Vector Calculus and Applications

Transcription

1 Unit 1 Vector Calculus and Applications It is well known that accelerating electric charges produce time-varying fields which have an electric and magnetic component thus the name electromagnetic fields. On the other hand, stationary charges produce static electric fields, while moving, but non-accelerating, charges give rise to static magnetic fields. Vector calculus is an extremely useful tool in characterizing and applying the properties of these fields. In as far as is feasible, in this section and others, we shall seek to illustrate the rudiments of vector calculus as they may be applied in the study of engineering electromagnetics. That is, rather than treating the subject from a purely mathematical viewpoint, we wish to investigate some of the more prominent features of the vector calculus in the context of the development of the fundamentals of electromagnetics. Some of the mathematics, in its most general application, proves to be formidable, but fortunately many of the important features may be developed in the study of basic electromagnetics problems. 1.1 Review - A Few Comments on the Properties of Vectors The following material has been encountered much earlier in the programme and, with the exception of the first one or two tutorials, will not receive any detailed treatment here: 1

2 1. Definition of a scalar: a quantity possessing magnitude only (eg., potential difference); 2. Definition of a vector: a quantity possessing magnitude and direction (eg., electric field intensity); in these notes, non-time-varying vectors are indicate by an arrow (eg., E) and time-varying vectors will be indicated by a tilde underscore such as Ẽ. Ordinary phasors will be represented by a tilde overbar such as in Ṽ and time-harmonic vector quantities will be represented in the same manner (eg., Ẽ). The text uses boldface letters for vectors, but this is not convenient for note-taking. 3. Definition of a unit vector: a vector whose magnitude is unity. For example, ˆr = r r is a unit vector in the direction of a vector, r, whose magnitude is r. 4. We will also consider the following operations: (a) addition (or subtraction) of vectors (b) the scalar (or dot) product (c) the vector (or cross) product (d) triple products A little consideration will be given to the specification of vectors in each of the three most commonly used coordinate systems: Cartesian (or rectangular), cylindrical and spherical Vectors and Vector Space Discussion Based on the Cartesian Coordinate System Consider the right-handed rectangular or Cartesian coordinate system. A vector r may be specified in this system by its components along the respective axes as shown: 2

3 The x, y and z axes are mutually perpendicular. The position vector r is from the origin, O, to some arbitrary point P (x, y, z) and is given by r = x + y + z. Thus r may be given as an ordered triple (x, y, z). A collection of all such vectors which has the following five properties constitute a vector space: Let r = (x, y, z), r 1 = (x 1, y 1, z 1 ), and r 2 = (x 2, y 2, z 2 ). 1. Vector equality: r 1 = r 2 means x 1 = x 2, y 1 = y 2, and z 1 = z Vector addition: r = r 1 + r 2 means x = x 1 + x 2, and so on. 3. Scalar multiplication: For a real number a, a r = (ax, ay, az) and vice versa. 4. Negative of a vector: r = ( x, y, z) and vice versa. 5. Null vector: A vector 0 implies x = 0, y = 0, and z = 0 and vice versa. Since the vector components are real numbers, the following properties also hold: 1. Vector addition is commutative: r 1 + r 2 = r 2 + r Vector addition is associative: r 1 + ( r 2 + r 3 ) = ( r 1 + r 2 ) + r Scalar multiplication is distributive: a( r 1 + r 2 ) = a r 1 +a r 2 and (a+b) r = a r+b r. 4. Scalar multiplication is associative: a(b r) = (ab) r. 3

4 Furthermore, the null vector 0 and the negative of a given vector r are unique. For various topics in engineering electromagnetics, the concepts presented above may be extended to include (1) complex vectors, (2) functions of vectors, and (3) any number of vector components Miscellaneous Vector Characteristics and Elementary Operations Specification of r in Terms of the Standard Unit Vectors Let ˆx, ŷ, and ẑ be unit vectors pointing along the positive direction of the respective coordinate axes. Then, since x = xˆx, etc., r = xˆx + yŷ + zẑ Note, too, that the unit vector ˆr in the direction of r is given simply as ˆr = where r may be determined from the dot or inner product between r and itself as shown below. Dot Product The dot product of two vectors r 1 and r 2 may be defined geometrically by r 1 r 2 = r 1 r 2 cos(θ 2 θ 1 ) where (θ 2 θ 1 ) is the angle between r 1 and r 2. This is a scalar product i.e. the result of the product is a scalar quantity. As claimed above, we note that r r = r r cos 0 = r 2 or r 2. Clearly, the dot product of r 1 and r 2 is simply the projection of r 1 onto the direction of r 2 multiplied by the magnitude of r 2 or vice versa. 4

5 The dot product may be specified using the unit vector notation as follows: r 1 r 2 = (x 1ˆx + y 1 ŷ + z 1 ẑ) (x 2ˆx + y 2 ŷ + z 2 ẑ) = x 1 x 2 + y 1 y 2 + z 1 z 2 since ˆx ˆx = 1 1 cos 0 = 1 etc. while ˆx ŷ = 1 1 cos 90 = 0 etc.. Again, using this idea r r = x 2 + y 2 + z 2 so that r = r = x 2 + y 2 + z 2. Cross Product Using the above notation recall that the cross product of two vectors denoted by R = r 1 r 2 implies R = r 1 r 2 sin(θ 2 θ 1 ) This is a vector product and R is specified (by definition) to be normal to the plane containing r 1 and r 2 and in the direction determine by the right hand rule as illustrated. Illustration: R n r r 1 That is, r 1, r 2, and R form a right-handed system. ˆn is a unit vector normal to the plane containing r 1 and r 2 and is in the direction of advance of a right-handed screw as r 1 is rotated into r 2. 5

