Fundamentals of Applied Electromagnetics 8th Edition Exercices Solutions Manual PDF

Exercices Solutions Fundamentals of Applied Electromagnetics 8th Edition By Fawwaz T.Ulaby,Umberto Ravaioli




Fundamentals of Applied Electromagnetics 8e
by
Fawwaz T. Ulaby and Umberto Ravaioli

Exercise Solutions

Chapters
Chapter 1 Introduction: Waves and Phasors
Chapter 2 Transmission Lines
Chapter 3 Vector Analysis
Chapter 4 Electrostatics
Chapter 5 Magnetostatics
Chapter 6 Maxwell’s Equations for Time-Varying Fields
Chapter 7 Plane-Wave Propagation
Chapter 8 Wave Reflection and Transmission
Chapter 9 Radiation and Antennas
Chapter 10 Satellite Communication Systems and Radar Sensors

,Exercices Solutions Fundamentals of Applied Electromagnetics 8th Edition By Fawwaz T.Ulaby,Umberto Ravaioli




Chapter 1 Exercise Solutions

Exercise 1.1
Exercise 1.2
Exercise 1.3
Exercise 1.4
Exercise 1.5
Exercise 1.6
Exercise 1.7
Exercise 1.8
Exercise 1.9
Exercise 1.10
Exercise 1.11
Exercise 1.12

,Exercices Solutions Fundamentals of Applied Electromagnetics 8th Edition By Fawwaz T.Ulaby,Umberto Ravaioli




Exercise 1.1 Given charges q1 = 10 mC, q2 = 10 mC, and q3 = 5 mC, all in free space, what is the direction of the
force acting on charge q3 ?
y
q3



2m 2m


q1 q2
x

2m
Figure E1.1


Solution:
y
Fe31

q3


Fe32



q1 q2
x


F e 3 = F e 31 + F e 3 2
Forces F e 31 and F e 32 are equal in magnitude, with F e 31 pointing along 45 above the x axis and F e 32 pointing along 45 below
the x axis. The yˆ components cancel.
Hence, F e 3 is along +x. ˆ

, Exercices Solutions Fundamentals of Applied Electromagnetics 8th Edition By Fawwaz T.Ulaby,Umberto Ravaioli




Exercise 1.2 Two parallel, very long, wires carry currents I1 and I2 . The magnetic field due to current I1 alone is B 1 .
What is the magnetic field due to both currents at a point midway between the wires if:
(a) I1 = I2 and both currents flow along the + yˆ direction?
(b) I1 = I2 , but I2 flows along the yˆ direction?

I1 I2

B=?




Figure E1.2

Solution:
y

z x

I1 I2

B=0




(a) B 1 is into the page ( z)
ˆ and B 2 is out of the page (+z).
ˆ Since I1 = I2 in magnitude, B = 0.

y

z x

I1 I2

B = 2B1




(b) Both magnetic fields are into the page. Hence,
B = 2B 1 = 2jB 1 j( z):
ˆ