electromagnetism

Advanced ElectricalEngineering 1

Electromagnetics

Prof. S. Peik

September 5, 2013

„The prerequisite forknowledge is curiosity“

Jacques CousteauFrench marine biologist

1910 – 1997

Prof. S. Peik

Foreword

Unlike other engineering disciplines, the complete theory of electrical engineeringcan be summarized in four fundamental equations known as Maxwell’s equations.This course gives an introduction to electric and magnetic field theory, leading toMaxwell’s equations. In addition, the theory is applied to wave propagation prob-lems and guided waves on transmission lines. This knowledge enables us to un-derstand the physics behind electrical signals traveling through lines and electronicdevices.

These lectures notes are supplemental material to the lectures. Parts of this docu-ment are taken from various sources.

Lecturer:

Prof. Dr. S. PeikDepartment of Electrical Engineering and Computer ScienceUniversity of Applied Sciences BremenNeustadtswall 3028199 Bremen

Office E E609Phone 5905-2437email [email protected]

The title page shows Faraday’s first induction coil.Using this coil Faraday discovered the law of induc-tion, now known as Faraday’s law.

Version: September 5, 2013 1

Contents

1 Introduction 8

1.1 The Phenomenon of Electricity . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2 Electrical Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.2.1 Separation of Charges . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3 Conservation of Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.4 Charge Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.5 Moving Charge, Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 Electric Fields 16

2.1 Coulomb’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.2 Permittivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.2.1 Analogy to Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.3 Vectorial Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.3.1 Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.4 Definition of the Electric Field . . . . . . . . . . . . . . . . . . . . . . . . 20

2.4.1 The Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.5 The Field Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.6 Field Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.6.1 Electric Field of a Point Charge . . . . . . . . . . . . . . . . . . . . 24

2.6.2 Vector Type Field Equation . . . . . . . . . . . . . . . . . . . . . . 24

2.6.3 General Description . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.7 Field Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.8 Electric Fields in Conductors . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.8.1 Homogeneous Field . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.9 Typical E-Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.10Summary E-Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

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3 Potential 33

3.1 Energy of the Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.2 Electric Potential V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.2.1 Constant-Potential Surfaces . . . . . . . . . . . . . . . . . . . . . 37

3.3 Independence of Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.4 Potential of a Point Charge . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.5 Electric Voltage V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.6 Potential from Charge Distributions . . . . . . . . . . . . . . . . . . . . . 41

3.7 The Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.8 Analogy to Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.9 The Conversion Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.10Electric Dipoles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3.11Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4 Capacitance and Capacitor 50

4.1 Capacitance of a Parallel Plate Capacitor . . . . . . . . . . . . . . . . . . 51

4.2 Recipe for Deriving the Capacitance . . . . . . . . . . . . . . . . . . . . . 52

4.3 Circuits with Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4.4 Energy in Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

5 Electric Flux and Flux Density 57

5.1 Definition of the D-Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.2 D-Fields in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5.3 Electric Flux Ψ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5.4 Relation between D and E . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5.5 Gauss’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

5.6 Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

5.7 Gauss’s Law in Differential Form . . . . . . . . . . . . . . . . . . . . . . 61

5.8 Influence and Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5.9 Partly Filled Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5.10E-Field and D-Field on Boundaries . . . . . . . . . . . . . . . . . . . . . . 63

5.11Stored Energy in Electrostatic Fields . . . . . . . . . . . . . . . . . . . . . 66

5.12Poisson’s and Laplace’s Equations . . . . . . . . . . . . . . . . . . . . . . 68

5.13Procedure for Solving Boundary Value Problems . . . . . . . . . . . . . . 68

5.14Solving Laplace with Finite Element Methods . . . . . . . . . . . . . . . . 70

5.14.1Relaxation Method . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

5.15Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73


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6 Moving Charges 74

6.1 Current I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

6.2 U-I-Relation at Capacitors in AC Circuits . . . . . . . . . . . . . . . . . . . 75

6.3 Current Density ~J . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

6.4 Current Density Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

6.5 Relation Between E and J . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

6.6 Resistance R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

6.7 Drift Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

6.8 Converted Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

6.9 Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

7 Magnetostatics 84

7.1 History of Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

7.2 Permanent Magnets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

7.3 Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

7.4 Magnetic Field Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

7.5 Magnetic Field of the Earth . . . . . . . . . . . . . . . . . . . . . . . . . . 86

7.6 The Magnetic Field H . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

7.6.1 Magnetic Field of a Current Through a Long Wire . . . . . . . . . . 87

7.7 Ampere’s Law, Definition of H . . . . . . . . . . . . . . . . . . . . . . . . 88

7.8 Magnetic Fields Due to Current in a Long Straight Wire . . . . . . . . . . 89

7.8.1 The Curl Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

7.9 Ampere’s Law in Point Form . . . . . . . . . . . . . . . . . . . . . . . . . 92

7.9.1 H-Field on a Sheet of Current . . . . . . . . . . . . . . . . . . . . . 95

7.10Solenoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

7.11Biot-Savart’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

7.12Magnetic Dipoles and Current Loop Analogy . . . . . . . . . . . . . . . . 99

7.13Definition of Magnetic Field by the Effect . . . . . . . . . . . . . . . . . . 100

7.14Magnetic Flux Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

7.14.1Permeability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

7.15Hysteresis Loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

7.16Magnetic Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

7.17Magnetic Vector Potential . . . . . . . . . . . . . . . . . . . . . . . . . . 105

7.17.1Gauge Transformations . . . . . . . . . . . . . . . . . . . . . . . . 106

7.17.2Poisson Equation for Magnetostatics . . . . . . . . . . . . . . . . . 108

7.18Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110


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8 Inductance and Magnetic Circuits 111

8.1 Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

8.2 Flux Linkage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

8.3 Inductance of a Long Solenoid . . . . . . . . . . . . . . . . . . . . . . . . 112

8.4 Circuits with Inductances . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

8.5 Energy in the Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . 114

8.6 Magnetic Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

8.7 Magnetic Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

8.7.1 Perfect Magnetic Conductor . . . . . . . . . . . . . . . . . . . . . . 118

9 Time-Varying Fields 119

9.1 Faraday’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

9.2 Lenz’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

9.3 Moving Loop in a Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . 121

9.4 Self- and Mutual Inductance . . . . . . . . . . . . . . . . . . . . . . . . . 122

9.4.1 Mutual Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

9.4.2 Self Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

9.5 V-I-Dependence in coils in AC-Circuits . . . . . . . . . . . . . . . . . . . 123

9.6 Transformer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

9.6.1 Ideal Transformer . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

9.7 Eddy Currents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

10Electromagnetic Forces 130

10.1Lorentz Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

10.1.1Force on a Current Carrying wire . . . . . . . . . . . . . . . . . . . 131

10.2Force on two Parallel Wires . . . . . . . . . . . . . . . . . . . . . . . . . . 132

10.3Force on an Air Gap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

11Electromagnetic Waves 134

11.1History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

11.2Maxwell’s Contribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

11.3Displacement Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

11.4Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

11.5Time Harmonic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

11.6Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

11.7Plane-Wave Propagation in Medium . . . . . . . . . . . . . . . . . . . . . 142

11.8EM-Wave Reflections at Normal Incidence . . . . . . . . . . . . . . . . . 144


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12Waves on Transmission Lines 147

12.1Time-Harmonic Signals on Lines . . . . . . . . . . . . . . . . . . . . . . . 149

12.2Solution of the Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . 150

12.3Propagation Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

12.4Wave Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

12.5Wave Length and Propagation Constant . . . . . . . . . . . . . . . . . . . 152

12.6Phase Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

12.6.1Lossless Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

12.7Determining the Constants by Boundary Conditions . . . . . . . . . . . . 156

12.8End Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

12.9Terminated Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

12.9.1Matched Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

12.9.2Arbitrary Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

12.10Reflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

12.11Input Impedance of a Line . . . . . . . . . . . . . . . . . . . . . . . . . . 161

12.11.1Impedance Transformation . . . . . . . . . . . . . . . . . . . . . . 162

12.11.2Short and Open Line Impedance . . . . . . . . . . . . . . . . . . . 163

A Vector Algebra 166

A.1 Vectors and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

A.1.1 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

A.1.2 Vector Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . 167

A.2 Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

A.2.1 Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 168

A.2.2 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 169

A.2.3 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 170

A.2.4 Relations of Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 171

A.2.5 The Poisonous Snake . . . . . . . . . . . . . . . . . . . . . . . . . 171

A.3 Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

A.3.1 Line, Surface and Volume Integrals . . . . . . . . . . . . . . . . . . 172

A.3.2 Del Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

A.3.3 Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

A.3.4 Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176

A.3.5 Curl and Stokes Theorem . . . . . . . . . . . . . . . . . . . . . . . 177

A.3.6 Laplacian Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

A.4 Converting Operators between Coordinate Systems . . . . . . . . . . . . 179


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B Useful Tables 182

B.1 Greek Alphabet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

B.2 Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183


Introduction 1During this course we develop a better understanding of the underlying physics ofelectric circuits, including properties of resistors, capacitors, inductors, and trans-mission lines. Firstly, we concentrate on the fundamental concepts of electrostatics(capacitors, resistors) and magneto-statics (inductors). Here we explore the ba-sic laws of electromagnetics such as Coulomb’s law, Gauss’ law, Ampere’s law andFaraday’s law. In addition to descriptive methods we acquire mathematical methodssuch as vector calculus in order to solve more complex electromagnetic problems.Secondly, we extend our knowledge to electrodynamic concepts which will enable usto investigate wave problems and transmission lines. The course concludes in theformulation of the complete Maxwell’s equations and the application of Maxwell’sequations. [1, 2, 3, 4, 5, 6]

1.1 The Phenomenon of Electricity

Electricity was first observed by the Greek philosopher Thales (577 b.c). He noticed,that a glass and an amber rod attract each other when rubbed with silk.

Figure 1.1: Thales of Miletus (ca. 635 BC-543 BC), also known as Thales the Mile-sian, was a pre-Socratic Greek philosopher and one of the Seven Sages of Greece.Many regard him as the first philosopher in the Greek tradition as well as thefather of science. Thales is credited with first popularizing geometry in ancientGreek culture, mainly that of spatial relationships. Herodotus cites him as hav-ing predicted the solar eclipse of 585 BC that put an end to fighting between theLydians and the Medes.

Two amber rods expel each other, whereas an amber and a glass rod attract eachother.

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Am

ber Rod

Glass Rod

Am

ber Rod

Amber Rod

Figure 1.2: Phenomena of Electricity

This phenomena can be explained with the existence of electric charges. Since weobserve attraction and repulsion, there must be two different kinds of charges. Wename them positive (+) and negative (-) charge. This definition goes back to Ben-jamin Franklin, who defined arbitrary, that positive charges are the charges on theglass. The word electricity is derived from the Greek word Eλεκτρων (electron)meaning amber.

