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

qq

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