1 Vector nalsis 1.1 VECTR NYSIS Introduction In general, electromagnetic field problem involves three space variables, as a result of which the solutions tend to become complex. This can be overcome b the use of vector analsis. So, knowledge of vector analsis is an essential pre-requisite to the stud of theor of electromagnetic fields. 1.1.1 Scalar and Vectors quantit is said to be a scalar if it has onl magnitude. uantities such as time, mass, distance, temperature, electric potential and population are scalars. quantit is called vector if it has both magnitude and direction. Vector quantities include velocit, force, displacement and electric field intensit. To distinguish between a scalar and a vector, it is customar to represent a vector b a letter with an arrow on the top, such as and. scalar is represented b a letter e.g.,, and U. Electromagnetic theor is essentiall a stud of some particular fields. We define a field as a function that specifies a particular quantit, everwhere in a region. Some examples of scalar fields are: temperature distribution in a building, sound intensit in a theatre and electric potential in a region. The gravitational force on a bod in space and the velocit of raindrops in the atmosphere are examples of vector fields. 1.1.2 Unit Vector vector has both magnitude and direction. The magnitude of is a scalar written as or. unit vector a along is defined as a vector whose magnitude is unit and its direction is along. In general, an vector can be represented b its magnitude and its direction as follows. = a,...(1.1) where magnitude of the vector a, direction of the vector. This a is called a unit vector.
2 Electromagnetic Fields From eqn. (1.1), unit vector a = =....(1.2) Note that a = 1. vector in cartesian (rectangular) co-ordinates ma be represented as dx,, i or a + a + a, where,, are called components of in x,, directions respectivel; x x ax, a and a are unit vectors in the x, and directions respectivel. The magnitude of is given b = = + + and the unit vector along is given b a = = a = a + a + a x x + +....(1.3) 1.1.3 Vector ddition and Subtraction Two vectors and can be added together to give another vector C. The vector addition is carried out component b component. Thus if = dx,, i and =,,, d i b g d i b g C = + a + + a + + a....(1.4) x x x Vector subtraction is similarl carried out as D= = + d i b x xg x d i b g....(1.5) D= a + a + a Graphicall, vector addition and subtraction are obtained b parallelogram rule as in Fig. 1.1 and Fig. 1.2.
Vector nalsis 3 C Fig. 1.1 C = + D Fig. 1.2 D = 1.1.4 osition and Distance Vectors point in cartesian coordinates ma be represented b (x,, ). The position vector r p (radius vector) of point is defined as the directed distance from the origin to, that is rp = = xax + a + a....(1.6) For example, the position vector for the point (3, 4, 5) is 3ax + 4a + 5a which is shown in Fig. 1.3. If two points and are given b (x p, p, p ) and (x,, ), the distance vector (or separation vector) between and is the displacement from to as in Fig. 1.4. r = r r d i d i d i = x x a + a + a x..(1.7)
4 Electromagnetic Fields Z (3,4,5) Z = 5 Y x = 3 = 4 X Fig. 1.3 r r r Fig. 1.4 Example 1.1 Given vectors = 2ax + 3a + 4a and = 5ax + 2a 6a, determine: (a) + (b) The component of along a (c) unit vector parallel to 3 +.
