INSTRUCTOR’S SOLUTIONS MANUAL FOR ELEMENTS OF ELECTROMAGNETICS
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Module
ELECTROMAGNETICS
Institution
ELECTROMAGNETICS
CHAPTER 1
P. E. 1.1
= A(a+) B + 1(,0, 3−) (=5,2, 6−) 6(,2, 3 )
A= +B +3 46 + 9= 7
(bA) B5= − −(5,0, 15 ) (5,2, −6 ) =0(, 2−, 21 )
(c) The component of A along ay is Ay = 0
(dA) B 3= + +(3,0, 9 ) (5,2, −6 ) = (8,2, 3 )
A unit vector parallel to this vector is
( )
( ) x y z a a...
, Sadiku & Kulkarni Principles of Electromagnetics, 6e
1
CHAPTER 1
P. E. 1.1
(a) A + B = (1,0,3) + (5,2,−6 ) = (6,2,−3)
A + B = 36 + 4 + 9 = 7
(b) 5 A − B = (5,0,15) − (5,2,−6) = (0,−2,21)
(c) The component of A along ay is Ay = 0
(d) 3 A + B = (3,0,9 ) + (5,2,−6 ) = (8,2,3)
A unit vector parallel to this vector is
a11 =
(8,2,3)
64 + 4 + 9
= ±(0.9117a x + 0.2279a y + 0.3419a z )
P. E. 1.2 (a) rp = a x − 3a y + 5a z
rR = 3a y + 8a z
(b) The distance vector is
rQR = rR − rQ = (0,3,8) − (2, 4, 6) = −2a x − a y + 2a z
(c) The distance between Q and R is
| rQR |= 4 + 1 + 4 = 3
P. E. 1.3 Consider the figure shown on the next page:
40
uZ = uP + uW = −350a x +
2
( −a x + a y )
= −378.28a x + 28.28a y km/hr
or
uz = 379.3∠175.72 km/hr
Where up = velocity of the airplane in the absence of wind
uw = wind velocity
uz = observed velocity
, Sadiku & Kulkarni Principles of Electromagnetics, 6e
22
N
y
up W E
x
uW
uz
S
P. E. 1.4
Using the dot product,
A B −13 13
cos θ AB = = =−
AB 10 65 50
θ AB = 120.66
P. E. 1.5
(a) E F = ( E ⋅ a F )a F =
(E ⋅ F )F =
− 10(4,−10,5)
2
F 141
= − 0.2837a x + 0.7092a y − 0.3546a z
ax a y az
(b) E × F = 0 3 4 = (55,16,−12 )
4 − 10 5
a E ×F = ± (0.9398,0.2734,−0.205)
P. E. 1.6 a + b + c = 0 showing that a, b, and c form the sides of a triangle.
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