Engineering Electromagnetics, 8th Edition Solution Manual

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CHAPTER 11.1.Given the vectorsM=10ax+ 4ay8azandN= 8ax+ 7ay2az, find:a) a unit vector in the direction ofM+ 2N.M+ 2N= 10ax4ay+ 8az+ 16ax+ 14ay4az= (26,10,4)Thusa=(26,10,4)|(26,10,4)|= (0.92,0.36,0.14)b) the magnitude of 5ax+N3M:(5,0,0) + (8,7,2)(30,12,24) = (43,5,22), and|(43,5,22)|= 48.6.c)|M||2N|(M+N):|(10,4,8)||(16,14,4)|(2,11,10) = (13.4)(21.6)(2,11,10)= (580.5,3193,2902)1.2.VectorAextends from the origin to (1,2,3) and vectorBfrom the origin to (2,3,-2).a) Find the unit vector in the direction of (AB): FirstAB= (ax+ 2ay+ 3az)(2ax+ 3ay2az) = (axay+ 5az)whose magnitude is|AB|= [(axay+ 5az)·(axay+ 5az)]1/2=p1 + 1 + 25 =3p3 = 5.20. The unit vector is thereforeaAB= (axay+ 5az)/5.20b) find the unit vector in the direction of the line extending from the origin to the midpoint of theline joining the ends ofAandB:The midpoint is located atPmp= [1 + (21)/2,2 + (32)/2,3 + (23)/2)] = (1.5,2.5,0.5)The unit vector is thenamp=(1.5ax+ 2.5ay+ 0.5az)p(1.5)2+ (2.5)2+ (0.5)2= (1.5ax+ 2.5ay+ 0.5az)/2.961.3.The vector from the origin to the pointAis given as (6,2,4), and the unit vector directed fromthe origin toward pointBis (2,2,1)/3. If pointsAandBare ten units apart, find the coordinatesof pointB.WithA= (6,2,4) andB=13B(2,2,1), we use the fact that|BA|= 10, or|(623B)ax(223B)ay(4 +13B)az|= 10Expanding, obtain368B+49B2+ 483B+49B2+ 16 +83B+19B2= 100orB28B44 = 0. ThusB=8±p641762= 11.75 (taking positive option) and soB= 23 (11.75)ax23 (11.75)ay+ 13 (11.75)az= 7.83ax7.83ay+ 3.92az1

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1.4.A circle, centered at the origin with a radius of 2 units, lies in thexyplane.Determine the unitvector in rectangular components that lies in thexyplane, is tangent to the circle at (p3,1,0),and is in the general direction of increasing values ofy:A unit vector tangent to this circle in the general increasingydirection ist=a. Itsxandycomponents aretx=a·ax= sin, andty=a·ay=cos. At the point (p3,1),= 150, and sot= sin 150axcos 150ay= 0.5(ax+p3ay).1.5.A vector field is specified asG= 24xyax+ 12(x2+ 2)ay+ 18z2az. Given two points,P(1,2,1)andQ(2,1,3), find:a)GatP:G(1,2,1) = (48,36,18)b) a unit vector in the direction ofGatQ:G(2,1,3) = (48,72,162), soaG=(48,72,162)|(48,72,162)|= (0.26,0.39,0.88)c) a unit vector directed fromQtowardP:aQP=PQ|PQ|= (3,1,4)p26= (0.59,0.20,0.78)d) the equation of the surface on which|G|= 60: We write 60 =|(24xy,12(x2+ 2),18z2)|, or10 =|(4xy,2x2+ 4,3z2)|, so the equation is100 = 16x2y2+ 4x4+ 16x2+ 16 + 9z41.6.Find the acute angle between the two vectorsA= 2ax+ay+ 3azandB=ax3ay+ 2azby usingthe definition of:a) the dot product: First,A·B= 23 + 6 = 5 =ABcos✓, whereA=p22+ 12+ 32=p14,and whereB=p12+ 32+ 22=p14. Therefore cos✓= 5/14, so that✓= 69.1.b) the cross product: Begin withA⇥B=axayaz213132= 11axay7azand then|A⇥B|=p112+ 12+ 72=p171. So now, with|A⇥B|=ABsin✓=p171,find✓= sin1p171/14= 69.11.7.Given the vector fieldE= 4zy2cos 2xax+ 2zysin 2xay+y2sin 2xazfor the region|x|,|y|, and|z|less than 2, find:a) the surfaces on whichEy= 0. WithEy= 2zysin 2x= 0, the surfaces are 1) the planez= 0,with|x|<2,|y|<2; 2) the planey= 0, with|x|<2,|z|<2; 3) the planex= 0, with|y|<2,|z|<2;4) the planex=⇡/2, with|y|<2,|z|<2.b) the region in whichEy=Ez: This occurs when 2zysin 2x=y2sin 2x, or on the plane 2z=y,with|x|<2,|y|<2,|z|<1.c) the region in whichE= 0: We would haveEx=Ey=Ez= 0, orzy2cos 2x=zysin 2x=y2sin 2x= 0. This condition is met on the planey= 0, with|x|<2,|z|<2.2

