Download - EM Course Module 2 for 2009 India Programs
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Fundamentals of Electromagnetics
for Teaching and Learning:A Two-Week Intensive Course for Faculty in
Electrical-, Electronics-, Communication-, andComputer- Related Engineering Departments in
Engineering Colleges in India
by
Nannapaneni Narayana RaoEdward C. Jordan Professor Emeritus
of Electrical and Computer EngineeringUniversity of Illinois at Urbana-Champaign, USADistinguished Amrita Professor of Engineering
Amrita Vishwa Vidyapeetham, India
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Program for Hyderabad Area and Andhra Pradesh FacultySponsored by IEEE Hyderabad Section, IETE Hyderabad
Center, and Vasavi College of EngineeringIETE Conference Hall, Osmania University Campus
Hyderabad, Andhra PradeshJune 3 June 11, 2009
Workshop for Master Trainer Faculty Sponsored byIUCEE (Indo-US Coalition for Engineering Education)
Infosys Campus, Mysore, Karnataka
June 22
July 3, 2009
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Module 2
Maxwells Equations
in Integral Form
2.1 The line integral
2.2 The surface integral
2.3 Faradays law
2.4 Amperes circuital law2.5 Gauss Laws
2.6 The Law of Conservation of Charge
2.7 Application to static fields
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Instructional Objectives
8. Evaluate line and surface integrals9. Apply Faraday's law in integral form to find the
electromotive force induced around a closed loop, fixed orrevolving, for a given magnetic field distribution
10. Make use of the uniqueness of the magnetomotive force
around a closed path to find the displacement currentemanating from a closed surface for a given currentdistribution
11. Apply Gauss law for the electric field in integral form tofind the displacement flux emanating from a closed
surface bounding the volume for a specified chargedistribution within the volume12. Apply Gauss law for the magnetic field in integral form
to simplify the problem of finding the magnetic fluxcrossing a surface
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Instructional Objectives (Continued)
13. Apply Gauss' law for the electric field in integral form,Ampere's circuital law in integral form, the law ofconservation of charge, and symmetry considerations, tofind the line integral of the magnetic field intensityaround a closed path, given an arrangement of point
charges connected by wires carrying currents14. Apply Gauss law for the electric field in integral form to
find the electric fields for symmetrical chargedistributions
15. Apply Amperes circuital law in integral form, without
the displacement current term, to find the magnetic fieldsfor symmetrical current distributions
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2.1 The Line Integral(EEE, Sec. 2.1; FEME, Sec. 2.1)
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a1a2E1E2
A l1
l2
B
The Line Integral:
Work done in carrying a charge from Ato Binan electric field:
WAB dWjj1
n
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(Voltage betweenAandB)
coscos
j j j j
j j j
j j
dW qE l
qE l
q
a
a
E l
1
W
n
AB j j
j
q E l
1
nWV
AB
AB j j
jq
E l
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In the limit ,n VAB E dlAB
= Line integral of EfromAtoB.
= Line integral of E
around the closedpath C.
Cd E l
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A
B
CC
R
L
is independent of
the path fromAtoB(conservative field)
If
then E lB
Ad
E l = 0
C
d
E l E l E lARBLA ARB BLA
d d d
0E l E lARB ALB
d d
E l = E lARB ALB
d d
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(1, 0, 0)
z
x
y
(1, 2, 3)
(1, 2, 0)
(0, 0, 0)
F yzax zxay xyaz ,along the straight line paths,
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).
(1,2,3)
(0,0,0)d F l
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From (0, 0, 0) to (1, 0, 0),
From (1, 0, 0) to (1, 2, 0),
1, 0 ; 0
z
x y z y
x z dx dz
yd dx dy dz dyF a
l a a a a
1,0,0
0,0,0
0; 0
0, 0
y z dy dz
d
F F l
1,2,0
0,0,0
0, 0d d
F l F l
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From (1, 2, 0) to (1, 2, 3),
x 1, y 2 ; dx dy 0F 2zax zay 2az, dl dzazF dl 2 dz, 2dz 6
(1,2,0)
(1,2,3) F dl 0 0 6 6
(0,0,0)
(1,2,3)
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In fact,
x y z
x y z
d yz zx xydx dy dz
yz dx zx dy xy dz
d xyz
F l a a a
a a a
1,2,3 1,2,3 1,2,3
0,0,00,0,0 0,0,0
1 2 3 0 0 06, independent of the path.
d d xyz xyz
F l
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Review Questions
2.1. How do you find the work done in moving a test chargeby 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 ofEbetween two pointsAandB?
