engineering electromagnetic fields and waves 2nd edition.pdf

_________________________________________ CHAPTER 1

Vector Analysis and Electromagnetic Fields in Free Space

The introduction of vector analysis as an important branch of mathematics dates back to the midnineteenth century. Since then, it has developed into an essential tool for the physical scientist and engineer. The object of the treatment of vector analysis as given in the first two chapters is to serve the needs of the remainder of this book. In this chapter, attention is confined to the scalar and vector products as well as to certain integrals involving vectors. This provides a groundwork for the Lorentz force effects defining the electric and magnetic fields and for the Maxwell integral relationships among these fields and their chargc and current sources. The coordinate systems employed are confined to the common rectangular, circular cylindrical, and spherical systems. To unifY their treatment, the generalized coordinate system is used. This timesaving approach permits developing the general rules for vcctor manipulations, to enable writing the desired vector operation in a given coordinate system by inspection. This avoids the rederivation of the desired operation for each new coordinate system employed. Next arc postulated the Maxwell integral relations for the electric and magnetic fields produced by charge and current sources in free space. Applying the vector rules developed earlier, their solutions corresponding to simple classes of symmetric static charge and current distributions are considered. The chapter concludes with a discussion of transformations among the three common coordinate systems.

1·1 SCALAR AND VECTOR FIELDS A field is taken to mean a mathematical function of space and time. Fields can be classified as scalar or vector fields. A scalar field is a function having, at each instant in

1

lJ

i'

2

VECTOR ANALYSIS AND ELECTROMAGNETIC FIELDS IN FREE SPACE

time, an assignable magnitude at every point of a region in space. Thus, the temperature field t) inside the block of material of Figure 1-1 (a) is a scalar field. To each point there exists a corresponding temperature T(x,]!, z, t) at any instant t in time. The velocity of a fluid moving inside the pipe shown in Figure 1-1 (b) illustrates a vector field. A variable direction, as well as magnitude, of the fluid velocity occurs in the pipe where the cross-sectional area is changing. Other examples of scalar fields are mass, density, pressure, and gravitational potential. A force field, a velocity field, and an acceleration field are examples of vector fields. The mathematical symbol for a scalar quantity is taken to be any letter: for example, A, T, Il, f. The symbol for a vector quantity is any letter set in boldface roman type, ff)!' A, H, a, g. Vector quantities are represented graphically by

(z)

6 200· 6

(x)

Temperature field at x 4 em

=

Heat source (a)

F

FIGURE 1-1. Examples of material. (b) Fluid velocity field ill,ide

fidd inside a block of

m-

Id. ny (b) ity lar ity for lce

3--

I

~

----...

1~!

~\ ~~\~I("'"

Unit vector a

B=C

y

FIGURE 1-2. Graphic representations of a vector, equal vectors, a uni t vector, and the representation of magnitude or length of a vector.

by means of arrows, or directed line segments, as shown in Figure 1-2. The magnitude or length of a vector A is written \A\ or simply A, a positive real scalar. The negative of a vector is tbat vector taken in an opposing direction, with its arrowhead on the opposite end. A unit vector is any vector having a magnitude of unity. The symbol a is used to denote a unit vector, with a subscript employed to specify a special direction. For example, ax means a unit vector having the positive-x direction. Two vectors are said to be equal if they have the same direction and the same magnitude. (They need not be collinear, but only parallel to each other.)

1·2 VECTOR SUMS The vector sum of A and B is defined in relation to the graphic sketch of the vectors, as in Figure 1-3. A physical illustration of the vector sum occurs in combining displacements in space. Thus, if a particle were displaced consecutively by the vector distance A and then by B, its final position would be denoted by the vector sum A + B = C shown in Figure 1-3 (a). Reversing the order of these displacements provides the same vector sum C, so that

A+B=B+A

( 1-1)

the commutative law of the addition of vectors. If several vectors are to be added, an associative law

(A + B)

+ D = A + (B + D)

follows £I'om the definition of vector sum and from Figure 1-3(b).

I

B

I I I

:A

+ B =C

I

I I I

(a)

(b)

FIGURE 1-3. (a) The graphic definition of the sum of two vectors. (b) The associative law of addition.

(1-2)

1·3 PRODUCT OF A VECTOR AND A SCALAR u and if B denotes a vector quantity, their produc a magnitude u times the magnitude of B, and having tht a positive scalar, or the opposite direction if u is negative same direction The f()lIowing laws hold IiII' the products of vectors and scalars. If a scalar

uB

(u

Bu

Commutative law

( 1-3)

(uv)A

Associative law

( 1-4)

o)A = uA

+ vA

Distributive law

(1-5)

uA

+ uB

Distributive law

( 1-6)

u(A +B)

1·4 COORDINATE SYSTEMS

The solution of physical problems often requires that the framework of a coordinate system be introduced, particularly if explicit solutions are being sought. The system most familiar to engineers and scientists is the cartesian, or rectangular coordinate system, although two other ii'ames of reference often used are the circular cylindrical and the spherical coordinate systems, The symbols employed for the independent coordinate variables of these orthogonal systems are listed as follows. 1. Rectangular coordinates: (x,y, z) 2. Circular cylindrical coordinates: (p, cj>, z) 3. Spherical coordinates: (r, 8, cj»

In Figure 1-4(a), the point P in space, relative to the origin 0, is depicted in terms of the coordinate variables of the three common orthogonal coordinate systems: as P(x,y, z) in the rectangular system, as P(p, cj>, z) in the circular cylindrical (or just "cylindrical") system, and as P(r, 8, cj» in the spherical coordinate system. In the cylindrical and spherical systems, it is seen that the rectangular coordinate axes, labeled (x), and ,are retained to establish proper angular references. You should observr that. the coordinate variable cj> (the azimuth angle) is common to both

: (zi I

I

P(x, y, z)

z

-(x)

y Rectangular

Circular cylindrical

Spherical

(a)

FIGURE 1-4. Notational convcnlions (a) Location of a point P in space, (Ii) The of a vector A into its orthogonal COmpOllt'nts.

in the three nnnmoll coordinate systems. p"im P Ie) The resolution

s·

~ ~ ~~ ~!a ~

~ &~ ~ ~ ("l)

("l)

-

,

,

i:

S '""

,

(.n

~

-

~

M- "'''I

0..

("l) •

...

:;:."'" :::r- (").: ("l)

(z) (z)

(z)

r= Constant

(sphere)

d> = Constant . (plane)

z= Constant (plane)

,

z = Constant

\

.-,1

d> = Constant (plane)

I

I I I

0

,

--(y) (x) - -

(y)

(yl

(x)

x=Constant

p=Constant (circular cylinder) (b)

_--®~ZA:---..,

,r::::"- ... __A _. ...P"",,-- I" t

I

,

'

I'

:(z)

:(z)

:(z)

I

I

'

~a;~;

: I

axAx~:-::=L------ ayAy

~ z

_-'-_

_- ----- 0

(x)

_-

' ,,/, .. A f-

I

_---

y

x

Rectangular

'

n!I r;

arA~P: a A

i

"

, --"" 4 ' '

I

I

a

:::,-_......

I

i : / q,A1 aA-~'biO p p z

----_ Jy)

_- -

- _---(x)

--p

Circular cylindrical (c)

I

I I

...

I

O/, aeAo': "-t': r~O

A

-----~y)



q,

,,"

"

::'>w'

-'--~;'

r--'/

c:

§

-

- _---- : (x)

Spherical

(y)

?:

I:

:2

...:>-to' u u ~

~

FIGURE 1-4 (continued)

to'

::

" ~

the cylindrical and tbe spherical systems, with the x-axis taken as the = 0 reference, generated in the positive sense from (x) toward (y). (By the "right-hand rule," if the thum b of the right hand points in the positive z-dircction, the fingers will indicate the sense.) The radial distance in the cylindrical system is p, measured perpendicularly from the to the desired point P; in the spherical system, the radial distance is 1, measured from the origin 0 to the point P, with denoting the desired declination angle measured positively from the reference z-axis to 1, as shown 1-4( a). The th ree coordinate systems shown are so-called "right-handed" properly definable after first discussing the unit vectors at P.



°

A. Unit Vectors and Coordinate Surfaces To enable expressing any vector A at the point P in a desired eoordinate system, three orthogonal unit vectors, denoted by a and suitably subscripted, are defined at P in the positive-increasing sense of each of the coordinate variables of that system. Thus, as noted in Figure 1-4(b), ax, a y, a z are the mutually perpendicular unit vectors of the rectangular coordinate system, shown at P(x,y, z) as dimensionless arrows of unit length originating at P and directed in the positive X,], and;;; senses respectively. Note that the disposition of these unit vectors at the point P corresponds to a right-handed coordinate system, so-called because a rotation from the unit vector ax through thc smaller angle toward a y and denoted by the fingers of the right hand, corresponds to the thumb pointing in the direction of a z . Similarly, in the cylindrical coordinate system of that figure, the unit vectors at P(p, , z) are a p ' aq,' a z as shown, pointing in the positive p, , and;;; senses; at P(r, 0, in the position ofay generates the diHcrential vector dan r a direction specified by the unit vector a.p and a magnitude given by dq, sin (; makes day (for r = constant, () = constant) become a,,> sin 8 dq, as shown, '"

By means of graphic techniques simila,o to those used in Example 1-1, 011 show for spherical coordinates that all the spatial partial derivatives of the unit v in that system are zero except for Ja,o

°

J¢ = aq, 3m

[)

tJ

a y sin () while in the circular cylindrical system, all are zero except for

1·7 SCALAR AND VECTOR PRODUCTS OF VECTORS Besides the simple product of a vector with a scalar quantity discussed in Secli( two other kinds of products involving only vector quantities are now discussel lirst of these, called the scalar product (or dot product), is defined as followso

A· B == AB cos () in which () signifies the angle between the vectors A and B. Noting from (1-3 A· B may be written either (A cos 8)B or A(B cos 0) makes it evident that th, product A . B denotes the product of the scalar projection of either vector 0' other, times the magnitude of the other vector. The definition of A . B makes th,

1-7 SCALAR AND VECTOR PRODUCTS OF VECTORS

15

the angle unity), to

useful, tor example, in computing the work done by a constant force acting a distance expressed as a vector. A generalization of this idea extended to the expression for work is taken up in the next section. Definition (1-34) permits the conclusion that if A and B are perpendicular, cos () zero, making their scalar product zero. Again, if A and B happen to lie in the same then A • B denotes the product of their lengths. These observations lead to results involving the scalar products of the orthogonal unit vectors a l , a2, and 83 coordinate systems illustrated in Figure 1-5. For example, a l • a2 a2 • a3 = 8:\ • a l = 0, while at • at = a z . a2 = a3 • a3 = l. From the definition (1-34), and since B· A means BA cos 0, the commutative fhr the dot product follows.

. Allowing having sin O. This Il, whence

A·B=B·A

(1-35)

Ian

distributive law for the dot product of the sum of two vectors with a third vector

A . (B

+ C) = A . B + A . C

( 1-36)

also be proved. , one can lit vectors

IXAMPLE 1·2. Vector analysis can be used to shorten a number of proo[~ of g-eometry. Suppose one is to show that the diagonals of a rhombus arc perpendicular. Represent its sides and diagonals by means of the veetors shown in the diagram. The diagonals are A + B C and A B D. Form the dot product of C and D.

(A ( 1-32)

+ B)

. (A - B)

which must equal zero because A

=

A· A - B . B

B for a rhombus.

=

A2 - B2

Thercf()[(~

C and D are perpendicular.

1f rhe vectors A and B are expressed in terms of their generalized orthogonal c:omponents in the manner of (1-9), their scalar product can be written ( 1-33)

expanding this expression by means of the distributive law (1-36) and applying results obtained earlier fix the dot products of the unit vectors, one obtains

eclioll 1-3, ussed. The

(1-37a)

(1-34)

(1-34) that the scalar )r onto the s the scalar

t

EXAMPLE 1-2

if

16

VECTOR ANALYSIS AND ELECTROMAGNETIC FIELDS IN FREE SPACE

For example, the expansion of the dot product of two vectors in rectangular nates is

COOl

( 1-3' and in circular cylindrical coordinates (1-3

EXAMPLE 1-3. (a) At the point P(3, 5, 6), shown in (a) of the figure, are given the two veet D = - 50a x + 60a y + 100az and E = 12ax - 24ay- Find the vector magnitudes and dot product D . E. Use these to determine the projection D cos of D onto E, and angk between the vectors. (b) In (b) of the figure, at point P(5, 60°, 9) are given two vectors F = IOa p + Ba", 4a z and G = - 20ap + BOa z in cylindrical coord ina Find the vector magnitudes and F . G as well as the angle 0 between the vectors.

e

e

(2)

D=

50a.,

+603.,+ 1003,

(xl

~-

5

'~

(yl

(al

G

F F II.

(xl

EXAMPLE I

17

1-7 SCALAR AND VECTOR PRODUCTS OF VECTORS

coordi-

(a) By usc of (I

the vector magnitudes are

1-37b) while the dot product is found from expansion (1-37b) 50(12)

+ 60( -24)

2040

The latter, by (1-34), also means DE cos 0, whence the projection D cos 8 becomes vectors and the md the ven the :Iinates.

J)

D·E E

cos (J

-2040 26.833

76.03

This nCl-iative result shows that the projection D cos 0 alonl-i E is in the negative-E sense (meaninl-i that 0 exceeds 90°). The value of 0 is found from the definition (1-34), yieldinl-i

0= cos

.. 1

D .E

~- =

DE

cos

.. 1

2040

-~-..- - - - - =

126.886(26.883)

(b) The mal-iniludcs and dot product, from (1-7) and

, . ,0 126.82 in circular cylindrical

coordinates, arc

G= F' G

+ F~ + 1';]112 [10 2 + 8 2 + 4 2 ] [20 2 + 80 2FI 2 = 86.462 [F~

F

=

= 13.416

10(-20) - 4(80» = -520

The anl-ik () between F and G is found from definition (1-34), obtaining

0= cos

.. 1

F.G .. 1 520 , = cos --~"'-'-- - = 117.93 FG 13.416(82.462)

From this result you may determine that the projection of F ncgativc-G sellSe.

0

G is in the

The second kind or product of one vector with another is called the vector product cross product), defined as l()Uows

A x B = a"AB sin 0

( 1-38)

e

is the angle measured between A and B, and a" is a unit vector taken to be perpendicular to both A and B and having a direction determined {i-om the righthand rule provided that the rotation is taken {i'om A to B through the angle O. The vector product A x B is illustrated graphically in Figure 1-9. One may show from the diagram that in which

A

X

B

-B

X

A

(I

18

VECTOR ANALYSIS AND ELECTROMAGNETIC FIELDS IN FREE SPACE

AxB

AxB

,f.

--",

A~--~

A

()

B

Positiv€~""­

(J sense

from A to B

FIGURE 1-9. Illustrating the cross product.

which means that the vector product does not obey a commutative law. In forming t1: cross product, the ordering of the vectors, therefore, is an important consideratiOl If A and B are parallel vectors, sin is zero to make their cross product zen If A and B happen to be perpendicular vectors, then A X B is a vector having a lengt AB and a direction perpendicular to both A and B, with the ambiguity in the directio resolved by means of the right-hand rule. These observations applied to the crm products of the orthogonal unit vectors of Figure 1-5, for example, lead to the sped" results: al X a l = az X az = a3 X a3 = 0; a l X az = a 3 , az X a 3 = al, and a3 > a l = az. However, note that a i X a3 az. A distributive law can be shown to hold for the cross product

e

A

X

(B + C)

=

A

X

B +A

X

C

(1-40

Because of the noncommutativity of the cross product as expressed by (1-39), the orde of the factors in (1-40) is important. If the vectors A and B are given in terms of their orthogonal components il the manner of (1-9), then their vector product is written

The use of the distributive law (1-40) and the special results obtained for the cros products of the orthogonal unit vectors provides the following expansion.

which can alternatively be put into the compact determinentaI form

A

X

B

=

al

az

a3

Ai

Az B2

A3

Bl

B3

(I -41)

19

1-7 SCALAR AND VECTOR PRODUCTS OF VECTORS

Pivot

P EXAMPLE 1-4

EXAMPLE 1·4. The definition of the cross product can be used to express the moment of a force F about a point P in space. Suppose R is a vector connecting the point P with the point of application Qofthe force vector F, as shown in the diagram. Then the vector moment M has the magnitude M = RF sin (} = X Fl. The turning direction of the moment, as well as its magnitude, are thus expressed by the vector product

IR

M

Ig the

Hion.

RxF

(1-42)

EXAMPLE 1·5. A force F = !Oay N is applied at a point Q(O, 3, 2) in space. Find the moment ofF about the point P(2, 0, 0). The vector distance R between P and Q)s

zero. ~ngth

'ction cross )ecial

The vector moment at P is found by means of (l-42) and the determinant (1-41).

a3 X

ax

1-40)

ay az

M=RxF= -2

3

2

o

10

0

=

-20a x -20a z N-m

M, shown at P in the sketch, is a vector perpendicular to the plane formed by F and R.

)rder Its

in

EXAMPLE 1·6. Given the two vectors F and G in (b) of the figure in Example 1-3, determine their vector cross product F x G, as well as the magnitude of the latter. Find the unit veetor an in the direction of the vector F X G. Verify that an is perpendicular to F and to G.

cross 1 1(z)

21

/1d-,. ---_9(0,3,2) I

/

P(2 0 0) I , ,

/

I

I

R

---0 --_

_(-)---M

-41 )

I

I

x EXAMPLE 1-5

1

F = lOay

I

---3'--_ "1

(y)

20

VECTOR ANALYSIS ANI) ELECTROMAGNETIC FIELDS IN FREE SPACE

From (I

in circular cylindrical coordinates, F x G becomes

I)

FxG

ap

a",

az

IO

8

-4

20

0

80

+ a",[ -4( -20)

- 10(80)]

+ a z [IO(O)

8(

20)]

160az The F G is IF X GI = [640 2 + 720 2 + 160 2 ] 1/2 vector an in the directioll of the vector F x G is given by

FxG

a n

The dot

= 'iF x

0.655ap

Gi an'

= 976.5,

0.737a",

while the uni

+ 0.1638a z

F [wcnmes, from (l-37b), the zero result 10

0.737(8)

+ 0.1638( -4)

= 0

verifying frorn til!' definition (I that an and F are perpendicular vectors. You ma) similarly show that an and G are perpendicular.

1·8 VECTOR INTEGRATION Vector integration, f()f the purposes of field theory, encompasses integrals in space along lines, over surfaces, or throughout volume regions, as well as integrals in the time domain and the domain. The subject of the present discussion concerns only integrations in space. Tne vector notation embodies compactness as an important feature, so it is always worthwhile to examine the integrand ofa vector integral carefully. The integrand may be either a scalar or a vector Thus, the integrals

[ A' Bdt

Line integral

~I'

J, (C J: F'· possess scalar hand, the

Surface integral Volume integTal

Gdll

produce scalar results on integration. On the other

amI

G

Line integral

Hx

Surfilce in tegral

J and

D) • ds

X

X

K

Volume vector results. In the last three examples, acroullt the different directions assumed by the on the surhce ,,)', or in the volume V defined.

t1H'IT/ill'C

1-8 VECTOR INTEGRATION Typical di (Vector displacement) ' "

dt (Scalar displacement) P2

Patht

21

""

P2

~~-;;J

.l-----R Pl

(b) ~:XAMPLE

\-7. (a) Integration of the scalar dt over a path t. (b) Integration the vector dt over the path t.

, unit EXAMPLE 1·7. The difterent results provided by scalar and vector integrands is exemplified by simple integrals of scalar and vector displacements dt or dt along some prescribed path in space. The integral

may

summed over the path t shown in (a) of the figure, provides its true scalar length d. On the other hand, the integral of the vector displacement dt on the same path

R=

)ace the erns vays nay

Jtr dt

produces quite a diftercnt answer, a vector result R determined only by the endpoints P l and P2 of that path rather than by the form of the path between the endpoints. This vector R is illustrated in (b) of the accompanying figure. So the line integral of dt about a closed path is zero, whereas if dt is the integrand, the perimeter of the closed pa th is the result.

An integral flllding extensive utility in work or energy calculations is the scalar line integral

LF . dt == LF dt

cos

e

(1-43)

This integral sums the scalar product F . dt over the path t, as suggested by Figure 1-10. Only the projection ofF along de at each point on the path contributes to the integral result. The line integral (1-43) can be expressed in terms of the generalized orthogonal components of F and of de in the following way, making use of (l-9), (1-21), and her

1-37 a)

(1-44)

es, he ~d.

In the rectangular coordinate system, in which hi = h2 = h3

= 1,

(1-44) is written (1-45)

22

VECTOR ANALYSIS AND ELECTROMAGNETIC FIELDS IN FREE SPACE

- --

11----""'- F

(a)

(b)

(c)

FIGURE J-lO. A palh and the field F in space. (0) Division of t into vector elements dt. (c) product F· dt (to be summed over the path) shown at the typical point P on the path.

assuming (Xi,_Vl' of the path t.

.::tl

are the coordinates of the endpoints P 1 and P

and

EXAMPLE 1·8. Evaluate the line integral (1-43) between the points PI(O, 0,1) and P 2 (2, 4,1 ovcr a path t defined the intersection of the two surfaces y = x 2 and z = 1, if F is thl v(,ctor fidd

(1 The path t is illustrated ill the 2 Inserting = lOx, .')x y, and f~ it {()llows that dz = 0 from the definitiD!!

£

F . dt

the desired resnlt.

j'

(2,0,

( EXAMPLE I

fx2~O

lOx dx -

20

106.7 =

into (1-45) and since x 2 = y all(

fy4=O 5y2 dy + 0 86.7

1-9 ELECTRIC CHARGES, CURRENTS, AND THEIR DENSITIES

23

This answer can also be obtained by expressing the dificrential displacement dx along the path in terms of From the definition of l, dy = 2x dx and dz O. Thus 2 Jtr F . dt: = Jor

lOx dx

'4

j o 5y2 dy

4y =

-36.7

IXAMPlE 1·9, A line integral such as (1-4·3) in gcncral has a value depending on the shape of the path connecting the endpoints PI and P2 . Evaluate the integral of Example 1-3 for the same function F and the same endpoints PI(O, 0,1) and P2(2, 4,1), but deform t: into the straight-line path given by the intersection of the surElCes y = 2x and z = I. Integral (1-43) now becomes dy

+0

60

obviously dilll'rent from the result obtained over the parabolic path in the last example. F is f()r tbis reason called a nonconservalive field. A vector field fell' which the line integral (1-43) is independent of the shape of the path connecting a fixed pair of emlpoints is said to be conservative. More is said later of such fields in connection with static electric charge distributions in Chapter 4.

1·9 ELECTRIC CHARGES, CURRENTS, AND THEIR DENSITIES 4, I) is the

(I)

, and

The physical and the chemical properties of matter are known to be governed by the eitcctric and magnetic forces that act among the particles comprising all material sub!ltalH~es, whether inorganic or living cells. The fundamental electric panicles of matter of two varieties, commonly called positive and negative electric charges. Many experiments have provided the following conclusions concerning electric charges. 1. The algebraic sum oCthe positive and negative electric charges in a closed system never changes; that is, the total electric charge of a defined aggregate of matter is consewed. 2. Electric charge exists only in positive or negative integral multiples of the magnitude of the elect mnic charge, e = 1.60 X 10 - 19 C; this implies that electric charge is quantized. From the viewpoint of classical electromagnetic theory, an electric charge aggregate will be treated as though it were capable of being indefinitely divisible, such that a volume electric-charge density, denoted by the symbol Pv is defined as follows 3

Pv

Aq ,

=--

Ali

elm

3

( 1-46a)

This limit of this ratio is taken such that the volume-element in space does not become so small that it contains so few charged particles that the relatively smooth property of the density quantity p" is lost, although Ali is kept small enough thal thl' integration the quantities containing Av becomes a meaningful process. I-II (a) illustrates the meaning of these quantities relative to a volume eiemellt

or

3It is dear thaI Ihe symbol p, for volume ('haq;;" density should not be confused with the lIn.l11/" , the radial variahle of the circular cylindrical coordinales (p, 4>, 'c).

:!i

I 24

VECTOR

ANn ELECTROMAGNETIC FIELDS IN FREE SPACE

de

.. ex.· '~

dq = p{ dt on dt

(a)

FIGURE I-IL

(c)

(b)

used in ddining volume, surface, and line charge densities in space. Qualltit;t·, defining Ps' (el Quantities defining pt.

Aq residing within any element Av may vary from pOil point in a it is evident from (1-46a) that charge density function of space as possibly of time. Thus Pv is a .field, written in ger Pv(ur, U2' U 3 , t) or p,,(r, In some physical the charge I1q is identified with an element of su or line instead of a volume. The limiting ratio (1-46a) should then be defined as foil Aq

pS=AC/m

2

( 1-'

L.l.S

Aq

Pc

= 111' C/m

(1-

The quantltles associated with these definitions of volume, surface, and line ch. densities are illustrated in 1-11. In some systems aggregates, two species of positive and negative ch: densities may be simultaneously. A net charge density p" (volume, sud such an instance defined or line density) is

p"

p,;

+ pv

C/m 3

(1

P:

in which and denote limiting ratios defined due to the positive negative charges + ami Aq respectively in Av. occurrence of both pos: metallic ions and mobile electrons in a conductor is an example to which (1-47) rna applied. The ill this being of eqnal magnitudes but opposite s 41n some physical ent simultaneously characterized by

discharge, electrons and several kinds of ions maybe Their net density at any point in the region may th,

(I if a total of

to be (lUnd there.

25

1-9 ELECTRIC CHARGES, CURRENTS, AND THEIR DENSITIES

P;; = - p;;), cancel, providing the net density Pv 0 in such a compensated charge system. The total amount of charge contained by a volume, surface, or line region is . obtained from the integral of the appropriate density function (1-46a), (1-46b), or 1-46c). Thus in some volume region, each element dv contains the charge dq = Pvdv, making the total charge in 1) the integral

Iv dq Iv Pv du C

q=

=

Similar integral expressions may be constructed to yield the total charge on a given surface or a line in space.

EXAMPLE 1·10. (a) The radially dependent volume charge density Pv = 50r2 C/m 3 exists within a sphere of radius r 5 cnL Find tlfe total charge if contained by that sphere. (b) The same sphere of is now covered with the angularly dependent surface charge density Ps 2 x 1O~ 3 0 C/m 2 Find the total charge on the spherical surface.

point to lsity is a general

(a) Making usc of( 1-47) and dv of (1-17) obtains

q=

Sv p" dv SSS (50r2)r2 sin 0 dr dO dcjJ = 50 s:n d(p S: sin 0 dO S:·os r dr 4

,5 JO.os =

)f surface follows.

= 50(2n)2 -5

(1-46b)

(1-46e) ~

charge

°

3.927 x 10 - 5 = 39.27 j1C.

Attention is called to the "product separability" of the integrand in this example, enabling the expression of the triple integrand as the product of three separate integrals in r, 0, and cjJ.

Is

(b) Using q = Ps lis in this case, along with the scalar surhrce clement ds = r2 sin 0 dO dcjJ on this sphere or radius r = OJ)5 m, as suggested by ds shown in Figure 1-7(b), yields on the complete sphere If

=

e charge surface,

f p"d.1 = ff(2 S

X

2 x 1O~ 3(0.05)2 =

10- 3

cos 20)r2 sin OdO r/cjJ] ~

r.

In (OS2 0 sin 0 dO = Jo

2n

.0

dcjJ

r-O.OS

5 x 10-

[

_cos~Jn 3

0

20.9 /lC

( 1-47) rive and positive maybe te signs ly be presy then be

(1-47a)

A vector field F(Ul' U2, U,' t) at some given instant t, can be represented graphiby use of a myriad of vectors of appropriate lengths and directions at many in a region of space. A vector field plotted in this way is shown in Figure 1-12 (a). is, however, a cumbersome way to graph a vector field; usually a much more representation is by use of a/lux plot, a method replacing the vectors with lines (called jlux lines) drawn in accordance with the i()llowing rules. 1. The directions of the flux lines agree with the directions of the field vectors. The transverse densities of the flux lines are the same as the magnitudes of the fidd vectors. The flux plot of the vector field of Figure 1-12 (a), sketched in accordance with these is noted in (b) of that figure. If a surhlce S is, moreover, drawn in the region

26

VECTOR

(a)

ELECTROMAGNETIC FIELDS IN FREE SPACE

(b)

(c)

FIGURE 1-12. A veCWr field F, its flux and the flux through typical surfaces (a) A vector field F, denoted by "farrows. The flux map of the vector field F, showing an open surface S through a net flux passes. (e) A closed surface S, showing zero net flux emergent from it.

of space embracing that flux, then the net lines of flux r/J passing through S can be a measure of some physical quantity (such as charge,current, or power flow), depending on the physical meaning ofF. The differential amount of flux dr/J passing through any surface-element ds in space is defined by the scalar dr/J = F ds cos = F • ds, a positive or negative l-esuit, depending on the angle between F and ds. The net (positive or negative) flux of F through S is therefore the integral of dr/J over S

e

Is F' ds

(1-48)

in which ds is taken to emerge from that side of S assumed positive, as shown in Figure 1-12 (b). If S is a dosed surface, the net flux through it is given by

(1-49) as noted in Figure \-12 The la Her will integrate to zero (an indication that just as many flux Jines leave S' as enter it) unless the interior volume of S contains sources or sinks offlux lines. This view will be amplified later in the discussion of the divergence of a vector field. The current flow through a surb.ce embodies a good illustration of the flux COIlcept. Supr:iose that there are electric charges of density Pv(Ul, U2 , U3, t) in a region, and imagine that the cllarges have velocities averaging to the function v(ul' U2, U3, t) within the elements dv with which the densities Pv are identified. A current density function J may then be defined at any point P in the region by or

C/sec/m 2

( 1-50a)

This function is a measure, in the vicinity of any point P in space, of the instantaneous rate of flow of charge per unit cross-sectional area. If two species of charge density

l-9 ELECTRIC CHARGES, CURRENTS, AND THEIR DENSITIES

27

of opposite kinds, designated by P;; and Pv , exist simultaneously in a region of space, then their total current density J at each point is written

( 1-50b) In general, for n species with densities Pi and velocities Vi (e.g., electrons plus a mixture of ions)

( 1-50c) The differential current flux di flowing through a surface element ds at which the current density J exists, is di J . ds amperes, to make the net current i (current through S

i

be a :ling any ltive e or

48)

= S~ J . ds C/sec

or

(I-51 )

A

IXAMPLE 1·11. An electron bcam of circular cross-section 1 mm in diameter in a cathode ray tube (CRT) has a measured current of I itA, and a known average electron speed of 106 m/sec. Calculate the average current density, charge density, and rate of mass transport in the beam. Assuming a constant current density J = azJz in the cross-section (I-51), yields the following current through any cross-section.

in which A denotes the cross-sectional area of the beam. Thus the average current density is

Jz=

-ure

i A

=

10- 6 4 2 n(1O-3)2 = n Aim ----~

4

49)

The charge density in the beam, from (1-50a) in which becomes

ust ces Ice

J

a z 4/n imd v - - az I0 6 ,

Jz

)0-

)n, t) ds = azds

IC-

1 mm

a) us ty

Cross-section A I':XAMPLE I-II

i = l/1A

-----------l>-'(z)

28

VECTOR ANALYSIS AND ELECTROMAGNETIC FIELDS IN FREE SPACE

The rate of mass transport in the heam is the current times the electronic mass-to-charge ratio; this yields 5.7 x 10- 18 kg/sec, assuming an electron mass of9.! x 10 31 kg.

1·10 ELECTRIC AND MAGNETIC FIELDS IN TERMS OF THEIR FORCES Electric and magnetic fields are fundamentally fields of force that ongmate from electric charges. Whether a force field may be termed electric, magnetic, or electromagnetic hinges on the motional state of the electric charges relative to the point at which the field observations arc. made. Electric charges at rest relative to an observation point give rise to an electrostatic (time-independent) field there. The relative motion of the charges provides an additional force field called magnetic. That added field is magnetostatic if the charges are moving at constant velocities relative to the observation point. Accelerated motiolls, on the other hand, produce both time-varying electric and magnetic fields termed electromagnetic fields. The connection of the electric and magnetic fields to their charge and current sources is provided by an elegant set of relations known as Maxwell's equations, attributed historically to the work of many scientists and mathematicians well before Maxwell's time,5 but to which he made significant contributiohs. They are introduced in the next section. Suppose that electric and magnetic fields have been established in some region of space. The symbol for the electric field intensity (or just electric intensity) is the vector E; its units are force per unit charge (newtons per coulomb). The magnetic field is represented by means of the vector B called magnetic flux density; it has the unit weber per square meter. If the fields E and B exist at a point P in space, their presence may be detected physically by means of a charge q placed at that point. The force F acting on that charge is given by the Lorentz force law F = q(E

+v

x B)

(1-52a) (1-52b)

=FE+FBN in which

q is the charge (coulomb) at the point P v is the velocity (meter per second) of the charge

q

E is the electric intensity (newton per coulomb) at P B is thc magnetic flux density (weber per square meter or tesla) at P

FE

=

qE, the electric field force acting on q

F B = qv x B, the magnetic field force acting on q In Figure 1-13, these quantities are illustrated typically in space. The force FE has the same direction as the applied field E, whereas the magnetic field force F B is at right angles to both the applied field B and the velocity v of the charged particle. The Lorentz force expression (1-52) may be used for discussing the ballistics of charged particles traveling in a region of space on which the electric and magnetic fields E and B are imposed. The deflection or the focusing of an electron beam in a cathode ray tube are common examples. 5James Clerk Maxwell (1331-1379).

I

il

29

I-II MAXWELL'S INTEGRAL RELATIONS FOR FREE SPACE

FE

B flux ---

--".,-I

I I

,I

I

I

,~ :

B

I I

I

t

FB (b)

fa)

(c)

:FIGURE 1-13. Lorentz forces acting on a moving charge q in the presence of (a) only an E field, (b) only a B Geld, and (e) both electromagnetic Gelds,

EXAMPLE 1·12. An electron at a given instant has the velocity v (3) I05 ay + 105 az m/see at some position in empty space. At that point, the electric and magnetic fields are known to be E = 400a z V/m and B = O.005ay WbJm 2 , Find the total force acting on the electron. The total force is found from the Lorentz reaction (1-52a)

F = q[E

+v

X

B) = -1.6(10-19)[a z400

+ (a y 3' 105 + a z4'

105)

X

a y O.005)

= (a x 32 - a z 6.4)IO-17 N

Although this is quite a small force, the very small mass of the electron charge provides a tremendous acceleration to the partiele, namely a F 1m = (a x 3.51 - azO. 7) 10 14 m/sec 2 .

1·11 MAXWELL'S INTEGRAL RELATIONS FOR FREE SPACE The relationships among the electric and magnetic force fields and their associated charge and current distributions in space are provided by Maxwell's equations, postulated here in integral form for the fields E and B in free space.

~s (EoE) . ds

Iv pv dv C

J. B· ds = 0 Wb :Vs J, E. dt = -~

'Ji

dt

( 1-53) (1-54)

r B . ds V

Js

(1-55)

( 1-56)

30

VECTOR ANALYSIS AND ELECTROMAGNETIC FIELDS IN FREE SPACE

in which E B

=

E(Ul'

U2 , U3 ,

t) is the electric intensity field

B(ul'

U2, U3,

t) is the magnetic flux density field

Iv p" do = q(t) is the net charge inside any dosed surface S itt) is the net current flowing through any open surface S bounded by the closed line t Eo

is the permittivity offi-ee space (~10 9/36n F/m)

110

is the permeability of free space (= 4n x 10

7

Him)

The Maxwell equatious 6 (I-53) through (1-56) must be simultaneously sati~fied by the field solutions E and B for all possible closed paths t and surfaces S in the region of space occupied by these fields. This strict requirement might appear to limit severely the number of practical problems that can be solved by means of these integrals. Indeed, their application to the discovery offield solutions E(u l , U 2 , U 3 , t) and B(ul' 11 2 , 11 3 , t) is restricted, in the present treatment, to problems in which the charge or current distributions have particular symmetries that serve t~ simplify the solutions. The equivalent differential forms of Maxwell's equations, developed in the next chapter, have a somewhat wider range of application in problem solving at the introductory level. The reader is to be assured that only a low-level introduction to methods for obtaining electric and magnetic field solutions of Maxwell's integral relations (1-53) through (I-56) is attempted here. For the purposes of this introductory treatment, the Maxwell relations are simplified by considering only the field solutions of a few simple, symmetrical geometries of static charge or current distributions. In Examples 1-13 through 1-17 that follow, these simplifications are shown to enable, in one or two steps, solving for the electric or magnetic field of a given charge or current distribution. The symmetry of the distribution will be seen to be the key to providing quick solutions for the desired field. Symmetries about a point, a line, or a plane are considered.

