Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Willian Hayt wkh tusgkwsi f asjv gsgsd gasff, Study Guides, Projects, Research of Electromagnetic Engineering

Xyzedfbajeg wgsg sdgtgsr kgabfjkagb abfjkkbadjgus ajhfuafjk ahfaefbkafkafbuffafgafvnsafgsd shhsgf shgfshfaffhsfdshvfsgfsd.asfsahfd

Typology: Study Guides, Projects, Research

2016/2017

Uploaded on 12/09/2017

nisarg-patel
nisarg-patel 🇮🇳

1 document

1 / 26

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
CHAPTER
1
VECTOR
ANALYSIS
Vector analysis is a mathematical subject which is much better taught by math-
ematicians than by engineers. Most junior and senior engineering students, how-
ever, have not had the time (or perhaps the inclination) to take a course in vector
analysis, although it is likely that many elementary vector concepts and opera-
tions were introduced in the calculus sequence. These fundamental concepts and
operations are covered in this chapter, and the time devoted to them now should
depend on past exposure.
The viewpoint here is also that of the engineer or physicist and not that of
the mathematician in that proofs are indicated rather than rigorously expounded
and the physical interpretation is stressed. It is easier for engineers to take a more
rigorous and complete course in the mathematics department after they have
been presented with a few physical pictures and applications.
It is possible to study electricity and magnetism without the use of vector
analysis, and some engineering students may have done so in a previous electrical
engineering or basic physics course. Carrying this elementary work a bit further,
however, soon leads to line-filling equations often composed of terms which all
look about the same. A quick glance at one of these long equations discloses little
of the physical nature of the equation and may even lead to slighting an old
friend.
Vector analysis is a mathematical shorthand. It has some new symbols,
some new rules, and a pitfall here and there like most new fields, and it demands
concentration, attention, and practice. The drill problems, first met at the end of
Sec. 1.4, should be considered an integral part of the text and should all be
1
| | | |
e-Text Main Menu
Textbook Table of Contents
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a

Partial preview of the text

Download Willian Hayt wkh tusgkwsi f asjv gsgsd gasff and more Study Guides, Projects, Research Electromagnetic Engineering in PDF only on Docsity!

CHAPTER

VECTOR

ANALYSIS

Vector analysis is a mathematical subject which is much better taught by math- ematicians than by engineers. Most junior and senior engineering students, how- ever, have not had the time (or perhaps the inclination) to take a course in vector analysis, although it is likely that many elementary vector concepts and opera- tions were introduced in the calculus sequence. These fundamental concepts and operations are covered in this chapter, and the time devoted to them now should depend on past exposure. The viewpoint here is also that of the engineer or physicist and not that of the mathematician in that proofs are indicated rather than rigorously expounded and the physical interpretation is stressed. It is easier for engineers to take a more rigorous and complete course in the mathematics department after they have been presented with a few physical pictures and applications. It is possible to study electricity and magnetism without the use of vector analysis, and some engineering students may have done so in a previous electrical engineering or basic physics course. Carrying this elementary work a bit further, however, soon leads to line-filling equations often composed of terms which all look about the same. A quick glance at one of these long equations discloses little of the physical nature of the equation and may even lead to slighting an old friend. Vector analysis is a mathematical shorthand. It has some new symbols, some new rules, and a pitfall here and there like most new fields, and it demands concentration, attention, and practice. The drill problems, first met at the end of Sec. 1.4, should be considered an integral part of the text and should all be

1

s s

worked. They should not prove to be difficult if the material in the accompany- ing section of the text has been thoroughly understood. It take a little longer to ``read'' the chapter this way, but the investment in time will produce a surprising interest.

