• No results found

Fundamental Electromagnetics [ Chapter 2: Vector Algebra ]

N/A
N/A
Protected

Academic year: 2021

Share "Fundamental Electromagnetics [ Chapter 2: Vector Algebra ]"

Copied!
46
0
0

Loading.... (view fulltext now)

Full text

(1)

F undamental E lectromagnetics

[ Chapter 2: Vector Algebra ]

Prof. Kwang-Chun Ho

[email protected]

Tel: 02-760-4253 Fax:02-6919-2160

(2)

Basic concept of scalars and vectors What is unit vector?

Vector addition and subtraction Position and distance vectors Vector multiplication

Dot product Cross product

Components of a vector

Key Points

(3)

Scalar quantities:

Time, Mass, Distance, Temperature, etc Vector quantities:

Velocity, Force, Acceleration, etc Field:

A function that specifies a particular quantity everywhere in a region

Scalars and Vectors

Scalars Vectors

Definition Magnitude only Magnitude & Direction

(4)

Scalar field:

 Is a function that specifies a scalar quantity everywhere in a region

 Height of a mountain

 Sound intensity in a theater

 Is just one where a quantity in “space” is represented by numbers, such as this temperature map

Scalars and Vectors

Can express

as a function T x y ( , )

(5)

Vector field:

 Is a function that specifies a vector quantity everywhere in a region

 Gravitational force on a body in a space

 Wind map in the atmosphere

Scalars and Vectors

(6)

Vectors show magnitude and displacement, drawn as a ray :

Vector Notation

A

(7)

Vector :

Magnitude of : or

Direction of : unit vector whose magnitude is unity.

That is,

In Cartesian(or rectangular) coordinates, it represents as or

where are components of in x, y and z directions, respectively.

a

A

A

,

A

a A A A a

A

 

  

 

A A A

x

,

y

,

z

A A

A

x x y y z z

A a   A a   A a  , , and

x y z

A A A

  

Or can use Bold-Faced Type like

AA A

A

Unit vector

(Graphical Representation)

A

a

A

a

A

A   Aa

A

(8)

The magnitude and unit vector of vector are then

2 2 2

x y z

AAAA

2 2 2

x x y y z z

A

x y z

A a A a A a a

A A A

 

  

  

A

Unit vector

a

x

a

y

a

z

A

z z

A a

x x

A a

y y

A a

(9)

Example 2.1:

Describe the vectors and , shown in the following figures

P

R

  

Unit vector

 

x

y z

P

x

y z

R  3 a

x

4 a

y

(10)

Example 2.2:

Describe the vectors and , shown in the following figures

R

2

R

1

1

2

x y

,

x y z

R aa ba

R aa ba ca

 

  

  

   

Unit vector

R

1

R

2

(11)

Two vectors and add together to obtain a new vector :

if , then

Vector subtraction is similarly carried out as C    AB

x

,

y

,

z

and

x

,

y

,

z

A   A A A B   B B B

x x

x

y y

y

z z

z

C   AB a   AB a   AB a

x x

x

()

y y

y

z z

z

D A B A B

A B a A B a A B a

    

     

 

  

  

A

BC

Vector addition and subtraction

(12)

Vector addition and subtraction

Vectors may be added graphically,

“head to tail.”

(13)

Vector addition and subtraction

Special case of vector addition

Add the negative of the subtracted vector

Continue with standard vector addition procedure

 

A B       AB

(14)

Example 2.3:

If and , find

 (a) the component of along ,

 (b) the magnitude of ,

 (c) a unit vector along

Solution:

 (a)

 (b) Since , we have

 (c) Letting , a unit vector along is

10

x

4

y

6

z

A   a   a   a

2

x y

B   a   a3A B   

y

4 A  

3 A B     3(10, 4, 6) (2,1, 0)    (28, 13,18) 

2 2 2

3 A B     28   ( 13)  18  1277

2 (10, 4, 6) (4, 2, 0) (14, 2, 6) C    AB      

2 2 2

(14, 2, 6) 14 ( 2) 6

A

a C

C

  

  

 

A

a

y

2 A   B

C

Vector addition and subtraction

(15)

