Engineering_Electromagnetics___7th_Edition___William_H._Hayt___Solution_Manua

Your text here 1 SOLUTION MANUAL ENGINEERING ELECTROMAGNETICS 7TH EDITION BY WILLIAM
H. HAYT WITH 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. The three vertices of a triangle are located at A(−1, 2, 5), B(−4, −2, −3), and C(1, 3, −2).
a) Find the length of the perimeter of the triangle: Begin with AB = (−3, −4, −8),√ BC = (5, 5, 1),
and
√ CA = (−2, −1,
√ 7). Then the perimeter will be
= |AB| + |BC| + |CA| = 9 + 16 + 64 +
25 + 25 + 1 + 4 + 1 + 49 = 23.9.
b) Find a unit vector that is directed from the midpoint of the side AB to the midpoint of side
BC: The vector from the origin to the midpoint of AB is MAB = 12 (A + B) = 12 (−5ax + 2az ).
The vector from the origin to the midpoint of BC is MBC = 12 (B + C) = 12 (−3ax + ay − 5az ).
The vector from midpoint to midpoint is now MAB − MBC = 12 (−2ax − ay + 7az ). The unit
vector is therefore

MAB − MBC (−2ax − ay + 7az )
aM M = = = −0.27ax − 0.14ay + 0.95az
|MAB − MBC | 7.35

where factors of 1/2 have cancelled.
c) Show that this unit vector multiplied by a scalar is equal to the vector from A to C and that the
unit vector is therefore parallel to AC. First we find AC = 2ax + ay − 7az , which we recognize
as −7.35 aM M . The vectors are thus parallel (but oppositely-directed).

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 − 83 B + 49 B 2 + 16 + 83 B + 19 B 2 = 100
√
or B 2 − 8B − 44 = 0. Thus B = 8± 64−176
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
1

,1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determine
√ the unit
vector in rectangular components that lies in the xy plane, is tangent to the circle at ( 3, 1, 0), and
is in the general direction of increasing values of y:
A unit vector tangent to this circle in the general increasing y direction is t = √
aφ . Its x and y
components are tx = aφ · ax = − sin φ, and ty = √ aφ · ay = cos φ. At the point ( 3, 1), φ = 30◦ ,
and so t = − sin 30◦ ax + cos 30◦ ay = 0.5(−ax + 3ay ).

1.5. A vector field is specified as G = 24xyax + 12(x2 + 2)ay + 18z 2 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(x2 + 2), 18z 2 )|, or
10 = |(4xy, 2x2 + 4, 3z 2 )|, so the equation is

100 = 16x2 y 2 + 4x4 + 16x2 + 16 + 9z 4


1.6. If a is a unit vector in a given direction, B is a scalar constant, and r = xax + yay + zaz , describe
the surface r · a = B. What is the relation between the the unit vector a and the scalar B to this
surface? (HINT: Consider first a simple example with a = ax and B = 1, and then consider any a
and B.):
We could consider a general unit vector, a = A1 ax + A2 ay + A3 az , where A21 + A22 + A23 = 1.
Then r · a = A1 x + A2 y + A3 z = f (x, y, z) = B. This is the equation of a planar surface, where
f = B. The relation of a to the surface becomes clear in the special case in which a = ax . We
obtain r · a = f (x) = x = B, where it is evident that a is a unit normal vector to the surface
(as a look ahead (Chapter 4), note that taking the gradient of f gives a).

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.



2

, 1.8. Demonstrate the ambiguity that results when the cross product is used to find the angle between
two vectors by finding the angle between A = 3ax − 2ay + 4az and B = 2ax + ay − 2az . Does this
ambiguity exist when the dot product is used?
We use the relation A × B = |A||B| sin θn. With the given vectors we find


√ 2ay + az √ √
A × B = 14ay + 7az = 7 5 √ = 9 + 4 + 16 4 + 1 + 4 sin θ n
5
  
±n


where n is identified as shown; we see that n can be positive or negative, as sin θ can be
positive or negative. This apparent sign ambiguity is not the real problem, however, as we
really want
√ the√ magnitude
√ of the angle anyway. Choosing the positive sign, we are left with
sin θ = 7 5/( 29 9) = 0.969. Two values of θ (75.7◦ and 104.3◦ ) satisfy this equation, and
hence the real ambiguity.
√
In using the dot
√ product, we find A · B = 6 − 2 − 8 = −4 = |A||B| cos θ = 3 29 cos θ, or
cos θ = −4/(3 29) = −0.248 ⇒ θ = −75.7◦ . Again, the minus sign is not important, as we
care only about the angle magnitude. The main point is that only one θ value results when
using the dot product, so no ambiguity.

1.9. A field is given as
25
G= (xax + yay )
(x2 + 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  4 2  4
25 25 350
2 2
(xax + yay ) · ay dzdx = 2
× 7 dzdx = 2
dx
0 0 x +y 0 0 x + 49 0 x + 49

 
1 −1 4
= 350 × tan − 0 = 26
7 7


1.10. By expressing diagonals as vectors and using the definition of the dot product, find the smaller angle
between 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 be A = √ b(ax +
√ ay + az ) and B =
b(ax − ay + az ). Now use A · B = |A||B| cos θ, or b (1 − 1 + 1) = ( 3b)( 3b) cos θ ⇒ cos θ =
2

1/3 ⇒ θ = 70.53◦ . This result (in magnitude) is the same for any two diagonal vectors.




3

, 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 RM N : RM N = (−0.2, 0.1, 0.3) − (0.1, −0.2, −0.1) = (−0.3, 0.3, 0.4).
b) the dot product RM 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 of RM N on RM P :

(0.3, 0.2, 0.2) 0.05
RM N · aRM P = (−0.3, 0.3, 0.4) · √ =√ = 0.12
0.09 + 0.04 + 0.04 0.17

d) the angle between RM N and RM P :
 
−1 RM N · RM P −1 0.05
θM = cos = cos √ √ = 78◦
|RM N ||RM P | 0.34 0.17


1.12. Show that the vector fields A = ρ cos φ aρ + ρ sin φ aφ + ρ az and B = ρ cos φ aρ + ρ sin φ aφ − ρ az
are everywhere perpendicular to each other:
We find A · B = ρ2 (sin2 φ + cos2 φ) − ρ2 = 0 = |A||B| cos θ. Therefore cos θ = 0 or θ = 90◦ .

1.13. a) Find the vector component of F = (10, −6, 5) that is parallel to G = (0.1, 0.2, 0.3):

F·G (10, −6, 5) · (0.1, 0.2, 0.3)
F||G = G= (0.1, 0.2, 0.3) = (0.93, 1.86, 2.79)
|G|2 0.01 + 0.04 + 0.09

b) Find the vector component of F that is perpendicular to G:

FpG = F − F||G = (10, −6, 5) − (0.93, 1.86, 2.79) = (9.07, −7.86, 2.21)

c) Find the vector component of G that is perpendicular to F:

G·F 1.3
GpF = G − G||F = G − F = (0.1, 0.2, 0.3) − (10, −6, 5) = (0.02, 0.25, 0.26)
|F|2 100 + 36 + 25


1.14. Show that the vector fields A = ar (sin 2θ)/r2 +2aθ (sin θ)/r2 and B = r cos θ ar +r aθ are everywhere
parallel to each other:
Using the definition of the cross product, we find

sin 2θ 2 sin θ cos θ
A×B= − aφ = 0 = |A||B| sin θ n
r r

Identify n = aφ , and so sin θ = 0, and therefore θ = 0 (they’re parallel).




4