1.1 SCALARS AND VECTORS

The term scalar refers to a quantity whose value may be represented by a single (positive or negative) real number. The x; y, and z we used in basic algebra are scalars, and the quantities they represent are scalars. If we speak of a body falling a distance L in a time t, or the temperature T at any point in a bowl of soup whose coordinates are x; y, and z, then L; t; T; x; y, and z are all scalars. Other scalar quantities are mass, density, pressure (but not force), volume, and volume resistivity. Voltage is also a scalar quantity, although the complex representation of a sinusoidal voltage, an artificial procedure, produces a complex scalar, or phasor, which requires two real numbers for its representation, such as amplitude and phase angle, or real part and imaginary part. A vector quantity has both a magnitude^1 and a direction in space. We shall be concerned with two- and three-dimensional spaces only, but vectors may be defined in n-dimensional space in more advanced applications. Force, velocity, acceleration, and a straight line from the positive to the negative terminal of a storage battery are examples of vectors. Each quantity is characterized by both a magnitude and a direction. We shall be mostly concerned with scalar and vector fields. A field (scalar or vector) may be defined mathematically as some function of that vector which connects an arbitrary origin to a general point in space. We usually find it possible to associate some physical effect with a field, such as the force on a compass needle in the earth's magnetic field, or the movement of smoke particles in the field defined by the vector velocity of air in some region of space. Note that the field concept invariably is related to a region. Some quantity is defined at every point in a region. Both scalar fields and vector fields exist. The temperature throughout the bowl of soup and the density at any point in the earth are examples of scalar fields. The gravitational and magnetic fields of the earth, the voltage gradient in a cable, and the temperature gradient in a soldering- iron tip are examples of vector fields. The value of a field varies in general with both position and time. In this book, as in most others using vector notation, vectors will be indi- cated by boldface type, for example, A. Scalars are printed in italic type, for example, A. When writing longhand or using a typewriter, it is customary to draw a line or an arrow over a vector quantity to show its vector character. (CAUTION: This is the first pitfall. Sloppy notation, such as the omission of the line or arrow symbol for a vector, is the major cause of errors in vector analysis.)

2 ENGINEERING ELECTROMAGNETICS

(^1) We adopt the convention that magnitude'' infersabsolute value''; the magnitude of any quantity is therefore always positive.

s s

tion when multiplied by a negative scalar. Multiplication of a vector by a scalar also obeys the associative and distributive laws of algebra, leading to …r ‡ s†…A ‡ B† ˆ r…A ‡ B† ‡ s…A ‡ B† ˆ rA ‡ rB ‡ sA ‡ sB Division of a vector by a scalar is merely multiplication by the reciprocal of that scalar. The multiplication of a vector by a vector is discussed in Secs. 1.6 and 1.7. Two vectors are said to be equal if their difference is zero, or A ˆ B if A ÿ B ˆ 0. In our use of vector fields we shall always add and subtract vectors which are defined at the same point. For example, the total magnetic field about a small horseshoe magnet will be shown to be the sum of the fields produced by the earth and the permanent magnet; the total field at any point is the sum of the indivi- dual fields at that point. If we are not considering a vector field, however, we may add or subtract vectors which are not defined at the same point. For example, the sum of the gravitational force acting on a 150-lbf (pound-force) man at the North Pole and that acting on a 175-lbf man at the South Pole may be obtained by shifting each force vector to the South Pole before addition. The resultant is a force of 25 lbf directed toward the center of the earth at the South Pole; if we wanted to be difficult, we could just as well describe the force as 25 lbf directed away from the center of the earth (or ``upward'') at the North Pole.^2

1.3 THE CARTESIAN COORDINATE SYSTEM

In order to describe a vector accurately, some specific lengths, directions, angles, projections, or components must be given. There are three simple methods of doing this, and about eight or ten other methods which are useful in very special cases. We are going to use only the three simple methods, and the simplest of these is the cartesian, or rectangular, coordinate system.

4 ENGINEERING ELECTROMAGNETICS

FIGURE 1. Two vectors may be added graphically either by drawing both vectors from a common origin and completing the parallelogram or by beginning the second vector from the head of the first and completing the triangle; either method is easily extended to three or more vectors.