Example 2.4:

Given vector and , determine

 (a) ,

 (b) ,

 (c ) The component of along ,

 (d) A unit vector parallel to

Solution:

 (a)

 (b)

 (c)

 (d)

A unit vector parallel to this vector is

x

3

z

A   a   a

5

x

2

y

6

z

B   a   a   aA   B

5A B   

A

a

y

3A   B

Vector addition and subtraction

(16)

MatLab Lesson:

MatLab scripts to calculate a vector magnitude

MatLab scripts to calculate the unit vector

Find the unit vector of vector using MatLab

function y=magvector(R)

% Calculates the magnitude of a Cartesian vector R y=sqrt(R(1)^2+R(2)^2+R(3)^2);

function y=unitvector(R)

% Calculates the unit vector of a Cartesian vector R y=R/magvector(R);

>> A=[10 -4 6];

>> unitvector(A)

10

x

4

y

6

z

A   a   a   a

Vector addition and subtraction

(17)

Show the graphical representation of addition and subtraction of two vectors in Example 1.3

Run vectoralg.m!

Visual EMT using MatLab

(18)

A point P may be described by (x, y, z)

Then, the position vector of point P is the directed distance from the origin to P, that is,

As example, a position vector is shown as

p x y z

r   OPxa   ya   za

3 a

x

 4 a

y

 5 a

z

r

p

Position and distance vectors

, ,

p x y z

r xa ya za

x y z

  

   

a

x

a

y

a

z rp

(19)

Distance vector: displacement from one point to another

if two points P and Q are given as and , the distance vector is

x

P

, y

P

, z

P

x

Q

, y

Q

, z

Q

     

PQ Q P Q P x Q P y Q P z

r     rrxx a   yy a   zz a

Position and distance vectors

  

   

r

P

r

Q

r

PQ

QP P Q

r    rrr

P

r

Q

P

(20)

Example 2.5:

Given points and , find:

(a) the position vectors of P and R,

 (b) the distance vector ,

(c) the distance between P and Q

Solution:

 (a)

 (b)

 (c)

1, 3, 5 ,   2, 4, 6

PQ R0, 3,8

r

QR

Position and distance vectors

(21)

Example 2.6:

Woman walks with a velocity of 1.0 m/s along the

aisle of a train that is moving with a velocity of 3.0 m/s.

What is the woman’s velocity?

Solution:

 For passenger sitting in a train: 1.0 m/s

 For bicyclist standing:

Relative velocity in 1-D

(22)

Cyclist: frame of reference A

Moving train: frame of reference B

In 1-D motion, position of P relative to frame of reference A is given by distance x

P/A

Position of P relative to frame of reference B is given by distance x

P/B

Distance from origin A to origin B is given by x

B/A

 Thus,

Velocity v

P/A

of P relative to frame A is the derivative of x

P/A

with respect to time

Relative velocity in 1-D

/ / /

P A B A P B

xxx

/ / /

1.0 / 3.0 / 4.0 /

P A P B B A

v v v

m s m s m s

 

 

dt

dx dt

dx dt

dx

P/A P/B B/A

(23)

Example 2.7:

A river flows SE at 10 km/hr, and a boat flows upon it.

A man walks upon the deck at 2 km/hr to the perpendicular direction.

Find the velocity of the man with respect to

the earth.

Solution:

 Since the velocity of boat is

 

10 cos 45 sin 45 7.071( ) km/hr

b x y

u a a

a a

 

 

  

 

Relative velocity in 2-D

u

m

u

ab

u

b

(24)

, and the velocity of the man with respect to the boat (relative velocity) is

the absolute velocity of the man is

that is, 10.2 km/hr at 56.3

o

south of east

56.3

5.657 8.485 km/hr 10.2

ab m b x y

j

u u u a a

e

   

    

 

2 cos 45 sin 45 1.414( ) km/hr

m x y

x y

u a a

a a

  

  

  

 

Relative velocity in 2-D

(25)

Example 2.8:

The velocity of the boat relative to the water is 4.0 m/s, directed

perpendicular to the current

The river is 1.8 km wide and the velocity of the water relative to the shore is 2.0 m/s

How far upstream is the boat when it reaches the opposite shore?

