Elements of Electromagnetics Third Edition E Book Chapter 02

8/12/2019 Elements of Electromagnetics Third Edition E Book Chapter 02

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Chapter 2

COORDINATE SYSTEMSAND TRANSFORMATIONEducation makes a people easy to lead, but difficult to drive; easy to govern butimpossible to enslave.

HENRY P. BROUGHAM

2.1 INTRODUCTIONIn general, the physical quantities we shall be dealing with in EM are functions of spaceand time. In order to describe the spatial variations of the quantities, we must be able todefine all points uniquely in space in a suitable manner. This requires using an appropriatecoordinate system.

A point or vector can be represented in any curvilinear coordinate system, which m aybe orthogonal or nonorthogonal.An orthogonal system iso n ein which the coordinates arc mutually perpendicular.

Nonorthogo nal systems are hard to work with and they are of little or no practical use.Exam ples of orthogonal coordinate systems include the Cartesian (or rectangular), the cir-cular cylindrical, the spherical, the elliptic cylindrical, the parabolic cylindrical, theconical, the prolate spheroidal, the ob late spheroidal, and the ellipsoidal.1A considerableamount of work and time may be saved by choosing a coordinate system that best fits agiven problem. A hard problem in one coordi nate system m ay turn out to be easy inanother system.In this text, we shall restrict ourselves to the three best-known coordinate systems: theCartesian, the circular cylindrical, and the spherical. Although we have considered theCartesian system in Chap ter 1, we shall consider it in detail in this chapter. We should be arin mind that the concepts covered in Chapter 1and dem onstrated in Cartesian coordinatesare equally applicable to other systems of coordinates. For example, the procedure for

'For an introductory treatment of these coordinate systems, see M. R. Spigel,MathematicalHand-bookofFormulas andTables. NewYork:M cGraw-Hill,1968,pp.124-130.28


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2.3 CIRCULAR CYLINDRICA L COO RDINATE S (R, F, Z) 29finding dot or cross product of two vectors in a cylindrical system is the same as that usedin the Cartesian system in Chapter 1.Sometimes, it is necessary to transform points and vectors from one coordinate systemto another. The techniques for doing this will be presented and illustrated with examples.

2.2 CARTESIAN CO OR DINA TES (X, Y, Z)As mentioned in Chapter 1, a point P can be represented as (x, y, z) as illustrated inFigure 1.1. The ranges of the coordinate variablesx, y, andzare

-0 0 < X < 00

< y (2.1) 00 ,z)The circular cylindrical coordinate system is very convenient whenever we are dealingwith problems having cylindrical symm etry.A point P in cylindrical coordinates is represented as (p ,,z) and is as shown inFigure2.1.Observe Figure 2.1 closely and note how we define each space variable:pis theradius of the cylinder passing through Por the radial distance from the z-axis:,called the

Figure 2.1 PointPand unit vectors in the cylindricalcoordinate system.


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30 Coordinate Systems and Transformationazimuthal angle, is measured from the x-axis in the xy-plane; and z is the same as in theCartesian system. The ranges of the variables are

0


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2.3 CIRCULAR CYLIND RICAL COO RDINA TES (p, 0, z) 11 31

Figure 2.2 Relationship between (x, y, z) and(P,* . z).

The relat ionships between (a x , a y , a z) and (a p , a^, a2) are obtained geometrical ly fromFigure 2 .3 :

or

= cos 0 a p - sin

a p = cos (j>ax+ sin= - s i n= a7

co s

(b)Fig ure 2 .3 Unit vector transformation: (a ) cylindrical comp onents of ax, (b) cylin-drical components of a r

(2.9)

(2 .10)


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32 Coordinate Systems and TransformationFinally, the relationships between (Ax, Ay, A z) and(Ap,A0,Az) are obtained by simplysubstituting eq. (2.9) into eq. (2.2) and collecting terms. Thus

A = (Ax cos + Ay sin )ap + (~AX si n + Ay cos 0 )a0 + Azaz (2or

Ap = Ax cos + Ay sinA,/, = ~AX sin + Ay cos tj> (2.12)