Vector nalsis 5 (a) b g b g + = 2, 3, 4 + 5, 2, 6 = (7, 5, 2) b g + = 7 + 5 + 2 = 78 = 8.83 (b) The component of along a is 3. (c) unit vector parallel to 3 +. 3+ = 3 (2, 3, 4) + (5, 2, 6) = (6, 9, 12) + (5, 2, 6) = (11, 11, 6) et C = 3+ unit vector parallel to C is C a ac =± C =± 11, 11, 6f 11 + 11 + 6 d =± 0. 659 a + 0. 659 a + 0. 359 a Note that a c = 1 as expected. Example 1.2 oints and are located at (1, 2, 4) and ( 5, 4, 2). Determine (a) The position vector of. (b) The distance vector from to. (c) The distance between and. (d) vector parallel to with magnitude of 10. (a) r = a 2a + 4a (b) r = r r = 5, 4, 2 1, 2, 4 a f a f = ( 6, 6, 2) r = 6a + 6a 2a (c) Distance between and is the magnitude of vector r. d = = 6 + 6 + 4 r = 9.381. i
6 Electromagnetic Fields (d) et the required vector be. = a, where is the magnitude which is given as 10. Since is parallel to, it must have the same unit vector as r or r. Hence, af a r =± =± 6, 6, 2 r 9381. b = 0639. a + 0639. a 0213. a d d 639. a x 639. a 213. a i. =± 10 0639. a + 0639. a 0213. a =± + g 1.1.5 Vector Multiplication When two vectors and are multiplied, the result is either a scalar or a vector depending on how the are multiplied. There are two tpes of vector multiplication: (a) Scalar (or dot) product: (b) Vector (or cross) product:. Multiplication of three vectors, and C can result in either (c) Scalar triple product: C or (d) Vector triple product: d d i C i (a) Dot roduct The dot product of two vectors and, written as, is defined geometricall as the product of the magnitude of and the projection of onto (or vice versa). Thus = cosθ,...(1.8) where θ is the smaller angle between and. The result is called either the scalar product because it is scalar, or the dot product due to the dot sign. If =,, and =,,, d i d i then = x x + +,...(1.9) which is obtained b multipling and, component b component. i
Vector nalsis 7. = (rojection of on ) = ( cos θ ΑΒ) θ The dot product obes the following: (i) Commutative law: = (ii) Distributive law: + C = + C (iii) d i Fig. 1.5...(1.10)...(1.11) = bg 2 = 2....(1.12) lso note that a a = a a = a a = 0 x x a a = a a = a a =1...(1.13) x x (b) Cross roduct The cross product of two vectors and is defined as = sin θ a,...(1.14) n where a n is the unit vector normal to the plane containing and. The direction of a n is taken as the direction of right thumb when the fingers of the right hand rotate from and. lternativel, the direction of a n is taken as that of advance of right-handed screw as is turned into as shown in Fig. 1.6. The vector multiplication is called cross product due to the cross sign. It is also called as vector product because the result is a vector. If =,, and =,,, then d i d i ax a a = d i b g d i = a + a + a....(1.15) x x x x x
8 Electromagnetic Fields a n θ Fig. 1.6 Cross product has the following properties (i) It is not commutative:...(1.16) It is anti-commutative: =...(1.17) (ii) It is not associative: C C d i d i...(1.18) (iii) It is distributive: + C = + C d i d i d i...(1.19) a x a x a a -a -a a a a a a x -a x Fig. 1.7 Fig. 1.8 (a) Moving clockwise leads (b) Moving counterclockwise results to positive results in negative results.
Vector nalsis 9 (iv) =0...(1.20) lso note that ax a = a a a = ax...(1.21) a a = a x which are obtained in cclic permutation and illustrated in Figs. 1.7 and 1.8. (c) Scalar triple product Given three vectors, and C, we define the scalar triple product as d Ci= dc i= C d i...(1.22) obtained in cclic permutation. If = d x,, i, = dx,, i and C = dcx, C, Ci, then d Ci is the volume of a parallelopiped having, and C as edges and is easil obtained b finding the determinant of 3 3 matrix formed b, and C, that is, C = C C C d i....(1.23) Since the result is scalar, this is known as scalar triple product. (d) Vector triple product We define vector triple product for vectors, and C as C = C C = C C Note: d i d i d i d i d i...(1.24) e j C e Cj...(1.25) e j C = Ce j 1.1.6 Components of a Vector direct application of vector product is its use in determining the projection (or component) of a vector in a given direction. The projection can be scalar or vector. Given a vector, we define the scalar component of along vector as in Fig. 1.9. = cosθ = a cosθ...(1.26)
10 Electromagnetic Fields = a....(1.27) The vector component of along is the scalar component multiplied b a unit vector along, that is, = a = a a d i....(1.28) Notice from Fig. 1.9 and Fig. 1.10 that the vector can be resolved into two orthogonal components; one component parallel to, another d i perpendicular to. - θ θ Fig. 1.9 Fig. 1.10 (a) Scalar component (b) Vector component Example 1.3 Given vectors = 3ax + 4 a + a and = 2ax 3a + 5 a, find the angle between The angle θ can be found b either dot or cross product. = 341,, 2, 35, b gb g and. = 6 12+ 5 = 1. 2 2 = 3 + 4 + 1 = 26 b g = 2 + 3 + 5 = 38 1 cos θ = = = 0032. 26 38 θ = 91.834.
