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Solution manual electromagnetics engineering 8th edition
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Solution manual electromagnetics engineering 8th edition
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By: knoowy00785 • 6 days ago
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CHAPTERff 11.1. Givenf thef vectorsf Mf =f −10axf +f4ayf −f8azf andf Nf =f 8axf +f7ayf −f2az,f find:
a) af unitf vectorf inf thef directionf off −Mf+f2N.
−Mf+f2Nf =f 10axf −f4ayf +f8azf +f16axf +f14ayf −f4azf =f (26,f10,f4)
Thus
(26,f10,f4)
a f =f =f (0.92,f0.36,f0.14)
|(26,f10,f4)|
b) thef magnitudef off 5axf+fNf−f3M:
(5,f0,f0)f+f(8,f7,f−2)f−f(−30,f12,f−24)f =f (43,f−5,f22),f andf |(43,f−5,f22)|f =f 48.6.
c)f |M||2N|(Mf+fN):
|(−10,f4,f−8)||(16,f14,f−4)|(−2,f11,f−10)f =f (13.4)(21.6)(−2,f11,f−10)
=f (−580.5,f3193,f−2902)
1.2. VectorfAfextendsffromftheforiginftof(1,2,3)fandfvectorfBffromftheforiginftof(2,3,-2).
a) Findf thef unitf vectorf inf thef directionf off (Af−fB):f First
Af−fBf =f (axf+f2ayf +f3az)f−f(2axf+f3ayf −f2az)f =f (−axf −fayf +f5az)
√
w√ hoseff magnitudeff isff |Aff−ffB|fff=fff[(−axf −fayf +f5az )f·f(−axf −fayf +f5az )]1/2ff =fff 1f+f1f+f25fff=
3f 3f =f5.20.f Thef unitf vectorf isf therefore
aABf =f(−axf−fayf +f5az)/5.20
b) findf thef unitf vectorf inf thef directionf off thef linef extendingf fromf thef originf tof thef midpointf off theflin
ef joiningf thef endsf off Af andf B:
Thef midpointf isf locatedf at
Pmpf =f [1f+f(2f−f1)/2,f 2f+f(3f−f2)/2,f 3f+f(−2f−f3)/2)]f =f (1.5,f2.5,f0.5)
Thef unitf vectorf isf then
f (1.5axf+f2.5ayf+f0.5az)f +f2.5a +f0.5a )/2.96
a = =f(1.5a
x y z
p
f
mp
(1.5)2f +f(2.5)2f +f(0.5)2
1.3. ThefvectorffromftheforiginftofthefpointfAfisfgivenfasf(6,f2,f4), — fand
− fthefunitfvectorfdirectedffromfthef or
iginf towardf pointf Bf isf (2,f 2,f1)/3.−
fIff pointsf Af andf Bf aref tenf unitsf apart,f findf thef coordinatesfoff pointf B.
Withf Af =f (6,f−2,f−4)f andf Bf =f 13fBf (2,f−2,f1),f wef usef thef factf thatf |Bf−fA|f =f 10,f or
2 2 1
|(6 − 3fB)ax − (2 − 3fB)ay − (4f+f 3fB)azf = | f 10
Expanding,4f obtain
36f−f8Bf+f fB2f+f4f−f 8fBf+f 4fB2f+f16f+f 8fBf+f 1fB2f=f100
9 3 9 √ 3 9
orf B2f−f8Bf−f44f=f0.f Thusf Bf=f 8±f 64−176
2 f =f11.75f (takingf positivef option)f andf so
2 2 1
B
=f (11.75)axf −f (11.75)ayf +f (11.75)azf =f 7.83axf −f7.83ayf +f3.92az
3f 3f 3f
1
,1.4. Affcircle,ffcenteredffatffthefforiginffwithffaffradiusffofff2ffunits,ffliesffinfftheffxyffplane.fffDetermine √theffunit
vectorf inf rectangularf componentsf thatf liesf inf thef xyf plane,f isf tangentf tof thef circlef atf ( −
3,f1,f0),f
andf isf inf thef generalf directionf off increasingf valuesf off y:
Affunitffvectorfftangentfftoffthisffcircleffinfftheffgeneralffincreasingf yf directionffisfftf =f −aφ .ff Its√xffand
yf componentsf aref txf =f −aφf ·faxf =f sinfφ,f andf tyf =f −aφf ·fayf =f −√
fcosfφ. f Atf the f point f (− 3,f1),
φf=f 150 ,ffandf sof tf =f sinf150 axf −fcosf150 ayff=f 0.5(axf +
◦ ◦ ◦
3ay ).
