EM Course Module 2 for 2009 India Programs

8/12/2019 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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2-2

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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2-3

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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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 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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2-5

2.1 The Line Integral(EEE, Sec. 2.1; FEME, Sec. 2.1)


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2-6

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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2-7

(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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2-8

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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2-92-9

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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2-102-102-10

(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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2-112-11

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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2-12

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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2-13

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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2-14

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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2-15

Problem S2.1. Evaluation of line integral around a closed

path in Cartesian coordinates


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2-16

Problem S2.2. Evaluation of line integral around a closed

path in spherical coordinates


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2-17

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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2-192-19

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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2-202-20

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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2-21

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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2-22

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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2-23

(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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2-24

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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2-25

Problem S2.3. Evaluation of surface integral over a closed

surface in Cartesian coordinates


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2-26

2.3 Faradays Law(EEE, Sec. 2.3; FEME, Sec. 2.3)


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2-27

Faradays Law

E dl ddtC B dSS

dS

SC

B


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2-28

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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2-29

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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2-30

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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2-31

(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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2-32

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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2-33

where is the magnetic flux linked by one turn.

emf Nddt


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2-34

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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2-352-35

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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2-372-372-37

(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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2-38

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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2-39

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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2-40

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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2-41

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?

2 42


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2-42

Problem S2.4. Induced emf around a rectangular loop of

metallic wire falling in the presence of a magnetic field

2 43


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2-43

Problem S2.5. Induced emf around a rectangular metallic

loop revolving in a magnetic field

2 44


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2-44

2.4 Ampres Circuital Law

(EEE, Sec. 2.4; FEME, Sec. 2.4)

2 45


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2-45

Ampres Circuital Law

H dlC J dS ddtS D dSS

dS

SC

J, D

2 462 46


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2-462-46

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.

2 47


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2-47

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.

2 48


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2-48

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

2 49


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2-49

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)

2 50


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2-50

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

2 51


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2-51

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

2 52


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2-52

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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2-532-53

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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2-542-54

(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

2-55


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2 55

E dl = C

d

dtB dS

S

H dl=C J dS+

d

dtS D dSS

Interdependence of Time-Varying Electric and

Magnetic Fields

2-56


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2 56

I(t)

I(t) H(t) E(t)

Hertzian Dipole

2-57


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Radiation from Hertzian Dipole

2-58


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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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2 59

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.

2-60


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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)

2-63


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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

2-66


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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

2-67


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P2.23

B dSS

B dS + B dS2S2S1

B dS3 B dS4S4S3 0

z

y

x

dS3

dS1dS2

dS4 1

2-68


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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.

2-70


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Problem S2.8. Finding the displacement flux emanating

from a surface enclosing charge

2-71


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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)

2-73


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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)

2-742-74


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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

2-77


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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

2-78


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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)

2-81


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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.

2-93


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Problem S2.11. Application of Gauss law for the electric

field in integral form and symmetry

2-94

P bl S2 12 A li ti f A i it l l i


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Problem S2.12. Application of Amperes circuital law in

integral form and symmetry


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The End

Documents

EM Course Module 2 for 2009 India Programs