1.2 Electrical Charge

The origin of charges can be pinpointed inside the atom. Using Bohr’s model wecan associate positive charges with the protons in the atom’s nucleus, and negativecharges with the electrons, that form the atom’s hull. Electrons can be removedfrom the atom.

+

+

++

Elektron

Proton

Neutron

Figure 1.3: Bohr’s Model

As observed with the amber and the glass rod, charges of charges of opposite signattract each other, whereas charges of equal sign repel each other.

We use the letter Q for the quantity of charge. The unit of Q is Coulomb (or C inshort) named after Charles Augustin de Coulomb.

1.2.1 Separation of Charges

When we feed energy to the atom we can extract an electron from the atom. Theremaining positively charged atom is now called an ion. We can also add electrons


Prof. S. Peik

to atoms; in that case the atom is negatively charged by one electron charge. Thisis also called an ion.

Generally, matter is always composed of negative charges (electrons) and positivecharges (protons). When both charges are balanced within an object, the object iselectrically neutral outside. However, we can always separate positive and negativecharges, such that we create a positively charged section and a negatively chargedsection.

+

+

+

+

+

+ - -

-

-

-

-

+

+

+

+

+

+- -

---

Attraction F

Voltage V

Separation

by adding energy

Q=Q-+Q+=0

Q+ Q-

Figure 1.4: Separation of Charges

We will observe the following effects:

1. Force: On the charged bodies acts a force

• attraction on like charges

• repulsion on unlike charges

2. Voltage: Between the two charged bodies exists a voltage V = ϕ1 − ϕ2.

1.3 Conservation of Charge

By experiment we can show, that there is an interesting proposition about charge

Conservation of Charge:The total charge Q of a closed system is always constant. No physical or

chemical process is able to change it.

The proposition is based on the theory, that the charge of a proton and electronis fixed and electrons and protons cannot vanish and that opposite equal chargesnullify each other.

There are a number of other conservation laws in physics, e.g. energy conservationand mass conservation. These propositions can be confirmed by experiment butcannot be proved directly.

1.4 Charge Distributions

Charges can be be placed either in a point or can be distributed spatially. The maincharge distributions are


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• Point charges

• Line charges

• Surface charges

• Space or volume charges

Figure 1.5 shows common possible charge distributions. For line charges, the chargeis usually given by Charge per length, i.e. λ = Q

l , with the unit Coulombs per meter.

Similarly, this applies to surface charges σ = QA and volume charges ρ = Q

V .

Volume Charge ρLine Charge λ

Surface Charge σ

Figure 1.5: Charge Distributions

As mentioned earlier those charges are denoted usually by their densities, e.g. ρ inC

m3 .

In a cloud or line of charge we define dQ as the infinitesimal small charge element.The total charge is calculated by

Q =∫

VoldQ =

∫

Vρ(x, y, z) dV (1.1)

Similarly for line and surface charges by

Q =∫

Lengthλ(x, y, z) dl (1.2)

Q =∫

Areaσ(x, y, z) dA (1.3)

The charge per length, area, or volume is called the charge density. When thecharge density is equal everywhere in space, we say the charge distribution is ho-mogeneous. When the charge density changes spatially the charge distribution isinhomogeneous. In that case the charge density must be specified for all spatialpoints as a function of the spatial components x, y and z, that is ρ(x, y, z). In general

• Line charge density λ(~r) = dQdl with the unit C

m .

• Surface charge density σ(~r) = dQdA with the unit C

m2 .

• Space charge density ρ(~r) = dQdV with the unit C

m3 ,


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where~r = (x, y, z) is the vector leading to the point where the charge is located, theso called source points .

The total charge over a volume can be computed by integration of the charge den-sity. For example

Q =∫

Volρ(x, y, z) dx dy dz (1.4)

This integration means, that we sum up all charge densities over the complete vol-ume Vol. We can imagine the integration as a summation of charges in very small(infinite small) cubes filling the complete volume. Within each cube the charge ishomogeneous.

=⇒ See Also: Definition of Coordinate Systems Appendix A.2

The total charge is usually specified by the capital letter Q. In contrast, point chargesare specified by the small letter q.

Example 1: A cube of 1cm×1cm×1cm has a homogeneous volume charge of ρ = 12 µCm3 .

What is the total charge Q of the cube?

Q =∫ 1cm

0

∫ 1cm0

∫ 1cm0 ρ dx dy dz = 12 · 10−12C

Example 2:A square plate of size 2a× 2a is located in the x-y plane centered at x = y = 0 is charged

with a surface charge density of σ(x, y) = (1− x2

a )(1−y2

a ) . Find the total charge Q of theplate?

Q =∫ a

−a

∫ a

−aσ(x, y) dx dy =

∫ a

−a(1− y2

a)∫ a

−a(1− x2

a)dx dy (1.5)

=∫ a

−a(1− y2

a)(2a− 2a2

3)dy (1.6)

= (2a− 2a2

3)2 = 4a2 − 8a3

3+

4a4

9(1.7)

you may use ipython or a similar program, try:

x,y,a, sigma=symbols('x y a sigma')

sigma = (1-x**2/a)*(1-y**2/a)

Q=Integral(sigma, (x, -a, a), (y, -a, a))

Qresult=expand(Q.doit())

Eq(Q,Qresult)

∫ a−a∫ a−a

(1− x2

a

) (1− y2

a

)dx dy = 4

9 a4 − 83 a3 + 4a2


Prof. S. Peik

Example 3:Find the total charge Q in a sphere with volume charge ρ =

ρ0R r/3 and radius R

Q =∫∫∫

ρ(r)dV (1.8)

=ρ0R

∫ 2π

0

∫ R

0

∫ π

0

r3

sin θ r2 dθdrdφ (1.9)

=ρ0R

2π∫ R

0

∫ π

0

r3

3sin θ dθdr (1.10)

=ρ0R

2π∫ π

0

R4

12sin θ R dθ (1.11)

=ρ01

πR3

6

∫ π

0sin θ dθ (1.12)

= ρ0πR3

6[− cos θ]π0 (1.13)

= ρ0πR3

3(1.14)

Maxima Code for the Integration:

r,phi,theta,rho,rho0,R=symbols('r phi theta rho rho_0 R')

display('Volume Charge Density:',Eq(rho,rho0/R*r/3))

rho=rho0/R*r/3

dV=sin(theta)*r**2

Q=Integral(rho*dV, (theta, 0, pi), (r, 0, R), (phi,0,2*pi))

Eq(Q, Q.doit())

1.5 Moving Charge, Current

The energy used to separate charges is released when the charges are joined again.For example, in a flash light the separated charges inside the battery are joinedthrough the lamp. The energy is released as light and heat in the lamp. Electricalengineering makes use of this retaining and releasing of energy, for transmission ofenergy (power systems) or signal transmission (communications).

The flow of electrons through a wire is called electrical current. The current is alwayscaused by unequal charges at the end of the wire. This tension1 caused by theunequal charges is called the voltage.

The current I inside the wire is defined as the ratio of the charge Q flowing throughthe the time period t trough the cross section of the wire divided by the time periodt.

current I =Charges Q flowing through the cross section

time period t(1.15)

or using differentials

I =dQdt

⇔ dQ = I · dt (1.16)

1Note, in German the word for tension and voltage is the same called Spannung


Prof. S. Peik

Interestingly, the area of the cross section is irrelevant.

The unit of the electric current is consequently Cs . This unit is abbreviated Ampere

or A in short. For historic reasons, the Ampere is a fundamental SI unit whereas theCoulomb is derived from the Ampere. That is why the Coulomb is often referred toas Ampere-Seconds or As.

By counting electron flowing through a wire, we can derive the amount of chargeone single electron is carrying2. We get as the electron charge

qelectron = e− = −1, 602 · 10−19C = −qe (1.17)

where obviously qe = 1, 602 · 10−19C. The amount qe is the so called elementarycharge. This is the smallest quantity charges come in. We cannot generate half anelementary charge, as electrons cannot be split. Note, that the electron charge isthe negative elementary charge.

Example 4: One electron per second flows through a wire. Then, the current through thewire isI = −qe · 1s = −1, 6 · 10−19 AUsing very sensitive Ampere meters we can measure currents as small as 10−14 A. Thiscorresponds to 62.500 electrons/s.'

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Cell Phone Battery and Lightning

A cell phone battery with a capacity of 1000 mAh at 3.7V can deliver 1 Ampere current forone hour.The charge on the battery isQ = 1A · 3600s = 3600CThe stored energy is E = V · I · t︸︷︷︸

Q

= 3.7V · 1A · 3600s = 13320J

In comparison a cloud in a thunderstorm holds up to 200 Coulomb with voltages around 30Million Volts.The stored energy is E = V ·Q = 3 · 107V · 200C = 6, 000, 000, 000J

A detailed discussion of currents and current densities is performed in Chapter 6. Inthe next chapter we will concentrate on static (i.e. fixed) charges.

1.6 Summary

• There are positive and negative charges

2As the elementary charge is very small, we cannot just “count” electrons in a wire. That’s why theelectron charge is in reality measured with different methods.


Prof. S. Peik

• equal charges repel; unequal charges attract

• Separation of charges cause a voltage and mechanical forces

• The sum of all charges in a closed system is constant

• Charge only come in multiple of the elementary charge qe

• Charge of an electron is −qe =-1,602 10−19 C

• There are point, line, surface, and volume charges

• Electric current describes moving charges with I = dQdt

• The unit of charge is the Coulomb and the unit of current is the Ampere


Electric Fields 2In this chapter we will discuss the effects due to static electric charges only. Thismeans, that there are no currents flowing. This section of electromagnetism is calledelectrostatics.

2.1 Coulomb’s Law

In the last Chapter we discovered that there are the forces between charged bodies.However, we described the phenomena, qualitatively only. The exact strength offorces between charges can be found experimentally. We will perform the followingexperiments as shown in Figure 2.1. Two charges of strength q1and q2 are separatedby the distance r. On charge 1 acts the force F1, on charge 2 acts the force F2.Alternatively, we place the setup into a medium.

q1 q2

q1 q2

r

r

q1 q2r

F1 F2

F1 F2

F1 F2

Medium

Figure 2.1: Experiment to Coulomb’s Law

By accurate measurements we can find the following proportionalities:

• F1 = F2 = F always

• F ∝ q1

• F ∝ q2

16

Prof. S. Peik

• F ∝ 1r2

• F is a function of the medium, e.g. in water F = 181 Fvacuum

We can establish the following proportionality

F ∝q1 · q2

r2 (2.1)

assuming the charges are very small compared to the distance r. This first knownquantitative law of electricity was found in 1785 by Augustine Coulomb, see Figure2.2. Hence it is called Coulomb’s law.

