, 0, 0) to
(1, 2, 0) and then from (1, 2, 0) to (1, 2, 3).
z
(0, 0, 0)
(1, 0, 0)
x
(1, 2, 3)
y
(1, 2, 0)
2-11
From (0, 0, 0) to (1, 0, 0),
y  z  0; dy  dz  0
F  0,
1,0,0 

0,0,0 
F dl  0
From (1, 0, 0) to (1, 2, 0),
x  1, z  0 ; dx  dz  0
F  yaz
dl  dx ax  dy ay  dz az  dy ay
F d l  0,
1,2,0 

0,0,0 
F dl  0
2-12
From (1, 2, 0) to (1, 2, 3),
x  1, y  2 ; dx  dy 0
F  2za x  za y  2a z , dl  dz a z
F • dl  2 dz ,

(1,2,3)
(1,2,3)
(1,2,0) 2 dz  6
(0,0,0) F • dl  0  0  6  6
2-13
In fact, F d l   yzax  zxa y  xyaz 
 dx a
x
 dy a y  dz az 
 yz dx  zx dy  xy dz
 d  xyz 
1,2,3

0,0,0 
F d l 
1,2,3
 0,0,0 
d  xyz    xyz 
1,2,3
 0,0,0 
 1 2  3  0  0  0
 6, independent of the path.
2-14
Review Questions
2.1. How do you find the work done in moving a test charge
by an infinitesimal distance in an electric field?
2.2. What is the amount of work involved in moving a test
charge normal to the electric field?
2.3. What is the physical interpretation of the line integral of
E between two points A and B?
2.4. How do you find the approximate value of the line
integral of a vector field along a given path? How do you
find the exact value of the line integral?
2.5. Discuss conservative versus nonconservative fields,
giving examples.
2-15
Problem S2.1. Evaluation of line integral around a closed
path in Cartesian coordinates
2-16
Problem S2.2. Evaluation of line integral around a closed
path in spherical coordinates
2-17
2.2 The Surface Integral
(EEE, Sec. 2.2; FEME, Sec. 2.2)
2-18
The Surface Integral
Flux of a vector crossing a surface:
B
aann
Flux = (B)(S)
S
B
S
B
an
a
S
an
Flux = 0
Flux  (B cos a ) S
 B S cos a
 B • S a n
 B • S
2-19
Normal
anj
aj
B
Bj j
Flux 
n
  j
j1
Sj
S

In the limit n   ,
Flux,  =
n
 B j • S j
j1
S B • dS
= Surface integral of B over S.
2-20
S B • dS
D2.4
(a)
= Surface integral of B
over the closed surface S.
A  x  ax  a y 
z
x  0, an   a x
2
A  0, A • dS  0
A
A
2
dS  0
dS  0
x
y
2-21
(b)
z
2
x  2, an   ax
A  2  ax  a y 
d S   dy dz ax
A • dS   2 dy dz
2
2
x
2
2
 A • dS   y 0 z 0 2 dy dz   8
 A • dS  8
y
2-22
z
(c)
2
y  0, an   a y
A  x  ax  a y 
d S   dz dx a y
y
2
x
A d S   x dx dz
2
2
 A • dS   x 0 z 0 x dx dz   4
 A • dS
4
2-23
(d) From (c),
A • dS   x dx dz
2– x
2
 A • dS   x0 z 0 x dx dz
2
  0 x(2 – x) dx
4

3
4
 A • dS  3
z
2
x +z =2
y
x
2
2-24
Review Questions
2.6. How do you find the magnetic flux crossing an
infinitesimal surface?
2.7. What is the magnetic flux crossing an infinitesimal
surface oriented parallel to the magnetic flux density
vector?
2.8. For what orientation of an infinitesimal surface relative
to the magnetic flux density vector is the magnetic flux
crossing the surface a maximum?
2.9. How do you find the approximate value of the surface
integral of a vector field over a given surface? How do
you find the exact value of the surface integral?
2.10. Provide physical interpretation for the closed surface
integrals of any two vectors of your choice.,
2-25
Problem S2.3. Evaluation of surface integral over a closed
surface in Cartesian coordinates
2-26
2.3 Faraday’s Law
(EEE, Sec. 2.3; FEME, Sec. 2.3)
2-27
Faraday’s Law
d
C E • dl  – dt S B • dS
B
S
C
dS
2-28
C E • dl = Voltage around C, also known as
electromotive force (emf) around C
(but not really a force),
 V m  m, or V.
S B • dS = Magnetic flux crossing S,
2
2
Wb
m

