em course module 1 for 2009 india programs

8/12/2019 EM Course Module 1 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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1-2

Maxwells Equations

Electric field

intensity

Magnetic

flux density

Charge density

Magnetic

field intensity

Current

densityDisplacement

flux density

V m Wb m2 C m3

A m A m

2

C m

2

Edl=ddt BdSSC DdS= rdvVS

Hdl= J

dS+ d

dtSC

DdS

S

BdS=0

S


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1-3

Module 1Vectors and Fields

1.1 Vector algebra1.2 Cartesian coordinate system1.3 Cylindrical and spherical coordinate systems1.4 Scalar and vector fields1.5 Sinusoidally time-varying fields

1.6 The electric field1.7 The magnetic field1.8 Lorentz force equation


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1-4

Instructional Objectives

1. Perform vector algebraic operations in Cartesian,cylindrical, and spherical coordinate systems

2. Find the unit normal vector and the differential surface ata point on the surface

3. Find the equation for the direction lines associated with a

vector field4. Identify the polarization of a sinusoidally time-varying

vector field5. Calculate the electric field due to a charge distribution by

applying superposition in conjunction with the electric

field due to a point charge6. Calculate the magnetic field due to a current distribution

by applying superposition in conjunction with themagnetic field due to a current element


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1-5

Instructional Objectives (Continued)

7. Apply Lorentz force equation to find the electric andmagnetic fields, for a specified set of forces on a charged

particle moving in the field region


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1-6

1.1 Vector Algebra(EEE, Sec. 1.1; FEME, Sec. 1.1)

In this series of PowerPoint presentations, EEE refers toElements of Engineering Electromagnetics, 6th Edition,

Indian Edition (2006), and FEME refers to Fundamentals of

Electromagnetics for Engineering, Indian Edition (2009).

Also, all D Problems and P Problems are from EEE.


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1-7

(1) Vectors(A) vs. Scalars(A)

Magnitude and direction Magnitude only

Ex: Velocity, Force Ex: Mass, Charge


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1-8

(2) Unit Vectorshave magnitude

unity, denoted by symbol awith subscript. We shall use

the right-handed system

throughout.

Useful for expressing vectors in terms of their

components.

aA = AA

= A1a1 A2a2 A3a3A

12 A

22 A

32


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1-9

(3) Dot Productis a scalar

AA B=ABcos a

B

Useful for finding angle between two vectors.

cos a = A BAB

A = A1a1 A2a2 A3a3B = B1a1 B2a2 B3a3

= A1B1 A2B2 A3B3A

12 A

22 A

32 B

12 B

22 B

32

A

B

a


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1-10

(4) Cross Productis a vector

AA B=ABsin a

B

is perpendicular to both Aand B.

Useful for finding unit vector perpendicular to

two vectors.

an = A BAB sina

= A BA B

ana

right hand

screw A

B

an


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1-11

where

(5) Triple Cross Product

in general.

A B =a1 a2 a3

A1 A2 A3

B1 B2 B3

A (B C) is a vectorA (B C) B (C A) C (A B)


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1-12

(6) Scalar Triple Product

is a scalar.

AB C = B C A = C A B

= A1 A2 A3B1 B2 B3C1 C2 C3


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1-13

Volume of the parallelepiped

an C

B

A

Area of base Height

n

=

=

=

=

=

A B C a

A BA B C

A B

C A B

A B C


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1-14

D1.2 (EEE) A= 3a1+ 2a2+ a3

B= a1+ a2a3

C= a1+ 2a2+ 3a3

(a) A+ B4C

= (3 + 14)a1+ (2 +18)a2

+ (1112)a3

= 5a212a3

A B4C = 25 144 = 13


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1-15

(b) A+ 2BC

= (3 + 21)a1+ (2 + 22)a2

+ (123)a3

= 4a1+ 2a24a3

Unit Vector

=

=

4a1 2a24a34a1 2a24a31

3(2a1 a22a3)


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1-16

(c) A C= 3 1 + 2 2 + 1 3

= 10

(d)

=

= 5a14a2+ a3

B C = a1 a2 a31 1 11 2 3

(3 2)a1 (13)a2 (21)a3


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1-17

(e)

= 158 + 1 = 8

Same as

A (B C) = (3a1+ 2a2+ a3) (5a14a2+ a3)

= 3 5 + 2 (4) + 1 1

= 158 + 1

= 8

AB C =3 2 1

1 1 1

1 2 3


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1-18

P1.5 (EEE)

D= BA ( A+ D= B)

E= CB ( B+ E= C)

Dand Elie along a straight line.