This proportionality can be transformed into an equality by introducing a constant k

F = kq1 · q2

r2 (2.2)

By experiment we can figure out, that k depends on the media surrounding thecharges. The medium around charges is called the dielectric. The factor k has aunit. We get

[k] =[F][r]2

[Q]2(2.3)

[k] =kg ·m

s2 m2 1A2 · s2 (2.4)

with

[U] = 1V =kg ·m2

s3 ·A (2.5)

wet get

[k] =VmAs

(2.6)

By measurements, we can find the value of k in vacuum to be

k =1

4πε0(2.7)

with ε0=8.854·10−12 As/Vm

When introducing a surrounding medium, the k changes by a factor εr, where εr isa dimensionless constant which depends on the dielectric. Together with ε0 we canintroduce a combined constant

ε = ε0εr (2.8)


Prof. S. Peik

2.2 Permittivity

This dependence of the forces from the medium is called the permittivity. The Con-stant ε is the permittivity constant or number, with ε0 being the absolute permittivitynumber of vacuum and εr being the relative permittivity constant or number. A moredetailed description of permittivity is given in Chapter 5. The relative permittivitiesof some materials is given in Table 2.1.

Using the permittivity definition we can write

F =1

4πε0εr

q1 · q2

r2 (2.9)

This is the form of Coulomb’s law used in many textbooks.

Figure 2.2: Coulomb is distinguished in the history of mechanics and of electric-ity and magnetism. In 1779 he published an important investigation of the lawsof friction (Théorie des machines simples, én ayant egard au frottement de leurs partieset a la roideur des cordages), which was followed twenty years later by a memoiron viscosity. In 1785 appeared his Recherches théoriques et expérimentales sur la forcede torsion et sur l’élasticité des fils de metal. This memoir contained a description ofdifferent forms of his torsion balance, an instrument used by him with great suc-cess for the experimental investigation of the distribution of charge on surfacesand of the laws of electrical and magnetic force, of the mathematical theory ofwhich he may also be regarded as the founder.

Material εr Material εr

Vacuum 1 Quartz 3,8...5Air 1,00059 Glass 5...7

Polyethylene 2,3 Ceramic 9,5...100Rubber 2,5...2,5 dest. Water 81Amber 2,8 Diamond 16,5

Table 2.1: Permittivity of Some Materials

2.2.1 Analogy to Gravitation

Coulomb’s law is very similar to Newton’s law of gravitation:

F ∝m1 ·m2

r2 (2.10)

In contrast to Newton’s law, Coulomb’s law has some differences:


Prof. S. Peik

• There are two types of charges (+ and -)

• even charges repel, an effect unknown in classical gravitation theory

• Forces are much, much, much stronger

Example 5: Two students stand 1m apart. One student lacks 1% of his electrons. Theother student has 1% too much.

F F

++

+__

_

The force F is equal to the gravitational force of the earth. Normally, we cannot gener-ate such big electrostatic forces. Positive and negative charges are generally very nicelybalanced.

2.3 Vectorial Description

When the charges are positioned into the three-dimensional space at the positionvectors~r1and ~r2, we can rewrite Coulomb’s law as

~F21 =1

4πε0εr

q1q2

R2 ~a21 (2.11)

with

R = |~r2 − ~r1| (distance) (2.12)

~a21 =~r2 − ~r1

R(2.13)

Here we say, ~F21 is the force on charge 2 due to charge 1.

q1

q2

r1

r2

F12

F21

r2 - r1

Figure 2.3: Vectorial Coulomb Forces on Two Charges

We may write directly

~F21 =1

4πε0εrq1q2

(~r2 − ~r1)

|~r2 − ~r1|3 (2.14)


Prof. S. Peik

2.3.1 Superposition

The principle of superposition applies to Coulomb forces. That means that the forcesdo not interfere with each other and the resulting force is added linearly.

If we have a number of charges q1, q2, . . . , qN the total force on charge qj is

~Fj =qj

4πε0

N

∑k=1

qk(~rj −~rk)

|~rj −~rk|3(2.15)

or for charge distributions the force on charge qj is

~Fj =qj

4πε0

∫

V

ρ(~rj −~rρ)

|~rj −~rρ|3dV (2.16)

Example 6:Three charges q1 = 1C, q2 = 1 C and q3 = 2C are located at ~r1 = −x, ~r2 = x and~r3 = 2x+ 3yin vacuum.Find the force ~F1, ~F2,~F3 on the charge q1to q3.First we find

~r1 −~r2 = −2x (2.17)

~r1 −~r3 = −3x− 3y (2.18)

~r2 −~r3 = −x− 3y (2.19)

The force on q1 is

~F1 =1C

4πε0

(1C−2x|2|3 + 2C

−3x− 3y√

32 + 323

)=

1C2

4πε0

(−2x|2|3 +

−6x− 6y√

32 + 323

)(2.20)

=1C

4πε0

(−1

4x− 2√

2x− 2√

2y)

(2.21)

=1C

4πε0

((−1

4− 2√

2)x− 2√

2y)

(2.22)

Similar for the other charges. (Left to the reader)

2.4 Definition of the Electric Field

Coulomb’s law opens some questions: The force acts on separated charges that areplaced in a vacuum. How can a force act on something that is not connected to thecause by any medium. This problem is similar to the forces acting in Newton’s lawor on a magnetic compass.

To answer this question we have to abandon the idea of cause and effect are joinedthrough a physical touchable carrier. In our new understanding the space itself (evenempty space) becomes the carrier of forces. The physicians M. Faraday (1873) andJ.C Maxwell developed this idea. The property of space to act with forces on chargesetc. is called a field.


Prof. S. Peik

2.4.1 The Field

A field, therefore, describes a special property of the space. In an electric field, thisspecial state is created by the presence of charges somewhere in the field as seenfrom Coulomb’s law.

In order to understand, how fields look like, we have to answer the question howthe properties of the space changes due to the presence of charges? Let us walkthrough an example first.

We assume two electric point charges, one positive one negative, in space as seenin Figure 2.4.

Now, we position a small test charge somewhere in the space. The test chargeshould be very small compared to the other charges. How does the force act onthis test charge due to the presence of the two other charges? If we can answerthis question, we found the field description for the point where the test charge isplaced.

q1 q2

+ -

Test charge

Figure 2.4: Experiment for Field Determination

On the test charge acts a force from the two other charges. As we assume a positivetest charge, one force pushes the test charge away from the left charge And oneforce pulls it towards the right charge. The second force is stronger as the distancebetween test and right charge is smaller. The resulting Force Ftot is the vectorialaddition of the two forces as shown in Figure 2.5.

Figure 2.5: Total Force on Test Charge

Now the field is determined for the location of the test charge: In that location thefield creates a force down-right of certain strength. After determining the field, wenot not need to know the cause of the field (here two charges) anymore. The field isa complete description of the physical phenomenon.

However, we need to derive the field for the other locations in space as well. There-fore, we place the test charge at different locations in space and repeat our forcecalculation.


Prof. S. Peik

Now, we can describe the complete space in form of an electric field. The elec-tric field—describing a force—has a direction (field direction) and magnitude (fieldstrength) everywhere.

We can envisage the field best by drawing imaginative force lines in space. Theforce lines describe the path of the test charge traveling through space due to thepulling-pushing forces of the field. Some force lines are shown in Figure 2.6. Theselines are also known as field lines.

Figure 2.6: Force Lines of Example Field

2.5 The Field Concept

For a detailed discussion of fields we first define the field in general:

Fields describe a special property of space

Fields can either describes a cause or an effect in space.

For example, elevation lines describe the property of the surface of earth, that areat the same elevation. This is a field.

Generally, we have two types of fields: Scalar and vectorial fields. Vectorial fieldscan be either irrotational (divergence) fields or rotational fields.


Prof. S. Peik

Scalar Fields

Fields

Vectorial fields

describe a scalar property

e.g.. temperature, pressure,

elevation

describe a

property with magnitude and direction

e.g. wind, gravity

Conservative Fields

(Source Fields, Irrotaional Fields) Rotational Fields

Field lines have a source

and a sink

e.g. electric fields

field lines circle around

e.g. hurricanes

As mentioned before, electrical fields have a direction and a magnitude. The forcelines start a one charge and end on another charge (or at infinity). Hence, electricalfields are vectorial source fields.

2.6 Field Strength

We can now define a quantitative description of the space filled with charges. Wecall this description the electric field. The electric field describes the effect on acharge placed in the field.

However, this force depends on the test charge qt. Therefore, the force effect doesnot describe the electric field properly. The field would depend on the strength of thetest charge. We need to find a field description, that is independent of the charge qtand only describes the properties of space due to the placed charges.

Applying Coulomb’s law we find that the force on the test charge is doubled whenthe test charge qt is doubled. The charge qt is linearly proportional to the force thatact on it. Consequently, we can eliminate the effect of qt by defining the field as theratio of force to charge qt.


Prof. S. Peik

~E =~Fqt

(2.23)

with the unit

[E] =[F][Q]

=NAs

=VAsm

As=

Vm

(2.24)

With ~F(x, y, z) being the force acting on Q at the point x, y, z. As ~F is a vector, the

electric field ~E(x, y, z) is a vector, as well. Typical values of electric fields are givenin Table 2.2.

Field Type Field Strength EAtmosphere (clear weather) 100...200 V/m

Discharge level of Air 30 kV/cmSurface of cellular phone antenna (reception) 1...103 µV/mSurface of cellular phone antenna (transmit) 1000 V/m

Inside capacitor 106 V/mInside conducting wire 0,1 V/m

Inside depletion layer of semi conductor 104...108V/m

Table 2.2: Typical Values of Electric Fields

2.6.1 Electric Field of a Point Charge

From Coulomb’s law we can directly derive the electric field of a point charge. Withthe absolute value of the force F between a point charge q and a test charge qt:

|F| = 14πε

qt qr2 (2.25)

follows an electric field |E| as a function of r and the charge q of

|~E| = |~F|qp

=|F|qp

=1

4πε

qr2 (2.26)

The direction of the electric field is always radial symmetric.

2.6.2 Vector Type Field Equation

The above equation describes the absolute value of E only. The electric field, how-ever, is a vector quantity. We have to define the vector quantity ~E now.