1.5. AfvectorffieldfisfspecifiedfasfGf=f24xyaxf+f12(x2f+f2)ayf+f18z2az.ffGivenfftwoffpoints,ffP(1,f2,f−1)fandf Q(
−2,f1,f3),f find:
a)f Gf atf P: f G(1,f2,f−1)f =f (48,f36,f18)
b) af unitf vectorf inf thef directionf off Gf atf Q:f G(−2,f1,f3)f =f (−48,f72,f162),f so
f (−48,f72,f162)f
a = =f (f —
0f 26
f .
f 0f 39f 0f 88)
,ff. ,f .
G
|(−48,f72,f162)|
c) af unitf vectorf directedf fromf Qf towardf P:
f Pf−fQff
(3 1f 4) f
a
QP = = ,√
f−f ,f
=f (0.59,f0.20,f−0.78)
|Pf −f Q| 26
d) thef equationf off thef surfacef onf whichf|G|f =f 60:f Wefwritef 60f =f |(24xy,f12(x2f +f 2),f18z2)|,forf10f =f|
(4xy,f2x2f+f4,f3z2)|,f sof thef equationf is
100f=f16x2y2f+f4x4f+f16x2f+f16f+f9z4
1.6. Findf thef acutef anglef betweenf thef twof vectorsf Af =f 2axf+fayf +f3azf andf Bf =f ax −
3ayf+f2azfbyfusingfthef definitionf of:
√ √
a) theffdotffproduct√ 2f −f3f +f6ff=ff5ff=ffABfcosfθ,ffwhereffAf =ff 22ff+f12ff+f32ff =ff 14,
:ff First,ffAf ·f Bff=f√
andf wherefBf= 12f +f32f +f22f = 14.f Thereforef cosfθf =f 5/14,f sof thatf θf =f 69.1◦.
b) thef crossf product:f Beginf with
ax ay az
A×B = Ø 2 1 3 Ø =f11 a xf− ayf− 7az
Øf 1 −3 2f Ø
√ ¢f √ √
and
find f thenf |A
f θf=fsin−
1 f× B
√|ff=f 11=f 69.1
°171/14
2ff+f12◦+f72ff=f 171.ffSof now,f withf |Af×fB|f =f ABfsinfθf=f 171,
ff
1.7. Givenf thef vectorf fieldf Ef =f 4zy2fcosf2xaxf +f2zyfsinf2xayf +fy2fsinf2xazf forf thef regionf xf,ff y|f ,ff and
|f |f |ff z |f |
lessf thanf 2,f find:
a) thef surfacesf onf whichf Eyf=f 0.f Withf Eyf =f 2zyfsinf2xf =f 0,f thef surfacesf aref 1)f thef planef zf =f 0,fwithf|
x|f<f2,f |y|f<f2;f 2)fthefplanefyf=f0,f withf|x|f<f2,f |z|f<f2;f 3)f thefplanefxf=f0,f withf|y|f<f2,
|z|f<f2;f 4)f thef planef xf=fπ/2,f withf |y|f <f2,f |z|f <f2.
b) thef regionf inf whichf Eyff=f Ez:f Thisf occursf whenf 2zyfsinf2xf =f y2fsinf2x,f orf onf thef planef 2zf =f y,fwithf
|x|f <f2,f |y|f <f2,f |z|f <f1.
c) thef regionf inf whichf Ef =f 0:f Wef wouldf havef Exff=f Eyff =f Ezff =f 0,f orf zy2fcosf2xf =f zyfsinf2xf =
y2fsinf2xf =f 0.f Thisf conditionf isf metf onf thef planef yf =f 0,f withf |x|f <f 2,f |z|f <f 2.