D

A

B

E

C

CommonPoint


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1-19

What is the geometric interpretation of this result?

=DE 0

=B A C B 0

+ =BC AC BB AB 0

A B + B C + C A = 0


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1-20

E1.1 Another Example

Given

Find A.

1 2 3

2 1 3

2 (1)

2 (2)

= +

=

a A a a

a A a a

2 3 1 3

31 2

1 2 3

= 2 2

0 2 21 201 2

C

C C

+

= = + +

A a a a a

aa a

a a a


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1-21

To find C, use (1) or (2).

1 2 32 2 = + +A a a a

1 1 2 3 2 3

3 2 2 3

2 2 2

2 21

C

CC

+ + = +

= +=

a a a a a a

a a a a


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1-22

Review Questions

1.1. Give some examples of scalars.1.2. Give some examples of vectors.

1.3. Is it necessary for the reference vectors a1, a2, and a3

to be an orthogonal set?

1.4. State whether a1, a

2, and a

3directed westward,

northward, and downward, respectively, is a right-

handed or a left-handed set.

1.5. State all conditions for which ABis zero.

1.6. State all conditions for which ABis zero.

1.7. What is the significance of ABC=0?1.8. What is the significance of A(BC)=0?


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1-23

Problem S1.1. Performing several vector algebraic

manipulations


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1-24

Problem S1.1. Performing several vector algebraic

manipulations (continued)


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1-25

1.2 Cartesian

Coordinate System(EEE, Sec. 1.2; FEME, Sec. 1.2)


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1-26

Cartesian Coordinate System


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1-27

Cartesian Coordinate System

x

yx

y

z

Oaz z

az

ay

ayax

ax


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1-28

Right-handed system

xyz xy

ax, ay, azare uniformunit vectors, that is, the

direction of each unit vector is same everywhere inspace.

ax ay = azay az= axaz ax = ay


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1-29

1 12 2

12 2 1

+ =

=

r R r

R r r

zP2

P1R12

r1 r2

y

x

O

2 2 2

1 1 1

2 1 2 1 2 1

x y z

x y z

x y z

x y z

x y zx x y y z z

= + +

+ += + +

a a a

a a aa a a

Vector drawn from one point to another: From

P1

(x1

, y1

, z1

) toP2

(x2

, y2

, z2

)


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1-30

x

x2

x1

O

(x2 x

1)a

x r

1 z

1

r2

z

P1

P2R12

(z2z1)az

(y2y1)ay

y1

z2

y2y


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1-31

P1.8 A(12, 0, 0), B(0, 15, 0), C(0, 0,20).

(a)Distance fromBto C

=

=

(b)Component of vector fromAto Calong vector

fromBto C

= Vector fromAto C

Unit vector along vector fromBto C

(00)ax (015)ay (200)az152 202 = 25


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1-32

(c) Perpendicular distance fromAto the line throughB

and C

= (Vector fromAto C) (Vector fromBto C)BC

12 20 15 20

25

x z y z =

a a a a

15 20

12 20

15 20

40016

25

y z

x z

y z

=

= =

a aa a

a a


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1-331-33

(2) Differential Length Vector(dl)

180 240 300

25=

a a az y x

12 2=

dl = dxax

dyay

dzaz

, ,Q x dx y dy z dz + + +

dzdx

, ,P x y z

dl

dyya

xa

za


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1-34

dl= dx ax+ dy ay

= dx ax+f(x) dx ay

Unit vector normal to a surface

andl2

dl1

Curve 2

Curve 1an = dl1 dl2dl1 dl2

dldx

y =f(x)

dy=f(x) dxz= constant plane

dz= 0


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1-35

D1.5 Find dlalong the line and having the projection dzon

thez-axis.

(a)

(b)

x = 3,y =4dx = 0,dy = 0dl = dzazx y = ,y z=dx dy = 0,dy dz= 0

dy=dz,dx

=dy

=dz

x y z

x y z

d dz dz dz

dz

= +

= +

l a a a

a a a


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1-36

(c) Line passing through (0, 2, 0) and (0, 0, 1).

x = 0, dy02

= dz10

dx = 0,dy = 2dz

2

2

y z

y z

d dz dz

dz

= +

= +

l a a

a a


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1-37

(3) Differential Surface Vector(dS)

Orientation of the surface is defined uniquely by thenormal anto the surface.

