by nannapaneni narayana rao edward c. jordan professor emeritus of electrical and computer...
DESCRIPTION
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 - PowerPoint PPT PresentationTRANSCRIPT
Fundamentals of ElectromagneticsFundamentals of Electromagneticsfor Teaching and Learning:for Teaching and Learning:
A Two-Week Intensive Course for Faculty inA Two-Week Intensive Course for Faculty inElectrical-, Electronics-, Communication-, and Electrical-, Electronics-, Communication-, and
Computer- Related Engineering Departments in Computer- Related Engineering Departments in Engineering Colleges in IndiaEngineering Colleges in India
byby
Nannapaneni Narayana RaoNannapaneni Narayana RaoEdward C. Jordan Professor EmeritusEdward C. Jordan Professor Emeritus
of Electrical and Computer Engineeringof Electrical and Computer EngineeringUniversity of Illinois at Urbana-Champaign, USAUniversity of Illinois at Urbana-Champaign, USADistinguished Amrita Professor of EngineeringDistinguished Amrita Professor of Engineering
Amrita Vishwa Vidyapeetham, IndiaAmrita Vishwa Vidyapeetham, India
Program for Hyderabad Area and Andhra Pradesh FacultySponsored by IEEE Hyderabad Section, IETE Hyderabad
Center, and Vasavi College of EngineeringIETE Conference Hall, Osmania University Campus
Hyderabad, Andhra PradeshJune 3 – June 11, 2009
Workshop for Master Trainer Faculty Sponsored byIUCEE (Indo-US Coalition for Engineering Education)
Infosys Campus, Mysore, KarnatakaJune 22 – July 3, 2009
1-3
Maxwell’s Equations
Electric fieldintensity
Magnetic flux density
Charge density
Magneticfield intensity
Current density
Displacementflux density
V m Wb m2 C m3
A m A m2 C m2
Ed l – ddt BdS
SC DdS dvVS
Hdl JdS ddtSC DdS
S BdS 0S
1-4
Module 1Vectors and Fields
1.1 Vector algebra1.2 Cartesian coordinate system1.3 Cylindrical and spherical coordinate systems1.4 Scalar and vector fields1.5 Sinusoidally time-varying fields1.6 The electric field1.7 The magnetic field1.8 Lorentz force equation
1-5
Instructional Objectives1. Perform vector algebraic operations in Cartesian,
cylindrical, and spherical coordinate systems2. Find the unit normal vector and the differential surface at
a point on the surface3. Find the equation for the direction lines associated with a
vector field4. Identify the polarization of a sinusoidally time-varying
vector field5. Calculate the electric field due to a charge distribution by
applying superposition in conjunction with the electric field due to a point charge
6. Calculate the magnetic field due to a current distribution by applying superposition in conjunction with the magnetic field due to a current element
1-6
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
1-7
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.
1-8
(1) Vectors (A) vs. Scalars (A)
Magnitude and direction Magnitude only Ex: Velocity, Force Ex: Mass, Charge
1-9
(2) Unit Vectors have magnitudeunity, denoted by symbol awith subscript. We shall usethe right-handed systemthroughout.
Useful for expressing vectors in terms of their components.
aA AA
A1a1 A2a2 A3a3
A12 A2
2 A32
1-10
(3) Dot Product is a scalar A
A • B = AB cos B
Useful for finding angle between two vectors.
cos A • B
ABA A1a1 A2a2 A3a3B B1a1 B2a2 B3a3
A1B1 A2B2 A3B3
A12 A2
2 A32 B1
2 B22 B3
2
A
B
1-11
(4) Cross Product is a vector
AA B = AB sin
B
is perpendicular to both A and B.
Useful for finding unit vector perpendicular to two vectors.
an A B
AB sin
A BA B
an
right handscrew A
B
an
1-12
where
(5) Triple Cross Product
in general.
A B a1 a2 a3A1 A2 A3B1 B2 B3
A (B C) is a vector
A (B C) B (C A) C (A B)
1-13
(6) Scalar Triple Product
is a scalar.
