Section 1 1 TRƯỜNG ĐIỆN TỪ

No Slide Title Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jor[.]

to supplement “Elements of Engineering

Electromagnetics, Sixth Edition”

by

Nannapaneni Narayana Rao

Edward C Jordan Professor of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign, Urbana, Illinois, USA

Distinguished Amrita Professor of Engineering Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India

1.1 Vector Algebra

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

Magnitude and direction Magnitude only Ex: Velocity, Force Ex: Mass, Charge

(2) Unit Vectors have magnitude unity denoted by

symbol a with subscript

Useful for expressing vectors in terms of their components

aA A

A 

A1a1  A2a2  A3a3

A12  A22  A32

(3) Dot Product is a scalar

A

A • B = AB cos 

B

Useful for finding angle between two vectors

cos   A • B

AB A A1a1  A2a2  A3a3

B B1a1  B2a2  B3a3

 A1B1  A2B2  A3B3

A12  A22  A32 B12  B22  B32

A B



(4) Cross Product is a vector

A

A B = AB sin 

B

is perpendicular to both A and B.

Useful for finding unit vector perpendicular to two vectors



an  A B

A B

A B

right hand screw A

B

an

where

(5) Triple Cross Product

in general.

A B 

a1 a2 a3

A1 A2 A3

B1 B2 B3

A (B C) is a vector

A (B C) B (C A) C (A B)

(6) Scalar Triple Product

is a scalar

A • B C B • C A C • A B



A1 A2 A3

B1 B2 B3

C1 C2 C3

Volume of the parallelepiped

an C B

A

   

 

Area of base Height

n









A × B C a

A × B

A × B C

A × B

C A × B

A B ×C

D1.2 A = 3a1 + 2a2 + a3

B = a1 + a2 – a3

C = a1 + 2a2 + 3a3

(a) A + B – 4C

= (3 + 1 – 4)a1 + (2 + 1 – 8)a2

+ (1 – 1 – 12)a3

= – 5a2 – 12a3

A  B – 4C  25  144 13

(b) A + 2B – C

= (3 + 2 – 1)a1 + (2 + 2 – 2)a2

+ (1 – 2 + 3)a3

= 4a1 + 2a2 – 4a3

Unit Vector

=

=

4a1  2a2 – 4a3 4a1  2a2 – 4a3

1

3(2a1  a2 – 2a3)

(c) A • C = 3 1 + 2 2 + 1 3

= 10

(d)

=

= 5a1 – 4a2 + a3



B C 

a1 a2 a3

1 1 –1

(3  2)a1  (–1 – 3)a2  (2 –1)a3

(e)

= 15 – 8 + 1 = 8 Same as

A • (B C) = (3a1 + 2a2 + a3) • (5a1 – 4a2 + a3)

= 3 5 + 2 (–4) + 1 1

= 15 – 8 + 1

= 8

A • B C 

3 2 1

1 1 –1

1 2 3



D = B – A ( A + D = B)

E = C – B ( B + E = C)

D and E lie along a straight line.

D

E C

Common Point









What is the geometric interpretation of this result?

D× E 0

 B A × C B     0

B ×C A ×C B × B A × B 0

A × B + B ×C + C× A = 0

Another Example

Given

Find A

2 (1)

2 (2)

a × A a a

a × A a a

2 3 1 3

3

1 2

1 2 3

0

C



a

a a

1

C

C

C



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