For example, in Cartesian coordinates, dSin any planeparallel to thexyplane is

dS = dSan = dl1 dl2an = dl1 dl2

dx dyaz = dxax dyay

adS

dl1

dl2

an

x

ydSdx

dy

az

1 2

1 2

sindS dl dl

d d

a== l l


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1-38

(4) Differential Volume(dv)

In Cartesian coordinates,

dv = dl1dl2 dl3

dv = dxax dyay dzaz= dx dy dz dz dy

dx

z

y

x

dl2

dl1

dl3 dv


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1-39

Review Questions

1.9. What is the particular advantageous characteristicassociated with unit vectors in the Cartesian coordinatesystem?

1.10. What is the position vector?1.11. What is the total distance around the circumference of a

circle of radius 1 m? What is the total vector distancearound the circle?

1.12. Discuss the application of differential length vectors tofind a unit vector normal to a surface at a point on thesurface.

1.13. Discuss the concept of a differential surface vector.1.14. What is the total surface area of a cube of sides 1 m?

Assuming the normals to the surfaces to be directedoutward of the cubical volume, what is the total vectorsurface area of the cube?


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1-40

Problem S1.2. Finding the unit vector normal to a surface

and the differential surface vector, at a point on it


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1-41

1.3 Cylindrical and Spherical

Coordinate Systems(EEE, Sec. 1.3; FEME, Appendix A)

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1-42

Cylindrical Coordinate System

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1-43

Spherical Coordinate System

1 441 44


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1-441-44

Cylindrical (r, f,z) Spherical (r, q, f)

Only azis uniform. All three unit

arand afare vectors are

nonuniform. nonuniform.

Cylindrical and Spherical Coordinate Systems

af

f

x

x

y

yr

z

z ar

az

9090

z

r

y

ar

90

x

af

aq

f

q90

1 451 45


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1-451-45

x = rcos f x = rsin qcos fy = rsin f y = rsin qsin f

z = z z = rcos qD1.7 (a) (2, 5p/6, 3) in cylindrical coordinates

2 cos 5 6 33 1 2

2 sin 5 6 1

3

p

p

= = + =

= = =

x

y

z

x

z

3

2 y

5p/65 6p

1 461 46


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1-461-46

(b)

x = 4 cos 4p 3 =2y = 4 sin 4p 3 =2 3 4 12 = 4z=1

(4, 4 3, 1) in cylindrical coordinatesp

1 4

x

y4p/3

z

4 3p

1 471 47


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1-471-47

(c)

24 sin cos 3

3 624 sin sin 3 9 3 4 4

3 6

4 cos 23

p p

p p

p

= =

= = + + == =

x

y

z

(4, 2 3, 6) in spherical coordinatesp p

x

y

z

4

6p

2 3p

1 481 48


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1-481-48

x = 8 sinp4

cosp

3= 1

= 8 sinp4

sinp

3= 3

z= 8 cosp4

= 2

1 3 4 = 8

(d) 8, 4, 3 in spherical coordinates.p pz

y

x

8

3p

4p

1 49


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1-49

Conversion of vectors between coordinate systems

arc

af

az

=

cos f sinf 0

sinf cosf 0

0 0 1

ax

ay

az

a

rsaq

af

=

sinqcos f sin qsinf cos q

cos qcosf cos qsinf sin q

sinf cos f 0

a

xay

az

axarc

az ay

af

f

az ars

arcafa

q

q

1 501 50


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1-501-50

P1.18 A= ar at (2, p/6, p/ 2)

B= aq at (1, p/3, 0)

C= af at (3, p/4, 3p/2)

A = sinp6

ay cosp6

az

= 12

ay 3

2az

1

4 3

4= 1

x

z A

y

2p

6p

1 511 51


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1-511-51

B sinp

6 ax

cosp

6 az

1

2ax

3

2az

1

4 3

4= 1

C = ax

x

y

z

B

13p

C

x

y

z

p/4

3

3p/2

1 521 52


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1-521-52

(a)

(b)