Solution:

Relative velocity in 2-D

(26)

There are two types of vector multiplication:

Scalar (or dot) product:

Vector (or cross) product:

Dot product:

Here is the smaller angle between two vectors, and the result is scalar.

If and , then A B  

A B    cos

AB

A B    A B   

AB

x

,

y

,

z

A   A A A

x x y y z z

A B    A BA BA B

0 (Orthogonal) 1

x y y z z x

x x y y z z

a a a a a a a a a a a a

  

  

     

  

     

  

x

,

y

,

z

B   B B B

using

% Matlab script

>> dot(A,B)

Vector multiplication

(27)

Example 2.9:

Work of a force acting on a body when the body is moved by a small distance

Work done by the force is

x

 

cos

F    xF   x

( Product between the components of the same direction )

(Graphical Representation)

AB

B

Aa

a B

 

(Projection= )

B cos 

AB

Vector multiplication

(28)

Example 2.10:

Find the work done against gravity to move a 10 kg baby from the point (2,3) to the point (5,7) ?

Solution:

 We have that the force vector is

 And the displacement vector is

 The work is the dot product

 Notice the negative sign verifies that the work is done against gravity

 Hence, it takes 392 J of work to move the baby

 

10 9.8

y

F   mg    a

5 2

x

7 3

y

3

x

4

y

x a a a a

           

98

y

  3

x

4

y

392

F     xaa   a   

 

Vector multiplication

x y

(2, 3)

(5, 7)

(29)

Example 2.11:

If and , find Solution:

 Using the dot product,

 Thus,

AB

x

3

z

A   a   a

5

x

2

y

6

z

B   a   a   a

Vector multiplication

% Matlab script A = [1 0 3];

B = [5 2 -6];

Num = dot(A,B);

Den=sqrt(sum(A.^2))*sqrt(sum(B.^2));

Theta_AB = (180/pi)*acos(Num/Den)

(30)

There are instances where the product of two vectors is another vector

Torque:

The torque (turning force) vector lies in a direction perpendicular to the plane formed by the

position vector and the force vector

The torque is the vector (or cross) product of the position vector and the force vector

Vector multiplication

(31)

Cross Product:

where is a unit vector normal to the area of parallelogram, and the result is vector

a

n

A B     AB sin 

AB

a

n

% Matlab script

>> cross(A,B)

Vector multiplication

(Right‐hand rule) (Right‐handed screw rule)

sin

AB

A B     AB

= area of parallelogram

BA B 

A B 

B

a

n

a

n

(32)

If and , then

Basic properties:

   

x y z

x y z

x y z

y z z y x z x x z y x y y x z

a a a A B A A A

B B B

A B A B a A B A B a A B A B a

 

     

  

 

  

x

,

y

,

z

A   A A A

x

,

y

,

z

B   B B B

(More easily remembered form)

   

 

A B B A

A B C A B C

A B C A B A C

  

    

     

   

     

      

, , ,

x y z y x

y z x z y

z x y x z

a a a a a

a a a a a

a a a a a

     

     

      

    

    

    

(Commutative Law)

(Associative Law)

(Distributive Law)

Vector multiplication

a

x

a

y

a

z

a

z

a

y

a

x

(33)

Example 2.12:

Let and ,

 (a) Find the vector component of along

 (b) Determine a unit vector perpendicular to both and

Solution:

 (a) Since , using , we have

 (b) Using

3

y

4

z

A   a   a

4

x

10

y

5

z

B   a   a   a

 

B B B B B

A   A a   A a   a

B

B

aB

 

    

2

10 4, 10, 5

0.2837 0.7092 0.3546

B

141

x y z

A B B

A a a a

B

 

     

  

    

 

0 3 4 55,16, 12 ,

4 10 5

x y z

a a a

A B    

  

 

 

B

cos

AB B

A a    A   A

Vector multiplication

A

B

A

B

(34)

Example 2.13:

Show that vector , and form the sides of a triangle.

 Is this a right angle?