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.
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Problem S2.1. Evaluation of line integral around a closed
path in Cartesian coordinates
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Problem S2.2. Evaluation of line integral around a closed
path in spherical coordinates
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2.2 The Surface Integral(EEE, Sec. 2.2; FEME, Sec. 2.2)
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2-182-18
B
anS
The Surface IntegralFlux of a vector crossing a surface:
Flux = (B)(S)
Flux = 0
Flux (B cos a)S BScos a B San B S
B
an
S
a
Ban
S
an
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Flux jj1
n Bj Sj
j1n
In the limit n ,Flux, = B dS
S
= Surface integral of Bover S.
Normal
Bjanjaj
S
Sj
Bj
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z
y
x
2
2
B dS
S= Surface integral of B
over the closed surface S.
D2.4
(a)
x yx A a a
0,
0, 0
0
0
d
d
A S
A S
n xx
d
a a
A A S
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2,2
n x
x y
x
x
d dy dz
a a
A a a
S a
z
y
x
2
2
2
A dS 2dy dzA dS 2dy dz 8
z02
y02
A dS 8
(b)
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z
x
y
2
2
A dS x dx dz 4z02x02
A dS 4
(c)
0, n y
x y
y
y
x
d dz dxd x dx dz
a a
A a a
S a
A S
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(d) From (c),
A dS x dx dzA dS x dx dz
z02xx02
x(2 x)dx0
2 4
3
A dS 43
z
x
y
2
2
x+z= 2
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Review Questions
2.6. How do you find the magnetic flux crossing aninfinitesimal surface?
2.7. What is the magnetic flux crossing an infinitesimalsurface oriented parallel to the magnetic flux densityvector?
2.8. For what orientation of an infinitesimal surface relativeto the magnetic flux density vector is the magnetic fluxcrossing the surface a maximum?
2.9. How do you find the approximate value of the surfaceintegral 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.,
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Problem S2.3. Evaluation of surface integral over a closed
surface in Cartesian coordinates
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2.3 Faradays Law(EEE, Sec. 2.3; FEME, Sec. 2.3)
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Faradays Law
E dl ddtC B dSS
dS
SC
B
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2 2Wb m m , or Wb.
E dl =C
Voltage around C, also known as
electromotive force (emf) around C
(but not really a force),
B dSS
= Magnetic flux crossing S,
d
dt B dSS = Time rate of decrease of
magnetic flux crossing S,
Wb s, or V.
V m m, or V.
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Important Considerations
(1) Right-hand screw (R.H.S.) Rule
The magnetic flux crossing the
surface Sis to be evaluated toward
that side of Sa right-hand screwadvances as it is turned in the sense of C.
C
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z
R
C
yQ
Px
O
(2) Any surface Sbounded by C
The surface Scan be any surface bounded byC. For example:
This means that, for a given C, the values ofmagnetic flux crossing all possible surfaces
bounded by it is the same, or the magnetic flux
bounded by Cis unique.
z
C
y
R
x
P
QO
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(3) Imaginary contour Cversus loop of wire
There is an emf induced around Cin eithercase by the setting up of an electric field. A
loop of wire will result in a current flowing in
the wire.
(4) Lenzs Law
States that the sense of the induced emf is
such that any current it produces, if the closedpath were a loop of wire, tends to oppose the
changein the magnetic flux that produces it.
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Thus the magnetic flux produced by the
induced current and that is bounded by Cmustbe such that it opposes the changein the
magnetic flux producing the induced emf.
(5) N-turn coil
For anN-turn coil, the induced emf isNtimes
that induced in one turn, since the surface
bounded by one turn is boundedNtimes by
theN-turn coil. Thus
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where is the magnetic flux linked by one turn.
emf Nddt
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0sin cos
x y
B t t B a a
z
1
1
x
y
C
D2.5
0
0
sin
cos V
C
d
d B tdt
B t
E l
0= sinS
d B t B S(a)
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Lenzs law is verified.
emf < 0
emf > 0
0
emf
0
2 3
2 3t
t
dec.
inc.