1 3 1 32 2 2 2

3

4

1 3

2 2

0

A B a a a a

A C a a a

y z x z

y z x

= +

=

= +

=

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1-53

(c)

(d) A B C C A B

=

1 0 0

01

2

3

2

12

0 32

3

4

1 3

2 2

1

2

x z x

=

=

B C a a a

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1-54

Differential length vectors:

Cylindrical Coordinates:

dl= drar+ rdfaf +dz az

Spherical Coordinates:

dl= drar+ rdqaq+ rsin qdfaf

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1 55

Review Questions

1.15. Describe the three orthogonal surfaces that define thecylindrical coordinates of a point.1.16. Which of the unit vectors in the cylindrical coordinate

system are not uniform? Explain.1.17. Discuss the conversion from the cylindrical coordinates

of a point to its Cartesian coordinates.1.18. Describe the three orthogonal surfaces that define thespherical coordinates of a point.

1.19. Discuss the nonuniformity of the unit vectors in thespherical coordinate system.

1.20. Discuss the conversion from the cylindrical coordinatesof a point to its Cartesian coordinates.

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1 56

Problem S1.3. Determination of the equality of vectors

specified in cylindrical and spherical coordinates

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1 57

Problem S1.4. Finding the unit vector tangential to a

curve, at a point on it, in spherical coordinates

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1 58

1.4 Scalar and Vector Fields

(EEE, Sec. 1.4; FEME, Sec. 1.3)

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59

FIELDis a description of how a physical quantityvaries from one point to another in the region of the

field (and with time).

(a) Scalar fields

Ex: Depth of a lake, d(x,y)

Temperature in a room, T(x,y,z)

Depicted graphically by constant magnitudecontours or surfaces.

y

x

d1

d2d3

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(b)Vector Fields

Ex: Velocity of points on a rotating disk

v(x,y) = vx(x,y)ax+ vy(x,y)ay

Force field in three dimensions

F(x,y,z) =Fx(x,y,z)ax+Fy(x,y,z)ay

+Fz(x,y,z)az

Depicted graphically by constant magnitudecontours or surfaces, and direction lines (or

stream lines).

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Example: Linear velocity vector field of points

on a rotating disk

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(c) Static FieldsFields not varying with time.

(d) Dynamic Fields

Fields varying with time.

Ex: Temperature in a room, T(x,y,z; t)

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2 2 2

0

2 2

0

, , , 0 1 0 2 1 1 4

4

T x y z T x y z

T x z

= + + +

= +

2 24 const.x z+ =

2 2 20= 1 sin 2 1 cos 4T x t y t z p p+ + +

D1.10 T(x, y, z, t)

Constant temperature surfaces are elliptic cylinders,

(a)

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(b)

Constant temperature surfaces are spheres

(c)

Constant temperature surfaces are ellipsoids,

22 2

0

2 2 20

, , , 0.5 1 1 2 1 0 4

4 4 4

T x y z T x y z

T x y z

= + + +

= + +

2 2 2 const.x y z+ + =

2 2 2

0

2 2 2

0

, , , 1 1 0 2 1 1 4

16 4

T x y z T x y z

T x y z

= + + + +

= + +

2 2 216 4 const.x y z+ + =

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Procedure for finding the equation for the

direction lines of a vector fieldThe field Fis

tangential to the

direction line at

all points on a

direction line.

dl F =ax ay az

dx dy dz

Fx Fy Fz

= 0

dx

Fx= dy

Fy= dz

Fz

dlF

F

F

dl

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Similarly

dr

Fr= r df

Ff= dz

Fz

dr

Fr= r dq

Fq= rsinqdf

Ff

cylindrical

spherical

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P1.26 (b)xax yay zaz(Position vector)

dx

x= dy

y= dz

z

lnx = lny lnC1 = lnz lnC2lnx = lnC1y = lnC2z

x = C1y = C2z

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Direction lines are straight lines emanating

radially from the origin. For the line passingthrough (1, 2, 3),

1= C1(2) = C2(3)C1 = 1

2,C2 = 1

3

x = y2= z

3

or, 6 3 2= =x y z

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Review Questions

1.21. Discuss briefly your concept of a scalar field andillustrate with an example.

1.22. Discuss briefly your concept of a vector field and

illustrate with an example.

