Fundamentals of Applied Fundamentals of Applied Electromagnetics Electromagnetics Chapter 2 - Vector Analysis Chapter 2 - Vector Analysis
Fundamentals of Applied Fundamentals of Applied ElectromagneticsElectromagneticsChapter 2 - Vector AnalysisChapter 2 - Vector Analysis
Chapter ObjectivesChapter Objectives
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 OutlineChapter Outline
Basic Laws of Vector AlgebraOrthogonal Coordinate Systems
Transformations between Coordinate SystemsGradient of a Scalar FieldDivergence of a Vector FieldCurl of a Vector FieldLaplacian Operator
2-1)2-2)2-3)2-4)2-5)2-6)2-7)
2-1 Basic Laws of Vector Algebra2-1 Basic Laws of Vector Algebra
• Vector A has magnitude A = |A| to the direction of propagation.
• Vector A shown may be represented as
zyx AzAyAx ˆˆˆA
2-1.1 Equality of Two Vectors2-1.1 Equality of Two Vectors
• A and B are equal when they have equal magnitudes and identical unit vectors.
• For addition and subtraction of A and B,
zzyyxx
zzyyxx
BAzBAyBAx
BAzBAyBAx
ˆˆˆBAD
ˆˆˆBAC
2-1.2 Vector Addition and Subtraction2-1.2 Vector Addition and Subtraction
2-1.3 Position and Distance Vectors2-1.3 Position and Distance Vectors
• Position vector is the vector from the origin to point.
122112 RRPPR
2-1.4 Vector Multiplication2-1.4 Vector Multiplication
• 3 different types of product in vector calculus:
1. Simple Product with a scalar2. Scalar or Dot Product
where θAB = angle between A and B
ABAB cosBA
+ve-ve
2-1.4 Vector Multiplication2-1.4 Vector Multiplication
3. Vector or Cross Product
• Cartesian coordinate system relations:
• In summary,
ABAB sinnBA
0ˆˆˆˆˆˆ zzyyxx
zyx
zyx
BBB
AAA
zyx ˆˆˆ
BA
2-1.5 Scalar and Vector Triple 2-1.5 Scalar and Vector Triple ProductsProducts
• A scalar triple product is
• A vector triple product is
known as the “bac-cab” rule.
BACACBCBA
BACCABCBA
Example 2.2 Vector Triple ProductExample 2.2 Vector Triple Product
Given , , and , find (A× B)× C and compare it with A× (B× C).
A similar procedure gives
2zyxA zyB 3z2xC
SolutionSolution
zy3x
110
211
zyx
BA 2z7y3x
302
113
zyx
CBA
z4y2xCBA
2-2 Orthogonal Coordinate Systems2-2 Orthogonal Coordinate Systems
• Orthogonal coordinate system has coordinates that are mutually perpendicular.
• Differential length in Cartesian coordinates is a vector defined as
2-2.1 Cartesian Coordinates2-2.1 Cartesian Coordinates
dzzdyydxxd ˆˆˆl
2-2.2 Cylindrical Coordinates2-2.2 Cylindrical Coordinates
• Base unit vectors obey right-hand cyclic relations.
• Differential areas and volume in cylindrical coordinatesare shown.
ˆˆˆ ,ˆˆˆ ,ˆˆˆ rzrzzr
Find the area of a cylindrical surface described byr = 5, 30° ≤ Ф ≤ 60°, and 0 ≤ z ≤ 3
For a surface element with constant r,the surface area is
Example 2.4 Cylindrical AreaExample 2.4 Cylindrical Area
SolutionSolution
2
55
3
0
3/
6/
3
0
60
30
zdzdrSz
2-2.3 Spherical Coordinates2-2.3 Spherical Coordinates
• Base unit vectors obey right-hand cyclic relations.
where R = range coordinate sphere radius
Θ = measured from the positive z-axis
ˆˆˆ ,ˆˆˆ ,ˆˆˆ RRR
A sphere of radius 2 cm contains a volume chargedensity ρv given by
Find the total charge Q contained in the sphere.
Example 2.6 Charge in a SphereExample 2.6 Charge in a Sphere
SolutionSolution
32 C/m cos4 v
C 68.443
cos10
3
32
cossin3
4
sincos4
2
0 0
36
2
0 0
2
102
0
3
22
0 0
102
0
2
2
2
d
ddR
ddRdRdvQRv
v
2-3 Transformations between 2-3 Transformations between Coordinate SystemsCoordinate Systems
Cartesian to Cylindrical Transformations• Relationships between (x, y, z) and (r, φ, z)
are shown.• Relevant vectors are defined as
sinˆcosˆˆ yxr
cosˆsinˆˆ yx
cosˆsinˆˆ
,sinˆcosˆˆ
ry
rx
2-3 Transformations between 2-3 Transformations between Coordinate SystemsCoordinate Systems
Cartesian to Spherical Transformations• Relationships between (x, y, z) and (r, θ, Φ)
are shown.• Relevant vectors are defined as
cosˆsinsinˆcossinˆˆ zyxR
sinˆsincosˆcoscosˆˆ zyx
cosˆsinˆˆ yx
sinˆcosˆˆ
,cosˆsincosˆsinsinˆˆ
,sinˆcoscosˆcossinˆˆ
Rz
Ry
Rx
Express vector in spherical coordinates.