6 Therefore, R = r 1 r 2 = r 1 r 2 sin(θ 2 θ 1 )ˆn Clearly, ( r 1 r 2 ) = ( r 2 r 1 ) for any pair of vectors. Note too that ˆx ˆx = ŷ ŷ = ẑ ẑ = 0 since sin 0 = 0. Also, ˆx ŷ = ẑ ; ŷ ẑ = ˆx ; ẑ ˆx = ŷ. Determinant Form for the Cross Product: Given: r 1 = x 1ˆx + y 1 ŷ + z 1 ẑ and r 2 = x 2ˆx + y 2 ŷ + z 2 ẑ. r 1 r 2 = Therefore, r 1 r 2 = ˆx ŷ ẑ x 1 y 1 z 1 x 2 y 2 z 2 Of course, there is nothing special about the notation. For any two vectors, A = A xˆx + A y ŷ + A z ẑ and B = B xˆx + B y ŷ + B z ẑ, A B = ˆx ŷ ẑ A x A y A z B x B y B z A and B may be any kind of vector quantities (eg., force F, electric field intensity, E, and so on). Scalar and Vector Fields A field may be defined as some function of a vector which is specified by a connection from an arbitrary origin to a general point in space. (i.e., the field is a function of a position vector, r). 6

7 A vector field is a vector function i.e., it has both magnitude and direction, which will, in general, vary throughout the region. A scalar field is a scalar function i.e., it is a function of r but it has only magnitude. It is important in this course to realize that the value of a field may vary, in general, with time as well as position. Examples: (1) Electric Field Intensity ( E( r)) The electric field intensity, E( r), of a positive point charge, q, (at (0, 0, 0), say) is a vector field. Recall that E( r) is defined as the force per unit charge experienced by a small positive test charge brought to position r. Illustration: E( r) = 1 q 4πɛ 0 r ˆr 2 The parameter, ɛ 0, measured in farads per metre is the permittivity of free space, and it has a value of π F/m. The E field is radially away from the positive point charge and the field unit is the volt per metre (V/m). (2) Electric Potential V ( r) We shall see later that the electric potential, V, which is defined as the work done per unit charge in bringing that charge to a position, r, from some arbitraily chosen zero reference, is strictly a scalar field. For example, for the above illustrated charge with the zero reference at infinity, it will be observed that the potential, measured in volts, is given by the scalar equation V ( r) = 1 q 4πɛ 0 r. 7

8 Complex Representation and Time-Averaging of Time-Harmonic Scalars and Vectors 1. Scalars A time-harmonic scalar quantity, V (t), with amplitude V 0, angular frequency ω = 2πf (f frequency in hertz (Hz)), and phase φ may be written V (t) = V 0 cos(ωt + φ). Recalling that e jθ = cos θ + j sin θ i.e. Euler s identity, V (t) may clearly be cast as V (t) = Re { Ṽ e jωt} with Ṽ = V 0 e jφ which is called a phasor. Note that the phasor is not a function of time. Thus, we may say V (t) Ṽ where may be read as is equivalent in phasor form to in other words indicates a kind of transformation. Addition It may be readily verified that if U(t) = Re { Ũe jωt} with Ũ = U 0e jφ, then V (t) + U(t) Ṽ + Ũ. The same is NOT true of multiplication. (Check it out). Time Derivative Noting that V (t) t = ωv 0 sin(ωt + φ) = Re { jωv 0 e jφ e jωt}, V (t) t jωṽ. That is, jω can replace the time derivative in phasor (complex) representation of time-harmonic scalars, a useful feature when considering the time-varying form of Maxwell s equations (see Unit 2). 8

9 2. Vectors Consider the time-harmonic vector V (t) = [V x cos(ωt + φ x )] ˆx + [V y cos(ωt + φ y )] ŷ + [V z cos(ωt + φ z )] ẑ = Re {[( V x e jφx ) ˆx + ( Vy e jφy ) ŷ + ( Vz e jφz ) ẑ] e jωt }. The equivalent phasor notation notation (Ṽ ) for V (t) is V (t) Ṽ = ( V x e jφx ) ˆx + ( Vy e jφy ) ŷ + ( Vz e jφz ) ẑ so that V (t) = Re { Ṽ e jωt}. 1. Time Averages In considering such issues as power requirements in electronic communications systems, computation of time-average values of combinations of time-harmonic quantities is important. It is trivial to show that the time-averaged value of a time-harmonic quantity is identically zero. For example, using to indicate time averaging over a period T (T = 1/f) of such a quantity, it is obvious that V (t) = 1 T T 0 V 0 cos(ωt + φ)dt = 0 The same is true for complex time-harmonic vector quantities, V (t). The importance of non-zero time averaging shows up in the averaging of products of time-harmonic quantities. Consider the following examples: Example 1. Determine the time average of V 2 (t) for V (t) = V 0 cos(ωt + φ). 9

10 Example 2. Consider two complex vectors Ẽ = ẼR + jẽi and H = H R + j H I (we ll meet these later as complex electric and magnetic field intensity, respectively). We wish to compute the time-averaged cross product Ẽ. Remember Ẽ E(t) H etc.. It is easy to similarly show that for the dot product, Ẽ H = 1 2 Re { Ẽ H }. 10

11 1.2 Coordinate Systems The geometry of many electromagnetics problems often dictates which coordinate system is more appropriate in facilitating a solution. For example, in examining problems associated with coaxial cables or line sources, (circular) cylindrical coordinates are usually useful. In other instances, certain problem symmetries may suggest the use of spherical coordinates and in some cases, the ordinary Carstesian or rectangular system may be all that is needed. The proper choice of coordinate system is usually critical in ensuring the quickest solution. Of the several such systems, we ll consider only those mentioned above Cartesian (or Rectangular) Coordinates We have already assumed knowledge of this system in the previous section. Here, we will consider specification of differential elements used in various kinds of integrations involving the (x, y, z) (i.e., Cartesian or rectangular) coordinates. 1. Specification of a Differential Length, dl: Consider a curve in space as shown: z dl r r + dr curve in space y x r = xˆx + yŷ + zẑ The quantity, d L is simply a differential vector of magnitude dl lying along the curve, while r and r + d r are position vectors specifying the beginning and ending of d L. Clearly, d L = d r. Therefore, d L = dxˆx + dyŷ + dzẑ and the magnitude is given by dl = dl dl = dx 2 + dy 2 + dz 2. 11

