Engineering Electromagnetics; William Hayt & John Buck

Dr. Naser Abu-Zaid; Lecture notes on Electromagnetic Theory(1); Ref:Engineering Electromagnetics; William Hayt& John Buck, 7th & 8th editions; 2012

Preliminary material (mathematical requirements) Vector Analysis

Vector algebra:

Vector Calculus:

Addition; Subtraction; Multiplication

Differentiation; Integration

Vector: A quantity with both magnitude and direction. (Force F  10N to the east). Scalar:A quantity that does not posses direction, Real or complex. (Temperature T  20o .

Vector addition: 1) Parallelogram: A

B

A

A B

B

2) Head to Tail:

B

A

A B

B Chapter One

A

1


Vector Subtraction: A B

A

B

B

A B

A B B

Multiplication by scalar: B  k A

A

2A

B  0.5A

B  3A

0.5A A

    

A

 3A

Commulative law: A  B  B  A Associative law: A  B  C  A  B  C Equal vectors: A  B if A  B  0 (Both have same length and direction) Add or subtract vector fields which are defined at the same point. If non vector fields are considered then vectors are added or subtracted which are not defined at same point (By shifting them)

Chapter One

B  2A

2


THE RECTANGULAR COORDINATE SYSTEM

y

x , y , z are coordinate

Right Handed System

variables (axis) which are mutually perpendicular.

x

z Out of page

z z 3 P 1,2,3

A point is located by its x , y and z coordinates, or as the intersection of three constant surfaces (planes in this case)

y 2

y

x 1

Chapter One

x

3


z z 3

P 1,2,3

surface (plane)

y

x 1 Surface (plane)

y 2

Three mutually perpendicular surfaces intersect at a common point

Chapter One

x

surface (plane)

4


Increasing each coordinate variable by a differential amount dx , dy , and dz , one obtains a parallelepiped. z

P' x  dx , y  dy , z  dz 

P x , y , z  dz

dy

dx

y

Differential volume: dv  dxdydz

Chapter One

x

5


Differential Surfaces: Six planes with dierential areas ds  dxdy ; ds  dzdy ;

ds  dxdz

dx 2  dy 2  dz 2

Differential length: from P to P’ dl 

VECTOR COMPONENTS AND UNIT VECTORS A general vector r may be written as the sum of three vectors;

z Projection of r into z-axis

z 3 C

Projection of r into x-axis

A

P 1,2,3 r B

y 2

y

x 1 Projection of r into y-axis

x

r  A BC A , B , and C arevector componentswith constant directions.

Unit vectorsâ x , â y , and â z directed along x, y, and z respectively with unity

Chapter One

length and no dimensions.

6


â x

z

â z

â y

â z

â z

â y â x

â y

y

â x

â y

â x

x

So, the vector r  A  B  C may be written in terms of unit vectors as:

r  A  B  C  Aâx  B ây  C âz vector components scalar components       A , B, C A , B ,C Where: A is the directed length or signed magnitude of A . B is the directed length or signed magnitude of B .

C is the directed length or signed magnitude of C . As a simple exercise, let rp (Position vector) point from origin (0,0,0) to P(1,2,3), then

rP  1âx  2ây  3âz

rP are:

rPx  A  1 ; rPy  B  2 ; rPz  C  3 . Vector components of

rP are:

Chapter One

Scalar components of

7


rPx  A  1âx ; rPy  B  2ây ; rPz  C  3âz . If Q(2,-2,1) then

rQ  2âx  2ây âz And the vector directed from P to Q,

rPQ  rQ  rP (displacement vector)

which is given by

rPQ  2  1âx   2  2ây  1  3âz  âx  4ây  2âz z rPO

rQ

rP  rP

y

x

The vector

rP is termed position vector which is directed from the origin toward

Chapter One

the point in quesion.