Let us assume a point charge q in the origin of the coordinate system. We are nowlooking at the field point at x = rx, y = ry, z = rz. It is easy to see, that the direction


Prof. S. Peik

|E|

r

~1/r2

q

r

Field point with E

Figure 2.7: Field of a Point Source

of ~E in the point~r = (rx, ry, rz) is the the direction of~r, where~r is the vector leading to

the field point. In order to get the vector quantity ~E from the already known absolutevalue E we just multiply the absolute value by the direction of~r of length one. Withthe unit vector of r being r = ~r

~|r| we get1

~E =1

4πε

qr2

~r|~r| (2.27)

as the distance r2 = |~r|2 we can also write

~E =q

4πε

~r|~r|3 (2.28)

2.6.3 General Description

An electric field at the point~r due to a point charge q at r′ can be obtained by usingthe difference vector~r−~r′ as the distance vector ~p from equation 2.28

as

~E =q

4πε0

~r−~r′|~r−~r′|3 (2.29)

Where~r points to the field point of interest and~r′ points to the source point asseen Figure 2.3.

Field Point

q

r

r’

r - r’

E

Figure 2.8: Vectorial E-Field Description

1We denote unit vectors with a hat, e.g. z, p


Prof. S. Peik

Example 7:Find the electric field ~E at the point (x = 1m, y = 3m, z = −4m) with a point charge of 2 C at(x = 3m, y = 2m, z = 2m).

the difference vector is~l = (−2, 1,−6) with the absolute value l =√

41, hence

~E =1

4πε0

2As41m2

1√41

−21−6

(2.30)

The same applies for a field due to N charges

~E =1

4πε0

N

∑k=1

qk(~r−~r′)|~r−~r′|3 (2.31)

or in case of charge distributions with the infinite small charges dQ per element weget

~E =1

4πε0

∫λ(~r−~r′)|~r−~r′|3 dl for line charges (2.32)

~E =1

4πε0

∫σ(~r−~r′)|~r−~r′|3 dS for surface charges (2.33)

~E =1

4πε0

∫ρ(~r−~r′)|~r−~r′|3 dV for volume charges (2.34)

Example 8:Find the E-Field of a uniform infinite line charge on the z-axis.Since the line is infinite long and on z-axis the E-field is shows some symmetries. The fieldalways points radial outwards from the line and is independent of z. Hence, we need tocalculate the E-field in the xy-plane (with z = 0) only.

Using cylindrical coordinates (ρ, φ, z) we use as field point and source pointField point:~r = Rρ + φφSource point~r′ = zzhence:~r−~r′ = Rρ+ φφ− zz (Note: Here is the poisonous snake lurking, see Appendix A.2.5)

The distance is seen from the figure and Pythagoras: |~r−~r′| =√

z2 + R2

Now using equation 2.32

~E =1

4πε0

∫λ(~r−~r′)|~r−~r′|3 dl =

λ

4πε0

∫ zz− Rρ− φφ√

z2 + R23 dz (2.35)

using symmetrie we know that ~Ehas ρ-component only: (2.36)

=λ

4πε0ρ∫ R√

z2 + R23 dz =λ

4πε0Rρ

[z

R2√

z2 + R2

]∞

−∞(2.37)

=λ

4πε0Rρ(

1R2 − (− 1

R2 )) (2.38)

=λ

2πε0

1R

ρ (2.39)

The field of an infinite line decays with 1R with the distance to the line. The field is always

oriented in ρdirection.


Prof. S. Peik

Example 9:A ring of radius a is charged with the uniform charge density λ and is placed in the xy−Planewith the axis coaxial with the z-axis as shown below. Find the E-field along the z-axis.

x

y

z

a

Solution:A charge element has the charge dQ = λ dl, with dl = a dφThe vector~r leading to the field point on the z-axis is just ~z = zz. The vector leading to thesource point is in cylindrical coordinates ~r′ = aρ + φ′φ.Be aware, that~r−~r′ 6= zz− aρ− φ′φ, because of the poisonous snake in cylindrical coordi-nates (see Appendix A.2.5)The Field expression is

~E =1

4πε0

∫ 2π

0λ

~r−~r′

|~r−~r′|3a dφ′ =

14πε0

∫ 2π

0λ

z z√

z2 + a23 a dl′ +1

4πε0

∫ 2π

0

· · · φ· · · dl′ +

14πε0

∫ 2π

0

· · · ρ· · · dl′(2.40)

Due to symmetry we get only a z-component of ~E, such that Ez, i.e. ~E = (0, 0,~Ez) and wecan omit the other two integrals (which cause poisonous snake problems)with

Ez =1

4πε0

∫ 2π

0λ

z√

z2 + a23 a dl′ =1

2ε0

zaλ

(z2 + a2)32

(2.41)

when using the total charge Q = 2πaλ we get

Ez =1

4πε0

z

(z2 + a2)32

Q (2.42)

Note, that for z → ∞ (a is negligible), the expression converges towards Ez = 14πε0

1z2 Q,

which is the expression for the point charge. Hence, far away, the ring field resembles apoint charge.

2.7 Field Lines

Field lines are imaginary lines in space to visualize the electric field. Their tangentialdirection is always identical to the direction of ~E. Field lines do not exist in reality butare a helpful tool to understand the effects of the field. Field lines have the followingproperties

• The lines are always pointing in the direction of ~E. The direction is the directionof a force on a positive test charge.

• The density of the lines are a gauge for the absolute value of the field.Thedenser the lines the stronger the field.


Prof. S. Peik

• Field lines are always smooth and continuous. The start at positive charges orinfinity and end at negative charges or infinity. Exception: The lines can be noncontinuous at boundaries.

q>0 q<0

Field Line

E-V

ecto

rs

Q<0q>0

Figure 2.9: Field Lines as Description of Fields

Experimentally, field lines can be made visible through metal chips. Figure 2.10shows some of these experiments.

Figure 2.10: Field Lines Made Visible through Metal Chips, from [?]


Prof. S. Peik

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Lightning

Clouds contain millions and millions of water droplets and ice particles suspendedin the air. As the process of evaporation and condensation occurs, these dropletscollide with other moisture that is condensing as it rises. The importance of thesecollisions is that electrons are knocked off of the rising moisture, creating a chargeseparation. The newly knocked-off electrons gather at the lower portion of the cloud,giving it a negative charge. The rising moisture that has lost an electron carries apositive charge to the top of the cloud.As the rising moisture encounters colder temperatures in the upper cloud regionsand begins to freeze, the frozen portion becomes negatively charged and the un-frozen droplets become positively charged. At this point, rising air currents have theability to remove the positively charged droplets from the ice and carry them to thetop of the cloud. The remaining frozen portion either falls to the lower portion of thecloud or continues on to the ground.

The charge separation has an electric field associated with it. Like the cloud, thisfield is negative in the lower region and positive in the upper region. The strengthor intensity of the electric field is directly related to the amount of charge build-upin the cloud. As the collisions and freezing continue to occur, and the charges atthe top and bottom of the cloud increase, the electric field becomes more and moreintense – so intense, in fact, that the electrons at the Earth’s surface are repelleddeeper into the Earth by the negative charge at the lower portion of the cloud. Thisrepulsion of electrons causes the Earth’s surface to acquire a strong positive charge.All that is needed now is a conductive path so the negative cloud bottom can conductits electricity to the positive Earth surface. The strong electric field creates this paththrough the air, resulting in lightning. The lightning is a high-voltage, high-currentsurge of electrons, and the temperature at the core of a lightning bolt is incrediblyhot. For example, when lightning strikes a sand dune, it can instantly melt the sandinto glass. The combination of the rapid heating of the air by the lightning and thesubsequent rapid cooling creates sound waves. These sound waves are what we callthunder. There can never be thunder without lightning.

from: http://science.howstuffworks.com/question646.htm, photo: NASA


Prof. S. Peik

2.8 Electric Fields in Conductors

Electrical conductors, e.g. a copper rod, are defined as objects with charges inside,that can move freely around. In contrast, insulator, e.g. a plastic sheet, have fixedcharges, that cannot be moved.

In ideal conductor the charges can move without any obstacles such that we cannotobserve any frictional loss (=resistivity). Real conductors, however, always possesa resistivity. As a result, the freely movable charges cannot move without friction.

In ideal conductors we cannot observe electric fields. The reason for this is explainedas follows: Let us assume we created an electric field inside an ideal conductor. Theagile charges now move due to the field (which creates a force F = qE). Positivecharges will gather at one end of the conductor. The negative charges will gather atthe other end. Now, the charges themselves create an inner field in the conductor.The inner field compensates the applied field. The charges move as long around,as long there is a field (=force on charges). At the end, when no charges moveanymore the field is compensated to zero. We memorize :

There is no electric field inside ideal conductors~E ≡ 0

Using the same idea, we can find out that the inside of any hollow and even perfo-rated ( a cage) conducting body is field free. We call this a Faraday’s cage. A car, forexample is a Faraday’s cage.

ideal conductor

no field

ideal conductor

Applied Field

Forceneg. Charge

pos. Charge

ideal conductor

Inner Field compensates Applied Field

E E

no field inside

Faraday’s Cage

Inner Field compensates Applied Field

E

no field inside

Figure 2.11: E-Fields inside ideal conductors are compensated

Figure 2.12: Faraday’s Cage in Action, Please, Don’t Try This at Home! (takenfrom www.tesladownunder.com)


Prof. S. Peik

Using the same argument, we can demonstrate that the electric field is always per-pendicular to the surface of ideal conductors. Tangential components cannot exist,as the would create compensation currents in the conductor.

The E-Field is always perpendicular to ideal conductor surfaces!

E

Ideal ConductorId

eal Conductor

Ideal Conductor

EE

E=0E=0

Figure 2.13: Electric Field and Ideal Conductors

2.8.1 Homogeneous Field

A field is called homogeneous, when the field vector is the same in magnitude anddirection everywhere in the region of interest.

homogenes Feld inhomogenes Feld

Figure 2.14: Homogeneous and non-Homogeneous Field

2.9 Typical E-Fields

Besides the point charge field we can easily derive the field patterns of most sym-metric charge configurations. The most common fields are given in Figure 2.15.


Prof. S. Peik

| ~E| = 1

4πε

Q

r2for r > a

| ~E| = 0 for r ≤ a

| ~E| = 1

4πε

q

r2Point

Sphere

| ~E| = 1

2πε

λ

rLine

PlateParallel

Plates| ~E| = 1

ε

Q

2A

| ~E| = 1

ε

Q

A

Figure 2.15: Common Charge Arrangements and Their Fields

2.10 Summary E-Fields

• Coulomb’s Law describes the Forces between two electric charges.

