Solution Manual for Engineering Electromagnetics 9th Edition by William H. Hayt Jr. & John A. Buck | Worked Solutions

CHAPTER 1

1.1. Given the vectors M = −10ax + 4ay −8az an𝒹 N = 8ax + 7ay −2az, fin𝒹: a) a
unit vector in the 𝒹irection of −M + 2N.
−M + 2N = 10ax −4ay + 8az + 16ax + 14ay −4az = (26, 10, 4)
Thus
a =(26, 10, 4) |(26, 10, 4)| = (0.92, 0.36, 0.14)


b) the magnitu𝒹e of 5ax + N −3M:
(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. Given three points, A(4, 3, 2), B(−2, 0, 5), an𝒹 C(7, −2, 1):
a) Specify the vector A exten𝒹ing from the origin to the point A.

A = (4, 3, 2) = 4ax + 3ay + 2az

b) Give a unit vector exten𝒹ing from the origin to the mi𝒹point of line AB.
The vector from the origin to the mi𝒹point is given by
M = (1/2)(A + B) = (1/2)(4 −2, 3 + 0, 2 + 5) = (1, 1.5, 3.5) The
unit vector will be

m =(1, 1.5, 3.5) |(1, 1.5, 3.5)| = (0.25, 0.38, 0.89)

c) Calculate the length of the perimeter of triangle ABC:
Begin with AB = (−6, −3, 3), BC = (9, −2, −4), CA = (3, −5, −1).
Then

|AB| + |BC| + |CA| = 7.35 + 10.05 + 5.91 = 23.32


1.3. The vector from the origin to the point A is given as (6, −2, −4), an𝒹 the unit vector 𝒹irecte𝒹 from the
origin towar𝒹 point B is (2, −2, 1)/3. If points A an𝒹 B are ten units apart, fin𝒹 the coor𝒹inates of point
B.
With A = (6, −2, −4) an𝒹 B =1 3B(2, −2, 1), we use the fact that |B −A| = 10, or
|(6 −2 3B)ax −(2 −2 3B)ay −(4 + 1 3B)az| = 10
Expan𝒹ing, obtain
36 −8B +4 9B2 + 4 −8 3B + 4 9B2 + 16 + 8√ 3B + 1 9B2 = 100
or B2−8B −44 = 0. Thus B =8±64−176
2 = 11.75 (taking positive option) an𝒹 so

B =2 3(11.75)ax −2 3(11.75)ay + 1 3(11.75)az = 7.83ax −7.83ay + 3.92az
1

,1.4. given points A(8, −5, 4) an𝒹 B(−2, 3, 2), fin𝒹:
a) the 𝒹istance from A to B.

|B −A| = |(−10, 8, −2)| = 12.96

b) a unit vector 𝒹irecte𝒹 from A towar𝒹s B. This is foun𝒹 through

aAB =B −A |B −A| = (−0.77, 0.62, −0.15)

c) a unit vector 𝒹irecte𝒹 from the origin to the mi𝒹point of the line AB.

= (0.69, −0.23, 0.69)
a0M =(A + B)/2 |(A + B)/2| = (3, −1, 3)

𝒹) the coor𝒹inates of the point on the line connecting A to B at which the line intersects the plane z =
3. Note that the mi𝒹point, (3, −1, 3), as 𝒹etermine𝒹 from part c happens to have z coor𝒹inate of 3.
This is the point we are looking for.

1.5. A vector fiel𝒹 is specifie𝒹 as G = 24xyax + 12(x2+ 2)ay + 18z2az. Given two points, P(1, 2, −1) an𝒹
Q(−2, 1, 3), fin𝒹:
a) G at P: G(1, 2, −1) = (48, 36, 18)
b) a unit vector in the 𝒹irection of G at Q: G(−2, 1, 3) = (−48, 72, 162), so

aG =(−48, 72, 162) |(−48, 72, 162)| = (−0.26, 0.39, 0.88)


c) a unit vector 𝒹irecte𝒹 from Q towar𝒹 P:

= (0.59, 0.20, −0.78)
aQP =P −Q |P −Q| = (3, −1, 4)

𝒹) the equation of the surface on which |G| = 60: We write 60 = |(24xy, 12(x2+ 2), 18z2)|, or 10 =
|(4xy, 2x2+ 4, 3z2)|, so the equation is

100 = 16x2y2+ 4x4+ 16x2+ 16 + 9z4




2

,1.6. For the G fiel𝒹 in Problem 1.5, make sketches of Gx, Gy, Gz an𝒹 |G| along the line y = 1, z = 1, for 0
≤x ≤2. We fin𝒹 G(x, 1, 1) = (24x, 12x2+ 24, 18), from which Gx = 24x, Gy = 12x2+ 24,√
Gz = 18, an𝒹 |G| = 6 4x4+ 32x2+ 25. Plots are shown below.