1.23. How do you depict pictorially the gravitational field of

the earth?1.24. Discuss the procedure for obtaining the equations for

the direction lines of a vector field.

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Problem S1.5. Finding the equation for direction line of a

vector field, specified in spherical coordinates

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1.5 Sinusoidally

Time-Varying Fields


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Sinusoidal function of time

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Polarizationis the characteristic which

describes how the position of the tip of the

vector varies with time.

Linear Polarization:

Tip of the vectordescribes a line.

Circular Polarization:

Tip of the vectordescribes a circle.

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Elliptical Polarization:

Tip of the vectordescribes an ellipse.

(i) Linear Polarization

Linearly polarized in thexdirection.

F1 = F1 cos (t f)axDirection remains

along thexaxis

Magnitude variessinusoidally with time

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Linear polarization

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F2 = F2 cos (t q)ayDirection remains

along the yaxisMagnitude variessinusoidally with time

Linearly polarized in the ydirection.

If two (or more) component linearly polarized

vectors are in phase, (or in phase opposition), then

their sum

vector is also linearly polarized.

Ex: 1 2cos cosx y f f = + + +( ) ( )F t F tF a a

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Sum of two linearly polarized vectors in phase

(or in phase opposition) is a linearly polarizedvector

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(ii) Circular Polarization

If two component linearly polarized vectors are

(a) equal in amplitude

(b) differ in direction by 90(c) differ in phase by 90,

then their sum vector is circularly polarized.

a= tan1F2 cos (t f)F

1

cos (t f)= tan1F2

F1

= constant

y

xaF1

F2 F

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Circular Polarization

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Example:

1 1

2 2

1 1

1

1 1

1

1

cos sin

cos sin

, constant

sintan

cos

tan tan

x yF t F t

F t F t

F

F t

F t

t t

a

= +

= +

=

=

= =

F a a

F

1F

2F

F

x

y

a

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(iii) Elliptical Polarization

In the general case in which either (i) or (ii) is not

satisfied, then the sum of the two component

linearly polarized vectors is an elliptically polarized

vector.

Example:F = F1 costax F2 sintay

1F

2FF

x

y

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Example: 0 0cos cos 4x yF t F t p= + +F a a

xF0

F0

F0

F0F1

F2 F

y

4p

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D3.17

F1and F2are equal in amplitude (=F0) and differ in

direction by 90. The phase difference (say f) depends

onzin the manner2pz(3pz) = pz.

(a) At (3, 4, 0), f= p (0) = 0.

(b) At (3,2, 0.5), f= p (0.5) = 0.5 p.

8

1 0

8

2 0

cos 2 10 2

cos 2 10 3

x

y

F t z

F t z

p p

p p

=

=

F a

F a

1 2 is linearly polarized.

+F F

1 2 is circularly polarized.+F F

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(c) At (2, 1, 1), f= p (1) = p.

(d) At (1,3, 0.2) = f= p (0.2) = 0.2p.

1 2 is linearly polarized.+F F

1 2 is elliptically polarized.+F F

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Review Questions

1.25. A sinusoidally time-varying vector is expressed interms of its components along thex-,y-, andz- axes.What is the polarization of each of the components?

1.26. What are the conditions for the sum of two linearlypolarized sinusoidally time-varying vectors to be

circularly polarized?1.27. What is the polarization for the general case of the sumof two sinusoidally time-varying linearly polarizedvectors having arbitrary amplitudes, phase angles, anddirections?

1.28. Considering the seconds hand on your analog watchto be a vector, state its polarization. What is thefrequency?

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Problem S1.6. Finding the polarization of the sum of two

sinusoidally time-varying vector fields

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1.6 The Electric Field


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The Electric Field

is a force field acting on charges by virtue of theproperty of charge.

Coulombs Law

R

F1 a21Q1

Q2

a12

F2 F1 = Q1Q2

4p0R

2a21

F2 = Q2Q14p0R

2a12

0 =permittivity of free space 109

36pF / m

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Q

Q Q

aQ

2a

D1.13(b)

From the construction, it is evident that the resultant

force is directed away from the center of the square.The magnitude of this resultant force is given by

Q2/4p0(2a2)

Q2/4p0(4a2)

Q2/4p0(2a2)

Q = 4p0

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2 2

2 2

0 0

22

2

2 cos 454 2 4 4

1 1

42

0.957N

Q Q

a a

aa

a

p p +

= +

=

1-91


92/141

Electric Field Intensity, E

is defined as the force per unit charge experienced by

a small test charge when placed in the region of thefield.