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1.8.Demonstrate the ambiguity that results when the cross product is used to find the angle betweentwo vectors by finding the angle betweenA= 3ax2ay+ 4azandB= 2ax+ay2az. Does thisambiguity exist when the dot product is used?We use the relationA⇥B=|A||B|sin✓n. With the given vectors we findA⇥B= 14ay+ 7az= 7p52ay+azp5|{z}±n=p9 + 4 + 16p4 + 1 + 4 sin✓nwherenis identified as shown; we see thatncan be positive or negative, as sin✓can bepositive or negative.This apparent sign ambiguity is not the real problem, however, as wereally want the magnitude of the angle anyway.Choosing the positive sign, we are left withsin✓= 7p5/(p29p9) = 0.969.Twovalues of✓(75.7and 104.3) satisfy this equation, andhence the real ambiguity.In using the dot product, we findA·B= 628 =4 =|A||B|cos✓= 3p29 cos✓, orcos✓=4/(3p29) =0.248)✓=75.7. Again, the minus sign is not important, as wecare only about the angle magnitude.The main point is thatonly one✓value results whenusing the dot product, so no ambiguity.1.9.A field is given asG=25(x2+y2) (xax+yay)Find:a) a unit vector in the direction ofGatP(3,4,2): HaveGp= 25/(9 + 16)⇥(3,4,0) = 3ax+ 4ay,and|Gp|= 5. ThusaG= (0.6,0.8,0).b) the angle betweenGandaxatP:The angle is found throughaG·ax= cos✓.So cos✓=(0.6,0.8,0)·(1,0,0) = 0.6. Thus✓= 53.c) the value of the following double integral on the planey= 7:Z40Z20G·aydzdxZ40Z2025x2+y2(xax+yay)·aydzdx=Z40Z2025x2+ 49⇥7dzdx=Z40350x2+ 49dx= 350⇥17tan1✓47◆0= 261.10.By expressing diagonals as vectors and using the definition of the dot product, find the smaller anglebetween any two diagonals of a cube, where each diagonal connects diametrically opposite corners,and passes through the center of the cube:Assuming a side length,b, two diagonal vectors would beA=b(ax+ay+az) andB=b(axay+az). Now useA·B=|A||B|cos✓, orb2(11 + 1) = (p3b)(p3b) cos✓)cos✓=1/3)✓= 70.53. This result (in magnitude) is the same foranytwo diagonal vectors.3

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1.11.Given the pointsM(0.1,0.2,0.1),N(0.2,0.1,0.3), andP(0.4,0,0.1), find:a) the vectorRM N:RM N= (0.2,0.1,0.3)(0.1,0.2,0.1) = (0.3,0.3,0.4).b) the dot productRM N·RM P:RM P= (0.4,0,0.1)(0.1,0.2,0.1) = (0.3,0.2,0.2).RM N·RM P= (0.3,0.3,0.4)·(0.3,0.2,0.2) =0.09 + 0.06 + 0.08 = 0.05.c) the scalar projection ofRM NonRM P:RM N·aRM P= (0.3,0.3,0.4)·(0.3,0.2,0.2)p0.09 + 0.04 + 0.04 =0.05p0.17 = 0.12d) the angle betweenRM NandRM P:✓M= cos1✓RM N·RM P|RM N||RM P|◆= cos1✓0.05p0.34p0.17◆= 781.12.Write an expression in rectangular components for the vector that extends from (x1, y1, z1) to(x2, y2, z2) and determine the magnitude of this vector.The two points can be written as vectors from the origin:A1=x1ax+y1ay+z1azandA2=x2ax+y2ay+z2azThe desired vector will now be the di↵erence:A12=A2A1= (x2x1)ax+ (y2y1)ay+ (z2z1)azwhose magnitude is|A12|=pA12·A12=⇥(x2x1)2+ (y2y1)2+ (z2z1)2⇤1/21.13.a) Find the vector component ofF= (10,6,5) that is parallel toG= (0.1,0.2,0.3):F||G=F·G|G|2G= (10,6,5)·(0.1,0.2,0.3)0.01 + 0.04 + 0.09(0.1,0.2,0.3) = (0.93,1.86,2.79)b) Find the vector component ofFthat is perpendicular toG:FpG=FF||G= (10,6,5)(0.93,1.86,2.79) = (9.07,7.86,2.21)c) Find the vector component ofGthat is perpendicular toF:GpF=GG||F=GG·F|F|2F= (0.1,0.2,0.3)1.3100 + 36 + 25 (10,6,5) = (0.02,0.25,0.26)4

