Elements of Electromagnetics 7th Edition By Sadiku (Solution Manual)

1 CHAPTER 1 P. E. 1.1 (a)     3,2,6 6,2,5 3,0,1 BA 794 36 BA (b)     21,2,0 6,2,5 15,0,5 5 BA (c) The component of A along ay is A y = 0 (d)     3,2,8 6,2,5 9,0,3 3 BA A unit vector parallel to this vector is 
z y x a a aa
3419.0 2279.0 9117.094 643,2,8
11
   P. E. 1.2 (a) 35px y z ra a a 8Ry zr3 a a (b) The distance vector is (0,3,8) (2,4,6) 2 2QR R Q x y z    rr r a aa (c) The distance between Q and R is ||4 1 4 3QRr   P. E. 1.3 Consider the figure shown on the next page: 40350
2
378.28 28.28 km/hr
or
379.3 175.72 km/hrZ PW x xy
xy
za    
 
uuu a a a
a
u Where up = velocity of the airplane in the absence of wind uw = wind velocity uz = observed velocity(Elements of Electromagnetics 7e Sadiku) (Solution Manual) 2 P. E. 1.4 Using the dot product, cosABAB

AB 13
10 6513
50 66.120AB P. E. 1.5  
z y xF F F
a a aFFFEaaE E
3546.0 7092.0 2837.01415,10,410 (a)2
   
 205.0, 2734.0, 9398.012,16,55
5 10 44 3 0 (b)
 

FEz y x
aa a a
FE P. E. 1.6 a + b + c = 0 showing that a, b, and c form the sides of a triangle. ,0ba hence it is a ri ght angle triangle. 
51.10 144 289921Area12,17,321
4311 04
21
2121
21
21Area

baac cb ba N
E
SW
uz up uW y
x 3
P. E. 1.7   22 2
12 2 1 2 1 2 1(a) 25 4 64 9.644PP x x y y z z    
   
  
 .83,22,518,2,5 3,2,1 (b)
1 2 1

P P P P r r r r (c) The shortest distance is  2.8 27,73,14
93182 553 6
931sin
21 31 31
 PPaPP PPd  Prob.1.1 45
(4, 5,1)0.6172 0.7715 0.1543|| (16 25 1)OPOP x y z
OP
x yz
OP
   
rra a a
raa a ar Prob. 1.2 Method 1: ,,AB B A BC C B CA A C
AB BC CA B A C B A Crr r rr r rr r
r r r rrrrrr  
       0 Method 2 (2 , 0 , 3 ) ( 4 ,6 , 2 ) (6 , 6 , 1 )
(10,1, 7) ( 2,0,3) (12,1, 10)
(4, 6, 2) (10,1, 7) ( 6, 7,9)
(0 ,0 ,0)AB B A
BC C B
CA A C
AB BC CArr r
rr r
rr r
rrr     
    
    
  0 Prob. 1.3 (a) 4
3 (4, 2,6) 3(12,18, 8) (4, 2,6) (36,54, 24)
( 32, 56, 30)     
  A B (b) 22 22 5 2(4, 2,6) 5(12,18, 8) (68,86, 28)
| | 12 18 8 532 23.065
(2 5 )/ | | (68,86, 28) / 23.065 2.948 3.728 1.214x yz    
   
  AB
B
A BB a a a (c ) 1006242 6x yz    aA a a (d) 12 18 881 8100
() 8x yz
xy   
 Ba a a
Ba a Prob. 1.4 (a) (10, 6,8) (1, 0, 2) 10 16 26   AB (b) 10 6 8( 12 0) (8 20) (0 6)10 2
-12 12 6x yz
xy z    
 ABa a a
aa a (c) 2 3 (20, 12,16) (3,0,6) 17 12 10x yz      A Ba a a Prob. 1.5 (a) ( 2,5,1) ( 1,0, 3) (4, 6,10) (1, 1,8)      ABC (b) 10 3(18, 2, 6)46 1 0
( ) ( 2,5,1) (18,2, 6) 36 10 6 32  
      BC
AB C (c) 203cos 0.05773 86.69
42 51 19o
AB ABAB    
 AB Prob. 1.6 (a) 1112012
() ( 1 , 0 , 1 ) ( 1 , 2 , 1 ) 1 0 1 0xy z   
    BC a a a
ABC