Thus

Units:

E = Limq0

F

q

Fe = qE

qE

q

E

q

qE

Sources: Charges;Time-varyingmagnetic field

N N m V

C C m m

= =

1-92


93/141

2

0

2

0

4

4

due to

R

R

Qq

R

QqR

q Q

p

p

=

=

=

F a

a

E

aR

q

R

Q

2

0

due to4

RQQ

Rp =E a

Electric Field of a Point Charge

(Coulombs Law)

1-93


94/141

Constant magnitude surfaces

are spheres centered at Q.Direction lines are radial lines

emanating from Q.

E due to charge distributions(a) Collection of point charges

Qn

Q3

Q2

Q1 R1

R2

R3Rn

aRnaR3aR2aR1

E = Qj4p0Rj

2aRj

j =1n

E

Q

aRR

1-94


95/141

E1.2

Q (> 0) d

x

e

d Q (> 0)

y

z

d2+ x2 d2+ x2a

Electron (charge eand mass m) is displaced from the

origin by D(


96/141

For any displacementx,

is directed toward the origin,

and x D d. F Q e x

2p0d3ax

2 2

0

3 22 2

0

2 cos4

2

x

x

Q e

d x

Q e x

d x

ap

p

= +

=

+

F a

a

1-961-96


97/141

The differential equation for the motion of the

electron is

Solution is given by

md2x

dt2= Qe x

2p0d3

d2xdt2

Qe2pm0d

3x = 0

30

cos2p

= +Q e

x A t Bm d

30

sin2p

Q e t

m d

1-97


98/141

Using initial conditions and at t= 0,

we obtain

which represents simple harmonic motion about the

origin with period

x = D dxdt

= 0

x = D cos Qe2pm0d

3t

30

22

Q e

m dp

p

1-981-98


99/141

(b) Line Charges

Line chargedensity,rL(C/m)

(c) Surface Charges

Surface chargedensity,rS(C/m

2)

(d) Volume Charges

Volume charge

density,r(C/m3)

P

dl

dS

dv

1-991-99


100/141

E1.3 Finitely-Long Line Charge

x

za

dz

a

ar

y

z +z2r2

Erf

a

04 C mr r p= =L L0

1-1001-1001-100

2 Ldz

dr

E


101/141

2 202 cos

4

Lrd

r z

ra

p=

+E a

0

0

,2

For r

p E aL ra r

0

3 20 2 20

0

2 2 20 0

0

2 2 2 2

0

2

4

2

2

2

E a

a

a a

aL

rz

a

Lr

z

L

r r

r dz

r z

r z

r r z

a a

r r a r r a

r

p

rp

r

p

=

=

=+

= +

= =+ +

1-1011-101


102/141

Infinite Plane Sheet of Charge

of Uniform Surface Charge Density

z

z

y

dyx

y

+z2

y2

a

0Sr

1-1021-102

r dy


103/141

0

2 2

0

0

2 20

2 cos2

ra

p

r

p

=+

=+

S

z

S

dydE

y z

z dy

y z

0

2 200

20

00

0

0

1

2

Sz

y

S

S

z dyEy z

zd

z

p

a

rp

ra

pr

=

=

=+

=

=

1-1031-103

r


104/141

0

0

0

0

0

0

for 02

2

2

zr

r

r

=

=

E a

a

a

S

z

S

n

S

z

+

+

+

+

+

z< 0

z= 0 z

z> 0

0rS

0

02

r

aS

z

0

02

r

aS z

1-1041-104


105/141

D1.16

Given

(3,5,1) 0 V m(1, 2,3) 6 V m

(3,4,5) 4 V m

==

=z

z

EE a

E a

z= 0 z= 2 z= 4

1Sr 2Sr 3Sr

1-105

1


106/141

21 04 C mSr =

1 2 3

0

1 2 3

0

1 2 3

0

10

2

16

2

1

42

S S S

S S S

S S S

r r r

r r r

r r r

=

+ =

+ + =

22 06 C mSr =

Solving, we obtain

2

3 02 C mSr = 2,1, 6 4 V mz = E a(d)

(a)

(c)

(b)

1-106

Review Questions


107/141

Review Questions

1.29. State Coulombs law. To what law in mechanics is

Coulombs law analogous?1.30. What is the value of the permittivity of free space?