A • B C B • C A C • A B
A1 A2 A3
B1 B2 B3
C1 C2 C3
1-14
Volume of the parallelepiped
an CB
A
Area of base Height
n
A×B C a
A×BA×B CA×B
C A×BA B×C
1-15
D1.2 (EEE) A = 3a1 + 2a2 + a3
B = a1 + a2 – a3
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
1-16
(b) A + 2B – C= (3 + 2 – 1)a1 + (2 + 2 – 2)a2
+ (1 – 2 – 3)a3
= 4a1 + 2a2 – 4a3
Unit Vector
=
=
4a1 2a2 – 4a34a1 2a2 – 4a3
13
(2a1 a2 – 2a3 )
1-17
(c) A • C= 3 1 + 2 2 + 1 3
= 10
(d)
=
= 5a1 – 4a2 + a3
B C a1 a2 a31 1 –11 2 3
(3 2)a1 (–1 – 3)a2 (2 –1)a3
1-18
(e)
= 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
A • B C
3 2 1
1 1 –1
1 2 3
1-19
P1.5 (EEE)
D = B – A ( A + D = B)
E = C – B ( B + E = C)
D and E lie along a straight line.
D
AB
EC
CommonPoint
1-20
What is the geometric interpretation of this result?
D×E 0
B A × C B 0
B×C A×C B×B A×B 0
A×B + B×C + C× A = 0
1-21
E1.1 Another Example
Given
Find A.
1 2 3
2 1 3
2 (1)2 (2)
a × A a aa × A a a
2 3 1 3
31 2
1 2 3
= 2 2
0 2 21 201 2
C
C C
A a a × a a
aa aa a a
1-22
To find C, use (1) or (2).
1 2 32 2 A a a a
1 1 2 3 2 3
3 2 2 3
2 2 2
2 2
1
C
C
C
a × a a a a a
a a a a
1-23
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 right- handed 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?
1-24
Problem S1.1. Performing several vector algebraic manipulations
1-25
Problem S1.1. Performing several vector algebraic manipulations (continued)
1-26
1.2 CartesianCoordinate System
(EEE, Sec. 1.2; FEME, Sec. 1.2)
1-27
Cartesian Coordinate System
1-28
Cartesian Coordinate System
xy
xy
z
Oaz z
azay
ayax
ax
1-29
Right-handed system
xyz xy…
ax, ay, az are uniform unit vectors, that is, the direction of each unit vector is same everywhere in space.
ax ay az
ay az ax
az ax ay
1-30
1 12 2
12 2 1
r R rR r r
zP2
P1R12
r1 r2
y
x
O
2 2 2
1 1 1
2 1 2 1 2 1
x y z
x y z
x y z
x y z
x y z
x x y y z z
a a a
a a a
a a a
Vector drawn from one point to another: From P1(x1, y1, z1) to P2(x2, y2, z2)
1-31
xx2
x1
O
(x2 – x1)ax r1 z1r2
z
P1
P2R12
(z2 – z1)az(y2 – y1)ay
y1z2
y2y
1-32
P1.8 A(12, 0, 0), B(0, 15, 0), C(0, 0, –20).
(a) Distance from B to C
=
=
(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
(0 – 0)ax (0 – 15)ay (–20 – 0)az
152 202 25
1-33
(c)Perpendicular distance from A to the line through B and C
=(Vector from A to C) (Vector from B to C)BC
12 20 15 20
25x z y z
a a × a a
15 2012 20
15 20
400 1625
y zx z
y z
a aa a
a a
1-341-34
(2) Differential Length Vector (dl)
180 – 240 – 30025
a a az y x
12 2
dl dx a x dy ay dz az
, ,Q x dx y dy z dz
dzdx
, ,P x y z
d l
dyya
xa
za
1-35
dl = dx ax + dy ay
= dx ax + f (x) dx ay
Unit vector normal to a surface
andl2
dl1
Curve 2Curve 1an
dl1 dl2dl1 dl2
dldx
y = f(x)
dy = f (x) dxz = constant plane
dz = 0
1-36
D1.5 Find dl along the line and having the projection dz on the z-axis.
(a)
(b)
x 3, y –4dx 0, dy 0dl dz az
x y 0, y z 1dx dy 0, dy dz 0dy – dz, dx – dy dz
x y z
x y z
d dz dz dz
dz
l a a a
a a a
1-37
(c)Line passing through (0, 2, 0) and (0, 0, 1).
x 0, dy0 – 2
dz1 – 0
dx 0, dy – 2 dz
2
2y z
y z
d dz dz
dz
l a a
a a
1-38
(3) Differential Surface Vector (dS)
Orientation of the surface is defined uniquely by the normal ± an to the surface.