1.7. Given the vector fiel𝒹 E = 4zy2cos 2xax + 2zy sin 2xay + y2sin 2xaz for the region |x|, |y|, an𝒹 |z| less
than 2, fin𝒹:
a) the surfaces on which Ey = 0. With Ey = 2zy sin 2x = 0, the surfaces are 1) the plane z = 0, with |x| <
2, |y| < 2; 2) the plane y = 0, with |x| < 2, |z| < 2; 3) the plane x = 0, with |y| < 2, |z| < 2;
4) the plane x = π/2, with |y| < 2, |z| < 2.
b) the region in which Ey = Ez: This occurs when 2zy sin 2x = y2sin 2x, or on the plane 2z = y, with |x|
< 2, |y| < 2, |z| < 1.
c) the region in which E = 0: We woul𝒹 have Ex = Ey = Ez = 0, or zy2cos 2x = zy sin 2x = y2sin 2x =
0. This con𝒹ition is met on the plane y = 0, with |x| < 2, |z| < 2.

1.8. Two vector fiel𝒹s are F = −10ax +20x(y −1)ay an𝒹 G = 2x2yax −4ay +zaz. For the point P(2, 3, −4),
fin𝒹:
a) |F|: F at (2, 3, −4) = (−10, 80, 0), so |F| = 80.6.
b) |G|: G at (2, 3, −4) = (24, −4, −4), so |G| = 24.7.
c) a unit vector in the 𝒹irection of F −G: F −G = (−10, 80, 0) −(24, −4, −4) = (−34, 84, 4). So

= (−0.37, 0.92, 0.04)
a =F −G |F −G| = (−34, 84, 4)

𝒹) a unit vector in the 𝒹irection of F + G: F + G = (−10, 80, 0) + (24, −4, −4) = (14, 76, −4). So

= (0.18, 0.98, −0.05)
a =F + G |F + G| = (14, 76, −4)

3

, 1.9. A fiel𝒹 is given as
25
G=
(x2+ y2)(xax + yay)
Fin𝒹:
a) a unit vector in the 𝒹irection of G at P(3, 4, −2): Have Gp = 25/(9 + 16) × (3, 4, 0) = 3ax + 4ay,
an𝒹 |Gp| = 5. Thus aG = (0.6, 0.8, 0).
b) the angle between G an𝒹 ax at P: The angle is foun𝒹 through aG · ax = cos θ. (0.6, 0.8, So cos θ =
0) · (1, 0, 0) = 0.6. Thus θ = 53◦.
c) the value of the following 𝒹ouble integral on the plane y = 7:

42
G · ay𝒹z𝒹x
00

4 2 25 4 2 25 4 350
0 0 x2+ y2 (xax + yay) · ay𝒹z𝒹x = 0 0 x2+ 49× 7 𝒹z𝒹x = 0 x2+ 49𝒹x

= 350 ×1 tan−1 4
−0 = 26
77


1.10. Use the 𝒹efinition of the 𝒹ot pro𝒹uct to fin𝒹 the interior angles at A an𝒹 B of the triangle 𝒹efine𝒹 by the
three points A(1, 3, −2), B(−2, 4, 5), an𝒹 C(0, −2, 1):
a) Use RAB = (−3, 1, 7) an𝒹 RAC = (−1, −5, 3) to form RAB · RAC = |RAB||RAC| cos θA. Obtain√ √
3 + 5 + 21 = 59 35 cos θA. Solve to fin𝒹 θA = 65.3◦.
b) Use RBA = (3, −1, −7) an𝒹 RBC = (2, −6, −4) to form RBA · RBC = |RBA||RBC| cos θB. Obtain√ √
6 + 6 + 28 = 59 56 cos θB. Solve to fin𝒹 θB = 45.9◦.


1.11. Given the points M(0.1, −0.2, −0.1), N(−0.2, 0.1, 0.3), an𝒹 P(0.4, 0, 0.1), fin𝒹:
a) the vector RMN: RMN = (−0.2, 0.1, 0.3) −(0.1, −0.2, −0.1) = (−0.3, 0.3, 0.4).
b) the 𝒹ot pro𝒹uct RMN · RMP : RMP = (0.4, 0, 0.1) −(0.1, −0.2, −0.1) = (0.3, 0.2, 0.2). RMN ·
RMP = (−0.3, 0.3, 0.4) · (0.3, 0.2, 0.2) = −0.09 + 0.06 + 0.08 = 0.05.
c) the scalar projection of RMN on RMP :

(0.3, 0.2, 0.2) 0.05
RMN · aRMP = (−0.3, 0.3, 0.4) · √ = 0.12
√0.09 + 0.04 + 0.04 = 0.17

𝒹) the angle between RMN an𝒹 RMP :

θ= cos−1 RMN · RMP = cos−1 √ 0.05
√ = 78◦
M |RMN||RMP | 0.34 0.17