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SOLUTION MANUAL ENGINEERING ELECTROMAGNETICS BY WILLIAM H. HAYT 8TH
EDITION COMPLETE CHAPTERS
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
(26, 10, 4)
a= = (0.92, 0.36, 0.14)
|(26, 10, 4)|
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
(1, 1.5, 3.5)
m= = (0.25, 0.38, 0.89)
|(1, 1.5, 3.5)|
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 = 13 B(2, −2, 1), we use the fact that |B − A| = 10, or
|(6 − 23 B)ax − (2 − 23 B)ay − (4 + 13 B)az | = 10
Expanding, obtain
36 − 8B + 49 B 2 + 4 − 38 B + 49 B 2 + 16 + 83 B + 19 B 2 = 100
√
8± 64−176
or B 2 − 8B − 44 = 0. Thus B = 2 = 11.75 (taking positive option) and so
2 2 1
B= (11.75)ax − (11.75)ay + (11.75)az = 7.83ax − 7.83ay + 3.92az
3 3 3
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,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
B−A
aAB = = (−0.77, 0.62, −0.15)
|B − A|
c) a unit vector directed from the origin to the midpoint of the line AB.
(A + B)/2 (3, −1, 3)
a0M = = √ = (0.69, −0.23, 0.69)
|(A + B)/2| 19
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(x 2 + 2)ay + 18z2 az . 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
(−48, 72, 162)
aG = = (−0.26, 0.39, 0.88)
|(−48, 72, 162)|
c) a unit vector directed from Q toward P :
P−Q (3, −1, 4)
aQP = = √ = (0.59, 0.20, −0.78)
|P − Q| 26
d) the equation of the surface on which |G| = 60: We write 60 = |(24xy, 12(x 2 + 2), 18z2 )|, or
10 = |(4xy, 2x 2 + 4, 3z2 )|, so the equation is
100 = 16x 2 y 2 + 4x 4 + 16x 2 + 16 + 9z4
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,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, 12x 2 + 24, 18), from which Gx = 24x, Gy = 12x 2 + 24,
Gz = 18, and |G| = 6 4x 4 + 32x 2 + 25. Plots are shown below.
1.7. Given the vector field E = 4zy 2 cos 2xax + 2zy sin 2xay + y 2 sin 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 = y 2 sin 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 zy 2 cos 2x = zy sin 2x =
y 2 sin 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 = 2x 2 yax − 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
F−G (−34, 84, 4)
a= = = (−0.37, 0.92, 0.04)
|F − G| 90.7
d) a unit vector in the direction of F + G: F + G = (−10, 80, 0) + (24, −4, −4) = (14, 76, −4). So
F+G (14, 76, −4)
a= = = (0.18, 0.98, −0.05)
|F + G| 77.4
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, 1.9. A field is given as
25
G= (xax + yay )
(x 2 + y2)
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 . So cos θ =
(0.6, 0.8, 0) · (1, 0, 0) = 0.6. Thus θ = 53◦ .
c) the value of the following double integral on the plane y = 7:
4 2
G · ay dzdx
0 0
4 2 25
4 2
25
4
350
(xax + ya y ) · a y dzdx = × 7 dzdx = dx
0 0 x2 + y2 0 0 x 2 + 49
0 x 2 + 49
1 −1 4
= 350 × tan − 0 = 26
7 7
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 :
−1 RMN · RMP −1 0.05
θM = cos = cos √ √ = 78◦
|RMN ||RMP | 0.34 0.17
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