What are its units?1.31. What is the definition of electric field intensity?

What are its units?1.32. Describe the electric field due to a point charge.1.33. Discuss the different types of charge distributions.

How do you determine the electric field due to a chargedistribution?

1.34. Describe the electric field due to an infinitely long linecharge of uniform density.1.35. Describe the electric field due to an infinite plane sheet

of uniform surface charge density.

1-107

Problem S1 7 Determination of conditions for three point


108/141

Problem S1.7. Determination of conditions for three point

charges on a circle to be in equilibrium

1-108

Problem S1.8. Finding the electric field due to an infinite


109/141

Problem S1.8. Finding the electric field due to an infinite

plane slab charge of specified charge density

1-109


110/141

1.7 The Magnetic Field


1-1101-110

The Magnetic Field


111/141

dFm

B

I dl

The Magnetic Field

acts to exert force on charge when it is in motion.

B= Magnetic flux density vectorAlternatively, since charge in motion constitutes

current, magnetic field exerts forces on current

elements.

FmB

vq

Fm = qv B

I=F l Bmd d

1-111


112/141

Units of B:

Sources: Currents;

Time-varying electric field

2

2

N Nm

=A m A m

Wb= = T

m

1-112

A L f F


113/141

Ampres Law of Force

Ra12

a21

dl1

dl2

I1 I2

0 2 2 211 1 1 2

1 1 2

0 1 1 122 2 2 2

2 2 1

4

4

I dd I dR

I d

I dd I dR

I d

p

p

=

=

=

=

l aF l

l B

l aF l

l B

1-1131-113

Magnetic field due to a current element


114/141

I dlaR

R P B

Magnetic field due to a current element

(Biot-Savart Law)

B = 04p

I dl aRR2

B right-circular to the axis of the current element0

7

Permeability of free space

= 4 10 H m

p

=

a

2

sin1

R

a

B

B

Note

1-114


115/141

E1.4

0 0

2 34 4

since

2 1 1 2 3 2

R

R

x y z

x y z

I d I d

R R

R

p p= =

=

+ + +

= + +

l a l R B

Ra

R = a a a

a a a

A situated at 1, 2, 2 .

Find at 2, 1, 3 .

l a a

B

x yI d I dx= +

1-115


116/141

0

12 3 x y

I dx

p= a a

0

3=

4 3

x y x y zI dx

p

+ + + a a a a a

B

1-116


117/141

JS

w

Current Distributions

(a)Filamentary CurrentI (A)

(b) Surface Current

Surface current density, JS(A/m)

JS = Iw max

1-117

( ) V l C


118/141

areaA

J

(c)Volume Current

Density, J(A/m2)

J = IA max

1-118


119/141

P1.44

x a1

r

y

rI

zz

z

a2

dz

z

aR

a1

a

a2

P(r, f,z)

0

22

0

22

4

sin

4

z Rdzdr z z

I dz

r z z

f

p

a

p

=

+

= +

a aB

a

1-1191-119

az z z=


120/141

2

2

1

2

1

2

1

0

2 2

2

02

0

01 2

cot

sin

4 1

cosec sin4 cosec

cos4

cos cos4

1

a

aB B

a

a

a

a

z

z

z zd

r

d zI

r z z r

I dr

I

r

I

r

af

a a

a

fa a

a

a a f

f

a

a

p

a a ap a

a

p

a a

p

=

=

=

=

=

= =

=+

=

=

=

1-120


121/141

For infinitely long wire,

a1 ,a2 ,a1 0 , a2 p

B 0I2pr af

1-121


122/141

Magnetic Field Due to an Infinite Plane Sheet of

Uniform Surface Current Density

This can be found by dividing the sheet into

infinitely long strips parallel to the current density

and using superposition, as in the case of finding the

electric field due to an infinite plane sheet of uniform

surface charge density. Instead of going through this

procedure, let us use analogy. To do this, we firstnote the following:

1-1221-122


123/141

Point Charge Current Element(a)

0

24

RI dR

p

= l aB204

RQ

Rp=E a

1-1231-123


124/141

z

r= 0

P

arB

r

I

(b) Line Charge Line Current

0

02

r

p=E a

L

rr

0

0

2

2

f

p

p

=

=

B a

a az r

I

rI

r

z

r= 0

P

ar Er0L

r

1-1241-124

Then,


125/141

JS

P

B

an

,

(c) Sheet Charge Sheet Current

0

02

r

=E aS n

0

2

=B J aS n

P

an E

0Sr

1-125

Review Questions


126/141

Q

1.36. How is magnetic flux density defined in terms of force

on a moving charge? Compare the magnetic force on a

moving charge with electric force on a charge.