2
, 1.8. Demonstratefthefambiguityfthatfresultsfwhenfthefcrossfproductfisfusedftof findfthef anglefbetweenftwofv
ectorsfbyffindingfthefanglefbetweenfAf=f3axf2ayf+f4a— zfandfBf=f2axf+fayf2az.fDoesfthisfambiguity
− f existf
whenf thef dotf productf isf used?
Wef usef thef relationf Af ×f Bf =f |A||B|fsinfθn.f Withf thef givenf vectorsf wef find
∑f ∏
√f 2ayf +fazf √ √f
Af×fBf=f14ayf+f7azf =f7f 5 √ 4f+f1f+f4f sinfθfn
| 5 }= 9f+f4f+f16
{z
±n
wheref nf isf identifiedf asf shown;f wef seef thatf nf canf bef positivef orf negative,f asf sinfθf canf be
positivef orf negative.fff Thisf apparentf signf ambiguityf isf notf thef realf problem,f however,f asf we
reallyffwa√ ntfffth√
effm√
agnitudeffoffftheffangleffanyway.fffChoosingfftheffpositiveffsign,ffweffareffleftffwith
sinfθf =f 7f f 5/(f f 29f f 9)f =f 0.969.f Twoffvaluesf off θf (75.7◦f andff104.3◦)f satisfyf thisffequation,f and
hencefthefrealfambiguity.
√
Inffusingfftheffd√ otffproduct,ffwefffindffAf ·f Bff=ff6f −f 2f −f 8ff=ff−4ff=ff|A||B|fcosfθff=ff3 29fcosfθ,ffor
cosfθf =f −4/(3f f 29)f =f −0.248f ⇒f θf =f −75.7◦.f Again,f thef minusf signffisf notf important,f asf we
caref onlyf aboutf thef anglef magnitude.f Thef mainf pointf isf thatf onlyf onef θf valuef resultsf whenfusin
gf thef dotf product,f sof nof ambiguity.
1.9. Af fieldf isf givenf as
Gf =
25f(x (xax +fyay)
2f +fy2)
Find:
a) afunitfvectorfinfthefdirectionfoffGfatfP(3,f4,f−2):fHavefGpf=f25/(9f+f16)f×(3,f4,f0)f=f3axf+f4ay,fandf |
Gp|f =f5.f Thusf aGf =f(0.6,f0.8,f0).
b) thef anglefbetweenf Gfandfaxf atfP: f f Thef anglef isf foundf throughf aGf ·f axf =f cosfθ.f Sof cosfθf =f(0.6,f0.8
,f0)f·f(1,f0,f0)f=f0.6.f Thusf θf=f53◦.
c) thef valuef off thef followingf doublef integralf onf thef planef yf=f7:
Zf 4fZf 2f
G · aydzdx
0 0
Zf 4f Zf 2ff Zf 4f Zf 2ff Zf 4ff
25 ffffff 25 ffff 350f
(fxafffx+ y fy)f ·a y dzdx =f × 7 dzdx = dx
ffffff
a
0 x +f49f 0 x +f49f
f
0 0 x +fy
2f 2
0
2f 2f
∑ µf ∂f ∏f
1f 4f
=f 350f×ff tan− f 1
−f0 =f 26
7 7
1.10. Byfexpressingfdiagonalsfasfvectorsfandfusingfthefdefinitionfoffthefdotfproduct,ffindfthefsmallerfanglefbet
weenfanyftwofdiagonalsfoffafcube,fwherefeachfdiagonalfconnectsfdiametricallyfoppositefcorners,fandf pa
ssesf throughf thef centerf off thef cube:
Assumingffaff sidefflength,ff b,ff twoff diagonalffvectorsffwouldffbeff Aff =f√ b(axff+ √fayff+f az )ff andffBff =
b(axf −fayf +faz ).ffNowf usef Af·fBf =f |A||B|fcosfθ,f orf b2 (1f−f1f+f1)f =f (ffff3b)(ffff3b)fcosfθff ⇒fffcosfθf =
1/3f ⇒f θf=f70.53◦.f Thisf resultf (inf magnitude)f isf thef samef forf anyf twof diagonalf vectors.