12 2. Specification of a Vector Differential Area, d S: Suppose ds is a differential element of a surface which encloses a volume. Then, if ˆn is the outward unit normal (positive) by definition d S = ˆndS Illustration: If the surface is not closed, the positive unit normal depends on the direction in which the perimeter is traversed. For a right-handed coordinate system this may be illustrated as below: Now, consider differential areas in each of the three coordinate planes as illustrated We have d S x = dydz ˆx d S y = dxdz ŷ d S z = dxdy ẑ is a differential surface vector along the ˆx direction. is a differential surface vector along the ŷ direction. is a differential surface vector along the ẑ direction. 3. Specification of a Differential Volume, dv : Consider the differential volume element, dv, in rectangular coordinates (Note that the text uses dν): dv = dxdydz. 12

13 1.2.2 (Circular) Cylindrical Coordinates We use the term circular cylindrical coordinates because, strictly speaking, there are other kinds of cylindrical coordinate systems such as elliptic cylindrical coordinates. Throughout this course, the term cylindical coordinates will be used in reference to the former. It is one of the two curvilinear coordinate systems we will encounter, the other being the spherical coordinate system of the next subsection. 1. Specification of the coordinate system: Consider the following illustration: Any point in the above illustrated space may be represented by three curvilinear coordinates which are here labelled (ρ, φ, z). Just as in the Cartesian system where the intersection of the planes x = constant, y = constant, z = constant describes any point (x, y, z), so the intersection of the surfaces ρ = constant, φ = constant, z = constant describes any point (ρ, φ, z). The three coordinate surfaces are (i.) right circular cylinders having the z-axis as the common axis. ρ = x 2 + y 2 = constant Note that ρ is the perpendicular distance from the z-axis while r is the general distance from the origin. 13

14 (ii.) half planes through the z-axis: ( ) y φ = tan 1 = constant x (φ is measured counterclockwise (ccw) from the positive x-axis as shown.) (iii.) planes parallel to the xy-plane as in the Cartesian system: z = constant The limits on ρ, φ and z are 0 ρ <, 0 φ 2π, < z < Inverting the equations under (i), (ii) and (iii) above, or going directly to the geometry of the above figure, gives the transformation from cylindrical to Cartesian cordinates as: x = ρ cos φ y = ρ sin φ z = z 2. Unit vectors in Cylindrical Coordinates: Consider the following illustration: ˆρ unit vector normal to the cylindrical surface and pointing in the direction of increasing ρ. ˆφ unit vector (tangential to the cylindrical surface), perpendicular to the half-plane φ = constant and pointing in the direction of increasing azimuth angle, φ. ẑ unit vector pointing in the direction of increasing z. while Note that ˆρ ˆρ = ˆφ ˆφ = ẑ ẑ = 1 ˆρ ˆφ = ẑ; ˆφ ẑ = ˆρ; ẑ ˆρ = φ. 14

15 3. Relationships Between Cartesian and Cylindrical Unit Vectors: It is obvious that ˆx and ŷ, being perpendicular to ẑ, have no component in the z direction. However, both ˆx and ŷ may have projections along both the ˆρ and ˆφ directions (and vice versa). Illustration: Because we are considering unit vectors, the following result: ˆx ˆρ = cos φ ˆx ˆφ = cos(90 + φ) = sin φ ŷ ˆρ = cos(90 φ) = sin φ ŷ ˆφ = cos(φ) To summarize in table form: ˆx ŷ ẑ ˆρ cos φ sin φ 0 ˆφ sin φ cos φ 0 ẑ On this basis, ˆx = (ˆx ˆρ)ˆρ + (ˆx ˆφ) ˆφ = cos φˆρ sin φ ˆφ ŷ = sin φˆρ + cos φ ˆφ ẑ = ẑ with cos φ = x x2 + y 2 and sin φ = y x2 + y 2. Similarly, it is clear from the table that ˆρ = cos φˆx + sin φŷ ˆφ = sin φˆx + cos φŷ ẑ = ẑ 15

16 4. Position Vector, r, in Cylindrical Coordinates: The above relationships between unit vectors may now be used to established the form of the position vector, r, in cylindrical coordinates. We simply need to determine the projections of the Cartesian components unto the new unit vectors: r = Of course, this result could have easily established geometrically as depicted below: It may be noted that any general vector given by A = A xˆx + A y ŷ + A z ẑ may be transformed to cylindrical coordinates as A = A ρ ˆρ + A φ ˆφ + Az ẑ by determining the projections of A x and A y unto the ˆρ and ˆφ directions. That is, A = ( A ˆρ)ˆρ + ( A ˆφ) ˆφ + ( A ẑ)ẑ. and A ρ = ( A ˆρ) = [A xˆx + A y ŷ] ˆρ A φ = ( A ˆφ) = [A xˆx + A y ŷ] ˆφ A z = A z. 16

17 5. Vector Differential Length in Cylindrical Coordinates: Recall, in general, if u is a function of v and w, then du = u u dv + v w dw It is left as an exercise to verify, using partial derivatives and the relationships between the unit vectors, that d r = d L = dxˆx + dyŷ + dzẑ transforms to d r = d L = dρˆρ + ρdφ ˆφ + dzẑ (Remember that both x and y are functions, in general, of ρ and φ.) Also, the above result may be quickly established from simple geometry. Finally, the magnitude of the differential length is given by dl = dl dl = (dρ) 2 + ρ 2 (dφ) 2 + (dz) 2 6. Vector Differential Area and Differential Volume in Cylindrical Coordinates: Consider the following diagram: We note that ds p = ρdφdz ˆρ (vector differential area in ˆρ direction) ds φ = dρdz ˆφ (vector differential area in ˆφ direction) ds z = ρdρdφẑ (vector differential area in ẑ direction) It is easily deduced from the above diagram that the differential volume element in cylindrical coordinates is given by dv = ρdρdφdz 17

18 1.2.3 Spherical Polar Coordinates 1. Specification of the coordinate system: Consider the following illustration in which a general point, P, in space is located on a spherical surface centred at (0, 0, 0) of the Cartesian coordinate system. Point P may be specified by the three spherical polar coordinates (we ll use, simply, spherical coordinates in our terminology) (r, θ, φ). A point is uniquely specified by the intersection of the following surfaces: (i.) Concentric spheres centres at the origin and given by r = x 2 + y 2 + z 2 = constant (ii.) Right circular cones centred on the z-axis (the polar axis) with vertices at the origin and which are specified by the equation of their polar angle as ( ) θ = cos 1 z = constant x2 + y 2 + z 2 (iii.) Half planes through the z-axis given by ( ) y φ = tan 1 = constant x (φ is the azimuth angle). 18