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1.14.Given thatA+B+C= 0, where the three vectors represent line segments and extend from acommon origin,a) must the three vectors be coplanar?In terms of the components, the vector sum will beA+B+C= (Ax+Bx+Cx)ax+ (Ay+By+Cy)ay+ (Az+Bz+Cz)azwhich we require to be zero.Suppose the coordinate system is configured so that vectorsAandBlie in thex-yplane; in this caseAz=Bz= 0. ThenCzhas to be zero in order for thethree vectors to sum to zero. Therefore, the three vectors must be coplanar.b) IfA+B+C+D= 0, are the four vectors coplanar?The vector sum is nowA+B+C+D= (Ax+Bx+Cx+Dx)ax+ (Ay+By+Cy+Dy)ay+ (Az+Bz+Cz+Dz)azNow, for example, ifAandBlie in thex-yplane,CandDneed not, as long asCz+Dz= 0.So the four vectors need not be coplanar to have a zero sum.1.15.Three vectors extending from the origin are given asr1= (7,3,2),r2= (2,7,3), andr3=(0,2,3). Find:a) a unit vector perpendicular to bothr1andr2:ap12=r1⇥r2|r1⇥r2|= (5,25,55)60.6= (0.08,0.41,0.91)b) a unit vector perpendicular to the vectorsr1r2andr2r3:r1r2= (9,4,1) andr2r3=(2,5,6). Sor1r2⇥r2r3= (19,52,37). Thenap=(19,52,37)|(19,52,37)|= (19,52,37)66.6= (0.29,0.78,0.56)c) the area of the triangle defined byr1andr2:Area = 12|r1⇥r2|= 30.3d) the area of the triangle defined by the heads ofr1,r2, andr3:Area = 12|(r2r1)⇥(r2r3)|= 12|(9,4,1)⇥(2,5,6)|= 33.35

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1.16.IfArepresents a vector one unit in length directed due east,Brepresents a vector three units inlength directed due north, andA+B= 2CDand 2AB=C+ 2D, determine the length anddirection ofC. (diculty 1)Take north as the positiveydirection, and then east as the positivexdirection. Then we may writeA+B=ax+ 3ay= 2CDand2AB= 2ax3ay=C+ 2DMultiplying the first equation by 2, and then adding the result to the second equation eliminatesD, and we get4ax+ 3ay= 5C)C= 45ax+ 35ayThe length ofCis|C|=⇥(4/5)2+ (3/5)2⇤1/2= 1Clies in thex-yplane at angle from due north (theyaxis) given by↵= tan1(4/3) = 53.1(or36.9from thexaxis). For those having nautical leanings, this is very close to the compass pointNE34E (not required).1.17.PointA(4,2,5) and the two vectors,RAM= (20,18,10) andRAN= (10,8,15), define atriangle.a) Find a unit vector perpendicular to the triangle:Useap=RAM⇥RAN|RAM⇥RAN|= (350,200,340)527.35= (0.664,0.379,0.645)The vector in the opposite direction to this one is also a valid answer.b) Find a unit vector in the plane of the triangle and perpendicular toRAN:aAN= (10,8,15)p389= (0.507,0.406,0.761)ThenapAN=ap⇥aAN= (0.664,0.379,0.645)⇥(0.507,0.406,0.761) = (0.550,0.832,0.077)The vector in the opposite direction to this one is also a valid answer.c) Find a unit vector in the plane of the triangle that bisects the interior angle atA:A non-unitvector in the required direction is (1/2)(aAM+aAN), whereaAM=(20,18,10)|(20,18,10)|= (0.697,0.627,0.348)Now12 (aAM+aAN) = 12 [(0.697,0.627,0.348) + (0.507,0.406,0.761)] = (0.095,0.516,0.207)Finally,abis=(0.095,0.516,0.207)|(0.095,0.516,0.207)|= (0.168,0.915,0.367)6