In matrix form, we have the transformation of vector A from (Ax,A y,A z) to(A p,A0, A,) as

(2.13),Az

=cos sin 0 0sincos 0 00 0 1

AxAyAz

The inverse of the transformation (Ap, A^, A z)> (Ax, Ay, A z) is obtained asAx cos sin $ 0-sin^ > cos ^ 00 0 1 A,

(2.14)

or directly from eqs. (2.4) and (2.10). Thusc o s < > s i n > 0s i n < j> c o s < j> 0

0 0 1VA . (2.15)

An alternative way of obtaining eq. (2.14) or (2.15) is using the dot product. Forexample:

(2.16) A /AyAz

= a ^ a pa z - ap

a ^ a 0a y a 0a z a 0

*x

az az aza z

AAA

The d erivation of this is left as an exercise.

2.4 SPHERICAL COORD INA TE S (r, 0, (/>)The spherical coordinate system is most appropriate when dealing with problems having adegree of spherical symmetry. A pointPcan be represented as (r,6,4>)and is illustrated inFigure 2.4. From Figure 2.4, we notice that r is defined as the distance from the origin to


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2.4 SPHERICAL CO OR DINA TES (r, e, 33pointPor the radius of a sphere centered at the origin and passing throughP; 6(called thecolatitude)is the angle between the z-axis and the position vector ofP ;and4> is measuredfrom the x-axis (the same azimuthal angle in cylindrical coordinates). According to thesedefinitions, the ranges of the variables are

O < 0 < i r (2.17)0


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34 Co ordin ate Systems and TransformationThe space variables (x,y, z) in Cartesian coordinates can be related to variables(r, 0,


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2 .4 SPHERICAL C O O R D I N A T E S (r, e, 35The components of vector A = (Ax, Ay, Az) and A = (Ar,Ae, A^) are related by substitut-ing eq. (2.23) into eq. (2.2) and collecting terms. Thus,

A = (Axsin 0 cos 4> + Aysin 0 sin 0 + Az cos 0)ar + (Axcos 0 cos 0+ Aycos 0 sin 0 Az sind)ae + {Ax sin 0 + Ay cos + Aycos 0 sin Az sinA^ = A* sin + Ay cos 0

(2.26)

A. =sin 6cos 0 sin 0 sin 0 cos 0cos 0 cos 0 cos 0 sin sin 0 sin 0 cos 4> 0

In matrix form, the (Ax, Ay, Az) -> (Ar,Ae, A ) vector transformation is performed accord-ing to

2.27)

The inverse transformation (A nAe, A^) > (Ax, Ay, A z) is similarly obtained, or we obtain itfrom eq. (2.23). Thus,

2.28)

Alternatively, we may obtain eqs. (2.27) and (2.28) using the dot product. For example,

\A X~Av =sinsincos

000

COS 0sin 0 cos 0 coscos 0 sin 0- s i n 0

sincosT)

0_A rA s

Ar a r ax ar ay ar az (2.29)

For the sake of com pleteness, it may be instructive to obtain the point or vecto r trans-formation relationships between cylindrical and spherical coordinates using Figures 2.5and 2.6 (whereis held constant since it is comm on to both systems). This will be left asan exercise (see Problem 2.9). Note that in point or vector transformation the point orvector has not changed; it is only expressed differently. T hus, for exam ple, the m agnitudeof a vector will remain the same after the transformation and this may serve as a way ofchecking the result of the transformation.

The distance between two points is usually necessary in EM theory. The distance dbetween two points with position vectors rl and r2 is generally given byd=\r2- 2.30)


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36 Coordinate Systems and TransformationFigure 2.6 Unit vector transformations forcylindri-ca landspherical co ordinates.

or d2 =(x2- x ,f + (y2- yxf + (z2- zif (Cartesian)d2 =p\ + p2- 2p,p2cos((^2- 00 + (z2- Z\f (cylindrical)

- 2r^r2cosd2cos0jsin 02sindxcos(2- 0i)(spherical)d2 = r\ + r\ - 2r^r2cosd2cos0j

(2.31)(2.32)(2.33)

