ELECTROMAGNETICS
CC431
Dr. Ahmed Said Eltrass
Electrical Engineering Department
Alexandria University, Alexandria, Egypt
Spring 2018
Email: ahmed4@vt.edu
Office hours: Saturday12:30 p.m. to 01:30 p.m.
Wednsday 12:00 p.m. to 01:30 p.m.
4th floor, Electrical Engineering Building
Chapter 1 : Vector Analysis and Coordinate Systems
Difference between scalar and vector quantity
• A scalar is a quantity defined by magnitude only.
Examples: Distance (x), Mass (m), Voltage (V ), Charge (Q)
• A vector is a quantity defined by magnitude and direction
Examples: Electric field (E), Force (F)
Vector Analysis
ax →: Unit vector in x-direction
ay →: Unit vector in y-direction
az →: Unit vector in z-direction
• |ax| = 1 : Magnitude of the vector ax
• Similarly |az| = |ay|= 1
• Unit Vector : is a vector has unit
magnitude and directed in certain
direction
• The vector rP is the vector extending
from the origin (0,0,0) to the point
P(1,2,3):
rP = (1-0) ax + (2-0) ay+ (3-0) az
= ax + 2ay + 3az
• The vector from P(1,2,3) to Q(2,-2,1)
RPQ = (2-1) ax + (-2-2) ay+ (1-3) az
= ax − 4ay − 2az
• The vector from Q(2,-2,1) to P(1,2,3)
RQP = (1-2) ax + (2+2) ay+ (3-1) az
= -ax + 4ay + 2az = - RPQ
Then RQP = - RPQ
• Vector A is expressed in terms of its component as:
• Magnitude :
• Scalar Multiplication:
• Unit Vector : is a vector has unit magnitude and directed in certain direction
Unit vector in A direction:
Example1:
Find the unit vector extending from point P(1,2,3) to point Q(3,-3,2)?
Rules of Vector Algebra
1- Vector addition and subtraction
 
A  B  ( Ax ax  Ay a y  Az az )  ( Bx ax  By a y  Bz az )
  Ax  Bx ax  Ay  By a y   Az  Bz az
2- Dot Product
 
A  B  A B cos  Scalar

A


where  is the angle between the two vectors A and B
 
A  B  ( Ax ax  Ay a y  Az az )  ( Bx ax  By a y  Bz az )
 Ax Bx  Ay By  Az Bz  scalar
Note that : ax  ax  a y  a y  az  az  1
ax  a y  ax  az  a y  az  0
(  0)
(  90)


B
Applications of Dot Product
 
1 - If A  B  0


then A  B


2 - To get the angle between two vectors A and B
 
A B
cos 
AB
 




A B

3 - The component of A on B  A cos     Scalar
 B 
A






 
B 
4 - Projection of A on B  A   a B  Vector


B




 

Component
of
A
on
B

B 
  A   2 B

B 
Example 2:




Given : A  3a y  4a z and B  4a x  10a y  5 az
 
1 - Find A  B


2 - Find the vector component of A in the direction of B



B
3- Cross Product
   

A  B  A B sin  an  vector



an is the unit vector perpendicular to both A and B


 is the angle between the two vectors A and B
Note that : a x  a x  a y  a y  a z  a z  0
(  0)
ax  a y  az , a y  az  ax , az  ax  a y
a y  ax   az , az  a y  - ax , ax  az  - a y
ay
az


If A  Ax a x  Ay a y  Az a z and B  Bx a x  B y a y  Bz a z
ax
 
A  B  Ax
Bx
ay
Ay
By
az
Az
Bz
 Ay Bz  Az B y a x   Ax Bz  Az Bx  a y   Ax B y  Ay Bx a z
ax
Applications of Cross Product
 
1 - If A  B  0
 
then A // B


2 - To get the angle between two vectors A and B
 
A B
sin  
AB
ay
az
ax
Example 3:


Given : A  2a x  3a y  az and B  4a x  2a y  5 az
 
1 - Find A  B


2 - Using the definition of cross product, find the angle between A and B
Chapter 1 Problems:


1. Given : A  10a x  4a y  8a z and B  8a x  7a y  2 a z


i - Find a unit vector in the direction of - A  2 B


ii - Find the magnitude of 5a x  B  3 A


2. Given : F  10a x  20 x( y  1)a y and G  2 x 2 yax  4a y  2 a z
For the point p ( 2,3,-4 ), Find :


i - F and G
 
ii - a unit vector in the direction of F  G
3. Given the points A( 1,  2,  1 ), B (  2,1,3 ), and C ( 4 ,0 ,1 ), find


i - the vectors R AB and R AC


ii - the component of R AB on R AC


iii - the projection vector of R AB on R AC


iv - the angle between R AB and R AC by two different methods
Coordinates Systems
1- Cartesian/Rectangular Coordinates
• Variables: P(x, y, z)
• Unit vectors:
ax  a y  az  1
ax  a y  az

