Section 1 1 TRƯỜNG ĐIỆN TỪ

Slide Presentations for ECE 329,
Introduction to Electromagnetic Fields,
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


1.1-3

(1) Vectors (A)

vs.

Magnitude and direction
Ex: Velocity, Force

Scalars (A)
Magnitude only
Ex: Mass, Charge

(2) Unit Vectors have magnitude unity denoted by
symbol a with subscript
A A1a1  A2 a 2  A3a 3
aA  
A
A12  A22  A32
Useful for expressing vectors in terms of their
components.


1.1-4

(3) Dot Product is a scalar
A
A

A • B = AB cos 
B
B
Useful for finding angle between two vectors.
A • B
cos  
AB

A A1a1  A2 a 2  A3 a 3
B B1a1  B2 a 2  B3a 3

A1 B1  A2 B2  A3 B3
 2
A1  A22  A32

B12  B22  B32


1.1-5

(4) Cross Product is a vector

B

A
A  B = AB sin  an

right hand
screw
A


B

an

is perpendicular to both A and B.
Useful for finding unit vector perpendicular to
two vectors.

A B
A B
an 

AB sin  A B

1.1-6

where

a1
A B  A1
B1

a2
A2
B2

a3
A3
B3

(5) Triple Cross Product

A (B C) is a vector
A (B C) B (C A) C (A B)

in general.


1.1-7

(6) Scalar Triple Product
A • B C B • C A C • A B

A1

A2

A3

 B1

B2

B3

C1 C2

C3

is a scalar.


1.1-8

Volume of the parallelepiped

 Area of base  Height
 A × B   C a n 
 A × B  C
C A × B
A B ×C

A×B

A×B

an

C

B

A


1.1-9

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


1.1-10

(b)

A + 2B – C
= (3 + 2 – 1)a1 + (2 + 2 – 2)a2
+ (1 – 2 + 3)a3
= 4a1 + 2a2 – 4a3
Unit Vector
4a1  2a 2 – 4a 3
=
4a1  2a 2 – 4a 3
1
= (2a1  a 2 – 2a 3 )
3


1.1-11

(c) A • C = 3 1 + 2  2 + 1  3
= 10
a1 a 2
(d) B C  1 1
1 2

a3
–1
3

= (3  2)a1  (–1 – 3)a 2  (2 – 1)a 3

= 5a1 – 4a2 + a3

1.1-12

3 2
(e)

1

A • B C  1 1 –1
1 2

3

= 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


1.1-13

P1.5

D
A

E

B

C

Common
Point

D= B–A

A + D = B)
(

E= C–B

B + E = C)
(

D and E lie along a straight line.


1.1-14

 D× E 0

 B  A  ×  C  B  0
B × C  A × C  B × B  A × B 0

A × B + B ×C + C× A = 0
What is the geometric interpretation of this result?

1.1-15

Another Example
Given a1 × A  a 2  2a3

a 2 × A a1  2a3

(1)
(2)

Find A.

A =C   a 2  2a3  ×  a1  2a3 
a1 a 2 a3
C 0  1 2 C  2a1  2a 2  a3 
1 0 2


1.1-16

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

a1 C  2a1  2a2  a3   a2  2a3
C  2a3  a2   a2  2a3
C 1

 A  2a1  2a2  a3 