19 The limits on r, θ and φ are 0 r <, 0 θ π, 0 φ 2π. The transformation from spherical to rectangular coordinates may be readily verified from the geometry as: 2. Unit vectors in Spherical Coordinates: Consider the following illustration: x = r sin θ cos φ y = r sin θ sin φ z = r cos θ ˆr unit vector normal to the spherical surface and pointing in the direction of increasing r. ˆθ unit vector tangential to the spherical surface, along a line of longitude as shown and pointing in the direction of increasing θ. ˆφ unit vector tangential to the spherical surface, perpendicular to the half-plane φ = constant and pointing in the direction of increasing azimuth angle, φ. It must be emphasized that the unit vectors ˆr and ˆθ vary in direction as the angles θ and φ vary and ˆφ varies with φ. Note that ˆr ˆr = ˆθ ˆθ = ˆφ ˆφ = 1 while ˆr ˆθ = ˆφ; ˆθ ˆφ = ˆr; ˆφ ˆr = ˆθ. 3. Relationships Between Cartesian and Spherical Unit Vectors: We ll consider the ideas involved for the specific case of the projection of ˆr onto ˆx, ŷ, and ẑ and leave the other cases as an exercise. 19

20 Illustration: Projections of ˆr unto the Cartesian unit Vectors: A consideration of the geometry shows that the projection of ˆr unto ˆx is given as ˆx ˆr = sin θ cos φ. Similarly, ŷ ˆr = sin θ sin φ and ẑ ˆr = cos θ. We have seen from the cylindrical coordinate discussion that the ˆφ projections unto the Cartesian unit vectors are given by ˆx ˆφ = sin φ ; ŷ ˆφ = cos φ ; ẑ ˆφ = 0 Furthermore, it is not too difficult to set up the geometry which shows that the ˆθ projections are given by To summarize in table form: ˆx ˆθ = cos θ cos φ ŷ ˆθ = cos θ sin φ ẑ ˆθ = sin θ ˆr ˆθ ˆφ ˆx sin θ cos φ cos θ cos φ sin φ ŷ sin θ sin φ cos θ sin φ cos φ ẑ cos θ sin θ 0 20

21 On this basis, ˆr = (ˆx ˆr)ˆx + (ŷ ˆr)ŷ + (ẑ ˆr)ẑ = sin θ cos φˆx + sin θ sin φŷ + cos θẑ ˆθ = cos θ cos φˆx + cos θ sin φŷ sin θẑ ˆφ = sin θˆx + cos θŷ It is left as an exercise to similarly determine ˆx, ŷ, and ẑ in terms of the spherical polar coordinate unit vectors. Note that by definition r = rˆr. This may be arrived at algebraically by using and the fact that r = xˆx + yŷ + zẑ i.e. this is a way of verifying the correctness of the exercise. As another exercise, may be used along with the appropriate partial derivatives to show that the vector differential length is given by d r = drˆr + rdθˆθ + r sin θdφ ˆφ 21

22 4. Vector Differential Area and Differential Volume in Spherical Coordinates: Consider the following diagrams: Differential Area We note from the geometry that ds r = r 2 sin θdθdφˆr ds θ = r sin θdrdφˆθ ds φ = rdrdθ ˆφ (vector differential area in ˆr direction) (vector differential area in ˆθ direction) (vector differential area in ˆφ direction) Differential Volume It is easily deduced from the above diagram that the differential volume element in cylindrical coordinates is given by dv = r 2 sin θdrdθdφ 22

23 1.3 Line Integrals with Applications In electromagnetics, it is common to encounter integrals whose integrands consist of one or more operations involving vector quantities. We shall meet some of these in this and subsequent sections of this unit. For scalar fields, V ( r) say, and vector fields, A( r) say, the following forms of line integrals may be encountered: In each case, one attempts to reduce the vector integral to scalar forms. The contour, C, is the path over which the integral is evaluated. It may be an open contour in which case the starting and end points are different or a closed contour which has the same staring and end points. For closed contours it is common to use the symbol or. C Initially, let s consider the second of the three types. Suppose, for example, that A( r) = E( r) which is the electric field intensity in V/m. Recall that d r = dl. Illustration: E d L is the projection of E onto d L multiplied by d L i.e. E d L = E d L cos θ = EL dl. Thus, AB. B... = C A E d L is simply the sum of all such projections along the curve Application 1. Energy, Potential Difference and Potential Energy Suppose that it is desired to move a charge, Q, over a displacement, d L, in a 23

24 region where the electric field intensity is E (which, for now, is a non-time-varying field). The force, F E, which the field exerts on the charge is by definition of E given by F E = QE (1.1) By definition of work, the differential amount of work, dw E, done by the field is given by dw E = F E dl and from (1.1), dw E = QE dl (1.2) Now, if an external source is to move Q in the field, the the force, F ext, required is equal but opposite to F E (i.e. F ext = F E ). The differential amount of external work, dw, must therefore be given by and using (1.2), dw = QE dl (1.3) From our definition of the line integral, it is clear that the total work done by an external force in moving a charge Q along a path, C, in a region where E exists is therefore or W = Q C E d L (1.4) Remember, equation (1.4) is the WORK DONE BY THE EXTERNAL SOURCE NOT BY THE FIELD ITSELF i.e. it is the ENERGY expended by an external source. This definite integral is basic to the study of field theory. If E dl is the C 24

25 same for all paths C connecting two specific points (say A and B) in a field, then the field is said to be conservative. Illustration: If the field is non-time-varying (i.e. static), then this property of conservation will exist. In general, for time-varying fields this property will NOT hold. It should be intuitive that, if conservation holds, then for any CLOSED path C E d L = 0 (1.5) Recall, Kirchhoff s voltage law here we have it without wires!! Example: Determine the work done in carrying a 4.0-C charge from B(1, 0, 2) to A(0.8, 0.6, 2) along the shorter arc of the circle given by C: x 2 + y 2 = 1; z = 2 if E = yˆx + xŷ + 2ẑ. 25