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1.18.A certain vector field is given asG= (y+ 1)ax+xay. a) DetermineGat the point (3,-2,4):G(3,2,4) =ax+ 3ay.b) obtain a unit vector defining the direction ofGat (3,-2,4).|G(3,2,4)|= [1 + 32]1/2=p10. So the unit vector isaG(3,2,4) =ax+ 3ayp101.19.a) Express the fieldD= (x2+y2)1(xax+yay) in cylindrical components and cylindrical variables:Havex=⇢cos,y=⇢sin, andx2+y2=⇢2. ThereforeD= 1⇢(cosax+ sinay)ThenD⇢=D·a⇢= 1⇢[cos(ax·a⇢) + sin(ay·a⇢)] = 1⇢⇥cos2+ sin2⇤= 1⇢andD=D·a= 1⇢[cos(ax·a) + sin(ay·a)] = 1⇢[cos(sin) + sincos] = 0ThereforeD= 1⇢a⇢b) EvaluateDat the point where⇢= 2,= 0.2⇡, andz= 5, expressing the result in cylindricaland cartesian coordinates: At the given point, and in cylindrical coordinates,D= 0.5a⇢.Toexpress this in cartesian, we useD= 0.5(a⇢·ax)ax+ 0.5(a⇢·ay)ay= 0.5 cos 36ax+ 0.5 sin 36ay= 0.41ax+ 0.29ay1.20.If the three sides of a triangle are represented by the vectorsA,B, andC, all directed counter-clockwise, show that|C|2= (A+B)·(A+B) and expand the product to obtain the law of cosines.With the vectors drawn as described above, we find thatC=(A+B) and so|C|2=C2=C·C=(A+B)·(A+B) So far so good. Now if we expand the product, obtain(A+B)·(A+B) =A2+B2+ 2A·BwhereA·B=ABcos(180↵) =ABcos↵where↵is the interior angle at the junction ofAandB. Using this, we haveC2=A2+B22ABcos↵, which is the law of cosines.7

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1.21.Express in cylindrical components:a) the vector fromC(3,2,7) toD(1,4,2):C(3,2,7)!C(⇢= 3.61,= 33.7, z=7) andD(1,4,2)!D(⇢= 4.12,=104.0, z= 2).NowRCD= (4,6,9) andR⇢=RCD·a⇢=4 cos(33.7)6 sin(33.7) =6.66.ThenR=RCD·a= 4 sin(33.7)6 cos(33.7) =2.77. SoRCD=6.66a⇢2.77a+ 9azb) a unit vector atDdirected towardC:RCD= (4,6,9) andR⇢=RDC·a⇢= 4 cos(104.0) + 6 sin(104.0) =6.79.ThenR=RDC·a= 4[sin(104.0)] + 6 cos(104.0) = 2.43. SoRDC=6.79a⇢+ 2.43a9azThusaDC=0.59a⇢+ 0.21a0.78azc) a unit vector atDdirected toward the origin:Start withrD= (1,4,2), and so thevector toward the origin will berD= (1,4,2).Thus in cartesian the unit vector isa=(0.22,0.87,0.44). Convert to cylindrical:a⇢= (0.22,0.87,0.44)·a⇢= 0.22 cos(104.0) + 0.87 sin(104.0) =0.90, anda= (0.22,0.87,0.44)·a= 0.22[sin(104.0)] + 0.87 cos(104.0) = 0, so that finally,a=0.90a⇢0.44az.1.22.A sphere of radiusa, centered at the origin, rotates about thezaxis at angular velocity⌦rad/s.The rotation direction is clockwise when one is looking in the positivezdirection.a) Using spherical components, write an expression for the velocity field,v, which gives the tan-gential velocity at any point within the sphere:As in problem 1.20, we find the tangential velocity as the product of the angular velocity andthe perperdicular distance from the rotation axis. With clockwise rotation, we obtainv(r,✓) =⌦rsin✓a(r < a)b) Convert to rectangular components:From here, the problem is the same as partcin Problem 1.20, except the rotation direction isreversed. The answer isv(x, y) =⌦[yax+xay], where (x2+y2+z2)1/2< a.1.23.The surfaces⇢= 3,⇢= 5,= 100,= 130,z= 3, andz= 4.5 define a closed surface.a) Find the enclosed volume:Vol =Z4.53Z130100Z53⇢d⇢ddz= 6.28NOTE: The limits on theintegration must be converted to radians (as was done here, but notshown).b) Find the total area of the enclosing surface:Area = 2Z130100Z53⇢d⇢d+Z4.53Z1301003ddz+Z4.53Z1301005ddz+ 2Z4.53Z53d⇢dz= 20.78