A. Gauss's Law for Electric Fields in Free Space Maxwell's integral law (1-53) [I-53 J is also known as Gauss's law for electric fields in free space. The meanings of the quantities are illustrated in Figure 1-14. Thus, suppose that there is in free space an electric field E(Ub U2, U3, t) (denoted by the E-field flux line distribution in that figure), plus some related electric charge distribution of density Pv(Ul> U2, U3, t) as shown. Construct in this region a closed surface S, with S having any desired shape and enclosing 6 Although given the collective name Maxwell's equations, historically they were in a gradual process of evolution over many years before Maxwell's time. For an enjoyable and first-rate account of the details, you are encouraged to read the historical surveys at the beginning of each chapter in R. S. Elliott, Electromagnetics. New York: McGraw-Hill. 1966.

1-11 MAXWELL'S INTEGRAL RELATIONS FOR FREE SPACE

31

FIGURE 1-14. Typical closed surface S in a region containing an electric field and a related electric charge. Gauss's law must hold [or all closed surfaces constructed in the region, whether charges are contained or not.

all or any part of the electric charge in the region, or no charge at all, as desired. Then the Maxwell-Gauss law (I-53) means that the integral of the quantity (EoE) . ds over that closed surface S (the net, outward flux of EoE emanating from S) is a measure of the amount of electric charge Pv dv = q that is contained only within the volume V bounded by that surface S. The dosed-surface integral of (EoE) . ds thus automatically excludes any charge that happens to lie outside S. (The surface element ds on S is by convention taken as positively outward from S, as shown in Figure 1-14, or away from its interior volume v.) The constant Eo in this Maxwell-Gauss law, called the permittivity oIfree space, is approximately 1O-9/36n F/m in the mks system of units. 7 To evaluate the amount of electric charge q within some volume V surrounded by the dosed surface S, Gauss's law (1-53) can be employed to do this two ways: (1) from the right side of (1-53), by use of the volume integral of the charge density Pv contained within the volume V; or (2) from the left side of (1-53), by integrating (EoE) • ds over the closed surhtee S that bounds the volume V of interest. If a known charge distribution is static (motionless) and happens to possess a particular symmetry in free space, then Gauss's integral law (1-53) can even be used to evaluate the electric field E produced by that charge. The small class of symmetric, static charge problems that can easily be solved by use of Gauss's law are illustrated in the following example.

Iv

EXAMPLE 1·13. Find the electric field intensity E of the following static charge distributions in free space: (a) a point charge Q; (b) a spherical cloud of radius 10 containing a uniform volume density Pu; (c) a very long line charge of uniform linear density PI; (d) a very large planar (surface) charge of density Ps'

These charge distributions arc illustrated in Figure 1-15. Closed surfaces S arc shown, appropriately chosen to permit solving for E by the usc of Gauss's law (I-53). (a) Field of a charge (symmetry about a point). To evaluate the field E of the static point charge Q, choose S in Gauss's law (I-53) to be the sphere with Qat its center, as in Figure l-15(a). To show that E has only a radial component about 7The significance or the units of Eo is clarified in Chapter 4 in the discussion of capacitance. A correct interpretation of the factor Eo in (1-53) is that it is a proportionality factor accounting for the proper units (mks) of the equation.

32

i i

VECTOR ANALYSIS AND ELECTROMAGNETIC FIELDS IN FREE SPACE Sphere 8 1 ,\

/

"

Sphere 82

-

---~ ,

/ Point / charge

l\

\

"- "-

~E = arEr

P,

Qr'JJI

ds

/,- \ ~ r \

Spherical charged cloud

-,, -

"' , , ___ -,' ~Spherical

'\

~E(r ro)

(a)

(b)

Uniform surface charge density Ps

-~--

I

,I

Circular cylindrical closed surface S (e)

(d)

FIGURE 1-15. Static charge distributions having symmetries such that Gauss's law applied to appropriate closed surfaces will lead to solutions lor E. (a) Static point charge; spherical surface S constructed to evaluate E(r). (b) Charged cloud of uniform density, showing SI and 8 2 used to evaluate E(r). (t) Uniform line charge. (d) Uniform suriace charge.

the charge, obsfTve that for this time-static problem (did! 0, for all fields), (I-55) reduces to E . de' = 0 for all closed Jines e'. Then integrating E • de' about any circumferential path of radius r over the sphere in Figure 1-15(a) yields the conclusion that Eo and E are zero. Furthermore, assuming Q positive, E must be directed radially outward if the integral of E" oE over S is to yield a positive answer. Thus (I-53) yields

f

Since a r • a r and from the symmetry Er is constant on S, Er may be extracted from the integral to obtain (1-57a) or, in vector form E

(1-57b)

\-\\ MAXWELL'S INTEGRAL RELATIONS FOR FREE SPACE

33

Coulomb's law {Cll' the force acting on another point charge Q: in the presence of Qis deduced by combining (1-57b) with the Lorentz force relation (1-52a). In the absence of a B field, the force on Q: when immersed in the E field (1-57b) of the charge Qis ,

FE = Q:E = a r

Q:Q

(I-58)

z 4nEor

--

(b) Field of a charged cloud (symmetry about a point). For the spherical cloud containing a uniform charge density p" C/m 3 , two cases arise. The field outside the cloud (r > TO) can be obtained from Gauss's law (I-53) applied to a concentric sphere S1 of radius r, as shown in Figure 1-15(b). That E bas only an Er component is shown as in part . Then the charge q enclosed by Sl is obtained by integrating p" dv throughout sphere, so (I-53) becomes

Solving lor Er (constant

011

S 1) yields (I-59)

an inverse-square result. It is ofthe form of the point-charge result (1-57a), assuming the field point outside the charge cloud (r> ro). inside the cloud (r < ro), applying (I-53) to the closed surface S2 of Figure 1-15 (b) yields

in which the volume integration is carried out only throughout the interior of S2, obtaining . With l!,~ constant on S2, E = p"r r 3Eo

r<

(1-60)

TO

E inside the uniformly charged cloud is theref(Jre zero at its center and varies linearly to the samc valuc as (1-59) at the sllrface r roo (c) Field o/a long line (symmetry about a line). Construct a closed right circular cylinder of length t and radius p concentric about the line charge as in Figure 1-15(c). From symmetry, E is radially directed (apEp) and of constant magnitude over the peripheral surface So. The left side of Gauss's law (1-53) is zero over the endeaps of S because E . ds is zero on them. Thus (I-53) becomes

in which the right side reduces to a line integral over the linear charge distribution. Solving for Ep on Sp yields, with Pt dl = ptt and ds = 2npt

Js

ft

Ep

Pc 2nEop

. (1-61)

Thus, the electric intensity of an inflflitely long, uniform line charge varies inversely with p. .

34

VECTOR ANALYSIS AND ELECTROMAGNETIC FIELDS IN FREE SPACE

(d) Field an planar charge (symmetry about a plane). A closed surface 8 is constructed in the form of a rectangular parallelepiped extending equally on both sides of the planar charge, as in Figure 1-15(d). The symmetry of the infinitely extensive charge requin:s that E be directed normally away from both sides of the charge as shown (E = ±a,h'x)' Flux emanates only from the ends 8 1 and 8 2 of the parallelepiped, whence Gauss's law becomes

A denoting the area of the ends of the parallelepiped. The two integrals over 8 1 and 8 2 provide exactly the same amount of outward electric flux, whence E = x

o'--------------~r

2Eo

~----~----~~r

ro

(b)

(a)

(c)

fd)

FIGURE 1-16. Flux plots of the fields of Example I-II. (a) Point charge. An inverse r2 field. (b) Uniformly charged spherical volume. The graph depicts variations with r. (c) Unilormly charged infinite line. An inverse p field. (d) Uniformly charged infinite plane. E is uniform everywhere.

1-11 MAXWELL'S INTEGRAL RELATIONS FOR FREE SPACE

35

Writing this in vector form to include the fleldB on both sides of the planar charge distribution gives

-~

E=a x

2Eo

Ps E =-a x

2Eo

x>o xa

B

p-directed as in Figure 1-21 (a), a result t()llowing {i'om symmetry and the application of the right-hand rnle to the positive current sense shown. Thus, inside the toroid, B = a",B"" exact if the winding is idealized into a current sheet. The application of the time-static Ampere's circuital law to the symmetric closed line t of radius p shown therefore yields §t (a",8",) • alp dt = Jion1, in which B,p, from the symmetry, is constant around t, and nl is the nct current passing through S bounded by t. Thus

Jion1 B",= ._-2np

(1-66)

40

VECTOR ANALYSIS AND ELECTROMAGNETIC FIELDS IN FREE SPACE

Closed path f

IndLl( abou I

/ /

1

(b)

(a)

FIGURE 1-21. Two coil configurations, the magnetic fields of which can be found using Ampere's circuital law. (a) Toroidal winding ofT! turns, showing symmetric path t. (b) Infinitely long solenoid, showing a typical rectangular closed path t.

Fl m,

(a e~

an inverse p-dependent field inside the region bounded by the current sheet. If the radius p of t in Figure 1-21 (a) were chosen to cause t to 'fall outside the torus (with p < a, for example), then S would no longer intereept any net current i. Then from the symmetry, the B field outside the idealized toroid must be zero.

(b) The infinitely long solenoid of Figure 1-21 (b) may be regarded as a toroid !if infinite radius; its magnetic field is thus also completely contained within the coil if the winding is idealized into an uninterrupted current sheet. The symmetry requires a z-directed field, B azB" independent of z. Ampere's circuital law (1-63) is applied to the rectangular elosed path shown in Figure 1-21 (b), two sides lying parallel to the z-axis. A nonzero contribution to the line integral is obtained only over the interior path parallel to the z-axis, whence

B z is constant over the path, whence n J1 a1d

( 1-67)

the ratio nld denoting the turns per meter length. B is thus constant everywhere inside the infinitely long solenoid.

c.

Faraday's Law

Maxwell's integral law (1-55)

!!..-

r B· ds =

dtJs

dl/l m dt

[1-55 ]

is attributable to tbe work of Faraday, and is called the induced electromotive force (emf) Law. The essence of this law of electro magnetics is expressed in the symbolism of Figure

a t

s

1-11 MAXWELL'S INTEGRAL RELATIONS FOR FREE SPACE

41

1/Jm enclosed by

e

by integration sense about f (b)

(a)

t such as that of (a) may be superposed anywhere on the example of (b), such that .Faraday's law must be trne for it. (a) Typical dosed line bounding a snrface S, relative to the Gelds in Faraday's law. (h) A symmetric example showing tbe E field indun:d by a time-varying magnetic field.

FlGURE 1-22. Indnced electric fields and Faraday's law. Any closed line

1-22(a). The relationship of the positive line-integration sense to the positive direction assumed for ds is the same as for Ampere's circuital law. Faraday's law (1-55) states that the time rate of decrease of the net magnetic flux t/I m passing any arbitrary surface S equals the integral of the E field around the dosed line bounding S. This is tantamount to saying that an E field is generated by a time-varying magnetic flux. The E field, in general, must also be time-varying if ( 1-55) is to be satisfied at every instant. Faraday's law tor strictly time-static fields is (I-55) with its right side reduced to zero

f.

E . dt

= 0 Faraday's law for time-static fields

(1-68)

which states that the line integral of a static E field about any closed path is always zero. A field obeying (1-68) is called a conservative field; all static electric fields are conservative. If the electric charges that produce an electric field are fixed in space, that electric field must obey Faraday's law in time-static form, (1-68). Several examples of the electric fields of charges at rest have been treated in Example 1-13. All static distributions of ekctric charges in space may be regarded as superpositions of point-charge concentrations = p" dv in the volume-elements dv in space. The electric field of a point charge Q., on the other hand, has been shown to be (1-57b)

[1-57b] It is easily shown that this electric field obeys Faraday'S law (1-68) for a time-static field. If any dosed path, such as t = ta + tb shown in Figure 1-23, is chosen in the

42

VECTOR ANALYSIS AND ELECTROMAGNETIC FIELDS IN FREE SPACE

B sat Closed path f

Direction of integration

j

can 1-17 rise imp law tha1 iter qU • A, Aq, = aq, • A and A z = a z • A, (1-74a) (1-74b)

(I-He)

(xl

(6) F[GUR~:

Geomelli", circular cylindrical cmmli",,'," 1-2(j,

a" a, of rectangular coordinate, (a) to a p of coordinates. II"

1-12 COORDINATE TRANSFORMATIONS

coorletails al coSince rector d the mjee)1' the d the d for From

-73a)

47

These are the desired relations that permit finding the cmnponents Ap> A

;;

r = ';x 2 +y2 + Z2

in which

x

=

AT sin () cos

=

cos () (1-79a) (1-79b)

A.p

rp

(1-77)

SPHERICAL TO RECTANGULAR

RECTANGULAR TO SPHERICAL

AT

z=z

(")

t'l

r sin () cos

(1-80)

cos

e=

cos

rp

=

sin

x

sinrp

e=

=-===

(1-82)

(1-82)

1-13 UNITS AND DIMENSIONS

With Fx = 3z, l'~ = (1-74a,b,c) yields

F

and

F~ =

49

5x, the use of the component transformations

+ al'~ + aJ:'~ + F~ sin l + a ( - f',; sin + Fy cos =30° semi~ir1finite plane //

(y)

(xl

Pl

S 1 If

PROBLEMS

53

(z)

-_ I ---_yl~--.L.(x)

,Y2

-

I

I I

--J..... ..... -

(y)

PROBLEM 1-6

SECTION 1-6 1-6. Shown is the "distance vector," R, a vector directed from the point PdXl,Yl' zd to P Z (x 2 ,)'z, zz) in space, the position vectors of the latter being r 1 and rz. Observing graphically that R = r z - rj, write the expression for Rand IRI in rectangular coordinate {arm. Also write the expression in rectangular coordinates for the unit vector a R directed along R, making use Ahc definition, aR = Rj R.

\.!.:z}'

From the geometry, it is readily seen that the relation among the radial unit vector a r of spherical coordinates and unit vectors of the circular cylindrical coordinate system is a r = a p sin 0 + a z cos O. Use this relation to show that (oa,/iJ b) it is Pv(b 3 - a3)/3Eor2. (Show an appropriately labeled sketch along with the details of your proof.) 1-29. Let the volume charge density within a spherical region of radius r a be given by Pv = Po(l + kr), in which Po denotes the density at the origin. Determine the k that will make the total charge in the sphere zero. For this k, why is the E field external to the sphere zero? Find E. as a function of r within the sphere, making use of Gauss's law. [Answer: k = -4/3a] 1-30. An infinitely long, cylindrical clond of radius p a in free space contains the static, uniform volume charge density p". With a suitably labeled sketch, make use of the symmetry and Gauss's law (1-53) to obtain the following. (a) The electric field outside the cloud (p> a). (b) The interior electric field (p < a). (c) Show that the exterior E field is the'same as that expected if the same total charge per length t were concentrated as a line charge along the z axis, as in Figure 1-15(c). [Answer: E = appva2/2Eop (b) appvp/2Eo] 1-31. Let an infinitely long, cylindrical charged cloud of radius a contain the static charge density Pv = po(p/a)2, varying parabolically to the density Po at the cloud surface. (a) Make use of (1-17) to determine the total charge q in any length t of this cloud. (b) Sketch a diagram as suggested by Figure 1-15(c), making use of the symmetry and Gauss's law (I-53) to find the E field outside the cloud (p > a), and then inside it (p < a). Label the Gaussian surfaces used. Show from your solution that the field oLltside the cloud is the same as that expected if the total clurge were concentrated along the z-axis. [Answer: (a) p onta 2 /2 (h) poa 2/4E op for p > a, pop 3 /4E oa2 for p < a]

1-32.

Two parallel, planar charges of the kind shown in Figure l-15(d) are located at x = d and - d, possessing the nniform, opposite surface charge densities - Ps and p" respectively. (a) Use the vector superposition (summing) of the fields of these two planar charges, as given by (1-62), to prove that the total E field between the planes ( - d < x < d) is Ps!E V /m, whereas that outside the planes > d) is zero. (Do not usc Gauss's law.) (b) Repeat (a) if both surface charge densities are positive.

(Ixl

SECTION I-llB 1-33. A hollow, circular cylindrical conductor in free space, assumed infinitely long to avoid end efiects, and having the inner and outer radii band c, respectively, carries the direct current 1. (a) Assuming a constant, z-directed current density in the conductor cross sectiou, show that the vector current density at any point therein is J = a z 1/n(c2 - b2 ). (b) Usc Ampere's law to show that the exterior magnetic field is the same as that of the solid conductor of Figure 1-19 carrying the same total current f. Show that B inside the hollow interior (p < b) is zero, whereas that within the conductor (b < p < c) is a",Jlo1(p2 - b2 )/2np(c 2 - b2). (c) Sketch a graph showing how B", varies with p. 1-34. A coaxial pair of circular cylindrical conductors, infinitely long in frec space, have the dimensions shown and carry the equal and opposite total currents f. (a) Show tha.t the current density in the inner conductor is a//na 2 , whereas in the outer conductor it is the negative of that iCJUnd in Problem 1-33 (a). (b) Show that the B fields within the inner conductor (p < a) and betweell the C'Onductors (a < p < b) are identical to those of the isolated wire of Example 1-15. Use Ampere's law, together with an appropriately labeled diagram showing the closed

PROBLEMS

59

s sphere. y Gauss's (r> ro). sur/aces '\lith that n. (c) If ;wee (a)

twecn its ; that for 2. (Show l,riven by fill make ~rc zero?

-4/3a] lC static, fmmetry (p> a). ~ as that ng the <

: chalge 1) Make diagram find the :es used. ~d if the )r p > a, at x = d ectively. as given whereas if both

PROBLEM \-34

assumed, to prove in detail that the B field within the outer conductor (b < p < c) is IIIP/loJ(C 2 p2)/2np(c 2 b2 ) and that it is zero for p > c. (e) Sketch a graph of BIP versus p over the (0, c) range, assuming a = 3 mm, b = 6 mm, c = 3 mm, 1 = 100 A. Find the current in each conductor, expressed in A/Cln 2 • .

1..35. Show that the static B fields of the coaxiallinc of Problem 1-34 arc the superposition of the fields of the hollow conductor of Problem 1-33' and those of the isolated conductor of Example 1-15.

1..36.

Two parallel, indefinitely thin eurrent sheets ofinfinite extent in free space are located at = +d, possessing the unii()rm but oppositely directed surface-current densities ±Jsv respectively. (The currents are assumed charge-compensated, making electric fields absent in this problem.) (a) Employ resulls ofExamplc 1-16 and superposition (not Amp(~rc's law) to show that B between the sheets ( -d

UZ, u 3 ,

t)

(~

AUI

with similar expressions for the partial derivatives with respect to the remain variables. Successive partial differentiations yield functions such as jJ2FIe 2 F/oul eJu z . If F has continuous partial derivatives of at least the second order, i permissible to differentiate it in either order; thus

a

(2·

The partial derivative of the sum or product combinations of scalar 'and vect functions sometimes is useful. In particular, one can use (2-3) to prove tbat t following expansions are valid

)Ns

2-2 GRADIENT OF A SCALAR FUNCTION

63

of + F at at (2-6) o(F x G)

-~-=Fx

at

DG

DF

+~xG

(2-7)

at

if f is any scalar function and F and G are vector functions of several variables, among which t denotes a typical variable.

in the intor increion of the in Figure liffercncc erivative on. tor funcld vector

(2-2) n which

2·2 GRADIENT OF A SCALAR FUNCTION The space rate ofehange ofa sealar fieldf(ull Uz, U3, t) is frequently of physical interest. For example, in the scalar temperature field T(ull Uz, U3, t) depicted in Figure I-I(a), one can surmise from graphical considerations that the maximum space rates of temperature change OCCllr in di rections normal to the constan t temperatu re surfaces shown. Generally, the maximum space rate of change of a scalar function, induding the vector direction in which the rate of change takes place, can be characterized by means of a vector di!li:?rential operator known as the gradient of that scalar function. It is developed here. If, al allY fixed time t, a single-valued, well-behaved scalar field f(u l , U2, U3, t) is set equal to any cons/ant fo so that f(UlJ uz, U3, t) = Jo, a surface in space is described, as depicted by .)1 ill Figurc 2-2. A physical example of such a surface is any of the constant temperature surfaces of Figure l-l(a). Another SllrfaCe, 8 2 , an infinitesimal distance from 81> is described by letting f(u 1 + dUI' Uz + dU2, U3 + dU3) = Jo + df, in which dl is taken to mean a very small, constant, scalar amount. Suppose that two nearby points, P and P', are located a vector distance dt apart on these two surfaces

deriva-

(2-3)

82 (defined

by f =

to + df) 82 (f= (o + dt)

laining 2F/ouI, ~r, it is

(2-4) -1 I

vector at the

(a)

(b)

FIGURE 2-2. Two nearby surfaccs 1=10 and 1 .f~ + dl" rdatiVt" [0 a discussion or grad (a) Points P and P' separated by dt and on snr/aces defincd by 1 =.f~ and 1 = 10 + til Points P and P' on the same slIJ'face 1 = 10' to show that grad 1 and dt are perpendicular.

64

VECTOR DIFFERENTIAL RELATIONS AND MAXWELL'S DIFFERENTIAL RELATIONS

as in Figure 2-2(a), recalling from (1-21) that one may express dt = aldt as (2-8)

df is

the amount by which f changes in going from P to P' from the first surface to the second, written as the total differential

(2-9)

curvil

from in S(

The presence of the components of dt in (2-9) permits expressing df as the dot product

in tl Calling the bracketed quantity the gradient of the function f, or simply grad f, as follows

and

or

(2-10)

tha one may write the total differential df of (2-9) in the abbreviated form

df = (grad I) . dt

(2-11 )

Two properties of grad I are deducible from (2-11); 1. That the vector function grad I defined by (2-10) is a vector perpendicular to any I = Io surface is appreciated if the points P and P', separated by a distance dt, are placed on the same surface as in Figure 2-2(b). Then the amount by which I changes in going from P to P' is zero, but from (2-11), (gradf) . dt 0, implying that grad I and dt are perpendicular vee tors. Grad I is therefore a vector everywhere perpendicular to any surface on which I constant. 2. If a displacement dt from the point P is assigned a constant magnitude and a variable direction, then from (2-11) and the definition (1-34) of the dot product it is seen that dI = Igrad dt cos 0, 0 denoting the direction between the grad f and dt. The magnitude of gradf is therefore df/(dt cos 8), but from Figure 2-2(a), dt cos 0 = dn, the shortest (perpendicular) distance from the point P on the surface SI to the adjacent surface S2 on which I Io + dj, whence

hoI Co

Po

Fr be

II

df IgradII = dn

OJ

p: 0'

(2-12)

d (~

2-2 GRADIENT OF A SCALAR FUNCTION

65

The vector grad j therefore denotes both the mag"nitude and direction of the maximal space rate of change of j, at any point in a region. Note that the magnitude ofgradj can also be expressed in terms of its orthogonal lfvilinear components, given in the definition (2-10) by \gradji=

/ OJ)2 - + (or)2 -"- + (OJ - -)2J1 2 [(hi OUI h2 AU h3 OU3

(2-13)

2

The expressions for grad j in a specific orthogonal coordi nate system are obtained rom (2-10) on substituting into it the appropriate symbols {CH" U; and hi as discussed n Section 1-5. Thus, in the rectangular system

grad j = ax III

the circular

oj

oj

ox + a oy + a z - -

(2-14a)

y

system (2-14b)

and in the .>jJherical coordinate system

_ grad)

oj

loj

1

oj

= a,-or + a o -r oe + a.p -'-e r 5111 (3'"' 'V

(2-14c)

An integral property of grail j, of considerable importance in field theory, is that its line integral over any dosed path t in space is zero. Symbolically

~ (gradj)

. dt = 0

(2-15)

holding fell' all well-behaved scalar functions j, and proved in the riJlIowing manner. Consider (2-15) integrated over an open path between the distinct endpoints po(u7, u~) and P(u 1 , 112'

ug,

C'P Jpo

(grad j) . dt

(2-16)

From (2-11) it is seen that (grad j) . dt denotes the lotal differential df, so that (2-16) becomes

fP

Jpo

(gradj) . dt =

fP df = IJP

Jpo

Po

(2-17)

or the difference of the values of the function I at the endpoints P and Po. Thus, any path connecting Po and P will provide the same result, (2-17). Carrying out (2-16) over some path A from Po to the point P and then back to Po once more over a different path B, the contributions of the two integrals would cancel exactly, making (2-15) the result. The integral property (2-15) of any vector field grad f is sometimes

66

VECTOR DIFFERENTIAL RELATIONS AND MAXWELL'S DIFFERENTIAL RELATIONS

called the conservative property of that field, from the applications of integrals of that type to problems involving certain kinds of energy, Any field gradf is a conservative field,

2·3 THE OPERATOR V (Del) Recall that the gradient of a scalar field (2-14a)

f

is expressed in rectangular coordinates by

[2-14a] The presence of the common function f in each term permits separating from this expression a vector partial differential operator represented by the symbol V (pronounced del) as follows (2-18) to permit writing gradf in an alternative symbolism, Vf

8f 8f of gradf=Vf=ax-+a - + a z -

ax

Yay

(2-19)

oz

The notations grad f and Vf will henceforth be considered interchangeable, It may be noted that the operator V defined by (2-18) in the rectangular coordinate system can be defined in other coordinate systems as well, including the generalized orthogonal curvilinear system. This is not done here because of its lengthy form and because it serves no particular need in connection with the objectives of this text. You may wish to consult other sources relative to extending (2-18) to other coordinate systems.}

EXAMPLE 2·1. Suppose a scalar, time-independent temperature field in some region of a space is given by

T(x,y) = 200x

+

100y deg

with x andy expressed in meters. Sketch a few isotherms (constant temperature surfaces) of this static thermal field and determine the gradient of T. The isotherms arc obtained by setting T equal to specific constant temperature values. Thn5, letting T = 100° yields 100 = 200x + 1O(!y, the equation of the tilted plane y = 2x + I. Th.is and other isothermie surfaces are shown in the accompanying figure. The temperature gradient of T(x,y) is given by (2-14a)

vT

== grad T

aT + a -aT + a -aT- = ox y oy z a.::;

= ax -

200ax

+

I OOay0 1m

example, sec M.,Javid, and P. M. Brown, Field AnalYsis and Elfictromagnetics. New York: McGraw-Hill, 1963, p.477.

1 For

:I... TI

:r

Ih

-

i

, II!

2-4 DIVERGENCE OF A VECTOR FUNCTION

\

,

67

(y)

(b)

(a)

EXAMPLE 2-1. (a) Graph of T

constant. (b) Side view of (a).

a vector everywhere perpendicular to the isotherms, as noted in (b) of the figure. The x andy components of the temperature gradient denote space rate of change of temperature along these coordinate axes. From (2-13), the magnitude is

denoting the maximal space rate of change of temperature at any pomt. One may observe that heat will flow in the direction of maximal temperature decrease; that is, along lines perpendicular to the isotherms and thus in a dir'cction opposite to that of the vector grad T at any poin t.

1-4 DNERGENCE OF A VECTOR FUNCTION

The flux representation of vector fields was described in Section 1-9. If a vector field r is representable by a continuous system of unbroken flux lines in a volume region as for example, in Figure 2-3(a), the region is said to be sourcefree; or equivalently, field F is said to be divergenceless. (The divergence ofF is zero.) On the other hand,

,

111l1li;;;;

fjl!"! /%

"~ 1/

l;r '- " (a)

Ijffk/"/.,

\-y / / / (b)

(e)

FIGURE 2-3. Concerning the divergence of flux fields. (a) A vector field F in a source-free As many flux lincs enter S as leave it. (b) A vector Geld F in a region containing sources posse,;sirlg net outgoing flux). (c) The meaning of div F: net outward flux per unit volume as -0.

68

VECTOR DIFFERENTIAL RELATIONS AND MAXWELL'S DIFFERENTIAL RELATIONS

if the flux plot of F consists of flux lines that are broken or discontinuous, as depicted in Figure 2-3(b), the region contains sources of the field flux; the field F is then said to have a nonzero divergence in that region. The characterization of the divergence of a vector field on a mathematical basis is described here. The divergence of a vectorfield F, abbreviated div F, is defined as the limit of the net outward flux ofF, F • ds, per unit volume, as the volume !lv enclosed by the surface S tends toward zero. Symbolically,

fs

div F

==

F· ds

lim

!lv

fl ux lines/m

3

(2-20)

Av-+ 0

Thus, as the closed surface S is made very small, as depicted in Figure 2-3(c), the limiting, net outward flux pCI' unit volume in the neighborhood of the point P defines the divergence of the vector field F there. The shape of S is immaterial in this limit, as long as the dimensions of !lv tend toward zero together. The definition (2-20) leads to partial diHerential expressions for div F in the various coordinate systems. For example, in generalized orthogonal coordinates, div F is shown to become

FIC the (:on

oPI

ter (2-21 )

The derivation of the differential expression (2-21) for div F in generalized orthogonal coordinates proceeds from the definition (2-20). Express the function F in terms of its generalized components as follows F(Ul' U2, U3,

I) = a1Fdul,

U2, U3,

t)

+ a 2 F 2 (u l , li2,

li 3 ,

t)

+ a 3 F 3 (u l ,

li2, U3,

I) (2-22)

The definition (2-20) requires that the net effiux ofF be found over the closed surface S bounding any limiting volume !lv, which from (1 II) or (1-13) is expressed

It in di 01

(2-23) The net, outward Hux ofF is that emanating from the six sides of !lv, designated by Llsi> !lS'I' and so forth, in Figure 2-4(a). The contribution !It/ll entering element !lSl is just F· Lls i = (alF I )' !lSI' or

T tJ

(2-24) (2-25 ) the negative sign being the consequence ofassuming a positively direeted Fl component the outward !lSI = - al !lt2 Llt3; that is, the flux Llt/ll enters !lSI' In the limit, as the separation Lltl between !lSl and Lls'! becomes sufficiently small, the flux Llt/l'l leaving !ls'! in Figure 2-4(b) differs from !It/ll entering !lSI by an amount given by the second

(

2-4 DIVERGENCE OF A VECTOR FUNCTION

69

FIGURE 2-4. A volume-clement L'iu in the generalized orthogonal coordinate system nsed in the development of the partial diflcrcntial expression for div F. (a) A volume-element !lv and I:omponents ofF in the neighborhood of 1'(u 1 , 112' U3)' (b) Flux contribntions entering and leaving opposite surfaces of !lv. The remaining four sides are similarly treated.

term of the Taylor's expansion of L\I/;~ about the point P; that is, A ./,

lJ.'f'1

=

+ 0(L\1/; tl '" UUl

F'1L\t2L\t3

L\ Ul

+ [~(F'lM2L\t3)]L\Ul OUI

Fl L\t2 L\t 3 +

[a~l (Flh2 h3) ] L\Ul L\uz L\U3

(2-26)

It is permissible to remove L\U2 and L\u3 from the quantity affected by the O/OUI operator in the foregoing because each is independent of Ul, in view of the orthogonal coordinate system being used. The net outgoing flux emerging from the sides As[ and L\S'l of Figure 2-4(b) is thus the sum of (2-24) and (2-26) (2-27a) The two remaining pairs of surface elements tribute outgoing flux in the amounts

L\s~, As~,

and L\S3',

As~

similarly con-

(2-27b)

(2-27c) seen to be obtainable from the symmetry and the cyclic permutation of the subscripts of (2-27a). Finally, putting (2-27a, b, c) into the numerator of (2-20) obtains the result

70

VECTOR DIFFERENTIAL RELATIONS AND MAXWELL'S DIFFERENT1AL RELATIONS

anticipated in (2-21)

whence

I n rectangular coordinates, div F is found from (2-28) by setting hI and

U1

=

X, U 2

= y,

U3

=

hz

h3

=

1

Z

.

oFx

oFy

i3F

z ox + -oy + --

dlV F = - -

Rectangular

(2-29a)

whereas in the circular cylindrical and the spherical coordinate systems, the expressions become

. (hv F

div F

I

=-

a

P i3p

, + 1 -of,,, + -p i3

(pF )

Circular cylindrical

(2-2% )

Spherical

(2-29c)

P

13 2 (r p".) Or

1

+. r 8m 8

a (Flj sin 8)

138

1

+ r SIn . 8

"",

U (0) is given by (I-59). Its divergence III spherical eoordinates is

/ /

divE

=

/

(2-32)

Or

/

/

/ /

This null result signifies a flux plot in the region' > ro consisting of unbroken lines, as noted in Figure 1-16(b). All inverse r2 radial ficlds behave this way. Inside the charged cloud (r < r0), the E field (1-60) being proportional to , has the divergence

p" Eo

divE

r

<

TO

(2-33)

a nonzero, eonstant result, proportiollal to the density p" of the e1oud. Note that bringing Eo inside the divergence operator puts (2-33) into the form div (EoE) = Pv C/m

3

,<

;' /

FIG!. them insid, V.

clos'

'0

making the divergence of (EoE) the same as the charge density Pv inside the cloud. It is shown in Section 2-4B that this result is true in general, even for nonuniform charge distributions in free space.

A. Divergence Theorem If F(uj, U Z , U3, t) is well-behaved m some regIOn of space, then the integral identity Sv(divF)du

~sF'ds

(2-34)

is true for the dosed surface S bounding any volume V. Equation (2-34) implies that the volume integral of (div F) dv taken throughout any V equals the net flux of F emerging from the dosed surface S bounding V. A heuristic proof of (2-34) proceeds as follows. Suppose that V is subdivided into a large number n of volume-elements, any of which is designated AUi with each endosed by bounding surfaces ,S; as in Figure 2-5(a). The net flux emanating from AUi is the surface integral ofF· ds over S;, but from (2-20), this is also (div F) Av; for Au; sufficiently small, that is,

~Si F· ds

(div F)

AVi

(2-35)

The fluxes contributed by every Si will sum up to yield the net flux through the exterior surface S bounding the volume V. Thus the left side of (2-35) summed over the

ge

2-4 DIVERGENCE OF A VECTOR FUNCTION

73

,, (b)

(a)

fIGURE 2-5. Geometry of a .typical closed surface S, used in relation to the divergence theorem. (a) A volume V bounded by 8, with a lypical volume-dement t.v, bounded by S, inside. (b) Surfaces 8 2 and S, constructed to eliminate discontinuities or singularities from

V. do~cd

surfaces .1s i inside S yields

i

;= 1

[rh 1s,

F.dsJ = 1srh F.ds

(2-36) to the right side of (2-35) summed over the n volume elements Llv; as the number n tends toward infinity (and as .1vi --+ dv)

rh

Ie;;

F . ds

=

lim

Avc-~O

f

i=l

(div F) dv

=

r

Jv

(div F) do

(2-37)

just (2-34), known as the divergence theorem. If the limiting process yielding (2-37) is to be valid, it is necessary that F, together its first derivatives, be continuous in and on V. IfF and its divergence V . Fare not continuous, then the regions in Vor on S possessing such discontinuities or possible must be excluded by constructing closed surfaces about them, as typified 2-5(b). Note that the volume V of that figure is bounded by the multiple surface S = SI + S2 + S3, with S2 and S3 constructed to exclude discontinuities or singularities inside them. The normal unit vectors an, identified with each vee tor surface element ds = an ds on Sl' S2, and S3, are assumed outward unit vectors pointing away from the interi'or volume V. The following examples illustrate the foregoing remarks concerning the diver. gence theorem.

EXAMPLE 2·4. Supposc the one-dimensional field H(x) = axKx of Examplc 2-2(a) exists in a region. Illustrate the validity of the divergence theorem (2-34) by evaluating its volume and surface integrals inside and on the rcctangular parallelepiped bounded by the coordinate surfaccs x = 1, x = 4, Y = 2, y = - 2, z = 0, and z = 3, for the given H.

ail 74

VECTOR DIFFERENTIAL RELATIONS AND MAXWELL'S DIFFERENTIAL RELATIONS

ds = - ax dy dz on 82

2 (y) (x)

EXAMPLE 2-4

Since div H = K, the volume integral of (2-34) becomes (Jl

Evaluating the surface integral requires summing the integrals ofH . ds over the six sides of the parallelepiped. Because H is x-directed, however, H . ds is zero over four of these sides, the surface integral reducing to the same result as (I)

J. H 1s

. ds = ('3

JFO

('2 Jy~

= 48K - 12K = 36K

(2)

EXAMPLE 2·5. Given the p-dependent field: E = R"K/pl I 2, with K a constant, illustrate the validity of the divergence theorem by evaluating both integrals of (2-34) within and on a right circular cylinder of length L, radius R, and centered about the

0

Ii [fi.OJP

P

8 ilz

J

-~.

2na 2

a

z

~ [p fi.olp2

I' elP

2na

J=

azfi.o

-!na

2

0

a result proportional to the current density ]z = flna 2 in the wire. This special case demollstrates the validity of a Maxwell's diflerential relation to be developed in Section 2-SB. You may fi.nthcr show from (2-5+) that curl B outside the wire is zero, in view of the inverse p dependence of B there.

