INSTRUCTOR’S SOLUTIONS MANUAL FOR PRINCIPLES OF ELECTROMAGNETICS ASIAN EDITION

Sadiku & Kulkarni Principles of Electromagnetics, 6e




INSTRUCTOR’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 = 379.3175.72∘ 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