• Permittivity describes the dependence of the Coulomb’s forces from the media.

• The absolute dielectric constant is ε0=8.854·10−12 As/Vm

• The relative dielectric constant captures the factor of forces

• Fields describe a special property of the space. The space surrounding electriccharges is called an electric field.

• We distinguish scalar Fields and vector fields (source- or curling fields)

• The electric field is a vector source field

• The electric field strength is defined as the ration of the force per charge.

• The direction of the E-field is the direction of the Coulomb’s force on a positivecharge

• Field lines illustrate the shape of the electric field

• In ideal conductor E = 0 , Field lines are always perpendicular to the surface ofan ideal conductor.


Potential 3The introduction of an electric field allows us to describes the the effects on chargesplaced into the space. The electric field is an ideal tool for calculating forces oncharges.

The rise of an electric voltage is the second effect of the separation of charges.Voltage can exist between isolators, conductors. Voltage always occurs betweentwo points of different potential.

Obviously, the potential offers a different method for describing the special prop-erty of space due to electric charges in that space. We can also find a connectionbetween the description through an electric field and the description through poten-tials.

3.1 Energy of the Electric Field

The electric potential can be defined directly from the energy levels in an electricfield. The concept can be best seen, when we take a look at the potentials in thegravitational field first as shown in Figure 3.1. In a gravitational field ~g the work Wis performed when moving the mass m against the field.

Similarly, work is performed in the electric field when we move an electrical chargeq against the electrical field ~E.

m

F

1

2

F=m·g

Moving mass m from 1 to 2

performs the work

W12=g·m·∆s

Gravitational Field

q

F

1

2

F=E·q

Moving charge q from 1 to 2

performs the work

W12=E·q·∆s

Electrostatic Field

+

g E

Figure 3.1: Analogy Gravitation and Electrostatic Field

33

Prof. S. Peik

Figure 3.2: Definition: Performed Work

Let us assume the charge q is moved about the distance ∆s exactly against the field~E. Now the work performed is

∆W = F · ∆l (3.1)

When moving the charge at an angle α with respect to the field ~E we perform thework

∆W = F · ∆l · cos(α) (3.2)

as seen in Figure 3.2.

Generally, we can phrase the the work as a scalar product of the two vectors of forceand direction

∆W = ~F · ∆~l (3.3)

where ~∆s contains the distance and direction of the movement.

Using this definition, we can easily determine the work performed, when a charge ismoved along a straight line in a homogeneous field.

In case of an inhomogeneous field we have to integrate along the path of the move-ment. Assume we move from P1 to P2 as shown in Figure 3.3. As a first approxima-tion we can sum up the work performed of small section of length ∆s of the path.The total work is then

W = ∑~F · ~∆l (3.4)

as shown in the left hand side of Figure 3.3.

When summing up an infinite number of infinite short distances of length ds we getthe exact result as an integration of

WP1,P2 =∫ P2

P1~F d~l (3.5)


Prof. S. Peik

Figure 3.3: Performed Work along an Inhomogeneous Field

This is shown on the right hand side of Figure 3.3.

The force ~F can now be replaced by ~F = ~E · q. The performed work due to a chargeof strength q is hence defined as :

WP1,P2 =∫ P2

P1~E · q d~l (3.6)

When we move along the field work is released (denoted positive), when movingagainst the field work has to be expended, as denoted negative.

The Integral

W =∫ P2

P1~E · q d~l (3.7)

is a charge dependent measure of the energy released when a charge q ismoved in the field from P1 to P2 .

Using this integral we can also use the energy released when moving a charge frompoint Pxto a reference point P0 to define the field. This energy is stored in the charge.We call this energy the potential energy of the charge at point Px .

the potential energy describes the energy that is released when moving q from pointPx to the reference point P0, that is

WPot(Px) =∫ P0

Px~E · q d~l = −

∫ Px

P0~E · q d~l (3.8)

The minus sign appears, because we reverse the path direction from (Px→ P0) to(P0→ Px).


Prof. S. Peik

P0

P1

Path

of I

nt. 5J

5J

5J5J

2J

2J

2J2J

E

Alte

rnat

e Pat

h o

f In

t.

Figure 3.4: Definition of a Potential Field from an E-Field

3.2 Electric Potential V

Now we can define an electric potential V1 , which is independent of the test chargeq. Again we normalize the energy with respect to the charge

V =Wpot

q(3.9)

The electric potential V is a scalar function of the location (x,y,z), i.e.

The Electric Potential V is a Scalar Field

The potential V is always referenced to a reference point P0 with the reference po-tential V0 = 0. The potential’s unit is Joules per Coulomb known as Volt (V).

[V] =JC

=VAsAs

= V (3.10)

Using eqn. 3.7 we get for the potential

V = −∫ Px

P0

~E · d~l (3.11)

1In the English literature the potential is usually abbreviated as V. However, try to avoid confusionwith the voltage V and the unit “Volt” V which use the same symbol!


Prof. S. Peik

where P0 is a fixed reference point with the the reference potential V0 = 0. If thepotential V is positive in a given point, we gain potential energy moving a positivecharge from the reference to that point. We need an external agent to perform thework.

The reference is usually set at the origin of the coordinate system or at infinity. Inprinciple it can be set anywhere in space, however.

3.2.1 Constant-Potential Surfaces

The potential remains constant, when we move perpendicular to the E-field, becauseno work is performed. When moving on a surface with constant potential, we callthis surface a constant potential surface. In two-dimensional field problems constantpotential lines are formed. As a general rule we can state

Constant potential lines or surfaces are always perpendicular to theE-field vector!

since the E-field vector is always perpendicular to ideal conducting surfaces we canalso state:

Ideal conducting surfaces are always constant potential surfaces; idealconductors are always on a constant potential

We can place a conducting foil on constant potential surfaces without disturbing thefield. Examples for E-fields with it’s constant potential surfaces and lines are givenin Figure 2.4.

Figure 3.5: E-Fields and Their Constant Potential Surfaces and Lines, Respec-tively

3.3 Independence of Path

When moving a charge in the electrostatic field or moving a mass in the gravitationalfield from point 1 to point 2, we have to perform the work W12. This means that theobject has now a potential energy of

Wpot = −∫ P2

P1G ds (3.12)


Prof. S. Peik

The path we choose to get from point 1 to point 2 is irrelevant. The total performedwork is always the same. The is a consequence of the law of conservation of energy.

The total work performed when moving a charge is independent of thechosen path

When moving a charge through the field and returning back to the starting point thetotal performed work must be zero. Compared to gravitational field, we obviouslyrecognize that we moved as much up as we moved down in the terrain.

This observation is part of the so called Maxwell’s equation for electrostatic fields.

E

Closed Path of

Integration

Ι

Figure 3.6: The Total Work W =∮

E ds on a Closed Path is Zero

Maxwell’s equation of electrostatic fields states:

∮~E d~s = 0 (3.13)

The total work performed on a closed path is zero!

The circle in the integral means, that we integrate over a closed path. Due to theproperty of eqn 3.13 the E-field is said to be conservative or irrotational. Thus, anelectrostatic field is a conservative field.

From this Maxwell’s equation we can also directly derive Kirchhoff’s voltage law:

∑Mesh

V = 0 (3.14)

3.4 Potential of a Point Charge

We can calculate the potential of a point charge q located at the origin with thereference potential set to zero at r0.

E =q

4πε0r2 r (3.15)

V = −∫ r

r0

q4πε0r′2

r dr′r (3.16)

V = − q4πε0

(1r0− 1

r) (3.17)


Prof. S. Peik

With the reference of V(∞) = 0, i.e. r0 = ∞ the potential between r and infinity is

V =q

4πε0r(3.18)

For a point charge placed at~r′ we get at the field point~r

V(~r) =q

4πε0

1|~r−~r′| (3.19)

Note, the potential decreases with 1r with the distance from the point charge.

3.5 Electric Voltage V

The potential is defined as an energy level compared to a reference point V0. How-ever, often we prefer to express the difference of potentials between two pointsdirectly. The difference of potential is the so called voltage. The voltage betweenthe points 1 and 2 is defined as

V12 = V1 −V2 (3.20)

where V1is the potential at point 1 and V2is the potential at point 2. A negativevoltage indicates, that the potential at point 2 is actually higher than at point 1.

The Voltage between two points can be obtained from the E-field directly

V12 = V1 −V2 =∫ P2

P1~E d~s (3.21)

Figure 3.7: Aleksandro Volta (1745-1827)


Prof. S. Peik

Example 10:

E

9cm

24cm

12cm

1

3 2

In the sketched homogeneous field we observe a voltage of 7,2 V between point 1 and 2.a) What is the electric field magnitude at point 1b) What is the electric field magnitude at point 2c) What is the voltage between point 2 and 3d) When fixing the potential at point 3 to V =0 V, find the potentials at point 1 and 2Solution:

a) V12 =∫ P2

P1~E d~x = |E| · (x2 − x1) following: |E| = 7,2V

24 cm = 30 Vm

b) E is constant: 30 Vm

c) V = E x = −30 · 0, 09 V = −2, 7 Vd)V1 = 0, 15 m · E = 4, 5 V and V2 = 0, 09 m · E = −2, 7 V


Prof. S. Peik

Example 11:Find the potential inside a parallel plate capacitor with constant E-field.

Assume a zero potential on the left plate.Solution:Between the plates we have

E = const =1ε

QA

(3.22)

Setting the potential on the left plate to V0=0V at xre f = 0, we get

V(x) = −∫ x

0~E d~s = −E · x = −1

ε

QA

x (3.23)

The integral reduces to a multiplication as the field is constant. The potential decades lin-early with the distance of x. The Equipotential surfaces are parallel to the plates

3.6 Potential from Charge Distributions

With the knowledge of the field from point charges, we can now calculate the po-tential field from any charge distribution by superposition similar to eqns 2.32 to2.34

V(~r) =1

4πε0

∫

Cλ(~r′)

1|~r−~r′|dl′ (3.24)

V(~r) =1

4πε0

∫

Sσ(~r′)

1|~r−~r′|dS′ (3.25)

V(~r) =1

4πε0

∫

Vρ(~r′)

1|~r−~r′|dV ′ (3.26)

The potential can be calculated in two different ways


Prof. S. Peik

• If the E-Field is known we use equation 3.11

• If the charge distribution is known, we use the integral formulas as sketchedabove

Example 12:A ring of radius a is charged with the uniform charge density λ and is placed in the xy−Planewith the axis coaxial with the z-axis as shown below

x

y

z

a

Find the potential field function along the z-axisSolution:

We use cylindrical coordinates. The field point is~r = zz and the source point is ~r′ = aρ + φφ,

The distance between the points is |~r−~r′| =√

a2 + z2

Again the length of the loop element is dl = a dφHence using eqn. 3.24:

V(~r) =1

4πε0

∮

Loopλ(~r′)

1|~r−~r′| dl (3.27)

=1

4πε0λ∫ 2π

0

1√a2 + z2

adφ (3.28)

=1

4πε0λa

1√a2 + z2

2π (3.29)

=λa

2ε0√

a2 + z2(3.30)

3.7 The Gradient

The electric field vector always points in the direction of the greatest alteration of thepotential. The change of potential per length (unit V

m ) is indicated by the magnitudeof the E-field vector.