 Calculate the area of the triangle

Solution:

 Since , it is a right angle triangle

 Area =

4, 0, 1 ,   1, 3, 4

a    b   c    5, 3, 3

Rectangle Area

% Matlab script a = [4 0 -1];

b = [1 3 4];

c = cross(a,b);

Area = 1/2*sqrt(sum(c.^2));

Vector multiplication

(35)

Plot the trajectory of a particle moving in Cartesian coordinate in terms of vector notations

Run rectCoord.m!

Visual EMT using MatLab

(36)

Let ’s and

(a) Plot the Dot-product (b) Plot the Cross-product

2

x

A   a

A B  

A B   

Run vecalgebra.m!

 

2 2

sin

x x y y

B   xy a   e

a

Visual EMT using MatLab

(37)

Scalar triple product:

Given three vectors , and , the scalar triple product is

if and , then ,

A B  

     

AB C     BC    ACA B   

  

x

,

y

,

z

,

x

,

y

,

z

A   A A A B   B B B

 

xx yy zz

x y z

A A A

A B C B B B

C C C

 

  

C

x

,

y

,

z

C   C C C

Vector multiplication

(38)

Vector triple product:

Given three vectors , and , the vector triple product is

     

A   B C     B A C     C A B   

 

, A B  

C

Vector multiplication

Height = A a  

n

Area = B C    a

n

A

C

B

(39)

A direct application of vector product is to determine the projection (or component) of a vector in a desired direction

Given a vector , the scalar component along vector is

The vector component of along vector is

cos cos

B AB B AB B

AA   A a     A a  

A

B

 

B B B B B

A   A a   A a   a

Components of a vector

A

B

A

A

A   A

B

(40)

Division of vectors does not consider because it is undefined

Coordinate components of a vector:

Components of a vector

A

y

: Component

along y-direction

a

y

a

y

a

x

a

x

A

x

: Component along x-direction

(41)

There are two methods of vector addition

Graphical : represent vectors as scaled-directed line segments; attach tail to head

Analytical : resolve vectors into x and y components;

add components

Components of a vector

,

x x x

,

y y y

R A B

R A B

R A B

 

 

 

  

cos , sin

cos , sin

x A y A

A A A A

B B B B

 

 

 

 

 

 

(42)

Example 2.14:

Derive the cosine formula Solution:

 From the triangle of figure, we know that

2 2 2

2 cos abcbc A

0, that is,

a b c

b c a

  

  

  

  

Thus, it becomes

   

2

2 2

2 2 cos

a a a b c b c b b c c b c b c bc A

   

  

  

 

   

 

     

  

Application of vectors

a

b

a

b

c

(43)

Example 2.15:

Derive the law of sines for a triangle using vectors Solution:

 We have

 Because , we can write or

 Thus,

 Similarly, we have

BC

180

 

B    CA  0 B B    

  0

B   C   A   B C      BA

 

sin sin

BC   BA    sin sin

A C

sin sin sin

A B C

Application of vectors

(44)

Problem 2.1:

Find the unit vector along the line joining point (2, 4, 4) to point (-3, 2, 2).

Problem 2.2:

Let , and

 (a) Determine

 (b) Calculate

(c) For what values of k is ?

 (d) Find

Problem 2.3:

Show that

Homework Assignments

2

x

5

y

3 ,

z

3

x

4

y

A   a   a   a B    a   a

x y z

C   a   a   a  2

A   B  5 A   C

2 kB  

A B    / A B

  A B

2

  A B

2

  AB

2

(45)

Problem 2.4:

Points , and form a triangle in space. Find the three angles of the triangle.

Problem 2.5:

and are vector fields given by and . Determine

 (a) at (1, 2, 3)

 (b) The component of along at (1, 2, 3)

 (c) A vector perpendicular to both and at (0, 1, -3) whose magnitude is unity.

Homework Assignments

   

1

1, 2, 3 ,

2

5, 2, 0

P PP

3

 2, 7, 3  

E

F

2

x y z

E   xa   a   yza

2

x y z

F   xya   y a   xyzaE

E

FE

F

(46)

Problem 2.6:

Find the vector using the component of vectors

Homework Assignments

R

References

Related documents