26 Potential Difference By definition, the potential difference (V AB ) between two points in an E-field is the work per unit charge required to move a charge from point B to point A in that field. From equation (1.4), V AB = W Q = A B E d L (1.6) The unit is the joule per coulomb (J/C) or volt (V). The potential difference, V AB, is positive if work is done (i.e., energy is expended by an external source) in moving the charge from B to A. Notice that, with this definition for the volt, together with equation (1.1) the E-field unit is indeed as we have said. Given equation (1.6), it is clear that the potential difference between points A and B of the last example is V AB = 1.92 J/4.0 C = 0.48 V. Potential (or Absolute Potential), V The potential V A at a point A in an electric field is the work per unit charge required to move a charge from some arbitrarily specified zero reference to point A. Two of the most commonly used zero references are and 1. the earth s surface i.e., ground 2. infinity. For example, if infinity is chosen as the zero reference, using the same ideas as led to equation (1.6), we have the potential at point A as and that at point B as 26

27 Thus, V A V B = That is V A V B = V AB (1.7) and hence the name potential difference for V AB. Example: For a zero reference chosen at infinity, determine the potential at a distance r A from the origin due to a point charge q at the origin. Assume free space. Finally, extend the idea to the case of n separate point charges each of which is at position r n i.e., not necessarily at the origin. 27

28 Application 2. The Biot-Savart Law and Ampère s Law The Biot-Savart Law proposed in 1820 by Ampère s colleagues states that for a differential length of filament, dl, located at a point, P 1, and carrying a current I (dc for now) produces at any field point P 2 a (differential) magnetic field intensity (dh) whose magnitude is proportional to the product of the current, the magnitude of dl and the sine of the angle lying between the filament and a line connecting the filament to point P 2. dh is inveresely proportional to the square of the distance, R 12, from the position of the filament to P 2. Illustration: Using SI units and a proportionality constant of 1/4π, this may be formalized as d H = Id L ˆR 12 4πR 2 12 ; (The H unit is, clearly, A/m.) (1.8) For the entire current loop, this may be written as H = Id L ˆR 4πR 2 (1.9) This is the static H field at P 2 due to the closed-loop current. ˆR is the unit vector from any dl toward P 2. (Notice that this is the third type of line integral initially mentioned in this section and that the result is a vector quantity). Example: Consider a square loop centred on the origin in the x-y plane and carrying a dc current I as shown. Determine the magnetic field intensity at (0, 0, 0). 28

29 Ampère circuital law states that the line integral of the magnetic field intensity, H, around any closed path, C, is exactly equal to the direct current, I e, enclosed by the path. Symbolically, C H d L = I e (1.10) where I e is the current cutting through the loop. It is, in fact, possible to derive this expression from the Biot-Savart law and some math that we have not yet encountered. (For now, we shall accept the law as being capable of experimental verification). The orientation of the integration contour, C, follows the right-hand rule with respect to the direction of the current flow as illustrated below (thumb in direction of the current and fingers curled in direction in which path is traversed positively). Illustration 1. Current Filament : Illustration 2. Thick Wire Carrying a DC Current (I): In this illustration, J is referred to as the (volume) electric current density and it has units of A/m 2. If the cross section of the wire is A, then for the situation shown, J = I A ˆr For contour C 1 enclosing area A, I e = I. 29

30 ( ) For contour C 2, I e = J A A = I. A For some special, yet often important, cases a contour may be chosen such that (1) H is everywhere tangent to the contour and (2) H is constant along the contour. As we shall see, this greatly simplifies the integration in equation (1.10) and Ampère s law may be used to find H if I e is known. Example: Consider an infinitely long thin wire along the z-axis which is carrying a dc current, I. Determine the value of H at an arbitrary point, P, at a radial distance ρ from the wire. The contour C should be chosen with care. Here, let s choose C parallel to the x-y plane as shown. We make the following arguments : (1) It can be easily seen from the fact that the wire stretches to infinity in either direction along the z-axis that H has no z dependence. (2) From the Biot-Savart law, since dl ˆR (= d z ˆR) always lies in the ˆφ direction for any point on the contour, H must also lie in the ˆφ direction. Also, H will clearly depend on ρ. (3) From symmetry, there is clearly no reason why H should change in magnitude for different values of φ. 30

31 Using these three observations, we may thus write H = H φ (ρ) ˆφ. For the contour chosen, d L = ρdφ ˆφ so that Ampère s law C H d L = I e becomes H(ρ) = I 2πρ ˆφ (1.11) Thus, for a dc current as specified (i.e. an infinite line current), the static magnetic field intensity, H, is in the ˆφ direction and is inversely proportional to the radial distance, ρ, from the line. This example may be extended to the important case of the coaxial line and we shall do this in a tutorial to follow shortly. 1.4 Surface and Volume Integrals with Applications We have already encountered the ideas of a differential surface area and differential volume in three coordinate systems. We will now carry out applied integrations involving these. Initially, because both surface and volume integrals may appear in a single equation in electromagnetics, we briefly consider both types before proceeding to applications. 1. Surface Integrals Analogous to the line integral forms, surface integrals may appear (for scalars, V ( r), and vectors, A( r)) as 31

32 where ds is a local normal to the surface S as discussed earlier. The dot product is the most commonly encountered form. In fact, this form may be interpreted as a flow or flux through the given surface (more later). If the surface is closed, it is common to use the construct to indicate this fact. S Illustration: Determine the surface area of a sphere of radius a. 2. Volume Integrals Volume integrals tend to be somewhat simpler than surface integrals because the volume element, dv, is a scalar quantity. For a vector A( r) (in Cartesian coordinates in this illustration) so that the vector integral is essentially reduced to a vector sum of scalar integrals. A similar situation exists for the other coordinate systems. Of course, we may also encounter the usual scalar volume integrals, say for example, V ( r)dv where V ( r) is a scalar field. Example: Suppose a continuous charge distribution lies in a spherical shell defined by 1 < r < 2. Let the volume charge density, ρ v ( r), (i.e. the charge per volume in vol 32