8


Other types of vectors (vector fields such as Force vector) are denoted:

F  F x âx  F y ây  F z âz Where

Fx , Fy , F z

are scalar components, and

F x âx , F y ây , F z âz

are the

vector components. The magnitude of a vector

B  B x âx  B y ây  B z âz is;

B B A unit vector in the direction of

B x 2  B y 2  B z 2 B is; B

B B x âx  B y ây  B z âz âB   B B x 2  B y 2  B z 2 Let

âB

B  B x âx  B y ây  B z âz and A  Ax âx  Ay ây  Az âz , then A  B  Ax B x âx  Ay  B y ây  Az  B z âz

A  B  Ax  B x âx  Ay  B y ây  Az  B z âz Ex:Specify the unit vector extending from the origin toward the point G(2,-2,-1). Ex: Given M(-1,2,1), N(3,-3,0) and P(-2,-3,-4) Find: (a) R MN (b) R MN  R MP (d) âMP (e) 2rP  3rN

Chapter One

(c) rM

9


THE VECTOR FIELD AND SCALAR FIELD

Vector Field: vector function of a position vector r . It has a magnitude and direction at each point in space. v r   v x r â x  v y r ây  v z r âz

 v x x , y , z â x  v y x , y , z ây  v z x , y , z âz

Chapter One

Velocity or air flow in a pipe

10


Scalar field: A scalar function of a position vector r . Temperature is an example T r   T x , y , z  which has a scalar value at each point in space.

T

T

3

2

T1

Ex:A vector field is expressed as S

x

125

 1  y  2  z  1 2

2

2

x  1â

x

 y  2ây  z  1âz



(a) Is this a scalar or vector field? (b) Evaluate S @ P 2,4 ,3 . (c) Determine a unit vector that gives the direction of S @ P 2,4 ,3 .

Chapter One

(d) Specify the surface f x , y , z  on which S  1 .

11


THE DOT PRODUCT

A  B  A B cos  AB  Which results in a scalar value, and  AB is the smaller angle between A and B . Projection of B into A

 B cos  AB 

A

 AB

B Projection of A into B

 A cos  AB 

A B  B A

since

A B cos  AB   B A cos  AB 

âx âx  âx âx cos 0  111  1

ây ây  ây ây cos 0  111  1 âz âz  âz âz cos 0  111  1

âx âz  âx âz cos 90o   110  0  âz âx ây âz  ây âz cos 90o   110  0  âz ây

Chapter One

âx ây  âx ây cos 90o   110  0  ây âx

12


Let B  B x âx  B y ây  B z âz and A  Ax âx  Ay ây  Az âz , then A  B  Ax B x  Ay B y  Az B z 2

A  A  Ax2  Ay2  Az2  A  A 2  A  A  A

The scalar component of B in the direction of an arbitrary unit vector â  is given by B â  B

B â   B â  cos   

â 

 B cos  

Scalar Projection of B into â 

 B cos    B â  The vector component of B in the direction of an arbitrary unit vector

Chapter One

â  is given by B â  â  .

13


B



â 

Vector Projection of B into â 

 B â  â 

Distributive property: A  B  C  A  B  A  C Ex: Given E  y âx  2.5x ây  3âz and Q(4,5,2) Find: (a) E @ Q. (b) The scalar component of E @ Q in the direction of

ân 

1 2âx ây  2âz . 3

(c) The vector component of E @ Q in the direction of

1 2âx ây  2âz . 3 (d) The angle  Ea between ErQ  and ân .

Chapter One

ân 

14


THE CROSS PRODUCT

A  B  A B sin AB aˆ n results in a vector A  B  A B sin AB  Direction of A  B  aˆ n

ân is a unit vector normal to the plane containing A and B . Since

there are two possible ân' s , we use the Right Hand Rule (RHR) to determine the direction of A  B . Cross product clearly results in a vector, and  AB is the smaller angle between A and B . B sin  AB  which is the height

A B

Of the parallelogram

B

ân

 AB A

A B sin  AB  Is the area of the parallelogram

Properties: A  B  B  A

A  B  C  A  B  A  C

Chapter One

 A B  B A

15


A  B  C  A  B  C âx âx  âx âx sin 0ân  0 ây ây  ây ây sin 0ân  0

âz âz  âz âz sin 0ân  0

y

z

RHR

x

Out of page

âx ây  âx ây sin 90o ân  111ân  âz

âx âz  âx âz sin 90o ân  111ân  ây

ây âx  ây âx sin 90o ân  111ân  âz ây âz  ây âz sin 90o ân  111ân  âx âz âx  âz âx sin 90o ân  111ân  ây