Hence, we can derive the E-field ~E from a given potential field function V by usingthe following derivation

From our definition of the potential as V = −∫~E d~l follows that

dV = −~Ed~l = −Exdx− Eydy− Ezdz (3.31)

Using the term

dV =∂V∂x

dx +∂V∂y

dy +∂V∂z

dz (3.32)


Prof. S. Peik

we can obtain by comparing eqn. 3.31 and 3.32:

Ex = −∂V∂x

Ey = −∂V∂y

Ez = −∂V∂z

(3.33)

or

~E = −grad V(x, y, z) = (3.34)

with the definition

grad f (x, y, z) =

d fdxd fdyd fdz

(3.35)

This is the so called gradient operator. Note, that the gradient translates a scalarfunction into a vector function. Often we can find the gradient denoted with the socalled ∇ (say:nabla) operator, defined as

∇ =

ddxd

dyddz

(3.36)

Then we can write the gradient as

grad f (x, y, z) = ∇ · f (x, y, z) =

ddxd

dyddz

· f (x, y, z) =

d fdxd fdyd fdz

(3.37)

leading to the often used notation

~E = −∇ ·V (3.38)


Prof. S. Peik

Example 13:

Find the E-field from the potential field V(x, y, z) = 3 V + 2 Vm · y + 1, 5 V

m z2

From eqn. 3.11 we get

~E =

0−2−3z

V

m

The electric field and potential at point (1,1,1) is

V = 3V + 2V + 1.5V = 6.5V (3.39)

~E =

0−2−3

V

m(3.40)

We can derive the potential in point (1,1,1) also from the E-field by integrating the field fromthe origin to the point (1,1,1). Since the potential at the origin is already 3V we have to add3V to the final solution, hence

V = 3V−∫ (1,1,1)

(0,0,0)~E · d~l (3.41)

Since we can set the path of integration freely we integrate along the path (0, 0, 0) →(1, 0, 0) → (1, 1, 0) → (1, 1, 1). In other words, we integrate along the x-axis first, then alongthe y-axis and then along the z-axis

V = 3V−∫ (1,0,0)

(0,0,0)~E · d~l −

∫ (1,1,0)

(1,0,0)~E · d~l −

∫ (1,1,1)

(1,1,0)~E · d~l (3.42)

= 3V−∫ x=1

x=0~E · xdx

∣∣∣y=0,z=0

−∫ y=1

y=0~E · ydy

∣∣∣x=1,z=0

−∫ z=1

z=0~E · zdz

∣∣∣x=1,y=1

(3.43)

= 3V−∫ x=1

x=0

−

020

· xdx−

∫ y=1

y=0

−

020

· ydy−

∫ z=1

z=0

0−2−3z

· zdz(3.44)

= 3V−∫ x=1

x=00dx−

∫ y=1

y=0−2dy−

∫ z=1

z=0−3z · zdz (3.45)

= 3V + 2V + 1.5V (3.46)

= 6.5V (3.47)

Example 14:Find the electric field along the z-axis from the potential of the last Problem with the chargedring

~E = −∇V =

ddxd

dyddz

λa

2ε0√

a2 + z2= −

ddx

(λa

2ε0√

a2+z2

)

ddy

(λa

2ε0√

a2+z2

)

ddz

(λa

2ε0√

a2+z2

)

(3.48)

=

00

λa2ε0

z√z2+a23

(3.49)

Compare with solution of last chapter in eqn 2.41.

Equation 3.38 is only valid in electrostatic fields where no time-variations occur. Withmathematical thoroughness we can show, that the gradient can only be applied toirrotational (=conservative) fields.


Prof. S. Peik

3.8 Analogy to Gravitation

The electric potential field is a scalar field, similar to the elevation of a terrain in agravity field. We can find many analogies between electrostatic fields and gravita-tional fields, e.g.

Elevation h ⇔ Potential VElevation Difference ⇔ Voltage U

Mass m ⇔ Charge QSlope ~G ⇔ Electric Field ~E

The direction of ~E (or ~G, respectively) is always perpendicular to the constant po-tential lines (elevation lines)!

Slope G

10m

20m30m

40m

50m

Constant Potential Contours

(Elevation Lines)

Figure 3.8: Elevation Lines of a Terrain, and Perpendicular Slope Direction

The absolute value of the E-Field ~E (Slope ~G) is proportional to the potential drift(slope)!

G1

G2

Figure 3.9: Bigger Slope means Bigger |~G|

3.9 The Conversion Triangle

The relation between source charges ρ, vector field ~E and potential V. in electro-statics can be summarized in the following triangle:


Prof. S. Peik

ρ=ε∇

~E

~E

Vρ

~E=

14πε

∫ ∫∫ ρ

r−r′

|r −r′ |3

dVV =

1

4πε

∫∫∫ρ

1

|r − r′|dV

ρ = ε∇2V

Sources P

oten

tial

Vector

Field

~E=−∇

VV=− ∫

~Ed ~lco

min

g up

soon

in a

thea

tre

near

you

coming up

soon in a theatre

near you

3.10 Electric Dipoles

An electric dipole is formed when two point charges of equal but opposite magnitudeare separated by a small distance. Consider a dipole as shown in Figure 3.10.

+Q

-Q

d

x

y

z p

r

r1

r2

Figure 3.10: Dipole

The potential generated by this dipole is

V =Q

4πε0

(1r1− 1

r2

)=

Q4πε0

(r2 − r1

r1r2

)(3.50)

where r1 and r2 are the distances of the positive and negative charge from the fieldpoint respectively. If d r then we can approximate

V =Q

4πε0

d cos θ

r2 (3.51)

in spherical coordinates. With d cos θ = ~d · r with ~d = d · z and the definition

~p = Q · ~d (3.52)


Prof. S. Peik

as the dipole moment we may write eqn 3.51 as

V =~p · r

4πε0r2 (3.53)

If the dipole is not centered at the origin but at~r′ we get

V(~r) =~p(~r−~r′)

4πε0|~r−~r′|3(3.54)

The electric field due to a dipole centered in the origin can be obtained from eqn.3.38

E = −∇ ·V =Qd cos θ

2πε0r3 r +Qd sin θ

4πε0r3 θ (3.55)

Or shortly

~EDipole =p

4πε0r3 (2 cos θ r + sin θ θ) (3.56)

where p = |~p| = Qd

Note, field strength decays with 1/r3 with distance.


Prof. S. Peik

Example 15:An electric dipole ~p = 5 nCm y is located at the point (3, 5, 0) in Cartesian space. Derive theelectric potential and electric field in the origin

V(0, 0, 0) =~p(~0−~r′)

4πε0|~0−~r′|3=

5y− (3x + 5y)

4πε0√

4 + 93nCm m

m3 =−25

4π|ε0|√

133 10−9 C

mVmAs

(3.57)

= − 25

4√

133

1π|ε0|

10−9V (3.58)

For calculating the E-field, we need the function V(~r) , hence (without units)

V(x, y, z) =~p(~r−~r′)

4πε0|~r−~r′|3=

5y ((x− 3)x + (y− 5)y + zz)

4πε0√(x− 3)2 + (y− 5)2 + z23 (3.59)

=5(y− 5)

4πε0√(x− 3)2 + (y− 5)2 + z23 (3.60)

~E = −gradV = difficult (3.61)

Easier if we move the origin to the dipole location with x′ = x− 3, y′ = y− 5, z′ = z , then

V(x, y, z) =5y′

4πε0√

x′2 + y′2 + z′23 (3.62)

Ex = 54πε0

−3x′y′√x′2+y′2+z′2

5 =5

4πε0

−45√

345 (3.63)

Ey = 54πε0

−(2y2−x2−z2)√x′2+y′2+z′2

5 =5

4πε0

−44√

345 (3.64)

Ez = 54πε0

−y′z′√x′2+y′2+z′2

5 = 0 (3.65)

for the position in the old origin x′ = −3, y′ − 5, z′ = 0.


Prof. S. Peik

Example 16:Mupad Code :>> V:=K*(y-5)/sqrt((x-3)^2+(y-5)^2+z^2)^3;

>> Efield:=linalg::grad(V,[x,y,z]);

>>z:=0;K:=1;

>>field:= plot::vectorfield([Efield[1],Efield[2]], x = -1..4, y = -1..6,

Grid=[20,20], Color = [Flat, RGB::Red]);plot(field):

3.752.51.250

5

3.75

2.5

1.25

0

x

y

x

y

3.11 Summary

• When moving a charge in a field we perform the work W = Q E ds

• E-fields behave similar to gravitational fields

• The integration of the force along a path yields the performed work

• The potential is defined as V=Wpot/Q and describes the potential energy percharge.

• The potential is always referred to a reference potential V0 = 0V at referencepoint p0

• The potential field is a scalar field

• The E-field can be derived from the potential through the gradient operator

• Locations with the same potential form constant potential surfaces

• The voltage describes a potential difference V12 = V2 −V1

• The total work performed when moving a charge through an electrostatic fieldalong a closed path is always zero


Capacitance and Ca-pacitor

4Capacitors are two conducting bodies of arbitrary geometry that are isolated againsteach other. Between the conductors we have a potential difference (voltage) ofV1 − V2 = V. A general capacitor is shown in Figure 2.5. The capacitor is a veryimportant component in most electronic circuits. Capacitors are used for AC-Circuitimpedances and as energy storage devices (e.g. in watches).

Q-Q

E

V

Figure 4.1: Capacitor

The geometry and distance of the conductor surfaces of the capacitor determine thework performed for separating the charges on the capacitor. For any given charge Qstored on the capacitor surfaces, we get a certain voltage V. The charge Q is alwaysproportional to the voltage V. We define as capacitance of the capacitor

C =QV

(4.1)

The capacitance C specifies the amount of charge Q required to create a voltage of1V.