33 C/m 3 ) be given by Find the total charge Q within the shell. ρ v ( r) = 2 r 3 C/m3. Application 1. Electric Flux, Flux Density and Gauss s Law (Electric) Before using the surface integrals in an electrostatics application, we need to consider a few related details on electric flux. In (approximately) 1837, Michael Faraday, being interested in static electric fields and the effects which various insulating materials (or dielectrics) had on these fields, devised the following experiment: Faraday had two concentric spheres constructed in such a way that the outer one could be dismantled into two hemispheres. With the equipment taken apart, the inner sphere was given a known positive charge. Then, using about 2 cm of perfect (ideal) dielectric material in the intervening space, the outer shell was clamped around the inner. Next, the outer shell was discharged by connecting it momentarily to ground. The outer shell was then carefully separated and the negative charge induced on each hemisphere was measured. Faraday found that the magnitude of the charge induced on the outer sphere was equal to the that of the charge on the inner sphere, irrespective of the dielectric used. He concluded that there was some kind of displacement from the inner to the outer 33

34 sphere which was independent of the medium. This is now referred to severally as displacement, displacement flux, or, as we shall use, electric flux. (Of course, the idea of electric flux lines as entities streaming away from electric charge is simply an invention to aid our conceptualization of the presence of an electric field). This flux, denoted by Ψ (unfortunately, the text uses F ), is in SI units related to the charge, Q, producing it via a dimensionless proportionality constant of unity; i.e. the electric flux in coulombs is given by Ψ = Q. The path of the flux lines is radially away from the inner sphere as shown: An important entity in electromagnetics is the idea of electric flux density, D. For example, in the above illustration, at the surface of the inner sphere, while at the outer surface D = Ψ area of surface ˆr = D = Q 4πb 2 ˆr. Q 4πa 2 ˆr Clearly, D is measured in coulombs/metre 2. If we think of the inner sphere as shrinking to a point charge, Q, then at a distance r from the charge D = From this last expression, it may be seen that, for free space, Q ˆr. (1.12) 4πr2 D = ɛ 0 E (1.13) Equation (1.13) is one of the important so-called constitutive relations which are essential in solving electromagnetics problems. 34

35 If the charge is distributed within a volume, such that the charge density is ρ v, then the ideas used in developing equation (1.13) lead to a volume integral ρ v dv D = V 4πR ˆR. (1.14) 2 Here R is the distance from the differential volume, dv, under consideration to the point of observation, and ˆR is the unit vector in that direction. If the region of interest is NOT effectively free space, then the permittivity, ɛ 0, must be replaced with the permittivity, ɛ, of the region. If the region is not electrically isotropic i.e. not same electrical properties in all directions then ɛ could be a matrix rather than a single number. In passing, we note that it is common to tabulate relative permittivities, ɛ R, where ɛ R = ɛ. (1.15) ɛ 0 [Aside 1. Without discussing the mathematical details, it is important to realize that when a material (not free space) is placed in an electric field, there is generallly motion of the bound charge within the material as it reacts to the field. A measure of the tendency of bound charge to do this is called the susceptibility, χ e (a dimensionless constant). If the material is isotropic and if this reaction effect is linearly related to the applied field, this effect adds directly to the applied electric field and is proportional to that field. The overall electric flux density, D, is therefore increased by an amount, P, called the polarization. D becomes D = ɛ 0 E + P = ɛ 0 E + χe ɛ 0 E = (1 + χ e )ɛ 0 E. On comparing the last expression with equation (1.15) and using the fact that D = ɛe, we see that ɛ R = (1 + χ e ). 35

36 It should be pointed out that ɛ R is dependent on frequency (in time-varying fields). For example, at HF (3 30 MHz), sea water has a relative permittivity of ɛ R 80 whereas at optical frequencies, the value is about 2. End of Aside 1.] [Aside 2. The idea of charge density is obviously very important in e-m. For line charges, Q = ρ L dl where ρ L is the linear charge density in C/m. L For surface charges, Q = ρ S ds where ρ s is the surface charge density in C/m 2. S Volume charge density was encountered in an example given earlier in this section. End of Aside 2.] The generalization of Faraday s experiments led to the following formalization known as Gauss s Law: The electric flux passing through any closed surface is equal to the total free charge enclosed by that surface. In general, the closed surface may take any form we wish to visualize some will obviously be more convenient to use in calculations than others. Illustration: (in terms of the vector differential area, ds). Since the differential flux, dψ, crossing the differential area must be the product of the normal component of D and the differential surface d S: dψ = D d S = D d S cos θ Therefore, Ψ = dψ = S D d S = Q Gauss s Law (1.16) 36

37 since Ψ = Q where Q is the charge enclosed by S. Thus, too, S D ds = ρ v dv = Q (1.17) V Of course, line or surface integrals may replace the volume integral as circumstances warrant. Facilitating appplication of Gauss s law is dependent on a suitable choice of the closed surface for integration. In fact, if the charge distribution is known, equation (1.17) can be used to obtain D in an easy manner if it is possible to choose a closed surface, S, which satisfies the following two properties: (1) D is everywhere either normal or tangential to S so that D ds = DdS or 0, respectively. AND (2) On that portion of S where D ds 0, the magnitude, D = D is constant. Example: An infinite uniform line charge with linear density ρ L lies along the z-axis. Determine D and E at a distance ρ from the z-axis (see p. 37a). 37

38 Application 2. Electric Flux Density and Gauss s Law (Magnetic) Before using the surface integrals in a magnetostatics application, we need to consider a few related details on magnetic flux density. In addition to the constitutive relationship, D = ɛe, relating electric flux density, D, to the electric field intensity, E, there is another such relationship which relates the magnetic flux density, B, to the magnetic field intensity, H. B is defined by B = µ H where µ in henrys per metre (H/m) is called the permeability of the material and, for free space, we write µ 0 = 4π 10 7 H/m. B is measured in webers/metre 2 (Wb/m 2 or tesla, T) and since H is measured in A/m, the weber is the product of henrys and amperes. A discussion of the value of µ for non-free-space parallels that which we briefly considered for ɛ in the previous class handout. Again without discussing the theoretical detail, we observe that orbiting and spinning electrons within a material may be thought of as current loops which produce B-fields without external excitation. When an external field is applied to the material, the interactions between it and the internal structure may cause the magnetic flux density, B, to increase or decrease from its free-space value. This is reflected in a value of µ which is either greater or smaller, respectively, than µ 0. The B-field is changed from what it would be in free space according to the expression B = µ 0 ( H + M) where M, called the magnetization or magnetic-moment density, is a measure of how the material reacts internally in a magnetic sense when an external H-field is applied. A dimensionless constant, χ m, called the magnetic susceptibility is defined such that M = χ mh 38