(^2) A few students have argued that the force might be described at the equator as being in a ``northerly'' direction. They are right, but enough is enough.

s s

In the cartesian coordinate system we set up three coordinate axes mutually at right angles to each other, and call them the x; y, and z axes. It is customary to choose a right-handed coordinate system, in which a rotation (through the smal- ler angle) of the x axis into the y axis would cause a right-handed screw to progress in the direction of the z axis. If the right hand is used, then the thumb, forefinger, and middle finger may then be identified, respectively, as the x; y, and z axes. Fig. 1.2a shows a right-handed cartesian coordinate system. A point is located by giving its x; y, and z coordinates. These are, respec- tively, the distances from the origin to the intersection of a perpendicular dropped from the point to the x; y, and z axes. An alternative method of inter- preting coordinate values, and a method corresponding to that which must be used in all other coordinate systems, is to consider the point as being at the

VECTOR ANALYSIS 5

FIGURE 1. (a) A right-handed cartesian coordinate system. If the curved fingers of the right hand indicate the direction through which the x axis is turned into coincidence with the y axis, the thumb shows the direction of the z axis. (b) The location of points P… 1 ; 2 ; 3 † and Q… 2 ; ÿ 2 ; 1 †. (c) The differential volume element in cartesian coordinates; dx, dy, and dz are, in general, independent differentials.

s s

If the component vector y happens to be two units in magnitude and directed toward increasing values of y, we should then write y ˆ 2 ay. A vector rP pointing from the origin to point P… 1 ; 2 ; 3 † is written rP ˆ ax ‡ 2 ay ‡ 3 az. The vector from P to Q may be obtained by applying the rule of vector addition. This rule shows that the vector from the origin to P plus the vector from P to Q is equal to the vector from the origin to Q. The desired vector from P… 1 ; 2 ; 3 † to Q… 2 ; ÿ 2 ; 1 † is therefore

RPQ ˆ rQ ÿ rP ˆ … 2 ÿ 1 †ax ‡ …ÿ 2 ÿ 2 †ay ‡ … 1 ÿ 3 †az ˆ ax ÿ 4 ay ÿ 2 az

The vectors rP; rQ, and RPQ are shown in Fig. 1.3c.

VECTOR ANALYSIS 7

FIGURE 1. (a) The component vectors x, y, and z of vector r. (b) The unit vectors of the cartesian coordinate system have unit magnitude and are directed toward increasing values of their respective variables. (c) The vector RPQ is equal to the vector difference rQ ÿ rP:

s s

This last vector does not extend outward from the origin, as did the vector r we initially considered. However, we have already learned that vectors having the same magnitude and pointing in the same direction are equal, so we see that to help our visualization processes we are at liberty to slide any vector over to the origin before determining its component vectors. Parallelism must, of course, be maintained during the sliding process. If we are discussing a force vector F, or indeed any vector other than a displacement-type vector such as r, the problem arises of providing suitable letters for the three component vectors. It would not do to call them x; y, and z, for these are displacements, or directed distances, and are measured in meters (abbreviated m) or some other unit of length. The problem is most often avoided by using component scalars, simply called components, Fx; Fy, and Fz. The com- ponents are the signed magnitudes of the component vectors. We may then write F ˆ Fxax ‡ Fyay ‡ Fzaz. The component vectors are Fxax, Fyay, and Fzaz: Any vector B then may be described by B ˆ Bxax ‡ Byay ‡ Bzaz. The mag- nitude of B written jBj or simply B, is given by

jBj ˆ

B^2 x ‡ B^2 y ‡ B^2 z

q … 1 †

Each of the three coordinate systems we discuss will have its three funda- mental and mutually perpendicular unit vectors which are used to resolve any vector into its component vectors. However, unit vectors are not limited to this application. It is often helpful to be able to write a unit vector having a specified direction. This is simply done, for a unit vector in a given direction is merely a vector in that direction divided by its magnitude. A unit vector in the r direction is r=

x^2 ‡ y^2 ‡ z^2

p , and a unit vector in the direction of the vector B is

aB ˆ

B

B^2 x ‡ B^2 y ‡ B^2 z

q (^) ˆ

B

jBj

h Example 1. Specify the unit vector extending from the origin toward the point G… 2 ; ÿ 2 ; ÿ 1 †.

