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INSTRUCTOR’S SOLUTIONS MANUAL FOR PRINCIPLES OF ELECTROMAGNETICS ASIAN EDITION
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INSTRUCTOR’S SOLUTIONS MANUAL FOR PRINCIPLES OF ELECTROMAGNETICS ASIAN EDITION
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Sadiku & Kulkarni Principles of Electromagnetics, 6eINSTRUCTOR’S SOLUTIONS MANUAL FOR
PRINCIPLES
ELEMENTS OFOF
ELECTROMAGNETICS
ASIAN EDITION
INTERNATIONAL SIXTH EDITION
Matthew N. O. Sadiku
Prairie View A&M University
Sudarshan R. Nelatury
Pennsylvania State University
S.V. Kulkarni
IITYork
New Bombay
Oxford
Oxford University Press
POESM_Ch01.indd 1 9/14/2015
, Sadiku & Kulkarni Principles of Electromagnetics, 6e
Copyright © 2015 by Oxford University Press
POESM_Ch01.indd 2 9/14/2015
, Sadiku & Kulkarni Principles of Electromagnetics, 6e
1
CHAPTER 1
P. E. 1.1
(a) + (1, 3) (25, −6 ) = (,62−3)
+ 6 +9 = 7
(b) 5 − (05, 15) (52, −6 ) = (,0−,21)
(c) The component of A along ay is Ay = 0
(d) 3 + (03, 9 ) (52, −6 ) = (28, 3)
A unit vector parallel to this vector is
a 11 =
(8 ,3)
4 9
(=0.9117a x + 0.2279a y + 0.3419a z )
P. E. 1.2 (a) rp = ax − 3ay + 5az
rR = 3ay + 8az
(b) The distance vector is
rQR = rR − rQ = (0, 3,8) − (2, 4, 6) = −2ax − ay + 2az
(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:
u = u + u = −350a +
40
(−a + a )
Z P W x x y
2
= −378.28ax + 28.28ay km/hr
or
uz = 379.3175.72∘ km/hr
Where up = velocity of the airplane in the absence of wind
uw = wind velocity
uz = observed velocity
POESM_Ch01.indd 1 9/14/2015
, Sadiku & Kulkarni Principles of Electromagnetics, 6e
Copyright © 2015 by Oxford University Press
POESM_Ch01.indd 2 9/14/2015
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