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Fundamentals of Applied Electromagnetics
Chapter 2 - Vector Analysis
Chapter Objectives
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Operations of vector algebra
Dot product of two vectors
Differential functions in vector calculus
Divergence of a vector field
Divergence theorem
The curl of a vector field
Stokes’s theorem
Chapter Outline
2-1)
2-2)
2-3)
2-4)
Basic Laws of Vector Algebra
Orthogonal Coordinate Systems
Transformations between Coordinate Systems
Gradient of a Scalar Field
2-5) Divergence of a Vector Field
2-6) Curl of a Vector Field
2-7) Laplacian Operator
2-1 Basic Laws of Vector Algebra
•
Vector A has magnitude A = |A| to the direction
of propagation.
•
Vector A shown may be represented as
A  xˆAx  yˆAy  zˆAz
2-1.1 Equality of Two Vectors
•
A and B are equal when they have equal
magnitudes and identical unit vectors.
2-1.2 Vector Addition and Subtraction
•
For addition and subtraction of A and B,

D  A  B  xˆ  A  B   yˆ A

 B   zˆ A
C  A  B  xˆ  Ax  Bx   yˆ Ay  B y  zˆ Az  Bz 
x
x
y
y
z
 Bz 
2-1.3 Position and Distance Vectors
•
Position vector is the vector from the origin to
point.
R12  P1P2  R2  R1
2-1.4 Vector Multiplication
• 3 different types of product in vector calculus:
1. Simple Product with a scalar
2. Scalar or Dot Product
A  B  AB cos AB
where θAB = angle between A and B
+ve
-ve
2-1.4 Vector Multiplication
3. Vector or Cross Product
A  B  n̂AB sin  AB
•
Cartesian coordinate system relations:
xˆ  xˆ  yˆ  yˆ  zˆ  zˆ  0
•
In summary,
xˆ
A  B  Ax
yˆ
Ay
zˆ
Az
Bx
By
Bz
2-1.5 Scalar and Vector Triple Products
•
A scalar triple product is
A  B  C  B  C  A  C  A  B
•
A vector triple product is
A  B  C  BA  C  CA  B
known as the “bac-cab” rule.
Example 2.2 Vector Triple Product
Given A  x̂  ŷ  ẑ2 , B  ŷ  ẑ , and C  x̂2  ẑ3 , find
(A× B)× C and compare it with A× (B× C).
Solution
x̂ ŷ ẑ
A  B  1  1 2   x̂3  ŷ  ẑ
0 1 1
x̂
ŷ ẑ
A  B C   3  1 1  x̂3  ŷ7  ẑ2
2 0 3
A similar procedure gives A  B C  x̂2  ŷ4  ẑ
2-2 Orthogonal Coordinate Systems
•
Orthogonal coordinate system has coordinates
that are mutually perpendicular.
2-2.1 Cartesian Coordinates
•
Differential length in Cartesian coordinates is a
vector defined as
dl  xˆdx  yˆ dy  zˆdz
2-2.2 Cylindrical Coordinates
•
Base unit vectors obey right-hand cyclic relations.
rˆ  ˆ  zˆ,
•
ˆ  zˆ  rˆ,
zˆ  rˆ  ˆ
Differential areas and
volume in cylindrical coordinates
are shown.
Example 2.4 Cylindrical Area
Find the area of a cylindrical surface described by
r = 5, 30° ≤ Ф ≤ 60°, and 0 ≤ z ≤ 3
Solution
For a surface element with constant r,
the surface area is
60
5
S  r  d  dz  5  / 6 z 0 
2
 30 z 0
3
 /3
3
2-2.3 Spherical Coordinates
•
Base unit vectors obey right-hand cyclic relations.
Rˆ  ˆ  ˆ,
ˆ  ˆ  Rˆ ,
ˆ  Rˆ  ˆ
where R = range coordinate sphere radius
Θ = measured from the positive z-axis
Example 2.6 Charge in a Sphere
A sphere of radius 2 cm contains a volume charge
density ρv given by
C/m 
 v  4 cos 2 
3
Find the total charge Q contained in the sphere.
Solution
Q    v dv 
v
2 
R
 4   
3
0 0
3
2
 2102
   4 cos  R
2
2
sin dRd d
 0  0 R 0
2102


