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Fundamentals of Electromagnetics
for Teaching and Learning:
A Two-Week Intensive Course for Faculty in
Electrical-, Electronics-, Communication-, and
Computer- Related Engineering Departments in
Engineering Colleges in India
by
Nannapaneni Narayana Rao
Edward C. Jordan Professor Emeritus
of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign, USA
Distinguished Amrita Professor of Engineering
Amrita Vishwa Vidyapeetham, India
Program for Hyderabad Area and Andhra Pradesh Faculty
Sponsored by IEEE Hyderabad Section, IETE Hyderabad
Center, and Vasavi College of Engineering
IETE Conference Hall, Osmania University Campus
Hyderabad, Andhra Pradesh
June 3 – June 11, 2009
Workshop for Master Trainer Faculty Sponsored by
IUCEE (Indo-US Coalition for Engineering Education)
Infosys Campus, Mysore, Karnataka
June 22 – July 3, 2009
1-2
Maxwell’s Equations
d
E  dl = –
B  dS
C
dt S
Electric field
intensity
V m
S
C m3 
Wb m2 
A m
Current
density
A m2
Displacement
flux density
C m2

V
r dv
Charge density
Magnetic
flux density
d
H
dl
J
dS
D
dS
=
+



C
S
dt S
Magnetic
field intensity
D  dS =

S
B  dS = 0
1-3
Module 1
Vectors and Fields
1.1 Vector algebra
1.2 Cartesian coordinate system
1.3 Cylindrical and spherical coordinate systems
1.4 Scalar and vector fields
1.5 Sinusoidally time-varying fields
1.6 The electric field
1.7 The magnetic field
1.8 Lorentz force equation
1-4
Instructional Objectives
1.
2.
3.
4.
5.
6.
Perform vector algebraic operations in Cartesian,
cylindrical, and spherical coordinate systems
Find the unit normal vector and the differential surface at
a point on the surface
Find the equation for the direction lines associated with a
vector field
Identify the polarization of a sinusoidally time-varying
vector field
Calculate the electric field due to a charge distribution by
applying superposition in conjunction with the electric
field due to a point charge
Calculate the magnetic field due to a current distribution
by applying superposition in conjunction with the
magnetic field due to a current element
1-5
Instructional Objectives (Continued)
7.
Apply Lorentz force equation to find the electric and
magnetic fields, for a specified set of forces on a charged
particle moving in the field region
1-6
1.1 Vector Algebra
(EEE, Sec. 1.1; FEME, Sec. 1.1)
In this series of PowerPoint presentations, EEE refers to
“Elements of Engineering Electromagnetics, 6th Edition,”
Indian Edition (2006), and FEME refers to “Fundamentals of
Electromagnetics for Engineering,” Indian Edition (2009).
Also, all “D” Problems and “P” Problems are from EEE.
1-7
(1) Vectors (A)
vs.
Magnitude and direction
Ex: Velocity, Force
Scalars (A)
Magnitude only
Ex: Mass, Charge
1-8
(2) Unit Vectors have magnitude
unity, denoted by symbol a
with subscript. We shall use
the right-handed system
throughout.
Useful for expressing vectors in terms of their
components.
A A1a1 + A2 a2 + A3a3
aA = =
A
A12 + A22 + A32
1-9
(3) Dot Product is a scalar
A
A
A • B = AB cos a
a
B
B
Useful for finding angle between two vectors.
A • B
cos a =
AB
=
A=
B=
A1a1 + A2a 2 + A3a3
B1a1 + B2a 2 + B3a3
A1 B1 + A2 B2 + A3 B3
A12 + A22 + A32 B12 + B22 + B32
1-10
(4) Cross Product is a vector
right hand
screw
A
A
A  B = AB sin a an
B
a
B
an
is perpendicular to both A and B.
Useful for finding unit vector perpendicular to
two vectors.