A. Theorem of Stokes If F(ub u2, U3, t) is well-behaved in some region, then the integral identity

1

(V x F)' ds

rf:

'Yt F

· dt

(2-56)

holds [i)r every closed line t in the region, if S is a surELee bounded by t. This is ealled the theorem (Jf Stokes. J\ heuristic proof follows along lines resembling the proof of the divergence theorem. Suppose the arbitrary S is subdivided into a large number n of surface-elements, typical ofwhieh is ~Si bounded by til as in Figure 2-9(a). The line integral ofF' de around Ii is inidt

(bl

FIGURE 2-9. Relative to Stokes's theorem. (a) Showing a typical interior surfaceelement ~Sl bounded by t't· (b) Closed lines t'2 and l3 comtructcd to eliminate discontinuities from S.

(2-57)

82

VECTOR DIFFERENTIAL RELATIONS AND MAXWELL'S DIFFERENTIAL RELATIONS

for ~Si sufficiently small. If the left side of (2-57) is surnmed over all closed contours t; on the surface S of Figure 2-9(a), the common edges of adjacent elements are traversed twice and in opposite directions to cause the integrations about t; to cancel everywhere on S except on its outer boundary t. Summing the left side of (2-57) over the n interior elements ~s; therefore obtains

f [rf-j{, F . dt]

;=1

=

rfF • dt j{

(2-58)

and equating to the right side of (2-57) summed over the same elements yields the result, as n approaches infinity

rf- F . dt ::Yt

=

lim

f

[(curl F) • ~s;]

As,->O t=1

=

r (curl F) . ds

Js

(2-59)

which is Stokes's theorem (2-56). As with the divergence theorem, It IS necessary in (2-56) that F together with its first derivatives be continuous. Ifnot, the discorHlnuities or singularities are excluded by constructing closed lines about them as in 2-9(b), causing S to be bounded by the closed line t = tl + t2 + t3' The connective strips, of vanishing widths as shown, are however, traversed twice so their integral contributions cancel. The positive sense of ds should as usual agree with the integration sense around t according to the right-hand rule.

w

E EXAMPLE 2·9. Given the vector field (1)

illustrate the validity of Stokes's theorem by evaluating (2-56) over the open surface S defined by the five sides of a cube measuring 1 m on a side and about the closed line t bounding S as shown.

(z)

(z)

Positive side of S

P4

____ Positive integration

s~: x= 0 ds =- axdydz

(x)

(x)

(b)

(a)

EXAMPLE 2-9. (a) l.ine elements on

(y)

t.

(b) Surface elements on S.

2-5 CURL OF A VECTOR FIELD

The line integral is evaluated first. The right side of (2-56) applied to making usc of figure (a)

= 0

+ Jz--o rl_ zdz + Jx-l ro_

5xdx

+ Jz-l ro_ zdz =

t:

83

becomes,

(2)

The surface integral of (2-56) is found next. From (2-52). ax

curl F =

ay

az

a

a a

ox

oy oz yZ yz

=

axz

+ a y5xy -

a z 5xz

(3)

whence the surface integral of (2-56) evaluate over S\, ... , '')5 yields, using figure (b),

r (curl F) • ds Js \

=

rl Jx~o r1

Jy=o

(4)

which agrees with (2).

EXAMPLE 2-10. Given the veetor field

F(ti) = a",K cot ti

(I)

in which K is a constant, illustrate the validity of Stokes's theorem evaluating (2-55) for the hemispherical surface S with a radius a, bounded by the closed jine t: ti = 90°, r = a as shown. There is a singularity in F on Sat () = 0°; it must be excluded to assure the validity of Stokes's theorem on the given surface. To accomplish this, a small circle t:3 at. (1 = til and r = a is constructed as in (b). Ifds is assumed positive outward on S, then the sense of the line integration is as noted, the integrals cancelling along (z and (4 oftlw connective strip

jr----Integration sense (a)

EXAMPLE 2-10. (a) Open hemispheric surface S. (h) Exclusion

(b)

or the singular point.

84

VECTOR DIFFERENTIAL RELATIONS AND MAXWELL'S DIFFERENTIAL RELATIONS

as its width vanishes, The line integral around t

r2~ F

J~-o

0

a",r

sin 8 d]

r=a

O~,,/2

= tl + t3

+ J O. The surface integral is evaluated using

Mal

ae

a4>

r sin

r

ar

a ao

a a

0

0

(r sin 8) K cot ()

curlF

-ar

K K - ao - cot 0 r r

whence

Jsr(curIF)ods=

r2:

l~

unit vect'

r1t~2 (-It'asinOdOd)

J",-o JO-Ol

=

~2naKeos81

(3)

which agrees with (2). You might consider how the results would have compared had one ignored the singularity.

B. Maxwell's Curl Relations for Electric and Magnetic Fields in Free Space In Section 2-4B, the divergence of a vector fUllction was put to use in deriving the differential Maxwell equations (2-39) and (2-41) from their integral versions (1-53) and (1-54). The definition of the curl may similarly be used to obtain the differential forms of the remaining equations (I-55) and (i-56), Because the latter are correct for closed lines of arbitrary shapes and sizes, one may choose t in the form of any small closed path bounding a j Lls 1 in the vicinity of any point, as in Figure 2-7. Taking the ratio of (I-55) to Lls 1 yidds, with the assignment of the vector sense a l to each side, d dt

r

Jl1s, Lls 1

the the Th sib] sati

reI;

It cu

aF fo]

Bods

(2-60)

(2-43), the left side, as AS l -40, becomes a1rcurlEl 1 . The right side denotes time rate of decrease of the ratio of the magnetic flux I1ljJm to I1S1> but this is just compon(~nt Bl at the point P. The limit of (2-60) therefore reduces to

£,

(~

(2-61 )

te

''''UIII: the at component of curl E to the time rate of decrease of the at component tnagneticJlux densilY B at any point. 4 The choice of the direction assigned by

2

al

[curl EJI

differentiation symbol alDt the lild that the field B is a a function of t only, I,)r a fixed

in (2-61) replaces the total differentiation did! in (2-60), of space as well as of time, whereas the volume integral

t.\,

C fl C

2-6 SUMMARY OF MAXWELL'S EQUATIONS: COMPLEX, TIME-HARMONIC FORMS

85

is arbitrary, implying that two similar results aligned with the directions of the unit vectors az and a 3 and independent of (2-61) are also valid. Combining these Ilectorially thus obtains the total curl of E at the point 1.1

Making use of the notation of (2-42) yields the more compact form

aB

VxE=--Vjm

2

at

(2-62)

the differential form of Faraday's law (I-55). Equation (2-62) states that the curl of the field E at any position is precisely the time rate of decrease of the field B there. This implies that the presence of a time-varying magnetic field B in a region is responsible for an induccd time-varying E in that region, such that (2-62) is cverywhere satisfied. A procedure similar to that used to derive (2-62) is applicable to the Maxwell relation (I-56), yielding the differential equation

V

B = Ilo

X -

a(EoE) J +- - A/m z

at

(2-63)

It states that the curl of B/llo at any point in a region is the sum of the electric current density J and the displacement current density a(EoE)/ot at that point. If the electric and magnetic fields in free space are static, the operator Ojat appearing in (2-62) and (2-63) should be set to zero. This restriction provides the following curl relations for time-static fields

VxE=O B V x-=J Ilo

(2-64 ) Curl relations for static E, B fields (2-65)

Equation (2-64) stales that any static E field is irrotational (conservative), whereas (2-65) specifies that the curl of a static B field at every point in space is proportional to the current density J there. 2·6 SUMMARY OF MAXWELL'S EQUATIONS: COMPLEX, TIME·HARMONIC FORMS

One may recall that in Sections 2-4B and 2-5 the differential Maxwell equations for free space were obtained Irom their integral forms, (1-53) through (I-56). These are collected for reference in Table 2-1, columns I and III. The integral Maxwell equations

~

TABLE 2-1 Time-dependent and complex time-harmonic forms of Maxwell's equations in free space

Differential forms

Integral forms TIME-DEPENDENT

~s EOE' ds

j:

~s EoE' ds

p,du

¢. B· ds = 0 v

~sB' ds

S

~

E. dt =

d

dl

v {

A: ~ . dt 'fr flo

i' B· ds

i' J' ds + r!.-, i' EoE' ds

Js

dt Js

J,B . dt ::rc fto

V' (EoE)

vx

Is B' ds

V' (EoE) =

V'B

0

E

cB

v x

fJt

i' j . ds + jw i' EoE' ds

Js

p,

V' B

0

A: E' dt = :Yt

[I

Js

Iv p,d"

IV, COMPLEX, TIME-HARMONIC

TIME-DEPENDENT

COMPLEX, TIME-HARMONIC

Js

V x

a

B

J + _ (EoE)

flo

Vx

ct

P" 0 r2-71]

E -jwB [2-72J ~

13 flo

~

J + jWEoE

[I-56]

,.,



a result decidedly not of the form of (2-83) with } + = constant', whereupon differentiating it to evaluate dz/dt yields the phase velocity

dz vp

= dt =

w

Po m/sec

(2-125a)

Because {jo = wji;~~, and with flo = 4n x 10 7, Eo ~ 1O- 9/36n, the phase velocity of a uniform plane wave in empty space is I

--- = c ~ JfloEo

3

X

10 8 m/sec

(2-125b)

-

the speed of light. 7 ._ A comparison of the complex expressions, (2-115) and (2-ll7), for Ex(z) and By(z) shows that their separate traveling wave terms are paired into ratios producing the sarne constant. Thus, write (2-115) and (2-117) in the forms = E~ e -

j{Joz

+ E;;' ei{loz

E; (z) + E; (z)

(2-126)

and -+

- Eme - j/1oZ B- (Z ) y

C

--

Em

_ilJoz

--I:""

C

(2-127) in which E;(Z), E;(z) and B;(x), fJ;(z) symbolically denote the positive z and negative Z traveling wave terms directly above them. Then the following complex ratios hold at any point in the region

c~ 3

X

10 8 m/sec

(2-128)

to provide j means for finding one of the fields whenever the other is known_ A more common variation of this technique is achieved by modifying the B field in empty 7Experiments have shown that the speed oflight, c, is more nearly 2.99792 x 10 8 m/sec. This value, together with the assumed permeability for free space fJ-o = 4n x 10- 7 H/m, inserted into (2-125b), is seen to a value for Eo that departs slightly from the approximate value 10 - 9/36n given.

DIFFERENTIAL REI"ATIONS AND MAXWELL'S DIFFERENTIAL RELATIONS (x)

t;:

--Z B/

-~

t) ""

Ei' (z, t) = Ei;,

cos (wi - f30z

+ 1>+) Equiphase surface

Motion

cE;:;

cos(wt - f30z

+ .... .,.. +)

(at t = 0) (z)

[nOURE 2-12. Vector plot ofthc fields ofa uniform plane wave along the z axis. Note the typical equiphase surface, depicting fluxes of

E;

and 13;.

through a division by /lo, defining a magnetic intensity field denoted by the symbol lor empty space as follows.

B

= H A/m

For empty space

(2-129)

/lo

Thus, denoting B: (Z)//lo by if: (z), and B; (Z)//lo by if; (z) the following ratios the traveling wave terms are valid for plane waves in empty space

E: (z) =~: (z)

/lo

(z)

fly

(z)

=

/loc

=~= J/loEo

r;;;, == 110 ~ 120n Q

V~

r;;;, == '10 ~ 120n Q

V~

(2-130a)

(2-130b)

2·

J

The real ratio, /lo/Eo the units volts per meter per ampere per meter, or ohms), ill called the intrinsic wave !11l;(JI!llran.ce empty space, and is denoted by the symbol 110' The advantage of (2-130) over ill that the ratio 110 is a usefully smaller number. The real impedal1ce ratio of 30) shows that the electric and magnetic fields of uniform plane waves in are in phase with one another, a condition evident on comparing tht~ the negative Z traveling solutions of (2-126) and (2-127). Each contains argument in the exponential factors, ample evidence of their 2-12 depicts the real-time electric and magnetic fields of in space at t = O.

T dl Ul

m E

li sl 81

p p

IXAMPLE 2·11. Suppose

p

empty space has the electric field (I)

a a

2-11 WAVE POLARIZATION

103

its frequency being 20 MHz. (a) What 1~ it§. direction of travel? Its amplitude? Its vector direction in space? (b) Find the associated B field and the equivalent H field. (e) Express E, :8, and H in real-time form. (d) Find the phase factor Po, the phase velocity, and the wavelength of this electromagnetic wave. (a) A comparison of (1) wit!: (2-115) 2r (2-126) reveals a positive z}raveling wave, whence the symbolism: E(z) = axE; (z). The real amplitude is E~ = 1000 Vim, with the vector field x directed in space. (b) Using either (2-127) or the ratio (2-128)

The use of (2-130a) obtains the magnetic intensity ~+

- Ex-(z) _1000 _ 2 65 -jPoz A/ z) -- e -jPoz . e m H~+( y '10 120n

(c) The real-time fields are obtained from (2-74) by taking the real part after multiplication by Jillt

E; (z, t) = Re [1000e- iPozJwt] =

B;

t) = 3.33

H;

t) = 2.65 cos (wt - Poz) A/m

X

Poz) V/m

1000 cos (wt 2

10- 6 cos (wt - Poz) Wb/m (or T)

(d) Using (2-118), (2-125), (2-124), and (2-122) yields

r--:-

150 =0 OJ,>! f.loEo vp = c = 3 ~

X

2n(20 x 10 6 )

c

3 x 10

.f

8

= 0.42

rad/m

10 8 m/sec 3 x 10 8

2n

Po

OJ

= - =

20

x 106

=15m

2·11 WAVE POLARIZATION

The vector orientation, or polarization, of an electromagnetic wave in space is usually described with reference to its electric field direction. Thus in Figure 2-12, the z traveling uniform plane wave shown with the field components Ex) Hy is said to be polarized in the x direction (or simply x-polarized). Similarly, the plane wave with the components Ey , Hx described in Problem 2-43 is polarized in the y direction. Both these waves are linearly polarized, because the electric field vector in any fixed z plane describes a straight-line path as time passes. Because Maxwell's equations are linear equations, a vector superposition, or summing, of the two linearly polarized uniform plane waves just introduced will also provide a valid field solution. The resultant vector sum will not necessarily be linearly polarized, however, depending on the phase condition between the x- and the ypolarized electric field components. For example, with Ex(::., t) Emx cos (wt Poz) and Ey(z) t) = Emy cos (wt - Poz) propagating in phase and at the same frequency along the z-axis, their sum, E = axEx + ayEy, would appear as depicted in Figure

104

VECTOR DIFFERENTIAL RELATIONS AND MAXWELL'S DIFFERENTIAL RELATIONS

l'

(x) I

Locus of E in z=O plane

I

.E)O.lOpz cos (f) K(r, 1» = a,100/r 2 + a820/r + a4>10r cos 1> [Answer: div F = 3z + 4x, fields A, G, and J are sourceless]

e,

2-9. Prove, by expansion in rectangular coordinates, that V . (F identity (12) in Table 2-2.

+ G) =

V • F + V . G, the

2-10. By expansion in the rectangular coordinate system, prove the identity (15) in Table 2-2, V' (fF) =F·Vf+f(V·F).

2-11. Show that the following fields are, divergen~eless (source-free). (a) The p-directed, inverse-p dcpendent field F = a,,/p, for p > 0; and (b) the r-directed, inverse-r 2 dependent field, G = aJ r2, for r > O. (By comparison with results found in Example 1-13, with what kinds of static-charge sources are these field-types identified?) (a) Given the class of electric fields E(p) = apK/ pn with K a constant and n a parameter, find div E. What choice of n yields a divergenceless (charge-free) field everywhere (excluding p = OJ? Comment on this conclusion relative to (1-61), applicable to the uniform line charge. (b) Given the class of electric fields E = a,K/rn , find div E for r > O. Which choice of the parameter n provides a divergenceless field? Comment on this conclusion with respect to (1-5 7b), the electric field of the point charge.

2-12.

PROBLEMS

107

SECTION 2-4A 2-13. Assuming the same six-sided closed surface S to bound the box-shaped interior volume as in Example 2-4, assume the field G(x,y, z) = a z lOxy 2 z3 exists in the region. Illustrate the validity of the divergence theorem (2-34) by evaluating its volume and surface integrals in and on the given parallelepiped. [Answer: 10,800] 2-14. Assuming the same right circular cylindrical region of radius p = a and length t as for Example 2-5, illustrate the correctness of the divergence theorem for this region, given the electric field E = appop3/4Eoa2, that corresponds to the nonuniform charge density of Problem 1-43. [Note for this case that no singularity exists within the given V or on S, thereby obviating any need for the exclusion surface S2 used in Example 2-5.} [Answer: nLpoa 2/2E o] 2-15. The first octant of a sphere centered at the origin is bounded by the four coordinate surfaces: r = a, 4> = 0, 4> = n/2, and on the bottom by the plane = n/2. Sketch it. Given that the field F(r, 4» arlO - a.,,30r sin () cos 4> exists in this region, illustrate the truth of the divergence theorem (2-34) by evaluating the volume and surface integrals within and on the defined region for the given field. [Answer: lOa 2 (a + n/2)]

e

e,

SECTION 2-48 2-16. In Problem 1-28, the electric field within the uniformly charged spherical shell (a < r < b) was found to consist of only thl! component Er = pv(r3 - a3 ) /3Eor2. Show that inserting this field into the Maxwell divergence relation (2-39) yields the charge density originally assumed. 2-17. It was found by use of Gauss's law in Problem 1-29 that the choice of the nonuniform charge density Pv = Po(l - 4r/3a) within a sphere of radius a yields the electric field therein given by Po EO

(r3

r2) 3a

Show that div (EoE) for this field yields the charge density originally assumed, thereby satisfying Maxwell's equation (2-39).

2-18. (a) In Problem 1-30(b) it was found, using Gauss's law, that the static electric field within the uniformly charged cylindrical cloud is E = a p pvp/2E o ' Determine div (EoE), to prove that Maxwell's divergence relation (2-39) is satisfied. (b) Show similarly, from the E-field solution of Problem 1-31 (b), that E inside the nonuniformly charged cylindrical cloud of that problem satisfies the Maxwell divergence relation (2-39). 2-19. By the application of (2-28) in the appropriate coordinate system, show that the Maxwell relation (2-41), div B = 0, is satisfied for each of the B fields given by (1-64) for the long, straight wire, by (1-65) for the current sheet, and by (1-66) and (1-67) lor the toroid and solenoid. What is the physical interpretation of the zero value of the divergence expected of each and every B field?

SECTION 2-5 2-20. With reference to a diagram resembling Figure 2-8 but adapted to the rectangular coordinate system, give the details of a proof of the curl expression (2-50) carried out in rectangular coordinate form. 2-21. (a) By the substitution of the appropriate coordinate variables and metric coefficients into the determinant (2-51) for the curl of a vector field, show to what result it expands in the rectangular coordinate system. (b) Similarly show that (2-54) and (2-55) are the results of expanding (2-51) in the circular cylindrical and the spherical coordinate systems. 2-22. Find the curl of each of the vector fields given in Problem 2-8. Which of those fields are irrotational (conservative)? [Answer: (c) -3ax - IOxa y - 3a., (e) -aplOp cos 4> + a.,,5p sin 4> + a z l5Z cos 4>]

108

VECTOR DIFFERENTIAL RELATIONS AND MAXWELL'S DIFFERENTIAL RELATIONS

2-23.

Find in detail the curl of the vcctors Vg, VG, and Vh generated in Problem 2-2, [These results exemplify the validity of the vector identity (20) in Table 2-2.]

2-24.

By use of expansions in rectangular coordinates, prove the vcctor identity (17) in Table 2-2, that V x (iF) = (Vf) x F + f(V x F).

2-25. Given the vector field F(x,y, z) 2xz + 5YZ 2, find the following.

=

3xy 3 ax

+ 4y 2 z 2 ay

and the scalar field f(x,y, z) =

(a) Vf (b) V' F (c) V x F (d) V· (iF) (e) V' (Vj) =- V 2f (f) V x (V x F) (g) V· (V X F) (h) V X (Vf) [Answcr: (a) 2zax + 5z 2 a y + + lOyz)a z (c) -3y 2zax - 9xy 2 a z (e) lOy (g) 0: also by identity (19) in Table 2-2]

2-26. Given are the fields G(p, , z) = aq,5p sin - a z 6p 2 z 2 and g(p, , z) = 3pz sin . Find the functions (a) Vg (b) V· G (c) V x G (d) V· (gG) (el V· (Vg) =- V 2 g (f) V X (V x G) (g) V(V·G) [Answer: (b) 5 cos 12p2z, (e) zero (f) ap(lOp-I cos - 24pz) + a z 24zZ]

2-27.

Given functions (a) Vh (b) (d) V· (Vil) [ Answer: (a) (f) 0]

thc fields H(r, 0) = arlOr cos 0

+ aq,20r 2

and

her,

0, ' ~"IOSinwt,mJo B = a Z --

L

Conductive, magnetic core (g,o)

at

sin wt

aB

at

--,

E and J flux ) (induced eddy currents)

aB

at (6)

(a)

FIGURE :1-14. Eddy current, in con CfJ

Dnl =

or

Bn2

D 2 ) = Ps

Case B: a2 Ps

[3-42]

Dn2

n' (Dl

n X (HI - H 2)

Case B: a 2 -> n x HI = Js n X (El - E2l = 0

Js CfJ

[3-72] [3-79]

148

MAXWELVS EQUATIONS AND BOUNDARY CONDITIONS

A boundary condition, comparing the tangential components of the E fields to either side of an interface, may he obtained from Faraday's integral law (3-78). The details of the derivation may be avoided if one recalls that Ampere's line-integral law (3-66) leads to the boundary condition (3-70a), lIt! - 1It2 = }s(n)' The boundary condition comparing the tangential components ofE can be analogously found by applying (3-78) to a similar thin rectangle, yielding the analog of (3-70a)

(3-79)

Thus the tangential component of the E field is alwOeYs continuous at an inte~face. The right side of (3-79) is evidently zero because no magnetic currents are physically possible. A summary of the four boundary conditions derived from Maxwell's integral laws for material regions in Sections 3-2C, 3-4B, and in the present section, is given in Table 3-2.

EXAMPLE 3·6. (a) Derive a refractive law for E at an interface separating two noneonductive regions. (b) Deduce from boundary conditions the direction ofE just outside a perfect conductor. (a) The boundary conditions for the tangential and the normal eomponents of E at an inter1i:lce separating nonconductive regions are (3-43) and (3-79); that is, EIEni = E2En2 and E'l = E'2' From the latter and the geometry of (a), one obtains

(3-80) a result analogous with (3-76) of Example 3-5 concerned with the refraction of B lines. (b) From (3-44), a perfectly conductive region 2 implies null fields inside it. Then (3-79), Etl in the adjacent region I must vanish also. The remaining normal component in region 1 is given by (3-45). ])n1 p" yielding Ps E lE"1 as shown in (b).

Region 1:

Cl.q. € 1. 0 being air. With H = 150ax Aim in the air region, what is the snrface current density on the perfect conductor? How much total current I flows in a 20-cm-wide x-directed strip of this conductor surlace? Sketch this system showing H, J" and a [(OW current flux lines. (b) Find the current density on the conductor surface of (a), this time assuming H = 30ax + 40a y A/m. Sketeh this system. (c) Suppose in the geometry of Figure 1-19(a) that the long, straight wire shown is a perfect conductor, and that surface eurrents totaling I flow on the conductor surface p = a. The B field for p > a is still given correctly (1-64). Use this field to deduce the surface current density Js on the wire. Formulate a vector integral relationship between J and J" showing a related sketch. 3-25. What two simultaneous boundary conditions arc being satisfied by the magnetic field refraction expression (3-76)? Establish that, if region I is air and region 2 is iron with 11,2 = 104 (a case of high contrast in permeabilities), the tilt angle 01 of Br from the normal in region I is very slllall for most values or0 2 . For example, find 1 if0 2 = 0, 45°, 89°, and then 89.9°. How lin from the normal must O2 be if 8 1 is to become as large as 10°? Sketch this example.

°

3-26. The toroidal iron core ofrectangular cross section partly fills the closely wound toroidal coil of!l turns and carrying the direct current 1 as shown. (a) Usc the right-hand rule (thumb in the sense or 1) to establish the direction of H inside the winding. (b) Use the static form of Ampt're's law C)-66) to deduce H at any radius p within the winding, and determine B for is being satisfied the two regions. Which boundary condition for magnetic fields Cfable at the air-iron interface? (c) From H deduce expressions for the magnetization density field M in the two regions. Sketch flux plots showing (in side views) the relative densities of H, B/l1o, and M in the two regions, assuming 11, » 1 for the iron. (d) Find Jm within the iron as well as J8m on the timr sides of the iron core. Sketch representative vectors or fluxes depicting these quantities. (c) If a = 1 em, b = 1.5 cm, C= 2 em, d = I cm, fl, = 1000, n = 100 turns, 1= 100 rnA, find t he values of Hand B at p = a + and b (just within the iron), at p b + and p = c.

3-27. As a simple exercise iu applying boundary conditions, an air space (region 1) defined for all > 0 and a magnetic substrate with 11, = 4 (region 2) occurring for all z < 0 are separated by the inllnitc pbmc interlace at z O. The constant, static magnetic field in region I is given to be BI = O.3a x + O.4ay + 0.5a z Wb/m 2 . Sketch BI (shown for convenience at the origin) and the Ilormal unit vcctor n at the interface (its direction taken as going from region 2 to region I). (a) Make use of the boundary conditions (Table 3-2), concerning the continuity of appropriate tangential or llormal field components at the interface, to deduce the vector fields HI> B 2 , and H2 in the as well as the field magnitudes. (Leave H expressions in terms of

PROBLEM 3-26

178

MAXWELL'S EQUATIONS AND BOUNDARY CONDITIONS

If"\!, ~----------

, f

\

~+-..,...-,

\-+4---:/ -11-+"""-,

(

----t--- (z)

1: (1-'0)

'-:1-+.......- ;

' -2;-(p) 3;

(1-'0)

PROBLEM 3-28

the symbolic 110.) (b) By use of the definition of n • B, find the angles (), and (}2 between n and B (or H) in the two regions. (Label (J, on the sketch.) Check your answers by use of (3-76). [Answer: (a) B2 = 1.2a x + 1.6ay + 0.5a z Wb/m 2 (b) (}2 = 76°]

3-28. A very long, nonmagnetic conductor (fl. I) of radius a carries the static current I as shown. The conductor is surrounded by a cylindrical sleeve of nonconducting magnetic material with a thickness extending from p = a to p = b and the permeability fl. The surrounding region is air. (a) Make use of symmetry and Ampere's law (3-66) to find Hand B in the three regions. (Label the closed lines employed in the proof, depicting H in the proper sense on each line.) (b) Find the M field in the magnetic region. If 1= 628 A, a = I em, b = 1.5 cm, fl. = 6 for the magnetic sleeve, sketch H"" B"" and M", versus p for this system. Comment on the continuity (or otherwise) of these tangential fields at the interfaces. (c) By use of (3-56) and (3-73b), find the volume magnetization current density 1m and the bound surface current densities J,m within and on the magnetic sleeve.

SECTION 3-5 3-29. Two semi-infinite regions, air (region I) for z > 0, and a dielectric (region 2, in which E = 4Eo) for z < 0, are separated by the interface at z = O. In the air region, the constant electric field El = l5a x + 20ay + 30az Vim is given. Sketch El for convenience at the origin. (a) Find D and E for both regions, making use of boundary conditions (Table 3-2). (Leave Eo explicitly in the D expressions.) (b) Find the refraction angles ()l and ()2 from the normal in both regions, making use of the definition ofn . E ifn is directed from region 2 to region 1. Use the refraction law (3-80) as a check. Answer: E2 = - 15ax + 20a y + 7.5a z V /m, ()2 = 73.30°]

r

SECTION 3-7 3-30. Prove the expressions (3-90a) and (3-9Gb) for the attenuation constant IX and the phase constant fJ associated with uniform plane waves in an unbounded, lossy region. 3-31. Assume uniform plane waves to be traveling at the frequency f= 100 MHz in a lossy region having the constitutive parameters fl = flo, E = 6Eo, (J = 10- 2 mho/m. (a) By direct substitution into (3-88), determine the value of the complex propagation constant associated with the waves, expressing y in its complex rectangular 101'm denoted by (3-89). From this result infer the values of the wave attenuation constant and phase constant. (b) Find the attenuation constant and the phase constant by use of (3-90a) and (3-90b). [Answer: IX 0.761 Np/m, f3 = 5.187 rad/m] 3-32. Repeat Problem 3-31, this time assuming the parameters of the lossy region to be fl = flo, E I.B Eo, (J = 10 mho/m, and in which uniform plane waves are traveling at the frequency f= 10 GHz. [Answer: y = 597.7 + j660.5 m- i ] 3-33. M.aking use of the free-space parameters fl flo, E Eo and (J 0, show that the expressions (3-90a), (3-90b) and (3-99a) reduce to the free-space results IX = 0, fJ = fJo of(2-11B), and I] = 1]0 of (2-13Gb). 3-34. Prove that the penetration of three skin depths by a plane wave into a conductive region produces an amplitude reduction to 5% of the reference value. Show that six skin depths yields 0.25'Yo'

PROBLEMS

179

3-35. Given the electric-field plane wave solution (3-91 b) in which the propagation constant is defined by (3-88), show by substitution into the Maxwell curl equation (3-83) that the corresponding magnetic field solution becomes (3-98c), if the intrinsic wave impedance q is defined by (3-99a). [Hint: Show that the coefficient y/jWfl reduces to q-I.] 3-36.

Show that the expression for intrinsic wave impedance

q,

defined by (3-97) as

can be re-expressed in complex polar form by the last expression given in (3-99a).

E;

H;,

3-37. A positive z-traveling, uniform plane wave has the field components and with the electric-field amplitude E; = 200 Vim, and operates at the frequency f = 100 MHz. It travels in a lossy region with the parameters given in Problem 3-31 (flo, 6Eo, (J = 10-2 ). The propagation constant in this region at 100 MHz was found to be y = 0.761 + jS.IS7 m- I ). (a) Determine the wavelength of this wave. Find its depth of penetration, b. What is the phase velocity of this wave? (b) Determine the intrinsic wave impedance q fo!, this region, at the (z) accompanying given frequency. Use this to obtain the expression for the magnetic field the given electric field. (c) Show a labeled sketch, patterned after Figure 3-18, showing the real-time E; t) and H; (z, t) fields of this uniform plane wave, at t = O. Label the depth of penetration as well as the wavelength on your diagram. [Answer: (a) b = 1.314 m (b) ~ = 150.6e iS . 4 ' il]

H;

3-38. A vehicle located far above the surface of the sea transmits an electromagnetic signal at the frequency j. Upon striking the air-sea interface, a transmitted wave penetrates the sea. The waves at the surface are presumed to be sufficiently far from the source that they,may, locally at least, be considered to be uniform plane waves. Supposing the net transmitted electric field amplitude is f-; = I Vim, how far will the wave penetrate before reaching of its surface value? Perform this calculation at two very low radio frequencies: 10 kHz (in the VLF range) and 1000 Hz (ELF), assuming sea water has the constants E, = 81 and (j = 4 U/m at these frequencies. Comment on the effectiveness of undersea radio communication, bascd on your results.

SECTION 3-8 3·39. Use the general expressions (3-90a, 90b, 99a) to derive (3-112), applicable to a good conductor (for which a/wE> 1).

3-40. for ex,

p,

Use the general exprcssions (3-109) through (3-111) to prove (3-113), approximations and q applicable to waves traveling in good dielectrics, for which EN/E' « 1.

3-41. An electromagnetic uniform plane wave is specified in some lossy region by the fields E; (z) = 3142e- Yz Vim, fI; (z) = fI;e- YZ A/m at the frequcncy f = 1000 MHz. The region has the parameters 11 = /-10, E = 24Eo, and (j = 48 mho/m. (a) Show that the loss tangent of this region at the givcn frequency is 36. Is the region classified as a "good conductor" or not? Explain. (b) Find the attenuation and phase constants of the region at this frequency. (Reasonable approximations are allowed.) Show that the "depth of penetration" of the wave into the region at this frequency is about 2 mm. Use a sketch of the real-time electric field (in the vicinity of the z-origin) to explain the meaning of "depth of penctration." (c) Find the wavelength of this wave, labeling it on the sketch of part (b). Compare this wavelength with that occurring in this region assuming now that it is completely lossless (samc fl and E, but now with a = 0). Comment. (d) Evaluate the complex amplitudc fI; of the magnetic field of this plane wave, making use of the intrinsic wave impedance of the region. (To what fact do you attribute the angle of q being close to 45°?)

CHAPTER 4 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __

Static and Quasi-Static Electric Fields

Electric fields of stationary charge distributions in space are considered in this chapter. Maxwell's equations, subjected to the time-static assumption, tJ/Dt = 0, provide an uncoupling of the static electric fields from the static magnetic fields. Gauss's law is applied to symmetrical systems; and the scalar potential field (]) is derived to supply an intermediate, often simplifying, step useful f(lr finding the static E field. Expressions fix the stored energy of an electrostatic system are derived and applied to twoconductor capacitance systems. Boundary-value problems of electrostatics are treated by means of Laplace's equation and extended to finite difIerence methods of solution for arbitrary, two-dimensional boundary shapes. Image and fIux-mapping techniques are discussed as alternative approaches to capaci lance problems, all(i. a capacitanceconductance analog is developed. The chapter is concluded with a consideration of the fc)rces of electric charge systems.

4·1 MAXWELL'S EQUATIONS FOR STATIC ELECTRIC FIELDS In Chapter 3, the Maxwell equations and boundary conditions f(lr time-varying electromagnetic fields in material media at rest were developed. Time-varying B (or H) and E (or D) fields are produced in a region whenever the charge and current sources of the fields are time-varying. For certain generic classes of field problems, it is advantageous to consider the sources to be non-lime-varying, that is, time-static (or just static). Then the charges and possibly currents responsible for the fields are stationary. The governing Maxwell equations for time-static fields are (3-24), (3-48), (3-59), and with the operator a/at set to zero, yielding V . D PI!' V x E 0, V' B = 0, and V x H = J. The static fields are designated D(u j , 11 2 , Pv(Ub U2, and so on.

180

4-2 STATIC ELECTRIC FIELDS OF FIXED-CHARGE ENSEMBLES iN FREE SPACE

181

An inspection of the static Maxwell relations reveals a new property not valid for their more general, time-varying forms. Thus, the static electric fields D and E are governed solely by the divergence and curl properties V' D

= Pv

VxE

(4-1 )

0

(4-2)

whereas the behavior of the static magnetic ,fields Band H is dictated by

V·B = 0

(4-3)

v x H=J

(4-4)

The coupling between the electric and magnetic field quantities, generally provided under time-varying conditions by the terms aB/at and aD/at appearing in (3-77) and (3-59), is seen to be missing in these pairs of equations. The sources of electrostatic fields are, from the divergence expression (4-1), static charges of density Pv' Magnetostatic fields, on the other hand, have static (direct) currents for sources, as noted in (4-4). In the present chapter, solutions of the electrostatic field equations (4-1) and (4-2) are considered from several points of view, whereas a detailed discussion of magneto statics by use of (4-3) and (4-4) is deferred until Chapter 5. The differential equations of electrostatics are, together with their integral forms and boundary conditions, given TABLE 4·1 Maxwell's Equations of Electrostatics DIFFERENTIAL FORM

V- D = Pv

vx

E

=

0

INTEGRAL FORM

[4-1]

~s D

[4-2]

1:

~

. ds = q

(4-5)

E - dt =O

(4-6)

BOUNDARY CONDITION

Etl

Et2 = 0

(4-3)

in Table 4-1. For a linear, homogeneous, and isotropic material, moreover, (3-30c) is applicable

D=EE

(4-9)

4·2 STATIC ELECTRIC FIELDS OF FIXED-CHARGE ENSEMBLES IN FyE SPACE

••

The Maxwell equations in Table 4-1 apply to fixed charges in free space, as well as to systems of rlielectrics and conductors into (or onto) which charges have been introduced such that static equilibrium of the charge distribution has been reached. Examples of the applications of the Gauss law (4-5) are given in Section 1-9. One of the results, (1-58), is Coulomb's force law (4-10a)

182

STATIC AND QUASI-STATIC ELECTRIC FlELDS

FIGURE 4-1. Illustrating quantities appearing in Coulomb's [()rcc law,

glvmg the force acting on q m the presence of the field E produced by a second charge q as shown in Figure 4-1, The symbol R is used instead of the spherical coordinate variable r because the source q is not necessarily located at the origin 0, The field of q was deduced from Gauss's law in Section 1-9 to be

E = aR--'---;;:-N/C

or

Vim

lOb)

Thus (4-10a) is a special case of the Lorentz [(lITe law (1-52) in the absence of a magnetic field; that is, F = q'E, Maxwell's equations (4-1) and are linear equations; therefore, any sum of their solutions in free space constitutes a solution, Suppose an aggregate of point charge of arbitrary posi tive or negative strengths is located at fixed points P' as in Figure 4-2, The total electrostatic field at the field point P is the sum of n terms like (4-lOb)

E Moreover, if a charge

Ij

10e)

is placed at P, the [()rce on it, from (1-52), becomes (4-10d)

If a system contains a large number of fixed charges, it is undesirable to use a surnmation like (4-lOc) or (4-10d), It is pre/tTable to replace the charge ensemble

(z)

FIGURE 4-2. Electrostatic fIeld of n discrete charges.