m

 , or Wb.
d
–  B • dS = Time rate of decrease of
dt S
magnetic flux crossing S,
Wb s, or V.
2-29
Important Considerations
(1) Right-hand screw (R.H.S.) Rule
The magnetic flux crossing the
surface S is to be evaluated toward
that side of S a right-hand screw
advances as it is turned in the sense of C.
C
2-30
(2) Any surface S bounded by C
The surface S can be any surface bounded by
C. For example:
z
R
z
R
C
O
C
Q
O
y
P
x
x
Q
y
P
This means that, for a given C, the values of
magnetic flux crossing all possible surfaces
bounded by it is the same, or the magnetic flux
bounded by C is unique.
2-31
(3) Imaginary contour C versus loop of wire
There is an emf induced around C in either
case by the setting up of an electric field. A
loop of wire will result in a current flowing in
the wire.
(4) Lenz’s Law
States that the sense of the induced emf is
such that any current it produces, if the closed
path were a loop of wire, tends to oppose the
change in the magnetic flux that produces it.
2-32
Thus the magnetic flux produced by the
induced current and that is bounded by C must
be such that it opposes the change in the
magnetic flux producing the induced emf.
(5) N-turn coil
For an N-turn coil, the induced emf is N times
that induced in one turn, since the surface
bounded by one turn is bounded N times by
the N-turn coil. Thus
2-33
d
emf  – N
dt
where  is the magnetic flux linked by one turn.
2-34

D2.5 B  B0 sin t ax  cos t a y
(a)
B
S

d S = B0 sin t
d
C E d l   dt  B0 sin t 
  B0 cos t V
z
1
C
1
x
y
2-35

B0
0
 dec.

2
3
t
 inc.
 B0
emf
 B0
0
 B0
emf < 0

2
emf > 0
Lenz’s law is verified.
3
t
2-36
(b)
B
dS
S
z
1
1
1
 B0 sin t – B0 cos t
2
2
1



B0 sin  t – 
4
2


E dl
C
d  1
 

– 
B0 sin  t –  
dt  2
4 

–
 B0


cos  t –  V
4
2

C
1
x
1 y
2-37
(c)
B
dS
z
S
1
 B0 sin t  B0 cos t
C


 2 B0 sin  t  
4


1
1
E dl
C
d 
 

 –  2 B0 sin  t   
dt 
4 



 – 2  B0 cos  t   V
4

x
y
2-38
E2.2 Motional emf concept
B
C
l
S
x
z
S
d S = B0ly
= B0l  y0  v0t 
v0 ay
conducting
rails
y
B = B0az
B
dS
conducting bar
y  y 0  v 0t
2-39
d
C E • dl  – dt S B • dS
d
   B0l  y0  v0t  
dt
  B0lv0
This can be interpreted as due to an electric field
F
E   v0 B0 a x
Q
induced in the moving bar, as viewed by an observer
moving with the bar, since
l
v0 B0 l  
v B a • dx a x
x0 0 0 x
l
 x0 E • dl
2-40
where
F  Qv  B
 Qv0 a y  B0 a z
 Qv0 B0 a x
is the magnetic force on a charge Q in the bar.
Hence, the emf is known as motional emf.
2-41
Review Questions
2.11. State Faraday’s law.
2.12. What are the different ways in which an emf is induced
around a loop?
2.13. Discuss the right-hand screw rule convention
associated with the application of Faraday’s law.
2.14. To find the induced emf around a planar loop, is it
necessary to consider the magnetic flux crossing the
plane surface bounded by the loop? Explain.
2.15. What is Lenz’ law?
2.16. Discuss briefly the motional emf concept.
2.17. How would you orient a loop antenna in order to receive
maximum signal from an incident electromagnetic wave
which has its magnetic field linearly polarized in the
north-south direction?
2-42
Problem S2.4. Induced emf around a rectangular loop of
metallic wire falling in the presence of a magnetic field
2-43
Problem S2.5. Induced emf around a rectangular metallic
loop revolving in a magnetic field
2-44
2.4 Ampére’s Circuital Law
(EEE, Sec. 2.4; FEME, Sec. 2.4)
2-45
Ampére’s Circuital Law
d
C H • dl  S J • dS  dt S D • dS
J, D
S
C
dS
2-46
C H • dl
= Magnetomotive force (only by
analogy with electromotive
force),
 A m  m, or A.
S J • dS
= Current due to flow of charges
crossing S,
2
2
A
m

m

 , or A.
S D • dS
= Displacement flux, or electric
flux, crossing S,
2
2
C
m

m

 , or C.
2-47
d
D • dS = Time rate of increase of

S
dt
displacement flux crossing S,
or, displacement current
crossing S,
C s, or A.
Right-hand screw rule.
Any surface S bounded by C, but the same surface
for both terms on the right side.
2-48
Three cases to clarify Ampére’s circuital law
(a) Infinitely long, current carrying wire
No displacement flux
S1 J • dS  S2 J • dS  I
C H • dl  I
S1
From 
I
C
S2
To 
2-49
(b) Capacitor circuit (assume electric field between
the plates of the capacitor is confined to S2)
S2
C
I(t)
S1
S1 J • dS = I but S2 J • dS = 0
S1 D • dS = 0 but S2 D • dS 
d
D • dS must be  I