AB
AB
an =
=
AB sin a A  B
1-11
where
a1 a 2
A  B = A1 A2
B1 B2
a3
A3
B3
(5) Triple Cross Product
A  (B  C) is a vector
A  (B  C)  B  (C  A)  C  (A  B)
in general.
1-12
(6) Scalar Triple Product
A•B C= B• C A = C• A  B
A1
A2
A3
= B1
B2
B3
C1 C2
C3
is a scalar.
1-13
Volume of the parallelepiped
= Area of base  Height
=  A × B  C a n 
=  A × B C
= C A×B
= A B×C
A×B
A×B
B
an
C
A
1-14
D1.2 (EEE)
A = 3a1 + 2a2 + a3
B = a1 + a2 – a 3
C = a1 + 2a2 + 3a3
(a)
A + B – 4C
= (3 + 1 – 4)a1 + (2 + 1 – 8)a2
+ (1 – 1 – 12)a3
= – 5a2 – 12a3
A + B – 4C = 25 + 144 = 13
1-15
(b)
A + 2B – C
= (3 + 2 – 1)a1 + (2 + 2 – 2)a2
+ (1 – 2 – 3)a3
= 4a1 + 2a2 – 4a3
Unit Vector
4a1 + 2a 2 – 4a 3
=
4a1 + 2a 2 – 4a 3
1
= (2a1 + a 2 – 2a 3 )
3
1-16
(c) A • C = 3  1 + 2  2 + 1  3
= 10
a1 a 2
(d) B  C = 1 1
1 2
a3
–1
3
= (3 + 2)a1 + (–1 – 3)a2 + (2 –1)a3
= 5a1 – 4a2 + a3
1-17
3 2
(e)
1
A • B  C = 1 1 –1
1 2
3
= 15 – 8 + 1 = 8
Same as
A • (B  C) = (3a1 + 2a2 + a3) • (5a1 – 4a2 + a3)
= 3  5 + 2  (–4) + 1  1
= 15 – 8 + 1
= 8
1-18
P1.5 (EEE)
E
D
B
A
C
Common
Point
D= B–A
(
A + D = B)
E= C–B
(
B + E = C)
D and E lie along a straight line.
1-19
D×E = 0
 B  A × C  B = 0
B×C  A×C  B×B + A×B = 0
A×B + B×C+C×A = 0
What is the geometric interpretation of this result?
1-20
E1.1 Another Example
Given a1 × A = a 2 + 2a3
a2 × A = a1  2a3
(1)
(2)
Find A.
A =C  a 2 + 2a3  ×  a1  2a3 
a1 a 2 a3
= C 0 1 2 = C  2a1 + 2a 2 + a3 
1 0 2
1-21
To find C, use (1) or (2).
a1 × C  2a1 + 2a2 + a3  = a2 + 2a3
C  2a3  a2  = a2 + 2a3
C =1
 A =  2a1 + 2a2 + a3 
1-22
Review Questions
1.1. Give some examples of scalars.
1.2. Give some examples of vectors.
1.3. Is it necessary for the reference vectors a1, a2, and a3
to be an orthogonal set?
1.4. State whether a1, a2, and a3 directed westward,
northward, and downward, respectively, is a righthanded or a left-handed set.
1.5. State all conditions for which A • B is zero.
1.6. State all conditions for which A × B is zero.
1.7. What is the significance of A • B × C = 0?
1.8. What is the significance of A × (B × C) = 0?
1-23
Problem S1.1. Performing several vector algebraic
manipulations
1-24
Problem S1.1. Performing several vector algebraic
manipulations (continued)
1-25
1.2 Cartesian
Coordinate System
(EEE, Sec. 1.2; FEME, Sec. 1.2)
1-26
Cartesian Coordinate System
1-27
Cartesian Coordinate System
z
az
O
ay
ax
y
x
az
ax
ay
z
x
y
1-28
Right-handed system
a x  ay = az
ay  az = a x
xyz xy…
az  a x = a y
ax, ay, az are uniform unit vectors, that is, the
direction of each unit vector is same everywhere in
space.
1-29
Vector drawn from one point to another: From
P1(x1, y1, z1) to P2(x2, y2, z2)
z
r1 + R12 = r2
P1
R12 = r2  r1
r2
r1
O
=  x2ax + y2a y + z2az 
  x1ax + y1a y + z1az 
P2
R12
x
=  x2  x1  ax +  y2  y1  a y +  z2  z1  az
y
1-30
z
R12
P1
(x 2 – x1)ax
r1
O
x1
x2
x
r2
z1
(y2 – y1)ay
P2
(z2 – z1)az
y1
y2
z2
y
1-31
P1.8 A(12, 0, 0), B(0, 15, 0), C(0, 0, –20).
(a) Distance from B to C
= (0 – 0)a x + (0 – 15)a y + (–20 – 0)a z
=
152 + 20 2 = 25
(b) Component of vector from A to C along vector
from B to C
= Vector from A to C
• Unit vector along vector from B to C
1-32
=  12a x  20a z 
 15a y  20a z 
15a y  20a z
400
=
= 16
25
(c) Perpendicular distance from A to the line through B
and C
= (Vector from A to C)  (Vector from B to C)
BC
=
 12ax  20az  ×  15ay  20az 
25
1-33
180az – 240a y – 300a x
=
25
= 12 2
(2)
Differential Length Vector (dl)
az
dl