EXAMPLE2.1 Given point P(2,6, 3) andvectorA =yax+ (x + z)ay,expressPand A incylindricaland spherical coordinates. Evaluate A atPin the Cartesian, cylindrical, and sphericalsystems.Solution:At pointP:x = - 2 ,y = 6, z = 3. Hence,

p= Vx2+ y2 = V 4 + 36 = 6.324>= tan 1- = tan 1 = 108.43x - 2

r= Vx2 +y

2 + z

2 = V4 + 36 + 9 = 7

= 64.62, Vx2 + y2 _ V40d = tan ' - * ' tanThus,

P(-2, 6, 3) = P(6.32, 108.43, 3) = P(7, 64.62, 108.43)In the Cartesian system, A at P is

A = 6ax+ a


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2.4 SPHERICAL COO RDINA TES r, e, 37For vector A,Ax = y, Ay = x + z, Az = 0. Hence, in the cylindrical system

ApA*Az

= cossin 4>0sincos 000

001y

X +0 z

orAp = ycos(j) + (x + z) sin 0A,/, = y sin 0 + (JC+ z)cos =

A =

sin =- 2

- 2V40 '

V40 V40 V V4040

6 / \ / - 2V40- + 3 1

- ~ 6 _ 38y ^p /V 4 0 V 4 0

Similarly, in the spherical system

V 4 0 / V 4 0 -= -0.9487a,, - 6.OO8a0

ArAeA*

or

sin 0cos sin9sin 0 cos 0cos6cos 0 cos6sin 0 sin6sin 0 cos 0 0

Ar = ysin 0 cos + (x + z)sin6sin 0A9 = ycos 0 cos 0 + (x + z)cos 0 sinA4, = ~y sin(j> + (x + z)cos 0

x +z


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38 I f Coordinate Systems and TransformationB ut x = rsin 6cos (j>, y = rsi n 6sin , and z = r cos 0. Subs titut ing the se y ield s

r 8 += r[s in 2 6cos $ sin 0 + (sin 0c os+ cos 6) sin 0 sin 4>]ar+ r[sin 0 cos 6sin cos 0 + (sin 0 cos+ cos 0) cos 6 sin

A t P

Hence,

+ r[ sin sin 2

r = 1,

-2

(sin6 cos 0 + cos 8 )cos

tan 0 = tan0 = 40

40cos


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2.4 SPHERICAL COO RDINA TES r,e, 39A n s w e r : (a) P (3 . 16 2 , 71 . 56 ,5),P(5 .9 16 , 32 .31 , 71 .56) ,T(4,270 ,3),

T(5, 53 .13 , 270) ,5(5,233 . 1 , - 1 0 ) ,5(11 .18 , 153.43 , 233 .1)(b) :(cos4>ap sina , z sinaz) ,sin9(sin 0 cos rc o s20sin)ar + sin 0 cos 0 (cos 0 + rsin 0 sin agsin 0 sin(c)0. 2.4az,0. 2.4az,1.44ar - 1.92a,,

EXAMPLE2.2 Express vec tor10B = ar + rcos6 ae + a,*

in Cartesian and cyl indrical coordinates. FindB ( 3,4, 0) andB(5,TT/2,2).

Solut ion:U si ngeq.(2 .28 ) :

si n0 cos -sincos0 sin 0 cos

- s i n 0 0

K)r

rco sI1

or10B x = sin 0 cos


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40 H Co ordi nate Systems and TransformationSubstituting all these gives

l o V x 2 + y2 x Vx2 + / + z2 z2x'x2 + y2

lOx + xzx2 + y2 + z2 V ?1 0 W + y2

y ~~ / 2 , .2 , 2-.

y2 + z2) f)Vx2 + y2 + z

2 x2 + y2 + z2y

lO y +B 7 =

x2 + y2 + z2 V ( ? + y2){x2 + y2 + z2)lOz

+>xA + /

X'x2 + y2

zV x2 + y2x 2 + y2 + z2

B = B A + Byay + BzazwhereBx, B y, andBzare as given above.At ( - 3 , 4 , 0) ,x = - 3 ,y = 4, andz = 0, so