1.37. How is magnetic flux density defined in terms of force

on a current element?

1.38. What are the units of magnetic flux density?1.39. State Amperes force law as applied to current elements.

Why is it not necessary for Newtons third law to hold

for current elements?

1.40. Describe the magnetic field due to a current element.1.41. What is the value of the permeability of free space?

What are its units?

1-126

Review Questions (continued)


127/141

Q ( )

1.42. Discuss the different types of current distributions.

How do you determine the magnetic flux density due to

a current distribution?

1.43. Describe the magnetic field due to an infinitely long,

straight, wire of current.

1.44. Discuss the analogies between the electric field due tocharge distributions and the magnetic field due to

current distributions.

1-127

Problem S1.9. Finding parameters of an infinitesimal


128/141

current element that produces a specified magnetic field

1-128

Problem S1.10. Finding the magnetic field due to a

ifi d t di t ib ti ithi i fi it l l b


129/141

specified current distribution within an infinite plane slab

1-129


130/141

1.8 Lorentz Force Equation


1-130

Lorentz Force Equation


131/141

For a given B, to find E,

E = Fq v B One force is sufficient.

Fm

q

B

E

v

Fe

e

m

e m

q

q

q

=

=

+

F E

F v B

F = F F

F = E + v B

1-131

D1.21

0 2 2B

B


132/141

0 2 23

x y z= + B a a a

0q= + =

F E v B

E = v B

Find E for which acceleration experienced by qis

zero, for a given v.

(a)

0

0 0

0 0

2 23

x y z

x y z x y z

y z

v

v B

v B

+

+ + = +

v = a a a

E = a a a a a a

a a

1-1321-1321-132

(b)


133/141

0

0 0

0 0

2 2

2 2 2 23

2 2

x y z

x y z x y z

x y z

v

v B

v B

+ +

= + + +

=

v = a a a

E a a a a a a

a a a

(b)

(c)

0

0

0 0

along 2

2 23

2 2 2 23

0

v =

a a a

E a a a a a a

x y z

x y z x y z

v y z x

v

v B

= =

= +

= + + +

=

1-133

For a given E, to find B,


134/141

For a given E, to find B,

One force not sufficient. Two forces are needed.

v B =F

q E

v1 B = F1q

E = C1v2 B = F2

qE = C2

1 2 1 2

1 2 1 2

1 2

==

C C v B v Bv B B v v B v B

= C v B

1-1341-134


135/141

B = C2 C1C1 v2

provided , which means v2and v1

should not be collinear.1 2 0 C v

1-135

P1.54 For v= v1 or v= v2, test charge moves with


136/141

constant velocity equal to the initial value. It is to be

shown that for

the same holds.

(1)

(2)

(3)

1

2

1 3

2 3

v v B = 0

v v B = 0

1 2 , where + 0,+

= +

m nm n

m n

v vv

1+ =q qE v B 0

2+ =q qE v B 0

+ =q qE v B 0

1-1361-136

Both and are collinear tov v v v B


137/141

Alternatively,

1 2

1 2

Both and are collinear to .

k

=

v v v v B

v v v v

(1) (2)+ + +m n

m n m n

1 2

1 2

1

for =

=

+= +

v vv

v v

k

k

m n nkm n m

1-137


138/141

1 2 0 +

+ = +

m nq q

m n

v vE B

1 2+ =+

m n

m n

v vv

1-138

Review Questions


139/141

1.45. State Lorentz force equation.

1.46. If it is assumed that there is no electric field, the

magnetic field at a point can be found from the

knowledge of forces exerted on a moving test charge

for two noncollinear velocities. Explain.

1.47. Discuss the determination of Eand Bat a point from theknowledge of forces experienced by a test charge at that

point for several velocities. What is the minimum

number of required forces? Explain.


140/141

1-140


141/141

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