For example, in Cartesian coordinates, dS in any plane parallel to the xy plane is
dS dS an dl1 dl2 an dl1 dl2
dx dy az dx ax dy ay
dS
dl1
dl2an
x
ydSdxdy
az
1 2
1 2
sin dS dl dl
d d
l × l
1-39
(4) Differential Volume (dv)
In Cartesian coordinates,
dv dl1 • dl2 dl3
dv dx ax • dy ay dz a z
dx dy dzdz dy
dx
z y
x
dl2
dl1
dl3 dv
1-40
Review Questions1.9. 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?
1-41
Problem S1.2. Finding the unit vector normal to a surface and the differential surface vector, at a point on it
1-42
1.3 Cylindrical and Spherical Coordinate Systems
(EEE, Sec. 1.3; FEME, Appendix A)
1-43
Cylindrical Coordinate System
1-44
Spherical Coordinate System
1-451-45
Cylindrical (r, , z) Spherical (r, , )
Only az is uniform. All three unit
ar and a are vectors are nonuniform. nonuniform.
Cylindrical and Spherical Coordinate Systems
a
x
x
y
yr
z
z ar
az
90Þ90
z
r
y
ar
90Þ
x
a
a
90
1-461-46
x = r cos x = r sin cos y = r sin y = r sin sin z = z z = r cos
D1.7 (a) (2, 5/6, 3) in cylindrical coordinates
2 cos 5 6 – 3 3 1 22 sin 5 6 13
xyz
x
z
3
2 y5/65 6
1-471-47
(b)
x 4 cos 4 3 – 2
y 4 sin 4 3 – 2 3
4 12 4
z – 1
(4, 4 3, –1) in cylindrical coordinates
1 4
x
y4/3
z
4 3
1-481-48
(c)
24 sin cos 33 624 sin sin 3 9 3 4 43 6
4 cos – 23
x
y
z
(4, 2 3, 6) in spherical coordinates
x
y
z
4
6
2 3
1-491-49
x 8 sin 4
cos 3
1
y 8 sin4
sin3
3
z 8 cos 4
2
1 3 4 8
(d) 8, 4, 3 in spherical coordinates. z
y
x
8
3
4
1-50
Conversion of vectors between coordinate systems
arcaaz
cos sin 0–sin cos 0
0 0 1
axayaz
ars
a
a
sin cos sin sin cos
cos cos cos sin – sin
–sin cos 0
ax
ay
az
axarc
az ay
a
az arsarca
a
1-511-51
P1.18 A = ar at (2, /6, 2)
B = a at (1, /3, 0)
C = a at (3, /4, 3/2)
A sin 6
ay cos 6
az
12
ay 3
2az
14
34
1x
z A
y
2
6
1-521-52
B sin 6
ax – cos 6
az
12
ax –3
2az
14
34
1
C ax
x
y
z
B
13
C
x
y
z
/43
3 /2
1-531-53
(a)
(b)
1 3 1 32 2 2 2
341 32 2
0
A B a a a a
A C a a a
y z x z
y z x
1-54
(c)
(d) A B • C C • A B
=
1 0 0
012
32
12
0 – 32
–3
4
1 32 2
12
x z x
B C a a a
1-55
Differential length vectors:
Cylindrical Coordinates:
dl = dr ar + r d a+ dz az
Spherical Coordinates:
dl = dr ar + r d a + r sin d a
1-56
Review Questions1.15. 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.
1-57
Problem S1.3. Determination of the equality of vectors specified in cylindrical and spherical coordinates
1-58
Problem S1.4. Finding the unit vector tangential to a curve, at a point on it, in spherical coordinates
1-59
1.4 Scalar and Vector Fields(EEE, Sec. 1.4; FEME, Sec. 1.3)
1-60
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
x
d1
d2d3
1-61
(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).
1-62
Example: Linear velocity vector field of points on a rotating disk
1-63
(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)
1-64
2 2 20
2 20
, , , 0 1 0 2 1 1 4
4
T x y z T x y z
T x z
2 24 const.x z
2 2 20= 1 sin 2 1 cos 4T x t y t z
D1.10 T(x, y, z, t)
Constant temperature surfaces are elliptic cylinders,
(a)
1-65
(b)
Constant temperature surfaces are spheres
(c)
Constant temperature surfaces are ellipsoids,
22 20
2 2 20
, , , 0.5 1 1 2 1 0 4
4 4 4
T x y z T x y z
T x y z
2 2 2 const.x y z
2 2 20
2 2 20
, , , 1 1 0 2 1 1 4
16 4
T x y z T x y z
T x y z
2 2 216 4 const.x y z
1-66
Procedure for finding the equation for the direction lines of a vector field
The field F istangential to thedirection line atall points on a direction line.