Vector nalsis 11 lternativel: ax a a = 3 4 1 2 3 5 a f a f a f = 20+ 3 a 15 2 a + 9 8 a = (23, 13, 17) bg b g b g. = 23 + 13 + 17 = 31 41 31. 41 sinθ = = 26 38 = 0.999 θ = 87.8. Example 1.4 Three fields are given b = 2ax a + a = 4ax a R= a a x Determine: (a) R (b) unit vector perpendicular to both and R (c) The component of along. (a) The onl wa R makes sense is ax a a dr i= a4, 0, 1f 1 1 0 2 1 1 = (4, 0, 1) ( 1, 1, 1) = 4 + 0 1 = 5. (b) unit vector perpendicular to both and R R a =± R is given b
12 Electromagnetic Fields ax a a R= 4 0 1 1 1 0 = ax a 4 a. R = 1 + 1 + 4 = 18 F HG ax a a a =± 4 18 a =± d 0. 235ax 0. 235a 0. 942ai. (c) The component of along is = cosθa d = a a =. = d i a 2 i I KJ b2, 1, 1gb 4, 0, 1g = 2 2 4 + 1 = 8+ 0 1 17 bg b g 40,, 1 b g b g b g = 679. a 169. a. x 7 4, 0, 1 = 17 4, 0, 1 Example 1.5 If = αax + 2a + 10a and = 4αax + 8a 2αa, for what values of α are and perpendicular? The condition for and to be perpendicular is =0. = α, 210, 4α, 8, 2α = 0 b gb g
Vector nalsis 13 2 = 4α + 16 20α = 0 α α 2 2 + 4 5α = 0 5α+ 4 = 0 b gb g α 1 α 4 = 0 α = 1, 4. Example 1.6 Find the scalar component of 6ax + 2a 3a along 3ax 4a. et = 6a + 2a 3a = 3a 4 a. Component of along is = a a d x = = i 3a 4a x 2 2 3 + a 4f d i a f F H = a = 6, 2, 3 4 3 5, 5, 0 18 8 = = 2. 5 5 1.2 CRDINTE SYSTEMS ND TRNSFRMTINS I K Introduction In electromagnetics, all the quantities are functions of space and time. In order to describe the spatial variations of the quantities, all the points in space must be defined using an appropriate co-ordinate sstem. lthough there are different co-ordinate sstems available, the three well-known co-ordinate sstems are Cartesian (or rectangular) co-ordinates, circular clindrical co-ordinates and spherical coordinates. 1.2.1 Cartesian Co-ordinates (X, Y, Z) point is represented in cartesian or rectangular co-ordinates as (X, Y, Z). From Fig. 1.11, it is known that an point in rectangular co-ordinate is the intersection of three plane (i) constant X-plane (ii) constant Y-plane and (iii) constant Z-plane which are mutuall perpendicular.
14 Electromagnetic Fields Z R x = Constant a = Constant a x a Y X = Constant Fig. 1.11 The range of the co-ordinate variables x, and are < x < < < < <. vector in cartesian co-ordinates can be written as ( x,, ) or x ax + a + a, where a, a and a are unit vectors along x, and directions respectivel. 1.2.2 Circular Clindrical Co-ordinates (,, Z) point in clindrical co-ordinate is represented as (,, Z). From Fig. 1.12, it is known that an point in the clindrical co-ordinates is an intersection of three planes vi. (i) constant plane (a clinder) (ii) constant plane (plane perpendicular to XY plane and displaced b angle from X-axis) (iii) constant Z-plane (parallel to XY plane). Each variable is defined as (i) is the radius of clinder passing through or the radial distance from Z-axis. (ii) is called the aimuthal angle and is measured from the X-axis in the XY plane: (iii) Z is the same as in cartesian sstem. The ranges of variables are 0 <
Vector nalsis 15 0 < 2π < <. = Constant Z R = Constant x = Constant Fig. 1.12 vector in clindrical co-ordinates can be written as (,, ) or a + a + a, where a, a and a are unit vectors in the,, and -directions as in Fig. 1.13. a a = a a = a a = 1...(1.29) a a = a a = a a = 0 a a = a a a a a = a...(1.30) a a = a Note: a, a, a are mutuall perpendicular and all point to the direction of increasing magnitude. a p Z x Fig. 1.13
16 Electromagnetic Fields 1.2.3 Relationship between Cartesian and Clindrical Co-ordinates Relationship between the variables (x,, ) of the cartesian co-ordinate sstem and those of the clindrical sstem (,, ) are easil obtained from Fig. 1.14 as = 2 2 1 x +, = tan, x =...(1.31) (or) x = cos, = sin, =...(1.32) Eqn. (1.31) is for transforming a point from cartesian (x,, ) to clindrical (,, ) co-ordinates, whereas Eqn. (1.32) is for (,, ) (x,, ) transformation. Z (X, Y, Z) = (,, Z) Z Y x = cos = sin X Fig. 1.14 1.2.4 Relationship between Unit Vectors of Clindrical and Rectangular Co-ordinates From the Figure 1.15 ax a = cos ax a = sin a a = sin...(1.33) a a = cos a = a
Vector nalsis 17 Y a a a 90 a x X Fig. 1.15 Transformation of the given vector ( x,, ) in cartesian into clindrical. = x ax + a + a The component along a is = a d d i x x i = a + a + a a = x cos + sin = cos+ sin a d x i. The component of along a is d i da x x a a i a d xsin cosi d xsin cosi. = a = + + = + = + a The component of along a is = a = a.