âz ây  âz ây sin 90o ân  111ân  âx Let B  B x âx  B y ây  B z âz and A  Ax âx  Ay ây  Az âz , then â x â y â z A  B  A B sin  AB ân  Ax Ay Az Bx By Bz

Chapter One

 Ay B z  Az B y â x  Ax B z  Az B x â y  Ax B y  Ay B x â z

16


CIRCULAR CYLINDRICAL COORDINATES

y

 ,  , z are coordinate variables

which are mutually perpendicular.

y

Remember polar coordinates (The 2D version)

P ( x , y ) or P( ,)



z

x

x

x   cos  

y   sin    is measured from x-axis

toward y-axis.

Including the z-coordinate, we obtain the cylindrical coordinates (3D version)

z

A point is located by its  ,  and z coordinates. Or as the intersection of three mutually orthogonal surfaces.

P ( x  1, y  2 , z  3) or P (   5 ,   63.4 o , z  3)

z 3



x 1

y 2 

y

Chapter One

x

17


z  z1

z

surface (plane)

P 1 , 1 , z 1 

â z

â

â

  1

z1

Surface (plane)

  1 surface (cylinder)

1

y

1

x

Chapter One

1) Infinitely long cylinder of radius   1 . 2) Semi-infinite plane of constant angle   1 . 3) Infinite plane of constant elevation z  z 1 .

18


The three unit vectors â z , â , and â are in the direction of increasing variables and are perpendicular  to the surface at which the coordinate variable is constant.

y â

â

â z



y



x

x Chapter One

z

19


Note that in Cartesian coordinates, unit vectors are not functions of coordinate variables. But in cylindrical coordinates â , and â are functions of  .

y

â 2

â1

2

2

z

â 1

â x

1

2

1

â y

x

2

z

â x

1

1

The cylindrical coordinate system is Right Handed: â â  âz .

x

Chapter One

â  2

y â y

20


Increasing each coordinate variable by a differential amount d , d , and dz , one obtains:

y

dd  area

d  arc length 

d 

d

x

z

Note that  and z are lengths, but  is an angle which requires a metric coefficient to convert it to length.



d

metric coefficient

Chapter One

arc length 

21


ds  ddz

z

ds  ddz

dz

ds  dd

y 

x

d

d

d

Differential volume: dv  dddz

Chapter One

Differential Surfaces: Six planes with dierential areas shown in the figure above. (Try it!)

22


Transformations between Cylindrical and Cartesian Coordinates

z

From cylindrical to cart: x   cos  

z  z1

y   sin  

z z

From cart. To cyl.:

x  x1

1

1

y  y1 y

  x2 y2 y    tan   x 

x

1

Chapter One

z z

23


Consider a vector in rectangular coordinates;

E  E x âx  E yây  E zâz Wishing to write E in cylindrical coordinates:

E  E  â  E â  E zâz From the dot product:

E   E â

E   E â

E z  E âz

E   E x âx  E yây  E zâz  â  E x âx â  E y ây â  E zâz â         

?

?

?

E   E x âx  E yây  E zâz  â  E x âx â  E y ây â  E zâz â         

?

?

?

E z  E x âx  E yây  E zâz  âz  E x âx âz  E y ây âz  E zâz âz         

?

?

Chapter One

?

24


y

900   â



â y

â

 â x



y



z

x

x â z

Clearly:

âx â  âx â cos    cos  

ây â  ây â cos 90o     sin   âx â  âx â cos 90o      sin  

ây â  ây â cos    cos   âz â  âz â cos 90o   0

âz â  âz â cos 90o   0 So:

E   E x cos    E y sin  

Or in matrix form

E    cos   sin    E x      E    sin   cos   E y 

Chapter One

E   E x sin    E y cos  

25


And, the inverse relation is:

E x   cos    sin   E       E y   sin   cos    E  

Chapter One

Note that the story is not finished here, after transforming the components; you should also transform the coordinate variables.