Q  x + dx, y + dy, z + dz 
P  x, y , z 
dz
dx

dy

ax
ay
dl = dx a x + dy a y + dz a z
1-34
dl
dx
y = f(x)
z = constant plane
dy = f (x) dx
dz = 0
dl = dx ax + dy ay
= dx ax + f (x) dx ay
Unit vector normal to a surface
dl1  dl 2
an =
dl1  dl 2
an
dl2
dl1
Curve 2
Curve 1
1-35
D1.5
Find dl along the line and having the projection dz on
the z-axis.
(a) x = 3, y = –4
dx = 0, dy = 0
dl = dz a z
(b) x + y = 0, y + z = 1
dx + dy = 0, dy + dz = 0
dy = – dz, dx = – dy = dz
d l = dz ax  dz a y + dz az
=  ax  a y + az  dz
1-36
(c) Line passing through (0, 2, 0) and (0, 0, 1).
dy
dz
x = 0,
=
0 – 2 1 –0
dx = 0, dy = – 2 dz
d l = 2 dz a y + dz az
=  2a y + az  dz
1-37
(3)
Differential Surface Vector (dS)
dS =  dl1  dl2  sin a
= d l1 × d l2

an
a
dl2
dS
dl1
Orientation of the surface is defined uniquely by the
normal ± an to the surface.
dS =  dS a n =  dl1  dl 2 a n =  dl1  dl 2
For example, in Cartesian coordinates, dS in any plane
parallel to the xy plane is
az
dx dy a z =  dx a x  dy a y
dy
dx dS
x
y
1-38
(4)
Differential Volume (dv)
dv = dl1 • dl 2  dl3
dl2
dl3
dv
dl1
In Cartesian coordinates,
dv = dx a x • dy a y  dz a z
= dx dy dz
dz
z
dy
dx
y
x
1-39
Review Questions
1.9. What is the particular advantageous characteristic
associated with unit vectors in the Cartesian coordinate
system?
1.10. What is the position vector?
1.11. What is the total distance around the circumference of a
circle of radius 1 m? What is the total vector distance
around the circle?
1.12. Discuss the application of differential length vectors to
find a unit vector normal to a surface at a point on the
surface.
1.13. Discuss the concept of a differential surface vector.
1.14. What is the total surface area of a cube of sides 1 m?
Assuming the normals to the surfaces to be directed
outward of the cubical volume, what is the total vector
surface area of the cube?
1-40
Problem S1.2. Finding the unit vector normal to a surface
and the differential surface vector, at a point on it
1-41
1.3 Cylindrical and Spherical
Coordinate Systems
(EEE, Sec. 1.3; FEME, Appendix A)
1-42
Cylindrical Coordinate System
1-43
Spherical Coordinate System
1-44
Cylindrical and Spherical Coordinate Systems
Cylindrical (r, f, z)
z
Spherical (r, q, f)
z
az
af
90Þ
90
z
x
ar
af
q r
ar
x f r
y
Only az is uniform.
ar and af are
nonuniform.
y
f
x
All three unit
vectors are
nonuniform.
90Þ
90
aq
y
1-45
x = r cos f
y = r sin f
z =z
x = r sin q cos f
y = r sin q sin f
z = r cos q
D1.7 (a) (2, 5p/6, 3) in cylindrical coordinates
z
3
2
x
55p/6
p 6
x = 2 cos 5p 6 = – 3 

y = 2 sin 5p 6 = 1

z=3
y
3+1 = 2
1-46
(b) (4, 4p 3, –1) in cylindrical coordinates
z
1
4
44p/3
p 3
y
x
x = 4 cos 4p 3 = – 2

 4 + 12 = 4
y = 4 sin 4p 3 = – 2 3 
z = –1
1-47
(c) (4, 2p 3, p 6) in spherical coordinates
z
2p 3
y
p
x
6
4
2p
p

cos = 3 
3
6

2p
p

y = 4 sin
sin = 3  9 + 3 + 4 = 4
3
6

p

z = 4 cos = – 2

3
x = 4 sin
(d)