S
dt 2
so that C H • dl is unique.
0
2-50
(c) Finitely long wire
 J • dS  I and  D • dS  0
S1
S1
S2 J • dS  0 and S2 D • dS  0
d
S1 J • dS  dt S1 D • dS must be
d
  J • dS 
D • dS

S2
S
dt 2
S1
Q1
I
C
Q2
S2
2-51
Uniqueness of C H • dl
dS1
S1
dS2
C
S2
potato
rubber band
d
C
S1
 D • dS1
dt S1
d
H
•
dl

–
J
•
dS
–
C
S2
2
 D • dS 2
dt S2
d
d
  J • dS1   D • dS1  –  J • dS 2 –  D • dS 2
S1
S2
dt S1
dt S2
d
d
D
•
dS


1
 D • dS 2  – S1 J • dS1 – S2 J • dS2
dt S1
dt S2
H • dl 
J • dS1 
2-52
d
D • dS  – 
J • dS

SS1 S2
dt SS1S2
Displacement current emanating from a closed
surface = – (current due to flow of charges
emanating from the same closed surface)
2-53
D2.9 (a) Current flowing from Q2 to Q3.
d
 I  I23 
dt
D
dS  0
S2
 I  I23  2 I  0
Q2
I23  3I A
S2
I23
I
Q1
3I
S1
Q3
S3
2-54
(b) Displacement current emanating from the
spherical surface of radius 0.1 m and centered at Q1.
d
I  3I 
dt
d
dt
D
D
dS  0
S1
d S  4 I A
S1
(c) Displacement current emanating from the spherical
surface of radius 0.1 m and centered at Q3.
d
3I  I23 
dt
d
dt
D
S3
D
dS  0
S3
d S  3I  I23  3I  3I  6 I A
2-55
Interdependence of Time-Varying Electric and
Magnetic Fields
d
C E • dl = – dt S B • dS
d
C H • dl = S J • dS + dt S D • dS
2-56
Hertzian Dipole
I(t)
I(t)
H(t)
E(t)
2-57
Radiation from Hertzian Dipole
2-58
Review Questions
2.18. State Ampere’s circuital law.
2.19. What is displacement current? Compare and contrast
displacement current with current due to flow of
charges, giving an example.
2.20. Why is it necessary to have the displacement current
term on the right side of Ampere’s circuital law?
2.21. Is it meaningful to consider two different surfaces
bounded by a closed path to compute the two different
currents on the right side of Ampere’s circuital law to
find the line integral of H around the closed path?
2.22. When can you say that the current in a wire enclosed by
a closed path is uniquely defined? Give two examples.
2-59
Review Questions (Continued)
2.23. Give an example in which the current in a wire enclosed
by a closed path is not uniquely defined.
2.24. Discuss the relationship between the displacement
current emanating from a closed surface and the current
due to flow of charges emanating from the same closed
surface.
2.25. Discuss the interdependence of time-varying electric
and magnetic fields through Faraday’s law and
Ampere’s circuital law, and, as a consequence, the
principle of radiation from a wire carrying time-varying
current.
2-60
Problem S2.6. Finding the displacement current emanating
from a closed surface for a given current density J
2-61
Problem S2.7. Finding the rms value of current drawn
from a voltage source connected to a capacitor
2-62
2.5 Gauss’ Laws
(EEE, Sec. 2.5; FEME, Secs. 2.5, 2.6)
2-63
Gauss’ Law for the Electric Field
 C

3
D
d
S

r
dv

m
,
or
C
 3

S
V
m

D
r•
V
S
dS
Displacement flux emanating from a closed surface S =
charge contained in the volume bounded by S = charge
enclosed by S.
2-64
Gauss’ Law for the Magnetic Field
 B • dS = 0
S
B
dS
S
Magnetic flux emanating from a closed
surface S = 0.
2-65
P2.21 Finding displacement flux emanating from a surface
enclosing charge

2
2
2
(a) r  x, y, z   r0 3  x  y  z

Surface of cube bounded by
x   1, y   1, and z   1

S
D d S   r dv
V

1
1
 
1
x 1 y 1 z 1
 8 r0 
1
1
 
r0  3  x2  y2  z 2  dx dy dz
1
x 0 y 0 z 0
2
2
2
3

x

y

z

 dx dy dz
 1 1 1
 8 r0  3    
 3 3 3
 16 r0
2-66
(b) r  x, y, z   r0  x y z 
Surface of the volume x > 0, y > 0, z > 0, and (x2 + y2 + z2) < 1.

