Electromagnetics Drill Solution Hayt8e Chapter 1to5 Drill problem Solution Manual

6/21/23, 6:12 AM 284428991 Electromagnetics Drill Solution Hayt8e Chapter 1to5




ELECTROMAGNETICS HAYT 8 EDITION
DRILL SOLUTION FROM CHAPTERS 1 TO 5
CHAPTER 1
D1.1. Given points M(–1, 2, 1), N(3, –3, 0), and P(–2, –3, –4), find: (a) RMN; (b) RMN + RMP; (c)
|rM|; (d) aMP; (e) |2rP – 3rN|.
(a) RMN = (3 – (–1))ax + (–3 – 2)ay + (0 – 1)az = 4ax – 5ay – az
(b) RMN + RMP = RMN + (–2 – (– 1))ax + (–3 – 2)ay + (–4 – 1)az
= RMN – ax – 5ay – 5az = 4ax – 5ay – az – ax – 5ay – 5az
= 3ax – 10ay – 6az
(c) |rM| = (1)2  22  12 = 6 = 2.45
ax  5ay  5az ax  5ay  5az
(d) aMP = = = –0.14ax – 0.7ay – 0.7az
(1)2  (5)2  (5)2 51
(e) 2rP – 3rN = 2(–2ax – 3ay – 4az) – 3(3ax – 3ay) = –4ax – 6ay – 8az – 9ax + 9ay
= –13ax + 3ay – 8az
|2rP – 3rN| = (13)2  32  (8)2 = 11 2 = 15.56
D1.2. A vector field S is expressed in rectangular coordinates as S = {125/[(x – 1)2 + (y – 2)2 +
(z + 1)2]}{(x – 1)ax + (y – 2)ay + (z + 1)az}. (a) Evaluate S at P(2, 4, 3). (b) Determine a unit vector
that gives the direction of S at P. (c) Specify the surface f(x, y, z) on which |S| = 1.
125
(a) S = {(2  1)ax  (4  2)ay  (3  1)az }
(2 1)2  (4  2)2  (3 1)2
125 125
= 2 2 2 (ax  2ay  4az ) = ( ax  2 a y  4 az)
1 2 4 21
= 5.95ax + 11.9ay + 23.8az
5.95ax  11.9ay  23.8az 5.95ax  11.9ay  23.8az
(b) aS = =
5.952  11.92  23.82 27.27
= 0.218ax + 0.436ay + 0.873az
(c) 1 = |S| =
2 2 2
 125(x  1)   125( y  2)   125(z  1) 
 ( 1)2 ( 2)2 ( 1) 2    ( 1)2 ( 2)2 ( 1) 2    ( 1)2 ( 2)2 ( 1) 2 
 x  y  z   x  y  z   x  y  z 
1252 [(x  1)2  ( y  2)2  (z  1)2 125 ( x 1) 2  ( y  2) 2  ( z 1) 2
1 = =
[( x 1)2  ( y  2)2  ( z 1)2 ]2 ( x 1)2  ( y  2)2  ( z 1) 2
Transposing,
( x  1)2  ( y  2)2  ( z  1)2
= 125
(x  1)2  ( y  2)2  (z  1)2
Conjugating,
( x  1)2  ( y  2)2  ( z  1)2 (x  1)2  ( y  2)2  (z  1)2
= 125
(x  1)  ( y  2)  (z  1)
2 2 2
(x  1)2  ( y  2)2  (z  1)2
( x  1)2  ( y  2)2  ( z  1)2 = 125
D1.3. The three vertices of a triangle are located at A(6, –1, 2), B(–2, 3, –4), and C(–3, 1, 5). Find:
(a) RAB; (b) RAC; (c) the angle θBAC at vertex A; (d) the (vector) projection of RAB on RAC.
(a) RAB = (–2 – 6)ax + (3 – (–1))ay + (–4 – 2)az
= –8ax + 4ay – 6az
(b) RAC = (–3 – 6)ax + (1 – (–1))ay + (5 – 2)az
= –9ax + 2ay + 3az
(c) Locating angle θBAC ,
RAB RAC = (–8)(–9) + 4(2) + (–6)3 = 72 + 8 – 18 = 62
Using dot product,
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,6/21/23, 6:12 AM 284428991 Electromagnetics Drill Solution Hayt8e Chapter 1to5

RAB RAC = |RAB||RAC| cos θBAC

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,6/21/23, 6:12 AM 284428991 Electromagnetics Drill Solution Hayt8e Chapter 1to5




R AB 62
θBAC = cos1 = cos1
RAB RAC  ( 8)2  42  ( 6)3  ( 9)2  22  33 
= 53.6°
(d) Analyzing the projection,
R 62
|RAB (L)| = RAB aAC = RAB = = 6.395
RAC (9)2  22  32
 RAC  9 ax  2 ay  3az
RAB (L) = |RAB (L)| aAC = |RAB (L)| 
 RAC 
= (6.395)
  ( 9)2  22  32
= –5.94ax + 1.319ay + 1.979az
D1.4. The three vertices of a triangle are located at A(6, –1, 2), B(–2, 3, –4), and C(–3, 1, 5). Find:
(a) RAB × RAC; (b) the area of the triangle; (c) a unit vector perpendicular to the plane in which the
triangle is located.
a x a y az
(a) RAB × RAC =  8 4  6 = [4(3) – (–6)(2)]ax + [(–6)(–9) – (–8)(3)]ay + [(–8)(2) – 4(–9)]az
9 2 3
= 24ax + 78ay + 20az
1 1
(b) A = base  height = R AB  R BC
2 2
Finding the direction of vector RBC
RBC = (–3 – (–2))ax + (1 – 3)ay + (5 – (–4))az = –ax – 2ay + 9az
ax a y az
RAB × RBC =  8 4  6 = 24ax + 78ay + 20az
1 2 9

|RAB × RBC| = 242  782  202 = 84
A = (1/2)(84) = 42
24ax  78ay  20az
(c) R AB R AC =
a = 0.286ax + 0.928ay + 0.238az
84
D1.5. (a) Give the rectangular coordinates of the point C(ρ = 4.4, ϕ = –115°, z = 2). (b) Give the
cylindrical coordinates of the point D(x = –3.1, y = 2.6, z = –3). (c) Specify the distance from C to D.
(a) Converting cylindrical to rectangular coordinates
Table 1 Cylindrical to rectangular coordinate systems
x = ρ cos ϕ
y = ρ sin ϕ
z = z
x = 4.4 cos –115° = –1.86
y = 4.4 sin –115° = –3.99
z = 2
C(x = –1.86, y = –3.99, z = 2)
(b) Converting rectangular to cylindrical coordinates
Table 2 Rectangular to cylindrical coordinate systems
ρ = x2  y 2
ϕ = tan–1 y/x
z = z
ρ = (3.1) 2  2.6 2 = 4.05
(ϕ is added with 180° which it lies on the second quadrant.)
ϕ = tan–1 y/x + 180° = tan–1 (2.6/–3.1) + 180° = 140°
z = –3
D(ρ = 4.05, ϕ = 140°, z = –3)
(c) CD = (–3.1 – (–1.86))ax + (2.6 – (–3.99))ay + (–3 – 2)az
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, 6/21/23, 6:12 AM 284428991 Electromagnetics Drill Solution Hayt8e Chapter 1to5
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= –1.24ax + 6.59ay – 5az

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