3
, 1.11.f Givenf thef pointsf M(0.1,f−0.2,f−0.1),f N(−0.2,f0.1,f0.3),f andf P(0.4,f0,f0.1),f find:
a)f thef vectorf RMNf:f RMNf =f (−0.2,f0.1,f0.3)f−f(0.1,f−0.2,f−0.1)f =f (−0.3,f0.3,f0.4).
b)ff thef dotf productf RMNf ·fRMPf:ff RMPf =f(0.4,f0,f0.1)f−f(0.1,f−0.2,f−0.1)f =f(0.3,f0.2,f0.2).ffRMNf ·
RMPf =f (−0.3,f0.3,f0.4)f·f(0.3,f0.2,f0.2)f =f −0.09f+f0.06f+f0.08f =f 0.05.
c) thef scalarf projectionf off RMNff onf RMPf:
(0.3,f0.2,f0.2) 0.05
RMNf ·faRMPf =f (−0.3,f0.3,f0.4)f·f √ =f √ =f 0.12
0.09f+f0.04f+f0.04f 0.17f
d) thef anglef betweenf RMNf andf RMPf: ∂f
µ ∂ µ
−1f fRMNf ·f RMPff 0.05 =f78◦
θ =f cos −1f
M
|RMNf ||RMPf | √ √
=f0.34 0.17
cos
1.12. Writef anf expressionf inf rectangularf componentsf forf thef vectorf thatf extendsf fromf (x1,fy1,fz1)f tof(x2,fy2,f
z2)f andf determinef thef magnitudef off thisf vector.
Thef twof pointsf canf bef writtenf asf vectorsf fromf thef origin:
A1f =f x1axf +f y1ayf +f z1az and
A2f =f x2axf +f y2ayf +f z2azfThefde
siredfvectorfwillfnowfbefthefdifference:
A12f=fA2f−fA1f=f(x2f−fx1)axf+f(y2f−fy1)ayf+f(z2f−fz1)az
whosef magnitudef is
p £ §1/2
|Afff |f =f
12 A
12 ·12fA =f 2(xf −1fxf )2f+f(y2f −fyf 1)2f+f(zf −2fzf )2 1
1.13. a)f Findf thef vectorf componentf off Ff =f (10,f−6,f5)f thatf isf parallelf tof Gf =f (0.1,f0.2,f0.3):
Ff ·f G (10,f−6,f5)f ·f (0.1,f0.2,f0.3)
F =f G f =f (0.1,f0.2,f0.3)f =f (0.93,f1.86,f2.79)
||G
|G|2 0.01f+f0.04f +f0.09
b) Findf thef vectorf componentf off Ff thatf isf perpendicularf tof G:
FpGf =f Ff −f F||Gf =f (10,f−6,f5)f −f (0.93,f1.86,f2.79)f =f (9.07,f−7.86,f2.21)
c) Findfthef vectorf componentf off Gfthatfisf perpendicularftof F:
Gf·fF 1.3
G =fGf−fG =fGf−f Ff=f(0.1,f0.2,f0.3)f− (10,f−6,f5)f =f (0.02,f0.25,f0.26)
pF ||F
|F|2 100f+f36f+f25
4
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