solutions manual for Engineering Electromagnetics, 6th Edition by William H. Hayt and John A. Buck.

Solutions of engineering electromagne
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tics 6th edition william
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h. hayt, john a. buck.pdf
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,CHAPTERc1

1.1. cGivencthecvectorscMc =c−10axc +c4ayc −c8azc andcNc =c8axc +c7ayc −c2az,cfind:
a) acunitcvectorcincthecdirectioncofc−Mc+c2N.
−Mc+c2Nc=c10axc−c4ayc+c8azc+c16axc+c14ayc−c4azc =c(26,c10,c4)
Thus
(26,c10,c4)
ac=c =c(0.92,c0.36,c0.14)
|(26,c10,c4)|c

b) thecmagnitudecofc5axc +cNc−c3M:
(5,c0,c0)c+c(8,c7,c−2)c−c(−30,c12,c−24)c =c (43,c−5,c22),c andc |(43,c−5,c22)|= c48.6.
c) |M||2N|(Mc +cN):
|(−10,c4,c−8)||(16,c14,c−4)|(−2,c11,c−10)c =c (13.4)(21.6)(−2,c11,c−10)
=c (−580.5,c3193,c−2902)

1.2. Givenc threec points,c A(4,c3,c2),c B(−2,c0,c5),c andc C(7,c−2,c1):
a) SpecifycthecvectorcAcextendingcfromcthecoriginctocthecpointcA.

Ac=c(4,c3,c2)c=c4axc +c3ayc +c2az

b) GivecacunitcvectorcextendingcfromcthecoriginctocthecmidpointcofclinecAB.c
Thecvectorcfromcthecoriginctocthecmidpointciscgivencby
M = (1/2)(A + B) = (1/2)(4 − 2,c3 + 0,c2 + 5) = (1,c1.5,c3.5)
Thecunitcvectorcwillcbe
(1,c1.5,c 3.5)
mc=c =c(0.25,c0.38,c0.89)
|(1,c1.5,c3.5)|c
c) CalculatectheclengthcofcthecperimetercofctrianglecABC:
BegincwithcABc =c(−6,c−3,c3),cBCc =c(9,c−2,c−4),cCAc =c(3,c−5,c−1).
Then

|AB|+c|BC|+c|CA|=c7.35c+c10.05c+c5.91c =c23.32

1.3. ThecvectorcfromcthecoriginctocthecpointcAciscgivencasc(6,c 2,c−4),c− andcthecunitcvectorcdirectedcfromcthecorigin
—
ctowardcpointcBc isc(2,c c 2,c1)/3. c If cpointscA c and cB c arectencunitscapart,cfindctheccoordinatescofcpointc B.


With2cAc =c(6,c−2,c−4)
2
candcBc =c B(2,c−2,c1),cwecusecthecfactcthatc|Bc−cA|= c10,cor
1c
3 1
|(6 − 3cB)ax − (2 − 3cB)ay − (4 + 3cB)az | = 10
Expanding,c4obtain
36c−c8Bc +c cB2c +c4c−c 8cBc +c 4cB2c +c16c+c 8cBc +c 1cB2c =c100
9 3 9 c √ 3 9
orcB −c8Bc −c44c=c0.c ThuscBc =c
2 c 8±c 64−176c
2 =c11.75c(takingcpositivecoption)candcso
2 2 1
Bc =c (11.75)axc −c (11.75)ayc +c (11.75)azc =c7.83axc −c7.83ayc +c3.92az
3c 3c 3c
1

,1.4. givencpointscA(8,c−5,c4)candcB(−2,c3,c2),cfind:
a) thecdistancecfromcActocB.

|Bc−cA|= c|(−10,c8,c−2)|= c12.96

b) acunitcvectorcdirectedcfromcActowardscB.c Thisciscfoundcthrough
c Bc−cAc
aAB = (c c 0.77,c0.62,cc c 0.15)
= − −
|Bc−cA|
c) acunitcvectorcdirectedcfromcthecoriginctocthecmidpointcofctheclinecAB.
c (Ac+cB)/2cc (3,c−1,c3)c
a =c =c =c(0.69,c−0.23,c0.69)
0M √
|(Ac+cB)/2| 19

=cN
d) theccoordinatescofcthecpointconctheclinecconnectingcActocBcatcwhichctheclinecintersectscthecplaneczc c 3.
otecthatcthecmidpoint,c(3,c c 1,—c3),cascdeterminedcfromcpartccchappensctochaveczccoordinatecofc3.c Thiscisct
hecpointcwecareclookingcfor.

