Engineering Electromagnetics – Solutions Manual (William H. Hayt & John A. Buck) | Complete Worked Solutions

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Problem Solutions for E-text

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

1.17. Point A(−4, 2, 5) and the two vectors, RAM = (20, 18, −10) and RAN = (−10, 8, 15), define a triangle.
a) Find a unit vector perpendicular to the triangle: Use

RAM × RAN (350, −200, 340)
ap = = = (0.664, −0.379, 0.645)
|RAM × RAN | 527.35

The vector in the opposite direction to this one is also a valid answer.
b) Find a unit vector in the plane of the triangle and perpendicular to RAN :

(−10, 8, 15)
aAN = √ = (−0.507, 0.406, 0.761)
389

Then

apAN = ap × aAN = (0.664, −0.379, 0.645) × (−0.507, 0.406, 0.761) = (−0.550, −0.832, 0.077)

The vector in the opposite direction to this one is also a valid answer.
c) Find a unit vector in the plane of the triangle that bisects the interior angle at A: A non-unit
vector in the required direction is (1/2)(aAM + aAN ), where

(20, 18, −10)
aAM = = (0.697, 0.627, −0.348)
|(20, 18, −10)|

Now
1 1
(aAM + aAN ) = [(0.697, 0.627, −0.348) + (−0.507, 0.406, 0.761)] = (0.095, 0.516, 0.207)
2 2
Finally,
(0.095, 0.516, 0.207)
abis = = (0.168, 0.915, 0.367)
|(0.095, 0.516, 0.207)|

, 2




1.27. The surfaces r = 2 and 4, θ = 30◦ and 50◦ , and φ = 20◦ and 60◦ identify a closed surface.
a) Find the enclosed volume: This will be

60◦
50◦
4
Vol = r2 sin θdrdθdφ = 2.91
20◦ 30◦ 2

where degrees have been converted to radians.
b) Find the total area of the enclosing surface:

60◦
50◦
4
60◦
Area = 2 2
(4 + 2 ) sin θdθdφ + r(sin 30◦ + sin 50◦ )drdφ
20◦ 30◦ 2 20◦

50◦
4
+2 rdrdθ = 12.61
30◦ 2


c) Find the total length of the twelve edges of the surface:

4
50◦
60◦
Length = 4 dr + 2 (4 + 2)dθ + (4 sin 50◦ + 4 sin 30◦ + 2 sin 50◦ + 2 sin 30◦ )dφ
2 30◦ 20◦
= 17.49

d) Find the length of the longest straight line that lies entirely within the surface: This will be from
A(r = 2, θ = 50◦ , φ = 20◦ ) to B(r = 4, θ = 30◦ , φ = 60◦ ) or
A(x = 2 sin 50◦ cos 20◦ , y = 2 sin 50◦ sin 20◦ , z = 2 cos 50◦ )
to
B(x = 4 sin 30◦ cos 60◦ , y = 4 sin 30◦ sin 60◦ , z = 4 cos 30◦ )
or finally A(1.44, 0.52, 1.29) to B(1.00, 1.73, 3.46). Thus B − A = (−0.44, 1.21, 2.18) and
Length = |B − A| = 2.53


2.5. Let a point charge Q1 25 nC be located at P1 (4, −2, 7) and a charge Q2 = 60 nC be at P2 (−3, 4, −2).
a) If  = 0 , find E at P3 (1, 2, 3): This field will be
 
10−9 25R13 60R23
E= +
4π0 |R13 |3 |R23 |3
√ √
where R13 = −3ax + 4ay − 4az and R23 = 4ax − 2ay + 5az . Also, |R13 | = 41 and |R23 | = 45.
So  
10−9 25 × (−3ax + 4ay − 4az ) 60 × (4ax − 2ay + 5az )
E= +
4π0 (41)1.5 (45)1.5
= 4.58ax − 0.15ay + 5.51az

b) At what point on the y axis is Ex = 0? P3 is now  at (0, y, 0), so R13 = −4ax + 
(y + 2)ay − 7az
and R23 = 3ax + (y − 4)ay + 2az . Also, |R13 | = 65 + (y + 2) and |R23 | = 13 + (y − 4)2 .
2

Now the x component of E at the new P3 will be:
 
10−9 25 × (−4) 60 × 3
Ex = +
4π0 [65 + (y + 2)2 ]1.5 [13 + (y − 4)2 ]1.5

, 3




To obtain Ex = 0, we require the expression in the large brackets to be zero. This expression
simplifies to the following quadratic:

0.48y 2 + 13.92y + 73.10 = 0

which yields the two values: y = −6.89, −22.11

2.19. A uniform line charge of 2 µC/m is located on the z axis. Find E in cartesian coordinates at P (1, 2, 3)
if the charge extends from
a) −∞ < z < ∞: With the infinite line, we know that the field will have only a radial component
in cylindrical coordinates (or x and y components in cartesian). The field from an infinite line
on the z axis is generally E = [ρl /(2π0 ρ)]aρ . Therefore, at point P :

ρl RzP (2 × 10−6 ) ax + 2ay
EP = = = 7.2ax + 14.4ay kV/m
2π0 |RzP |2 2π0 5

where RzP is the vector that extends from the line charge to point P , and is perpendicular to
the z axis; i.e., RzP = (1, 2, 3) − (0, 0, 3) = (1, 2, 0).
b) −4 ≤ z ≤ 4: Here we use the general relation


ρl dz r − r
EP =
4π0 |r − r |3

where r = ax + 2ay + 3az and r = zaz . So the integral becomes


(2 × 10−6 ) 4
ax + 2ay + (3 − z)az
EP = dz
4π0 −4 [5 + (3 − z)2 ]1.5

Using integral tables, we obtain:
 4
(ax + 2ay )(z − 3) + 5az
EP = 3597 V/m = 4.9ax + 9.8ay + 4.9az kV/m
(z 2 − 6z + 14) −4

The student is invited to verify that when evaluating the above expression over the limits −∞ <
z < ∞, the z component vanishes and the x and y components become those found in part a.

2.27. Given the electric field E = (4x − 2y)ax − (2x + 4y)ay , find:
a) the equation of the streamline that passes through the point P (2, 3, −4): We write

dy Ey −(2x + 4y)
= =
dx Ex (4x − 2y)

Thus
2(x dy + y dx) = y dy − x dx
or
1 1
2 d(xy) = d(y 2 ) − d(x2 )
2 2
So
1 2 1 2
C1 + 2xy = y − x
2 2