Solution. We first construct the vector extending from the origin to point G, G ˆ 2 ax ÿ 2 ay ÿ az We continue by finding the magnitude of G,

jGj ˆ

•••••••••••••••••••••••••••••••••••••••••• … 2 †^2 ‡ …ÿ 2 †^2 ‡ …ÿ 1 †^2

q ˆ 3

8 ENGINEERING ELECTROMAGNETICS

s s

ingly more complex and difficult to interpret. We shall come across many fields in our study of electricity and magnetism which are simpler than the velocity example, an example in which only the component and one variable were involved (the x component and the variable z). We shall also study more com- plicated fields, and methods of interpreting these expressions physically will be discussed then.

\ D1.2. A vector field S is expressed in cartesian coordinates as S ˆ f 125 =‰…x ÿ 1 †^2 ‡ …y ÿ 2 †^2 ‡ …z ‡ 1 †^2 Šgf…x ÿ 1 †ax ‡ …y ÿ 2 †ay ‡ …z ‡ 1 †azg. (a) Evaluate S at P… 2 ; 4 ; 3 †. (b) Determine a unit vector that gives the direction of S at P. (c) Specify the surface f …x; y; z† on which jSj ˆ 1 :

Ans. 5: 95 ax ‡ 11 : 90 ay ‡ 23 : 8 az; 0: 218 ax ‡ 0 : 436 ay ‡ 0 : 873 az; ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• …x ÿ 1 †^2 ‡ …y ÿ 2 †^2 ‡ …z ‡ 1 †^2

q ˆ 125

1.6 THE DOT PRODUCT

We now consider the first of two types of vector multiplication. The second type will be discussed in the following section. Given two vectors A and B, the dot product, or scalar product, is defined as the product of the magnitude of A, the magnitude of B, and the cosine of the smaller angle between them,

A  B ˆ jAj jBj cos AB … 3 †

The dot appears between the two vectors and should be made heavy for empha- sis. The dot, or scalar, product is a scalar, as one of the names implies, and it obeys the commutative law,

A  B ˆ B  A … 4 †

for the sign of the angle does not affect the cosine term. The expression A  B is read ``A dot B.'' Perhaps the most common application of the dot product is in mechanics, where a constant force F applied over a straight displacement L does an amount of work FL cos , which is more easily written F  L. We might anticipate one of the results of Chap. 4 by pointing out that if the force varies along the path, integration is necessary to find the total work, and the result becomes

Work ˆ

F  dL

Another example might be taken from magnetic fields, a subject about which we shall have a lot more to say later. The total flux  crossing a surface

10 ENGINEERING ELECTROMAGNETICS

s s

of area S is given by BS if the magnetic flux density B is perpendicular to the surface and uniform over it. We define a vector surface S as having the usual area for its magnitude and having a direction normal to the surface (avoiding for the moment the problem of which of the two possible normals to take). The flux crossing the surface is then B  S. This expression is valid for any direction of the uniform magnetic flux density. However, if the flux density is not constant over the surface, the total flux is  ˆ

B  dS. Integrals of this general form appear in Chap. 3 when we study electric flux density. Finding the angle between two vectors in three-dimensional space is often a job we would prefer to avoid, and for that reason the definition of the dot product is usually not used in its basic form. A more helpful result is obtained by considering two vectors whose cartesian components are given, such as A ˆ Axax ‡ Ayay ‡ Azaz and B ˆ Bxax ‡ Byay ‡ Bzaz. The dot product also obeys the distributive law, and, therefore, A  B yields the sum of nine scalar terms, each involving the dot product of two unit vectors. Since the angle between two different unit vectors of the cartesian coordinate system is 90 8 , we then have

ax  ay ˆ ay  ax ˆ ax  az ˆ az  ax ˆ ay  az ˆ az  ay ˆ 0

The remaining three terms involve the dot product of a unit vector with itself, which is unity, giving finally

A  B ˆ AxBx ‡ AyBy ‡ AzBz … 5 †

which is an expression involving no angles. A vector dotted with itself yields the magnitude squared, or

