Fundamentals of Applied Electromagnetics. Chapter 2 - Vector Analysis

Transcription

1 Fundamentals of pplied Electromagnetics Chapter - Vector nalsis

2 Chapter 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

3 Chapter Outline -) -) -) -4) -5) -6) -7) Basic Laws of Vector lgebra Orthogonal Coordinate Sstems Transformations between Coordinate Sstems Gradient of a Scalar Field Divergence of a Vector Field Curl of a Vector Field Laplacian Operator

4 - Basic Laws of Vector lgebra Vector has magnitude to the direction of propagation. Vector shown ma be represented as ˆ ˆ ˆ

5 -. Equalit of Two Vectors. Equalit of Two Vectors and B are equal when the have equal magnitudes and identical unit vectors. For addition and subtraction of and B, ( ) ( ) ( ) ( ) ( ) ( ) B B B B B B ˆ ˆ ˆ B D ˆ ˆ ˆ B C -. Vector ddition and Subtraction. Vector ddition and Subtraction

6 -. Position and Distance Vectors Position vector is the vector from the origin to point. R PP R R

7 -.4 Vector Multiplication different tpes of product in vector calculus:. Simple Product with a scalar. Scalar or Dot Product B B cosθ B where θ B angle between and B ve -ve

8 -.4 Vector Multiplication. Vector or Cross Product B nˆ Bsinθ B Cartesian coordinate sstem relations: ˆ ˆ ˆ ˆ ˆ ˆ 0 In summar, ˆ ˆ ˆ B B B B

9 -.5 Scalar and Vector Triple Products scalar triple product is ( B C) B ( C ) C ( B) vector triple product is ( B C) B( C) C( B) known as the bac-cab rule.

10 Eample. Vector Triple Product Given ˆ ŷ ẑ, B ŷ ẑ, and C ˆ ẑ, find ( B) C and compare it with (B C). Solution ˆ B ˆ ŷ ẑ 0 ŷ ẑ ( B) C ˆ ŷ7 ẑ ˆ ŷ 0 ẑ similar procedure gives ( B C) ˆ ŷ4 ẑ

11 - Orthogonal Coordinate Sstems Orthogonal coordinate sstem has coordinates that are mutuall perpendicular. -.. Cartesian Coordinates Differential length in Cartesian coordinates is a vector defined as dl d ˆ d ˆ d ˆ

12 -.. Clindrical Coordinates Base unit vectors obe right-hand cclic relations. rˆ ˆ φ ˆ, ˆ φ ˆ rˆ, ˆ rˆ ˆ φ Differential areas and volume in clindrical coordinates are shown.

13 Eample.4 Clindrical rea Find the area of a clindrical surface described b r 5, 0 Ф 60, and 0 Solution For a surface element with constant r, the surface area is S 60 φ 0 dφ r 0 d 5φ π / π / 6 0 5π

14 -.. Spherical Coordinates Base unit vectors obe right-hand cclic relations. R ˆ ˆ θ ˆ, φ ˆ θ ˆ φ Rˆ, ˆ φ Rˆ ˆ θ where R range coordinate sphere radius Θ measured from the positive -ais

15 Eample.6 Charge in a Sphere sphere of radius cm contains a volume charge densit ρ v given b Find the total charge Q contained in the sphere. Solution Q v ρ v 4cos θ ρ dv v π π φ 0 θ 0 0 R 0 ( C/m ) ( 4cos θ ) R sinθdrdθdφ 4 π π 0 0 R 0 π sinθ cos cos θ π 0 0 θdθdφ dφ ( µ C)

16 - Transformations between Coordinate Sstems Cartesian to Clindrical Transformations Relationships between (,, ) and (r, φ, ) are shown. Relevant vectors are defined as r ˆ ˆ cosφ ˆ sinφ ˆ φ ˆ sinφ ˆ cosφ ˆ rˆcos φ ˆsin φ φ, ˆ rˆsin φ ˆcos φ φ

17 - Transformations between Coordinate Sstems Cartesian to Spherical Transformations Relationships between (,, ) and (r, θ, Φ) are shown. Relevant vectors are defined as R ˆ ˆ sinθ cosφ ˆ sinθ sinφ ˆ cosθ ˆ θ ˆ cosθ cosφ ˆ cosθ sinφ ˆ sinθ ˆ φ ˆ sinφ ˆ cosθ ˆ ˆ ˆ Rˆ sinθ cosφ ˆcos θ θ cosφ ˆsin φ φ, Rˆ sinθ sinφ ˆcos θ θ sinφ ˆcos φ φ, Rˆ cosθ ˆsin θ θ

18 Eample.8 Cartesian to Spherical Transformation Epress vector ˆ ŷ ẑ in spherical coordinates. Solution ( ) ( ) Using the transformation relation, R sinθ cosφ sinθ sinφ ( ) sinθ cosφ ( ) sinθ sinφ cosθ Using the epressions for,, and, ( cos φ sin φ) R Rsin θ Rsin θ R cos θ R cosθ R cos θ