The unit of C is

[C] =[Q]

[V]=

AsV

= F (4.2)

The unit 1AsV = 1 F is called a Farad in honour of the physicist Michael Faraday.

50

Prof. S. Peik

Figure 4.2: Michael Faraday (1791-1867)

In practice, the capacitance is extremely small, such that we often use the sub-units

1µF = 10−6 F = 1 Microfarad

1nF = 10−9 F = 1 Nanofarad

1pF = 10−12 F = 1 Pikofarad

1fF = 10−15 F = 1 Femtofarad

4.1 Capacitance of a Parallel Plate Capacitor

The concept of capacitance can be explained with the parallel plate capacitor. InFigure 4.3 a parallel plate capacitor is shown wit plate area A and plate distance d .The capacitor is charged with the charge Q on the left plate and the charge −Q onthe right plate. Between the plates we have a homogeneous E-Field.

E=const

A

V

Figure 4.3: Parallel Plate Capacitor

The electric field strength is

|E| = const =1ε

QA

(4.3)

The potential along the x-axis is

V = −E · x = −1ε

QA

x (4.4)

and therefore the voltage between the plates is

V12 = V(x = 0)−V2(x = d) =1ε

QA

d (4.5)

We observe, that the voltage V is directly proportional to the charge Q on the plates.

We can directly derive the capacitance using the definition of eqn. 4.1


Prof. S. Peik

CPPC =QU

= εAd

(4.6)

We seen from the result, the capacitance of a parallel plate capacitor is only a func-tion of geometric quantities as the area A, and the plate distance d as well as themedium (with permittivity εr (remember: ε = ε0εr).

In practice, we often need large capacitances. This means, we have to maximizearea while minimizing the distance between the plates. Technical implementationsof capacitors use coiled up plates for example.

Figure 4.4: Various types of capacitors, from [?]

Example 17: A parallel plate capacitor (Area A= 1000 cm2, plate distance d= 1 mm, Airas Dielectricum) has a capacitance of C = ε0 A/d= 885,4 pF. A voltage of U= 10 V betweenthe plates gives rise to a chargeQ = CU = 885, 4 pF · 10 V = 8, 85 nCon the plates. This charge is very small; however, it includes n = Q/q = 5, 53 · 1010 elemen-tary charges already.

4.2 Recipe for Deriving the Capacitance

We can derive the capacitance of arbitrary structures with the following step-by-stepapproach.

1. We assume a charge Q on the capacitor surfaces

2. We derive the E-field due to these charges , if necessary with numerical meth-ods (computers)

3. We derive the voltage between the surfaces by V =∫ Pl2

Pl1E ds with the limits

being points on plate 1 and 2

4. We compute the capacitance C by the definition C = QV . The charge Q drops

out, such that only geometrical parameters are left


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Example 18:A spherical capacitor is a capacitor composed of two concentric spheres. with radius r1 andr2

If the distance d between the spheres is very small, i.e. d = r2 − r1 r1, we can derive thecapacitance from the equation for the parallel plate capacitor; now, the area is the surfaceof the sphere of A = 4 π r2 and d=r2-r1, such that we get

CSphere = 4 πεr2

r2 − r1(4.7)

For large distances of the spheres we use the recipe

1. The inner sphere has charge Q

2. The field is |E| = 14πε

Qr2 from the table

3. Th e voltage between the spheres is the integration from r1 to r2

UV =∫ r2

r1

14πε

Qr2 dr =

[− 1

4πεQr

]r2

r1= Q

4πε

(− 1

r2+ 1

r1

)= Q

4πε

(r2−r1r1 r2

)

4. Now the capacitance is

C =QV

= 4 πεr1 r2

r2 − r1(4.8)

Other capacitances of popular arrangements are given in Table 4.1:


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Arrangement Geometry Capacitance

Parallel Plates

E=const

A

V

C = ε Ad

Concentric spheres C = 4 πε r1 r2r2−r1

Two identical spheres

a

r r C = 2πε[1 + r(a2−r2)

a(a2−ar−r2)

]

Double cylinder co-axial

lr1

r2

C = 2πεlln(

r2r1

)

Two wire line a

l

r C = πεlln( a

r )

Table 4.1: Capacitances

4.3 Circuits with Capacitors

Capacitors in parallel are exposed to the same voltage V, whereas the charges onthe plates add up

Qtot = Q1 + Q2 + Q3 + · · · (4.9)Qtot

V=

Q1

V︸︷︷︸+

Q2

V+

Q3

V+ · · · (4.10)

Ctot = C1 + C2 + C3 + · · · (4.11)

We can summarize

Capacitors in Parallel:Ctot = C1 + C2 + C3 + · · ·

Capacitors in series will hold the same charge Q on every capacitor. Hence we getfor the total voltage

Vges = V1 + V2 + V3 + · · · (4.12)

using V = Q/C follows

QCges

=QC1

+QC2

+QC3

+ · · · (4.13)


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dropping Q :

Capacitors in Series:1

Cges= 1

C1+ 1

C2+ 1

C3+ · · ·

4.4 Energy in Capacitor

A capacitor can hold energy in form of an electric field. The field with the chargeshas the ability to perform work, as mentioned before. We can calculate the amountof stored energy by integration.

While charging a capacitor the current i(t) flows through into the capacitor. Thework dW, that is performed per time interval dt is hence

dW = v i dt (4.14)

where u and i are the instantaneous values of the voltage and current in the capac-itor.

Using the definition

v =QC

(4.15)

and

i =dQdt

(4.16)

follows

dW =QC

dQdt

dt =1C

Q dQ (4.17)

The total Energy is calculated by the summation of the energy increments

W =∫ 1

CQ dQ =

1C

Q2

2(4.18)

with Q = CV follows

W =12

CV2 (4.19)

as the stored energy within the capacitor.

4.5 Summary

• The capacitance C is defined as the charge Q per voltage V stored on a two-conductor body, i.e. C = Q/V

• The unit is Farad F and usually very small


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• The Capacitance is only a function of geometry and material

• The capacitance of a parallel plate capacitor is C = ε Ad . For many other ar-

rangements also formulas are given

• capacitors in parallel have a total capacitance of Cges = C1 + C2 + C3 + · · ·

• Capacitors in series have a total capacitance of 1Cges

= 1C1

+ 1C2

+ 1C3

+ · · ·

• The stored energy in a capacitor is W = 12 CV2


Electric Flux and FluxDensity

5The electric field is defined as the force F per charge Q. As shown before the forceis not only dependent from the sources in space but also from the material filling thespace. The electric field is a description of the effects of electrostatics and does notdescribe the cause of the field directly.

For describing the field by the cause we introduce the electric flux density ~D. TheD-field is defined by the cause, namely charges positioned in space.

5.1 Definition of the D-Field

The electric flux density is defined from the assumption, that all electric charges onsurfaces emit electric flux lines, usually visualized by field lines.

As seen from Figure 5.1 the flux lines are created by the charges. We can now define

|~D| = ChargesSurface Area

= lim∆A→0

∆Q∆A

=dQdA

(5.1)

D

Charges Q

Figure 5.1: Definition of the Electric Flux Density~D

The more charges per surface element, the bigger the D-field. The D-field directionis identical with the E-field direction, as the force lines follow the flux lines.

57

Prof. S. Peik

The unit of the electric flux density is

[D] =[Q]

[A]=

Asm2 (5.2)

5.2 D-Fields in Space

The D-field in not only defined for charges surfaces. The D-field fills the completespace as the E-field does. The D-field associated with an electrostatic field anywherein space can be derived by the influence experiment.

Figure 5.2: Influence Experiment

When placing two conducting touching plates into a field the charges in the platesare separated. When we now disconnecting the plates and removing them from theField we observe a field between the plates. The charges per surface on these platesare equal to the D-field strength.

The D-field is the same not matter what the medium in the causing field is. Due tothis definition the flux density is sometimes called displacement flux density.

5.3 Electric Flux Ψ

The total all of D-field lines form the electric flux Ψ. The exact amount of flux isdetermined by integration

Ψ =∫∫

~Dd~A (5.3)

Obviously, the electric flux is measured in Coulomb.

5.4 Relation between D and E

As mentioned before, the electric field ~E describes the effect of electrostatics inspace, namely the force on test charges. The D-field ~D describes the cause ofelectrostatic fields, i.e. charges placed in space. Using this definition the D-fieldis independent of the medium filling the space.

Fro Figure 5.2 see can derive the relation between D and E. In the small parallel platecapacitor with homogeneous field we have E = 1

εoεr

QA as known from Figure 2.15.

As ~D is always in the direction of ~E we can define


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~D = ε0εr · ~E (5.4)

where the permittivity ε is a constant factor describing the dependence of the medium.

This constant ε = ε0εr is the already defined permittivity constant ε . In Chapter 2we introduced the constant without explaining the exact definition. This definition isnow complete.

5.5 Gauss’s Law

When a charge distribution over a closed volume is given, we can find an interestingconnection between the amount of charge in the volume an the D-Field penetratingthe surface of the volume. This equation is known as Gauss’s law.

Gauss’s law states that the total electric flux Ψ through any closed surface is equalto the total charge enclosed by that surface. As an equation we can write

Ψtot =∮

Ψ =∮

S~D · d~A = ∑ Charges =

∫

Vρv dv (5.5)

or

∮

S~D · d~A =

∫

Vρv dv (5.6)

This equation is known as the integral form of Gauss’s law. A visual interpretation isgiven in Figure 5.3. This equation is also known as one of Maxwell’s equations.

Q

dA

D

Figure 5.3: Visual Interpretation of Gauss’s Law: The Sum of all D-lines Pene-trating the Hull A is Equal to the Total Charge Q Inside the Hull.

Another implied statement of Gauss’s law is, that electric field lines always start onpositive charges, so called sources, and end on negative charges, so called sinks.The electric field is hence a conservative field.

5.6 Divergence

We can obtain the total outward flow of the flux Ψ of the D-field ~D from a closedsurface S from the integral

∮~D · d~S. We can now define the divergence of ~D as the

net outward flow of flux per unit volume over a closed incremental surface


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Figure 5.4: Friedrich Gauss (1777-1855), German Mathematician, developed thetheorem known as Gauss’s Law. He was the first to measure electric and mag-netic fields in absolute quantities

div~D = ∇ · ~D = lim∆v→0

∮~Dd~A∆v

(5.7)

where ∆v is the volume enclosed by the surface A in which P is located .

An inward flow means negative divergence. An outward flow means positive diver-gence, as illustrated in Figure A.4.