A  A ˆ A^2 ˆ jAj^2 … 6 †

and any unit vector dotted with itself is unity,

aA  aA ˆ 1 One of the most important applications of the dot product is that of finding the component of a vector in a given direction. Referring to Fig. 1.4a, we can obtain the component (scalar) of B in the direction specified by the unit vector a as

B  a ˆ jBj jaj cos Ba ˆ jBj cos Ba

The sign of the component is positive if 0  Ba  908 and negative whenever 908  Ba  1808 : In order to obtain the component vector of B in the direction of a, we simply multiply the component (scalar) by a, as illustrated by Fig. 1.4b. For example, the component of B in the direction of ax is B  ax ˆ Bx, and the

VECTOR ANALYSIS 11

s s

\ D1.3. The three vertices of a triangle are located at A… 6 ; ÿ 1 ; 2 †, B…ÿ 2 ; 3 ; ÿ 4 †, and C…ÿ 3 ; 1 ; 5 †. Find: (a) RAB; (b) RAC ; (c) the angle BAC at vertex A; (d) the (vector) projection of RAB on RAC :

Ans. ÿ 8 ax ‡ 4 ay ÿ 6 az; ÿ 9 ax ÿ 2 ay ‡ 3 az; 53: 68 ; ÿ 5 : 94 ax ‡ 1 : 319 ay ‡ 1 : 979 az

1.7 THE CROSS PRODUCT

Given two vectors A and B, we shall now define the cross product, or vector product, of A and B, written with a cross between the two vectors as A  B and read A cross B.'' The cross product A  B is a vector; the magnitude of A  B is equal to the product of the magnitudes of A; B, and the sine of the smaller angle between A and B; the direction of A  B is perpendicular to the plane containing A and B and is along that one of the two possible perpendiculars which is in the direction of advance of a right-handed screw as A is turned into B. This direction is illustrated in Fig. 1.5. Remember that either vector may be moved about at will, maintaining its direction constant, until the two vectors have acommon origin.'' This determines the plane containing both. However, in most of our applications we shall be concerned with vectors defined at the same point. As an equation we can write

A  B ˆ aN jAj jBj sin AB … 7 †

where an additional statement, such as that given above, is still required to explain the direction of the unit vector aN. The subscript stands for ``normal.'' Reversing the order of the vectors A and B results in a unit vector in the opposite direction, and we see that the cross product is not commutative, for B  A ˆ ÿ…A  B†. If the definition of the cross product is applied to the unit

VECTOR ANALYSIS 13

FIGURE 1. The direction of A  B is in the direction of advance of a right-handed screw as A is turned into B:

s s

14 ENGINEERING ELECTROMAGNETICS

vectors ax and ay, we find ax  ay ˆ az, for each vector has unit magnitude, the two vectors are perpendicular, and the rotation of ax into ay indicates the posi- tive z direction by the definition of a right-handed coordinate system. In a similar way ay  az ˆ ax, and az  ax ˆ ay. Note the alphabetic symmetry. As long as the three vectors ax, ay, and az are written in order (and assuming that ax follows az, like three elephants in a circle holding tails, so that we could also write ay, az, ax or az, ax, ay), then the cross and equal sign may be placed in either of the two vacant spaces. As a matter of fact, it is now simpler to define a right-handed cartesian coordinate system by saying that ax  ay ˆ az: A simple example of the use of the cross product may be taken from geometry or trigonometry. To find the area of a parallelogram, the product of the lengths of two adjacent sides is multiplied by the sine of the angle between them. Using vector notation for the two sides, we then may express the (scalar) area as the magnitude of A  B, or jA  Bj: The cross product may be used to replace the right-hand rule familiar to all electrical engineers. Consider the force on a straight conductor of length L, where the direction assigned to L corresponds to the direction of the steady current I, and a uniform magnetic field of flux density B is present. Using vector notation, we may write the result neatly as F ˆ IL  B. This relationship will be obtained later in Chap. 9. The evaluation of a cross product by means of its definition turns out to be more work than the evaluation of the dot product from its definition, for not only must we find the angle between the vectors, but we must find an expression for the unit vector aN. This work may be avoided by using cartesian components for the two vectors A and B and expanding the cross product as a sum of nine simpler cross products, each involving two unit vectors,