0
2
sin  cos 2 dd

32
cos  
6 
 d  44.68 C 
 10   
3
3 0
0
3
2-3 Transformations between Coordinate Systems
Cartesian to Cylindrical Transformations
• Relationships between (x, y, z) and (r, φ, z) are
shown.
• Relevant vectors are defined as
rˆ  xˆ cos   yˆ sin 
ˆ   xˆ sin   yˆ cos 
xˆ  rˆ cos   ˆ sin  ,
yˆ  rˆ sin   ˆ cos 
2-3 Transformations between Coordinate Systems
Cartesian to Spherical Transformations
• Relationships between (x, y, z) and (r, θ, Φ) are
shown.
• Relevant vectors are defined as
Rˆ  xˆ sin  cos   yˆ sin  sin   zˆ cos 
ˆ  xˆ cos  cos   yˆ cos  sin   zˆ sin 
ˆ   xˆ sin   yˆ cos 
xˆ  Rˆ sin  cos   ˆ cos  cos   ˆ sin  ,
yˆ  Rˆ sin  sin   ˆ cos  sin   ˆ cos  ,
zˆ  Rˆ cos   ˆ sin 
Example 2.8 Cartesian to Spherical Transformation
Express vector A  x̂x  y   ŷ y  z   ẑz in spherical
coordinates.
Solution
Using the transformation relation,
AR  Ax sin  cos   Ay sin  sin   Az cos 
 x  y sin  cos    y  x sin  sin   z cos 
Using the expressions for x, y, and z,
AR  R sin 2  cos 2   sin 2    R cos 2 
 R sin 2   R cos 2   R
Solution 2.8 Cartesian to Spherical Transformation
Similarly,
A  x  y  cos  cos    y  x  cos  sin   z sin 
A  x  y sin    y  x  cos 
Following the procedure, we have
A  0
A   R sin 
Hence,
A  R̂AR  θ̂A  φ̂A  R̂R  φ̂R sin 
2-3 Transformations between Coordinate Systems
Distance between Two Points
• Distance d between 2 points is

d  R12  x2  x1    y2  y1   z2  z1 
•
2
2
2

1
2
Converting to cylindrical equivalents.

 r
d  r2 cos 2  r1 cos 1   r2 sin 2  r1 sin 1   z2  z1 
2
2
•
2
2
 r  2r1r2 2  1    z2  z1 
2
2
1

2

1
2
1
2
Converting to spherical equivalents.
d  R  R  2R1R2 cos  2 cos 1  sin 1 sin  2 cos2  1 
2
2
2
1
1
2
2-4 Gradient of a Scalar Field
•
•
Differential distance vector dl is dl  xˆdx  yˆdy  zˆdz .
Vector that change position dl is gradient of T, or
grad.
T
T
T
T  grad T  xˆ
 yˆ
 zˆ
x
y
z

•
The symbol ∇ is called the del or gradient
operator.



  xˆ  yˆ  zˆ
(Cartesian )
x
y
z

2-4 Gradient of a Scalar Field
•
•
Gradient operator needs dl  aˆl dl to be scalar
quantity.
dT
Directional derivative of T is given by dl  T  aˆl
2-4.1 Gradient Operator in Cylindrical and Spherical Coordinates
•
Gradient operator in cylindrical and spherical
coordinates is defined as
  rˆ
 ˆ1 


 zˆ
(cylindric al)
r
r 
z

1  ˆ 1

ˆ
ˆ
R


(spherical )
R
R 
R sin  
Example 2.9 Directional Derivative
2
2
T

x

y
z along the
Find the directional derivative of
direction xˆ 2  yˆ 3  zˆ 2 and evaluate it at (1,−1, 2).
Solution
Gradient of T :
 


T   xˆ  yˆ  zˆ x 2  y 2 z   xˆ 2 x  yˆ 2 yz  zˆy 2
y
z 
 x
We denote l as the given direction,
Unit vector is
and
dT
dl
aˆl 
I  xˆ 2  yˆ 3  zˆ 2
I
xˆ 2  yˆ 3  zˆ 2 xˆ 2  yˆ 3  zˆ 2