dl F ax ay azdx dy dzFx Fy Fz
0
dxFx
dyFy
dzFz
dl F
FF
dl
1-67
Similarly
drFr
r dF
dzFz
drFr
r dF
r sin d
F
cylindrical
spherical
1-68
P1.26 (b)xax yay zaz(Position vector)
dxx
dyy
dzz
ln x ln y ln C1 ln z ln C2
ln x ln C1y ln C2z
x C1y C2z
1-691-69
Direction lines are straight lines emanating radially from the origin. For the line passing through (1, 2, 3),
1 C1(2) C2 (3)
C1 12
, C2 13
x y2
z3
or, 6 3 2 x y z
1-70
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.
1-71
Problem S1.5. Finding the equation for direction line of a vector field, specified in spherical coordinates
1-72
1.5 SinusoidallyTime-Varying Fields(EEE, Sec. 3.6; FEME, Sec. 1.4)
1-73
Sinusoidal function of time
1-74
Polarization is the characteristic which describes how the position of the tip of the vector varies with time.
Linear Polarization:Tip of the vectordescribes a line.
Circular Polarization:Tip of the vector describes a circle.
1-75
Elliptical Polarization:Tip of the vectordescribes an ellipse.
(i) Linear Polarization
Linearly polarized in the x direction.
F1 F1 cos (t ) ax
Direction remainsalong the x axis
Magnitude variessinusoidally with time
1-76
Linear polarization
1-77
F2 F2 cos (t ) ay Direction remainsalong the y axisMagnitude varies
sinusoidally with time
Linearly polarized in the y direction.
If two (or more) component linearly polarized vectors are in phase, (or in phase opposition), then their sumvector is also linearly polarized.
Ex: 1 2cos cosx y ( ) ( )F t F tF a a
1-781-78
Sum of two linearly polarized vectors in phase (or in phase opposition) is a linearly polarized vector
1-79
(ii) Circular PolarizationIf 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.
tan–1 F2 cos (t )F1 cos (t )
tan–1 F2F1
constant
y
x
F1
F2 F
1-80
Circular Polarization
1-81
Example:
1 1
2 21 1
1
1 1
1
1
cos sin
cos sin
, constant sin tan cos
tan tan
x yF t F t
F t F t
FF tF t
t t
F a a
F
1F
2FF
x
y
1-82
(iii) Elliptical PolarizationIn 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 ax F2 sin t ay
1F
2FF
x
y
1-831-83
Example: 0 0cos cos 4x yF t F t F a a
x–F0
–F0
F0
F0F1
F2 F
y
4
1-841-84
D3.17
F1 and F2 are equal in amplitude (= F0) and differ in direction by 90˚. The phase difference (say ) depends on z in the manner –2z – (–3z) = z.
(a) At (3, 4, 0), = (0) = 0.
(b) At (3, –2, 0.5), = (0.5) = 0.5 .
81 0
82 0
cos 2 10 2
cos 2 10 3
x
y
F t z
F t z
F a
F a
1 2 is linearly polarized.F F
1 2 is circularly polarized. F F
1-851-85
(c) At (–2, 1, 1), = (1) = .
(d) At (–1, –3, 0.2) = = (0.2) = 0.2.
1 2 is linearly polarized.F F
1 2 is elliptically polarized. F F
1-86
Review Questions1.25. 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?
1-87
Problem S1.6. Finding the polarization of the sum of two sinusoidally time-varying vector fields
1-88
1.6 The Electric Field(EEE, Sec. 1.5; FEME, Sec. 1.5)
1-89
The Electric Fieldis a force field acting on charges by virtue of the property of charge.
Coulomb’s LawR
F1 a21Q1
Q2a12
F2 F1 Q1Q2
40R2 a21
F2 Q2Q1
40R2 a12
0 permittivity of free space
10–9
36F / m
1-90
Q
Q Q
aQ
2a
D1.13(b)
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
Q2/40(2a2)Q2/40(4a2)
Q2/40(2a2)
Q 4 0
1-911-91
2 2
2 20 0
22
2
2 cos 454 2 4 4
1 142
0.957 N
Q Qa a
aa
a
1-92
Electric Field Intensity, Eis defined as the force per unit charge experienced by a small test charge when placed in the region of the field.