• Vectors: A  Ax a x  Ay a y  Az a z
• Ranges:
   x  ,    y  ,    z  
• Coordinate Planes (surfaces):
1- Plane (x=constant): // yz plane
y and z are variable
2- Plane (y=constant): // xz plane
x and z are variable
3- Plane (z=constant): // xy plane
x and y are variable
2- Cylindrical Coordinates
 Variables : P (  ,  , z )
 : radius of cylinder
 : angle measured from  ve x - axis
z : height from xy - plane
• Unit vectors: a   a  a z  1
a   a  a z

• Vectors: A  A a  A a  Az a z
• Ranges:
  0, 0    2 ,    z  
• Coordinate surfaces:
1 -   constant :  and z are variables
2 -   constant :  and z are variables
3 - z  constant :  and  are variables
Relations to transform a point:
1- From Cylindrical to Cartesian:
x   cos 
y   sin 
zz
2- From Cartesian to Cylindrical:
  x2  y2
  tan 1
y
x
Will be given in exams
zz
Example 4:
Transform the point C(3, 2,−7) to Cylindrical Coordinates?
Answer: C(3, 2,−7) → C(ρ = 3.61, φ = 33.7◦, z = −7)
3- Spherical Coordinates
 Variables : P (r ,  ,  )
r : radius of sphere
 : angle measured from  ve z - axis
 : angle measured from  ve x - axis
•Unit vectors: ar  a  a  1
ar  a  a

• Vectors: A  Ar ar  A a  A a
• Ranges:
r  0, 0     , 0    2
• Coordinate surfaces:
1 - r  constant :  and  are variables
2 -   constant : r and  are variables
3 -   constant : r and  are variables
Relations to transform a point:
1- From Spherical to Cartesian:
x  r sin  cos
y  r sin  sin 
z  r cos
2- From Cartesian to Spherical :
r  x2  y2  z 2

z
  cos  2
 x  y2  z2

y
  tan 1
x
1




Will be given in exams
Example 5:
Transform the point C(3, 2,−7) to Spherical Coordinates?
Vector Transformation
1- Cylindrical Vector Transformation
Cylindrical to Cartesian
Cartesian to Cylindrical
Will be given in exams
Example 6:




Transform the vector A  ya x  xa y  za z into Cylindrical Coordinates?
Recall : x   cos  , y   sin  , z  z
2- Spherical Vector Transformation
Spherical to Cartesian
Cartesian to Spherical
Will be given in exams
Example 7:
 xz 
Transform the vector A 
ax into Spherical Coordinates?
y
Recall : x  r sin  cos  , y  r sin  sin  , z  r cos 
Problems (Continue):
8. Find the distance between the two points



A   3,   100 , z  3 and B   5,   130 , z  4.5
 1
sin  
9. Given : A  2  cosar 
a  find :
r 
sin  

i - A at P(r  0.8,   30 ,   45 )

ii - a unit vector in the direction of A at P

Differential Elements
z
1- Cartesian Coordinates
dz
dl
Differential Length
dx
If we have charge distribution over length,
the differential length in general direction:



dl  dxa x  dya y  dza z
y
dy
x
Differential Area
If we have charge distribution over area(sheet), the differential area is:


Plane with constant z : ds1  dxdya z


Plane with constant y : ds 2  dxdza y


Plane with constant x : ds3  dydza x
Differential area vector is
perpendicular to the area
z
ds3
ds 2
x
ds1
y
Differential Volume
If we have charge distribution over volume, the differential volume is:
dv  dx dy dz 
Scalar
2- Cylindrical Coordinates
z
Will be given in exams
Differential Length
dz

In   direction : dl  da ( and z  constant)

In   direction : dl  da (  and z  constant)

In z  direction : dl  dza z (  and   constant)



x
In general direction : dl  da  da  dza z
y
d
d
d 
Differential Area


Plane with constant z : ds1  dda z


Plane with constant  : ds 2  ddza


Plane with constant  : ds3  ddza
Differential Volume
dv  d d dz

Scalar
ds3
ds 2
ds1
2- Spherical Coordinates
Differential Length

In r  direction : dl  drar ( and   constant)

In   direction : dl  rda ( r and   constant)

In   direction : dl  r sin da ( r and   constant)



In general direction : dl  drar  rda  r sin da
Differential Area
Will be given in exams


2
Plane with constant r : ds1  r sin ddar


Plane with constant  : ds 2  rdrda


Plane with constant  : ds3  r sin drda
ds3
ds 2
Differential Volume
dv  dr rd r sin d

Scalar
ds1
2
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