19 Solution.8 Cartesian to Spherical Transformation Similarl, θ φ ( ) cosθ cosφ ( ) ( ) sinφ ( ) cosθ cosθ sinφ sinθ Following the procedure, we have θ φ 0 Rsinθ Hence, Rˆ R θˆ θ φˆ φ Rˆ R φˆ Rsinθ

20 - Transformations between Coordinate Sstems Distance between Two Points Distance d between points is d R [ ( ) ( ) ( ) Converting to clindrical equivalents. d Converting to spherical equivalents. d [( ) ( ) ( ) ] r cosφ r cosφ r sinφ r sinφ [ r r r r ( φ φ ) ( ) ] [ cosθ cosθ sinθ sinθ ( φ )] { } R R R φ R cos ]

21 -44 Gradient of a Scalar Field Differential distance vector dl is dl d ˆ d ˆ d ˆ. Vector that change position dl is gradient of T, or grad. T T grad T ˆ T ˆ T ˆ The smbol is called the del or gradient operator. ˆ ˆ ˆ (Cartesian)

22 -44 Gradient of a Scalar Field Gradient operator needs dl aˆ ldl to be scalar quantit. Directional derivative of T is given b Gradient operator in clindrical and spherical coordinates is defined as dt dl T ˆ -4. Gradient Operator in Clindrical and Spherical Coordinates rˆ r ˆ φ r ˆ φ (clindrical) a l Rˆ R ˆ θ R ˆ φ θ R sinθ φ (spherical)

23 Eample.9 Directional Derivative Find the directional derivative of T along the direction ˆ ˆ ˆ and evaluate it at (,, ). Solution Gradient of T : We denote l as the given direction, ( ) ˆ ˆ ˆ T ˆ ˆ ˆ I ˆ ˆ ˆ Unit vector is ˆ a l I I ˆ ˆ ˆ ˆ ˆ 7 ˆ and dt dl T aˆ l (,,) 7 ( ) 7,,

24 -55 Divergence of a Vector Field Total flu of the electric field E due to q is Total Flu E S ds Flu lines of a vector field E is E div E E E E

25 -5. Divergence Theorem The divergence theorem is defined as v Edv E ds S (divergence theorem) E stands for the divergence of vector E.

26 Eample. Calculating the Divergence Determine the divergence of each of the following vector fields and then evaluate it at the indicated point: ( a) E ˆ ŷ ẑ at (,-, 0) ( ) ( b a θ R ) ˆ θ ( E Rˆ cos / a sinθ / R ) at ( a /, 0, π ) Solution ( a) E Thus E E E E 6 (,,0) R R sinθ θ ( b) E ( R E ) ( E sinθ ) R R θ R sinθ E φ φ a cosθ R Thus, E 6 ( a /,0, π )

27 -66 Curl of a Vector Field Circulation is ero for uniform field and not ero for aimuthal field. The curl of a vector field B is defined as B curl B lim s 0 s nˆ C B dl ma

28 -6. Vector Identities Involving the Curl Vector identities: () ( B) B, () ( ) 0 for an vector, () ( V ) 0 for an scalar function V. -6. Stokes s s Theorem Stokes s theorem converts surface into line integral. S ( B) ds B dl (Stoke's theorem) C

29 Eample. Verification of Stokes s s Theorem vector field is given b B ẑ cosφ / r. Verif Stokes s theorem for a segment of a clindrical surface defined b r, π/ φ π/, and 0, as shown. Solution Stokes s theorem states that Left-hand side: Epress in clindrical coordinates B B φ B B rˆ φˆ r φ sinφ cosφ rˆ φˆ r r r B r S ( B) ds ẑ r ( rb ) φ C B dl Br φ

30 Solution. Verification of Stokes s s Theorem The integral of B over the specified surface S with r is S ( B) ds 0 φ π π 0 π π rˆ sinφ φˆ r sinφ dφd r Right-hand side: Definition of field B on segments ab, bc, cd, and da is cosφ r r rˆ rdφd 4 B dl C b a B ab dl c b B bc dl d c B cd dl a d B da dl

31 Solution. Verification of Stokes s s Theorem t different segments, ( cosφ) / 0 where dl φˆ 0 B B ˆ rdφ ab cd B bc ẑ ( cosπ ) / where φ π da ( cos / ) / ẑ 4 where dl ẑd B ẑ π Thus, C B dl a d ẑ 4 ẑd 0 4 d 4 which is the same as the left hand side (proved!)

32 -7 Laplacian Operator 7 Laplacian Operator Laplacian of V is denoted b V. For vector E given in Cartesian coordinates as the Laplacian of E is defined as ( ) V V V V V E E E E E E ˆ ˆ ˆ E E ˆ ˆ ˆ E

33 -77 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 simplif it as E ( E) ( E)