Thus,, = 0 - 0 = 0

BFor spherical to cylindrical vector transformation (see Problem 2.9),

sin ^ cos6 00 0 1

orcosd - s i n 0 0

10 2= sin6 + rcos

H)rcos

107 = cosdr - rsin6cos6

Butr = V p z + zl and6 = tan ' -

V + y2

^ + 7


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2.5 CONSTANT-COORDINATE SURFACES 41

Thus,sin V = /7T7 cos C =

7 2+z2 VTT?- 2 0 1 0+29 V29

= 2.467apNote that at ( - 3 , 4, 0),

\B(x,y,z)\ = |B(p,


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42 M Coo rdinate Systems and Transformat ionCartesian system, if we keepxconstant and allowyandz to vary, an infinite plane is gen-erated. Thus we could have infinite planes

x = constanty = constantz = constant

(2.34)

which are perpendicular to thex-, y-,and z-axes, respectively, as shown in Figure 2.7. Theintersection of two planes is a line. For example,x = constant, y = constant (2.35)

is the line RPQ parallel to the z-axis. The intersection of three planes is a point. Forexample,x = constant, y = constant, z = constant (2.36)

is the pointP (x, y, z). Thus we may define point P as the intersection of three orthogonalinfinite planes. IfPis (1, - 5 , 3), thenP is the intersection of planesx = 1,y = - 5 , andz = 3.Orthogonal surfaces in cylindrical coordinates can likewise be generated. The sur-facesp = constant = constantz = constant

(2.37)

are illustrated in Figure 2.8, where it is easy to observe that p = constant is a circularcylinder, = constant is a semiinfinite plane with its edge along the z-axis, andz = constant is the same infinite plane as in a Cartesian system. Where two surfaces meetis either a line or a circle. Thus,z = constant, p = constant (2.38)

z = constant

x= constant Figure 2.7 Constantx, y,andzsurfaces.

.


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p= constant

z = constant

2. 5 CONS T ANT- COO RDI NATE S URF ACES H 43

Figure2.8 Constantp,(j>,andzsurfaces.

*-y

= cons tan t

= constant

(2.41)

Figure2.9 Constantr,9,andsurfaces.


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44 Coordinate Systems and Transformationis a semicircle passing through Qa ndP.The intersection of three surfaces gives a point.Thus,

r = 5, 0 = 30, 0 = 60 (2.42)is the pointP(5,30, 60 ). We notice that in general, a point in three-dimensional space canbe identified as the intersection of three mutually orthogonal surfaces. Also, a unit normalvector to the surface n = constant is an,wherenisx, y, z, p, , r,or6.For ex ample, toplane* = 5,aunit normal vectorisax and to pla ned = 20,aunit normal vectorisa^.

EXAMPLE 2.3 Two uniform vector fields are given by E = - 5 a p +23^, - 6az. Calculate+ 3az and F = ap

(a) |E X F(b) The vector compo nent of E atP(5, TT/2,3) parallel to the line x = 2, z = 3(c) The angle E mak es with the surfacez = 3 atPSolution:(a) E XF = - 5 101 2 - 6= (- 6 0 - 6)a, + (3 - 3O)a0 + ( - 1 0 - 10)3,= ( - 6 6 , - 2 7 , - 2 0 )

|E XF| = V 6 62 + 272 + 202 = 74.06(b) Linex = 2,z = 3 is parallel to the y-axis, so the compon ent of E parallel to the givenline is

(E av)a vBut atP(5, TT/2,3)

Therefore,= sin ap + c o s a= sin T T / 2ap + cos nil a^, = ap

(E di y= (E ap)ap = - (or -5a y )(c) Utilizing the fact that the z-axis is norm al to the surface z =3, the angle between thez-axis and E , as shown in Figure 2 .10, can be found using the dot product:

E az = |E|(1) cos6Ez- > 3 = V l 3 4 cos Ez3cosoEz = '134 = 0.2592 - BEz= 74.98

Hence, the angle betweenz = 3 and E is90 - BEz= 15.02 J


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46 SB Co ord inate Systems and Tran sform ation(b) Any vec tor D can a lways be reso lved in to two or thogonal com ponen t s :