0B
0B
0B
0B
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2-362-362-36
(b)
z
C
y
x
1
1
1
0 0
0
1 1sin cos
2 2
1sin
42
B t B t
B t
0
0
1 sin42
cos4
2
V
d B tdt
Bt
S
d B S
C
d E l
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(c)z
C
y
x
1
11
0 0
0
sin cos
sin
4
B t B t
B t2
0
0
2 sin4
2 cos4
V
d B tdt
B t
S
d B S
C
d E l
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E2.2 Motional emf concept
conducting bary0 v0t
C
lx
S dS
B
v0ay
yz conducting
rails
0
0 0 0
=
=
Sd B ly
B l y v t
B S0
=z
BB a
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E dl ddtC
B dSS
This can be interpreted as due to an electric field
induced in the moving bar, as viewed by an observer
moving with the bar, since
E FQ v0B0ax
v0B0l v0B0ax dxaxx0l E dl
x0l
0 0 0
0 0
dB l y v t
dt
B lv
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where
is the magnetic force on a charge Qin the bar.
Hence, the emf is known as motional emf.
F Qv B Qv0ay B0az Qv0B0ax
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Review Questions
2.11. State Faradays 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 Faradays law.
2.14. To find the induced emf around a planar loop, is itnecessary 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 receivemaximum signal from an incident electromagnetic wavewhich has its magnetic field linearly polarized in thenorth-south direction?
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Problem S2.4. Induced emf around a rectangular loop of
metallic wire falling in the presence of a magnetic field
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Problem S2.5. Induced emf around a rectangular metallic
loop revolving in a magnetic field
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2.4 Ampres Circuital Law
(EEE, Sec. 2.4; FEME, Sec. 2.4)
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Ampres Circuital Law
H dlC J dS ddtS D dSS
dS
SC
J, D
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A m m, or A.
H dlC
J dSS
D dSS
= Magnetomotive force (only by
analogy with electromotiveforce),
= Current due to flow of chargescrossing S,
= Displacement flux, or electricflux, crossing S,
2 2C m m , or C.
2 2A m m , or A.
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d
dtD dS
S = Time rate of increase ofdisplacement flux crossing S,
or, displacement current
crossing S,
C s, or A.
Right-hand screw rule.
Any surface Sbounded by C, but the same surfacefor both terms on the right side.
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Three cases to clarify Ampres circuital law
(a) Infinitely long, current carrying wire
No displacement flux
J dSS1 J dS I
S2
H dl IC
C
I To From
S1
S2
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C
S1
S2
I(t)
J dS=S1
Ibut J dS=S2
0
D dS=S1 0 but D dSS2 0
d
dtD dSmust be I
S2
so that H dlC is unique.
(b) Capacitor circuit(assume electric field between
the plates of the capacitor is confined to S2)
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I
CQ1Q2
S1S2
(c) Finitely long wire
J dS I andS1 D dS 0S1J dS 0and
S2 D dS 0
S2
J dS ddtS1 D dSmust beS1 J dS d
dtS2D dS
S2
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Uniqueness of H dlC
dS1 dS2
CS2S1
rubber bandpotato
H dl J dS1 ddtS1C
D dS1S1
H dl J dS2
d
dtS2CD dS
2S2
J dS1 ddtS1
D dS1 J dS2 ddtS2
D dS2S2S1
d
dtD dS1S1
ddt
D dS2S2 J dS1S1 J dS2S2
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d
dtD dS J dS
SS1S2SS1S2
Displacement current emanating from a closedsurface =(current due to flow of charges
emanating from the same closed surface)
2-532-53
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Q2
Q3Q1
II23
3I
S1 S3
S2
D2.9(a) Current flowing from Q2to Q3.
2
23
23
23
0
2 0
3 A
D S
S
dI I dd t
I I I
I I
2-542-54
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(b)Displacement current emanating from the
spherical surface of radius 0.1 m and centered at Q1.
(c)Displacement current emanating from the spherical
surface of radius 0.1 m and centered at Q3.
1
1
3 0
4 A
D S
D S
S
S
dI I d
dt
dd I
d t
3
3
23
23
3 0
3 3 3 6 A
D S
D S
S
S
d
I I ddt
dd I I I I I
dt
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E dl = C
d
dtB dS
S
H dl=C J dS+
d
dtS D dSS
Interdependence of Time-Varying Electric and
Magnetic Fields
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I(t)
I(t) H(t) E(t)
Hertzian Dipole
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Radiation from Hertzian Dipole
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Review Questions
2.18. State Amperes 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 Amperes 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 Amperes circuital law to
find the line integral of Haround 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.