S
D d S   r dv
V

1
x 0

r0
2
r0

1 x2
y 0
1
 
x 0
1 x2  y2

z 0
1 x2
y 0
r0 xyz dx dy dz
xy 1  x2  y2  dx dy
1 x2
 xy
x y
xy 



dx



x

0
2
2
4  y 0
 2
r0 1 
1
3
3
5
3

1
2
3
2
4
5 
x

x

x

x

x

2
x

x

 dx
4 x 0 
2
r0
1
r0  x2 x4 x6 
x 3 1 5
    x  x  dx 
  

x

0
4
2 
4  4 4 12 0
2

r0
48
1
2-67
P2.23
z
dS4
1
dS1
dS2
y

x
dS3
S B • dS

S1 B • dS + S2 B • dS2
 S B • dS3  S B • dS 4
3
4
0
2-68
  B d S1
S1
   B d S2   B d S3   B d S4
S2
 
1
S3




z 0 x 0
B0  yax  xa y 
S4
y 0
0  0
 
1
z 0 x 0
B0 x dx dz
B0 2

2
B0  2
Absolute value =
Wb
2
 dx dz a 
y
2-69
Review Questions
2.26. State Gauss’ law for the electric field.
2.27. How do you evaluate a volume integral?.
2.28. State Gauss’ law for the magnetic field.
2.29. What is the physical interpretation of Gauss’ law for the
magnetic field.
2-70
Problem S2.8. Finding the displacement flux emanating
from a surface enclosing charge
2-71
Problem S2.9. Application of Gauss’ law for the magnetic
field in integral form
2.6 The Law of
Conservation of Charge
(EEE, Sec. 2.6; FEME, Sec. 2.5)
2-73
Law of Conservation of Charge
d
S J• dS + dt V r dv  0
J
r(t)
V
S
dS
Current due to flow of charges emanating from a closed surface S
= Time rate of decrease of charge enclosed by S.
2-74
Summarizing, we have the following:
Maxwell’s Equations
d
C E d l   dt S B d S
d
C H d l  S J d S  dt S D d S


S
S
(1)
(2)
D dS   r dv
(3)
B dS  0
(4)
V
2-75
Law of Conservation of Charge

d
J dS 
S
dt

r dv  0
(5)
V
(4) is, however, not independent of (1), whereas (3) follows
from (2) with the aid of (5).
2-76
E2.3 Finding

C
H d l around C.
dS
I1
S
Q(t)

d
H d l  I2 
C
dt

1
D dS  Q
S
2

D dS
I2
C
(Ampére’s Circuital Law)
S
(Gauss’ Law for the electric
field and symmetry
considerations)
2-77
dQ
I2 – I1 
0
dt
(Law of Conservation
of Charge)
dQ
 I1 – I2
dt
d 1 
  H d l  I2   Q 
C
dt  2 
1
 I2   I1  I2 
2
1
  I1  I2 
2
2-78
Review Questions
2.30. State the law of conservation of charge..
2.31. How do you evaluate a volume integral?.
2.32. Summarize Maxwell’s equations in integral form for
time-varying fields.
2.33. Which two of the Maxwell’s equations are independent?
Explain..
2-79
Problem S2.10. Combined application of several of
Maxwell’s equations in integral form
2-80
2.6 Application to Static Fields
(EEE, Sec. 2.7)
2-81
2-82
2-83
2-84
2-85
2-86
2-87
2-88
2-89
2-90
2-91
2-92
Review Questions
2.34. Summarize Maxwell’s equations in integral form for
static fields.
2.35. Are static electric and magnetic fields interdependent?
2.36. Discuss briefly the application of Gauss’ law for the
electric field in integral form to determine the electric
field due to charge distributions.
2.37. Discuss briefly the application of Ampere’s circuital
law in integral form for the static case to determine the
magnetic field due to current distributions.
2-93
Problem S2.11. Application of Gauss’ law for the electric
field in integral form and symmetry
2-94
Problem S2.12. Application of Ampere’s circuital law in
integral form and symmetry
The End