Using the transformation relation,
Using the expressions for x, y, and z,
Example 2.8 Cartesian to Spherical Example 2.8 Cartesian to Spherical TransformationTransformation
SolutionSolution
zzyyxA zyx
cossinsincossin
cossinsincossin
zxyyx
AAAA zyxR
RRR
RRAR
22
2222
cossin
cossincossin
Similarly,
Following the procedure, we have
Hence,
Solution 2.8 Cartesian to Spherical Solution 2.8 Cartesian to Spherical TransformationTransformation
cossin
sinsincoscoscos
xyyxA
zxyyxA
sin
0
RA
A
sinφRφθRA RRAAAR
2-3 Transformations between 2-3 Transformations between Coordinate SystemsCoordinate Systems
Distance between Two Points• Distance d between 2 points is
• Converting to cylindrical equivalents.
• Converting to spherical equivalents.
21
212
212
21212R zzyyxxd
2
1
21
2121221
21
22
212
21122
21122
2
sinsincoscos
zzrrrr
zzrrrrd
21
122112212
122 cossinsincoscos2 RRRRd
2-4 Gradient of a Scalar Field2-4 Gradient of a Scalar Field
• Differential distance vector dl is .
• Vector that change position dl is gradient of T, or grad.
• The symbol ∇ is called the del or gradient operator.
dzzdyydxxd ˆˆˆl
z
Tz
y
Ty
x
TxTT
ˆˆˆ grad
)(Cartesian ˆˆˆz
zy
yx
x
2-4 Gradient of a Scalar Field2-4 Gradient of a Scalar Field
• Gradient operator needs to be scalar quantity.
• Directional derivative of T is given by
• Gradient operator in cylindrical and spherical coordinates is defined as
dlad lˆl
laTdl
dTˆ
2-4.1 Gradient Operator in Cylindrical and Spherical 2-4.1 Gradient Operator in Cylindrical and Spherical CoordinatesCoordinates
al)(cylindric ˆ1ˆˆ
zz
rrr
)(spherical sin
1ˆ1ˆˆ
RRR
R
Find the directional derivative of along the direction and evaluate it at (1,−1, 2).
Gradient of T :
We denote l as the given direction,
Unit vector is
and
Example 2.9 Directional DerivativeExample 2.9 Directional Derivative
SolutionSolution
zyxT 22 2ˆ3ˆ2ˆ zyx
222 ˆ2ˆ2ˆˆˆˆ yzyzyxxzyxz
zy
yx
xT
2ˆ3ˆ2ˆ zyxI
17
2ˆ3ˆ2ˆ
232
2ˆ3ˆ2ˆˆ
222
zyxzyx
I
Ial
17
10
17
264ˆ
2,1,1
2
2,1,1
yyzxaT
dl
dTl
2-5 Divergence of a Vector Field2-5 Divergence of a Vector Field
• Total flux of the electric field E due to q is
• Flux lines of a vector field E is
S
dsEFlux Total
z
E
y
E
x
E zyx
E divE
2-5.1 Divergence Theorem2-5.1 Divergence Theorem
• The divergence theorem is defined as
• ∇ ·E stands for the divergence of vector E.
theorem)e(divergenc EE
Sv
dsdv
Determine the divergence of each of the following vector fields and then evaluate it at the indicated point:
Example 2.11 Calculating the Example 2.11 Calculating the DivergenceDivergence
SolutionSolution
,,aRaRab
,,-zxzxa
02/at /sinˆ/cosRE
022at z2y3xE 2323
22
16 Thus
606
0,2,2
22
E
xxxxz
E
y
E
x
EEa zyx
16 Thus,
cos2
sin
1sin
sin
11
,0,2/
3
32
2
a
R
E
R
aE
RE
RER
RREb
2-6 Curl of a Vector Field2-6 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
max0
lBn1
limB curlB
Cs
ds
2-6.1 Vector Identities Involving the 2-6.1 Vector Identities Involving the CurlCurl
• Vector identities:(1) ∇ × (A + B) = ∇× A+∇× B,(2) ∇ ·(∇ × A) = 0 for any vector A,(3) ∇ × (∇V ) = 0 for any scalar function V.
• Stokes’s theorem converts surface into line integral.
2-6.2 Stokes’s Theorem2-6.2 Stokes’s Theorem
theorem)s(Stoke' lBsB C
S
dd
A vector field is given by . 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.
Example 2.12 Verification of Stokes’s Example 2.12 Verification of Stokes’s TheoremTheorem
r/coszB
SolutionSolutionStokes’s theorem states that
Left-hand side: Express in cylindrical coordinates
22
cosφ
sinˆ
1zφ
1ˆB
rrr
BrB
zrr
B
z
B
z
BB
rr rzrz
C
S
dd lBsB
The integral of ∇ × B over the specified surface S with r = 2 is
Right-hand side:Definition of field B on segments ab, bc, cd, and da is
Solution 2.12 Verification of Stokes’s Solution 2.12 Verification of Stokes’s TheoremTheorem
3
0
2
3
3
0
2
322
4
3
2
3sin
rcos
φsin
rsB
rdzd
r
dzrdrr
dzS
a
d
da
d
c
cd
c
b
bc
b
a
abCddddd lBlBlBlBlB
At different segments,
Thus,
which is the same as the left hand side (proved!)
Solution 2.12 Verification of Stokes’s Solution 2.12 Verification of Stokes’s TheoremTheorem
0φl where02/cosˆBB rdφdzcdab
2 where2/2coszB bc
dzdda zl where4z2/3/coszB
4
3
4
1z
4
1zlB
0
3
dzdzd
a
dC
2-7 Laplacian Operator2-7 Laplacian Operator
• Laplacian of V is denoted by ∇2V.
• For vector E given in Cartesian coordinates as
the Laplacian of E is defined as
2
2
2
2
2
22
z
V
y
V
x
VVV
zyx
zyx
EzEyExzyx
EzEyEx
2222
2
2
2
2
22 ˆˆˆEE
ˆˆˆE
2-7 Laplacian Operator2-7 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 EEE2