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1.23c) Find the total length of the twelve edges of the surfaces:Length = 4⇥1.5 + 4⇥2 + 2⇥30360⇥2⇡⇥3 +30360⇥2⇡⇥5= 22.4d) Find the length of the longest straight line that lies entirely within the volume: This will bebetween the points A(⇢= 3,= 100,z= 3) and B(⇢= 5,= 130,z= 4.5). Performingpoint transformations to cartesian coordinates, these become A(x=0.52,y= 2.95,z= 3)and B(x=3.21,y= 3.83,z= 4.5). Taking A and B as vectors directed from the origin, therequested length isLength =|BA|=|(2.69,0.88,1.5)|= 3.211.24.Two unit vectors,a1anda2lie in thexyplane and pass through the origin. They make angles1and2with thexaxis respectively.a) Express each vector in rectangular components; Havea1=Ax1ax+Ay1ay, so thatAx1=a1·ax= cos1. Then,Ay1=a1·ay= cos(901) = sin1. Therefore,a1= cos1ax+ sin1ayand similarly,a2= cos2ax+ sin2ayb) take the dot product and verify the trigonometric identity, cos(12) = cos1cos2+sin1sin2: From the definition of the dot product,a1·a2= (1)(1) cos(12)= (cos1ax+ sin1ay)·(cos2ax+ sin2ay) = cos1cos2+ sin1sin2c) take the cross product and verify the trigonometric identity sin(21) = sin2cos1cos2sin1: From the definition of the cross product, and sincea1anda2both lie in thex-yplane,a1⇥a2= (1)(1) sin(12)az=axayazcos1sin10cos2sin20= [sin2cos1cos2sin1]azthus verified.1.25.Given pointP(r= 0.8,✓= 30,= 45), andE=1r2✓cosar+ sinsin✓a◆a) FindEatP:E= 1.10a⇢+ 2.21a.b) Find|E|atP:|E|=p1.102+ 2.212= 2.47.c) Find a unit vector in the direction ofEatP:aE=E|E|= 0.45ar+ 0.89a9

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1.26.Express the uniform vector field,F= 5axina) cylindrical components:F⇢= 5ax·a⇢= 5 cos, andF= 5ax·a=5 sin. Combining, weobtainF(⇢,) = 5(cosa⇢sina).b) spherical components:Fr= 5ax·ar= 5 sin✓cos;F✓= 5ax·a✓= 5 cos✓cos;F= 5ax·a=5 sin. Combining, we obtainF(r,✓,) = 5 [sin✓cosar+ cos✓cosa✓sina].1.27.The surfacesr= 2 and 4,✓= 30and 50, and= 20and 60identify a closed surface.a) Find the enclosed volume: This will beVol =Z6020Z5030Z42r2sin✓drd✓d= 2.91where degrees have been converted to radians.b) Find the total area of the enclosing surface:Area =Z6020Z5030(42+ 22) sin✓d✓d+Z42Z6020r(sin 30+ sin 50)drd+ 2Z5030Z42rdrd✓= 12.61c) Find the total length of the twelve edges of the surface:Length = 4Z42dr+ 2Z5030(4 + 2)d✓+Z6020(4 sin 50+ 4 sin 30+ 2 sin 50+ 2 sin 30)d= 17.49d) Find the length of the longest straight line that lies entirely within the surface: This will befromA(r= 2,✓= 50,= 20) toB(r= 4,✓= 30,= 60) orA(x= 2 sin 50cos 20, y= 2 sin 50sin 20, z= 2 cos 50)toB(x= 4 sin 30cos 60, y= 4 sin 30sin 60, z= 4 cos 30)or finallyA(1.44,0.52,1.29) toB(1.00,1.73,3.46). ThusBA= (0.44,1.21,2.18) andLength =|BA|= 2.531.28.State whether or notA=Band, if not, what conditions are imposed onAandBwhena)A·ax=B·ax: For this to be true, bothAandBmust be oriented at the same angle,✓, fromthexaxis. But this would allow either vector to lie anywhere along a conical surface of angle✓about thexaxis. Therefore,Acanbe equal toB, but not necessarily.b)A⇥ax=B⇥ax: This is a more restrictive condition because the cross product gives a vector.For both cross products to lie in the same direction,A,B, andaxmust be coplanar. But ifAlies at angle✓to thexaxis,Bcould lie at✓orat 180✓to give the same cross product. Soagain,Acanbe equal toB, but not necessarily.10