1-48

8, p 4, p 3 in spherical coordinates.
z
p 4
8
y
p 3
x
p
p
= 1 

4
3

p
p

y = 8 sin sin = 3 1 + 3 + 4 = 8
4
3


p
z = 8 cos = 2

4

x = 8 sin
cos
1-49
Conversion of vectors between coordinate systems
a rc 
 
af =

a z 

cos f sin f 0  a x 
–sin f cos f 0  a 

  y 

0
1 
 0
 
a z 

a rs 
 
aq  =
 

af 

sin q cos f sin q sin f

cos q cos f cos q sin f

cos f
 – sin f
af
az
ay
f
ax arc
az
af
q
ars
arc
aq
cos q  a x 
  
– sin q  a y 
  
0  a z 
1-50
P1.18
A = ar at (2, p/6, p/2)
B = aq at (1, p/3, 0)
C = af at (3, p/4, 3p/2)
z
A
A = sin
p
6
x
p 2

y
p
6
a y + cos
1
3
= ay +
az
2
2
1 3
+ =1
4 4
p
6
az
1-51
z
B = sin
p 3
x
C
B
3
1
y
6
a x – cos
1
3
= ax –
az
2
2
z
1 3
+ =1
4 4
p /4
3p /2
x
p
y
C = ax
p
6
az
1-52
1
3 
(a) A B =  a y +
az 
2

2


3
=
4
1
3 
(b) A C =  a y +
az 
2

2


=0
1
3 
az 
 a x 

2
2


ax
1-53
(c)
(d)
1
3 
B C =  ax 
az  ax
2

2


1
=
2
AB•C=C•A B
1
= 0
1
2
0
1
2
0
3
=–
4
0
3
2
3
–
2
1-54
Differential length vectors:
Cylindrical Coordinates:
dl = dr ar + r df af+ dz az
Spherical Coordinates:
dl = dr ar + r dq aq + r sin q df af
1-55
Review Questions
1.15. Describe the three orthogonal surfaces that define the
cylindrical coordinates of a point.
1.16. Which of the unit vectors in the cylindrical coordinate
system are not uniform? Explain.
1.17. Discuss the conversion from the cylindrical coordinates
of a point to its Cartesian coordinates.
1.18. Describe the three orthogonal surfaces that define the
spherical coordinates of a point.
1.19. Discuss the nonuniformity of the unit vectors in the
spherical coordinate system.
1.20. Discuss the conversion from the cylindrical coordinates
of a point to its Cartesian coordinates.
1-56
Problem S1.3. Determination of the equality of vectors
specified in cylindrical and spherical coordinates
1-57
Problem S1.4. Finding the unit vector tangential to a
curve, at a point on it, in spherical coordinates
1-58
1.4 Scalar and Vector Fields
(EEE, Sec. 1.4; FEME, Sec. 1.3)
1-59
FIELD is a description of how a physical quantity
varies from one point to another in the region of the
field (and with time).
(a) Scalar fields
Ex: Depth of a lake, d(x, y)
Temperature in a room, T(x, y, z)
Depicted graphically by constant magnitude
contours or surfaces.
y
d3
d1
d2
x
1-60
(b) Vector Fields
Ex: Velocity of points on a rotating disk
v(x, y) = vx(x, y)ax + vy(x, y)ay
Force field in three dimensions
F(x, y, z) = Fx(x, y, z)ax + Fy(x, y, z)ay
+ Fz(x, y, z)az
Depicted graphically by constant magnitude
contours or surfaces, and direction lines (or
stream lines).
1-61
Example: Linear velocity vector field of points
on a rotating disk
1-62
(c) Static Fields
Fields not varying with time.
(d) Dynamic Fields
Fields varying with time.
Ex: Temperature in a room, T(x, y, z; t)
1-63
D1.10 T(x, y, z, t)

= T0  x 1 + sin p t   +  2 y 1  cos p t   + 4 z 2
(a)
2

2

T  x, y, z, 0  = T0  x 1 + 0   +  2 y 1  1  + 4 z 2
2
= T0  x 2 + 4 z 2 
Constant temperature surfaces are elliptic cylinders,
2
2
x
+
4
z

 = const.
2

1-64

(b) T  x, y, z, 0.5 = T0 x 1 + 1 +  2 y 1  0   + 4 z 2
2
2
= T0  4 x 2 + 4 y 2 + 4 z 2 
Constant temperature surfaces are spheres
x
2
+ y 2 + z 2  = const.