18 Electromagnetic Fields Similarl, M M N cos sin 0 = M sin cos 0 0 0 1 x = cos sin 0 sin cos 0 0 0 1 M x...(1.34)...(1.35) 1.2.5 Spherical Co-ordinates (r, θ, ) n point can be represented as (r, θ, ) i.e. intersection of three mutuall perpendicular planes (i) a sphere of radius r (centre at origin) (ii) cone having the Z-axis as its axis and vertex at origin. (iii) a constant plane which is perpendicular to XY plane and displaced b angle from X-axis. Z θ = Constant r = Constant Y X = Constant Fig. 1.16 From the Fig. 1.16, r is defined as the distance from the origin to the point or radius of the sphere, centred at origin, and passing through. θ (co-latitude) is the angle between the Z-axis and the position vector of, and is the angle measured from X-axis (same angle in clindrical coordinates). The range of r, θ, are 0 r < 0 θ< π
Vector nalsis 19 0 < 2π The three unit vectors ar, aθ and a are mutuall orthogonal and are shown in Fig. 1.17. Z ar aθ = a a θ a = a r...(1.36) a a = a a r r θ θ r a θ a Y X Fig. 1.17 1.2.6 Relationship between Cartesian, Clindrical and Spherical Co-ordinates From the Fig. 1.18, Z r = x + + = r sin θ 2 2 + θ= tan 1 x = tan 1....(1.37) Z = r cos θ (X, Y, Z) = (r, θ, ) = (,, Z) nd θ x = rsin θcos Z = rsin θsin...(1.38) Y = rcosθ. x = cos = sin X Fig. 1.18
20 Electromagnetic Fields 1.2.7 Relationship between Unit Vectors of Three Co-ordinates The transformation of vector requires the determination of the products of the unit vector in respective co-ordinates. The dot product of an spherical unit vector with an cartesian unit vector is the component of the spherical vector in the direction of the cartesian vector. The dot products are performed b first projecting the spherical unit vector on the XY plane and then projecting on to the desired axis. From the Fig. 1.19, dot products of unit vectors in spherical and cartesian co-ordinate sstems are as given in the Table 1.2.1. Z a a θ a r a 90-θ θ 90-θ θ a θ a θ a x Fig. 1.19 Table 1.2.1 a r aθ a a x. sin θcos cosθcos sin a. sinθsin cosθsin cos a. cosθ sin θ 0 Dot products of unit vectors in spherical and clindrical co-ordinate sstems are as given in Table 1.2.2.
Vector nalsis 21 Table 1.2.2 a r aθ a a. sin θ cosθcos 0 a. 0 0 1 a. cosθ sin θ 0 Using the Tables 1.2.1 and 1.2.2, the components of vector =,, related as given in the matrix form: r θ = sin θcos sinθsin cosθ cosθcos cosθsin sin θ sin cos 0 The inverse transformation ( r, θ, ) ( x,, ) is similarl obtained. M x = sin θcos cosθcos sin sin θsin cosθsin cos cosθ sin θ 0 M d i and = dr, θ, x r θ i are....(1.39)....(1.40) Transformation from spherical ( r, θ, ) to form clindrical (,, ) is also obtained from the Table 1.2.2. = sin θ cosθ 0 0 0 1 cosθ sin θ 0 Inverse transformation of clindrical to spherical is similarl written as r θ = sin θ 0 cosθ cosθ 0 sinθ 0 1 0 r θ...(1.41)....(1.42) Example 1.7 Given points (x = 2, = 3, = 1) and ( = 4, = 50, = 2), find the distance from (a) to the origin (b) to the origin (c) to. (i) to the origin. = x + + bg bg bg = 2 + 3 + 1 = 3.74.