1.5. AcvectorcfieldciscspecifiedcascGc =c24xyaxc +c12(x2c +c2)ayc +c18z2az.c Givenctwocpoints,cPc(1,c2,c−1)candcQ(
−2,c1,c3),cfind:
a) GcatcPc:c G(1,c2,c−1)c=c(48,c36,c18)
b) acunitcvectorcincthecdirectioncofcGcatcQ:c G(−2,c1,c3)c=c(−48,c72,c162),cso

c (−48,c72,c162)
aG = =c (−0.26,c0.39,c0.88)
|(−48,c72,c162)|

c) acunitcvectorcdirectedcfromcQctowardcPc:
cPc−cQcc (3,c−1,c4)c
a =c =c =c(0.59,c0.20,c−0.78)
QP √
|Pc−cQ| 26

d) thec equationc ofc thec surfacec onc whichc |G|c =c 60:c Wec writec 60c =c |(24xy,c12(x2c +c2),c18z2)|,c orc10
c=c|(4xy,c2x
2c +c4,c3z2)|,csocthecequationcis


100c =c16x2y2c +c4x4c +c16x2c +c16c+c9z4




2

, 1.6. Forcthec Gc fieldc inc Problemc 1.5,c makec sketchesc ofc Gxc,c Gyc,c Gzc andc |G|calongc thec linec yc =c 1,c zc =c 1,c forc0c
≤c xc ≤c 2.c Wec findc Gc√(x,c1,c1)c =c (24x,c12x2c +c24,c18),c fromc whichc Gxc =c 24x,c Gyc =c 12x2c +c24,cGzc
=c18,candc|G|= c6c 4xc +c32xc +c25.c Plotscarecshowncbelow.
4 2




1.7. GivencthecvectorcfieldcE = 4zy2ccosc2xax + 2zycsinc2xay + | | | | ||
y2csinc2xazc forcthecregionc xc ,c yc ,candc zc lesscthanc2,cfind:
a) thecsurfacesconcwhichcEyc =c0.c WithcEyc =c2zycsinc2xc =c0,cthecsurfacescarec1)cthecplaneczc =c0,cwith
|x|c<c2,c|y|c<c2;c 2)cthecplanecyc =c0,cwithc|x|c<c 2,c|z|c<c2;c 3)cthecplanecxc =c0,cwithc|y|c<c2,c|z|c<c 2;
4)cthecplanecxc =cπ/2,cwithc|y|c<c2,c|z|c<c2.
b) thecregioncincwhichcEyc =cEz:c Thiscoccurscwhenc2zycsinc2xc =cy2csinc2x,corconcthecplanec2zc =cy,cwith
|x|c<c 2,c|y|c<c 2,c|z|c<c 1.
c) thecregionc inc whichc Ec =c 0:c Wec wouldc havec Exc =c Eyc =c Ezc =c 0,c orc zy2ccosc2xc =c zycsinc2xc =
y2csinc2xc =c0.c Thiscconditionciscmetconcthecplanecyc =c0,cwithc|x|c<c 2,c|z|c<c 2.

1.8. TwocvectorcfieldscarecFc =c−10axc+c20x(yc−c1)ayc andcGc =c2x2yaxc−c4ayc +czaz.c ForcthecpointcPc(2,c3,c−4),
find:
a) |F|:c Fcatc(2,c3,c−4)c =c(−10,c80,c0),csoc|F|= c80.6.
b) |G|:c Gcatc(2,c3,c−4)c =c(24,c−4,c−4),csoc|G|= c24.7.
c) acunitcvectorcincthecdirectioncofcFc−cG:c Fc−cGc =c(−10,c80,c0)c−c(24,c−4,c−4)c =c(−34,c84,c4).c So
cFc−cGcc (−34,c84,c4)c
ac=c =c =c(−0.37,c0.92,c0.04)
|Fc−cG| 90.7

d) acunitcvectorcincthecdirectioncofcFc+cG:c Fc+cGc =c(−10,c80,c0)c+c(24,c−4,c−4)c =c(14,c76,c−4).c So
cFc+cGcc (14,c76,c−4)c
ac=c =c =c(0.18,c0.98,c−0.05)
|Fc+cG| 77.4

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