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1.28c)A·ax=B·axandA⇥ax=B⇥ax: In this case, we need to satisfy both requirements in partsaandb– that is,A,B, andaxmust be coplanar,andAandBmust lie at the same angle,✓, toax. With coplanar vectors, this latter condition might imply that both +✓and✓would thereforework. But the negative angle reverses the direction of the cross product direction. Therefore bothvectors must lie in the same planeandlie at the same angle tox; i.e.,Amustbe equal toB.d)A·C=B·CandA⇥C=B⇥CwhereCis any vector exceptC= 0: This is just the generalcase of partc.Since we can orient our coordinate system in any manner we choose, we canarrange it so that thexaxis coincides with the direction of vectorC. Thus all the argumentsof partcapply, and again we conclude thatAmustbe equal toB.1.29.Express the unit vectoraxin spherical components at the point:a)r= 2,✓= 1 rad,= 0.8 rad: Useax= (ax·ar)ar+ (ax·a✓)a✓+ (ax·a)a=sin(1) cos(0.8)ar+ cos(1) cos(0.8)a✓+ (sin(0.8))a= 0.59ar+ 0.38a✓0.72ab)x= 3,y= 2,z=1:First, transform the point to spherical coordinates.Haver=p14,✓= cos1(1/p14) = 105.5, and= tan1(2/3) = 33.7. Thenax= sin(105.5) cos(33.7)ar+ cos(105.5) cos(33.7)a✓+ (sin(33.7))a= 0.80ar0.22a✓0.55ac)⇢= 2.5,= 0.7 rad,z= 1.5: Again, convert the point to spherical coordinates.r=p⇢2+z2=p8.5,✓= cos1(z/r) = cos1(1.5/p8.5) = 59.0, and= 0.7 rad = 40.1. Nowax= sin(59) cos(40.1)ar+ cos(59) cos(40.1)a✓+ (sin(40.1))a= 0.66ar+ 0.39a✓0.64a1.30.Consider a problem analogous to the varying wind velocities encountered by transcontinental aircraft.We assume a constant altitude, a plane earth, a flight along thexaxis from 0 to 10 units, no verticalvelocity component, and no change in wind velocity with time.Assumeaxto be directed to theeast andayto the north. The wind velocity at the operating altitude is assumed to be:v(x, y) = (0.01x20.08x+ 0.66)ax(0.05x0.4)ay1 + 0.5y2a) Determine the location and magnitude of the maximum tailwind encountered: Tailwind wouldbex-directed, and so we look at thexcomponent only.Over the flight range, this functionmaximizes at a value of 0.86/(1 + 0.5y2) atx= 10 (at the end of the trip). It reaches a localminimum of 0.50/(1 + 0.5y2) atx= 4, and has another local maximum of 0.66/(1 + 0.5y2) atthe trip start,x= 0.b) Repeat for headwind: Thexcomponent is always positive, and so therefore no headwind existsover the travel range.c) Repeat for crosswind: Crosswind will be found from theycomponent, which is seen to maximizeover the flight range at a value of 0.4/(1 + 0.5y2) at the trip start (x= 0).d) Would more favorable tailwinds be available at some other latitude? If so, where? Minimizingthe denominator accomplishes this; in particular, the lattitude associated withy= 0 gives thestrongest tailwind.11

Engineering Electromagnetics, 8th Edition Solution Manual - Page 13 preview image