22 Electromagnetic Fields (ii) to origin. Transform (,, ) clindrical form into (x,, ) cartesian form. x = cos = sin. b Given = 4, = 50, = 2 b b g g x = 4cos 50 = 257. = 4sin 50 = 306. = 2 b g b g = 257. + 306. + 2 (iii) to = 4.47 = + + g b x xg d i b g dx,, i= b 257., 306., 2 g dx,, i= b 2, 3, 1 g = 6.79. Example 1.8 Transform each of the following vectors to clindrical co-ordinates at the point specified: (a) 5ax at b= 4, = 120, = 2 g (b) 5ax at b x = 3, = 4, = 1 g (c) 4a 2a 4a at x = 2, = 3, = 5 Transformation from cartesian to clindrical M N Given = (5, 0, 0) cos sin 0 = M sin cos 0 0 0 1 M N cos sin 0 = M sin cos 0 0 0 1 b M x N M 5 0 0. g
Vector nalsis 23 = 5cos = 5sin = 0 b g b g,, = 5cos, 5sin, 0. = 5cos a 5sin a b at = 4, = 120, = 2 = 25. a 433. a. g (ii) Transformation of 5 a x to clindrical is given in eqn. (5cos, 5sin, 0). Given point (x = 3, = 4, = 1). From the eqn. 2 2 1 = x + = 5, = tan x = 53.13. Given point in clindrical co-ordinate is b g b g b g = 5, = 5313., = 1 at = 5cos 5313. a 5sin 5313. a = 3 a 4a. (iii) et = 4a 2a 4a Using the transformation Eqn (1.34) M M M cos sin 0 4 = sin cos 0 2 0 0 1 4 = 4cos 2sin a + 4sin 2cos a 4a Coordinates of are given as (x = 2, = 3, = 5) b g b g. Clindrical co-ordinates of are = 36., = 563., = 5 at = 0555. a 444. a 4a b. Example 1.9 Given the points (x = 2, = 3, = 1) and (r = 4, θ = 25, = 120 ), find the distance from to. g
24 Electromagnetic Fields From eqn. (1.18) x = r sinθ cos = r sinθ sin = cosθ. The cartesian co-ordinates of are x = 0.845, = 1.464, = 3.63 b x xg d i b g = + + = 5.64. Example 1.10 Transform the vector = 4a 2a 4 a at (x = +2, = +3, = 4) to spherical co-ordinate. Transformation is given b eqn. (1.39) r θ = sin θcos sin θsin cosθ cosθcos cosθsin sin θ sin cos 0 = 4, = 2, = 4 b = 4sin θcos 2sin θsin 4cosθ Coordinates of is to be converted to spherical. r = x + + = 2 2 538. 1 x + θ= tan = 42. 03 1 = tan = 56. 3 x at 538., 4203., 563. b = 259. a + 311. a 443. a g r θ. M x g a r b4cos θ cos 2cos sin 4sin g θ b 4sin 2cos g + + a + a. 1.3 DIFFERENTI ENGTH, RE ND VUME Differential elements in length, area and volume are useful in vector calculus. The are defined in cartesian, clindrical and spherical coordinate sstems.
Vector nalsis 25 1.3.1 Cartesian Co-ordinates From the Fig. 1.20, if we move from to, dl = d a and if we move from to S, dl = d a + d a. Similarl, to move from D to would mean that dl = dx ax + d a + d a....(1.43) The differential surface (or area) element ds ma generall be defined as ds = ds a n,...(1.44) where ds is the area of the element and a n is unit vector normal to the surface. For example, for the surface CD in Fig. 1.20, ds = dda x, whereas for RS, ds = d d a x, because an = ax is normal to RS. Differential volume dv can be obtained from dl as the product of three components of dl. Note that dl and ds are vectors and dv is a scalar. Z d d S R dx a D C a Y a x X ds = d d a x = dx d a = dx d a dv Fig. 1.20...(1.45) = dx d d....(1.46)