Thus
Units:
E Limq 0
Fq
Fe qE
qEq
E
–q–qE
Sources: Charges;Time-varying magnetic field
N N m VC C m m
1-93
20
20
4
4
due to
R
R
QqR
QqR
q Q
F a
a
E
aRq
R
Q
20
due to 4 R
QQR
E a
Electric Field of a Point Charge
(Coulomb’s Law)
1-94
Constant magnitude surfaces are spheres centered at Q.Direction lines are radial lines emanating from Q.
E due to charge distributions(a) Collection of point charges
Qn
Q3
Q2
Q1 R1
R2R3
Rn
aRn aR3aR2aR1
E Q j
40R j2 a Rj
j 1
n
E
Q
aRR
1-95
E1.2
Q (> 0) d
x
e
d Q (> 0)y
z
d2 + x2 d2 + x2
Electron (charge e and mass m) is displaced from the origin by (<< 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.
1-96
For any displacement x,
is directed toward the origin,and x d.
F –Q e x
20d3 a x
2 20
3 22 20
2 cos 4
2
x
x
Q ed x
Q e x
d x
F a
a
1-971-97
The differential equation for the motion of theelectron is
Solution is given by
md2 xdt2 –
Qe x20d3
d2 xdt2
Q e2m0d3 x 0
30
cos2
Q e
x A t Bm d 3
0
sin2
Q e t
m d
1-98
Using initial conditions and at t = 0, we obtain
which represents simple harmonic motion about the origin with period
x dxdt
0
x cosQe
2m0d3 t
30
22
Q e
m d
1-991-99
(b) Line ChargesLine charge density, L (C/m)
(c) Surface ChargesSurface charge density, S (C/m2)
(d) Volume ChargesVolume charge density, (C/m3)
Pdl
dS
dv
1-1001-100
E1.3 Finitely-Long Line Charge
x
za
dz
–a
ar
y
z z2r2
Er
04 C m L L0
1-1011-1011-101
2 20
2 cos 4
Lr
dzdr z
E a
0
0, 2For
E aLra r
0
3 20 2 20
02 2 2
0 0
02 2 2 2
0
24
2
2
2
E a
a
a a
aL
rz
a
Lr
z
Lr r
r dz
r z
r z
r r za a
r r a r r a
1-1021-102
Infinite Plane Sheet of Chargeof Uniform Surface Charge Density
z
z
y
dyx
y
z2y2
0S
1-1031-103
0
2 20
02 2
0
2 cos2
Sz
S
dydE
y z
z dyy z
02 200
20
00
0
0
1
2
Sz
y
S
S
z dyEy z
z dz
1-1041-104
0
0
0
0
0
0
for 02
2
2
z
E a
a
a
Sz
Sn
Sz
+
+
+
+
+
z < 0
z = 0 z
z > 0
0S
0
02
aSz
0
02 aS
z
1-1051-105
D1.16
Given
(3,5,1) 0 V m(1, – 2,3) 6 V m(3,4,5) 4 V m
z
z
EE aE a
z = 0 z = 2 z = 4
1S 2S 3S
1-106
21 04 C mS
1 2 30
1 2 30
1 2 30
1 021 6
21 4
2
S S S
S S S
S S S
22 06 C mS
Solving, we obtain
23 02 C mS 2,1, 6 4 V mz E a(d)
(a)
(c)
(b)
1-107
Review Questions1.29. 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.
1-108
Problem S1.7. Determination of conditions for three point charges on a circle to be in equilibrium
1-109
Problem S1.8. Finding the electric field due to an infinite plane slab charge of specified charge density
1-110
1.7 The Magnetic Field(EEE, Sec. 1.6; FEME, Sec. 1.6)
1-1111-111
dFmB
I dl
The Magnetic Fieldacts to exert force on charge when it is in motion.
B = Magnetic flux density vectorAlternatively, since charge in motion constitutes current, magnetic field exerts forces on current elements.