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CHAPTER 22.1.Three point charges are positioned in thex-yplane as follows: 5nC aty= 5 cm, -10 nC aty=5 cm,15 nC atx=5 cm. Find the requiredx-ycoordinates of a 20-nC fourth charge that will produce azero electric field at the origin.With the charges thus configured, the electric field at the origin will be the superposition of theindividual charge fields:E0=14⇡✏015(5)2ax5(5)2ay10(5)2ay=14⇡✏0✓35◆[axay]nC/mThe field,E20, associated with the 20-nC charge (evaluated at the origin) must exactly cancelthis field, so we write:E20=14⇡✏0✓35◆[axay]=204⇡✏0⇢2✓1p2◆[axay]From this, we identify the distance from the origin:⇢=q100/(3p2) = 4.85.Thexandycoordinates of the 20-nC charge will both be equal in magnitude to 4.85/p2 = 3.43.Thecoodinates of the 20-nC charge are then (3.43,3.43).2.2.Point charges of 1nC and -2nC are located at (0,0,0) and (1,1,1), respectively, in free space. Determinethe vector force acting on each charge.First, the electric field intensity associated with the 1nC charge, evalutated at the -2nC chargelocation is:E12=14⇡✏0(3)✓1p3◆(ax+ay+az)nC/min which the distance between charges isp3 m. The force on the -2nC charge is thenF12=q2E12=212p3⇡✏0(ax+ay+az) =110.4⇡✏0(ax+ay+az)nNThe force on the 1nC charge at the origin is just the opposite of this result, orF21=+110.4⇡✏0(ax+ay+az)nN12

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2.3.Point charges of 50nC each are located atA(1,0,0),B(1,0,0),C(0,1,0), andD(0,1,0) in freespace. Find the total force on the charge atA.The force will be:F= (50⇥109)24⇡✏0RCA|RCA|3+RDA|RDA|3+RBA|RBA|3whereRCA=axay,RDA=ax+ay, andRBA= 2ax. The magnitudes are|RCA|=|RDA|=p2,and|RBA|= 2. Substituting these leads toF= (50⇥109)24⇡✏012p2 +12p2 + 28ax= 21.5axμNwhere distances are in meters.2.4.Eight identical point charges ofQC each are located at the corners of a cube of side lengtha, withone charge at the origin, and with the three nearest charges at (a,0,0), (0, a,0), and (0,0, a). Findan expression for the total vector force on the charge atP(a, a, a), assuming free space:The total electric field atP(a, a, a) that produces a force on the charge there will be the sumof the fields from the other seven charges.This is written below, where the charge locationsassociated with each term are indicated:Enet(a, a, a) =q4⇡✏0a226664ax+ay+az3p3|{z}(0,0,0)+ay+az2p2|{z}(a,0,0)+ax+az2p2|{z}(0,a,0)+ax+ay2p2|{z}(0,0,a)+ax|{z}(0,a,a)+ay|{z}(a,0,a)+az|{z}(a,a,0)37775The force is now the product of this field and the charge at (a, a, a). Simplifying, we obtainF(a, a, a) =qEnet(a, a, a) =q24⇡✏0a213p3 +1p2 + 1(ax+ay+az) = 1.90q24⇡✏0a2(ax+ay+az)in which the magnitude is|F|= 3.29q2/(4⇡✏0a2).2.5.Let a point chargeQ1= 25 nC be located atP1(4,2,7) and a chargeQ2= 60 nC be atP2(3,4,2).a) If✏=✏0, findEatP3(1,2,3): This field will beE= 1094⇡✏025R13|R13|3+ 60R23|R23|3whereR13=3ax+ 4ay4azandR23= 4ax2ay+ 5az. Also,|R13|=p41 and|R23|=p45.SoE= 1094⇡✏025⇥(3ax+ 4ay4az)(41)1.5+ 60⇥(4ax2ay+ 5az)(45)1.5= 4.58ax0.15ay+ 5.51azb) At what point on theyaxis isEx= 0?P3is now at (0, y,0), soR13=4ax+ (y+ 2)ay7azandR23= 3ax+ (y4)ay+ 2az. Also,|R13|=p65 + (y+ 2)2and|R23|=p13 + (y4)2.Now thexcomponent ofEat the newP3will be:Ex= 1094⇡✏025⇥(4)[65 + (y+ 2)2]1.5+60⇥3[13 + (y4)2]1.513

Engineering Electromagnetics, 8th Edition Solution Manual - Page 15 preview image