D = D, + Dnw here Dt is tangent ial to a given surface and Dn is normal to i t . In our case, since ar isnormal to the surface r = 10,

H e n c e ,D n = r sin 0 ar = 5ar

D , = D - Dn = 0.043a,, +(c) A vec tor atP perpendicular t o D and tangent i a l t o the cone 0 = 1 5 0 i s t he same as thevector perpendicular to both D and ae. H ence ,

D X afl =ar ae a0- 5 0.043 1000 1 0= - 1 0 0 a r - 5a*

Aunit vector along this is- 100a r - 5 3 Aa = , = - 0 .9 9 8 8 ar - 0.04993^VlOO2 + 52

PRAC TICE EXERC ISE 2.4IfA = 3ar + 2ae - 6a0andB = 4a,. + 33^,determine(a) A B(b) |A X B(c) The vector component of A alongazat(1, T T / 3 ,5ir/4)Answer: (a) - 6 , (b) 34.48, (c) -0 .1 16 ar + 0.201a,,.

S U M M A R Y 1 . The three common coordinate systems we shall use throughout the text are the Carte-sian (or rectangular), the circular cylindrical, and the spherical.2. A point P is represented asP(x,y,z), P(p,,z) , andP(r, 6,4>) in the Cartesian, cylin-drical, and spherical systems respectively. A vector field A is represented as(Ax, Ay, A z)orA ^ nx +Ayay + Azaz in the Cartesian system, as(Ap, A , A z) orApap + Aâ^ + Azazin the cylindrical system, and as (An Ae, A^) or Aâr + Aeae + Aâ^ in the sphericalsystem. It is preferable that mathematical operations (addition, subtraction, product,etc.) be performed in the same coordinate system. Thus, point and vector transforma-tions should be performed whenever necessary.

3. Fixing one space variable defines a surface; fixing two defines a line; fixing threedefines a point.4. A unit normal vector to surface n = constant is an.


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REVIEW QUESTIONS 47

PfVJEW QUtSTlONS2.1 The ranges of d and > as given by eq. (2.17) are not the only possible o nes. The followingare all alternative ranges of 6and, except

(a) 0


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48 B Coordinate Systems and Transformation2.7 Given G = 20a r + 50as + 4Oa0, at(1 , T/2, TT/6)the component of G perpendicular tosurface6= TT/2is

(a) 20ar(b)(c) 0(d) 20a r +(e) -40a r

2.8 Where surfacesp = 2andz = 1intersect is(a) an infinite plane(b) a semiinfinite plane(c) a circle(d) a cylinder(e) a cone

2.9 Match the items in the left list with those in the righ t list. Each answer can be used once,more than once, or not at all.(a) 0 = ?r/4(b) < > = 2 i r / 3(c ) JC = - 1 0(d ) r = 1 ,0 =(e) p = 5(f ) p = 3 ,


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P R O B L E M S 49PROBLEMS 2.1 Expressthefollowing points in Cartesian coordinates:( a )P( l ,60 , 2 )

(b) G(2, 90, -4)(c)R(,45, 210)(d)T(4,TT/2, TT/6)

2.2 Express the following points in cylindrical and spherical coordinates:(a) P(l, - 4 , - 3 )(b) g (3 , 0, 5)(c) R{-2, 6, 0)

2.3 (a) IfV= xz xy + yz,expressVincylindrical coordinates,(b) If U = x2 + 2>>2+ 3z2,expressUin spherical coordinates.

2.4 Transform the following vectors to cylindrical and spherical coordinates:(a) D = (x + z)ay(b) E = (y2 - x2)a x + xyzay + (x2 - Z2)a z2.5 Convert the following vectors to cylindrical and spherical systems:

xa x + yay + Aaz(a) F = Vx2(b) G = (x2 + y2) xa r

Vx2^

2.6 Express the following vectors in Cartesian coordinates:(a) A = p( z2 + l ) ap - pz cos a0(b) B = 2r sin 6c os ar + rcos 8c os 6 ae r sin 4>