FmB
vq
Fm qv B
IF l Bmd d
1-112
Units of B:
Sources: Currents;Time-varying electric field
2
2
N N–m=A m A m
Wb= = Tm
1-113
Ampère’s Law of Force
Ra12
a21
dl1
dl2
I1 I2
0 2 2 211 1 1 2
1 1 2
0 1 1 122 2 2 2
2 2 1
4
4
I dd I dR
I dI d
d I dR
I d
l ×aF l ×
l ×Bl ×aF l ×
l ×B
1-1141-114
I dlaR
R P B
Magnetic field due to a current element(Biot-Savart Law)
B 04
I dl a RR2
B right-circular to the axis of the current element
0
–7
Permeability of free space
= 4 10 H m
2
sin1R
B
B
Note
1-115
E1.4
0 02 34 4
since
2 1 1 2 3 2
R
R
x y z
x y z
I d I dR R
R
l ×a l × RB
Ra
R = a a a
a a a
A situated at 1, 2, 2 .
Find at 2, 1, 3 .
l a a
Bx yI d I dx
1-116
0
12 3 x yI dx
a a
03 =
4 3
x y x y zI dx
a a × a a aB
1-117
JS
w
Current Distributions
(a) Filamentary Current I (A)
(b) Surface CurrentSurface current density, JS (A/m)
JS Iw
max
1-118
area A
J
(c) Volume CurrentDensity, J (A/m2)
J IA
max
1-119
P1.44
xa1
r
yrI
z – z
z
a2
dz
z
aR
1
2P(r, , z)
022
022
4
sin
4
z Rdzdr z z
I dz
r z z
a ×aB
a
1-1201-120
2
2
1
2
1
2
1
02 2
20
2
0
01 2
cot
sin 4 1
cosec sin 4 cosec
cos 4
cos cos 4
1
a
aB B
a
a
a
a
z
z
z zdr
d zIr z z r
I drIrIr
1-121
For infinitely long wire,
a1 – , a2 ,1 0 , 2
B 0 I2r
a
1-122
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:
1-1231-123
Point Charge Current Element(a)
024
RI dR
l ×aB2
04 RQ
RE a
1-1241-124
z
r = 0
ParB
rI
(b) Line Charge Line Current
0
02E aL
rr0
0
2
2
B a
a az r
IrIr
z
r = 0
Par E
r0L
1-1251-125
JSP B
an
Then,
(c) Sheet Charge Sheet Current
0
02
E aSn
0
2
B J aS n
Pan E
0S
1-126
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?
1-127
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.
1-128
Problem S1.9. Finding parameters of an infinitesimal current element that produces a specified magnetic field
1-129
Problem S1.10. Finding the magnetic field due to a specified current distribution within an infinite plane slab
1-130
1.8 Lorentz Force Equation(EEE, Sec. 1.7; FEME, Sec. 1.6)
1-131
Lorentz Force Equation
For a given B, to find E,
E Fq
– v B
One force is sufficient.
Fm
q
B
E
v
Fe
e
m
e m
q
F EF v×BF = F FF = E + v×B
1-132
D1.21 0 2 2
3 x y zB
B a a a
0q
F E v×BE = v×B
Find E for which acceleration experienced by q iszero, for a given v.
(a)
0
0 0
0 0
2 23
x y z
x y z x y z
y z
v
v B
v B
v = a a a
E = a a a × a a a
a a
1-1331-1331-133
0
0 0
0 0
2 2
2 2 2 23
2 2
x y z
x y z x y z
x y z
v
v B
v B
v = a a a
E a a a × a a a
a a a
(b)
(c)
0
0
0 0
along 2
2 23
2 2 2 23
0
v =
a a a
E a a a × a a a
x y z
x y z x y z
v y z xv
v B
1-134
For a given E, to find B,
One force not sufficient. Two forces are needed.
v B Fq
– E
v1 B F1q
– E C1
v2 B F2q
– E C2
1 2 1 2
1 2 1 2
1 2
C × C v ×B × v ×B
v ×B B v v ×B v B
= C v B
1-1351-135
B C2 C1C1 • v2
provided , which means v2 and v1 should not be collinear.
1 2 0 C v
1-136
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
the same holds.(1)
(2)
(3)
1
2
1 3
2 3
v v ×B = 0
v v ×B = 0
1 2 , where + 0,
m nm n
m nv v
v
1 q qE v B 0
2 q qE v B 0
q qE v B 0
1-1371-137
Alternatively,
1 2
1 2
Both and are collinear to .
k
v v v v B
v v v v
(1) (2) m n
m n m n
1 2
1 2
–1 –
for = –
v vv
v v
kk
m n nkm n m
1-138
1 2 0
m nq q
m nv v
E B
1 2
m n
m nv v
v
1-139
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.
1-140
Problem S1.11. Finding the electric and magnetic fields from three forces experienced by a test charge
The EndThe End