2
2
2
I
17
2 3 2
4 x  6 yz  2 y 2
 T  aˆl 
17
1, 1, 2 

1, 1, 2 
 10
17
2-5 Divergence of a Vector Field
•
Total flux of the electric field
E due to q is
Total Flux   E  ds
S
•
Flux lines of a vector field E is
Ex E y Ez
  E  div E 


x
y
z

2-5.1 Divergence Theorem
•
The divergence theorem is defined as

   Edv   E  ds
v
•
(divergenc e theorem)
S
∇ ·E stands for the divergence of vector E.
Example 2.11 Calculating the Divergence
Determine the divergence of each of the following vector
fields and then evaluate it at the indicated point:
a  E  x̂3x 2  ŷ2 z  ẑx 2 z at 2,-2,0
b E  R̂ a3 cos / R 2  ˆa3 sin  / R 2  at a / 2,0, 
Solution
E x E y E z
a    E 


 6x  0  x2  x2  6x
x
y
z
Thus   E 2, 2, 0   16
3

E
1 
1

1
2
a
cos 

b    E  2
E sin   
R 2 ER 

R R
R sin  
R sin  
R3


Thus,   E a / 2,0,   16
2-6 Curl of a Vector Field
•
Circulation is zero for uniform field and not zero
for azimuthal field.
•
The curl of a vector field B is defined as

1 
  B  curl B  lim
n̂  B  dl

s 0 s
 C
 max

2-6.1 Vector Identities Involving the Curl
• Vector identities:
(1) ∇ × (A + B) = ∇× A+∇× B,
(2) ∇ ·(∇ × A) = 0 for any vector A,
(3) ∇ × (∇V ) = 0 for any scalar function V.
2-6.2 Stokes’s Theorem
•
Stokes’s theorem converts surface into line
integral.
   B ds   B  dl
C
S
(Stoke' s theorem)
Example 2.12 Verification of Stokes’s Theorem
A vector field is given by B  ẑ cos  / r . Verify Stokes’s
theorem for a segment of a cylindrical surface defined
by r = 2, π/3 ≤ φ ≤ π/2, and 0 ≤ z ≤ 3, as shown.
Solution
Stokes’s theorem states that
   B ds   B  dl
C
S
Left-hand side:
Express in cylindrical coordinates
 1 Bz B   Br Bz 
1 
Br
  φ̂



  B  rˆ



ẑ
rB



r



z

z

r
r  z



 
sin 
cos
 rˆ 2  φ̂ 2
r
r



Solution 2.12 Verification of Stokes’s Theorem
The integral of ∇ × B over the specified surface S with r
= 2 is
 2
cos 
 sin 
S   B ds  z0   3   r̂ r 2  φ̂ r 2   r̂rddz
3
3 2
sin 
3
3
  
ddz    
r
2r
4
0 3
Right-hand side:
Definition of field B on segments ab, bc, cd, and da is
b
 B  dl   B
C
a
ab
c
d
a
b
c
d
 dl   Bbc  dl   Bcd  dl   Bda  dl
Solution 2.12 Verification of Stokes’s Theorem
At different segments,
Bab  Bcd  zˆcos   / 2  0 where dl  φ̂rdφ  0
Bbc  ẑcos  2 / 2 where    2
Bda  ẑcos  / 3 / 2  ẑ 4 where dl  ẑdz
Thus,
1
3
 1
CB  dl  d  ẑ 4   ẑdz  3 4 dz   4
a
0
which is the same as the left hand side (proved!)
2-7 Laplacian Operator
•
Laplacian of V is denoted by ∇2V.
2
2
2

V

V

V
2
 V    V   2  2  2
x
y
z

•
For vector E given in Cartesian coordinates as
E  xˆE x  yˆE y  zˆE z
the Laplacian of E is defined as
 2
2
2 
 E   2  2  2 E  xˆ 2 E x  yˆ 2 E y  zˆ 2 E z
y
z 
 x
2
2-7 Laplacian Operator
•
•
In Cartesian coordinates, the Laplacian of a vector
is a vector whose components are equal to the
Laplacians of the vector components.
Through direct substitution, we can simplify it as
 2 E      E       E 