4-2 STATIC ELECTRIC FlELnS OF FIXED-CIIARGE ENSEMBLES IN FREE SPACE

183

with a jime/ion representing the average charge density in every volume-, surface-, or line-clement of the region. The symbols Pv, p" or Pc have been used to denote these density functions as discussed in Section 1-9 relative to Figure 1-1]. A continuum of charges distributed throughout some region with a density Pv thus possesses the charge dq = Pv dll in every dv element. Generally, Pv is a function of position and time, though for static Helds, the variable I is missing. With dq p"dv' located at the source point P' (x',y', the fidd dE at P due to dq is obta.ined from (4-10h), written

dE(x,y, a potential due to all the charges except qk' Denoting the bracketed factors by the absolute potential due to all the charges except the kth

If the assemblage of charges is not discrete, but rather a continuum of density Pv distributed throughout some volume region V, then (4-58a) becomes an integral on replacing qk with dq = Pvdv, obtaining

(4-58b)

wherein is the absolute potential at the position of Pv- For charge continua comprised ofswlace or line distributions as discussed in Sections 4-2 and 4-5, the following expressions are used in lieu of the preceding ones.

ull (4-58c)

(4-58d)

«II, In computing Ue £i-om one or a combinatioll of these four expressions, only the charge distributions and the potentials at the charge locations need to be known. The energy integrals (4-58), expressed in terms of the potential distribution accompanying static charge distributions in space, can also be written in terms of only the D and E fields that occupy the whole of space. The result becomes

(4-58e)

202

STATIC AND QUASI-STATIC ELECTRIC FIELDS

To prove the latter, suppose that surface charges of density Ps exist on the closed conducting surface S, where S may consist of n individual conductors such that S = S1 + S2 + ... + Sn, with the additional possibility of a volume charge density Pv occupying the region V enclosed by S.3 The two-conductor capacitor of Figure 4-6(b) or (c) represents such a system. The electrostatic energy of the system is the sum of (4-58b) and (4-58c) (4-59) in which S denotes the simply connected closed surface of the charged conductors, and V is the region between the conductors. Using the boundary condition (3-45) but with the unit vector n directed away from the volume V so that Ps = - n ' D, (4-59) becomes

Iv p/Pdv Iv V • (D) dv + ! Iv pv dll

Ue = -! ~s (D) • nds +!

(4-60)

in which the transformation of the dosed-surface integral to the volume integral4 is accomplished by use of the divergence theorem (2-34). The use of the vector identity (15) of Table 2-2, V- (fG) =IV'G+G-Vf, yields

Iv D • (V OCJ, choose the form (4-88a) (6)

The unknowns C 1 through C4 and kx are evaluated by use of the boundary conditions. Applying (2) to (6) yields

to obtain C 1 = O. Then (6) becomes

(7) Applying the boundary eondition (3) to the latter obtains

211

212

STATIC AND QUASI-STATIC ELECTRIC FIELDS

(x)

(a)

4> (x, 0) = V V I--L------,

o

a

1>

=V

o

(x)

4>

a

(x)

=- V

(b)

( c)

I

I

I Constant 100 V

I I

(x-h,y)

¢=V

Gap

QZ?1\vzzFlTil/?/A/

(x,y+h) (t)(x,y) '-C~-,

y

\

~-----,,"""-

__ +_

2' 3 -

I ¢(x,y-h) I I

!

-+ -

;6

'4 I

__ ¢=O

I o L - - - - - - ' - - ' - x - - L - - - - (x)

(6)

(a)

FIGURE 4-10. (a) Geometry relative to the finite-difference method. (b) Triangular-trough example of a two-dimensional electrostatic problem, showing a squaregridded overlay with a labeling of points at which the potentials , Co = up(L\C), yielding the total capacitance

c = up I1C ns

(4-113)

It remains to determine the field cell capacitance I1C. Assuming charges I1q, - L\q induced on the conducting-foil walls at the top and bottom of each cell as in Figure 4-I9(c), one obtains, fi'Orn (4-48), I1C = I1qlVo , in which the potential difference beE . dt, also expressed in terms of an average electween the boundaries is Vo tric field hy Vo = Eav L\hav> wherein L\hav is the median height of the typical cell; but

H;

230

STATIC AND QUASI-STATIC ELECTRIC FIELDS

Equipotential surfaces

Electric field flux

(a)

(b) ~

,

-

Constant ~

\

't

_\'---f_---P~

~-i~Wav

__}!I

- - - -; _-----.. ++ .. + ....

(d;

ac aWav -=ft ah av

-f1----~--P-I ~hav\ I

• :>:.I /

~ = Constant

/\-;1 .;-

Flux \ hnes

Vo

~

=

~

constan~"i

I

\ \

+~q

I

~Wav

.c

-= t

E

--i- __ I(~= Constant

(e)

I

I

~hav-+ I

(d)

FIGURE 4-19. Capacitance determination from a two-dimensional electric field map. (il) Insertion of conducting foil at equipotential surfaces, yielding scries capacitance equivalence. (b) Subdivision of region between equipotentials into parallel field cells. (c) Enlargement of field cell ~C of (b). (d) End view of field cells. A curvilinear rectangle and square.

Eav DaviE and Day equals Aq/Asav , As. v denoting the average area of the cell cross section: its length t times the average cell width Aw av as in Figure 4-19(c). Thus AC becomes

EAq

yielding the capacitance per meter depth of a field cell

(4-114) If the cells are sketched as cur1lilinear squares defined by Aw av = l1hav as shown in Figure

4-13 CAPACITANCE OF TWO-DIMENSIONAL SYSTEMS BY FIELD MAPPING

4-19(d), then

231

14) simplifies to

I1C

(4-115)

E

t

The incremental capacitance per meter depth of a eurvilinear square (tux cell thus equals the permittivity E of the dielectric filling the cell. In air, for example, eaeh square cell contributes Eo = 8.84 pF1m. The total capaeitance between the conductors, obtained from the series parallel comhination of all cells, is found from the substitution of (4-115) into (4-113)

C

-

t

ltp

= -E

Flm

(4-116)

lts

From the development of 113), it is evident that be integers, as noted in the following example.

ltp

and

lts

in (4-116) need not even

EXAMPLE 4-17. Sketch the electrostatic tlux plot of the coaxial capacitor of Figure 4--20, obtaining its capacitance per meter depth. Assume air dielectric and b/a = 2. Because of tile symmetry, a flux plot for only onc quadrant suffices. If the interval between the conduc'tors is subdivided, by trial, into two equal potential difference intervals as in Figure 4--20(b), a tlux map consisting of the curvilinear squares plus two leftover rectangks as shown is obtained. Then ns 2 and ltp = so (4--116) yields

c

-- (8.84 x 10- 12 ) = 79.5 pF/m

t

2

(1)

Another flux plot, dividing the quadrant into five flux tubes as shown in (e), yields ltp = and sketching in the equipotential surfaces to obtain the curvilinear squares as shown, lis = 2.3 to yield

c~ p

,

(a)

2.3

(8.84 x 10-

( b)

12

)

=

(2)

77 pFjrn

( c)

FIGURE 4-20. A coaxial capacitor and typical flux plots. (a) Coaxial capacitor: b!a = 2. (b) A flux plot using equal potential intervals. (c) A flux plot using five flux tubes per quadrant.

291

232 np

STATIC AND QUASI-STATIC ELECTRIC FIELDS

= 4 (3.33)

Ins = 2

I

(b)

(a)

(e)

FIGURE 4-21. Examples of flux plots for two-dimensional conductor systems. (a) Elliptical cylinder inside a pipe. (b) Rectangular cylinder inside a pipe. (c) Toothed structure above a plane.

The discrepancy between (I) and (2) is due to the unavoidable errors of estimation. It happens that this example can be checked by use of the exact (4-51), yielding

C

t

2rrEo

b

tn-

2rr(8.84 x 10-

0.093

= 80.3

pF/m

(3)

a

The chief merit of the flux-plotting method for two-dimensional electrostatic systems lies in its applicability to systems for which no analytical approach is feasible. In Figure 4-21 are shown such examples. Note that care must be exercised to assure the perpendicularity everywhere of the equipotential and flux lines; observe the tendency toward the compression of the flux lines at convex curves and corners because of the higher surface charge concentrations there. Advantage should always be taken of the symmetry, with no more equipotentials being employed than necessary to obtain satisfaetory curvilinear squares. A suitable procedure in Figure 4-21 (a), for example, is to begin at section A-A' by placing a trial equipotential surface at point C, inserting appropriate orthogonal flux lines while progressing toward the right, and checking continuously for the squareness of the flux cells that develop. Needless to say, an eraser is a valuable adjunct to these trial-and-error procedures. Further suggestions and examples are found in a number of sources. lO

4-14 CONDUCTANCE ANALOG OF CAPACITANCE A system is said to be analogous to another if a quantity in one system varies in the same way as some quantity in the other. An analogy may even exist between two quantities in the same system. If the quantities are vector fields, to be analogous they must satisfy comparable divergence and curl relationships as well as similar boundary conditions. I t is to he shown that the capacitance system of F'igure 4-6 in Section 4-6 leads to a conductance analog. In the capacitance system of Figure 4-22(a), applying a IOFor example, see S. S. Atwood, Electric and Magnetic Fields, 3,.d ed. New York: Wiley, 1949; S. Ramo,

J. Whinnery, and T. Van Duzcr. Fields and Waves in Communication Electronics. New York: Wiley, 1965, p. 159.

233

4-14 CONDUCTANCE ANALOG 01' CAPACITANCE

\

\

V

'----+-11\1--------' (b)

(a)

FlGURE 4-22. Analogous capacitance and conductance systems. (a) Capacitance system: conductors at potential difference V, separated by dielectric. (b) Conductance system: a small conductivity (J supplied to the dielectric.

voltage difference V between the conductors separated by a dielectric results in static charges +q and -q being deposited on the conductors. In the charge free dielectric, D = EE, obeying V' D = 0 and V X E = 0 of (4-1) and (4-2). These properties state that D between the conductors consists of uninterrupted flux lines, with the conservative E field implying a related potential field such that E = - V 0 and cosh u -> 1 as u -> 0.]

4-33.

For the long, covered conducting channel of Problem 4-32, assume V = 100 V and a = b (square cross section) and usc (4-164) to calculate the potential (x,y) in the cross section at the nine points detemdned by the intersections ofx = a/2, a/2 and 3a/4 withy = a/4, a/2, and 3a/4. Sketeh the cross section, labeling the potentials found at the indicated points. Usc the sketch as a basis for estimating the shapes of the = 25 V and 50 V equipotential contours in the cross sections.

4-34.

(al Make use of 164) to find the series expression for the E field at any P(x,y) in the conductive channel described in Problem 4-32. (b) Assmning V = 100 V and a = b I m, evaluate E(x,y) at the following points: (0, , (a/2, and (a, a/2). (c) Assuming air dielectric, use results of (b) to determine the surface charge density at the conductor locations (0, and a/2).

SECTION 4-10 4-35.

The very long rectangular conducting channel viewed sectionally in the figure has dimensions as indicated by the square-grid overlay. Thl' cover plate is at 100 V with the remaining sides at 0 V. Write the expression (4-9Id) for the potential at the three indicated interior points, taking advantage of the symmetry about the plane A. Solve the simultaneous linear equations for the potentials. Show a labeled sketch denoting the potential values obtained.

4-36.

Use the exact Fourier expression (4-164) to verify the potentials in the last column of the table at the end of Example 4-15; also calculate the values of 1, 2'

4-37.

Repeat Problem 4·35 for the square conducting channel shown, making use of the symmetry about the plane B. Find the six potentials at the indicated points (a) by matrix methods; A

B Insulated corners

II

(1)= 100V

q,= 100V \ I

I I

I

,.._1+_+_ I

2'

I

I

I

--..1-- .. - I

31 --.,.-

1-:

\

' -

a y cos

flat r 01

4>' + a y cos 4>') fa d4>'

(b) ,

4>') 1a d4>'

(1)

the' e oj

to provide a cancellation of the y components of the potential contributions of the pair of clements at P, leaving a net dA at P that is - x directed. Thus (5-28c) becomes (2) From the law of cosines applied to the triangle POP' in the figure, R2 = a2 + r2 2ar cos (l = a2 + (2 2ar sin 0 sin (//. If r »a, one can approximate, making use of the binomial theorem,2 R

~

r [ 1 - 2 ;a sin 0 sin

4>' J1

1

2~ r [ I -

;a sin 0 sin

4>' + ...

J

1

~

1

a.

.

--( + -sm esm 4> r2

2From the binomial theorem one may see that, in the expansion of (I

rg)

b) d en' Ill~

an OJ

la) :ll!

a ta

The reciprocal, for small a, is similarly approximated

R

eel,

iO ,

± b)", ifb« 1 then (I ± b)";?; 1 ± nh.

274

STATIC AND QUASI-STATIC MAGNETIC FIELDS

This puts (2) into the form

2/-lola .1:"/2 4n q,'=

A~--

q, -

n/2

[~r + a sin esin ¢'] sin ¢' d¢'

(3)

The integral of the (sin ¢')jr term is zero, so integrating the second term yields the answer

(5-32) Taking B = V

X

A in spherical coordinates therefore yields

(5-33) if a « r. The duality between the B field (5-33) of a small current-carrying loop and the electric field (4-44) of a small electrostatic dipole is noted. This gives rise to the name magnetic dipole, when reference is made to the field of a small loop earrying a steady current.

Taking the curl of (5-28a) leads to an alternative free-space integral expression for the B field of a static current distribution as follows.

B = V

X

A = V

X

r JicJ(U'l, u~, U3) dv'

Jv

4nR

(5-34)

One may note that the differentiations imposed by the V operator in this expression are with respect to the field point variables (u 1, Uz, U3), whereas the integration is performed within V with respect to the source point variables (U'l' . Thus R is a function of both the source point and field point variables, since R = so (5-34) becomes B

One can write V X

[J/R]

=

r

Jio V x

JV4n

[!J R

dv'

from the vector identity (17) in Table 2-2

The last term is zero because J is a function of only the source point variables; flll'thermore, the factor V (1/R) can be expressed

if aR is a unit vector pointing from P' to P. Thus

5-5 AN INTEGRAL SOLUTION FOR A IN FREE SPACE: BIOT-SAVART LAW

275

n

Field point P

R

(z)

iI

Source point P'

I I

------0----(x) (y) FIGURE 5-11. A volume distribution ofcurreuts, showing the dB contribution ofa typical current element J dv' from the Biot-Savart law.

obtaining (5-35a) This integral for B, expressed directly in terms of the static current distribution J in free space, is known as the Biot-Savart law. It provides an alternative approach for obtaining the magnetic fields of static current distributions in free space. Figure 5-11 shows the geometry relative to (5-35a), depicting a system of steady currents with densities J, and a typical field point P at which B is found by means of (5-35a). The differential contribution dB is given by the integrand of (5-35a)

hi o

b)

tht

~o

~el

meaning that dB contributed at P by J dv' is mutually perpendicular to both the current element vector J and the unit vector aR, as depicted in Figure 5-11. Specializations of the Biot-Savart law to surface or to line currents are readily obtained. Thus, if the volume current of Figure 5-11 is contracted to a thin filament of negligible cross section, putting J dv' --+ I dt' into (5-35a) obtains

rg~

ly 3; en in ar

:c (5-35b)

la

11l

)t;

EXAMPLE 5·6. Usc the Biot-Savart law to find the B field of the thin wire of length 2L and carrying a steady current, as given in Example 5-4. The form (5-35b) of the law is applicable. In the circular cylindrical system as shown in Figure 5-12, Idt' = azIdz', while a R is resolved into components as follows: azz'). With R = .}p2 + (zy, (5-35b) becomes a R = a p sin Ct: - a z cos tt. = Ir 1 (app

5(

'0

276

STATIC AND QUASI-STATIC MAGNETIC FIELDS

:(z')

£'

TIdf' R

Source point

P'

z'

z'

o ------------:.-:..----P

o

Field point

a -

",,~

I

--p----~~

---------

P

. . . --.,,'\

"',

P

-£ ,

-£ FIGURE 5-12. Geometry of'the straight wire of length 2L, using the Biot-Savart law to find B.

and integrating obtains

(5-36) Close to a wire of finite length (p« L), or for an infinitely long wire, (5-36) becomes B

(5-37)

results that agree with those of Example 5-4.

5-6 QUASI-STATIC ELECTROMAGNETIC FIELDS In previous sections of this chapter, only purely stalic magnetic fields, associated with steady current distributions, were considered. Such fields are required to satisfy the Maxwell integral laws (5-4) and (5-5) for all closed surfaGes or lines in the regions in question, or equivalently the differential laws (5-1) and (5-2) for all points in the regions. The boundary conditions, also to be satisfied at all interfaces, are (5-8) and (5-9). If the current sources are generalized to the time-varying case, their fields are then no longer purely magnetic but become electromagnetic, governed by all four Maxwell equations, (3-24), (3-48), (3-59), and (3-77), with the boundary conditions embracing the relations (3-42), (3-50), (3-70), and (3-79). For current sources that vary slowly in time, however, approximate methods, termed quasi-static, may sometimes be employed to advantage. An instance has already been given in Example 1-16. Quasi-static field solutions can be termed first-order solutions, because they do not satisfy Maxwell's equations exactly except in the zero frequency limit. Another view, bctter appreciated in Chapter 11 on radiation and antennas, is that the dimcnsions of the current-carrying system must be small compared with the wavelength AO in free space 3 if the system is to be amenable to a quasi-static method of attack. This 3Suppose one assumes that a device such as a coil or capacitor should not exceed 0.01,1,0 in its maximum dimension, adopted as a criterion for sufficient smallness to enable employing quasi-static analysis in the description of its fields. Operation of the device at a frequency of 100 MHz implies that its size should then not exceed 0.03 ill (3 em), since ,1,0 = 3 m at this frequency.

5-7 OPEN-CIRCUIT INDUCED VOLTAGE

277

91

constraint is equivalent to ignoring the finite velocity of propagation of the field from the sources to the nearby field points of interest, amounting to ignoring field radiation effects. A more sophisticated approach to quasi-static field solutions, using an appropriate power series representation of the fields, is described elsewhere. 4 The quasistatic approach to field problems is sometimes the only method that provides ready solutions to an otherwise difficult boundary-value problem. It has applications in the discussion of the voltages induced in stationary or moving coils immersed in magnetic fields that mayor may not be varying in time, as well as in the development of circuit theory, particularly regarding concepts of self- and mu tual inductance, to be discussed in subsequent sections.

EXAMPLE 5·7. Demonstrate that the approximate quasi-static fields of the long solenoid of Example 1-18 obey the Maxwell's equations (3-59) and (3-77) exactly only in the static field limit (}J -,> O. The quasi-static Band E fields inside the solenoid were found to be

wpBo E(p, t) = -a.,--cos wt

B(l) = azB o sin wt

2

Testing whether these fields satisfy (3-77), V ~ p

VxE=

a ap

0

X

a.,

az

0

0

pE.,

0

E =

(1)

-aB/at, one finds

p =

-azwB o cos wt

(2)

'!at oj

revealing that Band E of (I) do indeed satisfy (3-77). This is to be expected, because E was originally obtained using the integral form of (3-77), but Maxwell's equation (3-59), reducing to V x H = aD/at within the solenoid, is not satisfied by (I). This is evident on obtaining V X H = V x (B/Ilo) = 0, since B of (I) is independent of position inside the solenoid; whereas aD/at becomes

b), the ~

oj

:ell rg) b) j "

en'

a vanishing result only if w -> O. Thus (3-59) is satisfied only in the static field limit, though an approximate equality prevails if w is sufficiently small.

in~

an o la'

5·7 OPEN·CIRCUIT INDUCED VOLTAGE The transformer makes use of Faraday's law (3-77) to couple electromagnetic energy from one electric circuit to another through the time-varying magnetic field. Typical physical arrangements are diagramed in Figure 5-13. In (a) is shown the configuration of Figure 1-25(b): a primary coil consisting of a long solenoid, encircled by a secondary coil. Single-turn secondary coils are shown for simplicity; many turns are commonly

53

4See R. M. Fano, L. J. Chu, and R. B. Adler. Electroma.gnetic Fields, Energy and Forces. New York: Wiley, 1960, p. 221 If.

or

278

STATIC AND QUASI-STATIC MAGNETIC FIELDS B flux

B flux

B flux

Primary Secondary

(!Lo)

8;" Bj :' ':3 :3

~C-~;~

: ttttttt: I I ; I ! 11

1111111

(c)

(b)

(a)

FIGURE 5-13. Typical transformer configurations. (al Primary coil a long solenoid. (b) Short solenoid primary, secondary laterally displaced. (c) Gonfiguration of (b) with ferromagnetic core.

used to enhance the induced voltage V(t). A ferromagnetic or a ferrite core can also be used in a magnetic circuit arrangement as in f'igure 5-l3(c), to augmentsubstantially the magnetic flux intercepted by the secondary coil. The voltage V(t) developed at an open-circuit gap in the secondary coil 5 of a transformer is shown to be V(t)

dl/l m

--V dt

(5-38)

in which l/I m denotes the magnetic flux intercepted by the surface S bounded by the secondary winding. Suppose the coil shown in Figure 5-14(a) carries a time-varying current 1(/). In the surrounding region, the accompanying magnetic field B(Ut, U2, U3, t) induces an azimuthally directed, time-varying E field as described in Example 1-18 and depicted in the cross-sectional view of Figure 5-14(b). The secondary coil is shaped such that Hux orB passes through the surface S bounded by the coiL This assures the alignment of the conductor with the induced E field such that the free electrons in the conductor are urged by the E field forces to move along the conductor as noted in Figure 5-14(c). Thus an excess of electron charge accumulates at one end of the wire, while a dearth of electrons (a positive charge) is established at the other, producing about the gap another electric field denoted by Eo. Then the total E field about the system becomes E = El + Eo. Faraday's law (3-78) written about the closed path including the secondary coil and its gap thus becomes

J. E. dt == Jrconductor Yt

(E t

+ Eo)

. dt

+ Jrgap

(EI

+ Eo)

. dt =

dl/l m dt

(5-39)

The relationship between the total electric field El + Eo along the conductor and the current density J within it is given by (3-7), bccoming J = a(El + Eo) along the coil. SIt should be borne in mind that the designation secondary coil is arhitrary; either coil ora transformer may be designated as the primary coil, with the other coil taking the name secondary.