26


THE SPHERICAL COORDINATE SYSTEM

z

r ,  ,  are

coordinate variables.

P x  1, y  2 , z  3or



P r  14 ,   35.69o ,   63.4 o

z 3 

 is

measured from x-axis toward y-axis, and  is measured from the zaxis toward the xy plane.



r 

x 1

y 2 

y

x

Chapter One

A genral point is loacated by its coordinate variables r ,  ,  , or as the intersection of three mutually perpendicular surfaces.

27


z   1

P 1 , 1 , z 1 

surface (cone)

r  r1

â r

surface (sphere)

â

â

  1 Surface (plane)

y

x

increasing variables and are perpendicular  to the surface at which the coordinate variable is constant.

Chapter One

1) Sphere of radius r  r1 , centered at the origin. 2) Semi-infinite plane of constant angle   1 with it’s axis aligned with z-axis. 3) Right angular cone with its apex centered at the origin, and it axis aligned with z-axis, and a cone angle   1 . The three unit vectors âr , â , and â are in the direction of

28


z

P r1 , 1 , 1  â r â

1

1

x

â

y

Chapter One

r1

29


y

z

â

â r

z

r

â





z

x

x

xy plane



â z



y

Note that in spherical coordinates, unit vectors are functions of coordinate variables. â , â and âr are functions of  and  . y

z

aˆ  2

â

aˆ r1

2

r

1

r 

aˆ  1

xy plane

â z



y



z

x

x

Chapter One

aˆ r 2

30


y â  2

â 2

â1

2

2

z

â 1

â y

â x

1

2

1

y

â y

x

2

z

â x

1

1

x

The sphyrical coordinate system is Right Handed:

aˆ r  aˆ   aˆ  .

Chapter One

Increasing each coordinate variable by a differential amount dr , d , and d , one obtains:

31


Note that r is length, but and  are angles which requires a metric coefficient to convert them to lengths.

arc length 

d

r metric coefficient

arc length  rsin  

d

metric coefficient

Differential volume: dv  r 2 sin  dr d d Differential Surfaces: Six surfaces with differential areas shown in the figure. (Try itttttttttttt!)

Transformations between Spherical and Cylindrical Coordinates z

From spherical to cart: x  r sin cos 

z  z1

y  r sin sin 

1

z  r cos 

From cart. To spherical: r  x2  y2  z 2 ; ( √

 y x

;

x  x1

1

r1

1

y  y1 y

) x

Chapter One

  tan 1  

32


Consider a vector in rectangular coordinates;

E  Ex aˆ x  E yaˆ y  Ez aˆ z Wishing to write E in spherical coordinates:

E  Er aˆ r  E aˆ   E aˆ  From the dot product:

Er  E  aˆ r

E  E  aˆ 

E  E  aˆ 

Er  Ex aˆ x  E y aˆ y  Ez aˆ z   aˆ r

 E x aˆ x  aˆ r  E y aˆ y  aˆ r  E z aˆ z  aˆ r     ? ? ? E  Ex aˆ x  E yaˆ y  Ezaˆ z  aˆ 

 Ex aˆ x  aˆ   E y aˆ y  aˆ   Ez aˆ z  aˆ     ? ? ? E  Ex aˆ x  E y aˆ y  Ez aˆ z   aˆ 

Chapter One

 E x aˆ x  aˆ   E y aˆ y  aˆ   E z aˆ z  aˆ       ? ? ?

33




z

z

â z

r



aˆ r

90 0   aˆ 

 aˆ 



xy plane

From figure

aˆ z  aˆ r  aˆ z aˆ r cos   cos 









aˆ   aˆ r  aˆ  aˆ r cos 90o    sin 

aˆ z  aˆ   aˆ z aˆ  cos 90o     sin  And the rest is left to you as an exercise! So:

Er  Ex sin  cos   E y sin  sin   Ez cos 

Chapter One

Note that, after transforming the components; you should also transform the coordinate variables.

34

Chapter One


35

Engineering Electromagnetics; William Hayt & John Buck

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