(c) T  x, y, z, 1 = T0  x 1 + 0   +  2 y 1 + 1 + 4 z 2
2
= T0  x 2 + 16 y 2 + 4 z 2 
Constant temperature surfaces are ellipsoids,
2
2
2
x
+
16
y
+
4
z

 = const.
2


1-65
Procedure for finding the equation for the
direction lines of a vector field
The field F is
tangential to the
direction line at
all points on a
direction line.
F
F
dl
F
dl
ax ay
dl  F = dx dy
Fx Fy
az
dz = 0
Fz
dx dy dz
=
=
Fx Fy Fz
1-66
Similarly
dr r df dz
=
=
Fr
Ff
Fz
cylindrical
dr r dq r sin q df
=
=
Fr
Fq
Ff
spherical
1-67
P1.26 (b) xa x + ya y + za z
(Position vector)
dx dy dz
=
=
x
y
z
ln x = ln y + ln C1 = ln z + ln C2
ln x = ln C1y = ln C2 z
x = C1y = C2 z
1-68
 Direction lines are straight lines emanating
radially from the origin. For the line passing
through (1, 2, 3),
1 = C1(2) = C2 (3)
1
1
 C1 = , C2 =
2
3
y z
x= =
2 3
or, 6 x = 3y = 2z
1-69
Review Questions
1.21. Discuss briefly your concept of a scalar field and
illustrate with an example.
1.22. Discuss briefly your concept of a vector field and
illustrate with an example.
1.23. How do you depict pictorially the gravitational field of
the earth?
1.24. Discuss the procedure for obtaining the equations for
the direction lines of a vector field.
1-70
Problem S1.5. Finding the equation for direction line of a
vector field, specified in spherical coordinates
1-71
1.5 Sinusoidally
Time-Varying Fields
(EEE, Sec. 3.6; FEME, Sec. 1.4)
1-72
Sinusoidal function of time
1-73
Polarization is the characteristic which
describes how the position of the tip of the
vector varies with time.
Linear Polarization:
Tip of the vector
describes a line.
Circular Polarization:
Tip of the vector
describes a circle.
1-74
Elliptical Polarization:
Tip of the vector
describes an ellipse.
(i) Linear Polarization
F1 = F1 cos (t + f ) a x
Magnitude varies
sinusoidally with time
Direction remains
along the x axis
 Linearly polarized in the x direction.
1-75
Linear polarization
1-76
F2 = F2 cos (t + q ) a y
Magnitude varies
sinusoidally with time
Direction remains
along the y axis
 Linearly polarized in the y direction.
If two (or more) component linearly polarized
vectors are in phase, (or in phase opposition), then
their sum
vector is also linearly polarized.
Ex: F = F1 cos (t + f ) a x + F2 cos (t + f ) a y
1-77
Sum of two linearly polarized vectors in phase
(or in phase opposition) is a linearly polarized
vector
1-78
y
F2
a
F
a
F1
x
F2 cos (t + f )
–1
= tan
F1 cos (t + f )
F2
–1
= tan
F1
= constant
(ii) Circular Polarization
If two component linearly polarized vectors are
(a) equal in amplitude
(b) differ in direction by 90˚
(c) differ in phase by 90˚,
then their sum vector is circularly polarized.
1-79
Circular Polarization
1-80
Example:
F = F1 cos t ax + F1 sin t a y
F =
 F1
cos t  +  F1 sin t 
2
2
= F1 , constant
a = tan
= tan
1
1
F1 sin t
F1 cos t
 tan t  = t
y
F2
F
a
F1
x
1-81
(iii) Elliptical Polarization
In the general case in which either (i) or (ii) is not
satisfied, then the sum of the two component
linearly polarized vectors is an elliptically polarized
vector.
Example: F = F1 cos t a x + F2 sin t a y
y
F2
F
F1
x
1-82
Example: F = F0 cos t ax + F0 cos t + p 4 ay
y
F0
F2
F
p 4
F1 F0 x
–F0
–F0
1-83
D3.17
F1 = F0 cos  2p 108t  2p z  ax
F2 = F0 cos  2p 108t  3p z  a y
F1 and F2 are equal in amplitude (= F0) and differ in
direction by 90˚. The phase difference (say f) depends
on z in the manner –2pz – (–3pz) = pz.
(a) At (3, 4, 0), f = p(0) = 0.
 F1 + F2 
is linearly polarized.
(b) At (3, –2, 0.5), f = p(0.5) = 0.5 p.
 F1 + F2  is circularly polarized.
1-84
(c) At (–2, 1, 1), f = p(1) = p.
 F1 + F2 
is linearly polarized.
(d) At (–1, –3, 0.2) = f = p(0.2) = 0.2p.
 F1 + F2 
is elliptically polarized.
1-85
Review Questions
1.25. A sinusoidally time-varying vector is expressed in
terms of its components along the x-, y-, and z- axes.
What is the polarization of each of the components?
1.26. What are the conditions for the sum of two linearly
polarized sinusoidally time-varying vectors to be
circularly polarized?
1.27. What is the polarization for the general case of the sum
of two sinusoidally time-varying linearly polarized
vectors having arbitrary amplitudes, phase angles, and
directions?
1.28. Considering the seconds hand on your analog watch
to be a vector, state its polarization. What is the
frequency?
1-86
Problem S1.6. Finding the polarization of the sum of two
sinusoidally time-varying vector fields
1-87
1.6 The Electric Field
(EEE, Sec. 1.5; FEME, Sec. 1.5)
1-88
The Electric Field
is a force field acting on charges by virtue of the
property of charge.
Q1Q2
Coulomb’s Law
F1 =
a 21
F2
2
a12
4
p
R
0
Q
2
R
F1
a21
Q1
Q2 Q1
F2 =
a12
2
4p 0 R
 0 = permittivity of free space
10 –9