A  B ˆ AxBxax  ax ‡ AxByax  ay ‡ AxBzax  az ‡ AyBxay  ax ‡ AyByay  ay ‡ AyBzay  az ‡ AzBxaz  ax ‡ AzByaz  ay ‡ AzBzaz  az

We have already found that ax  ay ˆ az, ay  az ˆ ax, and az  ax ˆ ay. The three remaining terms are zero, for the cross product of any vector with itself is zero, since the included angle is zero. These results may be combined to give

A  B ˆ …AyBz ÿ AzBy†ax ‡ …AzBx ÿ AxBz†ay ‡ …AxBy ÿ AyBx†az … 8 †

or written as a determinant in a more easily remembered form,

A  B ˆ

ax ay az Ax Ay Az Bx By Bz

Thus, if A ˆ 2 ax ÿ 3 ay ‡ az and B ˆ ÿ 4 ax ÿ 2 ay ‡ 5 az, we have

s s

another plane (z ˆ constant). This corresponds to the location of a point in a cartesian coordinate system by the intersection of three planes (x ˆ constant, y ˆ constant, and z ˆ constant). The three surfaces of circular cylindrical coordi- nates are shown in Fig. 1.6a. Note that three such surfaces may be passed through any point, unless it lies on the z axis, in which case one plane suffices. Three unit vectors must also be defined, but we may no longer direct them along the ``coordinate axes,'' for such axes exist only in cartesian coordinates. Instead, we take a broader view of the unit vectors in cartesian coordinates and realize that they are directed toward increasing coordinate values and are per- pendicular to the surface on which that coordinate value is constant (i.e., the unit vector ax is normal to the plane x ˆ constant and points toward larger values of x). In a corresponding way we may now define three unit vectors in cylindrical coordinates, a; a, and az:

16 ENGINEERING ELECTROMAGNETICS

FIGURE 1. (aa) The three mutually perpendicular surfaces of the circular cylindrical coordinate system. (b) The three unit vectors of the circular cylindrical coordinate system. (c) The differential volume unit in the circular cylindrical coordinate system; d, d, and dz are all elements of length.

s s

The unit vector a at a point P… 1 ;  1 ; z 1 † is directed radially outward, normal to the cylindrical surface  ˆ  1. It lies in the planes  ˆ  1 and z ˆ z 1. The unit vector a is normal to the plane  ˆ  1 , points in the direction of increasing , lies in the plane z ˆ z 1 , and is tangent to the cylindrical surface  ˆ  1. The unit vector az is the same as the unit vector az of the cartesian coordinate system. Fig. 1.6b shows the three vectors in cylindrical coordinates. In cartesian coordinates, the unit vectors are not functions of the coordi- nates. Two of the unit vectors in cylindrical coordinates, a and a, however, do vary with the coordinate , since their directions change. In integration or dif- ferentiation with respect to , then, a and a must not be treated as constants. The unit vectors are again mutually perpendicular, for each is normal to one of the three mutually perpendicular surfaces, and we may define a right- handed cylindrical coordinate system as one in which a  a ˆ az, or (for those who have flexible fingers) as one in which the thumb, forefinger, and middle finger point in the direction of increasing ; , and z, respectively. A differential volume element in cylindrical coordinates may be obtained by increasing ; , and z by the differential increments d; d, and dz. The two cylinders of radius  and  ‡ d, the two radial planes at angles  and  ‡ d, and the two horizontal'' planes atelevations'' z and z ‡ dz now enclose a small volume, as shown in Fig. 1.6c, having the shape of a truncated wedge. As the volume element becomes very small, its shape approaches that of a rectangular parallelepiped having sides of length d; d and dz. Note that d and dz are dimensionally lengths, but d is not; d is the length. The surfaces have areas of  d d, d dz, and  d dz, and the volume becomes  d d dz: The variables of the rectangular and cylindrical coordinate systems are easily related to each other. With reference to Fig. 1.7, we see that

x ˆ  cos  y ˆ  sin  z ˆ z

VECTOR ANALYSIS 17

FIGURE 1. The relationship between the cartesian variables x; y; z and the cylindrical coordinate variables ; ; z. There is no change in the variable z between the two systems.