39 and B = µ 0 (1 + χ m ) H ; i.e. µ = µ 0 (1 + χ m ) and a relative permeability, µ R, may be defined as µ R = µ µ 0 = (1 + χ m ). Finally, we note that for magnetically anisotropic materials, the constitutive relationship B = µ H will be a matrix equation. (In this course, we will consider only magnetically isotropic materials so that µ will be an ordinary scalar). We are now in a position to discuss Gauss s law as it applies to magnetic fields. Unlike electric flux lines which terminate on negative charges and begin on positive charges, no such sources or sinks have been discovered for lines of magnetic flux. that is, no single magnetic charges or poles have ever been isolated. For this reason, Gauss s law for magnetic flux density B, which is analogous to equation (1.16) or equation (1.17) becomes S B d S = 0 (1.18) Equation (1.18) indicates that the magnetic charge contained within a closed surface is zero. If the integral is not closed S B d S = Φ (1.19) where Φ is the magnetic flux, measured in webers, linking the surface. Example: Find the magnetic flux, Φ, in a region of length d between the conductors of a coaxial cable if each conductor carries a current I. Take the radii of the inner and outer conductor to be a and b, respectively. 39

40 Summary Electric and Magnetic Field Results (Static) From Sections 1-3 and 1-4 From our application of line, surface and volume integrals, we have developed the following equations which govern static electric and (steady) magnetic fields. These are the non-time-varying integral forms of the famous, and essential, Maxwell Equations. We shall shortly find other forms of these equations using the so-called differential operators. Conservative E field: Ampère s Law: Gauss s Law (Electric): Gauss s Law (Magnetic): E dl = 0 (1.20) C H d L = I e = J ds (1.21) C S D d S = ρ v dv = Q (1.22) S v B ds = 0 (1.23) S The right hand side of equation (1.21) is a generalization of J A when J varies over the area, S, carrying the current and is not necessarily perpendicular to that area as illustrated. In addition to MAxwell s equations, we have two constitutive relationships: D = ɛ E (1.24) B = µ H (1.25) 40

41 1.5 Vector Differential Operators In the study of electromagnetics, as in many other physical sciences, the concept of vector operators provides a convenient means of writing in concise form what could otherwise be tedious and lengthy formulations of certain important physical relationships. In this section, we wish to (1) define, (2) interpret and (3) apply such operators in a variety of contexts The Gradient of a Scalar Definition: Consider a scalar point function ϕ(x, y, z). The del operator, symbolized, applied to this function is defined in cartesian coordinates as i.e. ϕ = ˆx ϕ x + ŷ ϕ y + ẑ ϕ z ; (1.26) ˆx x + ŷ y + ẑ z. (1.27) The expression ϕ is referred to as the gradient of the scalar function ϕ. Thus, the application of (1.27) to a scalar function results in a vector. At this point, we issue a caution that the del operator may be treated as a vector in the gradient (and other forms to follow) ONLY in cartesian coordinates. In the other coordinate systems we have studied, this idea will require modification. Illustration: If f(x, y, z) = x 2 yz, f = (2xyz)ˆx + (x 2 z)ŷ + (x 2 y)ẑ. Interpretation Initially, let us consider the dot product ϕ d r where d r = dx ˆx + dy ŷ + dz ẑ. Clearly, ϕ d r = ϕ x dx + ϕ y 41 ϕ dy + dz = dϕ (1.28) z

42 by definition (i.e. the dot product in (1.28) furnishes the change in the scalar function, ϕ, corresponding to a change in position d r). Now the expression, ϕ(x, y, z) = C where C is a constant describes some (arbitrary) surface in space. Consider two points, P and Q, separated by a displacement d r on such a surface. Then, moving from P to Q, the change in ϕ(x, y, z) = C is given, using (1.28), by dϕ = ( ϕ) d r = 0, (1.29) where the zero results because we stay on the constant surface. Equation (1.29) shows that ϕ d r. This is true for any direction of d r from P to Q as long as the two points are restricted to the surface. Thus, ϕ is seen to be normal to the surface ϕ = constant (A VERY IMPORTANT RESULT). Next, as illustrated, let us permit d r to extend from one surface ϕ = C 1 to an adjacent surface ϕ = C 2 so that invoking equation (1.28) dϕ = C 2 C 1 = C = ( ϕ) d r. (1.30) Now, for a given dϕ, d r is a minimum when it is chosen parallel to ϕ (since then cos θ = 1); or, for a given d r, the change in the scalar function ϕ is maximized by choosing d r parallel to ϕ. This identifies ϕ as a vector having the direction of the maximum space rate of change of ϕ. 42

43 Application: We have seen that in a conservative E field, the potential V at position P (x, y, z) is unique i.e. it is independent of the path of integration. We had that P V P = E d L. Because V P is a single-valued function, V (x, y, z), we may differentiate the latter equation to give or, equivalently, dv dl = E cos θ dv = E d L = E d r. (1.31) Comparing equations (1.30) and (1.31) immediately allows us to write that E = V (1.32) That is, the magnitude of E is given by the maximum space rate of change of V, and, based on the previous discussion, the direction of E is normal to the equipotential surface (a surface where V is constant) in the direction of decreasing potential (as indicated by the the minus sign). Equation (1.32) reads The electric field intensity, E, is the negative of the gradient of the potential, V. We now have a powerful way of determining the electric field once the potential field is known. GRADIENT IN CYLINDRICAL COORDINATES For a scalar function ϕ(ρ, φ, z) in cylindrical coordinates, dϕ = ϕ ρ ϕ ϕ dρ + dφ + φ z dz and d r = dρˆρ + ρdφ ˆφ + dzẑ. With a view to equation (1.28), we write dϕ = = ( ϕ ρ ˆρ + 1 ϕ ρ φ ˆφ + ϕ ) z ẑ dρˆρ + ρdφ ˆφ + dzẑ ( ϕ ρ ˆρ + 1 ϕ ρ φ ˆφ + ϕ ) z ẑ d r. 43