5-7 OPEN-CIRCUIT INDUCED VOLTAGE

279

11

B

",

I

~

~~~nductor ~ with gap ;I Eo field of . / displaced charges (b)

(a)

(c)

HGURE 5-14. Development of the open-circuit voltage V(t) ora transfOlmer. Transformer configuration nsed to prove (5-38). (b) Sectional view of E, induced by timevarying B of (a). (c) Showing charges displaced by El to produce Eo, canceling total E along wire.

Assuming the coil a jJe~lect conductor, E + Eo must tend toward zero if J is kept to a necessary finite value, making El + Eo = 0 along the conductor portion of the closed path t. This simplifies (5-39) to obtain

A: E' dt = r (El + Eo) . dt ~ Jgap

dt/lm dt

(5-40)

ne

tt

implying that the total E' dt generated by the time-varying magnetic flux t/lm embraced by t appears wholly at the gap. The closed-line integral of (5-40) is sometimes called the induced electromotive force (emf) abou t t, and is denoted by the voltage symbol V(t). Then (5-40) is written

V(t)

==

V ~ E· dt = --dl

dlPm

t

at of I),

(5-41)

of ell ~y

)y a nt :lg re or

ay Thus the induced emf: or equivalently the gap voltage V(l), depends only on the time rate of change of magnetic flux through the surface S bounded by the closed line t described by the wire. The explicit values ofE 1 and Eo are not required to be known on/he path. Furthermore, the wire path t may be distorted, if desired, into any arbitrary shape; for example, a square or a helix, in which case (5-4,1) is still valid. A h~lix­ shaped (many turn) conductor is useful for increasing the induced voltage across the gap, and it is commonly used in practical transfi)fmer and inductor designs. If in the foregoing discussions a finitely conducting wire had been assumed, the result (5-41) would have been modified only trivially if the conductivity (J were sufficiently large (of the order of 10 7 U/m, as for most good conductors).

llg a~

:a'

0: 3'

280

STATIC AND QUASI.STATIC MAGNETIC FIELDS

Long solenoid

(J turn/m)

FIGURE 5-15. Showing open-circuit coils ( and (' and the induced voltages Vet) obtained from the time-varying t/lm.

EXAMPLE 5·8. A thin wire is bent into a cirele of radius b and placed with its axis concentric with that of the solenoid in Example 1-16. Find Vet) induced across a small gap left in the conductor, for the two cases of Figure 5-15: (al b > a and (b) b < a. Include the polarity of Vet) in the answers. (a) If b > a, (5-41) combined with (1-67) yields, for the solenoid current 10 sin

Vet) =

-~ r B . ds = dt Js

d dt

r[

Js a z

llon10 sin d

wt,

wt] • (azds)

b>a

( 5-42)

since Ssds = na 2 • The polarity of Vet) is found by use of a right-hand-rule interpretation of the induced voltage law (5-41). Assuming, at a given t, that t/Jm through t is increasing in the positive z sense in Figure 5-15, aligning the thumb of the right hand in that direction points the fingers toward the terminal P2 at the gap, which at that moment is the positive terminal. The presence of the negative sign in the answer (5-42), however, requires that the true polarity of Vet) becomes the opposite of the indicated polarity in Figure 5-15, at that instant. (b) If b < a, the surface S' bounded by the wire t' is smaller than the solenoid cross section; (5-41) then becomes

Vet) =

-~ r B· ds = dt Js'

b=

H1> =

1

2np

-;Tir(~/-b2) (~ p)

O dz, the inductance contribution of a length (of conductor 2 only, from (5-73), is

IP) = .~1 12

j'

V

A . J dv - -Ao} 2

na2I

pzdpzde/>

The integral contribution of the third term in the integrand can be written

in which the second term integrates to zero (Peirce's integral 523 16 ), obtaining L(2)

= ito{ - 1

[ 8n

+

dJ

I (n-

2n

a2

A similar consideration of conductor I yields by analogy

making the total inductance L(l)

+ IPl

of the two-wire system

L= 16B.

/

O. Peirce, A Short 'fable oj Integrals. Boston: Ginn, 1910.

(5-84)

306

STATIC AND QUASI-STATIC MAGNETIC FIELDS

Comparison with (5-82) shows that the leading term of (5-84) is the internal inductance, making the last term the external inductance.

D. SeH-lnductance by the Method of Flux Linkages The resolution of the self·inductance of a circuit into the sum of internal and external self-inductances provided by (5-80) is closely related to another technique known as the method if flux linkages. This approach is based on the use of the energy definition, (5-78), but with the integration in all space replaced by a surface integral intercepting all the magnetic flux of the system, the self-inductance being thereby characterized by the linkage of that flux with the circuit current. The method is described here. For most circuits, the total magnetic flux generated by the current can be partitioned (exactly or approximately) into two amounts: that lying entirely outside the conductor, plus that flux wholly internal to the conductor. Such a flux division occurs precisely for the single round wire noted in Figure 5-27 (a), and very nearly so for the parallel two-wire line shown in the same figure,17 especially for wires with diameters small compared to their separation. Another example is the loop shown in Figure 5-27(e); for thin wire, the flux tubes can be separated into those wholly inside or outside the wire as shown. The volume occupied by the magnetic field (all space) is thus divisible into closed flux tubes that surround or are embedded in the current. The magnetic energy contained in all space has been given by (5-77)

u

m

=.1 2

r

Jv B· Hdv

[5-77]

Suppose the volume of the typical flux tube in Figure 5-27(b) is subdivided into elements dv = dsdt', in which dt' is aligned with the tube wall (and therefore with the B field) and ds denotes the cross-sectional area of the tube. Then B • H dv = (a,tB)· Hdsdt' = (a,tdt')· HBds = H· dt' dt/lm. Thus, if the integration (5-77) of the latter is to include all elements dv where Band H prevail, H • dt' should be integrated about the closed median line (' of the flux tube shown, with the remaining surface integration taken over an open surface S chosen to intercept all the flux tubes of the circuit. For the single-turn circuit of Figure 5-27 (b) or (e), the appropriate S intercepting all flux tubes is that bounded by a closed line { essentially coincident with the wire axis. Thus the energy integral (5-77) can be written

(5-85)

with S bounded by the circuit t. [Note: The last integral is the consequence off H . d{', integrated about any closed flux tube (I, being just the current i({I) enclosed by (I.] 170wing to the proximity effects of the low-frequency currents, the magnetic flux in the interior of parallel wires is not concentric about the centers of the wires, but about points moved slightly apart from the centers. This efleet is responsible for some of the magnetic flux being partly inside and partly outside the wire, as noted in Figure 5-27(a).

5-11 MAGNETIC ENERGY AND SELF-INDUCTANCE

307

, ,,

,,/ ,,

\

\

""\ \

,\

,, \ \ ,,

/

:

\

\ I

\

,

,// \ \

-----_ ..

\

\

--

/

,

,

, ,,

/

,,

\ \

,,

Parallel-wire line

/

Single wire (a)

dv = dsd/' Typical flux tube carrying dl/;m

I

dl/;m

dl/;m

Circuit

t

t' over which J,[H·dt=I ,

r

(b)

Flux tube {'

Conductor

, , '"

(c)

FIGURE 5-27. Concerning the method of flux linkages. (a) Examples of internal and external flux-partitioning. (b) Single-turn circuit (left) showing external flux tube linking lance, and a two-turn circuit (right) with a flux tube linking I twice (passing through Sex twice). (c) Wire loop, showing internal and external flux (lef!), and a typical internal flux tube (right) linking i(t'), a fraction of 1.

For all exterior flux tubes, passing through Sex as shown in Figure S-27(b), the total current I is linked by t', whereas a variable fraction itt') of I is linked by flux tubes t' located inside the conductor and passing through Sin, as shown in Figure S-27(c). In the event ofa circuit t having more than one turn as in Figure S-27(b), an exterior flux tube t' may even encompass I more than once (in general, as many as n times for

308

STATIC AND QUASI-STATIC MAGNETIC FIELDS

an n turn coil). It is thus evident in such cases that the same flux tuBe t' can contribute to (5-85) over the surface Sex several times, thereby increasing the magnetic energy and the self-inductance correspondingly. Whenever the magnetic flux of a circuit is separable into internal and external linkages passing through Sin and Sex as depicted in Figure 5-27 (e), it is convenient to separate (5-85) into the contributions (5-86)

In the latter, one is cautioned to observe that the quantity t/lrn.ex JSex B . ds appearing in the external energy term denotes a total flux through Sex, which can be the result of some or all the flux tubes passing through that surface more than once, for example, as in Figure 5-27 (b), or for the many turn coils illustrated in Figure 5-28. By use of (5-86), the self:'inductance of the circuit t is expressed in terms of internal and external contributions as follows

L

2Um

= J2 =

r . ,

I

Js z(t) dt/lm

I =

r

JSin itt') dt/lm +

(5-87)

I

such that the external inductance is given by

t/lrn.ex

L =

I

e

~

=

t

r

JSex

B· ds

(5-88a)

or just the total magnetic flux penetrating Sex divided by the current

t. The internal

n-turn coil (circuit f)

Circuit

t

BfIUX

t

-'-~' ~s~ ~ /

~ .---

l'::::;;: I

(a)

.1-.

,

,.exl

I' f"

B flux-...

I I~Y.

'~

I~el " ~

"'Y).!b~'< ~, , (b)

Core flux (c)

FIGURE 5-28. Examples of many-turn coils having negligible interual self-inductance. (a) Air core solenoid. (b) Toroidal winding. (c) Coil and iron core.

I

5-11 MAGNETIC ENERGY AND SELF-INDUCTANCE

309

inductance is

(5-88b)

a consequence of i(t') dljlm integrated over the appropriate internal strip Sin connecting with the wire axis, as depicted in Figure 5-27 (c). An illustration of the use of the latter for a long straight wire is taken up in Example 5-15. Although the internal conductor volume of a circuit may be small, the magnetic fields may be relatively large there; individual circumstances will dictate whether or not the internal inductance is negligible. For circuits having large external fluxes, such as those with iron cores, the total self-inductance is generally well approximated by (5-88a), the external selfinductance. II

EXAMPLE 5-15. Determine the internal self-inductance of every length t of the infinitely long wire shown, carrying the low-frequency current I. Use (5-38b), employing the method of flux linkages. A typical flux tube t' carrying dl/l m = B • ds through Sin is shown in the accompanying figure. With the internal B obtained from (1-64), the flux in the tube is

dl/l m = B· ds = Bq,dpdz =

/lIp --2

2na

dpdz

The current i(t') intercepted by dl/l m is the fraction I(np2jna 2 ) = I(p2ja 2), obtaining from (5-83b) (5-89) which agrees with (5-32).

~

iU')

=I ~ A, encompassed by dl/l m a

EXAMPLE 5-15

310

STATIC AND QUASI-STATIC MAGNt:TIC FIELDS

EXAMPLE 5-16. Determine the approximate self-inductance of a length t' of a long parallelwire line shown in (a), using the flnx linkage method. Assume the radii small compared to the spacing d. For well-separated conductors as in (b), the internal field is essentially that of an isolated conductor, making the internal self-inductance for both conductors just twice (5-89), that is, Li = JIt'/4n. The external inductance is found by using (5-88a), the ratio of the magnetic flux through Sex of (a), divided by I, but the total flux is just twice that through Sex due to one wire, given by

r

JSex

B' ds

i

t

z=O

J,d-a (JIOI -- a ) 2np 4> p=a

. a dp dz 4>

~

JIoIt' t' n d2n a

(1)

yielding for both wires

(2) The total self-inductance is the sum

L

=

Li

+ Le =

pt' 4n

+

JIot' n

d a

t'n - H

(5-90)

a result seen to agree with the exaet expression (5-84) on putting a 1 = a z into the latter and assuming nonmagnetic wire. For a nonmagnetic parallel-wire line with d = 12 in. and a = 0.1 in., one obtains

L/t'

=

lO-7

+ (4

x 10- 7 ) t'n 120

=

2.02 JIH/m

Neglecting the internal inductance would incur about 5% error' in this example, not a negligible amount.

In calculating self-inductance at low frequencies, the internal-inductance contribution is in some cases quite small; in others it should not be neglected. The internal inductance of a single-layer air core coil of several turns, .illustrated in Figure 5-28(a), contributes little to the self:inductance if the volume of the wire is small compared to

(a)

(b)

(c)

EXAMPLE 5-16. (a) Parallel-wire line and surfaces linked by internal and external magnetic fluxes. (b) Division of internal and external Huxes for thin wires. (c) Proximity effect for thick wires.

I 11.1

5-11 MAGNETIC ENERGY AND SELF-INDUCTANCE

311

EXAMPLE 5-17

,

the region where the significant fields are located. In the closely spaced toroid as in Figure 5-23(b), with every turn intercepting all the core flux, the self·inductance is proportional to the square of the turns, as seen in Example 5-17. The addi tion of an iron core in the form of the low-reluctance magnetic circuit of Figure 5-23(c) increases the selfinductance substantially more. In these cases, the added effect of the internal inductance is insignificant.

EXAMPlE 5·17. Find the self-inductance of an n turn toroid with a rectangular cross section as shown, for two cases: (a) with an air core, assuming closely spaced turns, and (b) with the core a linear ferromagnetic material (constant /l). (a) The magnetic flux in the air core, from Example S-2, is /lonld

b

I/Imcore=--tn , 2n a

An inspection of the sUlface Sex bounded by the circuit t, as given in Figure S-28(b), reveals that Sex intercepts the core flux n times,18 yielding t/lm,ex = nt/lm,core through Sex. Thus the self-inductance from (S-88a) becomes n!/lm, core ell

L~L

!/Im,ex

2

=--=-_._-=

/lon d 2n

b a

tn~

(S-9Ia)

with the internal inductance neglected. Thus, a 100-turn air core toroid with dimensions a = 1 em, b = 3 em, d = O.S cm has the inductance L = (4n x 10- 7 x 100 2 x O.OOS

tn 3)/2n = 11.0 /lH

Doubling the turns to 200 is seen to quadruple the inductance. (b) Inserting an iron core with the permeability /l, (S-91a) becomes L

(5-91b)

Using a linear ferromagnetic material with /lr = 1000 makes the inductance of the 100-tnrn toroid just 1000 times as large, yielding L = 11.0 mHo ISOr equivalently, every flux tube

dJ/!m encompasses the culTent

In times in this example.

I

I

312

STATIC AND QUASI-STATIC MAGNETIC FIELDS

t~eumann's Formula for External Inductance in Free Space An extension of the flux linkage expression (5-87) leads to Neumann's formula, applicable to circuits in free space. Equation (5-87) consists of internal and external self-inductance terms as follows

*E.

[5-87] Consider first only the external inductance term (5-88a) of (5-87), involving the flux t/lm,ex linked by the external surface Sex bounded by the circuit t. From (5-47), this is expressed

t/l m~ex =

r

JSex

B· ds =

r

(V

JSex

X

A) . ds =

~ A· dt Wb Yt

(5-92)

With (5-92), (5-88a) becomes L

e

t/lmex = = -'I

II

-

I

Sex

B • ds = -l~ A· dt I t

(5-93)

In free space, the vector magnetic potential A can be found by use of (5-28a) [5-28a] Applied to the circuit of Figure 5-29(a), (5-28a) obtains A at the typical field point P located on t bounding Sex. Another integration of A . dt about t in accordance with (5-93) then obtains the external self-inductance of the circuit. These steps are combined by inserting (5-28a) into (5-93), yielding

Le

= -1

I

f [i t

Vo

dV']

JloJ --R. dt H 4n

t (a)

(b)

FIGURE 5-29. A closed circuit in free space, relative to external self-inductance calculations. (a) Wire circuit, showing source and field points P' and P. (b) Simplification of (a), with sources I dt' concentrated on the wire axis.

(5-94a)

5-11 MAGNETIC ENERGY AND SELF-INDUCTANCE

313

This quadruple integration is simplified for a thin-wire circuit if I is considered concentrated on the wire axis as in Figure 5-29(b). Then J dv' becomes I dt', reducing (5-94a) to

floIdt'] L - -1 ~ [~ - -dte I { C' 4nR -

~~ t

('

flodt' - dt 4nR

(5-94b)

a result known as Neumann's formula for the external inductance of a thin circuit in free space. The order of the integrations relative to dt' and dt, and hence, relative to the source point and field point coordinates, is immaterial. From (5-87), the total self-inductance L is obtained by adding (5-94b) to the internal inductance term L i • Since the latter is a measure of the internal stored magnetic energy, Li is expressible using either (5-S1) or (5-8Sb); thus L = Li + Le becomes, in Fee space

,

i

L= 1

V;n

r JSin

~~

flo dt' - dt 4nR

+ J. J.

Il odt' - dt 4nR

B-Hdv + i(t') dl/tm

t

t'

'ft 'ft'

(5-95)

EXAMPLE 5·18. Find the self~inductance of a thin circular loop of wire in free space, with dimensions as in (a). Use the Neumann formula (5-94b). The current assumed concentrated on the wire axis as in (b) allows the use of (5-94b). In cylindrical coordinates, dt' = a",bdf/J' at the souree point P'(b, f/J', 0) on the

axis t'. From the circular symmetry, the location of the field point P on t is immaterial, so put P at f/J 0; that is, at P(b - a, 0, 0). The distance from 1" to P is given by the law of cosines

R=

(a)

(1)

(b)

EXAMPLE 5-UI. (a) Circular loop of round wire. (b) Axial, line current approximation of (a).

314

STATIC AND QUASI-STATIC MAGNETIC FIELDS

while dt' • dE in (5-94b), from part (b), means dE' dE cos cp' (implying that only the component of A parallel to dE at P is required in the integration.) Then (5-94b) becomes L e -

J.2l< J.2l< 1>= 0 1>' =0 ---;~================== (5-96)

This result is not integrable in closed form, though with numerical values of a and b it yields to computer solution. An alternative makes use of tabulated values of the complete elliptic integrals K(k) and E(k). A conversion of (5-96) in terms of such integrals is accomplished as follows. Change the variable cpt to 21X, making dcpt = 2 dlX and cos cpt = cos 21X = 2 cos 2 IX - 1, with the limits on IX going from 0 to n. Then R in (5-96) becomes

=

R

.jb 2

+ (b

- a)2 - 2b(b - a)(2 cos 2 IX - 1)

.j(2b - a)2 - 4b(b - a)

COSzlX

(2) if k 2

=

4b(b - a)/(2b - a)2. The complete elliptic integrals, defined by

de

J,fo"!2 -r===::;===

K(k)

(5-97)

are incorporated into (5-96) as follows. The integral in (5-96), making use of (2), becomes

21t fIt J.1>'=0 = Jo

2 cos 21X dlX (2b - a)

---;:===

S:

rk=co="s~2:=IX=d=lX=.::==

(3)

but the numerator of (3) is written

k cos 21X = k(2 cos 2 IX - I) = 2k cos 2 IX

2

2

k+k

k

to yield a further conversion of (3)

I

"

.jb(b - a) So

=

( -2 k ) dlX "k 2 - "7 { .jb(b - a) SO.jl - k 2 cos 2 IX k 1

" .j 1 - k 2 cos 2 IX dlX}

So

(4)

An inspection of the last integral shows that

which from (5-97) is just 2E( k). A similar consideration of the preceding integral in (4)

315

5-11 MAGNETIC ENERGY AND SELF-INDUG'TANCE

reveals that it is just 2K(k), so (5-96) becomes (5-98) The tabulated values 19 of K(k) and E(k) can be used in (5-97) to evaluate Le ofa circular loop with desired dimensions. For thin wires (a« b), the elliptic integrals are approx· imated by

E(k)

~

1

a«b

yielding the simplification

a«b

(5-100)

I

For example, a 2-mm diameter wire bent into a circle of 10-cm radius has the external inductance Le = (4n x 10- 7) (0. l)(tn 800 2) = 0.588/lH. The internal magnetic field of the loop is virtually that of a straight, isolated wire, making their internal inductances nearly the same. Applying the results of Example 5-12, the approximate internal inductance of the loop becomes

L. ~ /l(2nh) ,- 8n

ftb 4

(5-101)

With b = 10 cm and assuming nonmagnetic wire, L; = 0.031 /lH. Thus the self-inductance expressed by (5-93) becomes L = Le + L; = 0.619/lH, in which Le is seen to be the predominant term.

A summary of expressions for magnetic energy described in the foregoing discussion, together with expressions for the circuit inductance when the system is linear, is given in Table 5-1. *F. Kirchhoff VoHage Relation from Energy Considerations In concluding the remarks about the circuit of Figure 5-30(a), a Kirchhoff-type voltage equation resembling (5-63) can be obtained for it from the energy expression (5-68a)

VI dt = RI2 dt

+ Jvc r J' dA dv

[5-68a]

abbreviated [5-68b] Dividing by dt obtains

VI =

RP + dUm dt

19For example, sec E. Jahnke, and F. Emde. Tables oj Functions, 4th cd. New York: Dover, 1945.

(5-102)

316

STATIC AND QUASI.STATIC MAGNETIC FIELDS

TABLE 5-1 Summary of Magnetic Energy and Self-inductance Relations SELF-INDUCTANCE

MAGNETIC ENERGY

In terms of A and

J

integrated throughout conductor volume

In general (5-70)

Linear circuit (5-71 )

(5-73)

(5-74)

(5-75)

In free space

U =

m

11 1

fioJ" J du' dv

Jvc Jvc 4nR

In terms of Band H integrated throughout all space In general (5-76)

Linear circuit Um

= ~2 Jv 1 B· Hdv =1 1

JVin

(5-77)

B'Hdv+1 1

JVex

L

m 2U== -121 ~V B . H dv 12

(5-78)

B'Hdv(5-79)

Extension to method of flux linkages

Urn

=

1 Is itt') dt/lm

=

~2

I

Stn

itt') d,I,

'I'm

+ It/lm.ex 2

(5-85 )

L=

(5-86)

=

)2 Is itt') dt/lm ~ 12

I

Stn

itt') dt/l

m

+ t/lm.ex 1

(5-87)

In free space _

L= I

Is

in

i(t') dt/lm

+ rC, rC, j-t 'fe'

de" dt 4nR

(5-95)

5-11 MAGNETIC ENERGY AND SELF-INDUCTANCE

317

v

(a)

(b)

( c)

FIGURIfS-30. Development of circuit models of the circuit oLFigure 5-24. (a) Physical circuit driven by V(t). (Ii) Circuit model depicting terms of (5-106). (c) Circuit model using lumped elements.

signifying the instantaneous power delivered by V: the sum of the instantaneous heat loss plus dUm/dt, the power delivered to the magnetic field (rate of magnetic energy storage or release). Dividing by I produces a voltage relation 1 dUm V=RI+-I dt

(5-103)

in which Urn, the im;tantaneous magnetic stored energy, is specified by any of the expressions listed in Table 5-1, depending on whether the system is magnetically linear. For a linear circuit, a self~inductance L is attributable to the circuit energy by (5-78) (5-104 ) With L constant, the last term of (5-103) becomes (5-105) making (5-103) a voltage relation comparable to the Kirchhoff expression (5-61); that is,

V= RI + d(LJ) dt

(5-106)

Equation (5-106) states that the applied voltage V(t) supports two effects: (a) a voltage drop RI associated with the circuit resistance Rand (b) a back voltage d(LI)/dt or LdI/dt produced by the time-varying magnetic flux linking the circuit, a flux produced by 1. Because of the separation of these effects into two terms, one may properly lump the resistive voltage and the self-induced voltage to yield the series circuit model shown in Figure 5-30.

318

STATIC AND QUASI-STATIC MAGNETIC FIEI,DS

5-12 COUPLED CIRCUITS AND MUTUAL INDUCTANCE Besides the single circuit of Figure 5-24, also of physical interest is a pair of such circuits, coupled electromagnetically by the time-varying fields generated by their currents. Examples are the iron core and air core transformers of Figure 5-31 (a), which may have active sources in one or both windings. A generalization is illustrated in (b). The analysis of coupled circuits from the magnetic energy point of view closely parallels that for the single circuit. Consider the circuit pair of Figure 5-31 (b) with one driving source V(t) in circuit 1, producing the primary current 11 (t). The latter generates a field B 1 , the flux of which links not only circuit 1 but some fraction of that flux (governed by the geometry and the presence of ferromagnetic bodies) also links circuit 2, generating an emf about each circuit in accordance with the Faraday law, (3-78). The ensuing current 12 produces a field B2 reacting similarly on circuit 2 while also partly linking circuit 1, therehy establishing an additional back emf in each to modify 12 and 11 accordingly. The influence of these mutual coupling effects on current flow can conveniently be treated by use of Kirchhotf voltage equations, developed later in this section. The mutual magnetic coupling between the circuits leads to their mutual inductance parameters, developed in the following. A simple extension of the power integral (5-66) to the pair of circuits of Figure 5-31(b) yields

- J, (J) . ds j-'Sl

=

r

JVl

E· J dv

+

r

JV2

E· J dv

+

r aA. r. aA. at J dv + JV2 at J dv

JVl

(5-107) in which Vj , V2 denote the volumes inside the two conductors, with S] taken as the surbce enclosing VI exclusive of the driving source V(t). The left side of (5-107) denotes the instantaneous power Vi l delivered, whereas the two-volume integrals ofE· J are the ohmic losses Rili and R2I~ within the conductors. Multiplying (5-107) by dt

V«)~L._._(_fJ.)_~_.-...J I2

h

BI flux (of

h only)

RL

With iron core

V(t)

Vj (Conductor volume) With air core (a)

(b)

FIGURE 5-31. Magnetically coupled circuits. (a) Typical coupled circuits. (b) Generalized coupled circuits.

5-12 COUPLED CIRCUITS AND MUTUAL INDUCTANCE

319

yields (5-108a) abbreviated (5-108b) Integrating (5-1 08b) obtains

denotil)g the work done by V(t) in bringing the system up to the levels II and 12 at the instant t. The volume integrals in (5-109) represent the energy expended by V in establishing the magnetic fields of the coupled circuits; that is, the energy stored in the magnetic fields in the amount

(5-110)

The integrations are required only within the conductors, since no densities J exist outside them. Equation (5-110) is correct whether or not the system is linear. If the system of Figure 5-31 (b) is linear, one can assert that the contributions to the total A at any point in the region are proportional to the current densities J in the circuits. Then

Urn = 1 Iv! A" J dv

+ 1 IV2 A" J dv J

Linear system

(5-111 )

obtained analogously from (5-110) in the manner that (5-70) led to (5-71). I t is advantageous to rcexpress (5-111) in terms of the vector potential contributions of each current. Let the total vector potential at any field point P in either conductor be written (5-112) with Al and A2 denoting the potentials at P due to the currents in circuits 1 and 2, respectively. Then (5-111) splinters into the four contributions Um

=1'Jv!

A 1 "Jdv+1' JV2 A 2

·Jdv+1' JV2 A 1 ·Jdv+1'JVl A 2 "Jdv

(5-113a)

abbreviated as follows (5-113b)

320

STATIC AND QUASI-STATIC MAGNETIC FIELDS

Note that Urn 1 , for example, denotes the magnetic self~energy of circuit 1 taken alone (with circuit 2 open-circuited), with (5-71) revealing that Urn1 is the energy associated with the self-inductance of circuit 1, called L 1 . A similar remark applies to U m2 , leading to the self-inductance L2 of circ·uit 2. With J in conductor 1 as well as At both proportional to 11 , U rnl becomes proportional to Ii. Similarly, U m2 , U m12 , and U m21 are proportional to 1~, 1112, and 11 12 , respectively, yielding from (5-113a)

(5-1 14a) (5-1l4b) (5-114c) (5-114d)

The constants M12 and M21 appearing in (5-114c) and (5-114d) are known as the mutual inductances of the pair of circuits, related to the additional mutual magnetic energies associated with the magnetic coupling of the circuits. It is now shown that the mu tual inductances M 12 and M 21 are identical for linear systems, namely (5-115)

with the symbol M chosen to denote either parameter. That (5-115) is true for a linear system is demonstrated on reexpressing (5-113a) in terms of the volume integral of B . H by use of (5-77)

Um

= ~2 Jv f B· Hdv

[5-77]

This result, derived for the single circuit of Figure 5-24, is equally valid for the coupled circuits of Figure 5-31. Suppose Band H of the coupled system are expressed as the sums B = Bl

+ B2

H = HI

+ H2

(5-116)

in which Bl V X Al ,uH1' B2 V X A2 = ,uH 2 , whence B 1 , Hi are taken to be due to II in circuit I, while B 2 , H2 arc proportional to 12 in circuit 2. Then (5-77) expands into the four terms

in which the integrations are to be taken throughout all the space where the fields B and H exist. A comparison of the four integrals in (5-117) with those of (5-113a) reveals

5-12 COUPLED CIRCUITS AND MUTUAL INDUCTANCE

321

a one-to-one energy correspondence, implying that the self and mutual inductances defined in (5-114) can also be written

(5-118a)

(5-118b)

(5-118c)

, (5-118d)

but in the latter, the product Bl . H2 equals B2 • HI because (5-119) Thus (5-1 18c) and (5-118d) are identical, proving (5-115), that M12 = M 21 . Combining (5-114) and (5-115) into (5-1 13b) permits writing the total magnetic energy in the form

(5-120)

Hence, a knowledge of the inductance parameters and instantaneous currents determines the magnetic energy state of coupled circuits at any instant. Since the selfinductance expressions (5-118a, b) have already been considered in detail, the expressions (5-118c,d) concerning the mutual inductance M will occupy the attention of the remainder of this section. For coupled circuits in free space, M can be expressed by a volume integral in terms of the current sources, yielding a result resembling (5-75) for self-inductanee. Hence, substituting (5-28a) for A into (5-118c) or (5-118d) obtains

M21

= M=

I

1112

11 Vi

V2

lloJ" J v, dvH --d 4rcR

Free space

(5-121)

with primes again used to distinguish the souree point current element J'dv' from the unprimed field point element as in (5-75). In Figure 5-32(a) is shown the geometry

322

STATIC AND QUASI-STATIC MAGNETIC FIELDS

(a)

Current fiiamentt'2 linking 1{12

\

S2

(Cross- section) (b)

Circuit

(1

t~ (c)

FIGURE 5-32. Generalized coupled-circuit configurations pertaining to mutual energy and inductance calculations. (a) Linear coupled circuits in free space. (b) Linear coupled circuits in general (iron present or not), showing the portion tfr12(t'~) of the flux of I, liuking current filament t~. (e) Special case of (b): thin circuits. Depicting portions tfr12 (If:ft) and tfr21 (r(lflll) of the fluxes of I, and 12 ,

relative to the integrations. The Neumann integral (5-121) is not discussed further here; refer to other sources for applications. 20 More general expressions for M can be derived from magnetic flux and current linkage interpretations of (5-114c) and (5-114d), to include the effects of magnetic materials. Subdivide circuit 2 into closed current filaments t~ carrying the differential current di as in Figure 5-32(b), each linking a portion t/J12(t~) of the flux of circuit 1. 20See R. S. Elliott, Electromagnetics. New York: McGraw-Hill, 1966, p. 309.

5-12 COUPLED CIRCUITS AND MUTUAL INDUCTANCE

323

Equation (S-114c) for the mutual energy Um12 then becomes

a result that follows on noting that Ai . J dv = Ai . (atJl dt'ds Ai' dt" di, andobserving from (S-47) that Ai . dt" denotes t{! 12(t2), the portion of the flux of Ii linking t 2. Thus Um12 is found by integrating t{!12(t'2) di over the cross section S2 of wire 2, as depicted in Figure S-32(b). Similarly, (S-114d) becomes

f

I

(S-122b)

The use of the flux linkage expressions (S-122a, b) is facilitated by assuming Ii and 12 to be concentrated along the wire axes. Then t{!12(t2) and t{!21(t'1) in (S-122a) and (S-122b) become constants, yielding the simpler results

(S-122c) (S-122d)

in which

t{! 12 = the portion of the flux of Ii linked by circuit 2 t{! 21 = the portion of the flux of 12 linked by circui t 1 The simplifications (S-122c) and (S-I22d) are excellent approximations if the circuits are thin, as depicted in Figure S-32(c). The mutual inductance M is finally obtained by substituting the energies (S-122) into the definitions (S-llSc) and (S-11Sd), making use of M12 = M21 = M of (S-IIS); thus

M

Exact

(S-123a)

For thin circuits

(S-123b)

324

STATIC AND QUASI-STATIC MAGNETIC FIELDS

EXAMPLE 5-19

The latter approximations (5-123b) are usually acceptable in practical mutual inductance calculations.

EXAMPLE 5-19. Find M for the iron core toroidal transformer illustrated, the windings having n 1 and n2 turns and assuming no leakage flux. Compare M2 with the product LIL z . M for thin coils is conveniently found by use of (5-123b). For II in t l , the core flux obtained in Example 5-2 is

but t/I 12 linked by t2 (i.e., passing through obtaining from (5-123b)

Sex.2

bounded by

t 2) is

n2

times

t/lm,core,

(2)

The same answer is obtained using M = t/lzl/I2 , The se1f:'inductances of the coils, from Example 5-17, are (3)

Thus the product LIL2 equals the square of M given by (2). This is expected for coupled circuits whenever all the magnetic flux links each turn of the windings.

The idealization that all the magnetic flux produced by one circuit completely links the other, as in Example 5-19, is never quite attained in practice, even when high-permeability cores are used to minimize flux leakage. There is invariably some leakage, as depicted in Figure 5-33(a), causing M2 to be less than L I L 2 . This circumstance is expressed by the so-called coefficient of coupling between circuits, symbolized

5-12 COUPLED CIRCUITS AND MUTUAL INDUCTANCE

(a)

325

(b)

FIGURE 5-33. Magnetic coupling between circuits yielding high and low coupling coefficients.

(a) Iron core transformer with small leakage (k .... I). (h) Circuits coupled in air, Jar high-frequency applications.

by k and defined

M k=-.JL 1 L 2

(5-124)

The latter permits expressing M as a function of the self-inductance of each circuit whenever k is known; that is, (5-125) The maximum value attainable by k is unity, while for circuits totally uncoupled, k = O. If coils are coupled using high-permeability cores, k may have a value as high as 0.99 or better, though with air as the coupling medium as in Figure 5-33(b), a much smaller k is usual, in view of one circuit linking a correspondingly smaller fraction of the total self-flux of the other. The circuit model of coupled circuits can be deduced in the same manner as for single circuits. Since a pair of circuits is involved, two Kirchhoff voltage relations are desired. Three interrelated methods can be employed to obtain the Kirchhoff voltage equations: (a) a method based on the scalar and vector potentials I]) and A of the electromagnetic fields, described in Section 5-10; (b) a technique based on energy considerations, treated in Section 5-11, part F; and (c) an approach making use of the Faraday law, (3-78). The Kirchhoff voltage equations of coupled circuits are derived from application of the Faraday law, (3-78)

J: E. dt = _ dt/Jm ~

dt

[3-78]

to the closed paths tl and t2 defining the circuits. In (3-78), E denotes the total field existing at the elements dt of the paths tl and t 2 , with o/m the total flux intercepted by each circuit-flux generated by both 11 and 12 , To help visualize this process, in

326

STATIC AND QUASI-STATIC MAGNETIC FIELDS

(b)

(a)

FIGURE 5-34. Se!f- and mutual fluxes produced by 11 and 12 in coupled circuits. and emfs induced. (a) Flux of 11 only. The self-flux l/i I links t p inducing

~ ~tJ The mutual flux

1/112

=

IS

2

El 'dt

(1)

BI . ds links t 2 , inducing (2)

(b) Flux of 12 only. The self-flux l/i2links /:2, inducing

~ E . d/:

:Yt2 The mutual flux

t/121

=

= _dl/i2

2

Is! B2 . ds links t

1,

(3)

dt

inducing (4)

Figure 5-34 are shown the separate fluxes of 11 and 12 , Only one independent voltage source V(t) is used. The senses of 11 and 12 are arbitrary, being assumed as shown. Figure 5-34 shows that the total E generated along the closed path of circuit II and appearing in the left side of (3-78) consists of three contributions: a field El induced along II by -dl/1t/dt, in which 1/11 is the self~flux linking the circuit II and due to the current 11; another field E21 induced along II by -dl/121/dt, in which 1/121 is the "mutual flux" linking II and produced by 12 ; plus the generated field Eg produced only within the independent voltage source V(t). Thus, the total E· dl contribution to the integrand of the left side of the Faraday law (3-78) becomes E . dl = (E1

+ E21 + E 9 ) . dt = ! . dt + E 9 . dt (J

(5-126)

in which the current density J is that induced in the conductor via (3-7) and by the continuity of the tangential portion of the total electric field Econd = El + E21 appearing at the conductor surface.

5-l2 COUPLED CIRCUITS AND MUTUAL INDUCTANCE

(e)

(b)

(0)

327

FIGURE 5-35. Magnetically coupled circuits and circuit models. (a) The physical coupled circuits, with assumed current directions. (b) Circuit model showing elements corresponding to terms of (5-128). (e. Circuit model using symbolic convention to denote circuit self-indnctances.

The right side of the Faraday law (3-78) concerns the two magnetic flux contributions t/lm = t/ll + t/l21 linking the surface SI bounded by the circuit tl as shown in Figure 5-34(a) and (b). With this and (5-126), (3-78) finally becomes

~

:Yt,

l. dt + (J

j( +) E . dt J(-)

9

= _ dt/ll _ dt/l21 dt

dt

(5-127)

At low frequencies, the leftmost integral of (5-127) becomes Rllb Rl being the resistance of the conductive path by the arguments of Section 4-14B. Thus (5-127) may be written

The fluxes t/ll = Is, BI . ds and t/lzl = Is, B z . ds linked by tl are the positive quantities t/ll = LIIl and t/lZl = M12 , since those fluxes emerge from the positive side of S1 bounded by tl in Figure 5-34. With these substitutions one obtains (5-128a) the desired Kirchhoff voltage relation for the circuit t l . Applying a similar line of reasoning to the other circuit, one obtains the desired Kirchhoff voltage relation for t z

o

(5-128b)

These coupled diflcrential equations correspond to the circuit model in Figure 5-35. The use of this model makes it evident, without recourse to field theory, that on removing Rv tor example, the open-circuit voltage obtained across gap terminals at cod is just M dlddt. Other features of coupled circuits from the point of view of this model are treated in standard texts on circuit theory.21 liSee, for example, S. 1. Pearson, and G. J. Maler. introductory Circuit Analysis. New York: Wiley, 1965, pp. 54-63.

328

STATIC AND QUASI-STATIC MAGNETIC FIELDS

5·13 MAGNETIC FORCES AND TORQUES Although the force acting on a current-carrying circuit in the presence of an external magnetic field can often be obtained by use of the Ampere force law, (5-45a), frequently it is more expedient to obtain it from the stored magnetic field energy. It is shown how the force or torque acting on a current-carrying circuit or a nearby magnetic material region is deduced from an application of the conservation of energy principle to a virtual displacement or rotation of the desired body. This process is analogous to the determination of forces or torques exerted on charged conductors or dielectrics in the presence of an electrostatic field, discussed in Section 4-15. Suppose the magnetic circuit of Figure 5-36(a), having an air gap of variable width x, derives its energy from the source V supplying a direct current to the winding. If the armature were displaced a distance dt at the gap due to the magnetic field force F acting on it, the mechanical work done would be (5-129) This work is done by V at the expense of the energy in the magnetic field such that the following energy balance is maintained

+ dU dUs dUm ~~ '-M-a-g-n~etostatic\ Mechanical Work done work done by source V energy change

(5-130)