2.7 Convert the following vectors to Cartesian coordinates:(a) C = z si n ap - p cos a0 + 2pzaz

sin d cosd(b) D = - ar + r- ae

Vx2

2.8 Prove the following:(a ) ax ap = cos >a x a 0 = - s i n ^3 ^ - 3 ^ = sin


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50 Coordinate Systems and Transformationay- ae = cos 6sinaz ar = cos6a a s = sin6

2.9 (a) Show that point transform ation betwee n cylindrical and spherical coord inates is ob-tained using

p = r s in 9, z = r c o s 9, 4> = 4>(b) Show that vector transformation b etwe en cylindrical and spherical coordinates is ob-

tained using

or

ArAeA 0_

A,

=

=

si ncos0

si n0

cos

ee

9

0 cos0 - s i n1 0

cos80

sin8

99

010

Ar

(Hint: Make use of Figures 2.5 and 2.6.)2.10 (a) Exp ress the vector field

H = xy2za x + x2yzayin cylindrical and spherical coordinates,

(b) In both cylindrical and spherical coordinates, determine H at (3 , 4, 5).2.11 Let A = p cos 9 ap + pz2 si n az

(a) Transform A into rectangular coordinates and calculate i ts magnitude at point(3 , - 4 , 0 ).

(b) Transform A into spherical system and calculate its mag nitude at point (3 , 4, 0) .2.12 The transformation (Ap, A0 ,Az) > (Ax, A y, A z) in eq. (2.15) is not complete. Complete itby expressing cos4> and sinin terms ofx, y, an dz .Do the same thing to the transforma-tion (Ar, A e, A^) - ( A x , Ay, A z) in eq. (2.28).

2.13 In Practice Exercise 2.2, express A in spherical and B in cylindrical coordinates. Ev aluateA at (10 , TT/2, 3TI 74) and B at (2, TT/6, 1).


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PROBLEMS 51

2.14 Calculate the distance between the following pairs of points:(a) (2, 1,5) and (6, - 1 , 2 )(b) (3, T/2, -1) and (5, 3TT/2, 5)(c) (10, TT/4, 3TT/4) and (5, x/6, 7*74).

2.15 Describe the intersection of the following surfaces:

= 10a)b)c)d)e)f)

XXrp

r

= 2,= 2,= 10,= g= 60,

y =y = = t =z=0=

5- 1 ,30401090

2.16 At point 7(2, 3, 4), express azin the spherical system and arin the rectangular system.

*2.17 Given vectors A = 2a^ + 4ay + 10azand B = -5ap + a0 - 3az, find(a) A + BatP(0,2, -5)(b) The angle between A and B atP(c) The scalar component of A along B atP

2.18 Given that G = (x + y2)a x + xzay + (z2 + zy)az, find the vector component of Galong a0at point P(8, 30, 60). Your answer should be left in the Cartesian system.

*2.19 If J = rsin 0 cosar - cos26sin4> ae + tan - Inra0 atT(2, TT/2,3 12),determinethe vector component of J that is(a) Parallel to az(b) Normal to surface 4> = 37r/2(c) Tangential to the spherical surface r = 2(d) Parallel to the liney = - 2 ,z = 0

2.20 Let H - 5p sinap - pzcos a0 + 2paz. At pointP(2, 30, - 1), find:(a) a unit vector along H(b) the component of H parallel toax(c) the component of H normal top = 2(d) the component of H tangential to = 30


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52 11 Co ordina te Systems and Transformation*2.21 Let

andA = p(z2 - l)ap - pzcos a^, + /o2z2az

B = r2 cos 0 ar + 2r sin 0 a0At r(3,4, 1), calculate: (a) A and B, (b) the vector component in cylindrical coordi-nates ofAalong B atT (c) the unit vector in spherical coordinates perpendicular to bothA andBatT.

*2.22 Another way of defining a point P in space is(r, a, jS,7) where the variables are por-trayed in Figure2.11.U sing this definition, find (r,a, |8,7) for the following po ints:(a) ( - 2 , 3, 6)(b) (4, 30, - 3 )(c) (3, 30, 60)(Hint: ris the sphericalr,0 tyz ( x2 .G = a z + + ( 1 - - j Ia ,P p1 \ pzExpress G completely in spherical system.

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Elements of Electromagnetics Third Edition E Book Chapter 02

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