F/m
36p
1-89
D1.13(b)
Q
2a
a
Q
Q2/4p0(2a2)
Q2/4p0(4a2)
2/4p (2a2)
Q
0
Q
Q = 4p 0
Q
From the construction, it is evident that the resultant
force is directed away from the center of the square.
The magnitude of this resultant force is given by
1-90
2
Q
2
4p  0  2a
2

1
1
=
+ 2
2
2a 4a
0.957
= 2 N
a
cos 45 +
Q
2
4p  0  4a2 
1-91
Electric Field Intensity, E
is defined as the force per unit charge experienced by
a small test charge when placed in the region of the
field.
F
E = Lim
q0 q
Thus
E
qE
q
–q
–qE
Fe = qE
Units: N = N  m = V
C Cm m
Sources: Charges;
Time-varying
magnetic field
1-92
Electric Field of a Point Charge
(Coulomb’s Law)
Qq
F=
aR
2
4p  0 R
q


Q
= q
aR 
2
 4p  0 R

= q  E due to Q 
 E due to Q =
R
Q
Q
4p  0 R
2
aR
aR
1-93
Constant magnitude surfaces
are spheres centered at Q.
Direction lines are radial lines
emanating from Q.
E
aR
R
Q
E due to charge distributions
(a) Collection of point charges
Q1
E=
n

Qj
2
4
p
R
0 j
j =1
a Rj
aRn
R1
R2
R3
Q2
Q3
Qn
Rn
aR3
aR2
aR1
1-94
E1.2
z
Q (> 0)
d
d2 + x2
a
e
d
Q (> 0)
y
d2 + x2
x
Electron (charge e and mass m) is displaced from the
origin by D (<< d) in the +x-direction and released
from rest at t = 0. We wish to obtain differential
equation for the motion of the electron and its
solution.
1-95
For any displacement x,
F = 2
=
Qe
4p  0  d + x
2
2

Qex
2p  0  d 2 + x

2 32
is directed toward the origin,
and x  D  d.
Qe x
 F–
a
x
3
2p 0 d
cos a ax
ax
1-96
The differential equation for the motion of the
electron is
d2 x
Qe x
m 2 =–
dt
2p 0 d 3
d2 x
Qe
+
x=0
2
3
dt
2pm 0 d
Solution is given by
x = A cos
Qe
2p m 0 d
3
t + B sin
Qe
2p m 0 d
3
t
1-97
dx
Using initial conditions x = D and
= 0 at t = 0,
dt
we obtain
x = D cos
Qe
t
3
2pm 0 d
which represents simple harmonic motion about the
origin with period
2p
Qe
2p m 0 d
3
1-98
(b) Line Charges
Line charge
density, rL (C/m)
dl
P
dS
(c) Surface Charges
Surface charge
density, rS (C/m2)
(d) Volume Charges
Volume charge
density, r (C/m3)
dv
1-99
E1.3 Finitely-Long Line Charge
z
a
rL = rL 0 = 4p0 C m
dz
z
f
x
–a
r2 + z2
y
r
a
E
ar
dE = 2
E=
1-100
rL dz
4p  0  r + z
2
2
4p  0

2

cos a ar
rL 0 r dz
a
z =0
r
2
+z

2 32
ar
a

rL 0 r 
z
=
 2 2 2  ar
2p  0  r r + z  z =0
rL 0 a
2a
=
2p  0 r r + a
2
2
ar =
r r +a
2
rL 0
For a  , E 
ar
2p0r
2
ar
1-101
Infinite Plane Sheet of Charge
of Uniform Surface Charge Density
z
a
y2 + z 2
z
rS 0
x
dy
y
y
1-102
dEz = 2
rS 0 dy
2p0 y2 + z2
rS 0 z dy
=
p0 y2 + z2
rS 0 z
Ez =
p0


dy
y2 + z 2
y =0
rS 0 z 1
=
p0 z
rS 0
=
20

p 2
a =0
da
cos a
1-103
rS 0
E=
az for z
20
rS 0
=
an
20
rS 0

az
20
z<0
rS 0

az
20
0
rS 0
+
z>0
+
rS 0
a
20 z
+
+
+
z=0
z
1-104
D1.16
rS1
z=0
Given
E(3,5,1) = 0 V m
E(1, – 2,3) = 6az V m
E(3, 4,5) = 4az V m
rS 2
z=2
rS 3
z=4
1-105
1
 rS 1  rS 2  rS 3  = 0
2 0
1
 rS 1 + rS 2  rS 3  = 6
2 0
1
 rS 1 + rS 2 + rS 3  = 4
2 0
Solving, we obtain
2
r
=
4