The change in the magnetic energy, on changing the air gap in Figure 5-36(a), produces a corresponding inductance change. The magnetostatjc energy Urn is 1/2LP from (5-72), so the energy change occurring with 1 help constant becomes (5-131) Omitting the 12 R heat losses associated with the coil resistance in the equivalent circuit of this system depicted in Figure 5-30(c), the work dUs exerted by V to maintain (5-130) is done against the voltage induced by the flux change dl/l m in the time dt such

I

v-=-

(a)

(b)

FIGURE 5-36. Single circuits using magnetic cores subject to relative translation or rotation. (a) Armature translates. (b) Armature rotates.

5·13 MAGNETIC FORCES AND TORQUES

329

that V = -dljim/dt. With ljim = LI from (5-88a), and with I maintained at a constant value, the induced voltage becomes V = -dljimfdt = -ldL/dt. The work dUs done by the source in the time dt to overcome this voltage is therefore

dUs = - VI dt = 12 dL

(5-132)

which is just twice (5-131), the change in the stored energy. Combining (5-129), (5-131), and (5-132) into the energy balance, (5-130) thus yields (i)PdL+F'dt= [2 dL, reducing to F' dt = (1)P dL, or (5-133) The latter shows that the mechanical work just equals the change in the magnetostatic field energy. Thus, of thc electrical energy supplied by V, one-half goes to increasing the magnetic ener[jY of the system, whereas the other half is used up as mechanical work done by the magnetic force. The differential magnetostatic energy change dUm can be written in terms of the coordinate variations of Urn as the armature moves the distance dt = axdx + a y 4Y + a. dz if desired; that is, (5-134) gradient form allowable in view of (2-11). A comparison of (5-134) with (5-129), making use of (5-133), leads to the cartesian components ofF

(5-135a) Since Urn = (i)LP from (5-72), the force components with I constant can also be written in terms of the derivations of the self-inductance L as follows

I 2 0L

F=-y 2 oy

(5-135b)

To evaluate F, the magneLostatic energy Urn (or the self~inductance L) should be given in terms of the coordinates of the displaced element of the system. In Figure 5-36(a), for example, U m would be expressed in terms of the single coordinate x denoting the air-gap width. Suppose a portion of the iron core, instead of being translated, is constrained to rotation about an axis as in Figure 5-36(b). Then the differential work (with dU = dUm) done by the magnetic force in the angular displacement dO a1 dOl + a2 d0 2 + a3 d0 3 becomes (5-136) wherein T = a 1 T1 + a2 T2 + a 3 T3 denotes the vector torque due to the magnetic force. Then results analogous with (5-135a, b), in terms ofthe variations ofthe magnetic

330

STATIC AND QUASI-STATIC MAGN.ETIC FIELDS

energy with respect to angular changes, obtain as follows

(5-137a) and in terms of the variations in the circuit self-inductance with respect to the angular motions, one obtains

(5-137b)

EXAMPLE 5-20. A magnetic relay has a movable armature with two air gaps of width x as shown in the accompanying figure. The n turn coil carries a current 1 derived from the source V. The core and armature, both of permeability f-t, have the median lengths and cross-sectional areas t 1 , AI; t 2 , A 2 , respectively. (a) Find the expression for the magnetic flux, the magnetic energy stored, and the self-inductance of the system, expressed as functions of the gap width x. (b) Determine the force acting on the armature. Express this force in terms of magnetic flux in the air gap, and in terms of the air-gap Bav field.

(a) The core flux is obtained by use of the magnetic circuit methods in Section 5-3. The reluctances are 91 1 = tdf-tAl' 91 2 = tzlf-tAz, and that of the two air gaps in series is 2x/f-toA 1; whence

nl

t/lm,core =

------2-x91 1 + 91 2 + --_. f-toA l

(1)

L is well approximated by the extcrnal self-inductance (5-88a). The core flux passes n times through the surface Sex bounded by the coil, so that

L = nt/lm.core = 1 ----------2-x91 1 + 9l z +-f-toAl

(

I

I

1

I I L

I \

EXAMPLE 5-20

(2)

PROBLEMS

331

The magnetic energy of the system is therefore

(3)

It is evident that increasing the air gap results in a decrease in the core flux, the self-inductance, and the stored energy. (b) The force on the armature is obtained from (5-135a) or (5-135b); F has only an x component, as expected from the physical layout; thus

2x JloA [ [Jt 1 + [Jt 2 + JloA

J2

(4)

The negative sign means Fx is in the direction of deereasing gap width x, corresponding to an increase in magnetic energy. ''''ith the core flux expression (1), rewrite the air-gap force (4) as

F

x

With

!/Im,eore

=

BovA,

thi~

I = __

Jlo

A

.1, 2 'P m,core

(5-138a)

is also written (5-138b)

showing thc air-gap force to be proportional to the air-gap flux squared, as well as to the flux-density squared.

REFERENCES ELLIOTT, R. S. Electrornagnetics. New York:

McGraw~Hi1l,

1966.

LORRAIN, P., and D. R., CORSON. Electrornagnetic Fields and Waves, 2nd ed. San Francisco: Freeman, 1970. REITZ, R., and F. J. MILFORD. Foundations of Electrornagnetic Theory. Reading, Mass.: Addison~ Wesley, 1960.

PROBLEMS SECTION 5-1 5-1. From the divergence of the static diflerential Ampere law (5-2), show that the differential property of static current density (5-3) follows. Explain the physical meaning of (5-3). Show how (5-6) follows from (5-3), from an appropriate integration and by an application of the divergence theorem.

SECTION 5-2 5-2. In the figure is shown a toroid of permeability Jl

= JloJl" through which a long wire carrying the steady current I is coaxially threaded. (a) Making use of the symmetry, Ampere's and boundary conditions, argue why the same H field exists in the toroid as in the surrouuding air. Find B in the two regions. (b) With 1 = 10 A, Jl = 500Jlo, a = I cm, b = h = 2 cm,

332

STATIC AND QUASI-STATIC MAGNETIC FIELDS

I

I h

L PROBLEM 5-2

find Hand B to either side of the interface at the inner radius p in the toroid.

=

a. Determine the core flux

SECTION 5-3 5-3. A particnlar ferromagnetic core with an air gap is similar to that shown in Example 5-3. I t has a 5-cm 2 cross-sectional area, a median core length in the iron of 20 em, a 4-mm air-gap length, and is wound with a 200-turn coil carrying 0.1 A. The iron core has the constant permeability It = 5000lto. (a) Sketeh the analogous dc electric circuit and the equivalent magnetic circuit diagram, labeling the symbolic quantities that apply. (b) Calculate the reluctances of the iron path and the air gap. Find Bov and Hav values in each region. (c) At the iron-to-air-gap interface, which boundary condition (from Table 3-2) applies there? (d) Show that the Ampere integral law (5-5) is satisfied, by integrating H • dt about the closed median path. To which Maxwell law is the Ampere law analogous, but applicable to the electric-current analog? (e) If the air gap were missing and the applied mmfnI were the same, by what factor would Bov and the core flux increase? [Answer: (b.) Bav 6.22 mT]

5-4.

Suppose the toroidal magnetic circuit of Exam pie 5-3 had no air gap. With the dimensions and parameters as given, find the H field in the core along the median path (p = 5 cm) two ways: (a) using the magnetic circuit method; (b) using Ampere's law. Compare the answers. If this toroid has an air gap, explain why the Ampere law cannot be applied to find H directly.

5-5.

Given is the two-mesh magnetic circuit with an n-turn winding as shown in Figure 5-7 (b). Let the iron core /1 104 Ito and the coil wound about the middle leg carry 0.1 A with 80 tUfns. The median path length of the middle leg is t3 = 4 crn, whereas the outside legs have tl = 2 (2 = 12 em, with all cross-sectional areas fixed at 2 em . Sketch the schematic diagram of the magnetic circuit appropriately labeled, along with the analogous dc electric circuit. (a) Using the analogous circuit, employ simple circuit reduction methods borrowed from the analogous electric circuit to calculate the magnetic flux in each branch, neglecting leakage. Find Bav in each branch. (b) Find Hav in each branch. Check your solution by verifying whether Ampere's law is satisfied around one closed loop that includes the mmf source rd. [Answer: (a) ifJm3 = 0.201 mWbl

PROBLEMS

333

5-6.

Given is the same two-mesh magnetic circuit as in Problem 5-5, except that, additionally, a O.5-mm air gap is sawed through the middle branch t3' (a) What is the air-gap reluctance? Sketch the new analogous electric circuit, labeling appropriate quantities and their analogies. (b) Find the new value of current required in the n-turn coil to establish the same magnetic flux in each branch as was obtained for Problem 5-5. By what factor does the current need to be increased? Comment on the effect of the air gap. (c) If the air gap had instead been placed in the outer branch t 1 , comment qualitatively on its effects in this event. [Answer: (b) 1= 5.1 A]

5-7. Given is the two-mesh magnetic circuit of Figure 5-7(a), with the mmfsource nl wound on the outer leg t 1 • Sketch this system, along with a labeled schematic magnetic circuit. Assume the identical dimensions and parameters of Problem 5-3. (a) Repeat part (a) of Problem 5-5 for this new configuration. (b) Calculate Hap in each branch. Check your solution by verifying whether Ampere's law, of the form (5-20e), is satisfied around the closed loop defined by the branches tl and t 2. 5-8. A particular I %-silicon (Si) steel, useful in magnetic circuit applications, has the type of nonlinear B-H curve depicted in Figure 3-13(b). Only points on the virgin curve OP3 are considered lkre. (1'he hysteresis efleet is disregarded.) Tests on this steel show a curve having the (B, H) coordinates: (0.04,20), (0.13,40), (0.24,50), (0.39,60), (0.53,70), (0.63,80), (0.76,100), (0.87,125), (0.95,150), (1.06, 200), (1.19, 300), (1.25,400) in mks units. (a) Graph this B-H eurve on linear graph paper with reasonable care. (b) With fl, defined by B/floH, calculate the static fl and fl, values for each given point, and graph fl, as a function of Hover the given range. 5-9. The gapless toroidal ring shown is made of the Si steel described in Problem 5-8, with R = 10 em, r = 2 cm. Let the current in the 100-turn winding be 1.257 A. (a) Use (5-20e) to find Hav in this core. Find also Bap and the magnetic flux in the core. Employ the B-H characteristic given in Problem 5-8. [Answer: I/Im = 1.33 mWb] (b) Use answers obtained in (a) to deduce tbe values of fl and fl, of the core at its operating point. Find the reluctance of this magnetic core. Making use of the latter, check the value of the core flux obtained in (a). (c) Explain why the use of (5-20c) would have been unsuitable in part (a). 5-10. A toroidal magnetic circuit with an air gap has dimensions the same as those of Example 5-3. The core is made of the Si steel described in Problem 5-8. (a) Suppose that the maximum magnetic density Bav at which this device is to be operated is 1.06 T. Determine the corresponding core flux, the field Hav established in the steel core and in the air gap, and the mmf drops across the two regions. What mmf is required of the 100-turn coil to produce the desired Bav? What coil current? (b) If there were no air gap, what coil current would then be needed? Comment on the effect of the air gap on the required driving current to produce a desired Bap in the magnetic core. . 5-11. In the toroidal magnetic circuit with air gap of Problem 5-10, assume 1= 10 A flows in the 100-turn coil. Find Bav and the core flux. [Hint: Since neither (5-20c) nor (5-20e) is amenable to a direct solution for Bap , assume as a first approximation that the applied mmf due to nl is entirely across the air gap only, using successive approximations to find Bap from the B-H graph of Problem 5-8.]

PROBLEM 5-9

334

STATIC AND QUASI-STATIC MAGNETIC FlELDS

SECTION 5-4 5-12. Given a very long, round conductor of radius a carrying the static current I in free space and that its exterior B field is aq,1l01/2np, use (5-22) as the basis for finding the potential A outside the wire. [Hint: Expand (5-22), noting it has only a z-eomponent and that its %z operator is zero (why?). Integrate the resulting differential equation to obtain 1101

tnp+C

(p?:a)

2n

If desired, put the arbitrary potential reference (where A z

= 0)

at p

= a to eliminate C.J

5-13. Repeat Problem 5-12, but this time find A inside the wire, given that B there aq,llolp/2na2 Show that

IS

To what docs this result reduce, if the wire surface is taken as the potential reference?

SECTION 5-5 5-14. A finite length of this wire, in air, carries the static current I and lies on the z-axis as in Example 5-4, except it is displaced so that its lower end is at z = L J and its upper end is at L z . Sketch and label it. Find the vector magnetic potential A at any location P(p, 0, 0) on the p-axis by integrating (5-28c), showing that

5-15. (a) In Example 5-5 concerning the small curre'nt loop, show the details of inserting (5-32) into (5-22) to obtain B of (5-33). (b) Comment on the duality existing between the B field (5-33) of the current loop of Figure 5-10 and the E field (4-44) of the electric dipole charge of Example 4-8. How do their field sketches compare? The strength of the electric dipole moment in (4-44) is qd. Recalling the definition (3-53) of the magnetic moment of a current loop, what is the "magnetic dipole" moment inferred from (5-33)?

°

5-16. A square loop of thin wire centered in the z = plane and of sides 2a parallel to the X,] axes in air carries the current I flowing counterclockwise looking from the top. Sketch this geometry, and show details of how the Biot-Savart law (5-35b) is used to obtain the B field at P(O, 0, 0), yielding

1l0J21 B(O, 0, 0) = a z - - 11.a Make use of symmetry to show that integration along only one side of the loop is needed.

5-17.

(a) Show that B along the z-axis of the thin, square loop of Problem 5-16 is given by B(O, 0, z) = a z

11.(z

2

+

21loIa2 2 1/2 Z 2a) (z

2

+a )

\

[Hint: Make use of results of Example 5-4, if desired.] (b) To what result does this reduce at the center of the loop? (See Problem 5-16.) If 1= 10 A, a = 1 ern, find B(O, 0, 0). (c) Show, as z becomes sufficiently large, that B at great distances falls oft' as the inverse cube of the distance.

°

5-18. A thin, circular loop of thin wire centered in the z plane is of radius a and carries the current I (going couHterclockwise seen from the top) in air. Sketch it. (a) Use a direct inte-

/

PROBLEMS

335

: :

free Iltial

a/az

t

d[JU con~~~tlng b

I(t)

'~1

t

IS

I(t)

i-'- Viti I I

:e

t

(a)

I I I

I I

I I I

(b)

I'ROBLEM 5-19

ofthc Biot-Savan law (5-35b) to show that B along the

as is at the

IS

z axis

is given by

what result does this reduce at the loop center? Find B there iff = lOA, a = I em. Iff = lOA, 10 cm. (b) Show that this B field agrees, as the distance from the loop is made large, with f()!' the B field of a small loop.

SECTION 5-7 5--UI. ting 'ield e of lent

That the this j at

I

A highly conductive wire loop, of the rectangular dimensions as noted, is placed in the ('ommon plane of a nearby long wire carrying the current I(t) = 1m sin rot as shown in (a). What (quasi-static) B field is produced by the current'? (b) Use the Faraday law (5-41) to the open-circuit voltage V(t) at the loop gap. (Show on a sketch the direction orB on the bounded by the loop and the choice of a positive surface element.) What is the polarity at the gap? Explain. If 1m 10 A, f 20 kHz, d 4 mm, a b 10 em, find V(t), its polarity. (c) Repeat (b) for the parallel-wire system of figure (b), making use of ~ymmetry.

5--20.

A high-11 magnetic toroid has a rectangular cross section as shown, and is wound with n-turn coil carrying the current I(t) 1m sin rot. A one-turn secondary loop of wire embraces core as shown. (a) Use Ampere's law to deduce the quasi-static B(p, t) field in the toroidal Find the "core flux. (Sketch the flux in a side view of the system, noting its direction in relation to the positive current sense.) (b) Use Faraday's law (5-41) to deduce the open-circuit V(t) at the gap of the secondary loop t 2 . If a = 1 em, b = 3 em, d = 2 em, n1 = 150 turns, kHz, core 11 = 4000Jlo, and 1m = 2 A, find t/lm(t) and V2 (t). Label the polarity of V2 (t) thc gap, explaining your choicc.

by

-(z)

the be1ce. nes Ilte-

('ROBLEM 5-20

. E7

II

336

STATIC AND QUASI-STATIC MAGNETIC FIELDS

I

I I I 1'1

It

"I

T I l'--------'~

!

V(t)

p

1 I

I I 1 I

PROBLEM 5-21

SECTION 5-8 5-21. The long, straight, round wire shown carries the static current 1. The thin, rectangular loop shown is located with its nearest side at the distance p from the wire center. The rigid loop is moved radially away from the long wire, all points on the loop moving at the velocity v apv" relative to the wire. Use (5-44d) to determine V(t) induced at the loop gap, including its polarity and the reason for your choice. (Sketch the system, labeling typical v, B, and v X B symbols thereon, as required by the integration.)

SECTION 5-10 5-22. In Figure 5-22(c) assume, in the end view of the simple generator shown, that the radial magnetic field has the constant Eo magnitude in the gap over a ± 60° angular interval measured {i'om the vertical, and is zero outside the gap. Use (5-44d) to derive the motional voltage V(t) generated by the rotating coil across its open terminals, assuming the coil has n turns. (Show its polarity on a sketch, justifying your choice.) Show that V(t) = 2B odawn. If Bo = 0.3 T, d = 12 cm, a 4 cm, n 20 turns, and the r'otor is spinning at 50 revolutions per second, find V(t).

SECTION 5-11C 5-23.

(a) Make use of (5-77) to show that the magnetic energy stored in the toroid of Problem 5-20 is U m = (J.ldn 2 J2/4n) tn (b/a). Deduce its self inductance therefore to be L

/.ldn

2n

2

R b ,n-

a

and compare this wi th the result obtained in Exam pie 5-17 by the flux-linkage method. (b) For the toroid with dimensions as given in Problem 5-20(b), find its magnetic energy if J = 2 A, and its self-inductance. Under what condition would the self~inductance be a function of the current in the device? 5-24. (a) Find the magnetic energy stored in the toroidal inductor of Examplc 5-3, using average magnetic field values. What percentage of the total energy is stored in the air gap? What is the self-inductance? (b) Repeat the energy and inductance calculations of (a), but for no air gap in the core. Comment on the comparative results. 5-25. Determine, from results obtained in Example 1-17, the magnetic energy stored in a length d of a very long solenoid in air, with n/d closely spaced turns per meter. Show that its selfinductance per meter, L/d, is /.lonb 2 (n/d)2. For a long solenoid with b = 3 em and 10 turns per centimeter, find its inductance per meter.

PROBLEMS

337

PROBLEM 5-31

5-26.

(a) For the coaxial line of Example 5-13, verify the results (I), (2), and (3) obtained for its internal and external inductances, giving ample details. (b) The expression (5-83) is sometimes used for the inductance of a length t of the coaxial line. Under what condition(s) would this result be accurate?

SECTION 5-11D 5-27. For the toroidal inductor of Example 5-3, use the external flux linkage to lind its selfinductance. With no air gap, by what factor docs its inductance increase?

5-28.

Find, using the flux linkage method, the expression for the self-inductance of every length d of the very long solenoid in air of Example 1-17. Check the result with that given in Problem 5-25.

51.29.

For the two-mesh magnetic circuit with parameters as given in Problem 5-5, it was found that 0.1 A in its 30-turn coil produced 0.201 mWb of magnetic flux through the coil. Find its external self~inductancc, using the flux linkage method.

5-30.

In the two-mcsh magnctic circuit with an air gap, as described in Problem 5-6, it was found that the coil current of5.1 A produced the magnetic flux of 0.201 mWb through the coil. the flux linkage method to find the coil self-inductance. Neglect internal inductance.

5-31.

The toroidal magnetic core of circular cross section has a coil ofn turns as shown. Ncglectthe winding intcrnal inductance and the flux leakage and assuming the iron permeability p to be constant, use the flux-linkage expression (5-83a) to determine the approximate selfinductance. Use magnetic circuit methods to determine the core flux. Show that L = 1lT/ 2r2/2R. If Ilr 10 5 , n 50, r = .5 mm, R 3 em, find L. 5-32.

Rework Problem 5-31, this time employing Ampere's law to find the exact expression

H in the core, whence deduce the core flux from the integration ofB . ds over the core cross lIection. From this, deduce the external self~inductance by usc of (5-33a) and flux linkages. Calculate L for the values given in Problem 5-31. 5-33.

A wire circuit is threaded through a small toroidal low-loss ferrite bead of permeability shown. How much self-inductance is added to the circuit? [Hint: Reason that the H field or without the bead is essentially the same. The fields within the bead (sec enlarged figure) essentially those for the straight-wire Problem 5-2.]

I'ROBLEM .5-33

338

STATIC AND QUASI-STATIC MAGNETIC FIELDS

= 0.5 em

I

-107) for the power delivered to coupled circuits. [See thc proof of (5-66) for a single circuit.]

5-38.

From the expression (5-110) for the magnetic energy of coupled circuits, derive (5-111) for linear circuits. [Hint: Observe how the linear result (5-71) was obtained from the general expression (5-70) for a single magnetic circuit.]

5-39.

Usc (5-121) to deduce thc Neumann formula for two thin circuits in free space M=

11 {,

12

J1. o dt'· dt ...._ .4nR

Sketch a pair of circuits with labeling appropriate to the usc of this intcgral.

5-40.

Use the Neumann formula for thin circuits given in Problem 5-39 to derive the mutual inductance between two coaxial, circular loops with radii a and b, and separated by the distance

PROBLEM 5-40

PROBLEMS

339

free space as shown, obtaining

which K(k) and E(k) are the complete elliptic integrals (5-97), and k=

Proceed along lines suggested by Example 5-18, noting that the distance between a source P' and a field point P is

5-41. Given a fixed circuit tl in free space as shown, suggest, with respect to the flux-linkage definition (5-123b) of M, how the mutual inductance varies with respect to the second circuit on relocating it according to the three cases illustrated. Explain briefly, showing roughly the extent to which the flux of 11 (in t 1 ) links t z . 5-42. Suppose a second coil tz with Tt2 = 250 turns is wound on the iron core with an air gap, described in Example 5-3. Employ flux linkage methods to determine the self-inductance of each winding. Find the mutual inductance between these windings two ways: (I) by usc of the flux linkage result (5-123b); and (2) using (5-125), assuming zero leakage flux in this system.

~-43.

(a) In the coaxial coupled circuit system (in air) of Problem 5-40, assume the radius b of circuit t2 to be small compared to a, the radius of circuit t 1. Then the current 11 in tl would produce an essentially uniform B field over the smaller circuit t z . Using the solution to Problem 5-18 for Bl along the z-axis, show that the mutual inductance between these circuits is essentially J1.o1f.(ab)z/2(a 2 + d2 )3/2. (b) Find M between these circuits if a = 12 em, b 2 em for two cases: (I) if d = 20 cm (coaxial circuits), and (2) if d = 0 (coaxial and coplanar). Let the wire diameter be Imm. (c) If 11 = 10 A flows in circuit iI' find the magnetic flux 0/12 linking the second circuit. (If 12 were 10 A, then from (5-123b), how much flux 0/21 would link the first circuit?)

5-44. Make use of the inductance expressions (5-100) and (5-\ 0 1) for a circular wire loop to determine the self inductance of each of the two loops with dimensions as given in Problem 5-43(b). Usc these and the value of M to deduce the coupling coefficient k for both circuit separations d 0 and d 20 cm. 5-45. (a) For the same rectangular circuit near a long, straight wire in air as shown in figure (a) of Problem 5-19, find the expression for the mutual inductance between the two circuits. Sketch this labeled system. Find the value of M, using the dimensions given in Problem 5-19. (b) If ll(t) 10 sin Wi, make use of (5-123b), 0/12 MIll to find the amount of flux 0/12(t) linking the rectangular circuit having the given dimensions. (c) Use the Faraday law (5-41) to

(a)

(b)

(c)

PROBLEM 5-41 (a) Coaxial circuits. (b) Coplanar circuits. (e) Coaxial and coplanar circuits.

340

STATIC AND QUASI-STATIC MAGNETIC FIELDS

PROBLEM 5-46

deduce the open-circuit voltage V(t) (ineluding its polarity) appearing at the gap in the rectangular circuit at the frequency f= 20 kHz specified in Problem 5-19. (Identify the flux l/l", in (5-41) here as precisely 1/112' the flux produced by 11 and linking the circuit whcre t/1l2 = MIl' Evaluate V(t) making usc oftlle latteL) 5-46. Clamped firmly about the long, straight wire shown is a split toroidal core of permeability I' and the given dimensions, with n turns wound about it. The long wire carries the currentl j (t) 1m sill wt. (a) Based on thc flux produced in the toroidal core, obtain an expression for the mutual inductance betwecn circuits tj and t2 using the flux-linkage definition (5-123b). (Note tbat the flux t/112 linked by t z , that is, passing through the surface Sex,2 bounded by t z , is rl Limes the core flux.) (b) Find the value of Nt, if a 5 nnll, b = 1.5 cm, d = 3 cm, n = 200, 11 (t) = 50 sin wt A at the frequency f = 60 Hz, with I'r 5000. (c) For the values given in (b), use the Faraday law (5-41) to obtain the open-circuit voltage (including polarity) at the terminals of circuit l2' Do this two ways: (l) by usc of (5-41), or V -dl/l 12 /dt; and (2) making use of 123b) to express the flux t/l 12linked by l2 as t/1 12 M I], yielding V 2 (t) d(MItl/dt -Mdl l /dt.

SECTION 5-13 5-47. In Example 5-20, let n 150, 1 0.2 A, tl 10 cm, t2 5 em, 5 cm 2, A2 I cmz, gap x 1 mm, and I' = 80001'0' (a) Find the core flux, the densities Bav in the U-shaped stator and in the armature, and the force on the armature at the given gap width. (b) Repeat for the gap closed. 5-48. The hinged, movable iron armature provides a variable air gap oflength x with respect to the fixed iron U-shaped stator shown, both having the same cross-sectional area Ac- Assume that the small armature displacement x is linear translation. (a) Write the expression tor the core flux of this system, neglecting leakage. (b) Obtain an expression for the self-inductance of the coil, using the flux-linkage method. Find Ii'om this the expression for the magnetic stored

PROBLEMS

341

I I ~'-X

I I

t,.

r---------

I I I I

(Ill

I

.. I

(ILl

'- _ _ _ _ _ _ _ _

I I I I ---.I

PROBLEM 5-48

energy. (e) Determine the expression for the force on the armature, as a function of x. (d) If

t c = 12 em, Ac = 4 cm 2, x = 1.5 mm, I = 1.25 A, n = 200 turns, and J.l = 10 5J.lo (assuming linear iron), find the values of the core flux Eav and Hav in the iron and air-gap regions, the selfinductance, the stored magnetic energy, and the force on the armature. (e) [f the gap length x were reduced to 0.75 mm, by what factor would the force increase? If x were reduced to zero?

5-49. A magnetic relay has a rotating armature as in Figure 5-36(b). Label (as for the relay of Example 5-20) mean paths t 1 , t2 and cross-sectional areas A l , A z in the iron stator and armature, each of permeability J.l = J.lrPO' The air gap is produced by the small angle 0 = x/tz, x being the mean air gap length. Find expressions for the magnetic flux, self-inductance, stored energy, and torque, each in terms of the small angle 8.

,

(

CHAPTER 6 - - -_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __

Wave Reflection and Transmission at Plane Boundaries

This chapter is concerned with plane-wave boundary-value problems in one or two dimensions. The reflection from a perf(~ctly conducting' plane on which a uniform plane wave is incident is considered first. Replacing the perfect conductor with a lossy dielectric extends the problem into a two-region system, f()r which the wave transmitted into the dielectric is also of interest. The definition of wave impedance and reflection coefficient permits a systematic analysis of the multiple-layer problem, dealing with the reflected and transmitted waves excited by a normally incident wave. Next, a developmen1 of the Smith chart is discussed, with applications to the foregoing problems. Then the concept of standing waves and standing-wave ratio fiJr a lossless region is treated. The chapter concludes with a discussion of wave reflection and transmission at oblique incidence on a plane boundary.

6-1 BOUNDARY-VALUE PROBLEMS A boundary-value problem in electromagnetics is one involving two or more regions (separated by one or more intedaces) lew which solutions are desired such that (a) Maxwell's equations are satisfied by those held solutions in each of the regions, and (b) the boundary conditions discussed in Chapter 3 are satisfied at the interfaces. Examples are illust.rated in Figure 6-1. Figure 6-1 (a) shows a rudimentary boundaryvalue problem: a plane wave normally incident on a perfect conductor, yielding a reflected wave. In (b) is a two-region system separated by a plane interface. A given plane wave traveling in region I leads to the additional waves shown, such that the boundary conditions at the interf~tce are satisfied. In these problems, the given incident wave is presumed to originate hom an appropriate electromagnetic source (a generator) at the far left.

342

6-1 BOUNDARY-VALlIE PROBLEMS Region 2

Region 1

~

Reflected wave m~~n

{

- 7'

343

Region 2

Region 1 Incident

Transmitted

"-

/

-+-----,//

/

~

To sources of plane wave

/

Perfectly conducting plane boundary

-E-- To

sources of plane wave (b)

(a)

C~~

~ ~~\\ ;f\ \ \ \ \

(Region 2); Air

Monopolt;: Region 1

Voltage source

(d)

(c)

G;,,",'holo. . '

."

~z,

:vegUlde"" "'~~:'-"",Z

~ ~ 0'

-

~-

J:

"-

'"

-------tJli_____..... L

-~

Rectangular, hollow waveguide

(e)

Linear Biconical Biconical Spherical (thin) (fat)

(f)

I"IGURE 6-1. Examples of boundary-value problems in electromagnetic thCOIY, (Il) Reflection of a plane wave from a perfectly conducting plane. (b) Reflection of a plane wave from, and transmission into, a dielectric region 2. (el Monopole antenna at the earth's snrface. (d) Two types of conducting pairs, carrying waves from a generator to a load. (e) Two types of hollow waveguides, carrying waves from a generator to a load. (f) Four types of driven antennas in free spacc.

Whenever the source of electromagnetic energy is included in a boundary-value problem, you can say that you are discussing the complete boundary-value problem. If the reflected wave does not couple signiticantly with the generator, a discussion of the complete problem may not be necessary. [n Figure 6-1(c) is shown a three-region problem consisting of a driven monopole antenna source transmitting electromagnetic energy into the surrounding space (regio1l 2) and into the earth (region 3). In Figure 6-1 (d) and (e) are showIl other complete boundary-value problems involving generators (sources) driving waves down one- or two-conductor systems (waveguides or tr::msmission lines) to a load at the far end. Systems such as these are considered in Chapters 8 through 10.

(

344

WAVE REFLECTION AND TRANSMISSION AT PLANE BOUNDARIES

6-2 REFLECTION FROM A PLANE CONDUCTOR AT NORMAL INCIDENCE A fundamental boundary-value problem of electromagnetics involves the reflection of a normally incident uniform plane wave from a plane perfect conductor. Assuming a plane of infinite extent avoids edge (diffraction) effects, and with the simplification of normal incidence, the problem is reduced to two dimensions (t and z). The geometry is shown in Figure 6-2. The sources of the incident wave are assumed at the far left in lossless region 1. Assuming x polarization, the incident wave is given in the real-time domain by (2-121)

E; (z, t) = E:'

cos (wt

-/3z)

Vjm

(6-1)

letting the phase angle + = 0 for convenience, but the incident wave (6-1) alone cannot satisfy the tangential field boundary conditions (3-72) and (3-79) at the interface. One must add a reflected wave solution, its effect being such as to cancel the incident field everywhere on the perfect conductor at every instant t. This occurs only if the second solution has the same frequency and if its equiphase surfaces are parallel to the walL The only other independent solution of Maxwell's equations that meets these requirements is the negative Z traveling wave solution of (2-119)

E; (z, t) = E~ cos (wt

+ 13z + (P-)

(6-2)

The unknown amplitude E~ and phase - are found by applying the boundary condition (3-79). The details are more readily carried out in the complex time-harmonic form; hence, the sum of (6-1) and (6-2), in complex notation, takes the form of (2-115)

Ex(z)

+ E; (z) = E:'e- jpz + E~eiPz Vjm =

E; (z)

-

Sources

(z)

(or (:1z)

o

);

I I

I I

FIGURE 6-2. Reflection of normally incident planc wave from perfect conducting plane.

(6-3)

6-2 REFLECTION FROM A PLANE CONDUCTOR AT NORMAL INCIDENCE

345

The boundary condition (3-79), that the total tangential electric field must vanish at the surface of the perfect conductor, is written Ex(O) = 0; so (6-3) becomes 0 = +E;;', whence

E:.

E~ =

-E;;'

(6-4)

Thus total reflection occurs, with the reflected wave amplitude equaling the negative of the incident wave. Inserting (6-4) into (6-3), the total electric field at arry location to the left: of the conducting plane becomes (6-5) a result with a wave amplitude 2E~, just twice that of the incident wave. The dependence of (6-5) on z is unlike the traveling wave nature of either wave constituent in (6-3). 1t has instead a standing wave character, in view of the factor sin /lz. A graphical space-time sketch of this standing wave is facilitated on converting (6-5) to its real-time form by use of (2-74). Assuming the real amplitude E~, one obtains

[It(z)eiwtl = Re [ - j2E~ sin {3::: ejrot ] Re [e - j9002E~ sin {3::: ejwt 1 2E~ sin {h sin wt

EAz, t) = Re

(6-6)

A sketch depicting the dependence on Z at successive t is shown in Figure 6-3(a). The total magnetic field accompanying the electric field (6-5) is obtained directly by substituting (6-5) into Maxwell's curl relation (2-108). This was, in effect, already done in Section 3-6, however, in which it was shown in (3-98b) that magnetic field traveling waves are related to corresponding electric fields by the intrinsic wave impedanfi:e. Hence, to (6-3) correspond the two terms of the magnetic field

Hy(z) = II; (z) + H; (z) e- jf!z

Em

eifJ z A/m

(6-7)

1J in which 1J == (Il/E) 1/2 is, from (3-99a), the intrinsic wave impedance of the lossless region. If (6-4) is inserted into (6-7), the complex magnetic field reduces to

2£"+ m

1J

cos

IJz

(6-8)

The real-time form of (6-8) (with f;~ taken to be the pure real E~) becomes (6-9) another standing wave. It is plotted in Figure 6-3(b) for comparison with the electric field. A space phase shiji of 90° occurs hetween the peaks of the electric and magnetic field standing waves, with the maximum magnetic intensity appearing at the perfectly conducting surface z O. The magnetic field (6-9) cannot fall abruptly to zero on passing into the interior of the perfect conductor without inducing an electric swface current, predictable trom

346

WAVE REFLECTION AND TRANSMISSION AT PLANE BOUNDARIES (/J., t, "1

= 0)

(/J., t, "I = 0)

z

z

z

(a)

Region 1: (/J., t, "1

(b)

= 0)

( y)

(c)

FIGURE 6-3. Standing waves resulting from a plane wave normally incident on a pcrfect conductor. (a) Incident, reflected, and total electric fields. (b) Incident, reflected, and total magnetic fields. (e) Showing the vector electric and magnetic fields of (a) and (b).

the boundary condition (3-72). Observe that the induced surface current density J. is x directed and cophasal over the conducting plane as shown in Figure 6-3(c). One can see a close physical analogy between the electromagnetic standing waves of Figure 6-3 and the mechanical standing waves of displacements and tensions along a transversely oscillating string anchored at one end! as shown in Figure 6-4(a). In (b) is shown another example of standing waves resulting from the reflection of electro1 For example, sce D. Halliday, and R. Resnick. Physics for Studmls Wiley, 1962, p. 412.

~f

Science and Engineering. New York:

6-3 TWO-REGION REFLECTION AND TRANSMISSION

~ .. .'

Incident wave_ ~ Reflected wave

.....

Electromagnetic transmitting horn

Region far frorn horn: spherical waves nearly plane 1

1

1 1

I

------..

I

t=-~/ --~----. ~-.::-::~::;:-; l " Vibration source (wave generator)

347

):

,..+'"

I I ::;:::

I

J...--. .: _""'

.. , '\ ~ '. / Generator (a)

(b)

fIGURE 6-4. Experiments involving standing waves. (a) Standing waves on a string connected to a rigid body and a wave generator. Null locations are checked visually. (b) Electromagnetic standing waves ncar conducting plane. Waves may originate !i'om a distant source as shown. A neon bulb reveals maxima and nulls.

magnetic waves from a conducting plane. Although the waves emanating from the horn are essentially spherical in the vicinity of the horn, at suitable distances away and over a limited transverse region they are very nearly plane waves, so that the solutions (6-5) and (6-8) are applicable in the vicinity of the plane reflector. If sufficient power is available, a small neon bulb might be used for detecting the nulls in the electric-field standing waves, yielding a rongh measure of wavelength. 6-3 lWO-REGION REFLECTION AND TRANSMISSION

The wave problem of Figure 6-2 can be generalized by assnming region 1 conductive ((j, i= 0) instead oflossless, and region 2 with a finite conductivity instead of being a perfect reflector. The system is shown in Figure 6-5. An incident plane wave originating from the far left is given by the positive z traveling wave terms of (3-9Ib) and (3-98c)

j;+xl (7) = j;+ml e- Y'z ""

(6-10)

wherein fj 1 is specified by (3-99a) f()r conductive region I or equivalently by (3-111). The propagation constant of region 1 is 1'1> given by (3-89) (6-11 )

which ex and p are obtained from (3-90a,b), or equivalently from (3-109) and (3-110). The continuity of the tangential fields across the interface in Figure 6-5 (a) gives lise to another plane wave at the same frequency in region 2. This wave is not sufficient to satisfy the boundary conditions (3-71) and (3-79) at the interface, however. One more wave, reflected in region I, is required if the boundary conditions are to be met. The three waves are shown in Figure 6-5(a) in real-time, and as complex vectors ill Figure 6-5(b). Thus, in region 1, the reflected wave is required as follows

111

(6-12)

348

WAVE REFLECTION AND TRANSMISSION AT PLANE BOUNDARIES

-

Motion

Transmitted: A" +

'~+

Ex2

=

71c2Hy2

(a)

E;'(Z)

l

z

0

Wave • -- - -- motion

l

l

• - - -..-. Motion

, Motion ....;---- •

E:Z(Z)

o

EX](Zi

- - - (x)

if;;. (z) = (b)