s s

a as , and thus ax  a ˆ cos , but the angle between ay and a is 90 8 ÿ , and ay  a ˆ cos … 908 ÿ † ˆ sin . The remaining dot products of the unit vectors are found in a similar manner, and the results are tabulated as functions of  in Table 1. Transforming vectors from cartesian to cylindrical coordinates or vice versa is therefore accomplished by using (10) or (11) to change variables, and by using the dot products of the unit vectors given in Table 1.1 to change components. The two steps may be taken in either order.

h Example 1. Transform the vector B ˆ yax ÿ xay ‡ zaz into cylindrical coordinates.

Solution. The new components are B ˆ B  a ˆ y…ax  a† ÿ x…ay  a† ˆ y cos  ÿ x sin  ˆ  sin  cos  ÿ  cos  sin  ˆ 0 B ˆ B  a ˆ y…ax  a† ÿ x…ay  a† ˆ ÿy sin  ÿ x cos  ˆ ÿ sin^2  ÿ  cos^2  ˆ ÿ Thus, B ˆ ÿa ‡ zaz

\ D1.5. (a) Give the cartesian coordinates of the point C… ˆ 4 : 4 ;  ˆ ÿ 1158 ; z ˆ 2 †. (b) Give the cylindrical coordinates of the point D…x ˆ ÿ 3 : 1 ; y ˆ 2 : 6 ; z ˆ ÿ 3 †. (c) Specify the distance from C to D:

Ans. C…x ˆ ÿ 1 : 860 ; y ˆ ÿ 3 : 99 ; z ˆ 2 †; D… ˆ 4 : 05 ;  ˆ 140 : 08 ; z ˆ ÿ 3 †; 8.

\ D1.6. Transform to cylindrical coordinates: (a) F ˆ 10 ax ÿ 8 ay ‡ 6 az at point P… 10 ; ÿ 8 ; 6 †; (b) G ˆ … 2 x ‡ y†ax ÿ …y ÿ 4 x†ay at point Q…; ; z†. (c) Give the cartesian components of the vector H ˆ 20 a ÿ 10 a ‡ 3 az at P…x ˆ 5 ; y ˆ 2 ; z ˆ ÿ 1 †:

Ans. 12.81a ‡ 6 az; … 2  cos^2  ÿ  sin^2  ‡ 5  sin  cos †a ‡ … 4  cos^2  ÿ  sin^2 ÿ 3  sin  cos †a; Hx ˆ 22 : 3 ; Hy ˆ ÿ 1 : 857 ; Hz ˆ 3

VECTOR ANALYSIS 19

TABLE 1. Dot products of unit vectors in cylindrical and cartesian coordinate systems

a a az

ax cos  ÿ sin  0 ay sin  cos  0 az 0 0 1

s s

1.9 THE SPHERICAL COORDINATE SYSTEM

We have no two-dimensional coordinate system to help us understand the three- dimensional spherical coordinate system, as we have for the circular cylindrical coordinate system. In certain respects we can draw on our knowledge of the latitude-and-longitude system of locating a place on the surface of the earth, but usually we consider only points on the surface and not those below or above ground. Let us start by building a spherical coordinate system on the three cartesian axes (Fig. 1.8a). We first define the distance from the origin to any point as r. The surface r ˆ constant is a sphere.

20 ENGINEERING ELECTROMAGNETICS

FIGURE 1. (a) The three spherical coordinates. (b) The three mutually perpendicular surfaces of the spherical coordi- nate system. (c) The three unit vectors of spherical coordinates: ar  a ˆ a. (d) The differential volume element in the spherical coordinate system.

s s