C
m
(a) S 1
0
2
r
=
6

C
m
(b)
S2
0
2
r
=

2

C
m
(c) S 3
0
(d) E  2,1,  6 = 4az V m
1-106
Review Questions
1.29. State Coulomb’s law. To what law in mechanics is
Coulomb’s law analogous?
1.30. What is the value of the permittivity of free space?
What are its units?
1.31. What is the definition of electric field intensity?
What are its units?
1.32. Describe the electric field due to a point charge.
1.33. Discuss the different types of charge distributions.
How do you determine the electric field due to a charge
distribution?
1.34. Describe the electric field due to an infinitely long line
charge of uniform density.
1.35. Describe the electric field due to an infinite plane sheet
of uniform surface charge density.
1-107
Problem S1.7. Determination of conditions for three point
charges on a circle to be in equilibrium
1-108
Problem S1.8. Finding the electric field due to an infinite
plane slab charge of specified charge density
1-109
1.7 The Magnetic Field
(EEE, Sec. 1.6; FEME, Sec. 1.6)
1-110
The Magnetic Field
acts to exert force on charge when it is in motion.
Fm = qv  B
B
Fm
v
q
B = Magnetic flux density vector
Alternatively, since charge in motion constitutes
current, magnetic field exerts forces on current
elements.
dFm
B
d Fm = I d l  B
I dl
1-111
N
N–m
Units of B:
=
2
Am Am
Wb
= 2 =T
m
Sources: Currents;
Time-varying electric field
1-112
Ampère’s Law of Force
I1
a21 dl2
I2
a
dl1 12 R
 0 I 2 d l 2 × a 21 
d F1 = I1d l1 × 

2
4
p
R


= I1d l1 × B 2
 0 I1 d l1 × a12 
d F2 = I 2 d l 2 × 

2
4p R


= I 2 d l 2 × B1
1-113
Magnetic field due to a current element
(Biot-Savart Law)
0 I dl  a R
B=
4p
R2
a
Note B  sin a
I dl
aR
R
P
B
1
B 2
R
B right-circular to the axis of the current element
0 = Permeability of free space
= 4p  10–7 H m
1-114
E1.4 I d l = I dx  a x + a y  A situated at 1,  2, 2  .
Find B at  2,  1, 3 .
B=
0 I d l × a R 0 I d l × R
=
2
4p
R
4p
R3
R

since
a
=
R


R

R =  2  1 a x +  1 + 2  a y +  3  2  a z
= ax + a y + az
1-115
0 I dx  a x + a y  ×  a x + a y + a z 
B=
3
4p
3
 
0 I dx
=
ax  a y 

12 3p
1-116
Current Distributions
(a) Filamentary Current
I (A)
(b) Surface Current
Surface current density, JS (A/m)
JS
w
I 

JS =
w max
1-117
(c) Volume Current
Density, J (A/m2)
area A
J
I 

J =  
A max
1-118
P1.44
z
a2
P(r, f, z)
a2
a
dz
aR
z – z
z I
r
y
a1
x
a1
r
dB =
=
0 dz  a z × a R
2
2

4p r +  z  z   


0 I dz  sin a af
2
2

4p r +  z  z   


1-119
B=
z = a2
z  = a1
0 I
=
4p r 2
z  z
cot a =
r
dB
d z  sin a af
a2
a
= a1
1 +  z  z r  
2
0 I a cosec2a sin a da
=
af
2