44 Comparing this last expression with (1.28), we identify the gradient in cylindrical coordinates as ϕ = ( ϕ ρ ˆρ + 1 ϕ ρ φ ˆφ + ϕ ) z ẑ GRADIENT IN SPHERICAL COORDINATES (1.33) A procedure completely analogous to that which led to equation (1.33) may be used when the scalar function is given in spherical coordinates as ϕ(r, θ, φ). The result for the gradient is ϕ = ( ϕ r ˆr + 1 ϕ r θ ˆθ + 1 ) ϕ r sin θ φ ˆφ (1.34) The Divergence of a Vector The del operator,, is also used to operate on vector functions, say A. We shall first consider the dot product, A. Definition: The divergence of a vector function A (i.e. A(x, y, z) in cartesian coordinates) is defined as Therefore, A = ( ˆx x + ŷ y + ẑ ) (A xˆx + A y ŷ + A z ẑ). z A = A x x + A y y + A z z (1.35) Notice that the divergence of a vector is a scalar! As with the gradient (1.35) will not have such a simple form in cylindrical and spherical coordinates. Illustration: E = (x 2 yz)ˆx + (xyz 2 )ŷ + xẑ. The divergence of E ( sometimes written, div E), is E = 2xyz + xz 2. 44

45 Interpretation For the sake of simplicity, consider a vector function, A, that has only an ˆx component. From (1.35), A = A x x. With reference to the figure below and noting that the rectangular block is of differential volume, dv = lim x y z = lim v, the definition of partial ( ) x, y, z 0 v 0 differentiation gives A x x = lim A xright face A xleft face x 0 x or A x x = lim A xright face y z A xleft face y z v 0 v. (1.36) We next observe that lim y z = d S x. y, z 0 In (1.36), the first term in the numerator is the flow out while the second is the flow into the block through ds x. Therefore, (1.36) may be written as A x x = lim v 0 S x A d Sx v (1.37) while recalling that d S x points away from the block at both the right and left faces. Making identical arguments for Ay y Az and z, (1.37) may be generalized to include all 45

46 surfaces by writing A x x + A y y + A z z = lim v 0 S A d S v (1.38) where S is now the whole surface of the block. Thus, A = lim v 0 S A d S v (1.39) and the divergence of a vector field A represents the net outflow of the vector A per unit volume. For example, A could be the electric flux density, D, so that D would be the net flux per unit volume leaving a particular region. Hence the term divergence. A > 0 if there is a net outflow (i.e. a source region) and A < 0 if there is a net inflow (i.e. a sink region) DIVERGENCE IN CYLINDRICAL COORDINATES Starting with equation (1.39) and working backwards in cylindrical coordinates, it is not too difficult to show that the expression for divergence in cylindrical coordinates is: A = 1 (ρa ρ ) + 1 A φ ρ ρ ρ φ + A z z (1.40) DIVERGENCE IN SPHERICAL COORDINATES A similar procedure in spherical coordinates leads to A = 1 r 2 (r 2 A r ) r + 1 (A θ sin θ) + 1 A φ r sin θ θ r sin θ φ (1.41) 46

47 APPLICATIONS (1) Gauss Law We have seen the integral form of Gauss Law as S D ds = vol ρ vdv = Q (1.42) where Q is the charged enclosed by the surface, S. Now from equation (1.39) D S D = lim ds v 0 v Q = lim v 0 v (1.43) where we have indicated the volume, v, to be shrinking to infinitesimal proportions. In this case, lim v 0 Q v = ρ v with ρ v being the volume charge density. Therefore, equation (1.43) becomes D = ρ v (1.44) Equation (1.44) is referred to as the point form of Gauss Law, so called because the volume is becoming vanishingly small i.e. shrinking to a point. The magnetic Gauss Law (see equation (1.23)) in point form clearly becomes B = 0 (1.45) Example: Given D = ρz 2 sin 2 φ ˆρ + ρz 2 sin φ cos φ ˆφ + ρ 2 z sin 2 φ ẑ C/m 2, determine the volume charge density, ρ v, in the region where D exists. 47

48 (2.) Gauss Theorem or the Divergence Theorem Now, from equation (1.44), we may carry out the volume integration D dv = vol vol ρ v dv, which, on comparison with equation (1.42), gives the DIVERGENCE THEOREM D d S = D dv (1.46) S vol Equation (1.46) is true of any vector field, A, i.e. A d S = A dv, and we S vol see that the divergence theorem changes a surface integral to a volume integral. In words, Gauss theorem or the divergence theorem says that the integral of the normal component of a vector field over a closed surface is equal to the integral of the divergence of this vector field throughout the volume enclosed by that surface. (3.) Conservation of Charge The principle of conservation of charge states that charges can neither be created nor destroyed. Now, considering a region bounded by a closed surface, the current through that surface is I = S J d S. Recall that J is the current density while I is the current flowing outward through the surface, S. The rate of change of the charge, Q, inside the surface is dq dt ; i.e. or S I = dq dt J d S = dq dt, (1.47) which is the equation of conservation of charge (or continuity of current) in integral form. It implies that an outward flow of positive charge through the closed surface 48

49 must be balanced by a decrease of positive charge (or an increase in negative charge) within the region enclosed by that surface. Using the divergence theorem and the fact that Q = vol ρ v dv we get, from equation (1.47), J dv = d vol dt vol ρ v dv. d Then, if the surface around the volume is constant, may be replaced with a partial dt derivative to give J dv = vol vol ρ v t dv. Since the equation is true for any volume, a point form of the conservation of charge may be written as J = ρ v t. or J + ρ v t = 0 (1.48) In summary, equation (1.48) implies that the current, or charge per unit time, diverging from a small volume per unit volume is equal to the time rate of decrease of charge per unit volume at every point. (4.) Poisson s and Laplace s Equations Gauss law, in point form, says that D = ρ v which implies E = ρ v ɛ since D = ɛ E. However, as an example of the gradient of a scalar quantity we saw that E = V (V potential). 49