FIGURE 6-5. Plane wave normally incident on an interface separating two lossy regions. (a) Incident and reflected waves in region I, transmitted wave in region 2. (b) Vector representations denoting the fields of (0).

in which

iiI and Yl

are given by (3-99a) and (6-11). In region 2, the transmitted wave

IS

( ) - E"'+ -nz E"'+ ~'x2 Z m2 e

(6-13)

No reflected wave can exist in region 2, because that region is infinite in extent toward the right in Figure 6-5, whereas the only sources of the fields are to the far left in region 1. Satisfying the boundary conditions at the interface in Figure 6-5(b) requires setting the total tangential fields equal to each other at Z O. In region 1, the total electric and magnetic fields are given by the sums of (6-10) and (6-12) (6-14 )

349

0-3 TWO-REGION REFLECTION AND TRANSMISSlON

Tn region 2, they are simply (G-13) The boundary condition (3-79) requires the equality of the electric fields of (G-13) and (6-14) at z = 0; that is, + e- Y1Z [E~.. . . m1

+ E~-1111 e = £';+ e-Y,Z] z=o .1m2 i1Z

(6-15)

obtaining (6-16) The other boundary condition (3-71) requires the continuity of the magnetic fields there, obtaining (6-17) Thc linear results (6-16) and (G-17) involve the known impedances fil and fi2 of the regions, as well as the com plex amplitudes of the incident.:. the reflected, and the transmitted waves. Assuming the incident wave to be given (E;:;1 is known), the other amplitudes are pbtained from the simultaneous solution of (6-16) and (6-17). Rearranging them with £;:;! on the right yields ~-

~+

Em! - Em2 =

it;;'1

~+

+

Em2

~+

Eml

(6-18)

~+

Em! fi1

(6-19)

Their simultaneous solution obtains the complex amplitude of the reflected wave (6-20) Similarly, the transmitted wave has the amplitude (6-21) Additional confidence is gained in the results (6-20) and (6-21) on considering two special cases: (a) for which region 2 is a perfect conductor and (b) for which regions 1 and 2Jlave identical parameters (no interface exists). In case (a), with fi2 = 0, (6-21) yields E;:;2 = O,~a result ez:pected from the null fields within a perfect conductor; while (6-20) obtains E;;'1 = - E;:;I, agreeable with (6-4) as one should expect. In case ide~tjcal regions means fil = fi2' whence from (6-20) and (6-21), it;;'1 = 0 and = E;:;I, implying the reasonable conclusion that no reflection occurs if the region has no discontinuity.

IXAMPLE 6·1. A uniIorm plane wave with the amplitude £;1 = lOOeW Vim in air is normally incident on the plane surface of a losslcss dielectric with the parameters Ji2 = Jio, E2 = 4€o, and (J2 = O. Find the amplitudes of the reflected and transmitted fields.

350

WAVE REFLECTION AND TRANSMISSION AT PLANE BOUNDARIES

The geometry is sho""n in Figure 6-5. Region 1 is air, so iii = 110 = = 120n n. For region 2, liz = Jlo/4Eo = 60n n. The complex amplitudes of the reHected and transmitted waves are given by and (6-21)

J

60n 120n k';l1 = \00 ~~--~~60n + 120n

P;~2

=

I 00

_~~~On) 60n

These amplitudes into (6-13) and (6-

£.-(z)

z= d

_ ____ Q. ______ d ____ i>-(z) BCD

~

//'"

Incident field:

G

E:A = 111Hy~

-:.: I I I

I I

Interface 1

Interface 2

etc.

--E

"F'~H

:

....;--

etc:--

etc.

I

(a)

1

--

-J

etc.

(b)

FIGURE 6-6. Three-region system on which a uniform plane wave is normally incident. (a) Three-region system, showing the plane wave field incident on a thickness d of region 2. (b) Depicting the effects of the incident field on reflected and transmitted waves, with increasing time.

6-4 NORMAL INCIDENCE FOR MORE THAN TWO REGIONS

351

and 2, the ((lfWard wave C in region 2 striking the second interhce to produce a transmitted wave D, plus another reflected wave E returning to interl'ace 1. A continuation of this process, as time increases, produces an infinite sequence of reflected and transmitted waves, the linear sum of which obtains sinusoidal steady ,I-tate forward- and backward-traveling waves in the respective regions, Thus, in region I, the net positive z traveling electric field will consist only of the postulated x polarized incident wave A, denoted by

while the reflected wave in that region consists (Jfan infinite sequence ol' contributions of the waves B, G, ... ; that is,

Each wave term of the latter has a common factor eY1Z , so that the infinite sum, in the sinusoidal steady state, becomes

(6-22) reducing to a net reflected wave in region I designated by eY1Z E~"ml

(6-23)

in which £:;1 denotes its eomplex amplitude. Every term of (6-22) has an associated magnetic field related by the intrinsic wave impedance of region 1, yielding

E~l -- e YiZ

(6-24 )

it

The net, sinusoidal steady state f()rward and backward waves in region 1 are depicted in Figure 6-7. Similar arguments applied to the infinite sequences of waves in regions and 3 lead to the net field vectors shown. Region 1:

Region 2:

(ILl, fj, ITj) or ('n, 1/1)

(1L2, E2, tion

- +

Hyn

yJ =-;::~l

.....

i-~

EXl

Exl

Mot i~lE;i

E=--

Region n:

I ( J..1.ill fn, un

Mk, tk, Uk

-+

.

.l

\\

\\

"

\\

~l

..l

I I

353

= z (or (:Jz)

..:l

.. j

I

I

I

I

I I

xl

~l

--To

sources

I I

I I

I I

J:?IGURE 6-3. A multilayer syslem ofn layers, ou which a uniform plane wave is normally incident from the left.

and (E';, if;) except for the last (k = n) region, in which only the forward-traveling components f;;;n, H:~ appear. The total electric field for each region 2 becomes

in which fez) is caIled the reflection coefficient at any location z in the region, defined the complex ratio of the reflected wave to the incident wave as follows

(6-36)

The corresponding total magnetic field is

E,'-'"

~e

E;

2yz

J

E,'+

_ -'me -YZ[l

q

f(z) I

(6-37)

total-field impedance Z(z) is defined at any Z location by the ratio of the total electric (6-35) to the total magnetic field (6-37)

(6-38)

these results apply to any (kth) region, an additional k subscript should be applied to all quantities. simplicity, such subscripts have been dropped.

354

WAVE REFLECTION AND TRANSMISSION AT PLANE BOUNDARIES

A converse expression for ['(z) in terms of Z(z) is obtained from (6-38) by solving for ['(z)

['(z) =

~(z) Z(z)

fj

+ fj

(6-39)

a form convenient for finding ['(z) whenever Z(z) is known.~ Another useful expression is one that enables finding r at any location z' in a region in terms of that at another position z. At z', the reflection coefficient is expressed by use of (6-36): [,(z') = (E;;,/E:')e 2YZ '. Dividing the latter by (6-36) eliminates the wave amplitudes, yielding the desired result

(6-40)

In the application of (6-35) through (6-40) to the wave system of Figure 6-8, one should note the following properties of ['(z) and Z(z) at any interface separating two regIOns. 1. The total field impedance Z(Z) is continuous across the interface; that is, at an interface defined by z = a

(6-41 )

evident from the continuity of the tangential electric and magnetic fields appearing in the definition (6-38). 2. The reflection coefficient ['Jz) is discontinuous across the interface. This fol~ows from (6-39), for, because Z(z) must be continuous across the interface, r(z) cannot be if the wave impedance fj is different in the adjacent regions. The procedure for finding the complex amplitudes ofthe forward- and backwardtraveling waves in a multilayer system like that of Figure 6-8 is illustrated in two examples.

EXAMPLE 6·2. A uniform plane wave is normally incident in air on a slab of plastic with the parameters shown, a quarter-wave )hick at the operating frequency f = 1 MHz. The x polarized wave has the amplitude E~l = 100eN ' Vim. Use the concepts of reflection coeHicient and total field impedance to find the remaining wave amplitudes. To obviate carrying cumbersome phase terms across the interlaces, ass LIme separate Z origins 0 1 ,02> and 0 3 shown in (b) of the figure. The wave amplitudes are referred to these origins. First, values ofq for each region arcJ~und by using (3-99a); thus, fil fi3 .Jfto/Eo = 120n Q; in the plastic slab, q2 = .JJ4J4Eo = 60n Q. The propagation constants Y = rx + jf3 are computed from (3-90a,b) or (3-109) and (3-110); thus, in lossless region 2,

6-5 SOLUTION USING REFLECTION COEFFICIENT AND WAVE IMPEDANCE 1: Air (110, EO) I

!

Eii = 100e- Y1 " iI+

~~::n -+

yl

--..

(z)

~motion

Plastic

= Exl -

A

13 Air (110, EO)

I I

j:;+ x3

Ex2

L--

-+

HY2

l

0 1 O2

E;l

_

j,E;2 A

(~,

A

Motion -

......

Hy:

~l

M!i~jE~l HYI

I

2: Plastic (110, 4Eo)

355

03

-(z)

-- fry2

=-fI;

(b)

(a)

!EXAMPLE 6·2, (al Uniform plane wave normally incident on a plastic slab. (bl Side view wave components in the regions.

Then finding the complex wave amplitudes proceeds as follows. (a) One begins in region 3, containing no reflected wave. [3(Z), [rom (6-36), is therein zero, yielding the total field impedance from (6-38) Z3(Z) = ry3(1 + 0)/(1 - 0) = ry3 = 120n Q. By (6-4)), the t then the ratio sin OJsin ec of (6-94) will exceed unity. Thus, et becomes a complex angle, and the implications of tbis on the behavior of the field transmitted into region 2 may be deduced as follows. With sin t (6-94) exceeding unity, cos Ot is written

en

e or

(6-95) The negative root of the imaginary result is chosen here to preserve the physical realizability of the wave in region 2 (to be clarified momentarily relative to its producing

378

WAVE REFLECTION AND TRANSMISSION AT PLANE BOUNDARIES

a vanishing wave there as

z -+

co). Putting (6-94) into (6-95) yields

cos Ot = - j

(6-96)

and substituting the latter and (6-94) into the directed magnetic field in region 2

H2 (x, z) = =

=

H2 expression of (6-77)

provides they-

ay Et e- jP2(X sin 0, +. cos 0,) 112 a

Y

Et e!- i/J,[xJ. 1/'2 sin 6i -

Et

a y

jzJ(;, 1/'2) sin 2 iii - 1]

112 e -[1I2J(E-'/'2) sin 2 8i =1]ze - 11112./;'-0" sin O;]x

172

The form of (6-97a) shows that H2 is a wave attenuated in change in x in accordance with the form of

H~

-

2 -

Et

ay-e '12

z and

(6-97a)

exhibiting a phase

- az - jbx

e

(6-97b)

with attenuation and phase constants a and b defined by (6-98) T1ws, the wave function (6-97) represents the magnetic field transmitted in region 2 for angles of incidence 0i that exceed the critical value (6-93). It is a wave attenuated in the increasing z direction and shifted in phase in the increasing x sense in region 2, as shown in Figure 6-19. The reason for the choice of the negative root of the radical

I (x) I

Constant-phase plane

(a)

(b)

FIGURE 6-19. (a) A typical nonuniform plane wave produced in region 2 when the angle of incidence in region I \exceeds the critical angle. (b) Detail of Hy of (6-97) in region 2, at a fixed instant. ',--

PROBLEMS

379

in (6-96) is now evident; a positive root would make the z-dependent exponential factor in (6-97) grow indefinitely large as z -+ W, which is not sensible physically. The wave is thus "trapped" into traveling with pure phase change along x (parallel to the interlace), while being attenuated in amplitude as one moves away from the interface in region 2. This attenuation is clearly not associated with dissipation in region 2, which is a lossless region. In the foregoing discussions of parts C, D, and E of this section, only the obliqueincidence case involving two lossless regions was treated. If region 2 were made a conductive region, the penetration of the transmitted wave into region would be analyzed in much the same way as is done for the lossless case in part C, except for the replacement or E2 with the complex permittivity € defined in (3-103). This has the effect of injecting a "complex angle" interpretation into Snell's law (6-82a). Details of this case are found in Appendix A. Of special interest in Appendix A is the case for which region 2 is a good conductor. In that instance, it is shown that the transmitted wave enters region 2 with its direction of travel essentially normal to the interface, as illustrated by Figure A-2(b). This result has an important application, for example, to the penetration of electromagnetic fields into the conducting walls of rectangular hollow waveguides, a fact utilized later in Section 8-6.

REFERENCES LORRAIN, P., and D. R. CORSON. Electromagnetic Fields and Waves. San Francisco: Freeman, 1970. FANO, R. M., L. 1'. CHU, and R. P. ADLER. Electromagnetic Fields, Energy and Forces. New York: Wiley, 1960. RAMO, S., J. R. WHINNERY, and T. VAN DUZER. Fields and Waves in Commnnication Electronics, 2nd cd. New York: Wiley, 1984.

PROBLEMS

SECTION 6-2 6-1.

The total-reflection magnetic freld solution (6-8) was obtained by inserting the boundary result (6-4) into the general reflection result (6-7). Show that (6-8) can also bcfound by inserting the total-reflection electric field solution (6-5) into the appropriate time-harmonic Maxwell equation.

6-2.

Employ the boundary condition (3-72) to obtain the

expre&~ion

for the current density

Js induced by the magnetic field onto the perfectly reflecting plane of Figure 6-3(r). What depth of penetration of this curreut is expected into region 2? Explain.

"6-3.

Assume that the totally reflective syste!? of) • ds

(7-11 )

result is interpreted physically as follows. The total instantaneous generated power in V, given by the left side of (7-11), equals the sum of the time rate of increase in

388

THE POYNTING THEOREM AND ELECTROMAGNETIC POWER

electromagnetic energy in V, the ohmic losses in V, and the outgoing power flux passing through the surface enclosing V. This form has some interpretive advantages when applied to an antenna, for example, in which case the last term, the integral of [jJ • ds over any surface enclosing the antenna, denotes the power flux radiated into remote regions of space. A. The Static Poynting Theorem In a static electromagnetic system carrying only direct currents, the operator is zero, reducing Poynting's theorem (7-8) or (7-11) to

-

Iv J . E dv W

~s [jJ • ds

a/at

(7-12)

Time static

assuming V contains no generators. Thus, in a dc system, the net power flux entering a closed surface S constructed about the current-carrying conductors is a measure of the ohmic losses in those conductors. The application of (7-l2) to a dc-carrying wire is considered in an example.

EXAMPLE 7·1. From (7-12), evaluate the total power flux entering the closed surfl.tee S embracing a length t of a long round wire carrying a d.irect current 1 as in (a) of the accompanying figure. Compare the result with the volume integral of (7-12). I I

E

t

H

Closed surface S

Detail at an endcap

fa)

(b)

End view (c)

"

EXAMPLE 7-1. (al Long round wire carrying a static current 1. (b) The E and H fields on the surface S. (e) Inward power flux associated with direct current flow in a wire.

389

7-1 THE THEOREM OF POYNTING

dng hen °

ds

The closed surface Sis noted in (b). The Poynting vector iY' on the peripheral surface p = a is obtained from the known E and H fields, H being given by (5-11) of Example 5-1, whereas E is obtained from the currcnt dcnsity Jz ljA combined with (3-7)

lOte The Poynting vector at p = a on S is obtained from

a/at

·12)

iY'

=EX H=

(axl) erA

X

12

(a4>1)

-a

2na

p

2naAer

As seen in (b), iY' on the endcaps contributes nothing to thc inward powerflux, making the total inward power flux (7-9) over S p=

_J. iY" ds j"s

= -

ct Ch Jz=o J4>=o

(-a ~) p

2naAcr

°a p

adcpdz

rmg e of

NIre

em:om-

a result expressed in terms of the resistance (4-138) of the wire. From (7-12), the result j2 R is also obtainable from the volume integral ofJ ° E taken throughout the interior of S. Thus

C JoEdv= Jv C (erE)oEdv= Jv r erE;dv= Jz-o C_ C: r"_ er(~)2PdPdCPdZ Jv J4>-O Jp.-o erA 2

t

integrating to 12R as expected. The positive sign accounts for the actual inward sense of the power-flux P over S, as noted in (c).

B. Time-Instantaneous Poynting Theorem and Plane Waves Illustrations of the Poynting theorem in the time domain can be drawn from the of plane waves developed in Sections 2-10 and 3-7 .-Thus, the power-flux-density vector &' associated with a plane wave in a region is obtained by use of (7-1) applied the appropriate fields. In empty space, assume that a positive z traveling plane wave electric and magnetic fields inferred from (2-121 a) and (2-130a)

(7 -13) (7-14) Applying these to (7-1) obtains the time-instantaneous Poynting vector at any Z position

&'(z, t)

E XH =

[axE~ cos (wt -

Poz)]

X

[ay~: cos

(wt - Poz) ]

(E+)2

= a z ~- cos 2 '10

(wt - Poz)

(7-15a)

390

THE POYNTING THEOREM AND ELECTROMAGNETIC POWER

Inset: flux Plot} of &"

(z)

I I I I (Ito, fO)

__ ..---

--~ Wave

I __ J...._ 0 - - ___ _

(y)

motion

--------

----(z)

(a.)

\

\

-~

Motion

----(y)

(z) (1))

FIGURE 7-2. The Poynting vector associated with a plane .wave in empty space. (a) The vector.o/' = a/!i', versus . ds

391 (7-16)

signifying that the flux of:1' into a closed surface S in the lossless region is instantaneously a measure of the time rate of increase of the stored electromagnetic energy within S. In the example that follows, the validity of (7-16) is examined relative to a plane wave in free space. EXAMPLE 7-2. Given the plane wave defined by (7-13) and (7-14), determine the net power flux P( t) entering a closed box-shaped surface S having dimensions as in the accompanying figure. Show that the time rate of increase of the electromagnetic energy within the volume of the box provides the same answer. Because {YJ is everywhere z directed, the only contributions to power flux entering the box are on thc ends Sl and 5;2 shown, so (7-16) yields

1\ (I)

-

.- !is f .0/'. N,

= -

i

b

y~o

fa [a x=o

z

(E~)2 - cos 2 (wt '10 (7-17a)

'10 (7-17b) The net power flux entering S is therefore

'-Ps PI)· ds

1'(1) = 1\(1)

+

i-

ab[cos 2mt - cos 2(w/

=

flod)] W

2'10 the last being obtained by use of cos 2 0 =

EXAMPLE 7-2

! + (!) cos 20.

(7-18)

Ci

A

392

THE POYNTING THEOREM AND ELECTROMAGNETIC POWER

Equivalently, if the right side of (7-16) is integrated throughout the volume of the box, (7-18) should again be obtained. Substituting (7-13) and (7-14) yields

a '" vt

J. [1l0H2 EoE2] a {llo(E;:Y J.d J.b J.a [1 + cos 2(wl - Po,;:)] dxdyd,;: -2- + - dv = '" 2 J.d J.b J.. [1 + cos 2(wt - Po';:)].dxdyd,;: } + Eo(E;:Y 4 v

vi

--2-

41'/0

0

0

a {Eo(E;:Y = at 2 =

0

0

0

'

0

J.d J.b J.a [1 + cos 2(wt 0

-

0

'}

Po';:)] dx dyd,;:

0

wEo(E;:yab 2Po [-cos2(wt-Po,;:)1~ (7-19)

------ [cos 2wt - cos.2 (WI - Pod)] agreeing with (7-18) as expected. 2

For a plane wave traveling in a conductive (ohmic) region, the effects of the attenuation of E and H and the phase shift between them is expected to i~fluence the power-flux P(t) entering a closed surface S. For this case, the fields are'fven by real-time expressions inferred from (3-94) and (3-98c) \. (7-20) (7 -21 ) which (J is the angle of the wave impedance (3-99). The Poynting vector (7-1) thus becomes [#>(z, t)

=E

x H

=

(E+)2 a z _m_ e- 2az cos (Wi - (Jz) cos (wt

{Jz -

(J)

(7-22a)

1J

and the use of cos A cos B [#> = a z

2P;

=

(!)[eos (A

- B)] obtains

t)

e- 2az[cos

2P;

+ B) + cos (A

(J

+ cos (2wt -

2{Jz

(7-22b)

(z, t) versus Z at t = 0 is shown in Figure 7-3. Not only does the A graph of attenuation of and account for a doubly attenuated power-flux density but the effect of cos (J in (7 -22b), replacing the term unity in (7 -I5b) for the lossless case, to go negative over a portion of each cycle, ~an effect associated with the to cause

E:

H;

2P; ,

2P;

~In the course of obtaining (7-19), note that with the snbstitntion (Jio/'1l) = Eo, the two integrals in the first Jlep become identical, so they combine into one. The time differentiation is taken inside the integral to elimthe constant unity term, whereas in the last step, the identity (W€o/ Po) = 1/0 1 is used.

In

7-1 THE THEOREM OF POYNTING

393

(x)

At t = 0

(y) '~(z)

FIGURE 7-3. The instantaneous Poynting vector traveling plane wave in a conductive region.

(z, t) associated with a positive ::.

phase shift 0 between the electric and magnetic fields and detracting from the average power transmitted in the z direction. EXAMPLE 7·3. If a plane wave exists in a conductive region, evaluate the net instantaneous power flux entering the box-shaped closed surface of dimensions as shown. Integrating (7-22b) over the ends S1 at Z = 0 and 8 2 at z = d yields the instantaneous power fluxes

= -

f

b

y~O

fa {a z (E~)2 - [cos e + cos (2m{ X~O

2'1

e)]}. (-azdxdy)

abl cos 0 + cos (2m/ - e)]

P2(1) = -~~- e-2~dab[cos 2'1

e + cos (2mt -

(7-23a)

2{3d

(/J.,t, o)

..

'

(y) (z)

EXAMPLE 7-3

e) ]

(7-23b)

394

THE POYNTING THEOREM AND ELECTROMAGNETIC POWER

so that the net power flux entering the box is their sum P(t)

=

(E+ m

211

ah[(l

e- 2ad ) cos fJ

+ cos

(2M - fJ) - e- 2ad cos (2wt - 2fJd

fJ)] (7-24)

From the Poynting theorem (7-8) it is evident that (7-24) is a measure of the time rate of increase of the stored electromagnetic energy within the volume plus the instantaneous ohmic loss occurring therein. One can expect that (7-24) will reduce to (7-18) if a lossless region (IT = 0) is assumed.

7·2 TIME-AVERAGE POYNTING VECTOR AND POWER

In a consideration of the electromagnetic power delivered by sinusoidally time-varying fields to a region or system, one's interest from the point of view of practical measurements leans toward the time average of the power flux rather than its instantaneous value considered in the previous section. Time-average power in electromagrl~tic fields is important for the same reasons as in circuit theory. The time-average po"ter)entering the terminals of a passive network, found by use of an electrodynamomete~pe wattmeter or from the knowledge of the amplitude and phase of the input voltage and current, is a measure of the average power dissipated as heat in all the resistive elements of the network. From the electromagnetic viewpoint, the time-average power flux entering a closed surface containing no generators is a criterion of the same thing: the heat-producing ohmic losses in the region. In laboratory measurements, the time average of a time-harmonic function is customarily taken over a time interval embracing many cycles '01' periods. Since for steady state sinusoidal functions all periods are alike, an average over one period will yield the same result as that taken over many such periods. The time average of the Poynting vector ~(Ul' Uz , U3' f), denoted by ~av' is defined as the area under the function ~ over a cycle, divided by the duration T (period) of the cycle, that is,

f!lJav(Ul' Uz, U3)

=

Area under f!IJ over a cycle 1 Base ( T sec) = T

iT 0

~(Ul' Uz, U3, t) dt

(7-25a)

if t is chosen as the variable of integration. One may alternatively choose wt as the angular integration variable; then (7-25a) is written with 2n as the base-divisor

(7-25b)

It is evident that the time-average Poynting vector is a function solely of position in space, the time variable having been integrated out over definite limits (in t or wt) in the averaging process.

7-2 TIME-AVERAGE POYNTING VECTOR AND POWER

395

A. Time-Average Poynting Vector and Plane Waves Illustrations of the time-average Poynting vector can be drawn from examples in the last section. Equation (7-ISa) denotes a time-instantaneous Poynting vector ~(z, t) = a~z (z, t) attributed to the wave of (7-13) and (7-14). Applying (7-2Sb) obtains its time average

PI' av (z)

az

(E~

2

2n 2110 (E~)2

=a z

2110

P" d(wt) + a z (E~)2 f Z1t cos 2(wt

Jo

2n 2110

Jo

fJoz)d(w/)

(7-26)

Thus the time-average result (7-26) is attributable wholly to the constant first term of the time-instantaneous expression (7-ISb). The double frequency term contributes nothing on the time average because it possesses canceling positive and negative areas over a cycle, evident from the f!Jz(z, t) diagram of Figure 7-4(a), which is just an extension of Figure 7-2(b) to successive instants in time t. The inset in Figure 7-4(a), showing the wave at the fixed Z = 0 location, yields an average Poynting vector (area divided by the base) that is one-half the peak power density (E~)z/110' or (7-26). If the region is lossy, ~ av becomes a function of Z due to the wave attenuation produced by the losses. The time-instantaneous Poynting vector, in this case expressed by (7 -22b), is depicted in Figure 7-4( b), an extension of Figure 7-3. In the insets are shown variations off!Jz (z, t) with t at two fixed z locations (z = 0 and A). Making use of (7-2Sb) leads to the time-average Poynting vector

(7-27) The result is doubly attenuated in z; it also retains the factor cos 0 produced by the electric and magnetic fields being out of phase by an angle 0, a factor analogous to the power factor of a two-terminal impedance of circuit theory.

B. Time-Average Form of the Poynting Theorem If the total time-average power flux through some surface S (not necessarily a closed surface) is desired, one must integrate Pl'av over S by use of

(7-28a)

in which S denotes an arbitrary surface, open or closed. With S closed, (7-28a) yields the net (or total) power flux leaving that surface. A negative sign must be included with the integral of (7-28a) if Pay is to signify the net time-average power flux entering the

:1

396

THE POYNTING THEOREM AND ELECTROMAGNETIC POWER

z t

="

z (or (:iz)

(or wt)

o

\

(a)

(2wt - 2{:i - 8)1

~

"

Wave motion

z= "

(b)

j~~ and {YJ~~ associated with the ineident and reflected waves; and P a-v power flux passing through the normal surface S of area A = 4 m 2 , as well as passing through S. Find also the ratio of these positive powers and the return loss (in dB). What would the return loss be if 100%, reflection occurred? Zero reflection?

r::.,

406

THE POYNTING THEOREM AND ELECTROMAGNETIC POWER

1: air

r-81 1

--, I + 1;1'av<

~

I

I

o

~ .~ ..

, ,:52

---+

.'~a---:';ll

I ~-­

I ~.

2: (ila, 4'0)

1

1:1'.+vI

~..---- (z)

)

1

,

---4

d-o>-j

PROBLEM 7-15

7-15.

Given is the two-region lossless system of Example 6-1, with incident, reflected, and transmitted plane-wave solutions as noted. Construct a hypothetical closed rectangular box with opposite sides S't and S2 parallel to the interhce and protruding into Ihe two regions as shown. (a) Using the complex field solutions obtained in Exa mple 6-1, evaluate the corresponding timcaverage Poynting vector power densities &>,,',.1' &>~v.l' &>:V,2 as labeled on the diagram here. Given that the x and)) dimensions or the rectangular box arc a = b 50 em, determine t.he net time-average power flux entering the closed surface of the box. [s the time-average Poynting theorem (7-31) or satisfied? Explain. (c) It is seen that the conclusion orpart (b) is independent of the length d this closed box, Why is this so? [Answer: (a) 13.26, 1.47, 11.79 W/m2]

~

In the three-region lossless syste!p or Examrje 6-2, the s(~utions in 'lJe three regions revealed the electric field amplitudes E:;'1 = 100, E;d = 60, £:;'2 60, E;"2 = 20, and = ~ j80 V 1m. Sketch this system. (a) Find the magnetic-field amplitlldes (iI:;, 1, etc.) a('colIIpanying each of the given traveling-wave eleclric held amplitudes. Duermine the incident and reflecting time-average Poynting vector power dcnsilies (&'a~,l' etc.) in each regiou. (b) Add a dosed rectangular box, as for Problem 7-15, to this system, snch that the surfaces SI and S2 extend into I ancl2, respectively. Sketch it on yoU!' diagram. With the x and_y dimensions as a 2 m, fiud the time-average power fluxes, Pt, p~, pi, and Pi. passing through and 8 2 in the two regions. Calling the power flowing into the dosed box positive, determine net power flow into the box, whence conclude whether the time-average Poynting theorem ) or is satisfied. (c) Add a second dosed rectangnlar box to the this time with it frontal surbec S I still in region I, and with the opposing snrf;lCc extended into region 3. Sketch it OIl the diagram, Determine the net time-average power flow into the box. Is the Poynting theorem (7-56) satisfied?

7-16.

lc' 7-17.

(a) Do Problem 7-13(a), except assume this time a lossy region [i.e., llse the general Ibrms (6-35) and (6-37) instead of their lossless region versions]. Employ these in (7-47a) to show that the total (or net) time-average vector power density can be written &>av

az

(£:+)2 2az 2~ e[(1

··

(7-64)

in which f = Ir + jIi and fi = 11, + jl1; = firJo have been assumed. To what result does (7-64) reduce in a purdy rct1cctionlcss region? (b) Show that the general result (7-64) reduces to its lossless region version of Problem 7-13, if appropriate assnmptions concerning y and fi made. (e) In Problem 7-13, the result (7-59) shows, in a lossless regioll, that the total power can be dissociated into the contributions &':V + &'a--' provided by the incident and re!lee ted waves when considered individually, From the form of fi)!' this lossy region, argne to why no such equivalent statement can be made here.

PROBLEMS

407

2: (1'0, 6'0,

'---

0.03)

1: air Perfect conductor

j,;+ xl

/Waveguide

Direct ray ___ {

S --?-

iI.ii· ray

-~ ------

(z)

01 02 .:--

..... -

'0 4

(a)

(b)

PROBLEM 7-19

7-18. In Example 6-3, the f(mr-region problem involves a showing a hypothetical closed SurLUT S of rectangular box reglOn and the opposite such that its frontal surhec .)1 projects just within losslcss region 2 surlZtcc 8 2 is located at = 0 just inside region 4. 8 1 and S2 have the area A 4 m each. (a) With the known electric-field magnitudes E;:;2 = 85.7, E;;'2 37.0, and 47AVjm, find the time-average vector power densities on S 1 aryl 8 z . Label these vectors on the sketch. Find also lhe time-average power fluxes into (or out: of) S. How mueh power loss occurs within the region 3 bounded by the closed surl~lec S? 7-19. A microwave oven consists of a metal oven enclosure !Cd li'om a magnetron source, usually operating in the S-band (about 2.45 GHz), a hollow metallic by figure Power coupling from to the oven occurs (see Chapter B) as at the coupling aperture A, with microwaves illuminating the by direct rays /i'om the aperture, or from indirect rays produced as oblique reflections from the oven walls as shown. The sarnple is spaced above the oven wall to enhance the heating cfl(:cts. A simplilied model of the microwave heat ing process roughly produced by the direct is shown in figure (b). Let the operating frequency he 2.5 GHz. (a) Find the wavelength in air regions I and 3, the gap width and iX, /l, q and 'Az regi~.m 2. (b) Find the reflection coemcient at z d3 and Z 0 in region :l, as well as and r 2 (ri 2 l. (c) AssUfItc three cases of lossy slab thickness: d and Calculate lin' each the impedance ~.I' [Answer: 1 a z 13.25 kW/m2] Find the wave amplitudes R~2' E,;.2 in region 2. Sketch a labeled graph depicting only the magnitude versus of the incident and reflected electric lidds in the lossy 0 and ·c = slab region, labeling the values at

E:

7-20.

For the same microwave heating model of Problem 7-19, let d2 4.5Az and consider only the average power injected into a sample cube region of the lossy slab, with cross-sectional (x andy) dimensions a b = Sketch a cubical closed surLlec with these dimensions, jnst embracing that amount oflossy sample and having input and outpnt SUr!;lCCS S1 and la) Find the net time-average power Ilowing through 8 1 into the volume Why does zero power flow occur through Sz? Use (7-31) or (7-56) to illf(~r the timc-averageJoulc heat generated within the cubc. Answer: Pay 644 W] How do you know that heat is being generated nonunilormly in this lossy sample? Where the maximum density of heat generation? (e) The heat

r

408

THE POYNTING THEOREM AND ELECTROMAGNETIC POWER

generated within the lossy sample in t sec is given by

( (1)

if Fav is the net time-average power flux injected into the sample. Assuming that the heat is generated nearly uniformly in the cube, the heat required to raise the temperature of the sample mass m by /',. TOC becomes Ch /',. T, if Ch is the sample "specitic heat capacity." Assuming this particular sample to have the constant specific heat capacity Ch = 0.50 calorie/gOe (= 2.09 J/gOC) and a specific gravity dm = 1.3 (giving this sample thc mass m = drn V = 13.94 kg), calculate the amount of heat U h and the time t needed to raise the temperature of this particular sample ii'om ambient (20°C) to 170°C, assuming the same net power input as in (a). [Answer: U h = 4.37 MJ t 113 min]

.Ir 7-21.

At the distance from the sun to the earth, the sun produces the timc-average electromagnetic power flux density of about 1340 W /m 2 . Its power is contributed by frequency components ranging ii'om radio frequencies through the ultraviolet region and beyond. (a) Supposing that this power density arrived at a single sinusoidal frequency, what electric and magnetic field amplitudes would be required to produce this power density? (b) Use a suit.able surEtee integration to calculate the total time-average power radiated {l'om the sun. The distance from the sun to the earth is about 148 Gm.

__---------------------------------------CHAPTER8

Mode Theory of Waveguides

(

In this chapter, the wave reflection problems of Chapter 6 arc extended to the theory of waveguides, regions of uniform cross section bounded by conducting walls parallel to the propagation direction. 1 Typical waveguide configurations are shown in Figure 8-1. To simplify the analysis, perfectly conducting walls are assumed, except in Section 8-6 in which the attenuative eHects of wall losses are analyzed. The boundary effects of the conducting walls, producing only normal electric and tangential magnetic fields there, favors a z direction of energy How, so the waves are said to be guided in the z direction. In this sense, the wave transmission systems are said to be waveguides, though this term is usually restricted to the hollow, rectangular and circular cylindlical systems of Figures 8-1 (c) and (d). Two-conductor wave-guiding systems exemplified by the parallel-wire and coaxial lines of Figures 8-1 (a) and (b) arc commonly called transmission lines; in the strict sense they are also waveguides. The mode theory of uniform waveguides is considered in this chapter, with particular emphasis on the rectangular hollow wavegu'irles shown in Figure 8-1 (c). A boundary-value-problem approach is used, that is, solutions of Maxwell's equations, subject to boundary conditions, are obtained. The complex, time-harmonic forms of Maxwell's equations are used, time dependence of the fields being assumed according to the usual factor !1m!, but because of the invariance of the guide cross section with respect to the propagation direction z, an additional exponential Z dependence factor e+ Yz is assumed, with y identified as a z-direction propagation constant. With t and thus absorbed in the factor YZ, the wave equation in terms ofE or H reduces to iOptionally, you may elect to ddtT the study of Chapter 8 and direclly to Chapters 9 and 10 on transmission lines. Possible advantages of taking up Chapter 8 are that the study of velocity 8-5) and conductor attenuatiun losses (Section B-7) arc simpler ji)t rectangular than transmission lint'S.

409

410

MODE THEORY OF WAVEGUIDES

(a)

(b)

(c)

(d)

FIGURE 8-1. Uniform transmission line or waveguide structures of common occurrence. (a) Parallel-wire transmission line. (b) Circular cylindrical coaxial pair transmission line. (c) Rectangular hollow waveguide. (d) Circular hollow waveguide.

a dependence on only the transverse variables x,y in the case of the rectangular waveguide (or in terms of p, lc,mn; thus, I Np/m

Ymn

Ymn

jfimn

=jw~~

f-e 7Y c

rad/m

1< lc,mn

(8-54a)

I > J~,mn

(8-54b)

From (8-54b) one can inkr, lor a specified TErn» mode, a wavelength Amn and phase veloci ly vp,mn given by expressions identical with (8-45) and (8-4b) Ii)!' the comparable TMmn mode

I> lc,rnn

(8-55)

I > J~,m"

(8-56)

in which A(O) and 1J~0) an, the wavelength and phase velocity associated with plane waves propagating at the frequency I in an unbounded region filled with the same dielectric with the parameters J1 and E, A comparison of (8-21) with (8-23) shows that the intrinsic wave impedances of TE and TM modes are not the same; from (8-23) and (8-54b) one obtains for TErnn modes above cutoff

JWI'

1](0)

1]TE,mn = -:p-- = -;=======(;==;==)~2 n ) mn

1

_!c:

I > J~,mll

(8-57)

t

which deserves comparison with expression (8-47) for ~TM,m", If a TEmn mode is generated at a frequency below the cutoff value specified by (8-53), the propagation constant Ymn becomes the pure real amn of (8-54a), producing an evanescence of the field components (8-51) resembling that for TMmn modes below cutoff as shown in Figure 8-5(b), Although wavelength and phase velocity are undefined in the absence of wave motion for I < j~,m'" the intrinsic wave impedance lor a TErnn mode below cutotfis obtained from (8-54a) and (8-23), yielding

f < ,Ic,mn

(8-58)

432

MODE TREOR Y OF WAVEGUIDES

,50) 1 {:I«()

Allin

o Increasing

o

f

f Increasing

f

f= t~,mn FIGURE 8-8. Universal circle diagram (left) and qnantities plotted directly against freqnency (right), fix TM and TE modes.

From this result one may again sec, as from (8-48) for TMrnn modes, that whenever a mode evanesces (f < j~,mn) the wave impedance qTM or qTE becomes imaginary, showing that tor an evanescent mode, there is no time-average power How through a waveguide cross section. The common factor 1 - (!c,mnlf) 2 appearing in the various expressions (8-45) through (8-48) for TMmn modes, together with the comparable relations (8-54) through (8-57) for TErnn modes, permits graphing them as normalized quantities on the universal circle diagram shown in Figure 8-8. For example, the expressions (8-4·2b) and (8-54b) for the phase factor Pmn of TM or TE modes are normalized by dividing through by P(O) = wJjlE to obtain

J

Pmtl)2 + (!c.mtl)2 ( pt O) f the equation of a circle, considering PmnIP(O) and J~,mtllf as the variables. A discussion of the group velocity Vg noted in the diagram is reserved for Section 8-5. To the right in the figure is shown a graph of the same quantities plotted directly against frequency, which may have some interpretive advantages. Thus, the phase constant Pmn of a desired mode is seen to be zero at the cutoff frequency !c.mn while asymptotically approaching the unbounded space value P(O) = wJjlE represented by the diagonal straight line as f becomes sufficiently large. The expressions (8-51) feJr the five f-Ield components of the TE_ modes lead to flux plots of typical modes as seen in Figure 8-9. The electric field lines are entirely transverse in any cross section of the guide, as required for TE modes; they terminate normally at the perfectly conducting walls to satisfy the boundary conditions there. The magnetic lines, moreover, form closed loops and link electric flux (displacement currents) in the process, as required by Maxwell's equations. A comparison with Figure 8-6 points out the inherent diflerences between TM and TE mode field configurations in a rectangular guide. In Section B-3, the TM mode expressions (8-38) reveal that the lowest-order nontrivial mode of this group is the TM 11 mode. A similar inspection of the field expressions (8-51) shows that the lowest-order nontrivial TE modes are the TElO and TEO! modes, flux plots of which are depicted in Figure 8-9(a) and (b). Of these two,