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2.5b(continued) To obtainEx= 0, we require the expression in the large brackets to be zero.Thisexpression simplifies to the following quadratic:0.48y2+ 13.92y+ 73.10 = 0which yields the two values:y=6.89,22.112.6.Two point charges of equal magnitudeqare positioned atz=±d/2.a) find the electric field everywhere on thezaxis: For a point charge at any location, we haveE=q(rr0)4⇡✏0|rr0|3In the case of two charges, we would therefore haveET=q1(rr01)4⇡✏0|rr01|3+q2(rr02)4⇡✏0|rr02|3(1)In the present case, we assignq1=q2=q, the observation point position vector asr=zaz, andthe charge position vectors asr01= (d/2)az, andr02=(d/2)azThereforerr01= [z(d/2)]az,rr02= [z+ (d/2)]az,then|rr1|3= [z(d/2)]3and|rr2|3= [z+ (d/2)]3Substitute these results into (1) to obtain:ET(z) =q4⇡✏01[z(d/2)]2+1[z+ (d/2)]2azV/m(2)b) find the electric field everywhere on thexaxis: We proceed as in parta, except that nowr=xax.Eq. (1) becomesET(x) =q4⇡✏0xax(d/2)az|xax(d/2)az|3+xax+ (d/2)az|xax+ (d/2)az|3(3)where|xax(d/2)az|=|xax+ (d/2)az|=⇥x2+ (d/2)2⇤1/2Therefore (3) becomesET(x) =2qxax4⇡✏0[x2+ (d/2)2]3/2c) repeat partsaandbif the charge atz=d/2 isqinstead of +q: The field along thezaxis isquickly found by changing the sign of the second term in (2):ET(z) =q4⇡✏01[z(d/2)]21[z+ (d/2)]2azV/mIn like manner, the field along thexaxis is found from (3) by again changing the sign of thesecond term. The result is2qdaz4⇡✏0[x2+ (d/2)2]3/214

Engineering Electromagnetics, 8th Edition Solution Manual - Page 16 preview image

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2.7.A 2μC point charge is located atA(4,3,5) in free space. FindE⇢,E, andEzatP(8,12,2). HaveEP= 2⇥1064⇡✏0RAP|RAP|3= 2⇥1064⇡✏04ax+ 9ay3az(106)1.5= 65.9ax+ 148.3ay49.4azThen, at pointP,⇢=p82+ 122= 14.4,= tan1(12/8) = 56.3, andz=z. Now,E⇢=Ep·a⇢= 65.9(ax·a⇢) + 148.3(ay·a⇢) = 65.9 cos(56.3) + 148.3 sin(56.3) = 159.7andE=Ep·a= 65.9(ax·a) + 148.3(ay·a) =65.9 sin(56.3) + 148.3 cos(56.3) = 27.4Finally,Ez=49.4 V/m2.8.A crude device for measuring charge consists of two small insulating spheres of radiusa, one of whichis fixed in position. The other is movable along thexaxis, and is subject to a restraining forcekx,wherekis a spring constant.The uncharged spheres are centered atx= 0 andx=d, the latterfixed. If the spheres are given equal and opposite charges ofQcoulombs:a) Obtain the expression by whichQmay be found as a function ofx: The spheres will attract, andso the movable sphere atx= 0 will move toward the other until the spring and Coulomb forcesbalance. This will occur at locationxfor the movable sphere. With equal and opposite forces,we haveQ24⇡✏0(dx)2=kxfrom whichQ= 2(dx)p⇡✏0kx.b) Determine the maximum charge that can be measured in terms of✏0,k, andd, and state theseparation of the spheres then: With increasing charge, the spheres move toward each other untilthey just touch atxmax=d2a.Using the partaresult, we find the maximum measurablecharge:Qmax= 4ap⇡✏0k(d2a).Presumably some form of stop mechanism is placed atx=xmaxto prevent the spheres from actually touching.c) What happens if a larger charge is applied? No further motion is possible, so nothing happens.2.9.A 100 nC point charge is located atA(1,1,3) in free space.a) Find the locus of all pointsP(x, y, z) at whichEx= 500 V/m: The total field atPwill be:EP= 100⇥1094⇡✏0RAP|RAP|3whereRAP= (x+1)ax+(y1)ay+(z3)az, and where|RAP|= [(x+1)2+(y1)2+(z3)2]1/2.Thexcomponent of the field will beEx= 100⇥1094⇡✏0(x+ 1)[(x+ 1)2+ (y1)2+ (z3)2]1.5= 500 V/mAnd so our condition becomes:(x+ 1) = 0.56 [(x+ 1)2+ (y1)2+ (z3)2]1.515

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