4p r a = a
cosec a
0 I
a
=

cos
a

a = a af
4p r
0 I
=
 cos a1  cos a 2  af
4p r
2
1
2
1
1-120
For infinitely long wire,
a1  – , a2  ,
a1  0 , a 2  p
0 I
B
af
2pr
1-121
Magnetic Field Due to an Infinite Plane Sheet of
Uniform Surface Current Density
This can be found by dividing the sheet into
infinitely long strips parallel to the current density
and using superposition, as in the case of finding the
electric field due to an infinite plane sheet of uniform
surface charge density. Instead of going through this
procedure, let us use analogy. To do this, we first
note the following:
1-122
(a)
Point Charge
E=
Q
4p0 R
2
aR
Current Element
0 I d l ×aR
B=
4p R2
1-123
(b) Line Charge
z
rL 0
r
Line Current
z
I
P
r
ar E
r=0
rL 0
E=
ar
2p0r
P
B ar
r=0
0 I
B=
af
2p r
 0I
=
az  ar
2p r
1-124
Then,
(c)
Sheet Charge
rS 0
Sheet Current
JS
P
P
an E
rS 0
E=
an
20
B
an
B=
0
2
JS  an
1-125
Review Questions
1.36. How is magnetic flux density defined in terms of force
on a moving charge? Compare the magnetic force on a
moving charge with electric force on a charge.
1.37. How is magnetic flux density defined in terms of force
on a current element?
1.38. What are the units of magnetic flux density?
1.39. State Ampere’s force law as applied to current elements.
Why is it not necessary for Newton’s third law to hold
for current elements?
1.40. Describe the magnetic field due to a current element.
1.41. What is the value of the permeability of free space?
What are its units?
1-126
Review Questions (continued)
1.42. Discuss the different types of current distributions.
How do you determine the magnetic flux density due to
a current distribution?
1.43. Describe the magnetic field due to an infinitely long,
straight, wire of current.
1.44. Discuss the analogies between the electric field due to
charge distributions and the magnetic field due to
current distributions.
1-127
Problem S1.9. Finding parameters of an infinitesimal
current element that produces a specified magnetic field
1-128
Problem S1.10. Finding the magnetic field due to a
specified current distribution within an infinite plane slab
1-129
1.8 Lorentz Force Equation
(EEE, Sec. 1.7; FEME, Sec. 1.6)
1-130
Lorentz Force Equation
Fe = qE
Fm = qv × B
Fm
F = Fe + Fm
B
F = q E + v × B
q
For a given B, to find E,
F
E= –vB
q
 One force is sufficient.
E
v
Fe
1-131
D1.21
B0
B=
ax + 2a y  2az 

3
Find E for which acceleration experienced by q is
zero, for a given v.
F = q E + v × B = 0
E = v × B
(a) v = v0  ax  a y + az 
v0 B0
E=
ax  a y + az  ×  ax + 2a y  2az 

3
= v0 B0  a y + az 
1-132
(b)
v = v0  2ax + a y + 2az 
v0 B0
E=
2ax + a y + 2az  ×  ax + 2a y  2az 

3
= v0 B0  2ax  2a y  az 
(c)
v = v0 along y =  z = 2 x
v0
=  ax  2a y + 2az 
3
v0 B0
E=
ax + 2a y + 2az  ×  ax + 2a y  2az 

3
=0
1-133
For a given E, to find B,
F
vB= –E
q
One force not sufficient. Two forces are needed.
F1
v1  B =
– E = C1
q
F2
v2  B =
– E = C2
q
C1 × C2 =  v1 × B  ×  v2 × B 
=  v1 × B B  v2   v1 × B v2  B
=   C1 v2  B
1-134
C2  C1
B =
C1 • v 2
provided C1  v 2  0, which means v2 and v1
should not be collinear.
1-135
P1.54 For v = v1 or v = v2, test charge moves with
constant velocity equal to the initial value. It is to be
shown that for
mv1 + nv 2
v=
, where m + n  0,
m+ n
the same holds.
qE + qv1  B = 0
(1)
qE + qv2  B = 0
(2)
qE + qv  B = 0
(3)
1   3   v1  v  × B = 0
 2    3   v2  v  × B = 0
1-136
Both  v1  v  and  v2  v  are collinear to B.
  v1  v  = k  v2  v 
v1 – kv 2
v=
1– k
mv1 + nv 2
=
m+ n
n
for k = –
m
Alternatively,
m
n
(1) 
+ (2) 

m+n
m+n
1-137
 mv1 + nv 2
qE + q 
 m+ n
mv1 + nv 2
v =
m+n

B=0

1-138
Review Questions
1.45. State Lorentz force equation.
1.46. If it is assumed that there is no electric field, the
magnetic field at a point can be found from the
knowledge of forces exerted on a moving test charge
for two noncollinear velocities. Explain.
1.47. Discuss the determination of E and B at a point from the
knowledge of forces experienced by a test charge at that
point for several velocities. What is the minimum
number of required forces? Explain.
1-139
Problem S1.11. Finding the electric and magnetic fields
from three forces experienced by a test charge
1-140
The End