8-4 TE MODE SOLUTIONS OF RECTANGULAR WAVEGUIDES

2

433

2

Section 1-1

um \\I\\l\! !![ Section 2-2

(b)

(el

(d)

FIGURE 8-9. A few low-order TEmn modes of the rectangular waveguide. (a) TE!o mode. (b) TEo! mode. (e) TEll mode. (d) TE2! mode.

the mode I)aving the lowest cutofffi'equency is determined by whichever of the two trans~erse guide dimensions, a or b, is the larger. With m = I and n = 0 inserted into (B-5~»), the TE lo mode is seen to have a cutoff frequency

!c,10

=

a

2a

(8-59a)

a result independent of the b dimension because n = O. Thus (B-59a) states that the cutofHrequency of the TE 10 mode is the freq uency at which the width a is just one-half

434

MODE THEORY OF WAVEGUIDES

a frcc-space wavelength. Similarly, the TEol mode has a cutoff frequency (8-59b) a value larger than!c.10 if a > b, the dimensional condition depicted in Figure 8-1O(al. From the identical cutoff frequency expressions (8-53) and (8-40b), all higher-order TE and TM modes exhibit cutoff frequencies higher than (8-59a), assuming the standard convention of a> b, to make the TElO mode the dominant mode of that rectangular waveguide. For example, the so-called X-band rectangular waveguide, assumed airfilled and of interior dimensions a = 0.9 in. and b = 0.4 in. (0.02286 x 0.01016 m) has a cutoff frequency obtained from (8-59a), yielding 3 X 10 8 !c.lO = 2(0.02286) = 6.557 GHz

(8-60)

X-band guide

while the cutoff frequency of the next higher-order mode, TE 2o , becomes J~.20 = 13.12 GHz, from (8-53). The TEol mode, from (8-59b), yields !c.01 = 14.77 GHz, while using (8-53) or (8-40b) obtains cutoff frequencies for the TEll and TM 11 modes that are even higher (!c,ll = 16.10 GHz). Their positions on a frcquency scale are portrayed in Figure 8-10(a), showing why the propagation of electromagnetic power via the single dominant TElO mode in a rectangular waveguide is possible by kecping the generated frcquency f above the cutoff frequency of the TElO mode, but below the cutoff frequencies of all other modes. This choice assures a traveling wavc TElO mode and the evanescence of all other modes, thereby justifying the designation dominant for the propagating TE 10 mode. For example, the band 8.2 to 12.4 GHz is chosen as the X band; frequencies that propagate only in the dominant TElO mode in a 0.4 in. x 0.9 in. rcctangular waveguide.

ie, 10 = 01

6.557 GHz

For a = 0.9 in" b

=0,4 in,:

(GHz)

~

~ For

TE10

1- = 2.25:

[:j

.1

°

Q

r :}

TE21 TM21 I

tl

:)

3

i(>,rnn

fe, 10

(a)

I

1:

'f 2

TEJO TEoI

For~=

TEll TEoII TM[1 TE 20 I I I

I

TEll TMJl I

t

TE 20 TE02 I I

TE21 TEI2 TM21 TMI2 I I

r

~I__________-L 'f ____~_____ L_ _ _ __ L_ _ _ _~~~

°

1

't'

2

3

"

JO

(b)

}'IGURE 8-10, Cutoff frequencies oflower-order modes in rectangular and square waveguides, (a) }'or alb = 2.25, Cutoff frequencies shown relative to !c,IO on lower graph" (b) For alb = L

8-4 TE MODE SOLUTIONS OF RECTANGULAR WAVEGUIDES

435

It is evident that a square waveguide (a = b) will not possess only one dominant mode, for the TE 10 and TE01 modes then have identical cutofftrequencies from (8-59). Figure 8-1O(b) shows the positions of the cutoff frequencies oflower-order modes for a square waveguide on a relative frequency scale. A comparison ofF'igure 8-1O(a) with (b) reveals that the use of a rectangular waveguide, with a> b as in (a) of that figure, provides a desirable control over the E-field polarization of the propagated mode. Figure 8-11 illustrates the manner in which a microwave power source (a klystron, magnetron, etc.) is connected to a waveguide by using a small antenna wire protruding into the waveguide, such that the wire alignment agrees with the polarization of the dominant mode being launched. The power can similarly be extracted at the other end, if desired, by means of the center conductor of a coaxial line used as a receiving antenna (a waveguide-to-coax transition). The propagation of the energy down the waveguide via the dominant TElO mode thus assures the known field polarization necessary to the efficient launching and retrieval of the energy. Since signal power in a rectangular waveguide is commonly dispatched by use of the dominant TE 10 mode, its properties are for convenience collected separately· in the following. The expressions (8-51) for TEmn modes, with m = I, n = 0 inserted, reduce to three components ~±

;j{'z

(x)

.

~±

n

= Hz ,10 cos -a

~+

(8-6Ia)

X

[-JWlla ~+

]

.

n

"'+.

n

Si (x) = --n-- Hz-;l0 sm a x = Ei,10 sm -;; x

.m±( .:n x X ) --

-+ it -A-- -

'lTE.I0

±j2a ~ + 11.10

(8-61 b)

[±jf310a H~±z,10 ] Sln . ~x n a ]

.

n a

~+

.

n a

= [ - - 1 - Hz.10 sm - x = Hx.10 SlIl- X

(8-61c)

assuming J > !c.lO' The foregoing may for~some purposes be more conveniently expressed in terms of the complex amplitudes Ei,10 of the y-directed electric field (8-61 b), Microwave source (klystron, etc)

[W] TEiO mode

Coaxial-to -waveguide transition

FIGURE 8-11. Typical waveguide transmission system, showing launching of the dominant mode and a transition from waveguide to coaxial transmission line.

436

MODE THEORY OF WAVEGUIDES

yielding ~+

~+

n

"

a

gy- (x) = £;-;10 sin - x

(8-62a)

o . n A

1'/TE,1O ~+

SIn -

~ ± _. Ei:1O/i. yt'z(X)-J (0) 21'/ a

1(0)

a

.

~+

x

=

Hx.l0 sm

n a

(8-62b)

x

,n _ ~ ± n cos-x-Hz10cos X a . a

(8-62c)

The remaining properties of the TE10 mode are related to its cutoff frequency specified by (8-59a). From the latter, the ratio !c. I o/f is

.!c,10

viOl

J~.lO

(8-63)

-p-

2af

f

to permit writing the propagation constant, wavelength in the guide, and phase velocity for the TE IO mode as follows

YlO = 1J(1O == /3(0)

C(O)Y 2a

j/3(O)

Y1O=j/310

-

I Np/m

I - C(OIY rad/m

2a-

f < !c,10

(8-64a)

f > !c,10

(8-64b)

f > j~.10

(8-65)

f >

(8-66)

A(O) rn

)'10

vIOl

-(~:)Y

Pm/sec

j~,10

in which /3(0) = (JJ~, A(O) = 2n//3(0) = v~ol/I, and v~OI = (/lE) 1./2 as before. The intrinsic wave impedance obtained from (8-57) or (8-58) becomes 11(0)

fiTE, 1 0 = -;:::1=_='==(=A=(==01==)==2 Q

I > j~,lO

(8-67)

I

(8-68)

2a

fiTE,lO

-

A

Yf tn

(9-44)

bp a

The total magnetic field is given by (9-26), a superposition of (9-44) after multiplication by e- yz and eYz (9-45a) =

atjl

V+I

V-I]

_m _ _ e-Yz _ _ _ m_

bp Yftn~

[ in which

~

A

A

eYz

(9-45b)

bp

YJtn a

and yare given by (9-40) and (9-35).

(b) For a losslessdielectric, (9-35) yields y = jw.JP~ = jp and (9-40) yields ~ The real-time forms of (9-22) and (9-45), by use of (2-74), become

E(p,z, t)

a Re P

ap

[V~ .!. tJ(w,-pz) + V';; tn

bp a

tn

I

bp

'I

~.

tJ(W'+Pzl]

a

V+ I I .] cos(wt-fh+4>+)+ cos (wt+Pz+4>-) bp P In [ In a a _m_

(9-46) V+ I H(p,z,t)=a4> __ m_-cos(Wf bp [ 11 tn a

Pz+4>+)

_V_ m_

I cos (Wf bP 'I tna

+ pz + 4>-)] (9-47)

V;;

assuming complex amplitudes of the form V;; = tJ4> ±. A sketch of the positive z traveling fields is shown.

468

TEM WAVES ON TWO-CONDUCTOR TRANSMISSION LINES

--(z)

H; lines~~~~~~~=i~~~~~~:-::~~-~··~~:--:~~:-:-:·:-·;;~:~-:;; I

I

I

I

I

I

o

i3z =

I

I

Wave motion

I 7T

27T

EXAMPLE 9-2(b)

(c) Use is made of (9-25) to find the phasor current amplitudes given by (9-44)

~+

I;;; =

~

"(c.s)

~

+ • dt = ;H'-

I,!

from the

_ie±

fields

(a 4> - V;;;) ± i21t - - . a p d pfilna

+

v±

(9-48)

rn

fi b --In 2:n:

a

whence insertion into (9-27) yields the

+ z and

-

z traveling current waves of the

total line curTent

(9-49)

Normal E terminates on charges

Tangential H terminates on surface currents

~ ~~rface

~'-.(z)

Negative current: - I EXAMPLE 9-2(c)

(z)

9-2 CHARACTERISTIC IMPEDANCE

469

(d) In the accompanying sketch are shown the electric and magnetic fields (;?f only JlIe positive z traveling waves), along with their related voltage and current V+ and

r.

9·2 CHARACTERISTIC IMPEDANCE It is usually desirable to characterize TEM waves on a line in terms of their voltage and current waves rather than the electric and magnetic field quantities discussed in the foregoing sections. The advantages are evident from the fact that voltages and currents on a transmission line are readily measured scalar quantities at frequencies below I GHz or so, whereas the electric and magnetic fields must usually be inferred from such measurements. The comparison of the total line current (9-49) as evolved in Example 9-2, with the expression (9-27) for i(z), suggests writing the total line current i(z) in the equivalent form

(9-50)

In (9-50), the quantity be defined by

dz in the same cross section, as shown in Figure 9-3(b). The conductor at the assumed positive polarity is chosen for the constructioIl, where the positive 1 sense is taken to be z-directed. The right side of (3-82a) involves a surface charge increment I1q deposited at any t on the peripheral 8 3 shown, in view of the boundary condition (3-45). As seen from (9-5) and (9-3), the timevarying electric field of the TE M mode, in a fixed eross seetion of the line, satisfies the same Maxwell equations and boundary conditions as the static electric field between those conductors. The definition (4-47) of static capacitance can therefore be used to rdate I1q (0 the instantaneous voltage V between the conductors as follows I1q = (I1C) V

with I1C dcnoting the static capacitance ofthc I1z slice. Putting I1C

c

I1C I1z F/m

(9-60a) (cl1z), or

(9-60b)

474

TEM WAVES ON TWO-CONDUCTOR TRANSMISSION LINES

(9-60a) becomes

llq

= (c

llz) V

(9-60c)

in which e denotes the static capacitance per unit length, or distributed capacitance parameter of the uniform line. From (9-60b) it is seen that the distributed capacitance parameter c is also the static capacitance per unit length Cil as discussed in Chapter 4; so write (9-60b) as (9-60d) For example, e for a parallel-wire line is given by CIt obtained from (4-107). The left side of (3-B2a) denotes the net current flux emergent from S at any t. The contributions through S1 and S2 in Figure 9-3(b) yield the net amount

I

+J+

01

01 oz llz

llz

(9-61 )

An additional current increment llJ leaves the peripheral surface S3 and enters the region between the conductors, assuming the dielectric has a conductivity (J. From (4-119), llJ is proportional to V, obeying

111 = (llG) V

(9-62a)

in which llG denotes the eonductance of the llz slice. By putting llG = (g llz), implying

llG g= llzU/m

(9-62b)

III = (g llz) V

(9-62c)

(9-62a) becomes

in which g defined by (9-62b) denotes the conductance per unit length, or distributed eonductance parameter of the line. It is evident from (4-121), from which llG = ((J/E) llC, that g is not an independent quantity; it is related to the distributed capacitance parameter 3 c on making use of (3-IOB)

g

(J(

Elf) U 1m

(J

= -E e = -WE we = -E' we

(9-63)

Inserting (9-60c), (9-61), and (9-62c) into (3-B2a) yields

01 0 ozllz+ (gllz) V = -ot (ellz)V 3The last forms for g in (9-63) involve the frequency

ill

and so apply to tho; time-harmonic case only.

9-3 TRANSMISSION-LINE PARAMETERS, PERFECT CONDUCTORS ASSUMED

475

reducing to the differential equation

oJ OZ

0

= --(cV) -gV

ot

If the parameter c is not a function of time, the latter becomes

oJ

c

oz

oV -gV ot

(9-64)

or just (9-54). B. Line Constants ,)" Zo in Terms of Distributed Parameters Many dielectric materials used in transmission lines have parameters 11, E, and (J that may be functions of the sinusoidal frequency ill of the fields, as seen from Table 3-3. From this point of view, the time-harmonic forms of the transmission-line equations may be of greater interest than the real-time forms (9-53) and (9-54). Thus, if into the latter .

V(z, t) is replaced with v(z)ei wt l(z, t) is replaced with l(z)ei wt

(9-65)

one obtains the time-harmonic transmission-line equations

~

dV

- = -Jill I J

dz

e

(9-66a)

(9-66b) These are also written

dl

-=

dz

-if

(9-67a)

-yV~

(9-67b)

on taking i and y to mean

y = g + jwc t5/m

Series-distributed impedance

(9-68)

Shunt-distributed admittance

(9-69)

476

TEM WAVES ON TWO-CONDUCTOR TRANSMISSION LINES

With the substitution of (9-63), Y is written in terms of the dielectric loss tangent as follows

(9-70) For example, using a dielectric with a loss tangent of 0.00] yieldsy = (0.001 + j)wc, or very nearly jwc. The wave solutions for V(z) and i(z) of the transmission-line equat~ms (9-67) have been supplied by (9-20a) and (9-49). They yield expressions for y and :

t v

(9-117b) (9-117c)

Defining a real time domain rtiflection coefficient reflected to incident voltages

r

Vr(t, t) =

V+

at the load as the ratio of instantaneous

( +-;;t) t

(

t

(9-118)

t) v

permits writing (9-117a) and (9-117b) Vet, t)

= v+

(t - f) [I + r(t,

t)J

(9-119a)

[1 - r(t, t)]

(9-1 19b)

t)

V+ ( t-I(t, t) =

1/

Ro

From (9-117c), the ratio of the load voltage Vet, t) to the current I(t, t) is R, yielding from the ratio of (9-1 19a) to (9-119b) R-R 1 + ret) ' - () 1 - r(t)

V+(t -~)-'>oRg

Yg (t)

~

L

~V-(t+~)

(9-120)

I' 1

[(t, t)

r_(_R_O'_V_) _ _ _ _ _ _ _ _ _ V..:....(t..:....'....... t) c

~--- --~ z

z=O FIGURE 9-12, Resistive terminated line.

! z=t

~

R

9-7 WAVES OF ARBITRARY SHAPE ON LOSSLESS LINES

497

Solving for 1(t) obtains

(9-121)

1(t)

in which the notation 1(t, I) is altered to read just 1(t). Equation (9-121) is useful for finding V-(t + tlv) at the resistive load whenever the incident wave V+(t tlv) is known. It is emphasized that the pure real1(t) in (9-121) is a consequence ofthc output exprf:ssion (9-116a) being purely algebraic. 1\ time domain reflection cpefIicientjs undefined'll)r reactive loads corresponding to (9~116b, c,.aI~d d), except in the asymptotic limits for which the derivative terms in the load differential equations become negligible. EXAMPLE 9.6. If the line of Example 9-5 is terminated in a short circuit (R = 0), what reflected wave is produced by the incident trapezoidal voltage wave? Sketch the results. From (9-121), qt) = -1. 'fhen from (9-118)

(I) holding for all t> tic after V+ (I zlc) first appears at the load. To obtain the desired V" (t + ,:/c) reflected to theleft, the reflected wave (I) must be delayed in time by (t - z)le (a time delay incurred by wave motion from the load back to any Z location toward the generator), yielding

t

Rg ~(O,t) V(t,t) = 0 A:>--'-,.L£~-----"'''"'''''~--r . _/ --A ~------------------~ I

U

v. + g-

I Z

;"t

1=0

o t= 1. (

L - -_ _ _ _ _

o

L -_ _ _ _ _ _ _

--

-."j..J~ . .•.•i-Ir-:~~T~ __ I

o I=l! 2 (

...

~~= ~----~ --

•

I,.....f:.:'-_ ..

:~t:E;:J-:"'-" "..-----~---+--V---(-I....J:~:~ _v~,,-~ l--'f~m - .. -.

EXAMPLE 9-6

(z)

..,.... __

,b Mfl 498

TEM WAVES ON TWO-CONDUCTOR TRANSMISSION LINES

As a check, observe that (2) becomes V+ (t 2t/c) at the input z = 0, the reflection arriving there after a delay of 2t/c. These results, shown in the sketch, reveal the echo V- (t + z/c) as a mirrored replica of the incident V+, inverted by the effect of the minus sign in (2) (maintaining zero volts across the shorted load). The echo apparently originates from an image location z = 2t, due to the time-delay 2t/c needed for the reflection to reach to the input. The corresponding reflected current wave 1- (t + z/t) is obtained by substituting (2) into the second term of (9-1 lOb), yielding

V-

(t + ~)

=

0.02V+

Ro

(t + ~ _ c

2t) A c

(3)

In the foregoing example, a second echo (re-reflection) occurs when V- (t + z/v) reaches the input at z = O. It can be found by use of the arguments of that example applied to the generator circuit. With a generator internal resistance Rg R no Ol re-reflection occurs. In general, an internal resistance Rg provides the time-domain reflection coefficient at the line input (z = 0)

vt (t ~v) _ Rg -

Ro

0) - Rg + Ro

--~-.....;:-

V1- ( t

vi

(9-122)

+v

with denoting the forward wave re-reflected from the generator to the load. These processes repeat when the wave in its turn arrives at the load, causing a third echo V;:(t + t/1-», and so on. The total wave solution is the superposition of all waves obtained in this way.

vt

EXAMPLE 9-7. A lossless 50-Q eoaxial line 200 m long using dielectric with E, = 2.25 is terminated in 100 Q and fed from a l50-V dc source having Rg = 25 Q as shown. With the source switched on at t 0, Vg(t) 150u(t) V, a step function. Find the voltage and current waves on the line after t = O. The equivalent input circuit of (b) yields at A-A V(O, t)

J (0,

'I

V+

~

(t -~)

l50u(t) _5_:-- = 100u(t) V

50

r(, -~) ~ v.~: ~) ~

The incident waves are (1) and (2) delayed

''itl

A

(I)

(2)

z/v sec

~)V 1-'

t

(3)

t :.ered to t~e line? (d) Compute the load current and time-average power absorbed by ZL' (e) I1~ZI, 50 n, what is the input impedailce and how much average power is delivered to ZL?

ZO

3 (a) From Example B-4, = 50 n, rJ. = 1.97 X 10- Np/m, p = 0.595 rad/m. With A = 10.55 m, { in wavelengths becomes {IA = 6.33/10.55 = 0.6.

(b)

Zin is obtained by first finding f

Thcn

f

at the load using (10-5)

at A-A, by usc of (10-6), becomes 2(0.00197)6.33 e - j4n(O.6)

=O.2765ejI1l9'O.975e-J432' ~ 0.212+jO.177

n il?

II'

Vg = 100e V

50 Q

j (0)

cio."'()

A::= ~

i~(t)

-

I'-z-=-O----------z-=-(

~- . _ - - t = 6.33 m

---I

(a)

=

.lin i

(0)

=

70.5 + j28fl

(b)

'"

Vg =100e

10" .

V

(c)

EXAMPLE 10-1

(2)

516

PHASOR ANALYSIS OF REFLECTIVE TRANSMISSION LINES

TABLE 10-1. Transmission-line Analog of Plane Wave Propagation in

Multilayered Regions

Multilayer regions with plane waves

A:

Region 1

Region 2

:J - - - - --- --

+

A

Hy

• -.....

--..

---

---

-

Region 3

Region 4 (terminal region)

--;.-.

--;0..

------- ----- ..... ---

----

(ii 4 , 1'4) To source

I

I I

~--­

Interface A

I

I I

B

C

(z)

Total fields:

+

Ex(z) = E,;ie- YZ [l Hy(z)

1"(z)]

E,;i e-yz[l -

I\z)]

[6-29] [6-31]

with [6-30] Total transverse field impedance: Z(z)

== ~x(z)

=

Hy(z)

.ry 1 + 1"(z) 1 - 1"(z)

[6-32]

making 1"(z)

= ~(z)

- ~ + 1]

Z(z)

[6-33]

At another location z': 1"(z') = 1"(i)e 2Y (z, -

z)

[6-34]

Continuity of Z(z): Z(z-)

Z(z+)

[6-35]

Smith-chart use, normalizing (6-32): t(z)

== Z~z) TJ

=

+ i'(z) 1 - 1"(z)

1

[6-361

10-1 VOLTAGE AND CURRENT CIRCULATION ON LINES WITH REFLECTION

TABLE 10-1. continued

Cascaded transmission lines

Line 1

"--v-l~ Source or: generator

(201 ,1'1)

i I I

Line 2

Line 3

Line 4 or , a lumped : load

--;0..

~-~

I

(.2 02 ,1'2): (.2 03 ,1'3)

:

I I

' I

I

I

I

Junction A B C

Total voltage and current:

+

V';e- yz [1

V(z)

V';

~

I(z) = - , Zo

e-YZ[l

fez)]

,

r(z)]

-

[10-1] [10-2]

with

-

V,;

[10-3]

I'(z) == -~- e 2YZ

v.;i

Line impedance: Z(z) ==

~(z)

=

Z 1

l(z)

0

+

fez)

1 - f'(z)

[10-4]

making fez) =

~(z)

+

Z(z)

~o

[10-5]

Zo

At another location z':

fez')

=

f(z)e 2Y (Z'-Z)

[10-6]

Continuity of Z(z): Z(z-) = Z(z+)

[10-7]

Smith-chart use, normalizing (10-4): £(z) ==

Z~z) Zo

=

1 + fez) 1 - fez)

[10-8]

517

518

PHASOR ANALYSIS OF REFLECTIVE TRANSMISSION LINES

in which the I~lctor e = 0.975 is approximated as unity in what follows (line losses are ignored). into (10-4) then yields

75.8ei21.0' = 70.8

+ )27.1 n

~n is obtained from tbe equivalent input circuit of ------- =

(3)

(b)

O.813e- j12 . 6 ' A

(4)

Thus, the average input power becomes 1~*)-.1R,(7' 1~1A*)_1[> P P avjn --LR'(V~ 2 ( in in - 2 (. ". . . . in in in - 2 '"in in

=

1(70.H) (O.B!

=

23.4 W

(5 )

(d) I}y (I 7in 7(0) I,;;f°[I - f(O)L in which all quantities a~c known except 1:'. Solving f(H' it yields the l(lIward-travding currcnt-wave amplitude: O.813e - jI2,6'

7+ In

Then

7J"

1-0.212

(6)

)n.l77

written in terms ofJ,~ using (10-2), becomes

7L = itt) =

7:'e- jPt LI

- f(t))

=

I.Ok j216 l1

= 1.14e-j229.1" A

+ 0.103 -JO,257] (7)

The load average power is thell (8) agreeing with (e) With

cuit,

because of negligible line losses.

Z:L = ]:0 ~ 50!}, wave

a =5

0)

(

804 003

" 1 1 0 \ "" "01 =

h{~~=----7-n---n

~ot

Black screen1

Polar of farzone pattern: (

1

+ cos 2

0) I

Sin (5,.. Sin

5rrsinO

(1) 0)

~

0.".)

I

Cd)



a,p sin

4>]

(11-32)

components i£e and R,p in the farzone. the free-space version of (ll-S 7c). If only the u·',lThmeu. the results an~ irn.n.Wo'"''

(II thus related by the intrinsic wave impedance 110

574

RADIATION FROM ANTENNAS IN FREE SPACE

(d) Graphs of the farzone field pattern of the rectangular aperture of figurc (bJ arc usually desired in its "principal planes," the two symmetry planes that include the z axis and that slice normally through the aperture. Thus, the complete verticalfJrincipal filane of the aperture, as shown in figure (c), is defined by the = 0 and the 1) n semi-infinite planes, with the polar angle 0 having the nonnegative range 0 :s:; 0 ~ 0 90 in each. The taxzone E-fielel in that vertical plane is given by thc single expression

E(r, 0, , thereby concentrating the radiated power density in a desired direction even further in some point-to-point communication link. A property of an antenna indicating how effectively its radiation pattern concentrates its power density in a given angular direction (0',4>') is known as its directive gain, denoted by D(O', 4>'). The directive gain of a given antenna is defined as the ratio of the power density fYJaAr, 0' , ,) in all diredions (its power-density pattern is a perfect sphere), with Of, 4>') denoting the power density of the given antenna in the particular direction (0',4>') for which its directive gain is being defined. To illustrate this concept, Figure 11-13(a) shows how the directive gain of the half:wave linear dipole is obtained by use or (II-87b). Its axially symmetric power density pattern is independent of 4>, shown as the pattern H (in sectional view) in the figure, so suppose that one desires the directive gain D(O') in the particular direction 0= ()' shown. The spherical power density pattern of the comparison isotropic antenna, labeled f, is drawn so that its power density at 0., with the power density ;J}>av(r,O') of the given dipole. Then, as given by (II-87b), the ratio of the radiated power 4nr2;J}>av(r, 0') of the isotropic source to that radiated by the given dipole, Pay = ~s 'o/>av • ds, yields the desired directive gain D(()') in the direction 0'. The radius r denotes the radius of the integration sphere S', with r arbitrarily located in the f;irzone region of the given antenna. ]\Tore commonly, the directive gain of the given antenna is desired in the direction of its maximum power density, since this is the direction in which the maximum possible power can be captured by a distant receiving antenna. I n this case, the coincidence point of the power density patterns of the given antenna and the comparison

11-6 ANTENNA DIRECTIVE GAIN

577

Sphere S /'

I (z)

/'

/'

/

/

"\ \

I

/

/

/

\

/

!

I

I

I

\

\

\

\

\ \

(b)

(a)

antenna, (a) the direction 110 of its dipole; I is the power

It)r the example of a half-wave linear direction II and (b) the Ii is the power density field isotropic antenna, both at the distance r.

isotropic antenna yielding for lht'

0) of that given antenna maximum, the given antenna

FIGURE 11-13.

(II-87c)

in which

maximum of its power density pattern at a fixed Figure II-B(h) /()r the special case of the in the following example,

gain of the half,wave dipole antenna. maximum of the POW!'f density pattern of the half: souree is shown at (Lo, where 00 = 90°, as in density rJ'a,(r, ()o) in (11-El7c), at the pattern (II-52), with flol = 90° (rjJ being absent becanse fixed range r, (II-52) yields the maximum power

[COS (90" 90u~J2= 151~

(I)

Sill

use of (II-55) m the denominator of (11-87c) 4nl2 (151~/nr2) 36.51~ 9 Maximum

1.64

(2)

578

RADIATION FROM ANTENNAS IN FREE SPACE

the desired answer, showing that the comparison isotropic source of:Figure 11-13(b) would need to be driven with 1.64 times as much power as the given half-wave dipole to produce the same power density &'av at 90 0 • [Note that (2) can also be evaluated usiug the radiation resistance Read defined in (II if Pay !RradI; = t(73)1; = 36.51; is employed in (11-87e).] Expressed in decibels, (2) can be written

eo

D(900)dB

= =

10 log [D(900)]

lO log 1.64

= 2.15 dB

(3)

A second way of expressing the directive gain D(O', av(r, 0, 1 and bei w -'> w 2 /4 as w -'> O. Similarly, derivatives of the power series obtain ber w -'> w 3 /16 and bei w -'> w/2 as OJ -'> O. These into (B-21a) yield the dc resistance (B-22a) a result seen to agree with the static result (4-138)

The zero frequency inductance obtaincd from (B-2 I b) is

(B-22b) 4 agreeing with the static result Lit of (5-82).

q 609

B-1. CURRENT PENETRATION IN ROUND WIRE (SKIN EFFECT)

10 8 6

---

,-

---- c--

,/

,--- - -

--

--

4

ri

r:--;;-" ~r' t. /'

- --

2

I

V

0.8 - - - -0.6

V/ /

I

dc,----_

"

1-

1--- ..._- r---

0, 2

I I! I

--

--

JL~

Ii,

I

From (B 27b)

t--~~

~

From (B- 27a) (high-frequency approximation)

/~

i

1

1I

C

'"

1';

--".-

"

r----- 1--O. 1 0.2

2

0.4 0,6

=

4

6 810

~ 20

FIGURE B-2, Internal resistance and inductance parameters for an isolated ronnd wire.

A graph of ri and Ii for the isolated conductor, expressed as ratios to the de values (B-22), is shown as solid curves in Figure B-2. Also shown dashed are highfrequency approximations to the internal parameters, approaching the exact curves for alb sufficiently large as discussed in the following. Asympt9,tic approximatiolls can be found for Zj from (B-2), but it is convenient to reexpress :if", in terms of Jo(kp), instead of using the ber and bei functions. From (B-3) and (B-12) (B-23a) wherein

k -J -jWJ.l(J

112.j2IO and 0 is given by (B-9). Also

OA -;--- Jolkp) = up

0

A

o(kp)

~A-Jo(kp) -~:1 =

o(kp)

up

A koJ (kp) A,

in which the prime denotes the derivative with respect to the argument 1/2 .j2pll5. Then (B-23a) yields, at p = a

r

kp =

(B-23b) Using (B-23b) in (B-2), the internal distributed impedance becomes (B-24)

610

TRANSMISSION LINE PARAMETERS

in which the Bessel function of order unity Jl(V).5 The asymptotic (orm ten' It(ka) IS

obtained from the identity

IS

J~(u)

=

(B-25)

and the latter with (B-16b) into (B-24) yields

1/2 jill;,: -

~

~(allJ-1C18)

Zi --> ----.--

J2na

20"c

=-

1

2na

(1

.

+ J)

JWfJc - - !l/m 20"c

(B-26)

valid for sufficiently large a/b. The real and imaginary parts of (B-26) yield

ri

-->

1 2na

JWfJc

1 20", =2~aO"tb

a

a =

2b

ri(dc)

!lIm

J large

(B-27a)

a -large

(B-27b)

b

C

For instance, if alb = 5, the asymptotic expression (B-27a) can be used in lieu of (B-21a) with an error orabout 10'1,,, decreasing to zero error {()I' alb sufficiently larger. The asymptotic result (B-26) is seen to contain the quantity ~ of (3-112c), that is, the intrinsic wave impedance for a plane wave in a conductive region. Thus (B-26) yields tbe ratio lor a round wire a (j

large

(B-28)

One concludes tbat for current penetration small, the impedance ratio izj:it~ at the surface of a round wire becomesq, the same as the ratio of electric to magnetic plane wave fields in a conductor. B-2. DISTRIBUTED PARAMETERS OF A PARALLEL-WIRE LINE, CONDUCTOR IMPEDANCES INCLUDED The isolated wire internal impedance results of the previous discussion can be applied directly to the parallel-wire line of Figure 9-7(b), with the object of tin ding (9-98b), the distributed pararneter z. Proximity em~cts6 arc neglected, which assumes fields in each wire undisturbed from the axially symmetric configuration attained when isolated, a 5See S. Ramo, J. Whinnery, and T. van Duzer. Field, and 111alles in Communicatiolls Electronics, 2nd eeL New York: Wiley, 1934, p. 370. 6An analysis of the cf1(,cts of the proximity of the wires on the increase in internal rcsistdllcC is found in A. H. M. Arnold, "The alternating-current resistance of paralid conductors of circular cross-section," Jour. lEE., 77, 1935, p. 49.

611

B-2. DISTRIBUTED PARAMETERS OF A PARALLEL-WIRE LINE

reasonable assumption if the axial separation is greater than about 10 conductor diameters. With effects of proximity neglected, the internal parameters of the parallel-wire line of Figure 9-7(b) are double the results (B-21) obtained for the isolated conductor, in view of the impedance encountered twice along the edges Az of the rectangle t. The series parameter (9-98b) therefore becomes

z = 2ri + jw(2li + Ie)

(B-29)

= r + jwlOjm

'i

in which ri and are given by (B-2l) or (B-27). In (B-29) Ie is related to the magnetic field exterior to the wires, permitting the use of that obtained in Problem 9-12 (a) for the perfect conductor case

Ie = J.l t n h + d Hjm

(B-30)

a

1t

wherein a is the wire radius, 2h the separation, and d = --.lh 2 - a2 . The expression (9-100) for the shunt parameterj = g + jwe is the same whether or not the conductors are perfectly conducting; thus, from Problem 9-13

y

g+jwe

- + (Ell)

j we =

(Ell) + j

-

E'

E'

W1rE Ulm tnh+d' a

(B-3l)

For an air dielectric, the assumption g = 0 is appropriate. In telephone lines using poles for support, the insulator leakage is often reduced to an equivalent distributed loss effect along the line, yielding a parameter g determined by the number of poles used per mile or kilometer.

EXAMPLE B-3. A telephone line consists of 0.104 in. (0.264 cm) diameter hard-drawn copper wires (o-c 5.63 x 10 7 Vim) separated 12 in. (305 em) in air. Neglect leakage due to the supporting insulators. (a) Compute the distributed constants r, I, g, and c at 1 kHz. (b) Find Zo, iX, (3, .Ie, and 1/p at 1 kHz. (a) With 2a

= 0.104 in. and 2h 1CE

12 in., It/a

= 115.5, so (B-31) yields

10- 9 /36

c=-tn-[~-l+-J-::-(=~=y=···=I-] ~ t. [2311 ~ 5.10 pF(m ~ 0.00822 ,F(m; in whieh the conversion 1609 mimi is used. If leakage is neglected g = O.

612

TRANSMISSION LINE PARAMETERS

With z given by (B-29), evaluating r, and Ii requires first expressing the wire radius as a funetion of 0 given by (B-9) 0.00213 m 0.999. Thus, with

making alb 0.622. From Figure B-2, rjr i •de 1.003, Irl1i,de r j = L003r j • de , (B-22a) in (B-29) yields

2r j

r

or 11.47 Q/mi. Similarly, using (.de of (B-22b) in (B-29) obtains 2/i 8n = 0.999 ,uH/m = 0,161 mH/mi., and tl'om (B-30)

Ie

It + d 4n x 10 - 7 t" (231) = 2.18 ,uH/m nan

,uo

= ~tn~- =

=

2(0.999),u0l

3.51 mH/mi

so the total distributcd ind uetanee in (B-29) becomes

1= 21i

+ Ie =

2,28 flH/m

= 3.67 mH/mi

(b) Onc obtains /~o using (9-105) '7 _

,,"-0-

By usc of

~r + jwl _ "--.-g +Jwc

03a) y=

yielding

(X

+ jwl)(g + jwc) = 0.0083 + jO.035 rni I

= 0.0083 Np/mi, f3 = 0.035 A = 2n

f3 up =

w

2n 0.035

7i = fA =

=

rad/mi. From (9-37)

179 mi = 288 krn

179,000 rni/scc = 2.88 x 10 8 m/scc

from thc valuc of (x, a wavc 011 thc line attenuates to e- I = 36.8'/,) ol'its input valuc in d (X I = (0.0083) - I 120 mi at f 1 kHz.

B-3. DISTRIBUTED PARAMETERS OF A COAXIAL LINE, CONDUCTOR IMPEDANCES INCLUDED Figure 9-6 shows the current distributions obtained in the outer conductor ofa coaxial line at low, medium, and high fr·equencies. At the higher frequencies the currents concentrate toward the TEM fields responsible for those currents, namely, toward the outer wall of the inner conductor and toward the inner wall of the outer conductor. This latter property of the coaxial line, like hollow waveguides, makes it an excellent shielding device at high frequencies, the essentially zero currents on the outer wall eliminating the possibility of a small tangential electric field being coupled outside the outer conductor.

B-3. DISTRIBUTED PARAMETERS OF A COAXIAL LINE

613

Poynting vector

I,

;, (a-) Conductor 1 ;; flux

(a)

(b)

FIGURE B-3, Relative to the distributed internal impedance or a coaxial line. (a) Contin':!.ity of tangential magnetic fields into conductors. (b) Longitudinal electric field induced by .1f~ (~ +!) j~~J1.c + JOJ 2n a b 2a c

•

[_1,(~ +~) w2n

a

f0i. +~ tn ~J 2n a

b >J2~

= T + jwl n/m I ligh-freq uency approximations

(B-38)

The series distributed parameters of a coaxial line, from (B-38), have the following properties in the high-frequency approximation L The resistive part!" increases as the square root of the frequency, and it decreases inversely with c• 2. The inductive part has two contributions. The first is the inductive part of the internal impedance, behaving like T. The second is the external inductance (9-34), providing the major contribution to I in practical coaxial lines.

J'a

Finally, the shunt parameter (9-100), Y = g + jOJe, remains unaltered Irom (9-82), applicable to the line with perfect conductors. This conclusion follows as usual from the dependence of g and e on only the electric field in the dielectric region. Thus, A

y =g

(Elf.) We (Elf.) E' +.1 W2nE - - (JIm b

+ .1we = -? +.7 •

=

In

a

valid at all frequencies, and not just a high-frequency approximation.

(B-39)

B-3. DISTRIllUTED PARAMETERS (W A COAXIAL LINE

615

The general derivation of the series parameters of the coaxial line at an arbitrary frequency ill is omitted here, but at low frequencies they are readily obtained in the absence of the skin effect. Then the dc inductance results obtained in Example 5-13 are applicable, while the resistance parameters are obtained from an adaptation of 4-133), assuming uniform current densities over the conductor cross sections. It is left for you to carry out the details. EXAMPLE B·4. Assume the cable in Example 9-3 has the 1ielectric loss tangent of 0.0002, hut this time copper conductors are used. Recaleulate ZO, IX, and fj from the distributed

pal'ameters y and z, atf = 20 MHz. _._ Note first the plane wave b obtained from (3-114) is b -J2/CO~IPc = 1.48 x 10 - 5 m, = 0.0148 mm, sufficiently smaller than a = 0.05 in. 1.27 mm such that the high-frequency approximatioll assumed !()r (B-38) applies. The series r parameter then becomes r

I

(~+

2n

a

1,))(;12 b 2u c

=~, (}03. + 211

1.27

3

10 _) 4.14

r.;;(2~}~)411 ~li:-..~ 7

-V'"

2 x 5.8

X

10

(I)

= 0.191 Q/m

The inductance parameter has internal and external contributions

bJ.[

. .[r+w--tn" /Lo )col=) 2n a

7

l

,7 x 100.191+2n(2x 10 ) ..4n ·---tn3.26 2n

=)

= j29.8 Q/m

z

in which Ie contributes almost wholly to jwl. Thus, == r + jcoi 0.191 + j29.8 Q/m. The shunt parameter y is found from (B-39), whence c = 2nE/tn (bfa) =2n(2 x 10 9/36n)/1.18=94.2 pF/m, yielding y=g+jwc +j)wc (0.0002 + j)())c= ,i1.l84 x 10" 2 Vim. Dielectric losses are negligible in this example. The characteristic impedance is found by lise of (9-105) (3)

or essentially that of Example 9-3

f()f

the lossless case, expected since r« col

g « wc. The 191 + j29.8)j1.184 x 10-

obtained = 0.595ei 89 . 85 " m

(J = 0.595 sin 89.8,'j"

~

~wle

and

Jfi

I,

Ji'om (9-103a) IS ')' yielding from the imaginary part

0.595 rad/m

(4)

The latter is also obtainable using (9-103c), since r« col and g« wc are satisfied. To find IX, (9-103b) provides good accuracy

r

gZo

""- + -- = 2Zo 2 ~

1.97

X

10"3 Np/m

0.191

" -..~

2 x 50

2.37 x 10- 6

+ ~---,- .10 2

1.71 x 10- 2 dB/m

and while still a low value, it represents an increase of ovcr 30 times that obtained in Example 9-3 with conductors assumed. Waves thus attenuate to 35.8'1,) of the input value ofalcngth d=IX- 1 = (1.97 x 10- 3 ) I =508mofthislineatf=20MHz. At 20 MHz, (9-37) yields A 2n/f5 10.55 m and vp w/f5 2.12 x 10 8 results essentially those of Example 9-3, in view of the small losses.

_____________________________________________ APPENDIX C

Integration of the Inhomogeneous Wave Equation

The formal proof that (11-17) is a solution of (1l-16) is given here. It is convenient to expand (II 16) in rectangular coordinates into the scalar wave equations

(C-l) making use of (2-83). Denote any of (C-l) in the unsubscripted form (C-2) Recall the Green's symmetrical theorem (2-92) g

~~)dS

an

(C-3)

correct for any pair of well-behaved functions f and g in and on a volume V bounded by the closed surface S. It is shown that (C-3) leads directly to the solution (11-17) if suitable choices are made for the functions f and g. Choosef as the field A in (C-2), assuming A means any component of A located at a source point pt. Thus, A A(u'u u~, u~) = A (r'), in whicl.!c r' denotes the position vector ofP'o The wave equation (C-2) is certainly satisfied by A (r') at all such points in space, and the Green's theorem

616

INTEGRATION OF THE INHOMOGENEOUS WAVE EQUATION

617

P'(r')

on 81 Enlarged

view of 81

(b)

(a)

FIGURE C-1. Geometries relative to the derivation of (11-17). (a) Current distribution at source points P' and a fixed field point P enclosed by surface S. (b) Showing small sphere 8 1 excluding P( r) from the region of integration.

(C-3) is written (C-4)

in which dv and ds are primed, since they are identified with locations P' in V and on S. For reasons about to be clarified, the function g in (C-4) is chosen as

e- i/loR

(C-5)

g=--

R

In (C-5), flJr the free-space region to which the constants /10' Eo apply (C-6)

Po = w)/1oEo

and R = Ir - r ' \ denotes the d~tance between the source point P' (r') and any fixed field point P(r), at which A is desired to be expressed in terms of the sources. The geometry is shown in Figure C-l(a). The integration (C-4) in V and on S being taken over all points P' and with P an arbitrary fixed point in the region, it is seen that R = () at P (i.e., P is the origin of R). Moreover, a property of the function g is that it satisfies the scalar wave equation,

(C-7) With (C-5) inserted into (C-4), the latter becomes

rh [~ 0 = Ys A(r') on

(e-.iPOR) e- jPoR OA(r')] --R- - R --an- ds'

(C-3J

618

INTEGRATION OF THE INHOMOGENEOUS WAVE EQUATION

From (C-7), - p~ can replace V 2g in the first term of (C-8), obtaining for the volume integral

r[

Jv

A(r')p~

J

e - j(JoR R

=

e - j(JoR 2 R V A(r') dv'

r _ e-j(JoR [V A(r') + pf;A(r')]dv' = r J1.o](r')e- i (JOR dv' 2

Jv

Jv

R

(C-9)

R

the latter following from the use of (C-2). Next, an inspection of the right side of (C-8) reveals that if the field point per) is to be inside the volume region V of integration as in Figure Col (a), then the requirement of Green's theorem that the function g = e-j(JoR/R be well-behaved in V is violated at P (where R = 0), in view of the singularity in g there. The point P is excluded from the volume region of integration by constructing a small sphere S1 of radius R = R1 about P as shown in Figure Col (b). Then V is bounded by the closed surface S = S1 + S2 as noted in the figure, whence (C-8) is written symbolically

r [. ] dv' Jv

=

r JSl

[] ds' + JS2 r [] ds'

(C-lO)

in which the brackets denote the volume and surface integrands of (C-8). The contribution of the integral on S1 in (C-lO) is now shown to approach the value 4nA(r), or just 4n times the potential A at the field point P, as the sphere S1 vanishes. From Figure Col (b) it is evident that %n = -%R on S1, since the normal is directed outward with respect to the volume region V (meaning in the negative R sense). Th us ~ , 0 (e- j(JOR) A(r) -oR R

r [] ds' = JSl r [ JSl =

l { ~ (r), (I-Rl + . A

]Po

S,

e - j(JoR oA (r') ]

+-----

)e-i(JORl Rl

R

oR

R~Rl

+ e-j(JORlOA(r')J -R1

oR

ds

t

R=Rl

}' ds (C-II)

By definition A(r') is well-behaved in the vicinity of the fixed point P(rl;. Allowing the radius Rl of the sphere S1 to become arbitrarily small, the value of A(r') on S1 can be replaced by its mean value