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Review Questions (Continued)
2.23. Give an example in which the current in a wire enclosedby 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 Faradays law and
Amperes circuital law, and, as a consequence, the
principle of radiation from a wire carrying time-varyingcurrent.
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Problem S2.6. Finding the displacement current emanating
from a closed surface for a given current density J
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Problem S2.7. Finding the rms value of current drawn
from a voltage source connected to a capacitor
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2.5 Gauss Laws(EEE, Sec. 2.5; FEME, Secs. 2.5, 2.6)
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Gauss Law for the Electric Field
Displacement flux emanating from a closed surface S=
charge contained in the volume bounded by S= charge
enclosed by S.
V SdS
D
r
3
3
Cm , or CmS Vd dv
r
D S
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Gauss Law for the Magnetic Field
Magnetic flux emanating from a closed
surface S= 0.
B
dSS
B dS = 0S
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P2.21 Finding displacement flux emanating from a surface
enclosing charge
(a)
Surface of cube bounded by
1, 1, and 1x y z
2 2 2
0, , 3x y z x y zr r
1 1 12 2 2
01 1 1
1 1 12 2 2
0 0 0 0
0
0
3
8 3
1 1 18 3
3 3 3
16
S V
x y z
x y z
d dv
x y z dx dy dz
x y z dx dy dz
r
r
r
r
r
D S
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(b)
Surface of the volumex> 0,y> 0,z> 0, and (x2+y2+z2) < 1.
1
3 3 5 3 50
0
12 4 6
13 50 0
00
0
12
4 21
4 2 2 4 4 4 12
48
x
x
x x x x x x x dx
x x x xx x dx
r
r r
r
0, ,x y z x y zr r
2 2 2
2
2
1 1 1
00 0 0
1 12 20
0 0
12 3 2 4
10
00
1
2
2 2 2 4
S V
x x y
x y z
x
x y
x
xy
d dv
xyz dx dy dz
xy x y dx dy
xy x y xydx
r
r
r
r
D S
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P2.23
B dSS
B dS + B dS2S2S1
B dS3 B dS4S4S3 0
z
y
x
dS3
dS1dS2
dS4 1
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2
0Absolute value = Wb
2
B
2 3 42 3 4
1
0 00 0
1
00 0
2
0
0 0
2
S S S
x y yyz x
z x
d d d
B y x dx dz
B x dx dz
B
B S B S B S
a a a
11
Sd B S
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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.
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Problem S2.8. Finding the displacement flux emanating
from a surface enclosing charge
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Problem S2.9. Application of Gauss law for the magnetic
field in integral form
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2.6 The Law of
Conservation of Charge(EEE, Sec. 2.6; FEME, Sec. 2.5)
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Law of Conservation of Charge
Current due to flow of charges emanating from a closed surface S= Time rate of decrease of charge enclosed by S.
JdS +d
dtrdv 0
VS
VdS
J
S
r(t)
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Summarizing, we have the following:
Maxwells Equations
(1)
(2)
(3)
(4)0
E l B S
H l J S D S
D S
B S
C S
C S S
S V
S
dd d
dt dd d d
dt
d d v
d
r
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Law of Conservation of Charge
(4) is, however, not independent of (1), whereas (3) follows
from (2) with the aid of (5).
(5)0S V
dd dv
dtr J S
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I1
dS
I2
Q(t)C
S
(Ampres Circuital Law)
(Gauss Law for the electricfield and symmetryconsiderations)
E2.3 Finding around .C
d C H l
2C S
d
d I ddt H l D S
1
2Sd Q D S
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I2
I1dQ
dt 0(Law of Conservation
of Charge)
dQ
dt I1 I2
2 1 2
1 2
1
2
1
2
I I I
I I
2
1
2C
dd I Q
dt
H l
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Review Questions
2.30. State the law of conservation of charge..2.31. How do you evaluate a volume integral?.
2.32. Summarize Maxwells equations in integral form for
time-varying fields.
2.33. Which two of the Maxwells equations are independent?Explain..
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Problem S2.10. Combined application of several of
Maxwells equations in integral form
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2.6 Application to Static Fields(EEE, Sec. 2.7)
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Review Questions
2.34. Summarize Maxwells equations in integral form forstatic 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 Amperes circuital
law in integral form for the static case to determine the
magnetic field due to current distributions.
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Problem S2.11. Application of Gauss law for the electric
field in integral form and symmetry
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Problem S2.12. Application of Amperes circuital law in
integral form and symmetry
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The End