fundamentals of applied electromagnetics 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