P P P

div A<0 div A>0 div A=0

Figure 5.5: Definition of the Sign of the Divergence Operator

The divergence of any vector field ~A(x, y, z) is given by

div~A = ∇ · ~A =∂Ax

∂x+

∂Ay

∂y+

∂Az

∂z(5.8)

We note, that the result is a scalar field. The divergence can only be taken from avector. The divergence from a scalar does not make sense.

We can find a relation between the surface integral of an arbitrary volume V and itsvolume integral, by

∮~AdA =

∫

V∇~AdV (5.9)

This is known as the the divergence theorem.

The divergence theorem states that the total outward flow of a vectorfield ~A through a closed surface S is the same as the volume integral of

the divergence of ~A.


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Closed Surface S

dS

Volume V

dV

Figure 5.6: Divergence Theorem

5.7 Gauss’s Law in Differential Form

We can compare the divergence theorem and Gauss’s law of eqn. 5.6 and get

∇ · ~D = ρv (5.10)

This is the differential form of Gauss’s law. This law is another one of Maxwell’sequations. It states that the charge density at a given point is always the divergenceof the D-field at this point. Gauss’s law is directly derived from Coulomb’s law andhence just another way of stating Coulomb’s law.

Gauss law provides an easy way of calculating the electric field ~E quantitatively ifthe qualitative field distribution (e.g. in symmetric field) is known.

Using Gauss’s law we can derive the capacitance more generally by

C =QU

=

∮~D · d~A∫~E · d~l

(5.11)

In practice, the recipe of Section 4.2 is usually used.


Prof. S. Peik

Example 19:Uniformly charged dielectric sphereConsider a uniformly charged sphere of radius a with a charge density ρ.The total enclosed Charge is Q =

∫Sphere ρdV = 4

3 πa3ρ

The charge enclosed inside a smaller sphere of radius r inside is∫

ρdV = 43 πr3ρ

Now we can apply Gauss’s law. The right hand side of 5.6 is∫

ρdV = 43 πr3ρ. Since the

D-field is symmetric on the surface of the sphere and perpendicular to the surface, the lefthand side of Gauss’s law is ∮

~Dd~A = |~D|Asur f ace (5.12)

with Asur f ace = 4πr2 we get for Gauss’s law

|~D|4πr2 =43

πr3ρ (5.13)

|~D| =13

ρr (5.14)

Outside of the sphere for r > a the charge inside the volume stays constant Q = 43 πa3ρ and

we get

|~D|4πr2 =43

πa3ρ (5.15)

|~D| =13

a3ρ1r2 =

Q4π

1r2 (5.16)

|D|

ra

5.8 Influence and Polarization

Electrical conductors and isolators behave quite different in electrostatic fields. Theeffects are known as influence and polarization, respectively.

Influence:

Inside an ideal conductor the charges can move freely. When the force ~F = ~E · q actson the charges they separate. As explained before the charges accumulate on theconductor surface and compensate the outer field through their inner field.

Polarization:

Most molecule inside isolators are dipoles. When an electrostatic field is applied toan isolators, the molecules (dipoles) turn to get aligned with the field. This orienta-tion weakens the field. Hence the E-fields inside isolators is always smaller than invacuum. The weakening depends on the dipole character of the medium, i.e. εr.


Prof. S. Peik

E

EFreie Ladungen

Dipole

Isolator

Leiter

Figure 5.7: Influence and Polarization

5.9 Partly Filled Capacitors

The D-field in a parallel plate capacitor with layered dielectrics is constant, as noextra charge are located on the boundaries (Gauss’s law!). The E-field changes bythe ratio of the εr’s. We can use capacitors in series as equivalent circuit.

In capacitors with transverse layered dielectrics the E-field remains constant, as thevoltage between the plates is the same everywhere (V = E · d). The D-field is denserin the high material higher permittivity. The equivalent circuit for this capacitor arecapacitors in parallel.

5.10 E-Field and D-Field on Boundaries

On boundaries between two dielectric media with εr1and εr2, respectively, conditionsapply to E- and D-field. We call these conditions the boundary conditions.


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Figure 5.8: Partially filled Capacitor

Figure 5.9: Partially across filled Capacitor


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When we observe a tangential E-field on a surface with Et1 in medium 1 and Et2 inmedium 2 the following condition applies

Et1 = Et2 (5.17)

This condition can be easily proved by integrating the E-field along a short path with∮Edl as shown in Figure 5.11. Since the path is closed and the normal field is not

existent, the tangential field in both media must be the same. For the same reasontangential E-fields can never jump in value.

Et,1

Et,2

ε1

ε2

Int. Path

Figure 5.10: Integrating E-Field on Boundary

Using eqn. 5.4 we getεr2Dt1 = εr1Dt2 (5.18)

The tangential component of the E-field remains constant whereas the tangentialcomponent of the D-field changes by εr1

εr2.

For D-fields the normal component remains equal on both sides of the boundary, i.e.

Dn1 = Dn2 (5.19)

This statement can be proved by using Gauss’s law with a surface as shown in Figure. Since there are no charges in the surface, the total

∫DdS on the upper and lower

“lid” of the surface must be equal. Consequently, both D-fields must be equal.

Dn,1

Dn,2

ε1

ε2

Gaussian Surface

Figure 5.11: Integrating E-Field on Boundary

By using eqn. again we get for the E-field

εr1En1 = εr2En2 (5.20)

An overview is shown in Figure 5.12.


Prof. S. Peik

Et,1

Et,2

ε1

ε2

Et,1=Et,2

En,1

En,2

ε1

ε2

ε1En,1=e2En,2

Dt,1

Dt,2

ε1

ε2

ε2Dt,1=ε1Dt,2

Dn,1

Dn,2

ε1

ε2

Dn,1=Dn,2

Figure 5.12: Boundaries with E- and D- Fields

5.11 Stored Energy in Electrostatic Fields

Any static electric field stores energy, as it takes energy to separate charges, thatbuild up that field. We already calculated the energy stored inside the field of acapacitor resulting in eqn. 4.19

W =12

CV2 (5.21)

We can now calculate the total energy stored in any electric field, by the followingderivation. Let us assume an infinite small parallel plate capacitor of plate distancedl and with the plate area dA. This capacitor stores the energy dW = 1

2 CdV2, where

dV is the voltage across the capacitor. The capacitance of the capacitor is C = ε dAdl .

The voltage is determined by the E-field by dV = |E| dl. Hence the Energy is

dW =12

εdAdl

(|E| dl)2 =12

ε dA |E|2dl (5.22)

where dA dl is the Volume dV of the elementary capacitor. Hence the stored energyper volume element dV is

dW =12

ε|E|2 dV (5.23)

dW =12|D| |E| dV (5.24)

Now we can find the total stored energy in the System by integrating over the vol-ume of the system

W =12

∫

V|D| |E| dV (5.25)


Prof. S. Peik

Since ~D and ~E are always in the same direction we can also write |D| |E| = ~D · ~E =ε|~E|2 = 1

ε |~D|2 and

W =12

∫

V~D · ~E dV (5.26)


Prof. S. Peik

5.12 Poisson’s and Laplace’s Equations

Poisson’s and Laplace’s Equation are derived easily from Gauss’s law

∇~D = ∇ · ε~E = ρ (5.27)

and~E = −∇V (5.28)

Substituting the equations gives

∇ · (−∇V) =ρ

ε(5.29)

When our medium is homogeneous we can write

∇2V = −ρ

ε(5.30)

This is known as Poisson’s equation. In the special case of no volume charge ρ weget

∇2V = 0 (5.31)

known as Laplace’s equation.

Laplace’s equation is of significant importance in solving electrostatic boundaryvalue problems. If we are able to find a solution of Laplace’s satisfying all boundarycondition, we found a unique solution of the problem. This statement is phrased asthe uniqueness theorem as

The solution to the Laplace’s equation in some volume V is uniquely determined ifV is specified on the boundary surface S.

a visualization of the concept is sketched in Figure 5.13.

The solution to the multidimensional Laplace’s equation can be found in most casesusing the separation of variables.

5.13 Procedure for Solving Boundary Value Problems

The following procedure can be applied when solving Laplace’s or Poisson’s Equation

1. Solve the equation using either

• direct integration when V is a function of one variable or

• separation of variables if V is a function of more than one variable

The solution includes constants, that can be determined from the boundaryconditions


Prof. S. Peik

00.5

11.5

22.5

33.5

4 00.5

11.5

22.5

33.5

4

-1-0.8-0.6-0.4-0.2

00.20.40.60.8

1

Boundary Conditions

V(x,y)

x

y

Figure 5.13: Potential in a Closed Surface with Fixed Boundaries

2. Apply the boundary conditions, to determine a unique solution for V

3. With V we can obtain now ~E and ~D

4. If desired, find the charge Q induced on a conductor using Q =∫

ρ dS whereρ = Dn(D-component normal to surface)

5. If desired find the capacitance from C = Q/∆V

Example 20:Find the potential V on a dielectric rod (length l) charged with ρ0 and a voltage of V0 betweenthe ends of the rod.

z

V=V0

V=0V

ρ0

z=l

The potential depends only on z. Hence

d2Vdz2 =

−ρ0ε

(5.32)

The solution to this differential equation is (by integrating twice)

V = − ρ02ε

z2 + Az + B (5.33)

Applying the boundary conditions yield, when z = 0

V(0) = B = V0 hence B = V0 (5.34)

when z = l

V(l) = − ρ02ε

l2 + Al + V0 = 0 hence A =ρ02ε

l − V0l

(5.35)

The electric field is given by

~E = −∇V =

[V0l+

ρ0ε(z− l

2)

]z (5.36)


Prof. S. Peik

5.14 Solving Laplace with Finite Element Methods

Laplace’s equation is a kind of averaging instruction. The value of V(x, y, z) is theaverage of its surrounding points as we will show below.

Using finite differences, we can design a numerical technique, solving Laplace’sequation with arbitrary boundary conditions.

This can be sketched in two dimensions as follows

V0V1

V3

V2

V4

h

h

a b

c

d

Figure 5.14: Definitions for Finite Difference Method

With Laplace’s equation in two dimensions we have

∂2V∂x2 +

∂2V∂y2 = 0 (5.37)

using the finite, but very small distance h between two field points, we get

∂V∂x

∣∣∣∣a≈ V0 −V1

h(5.38)

Similar at the other distances:

∂V∂x

∣∣∣∣b≈ V2 −V0

h(5.39)

∂V∂y

∣∣∣∣c≈ V3 −V0

h(5.40)

∂V∂y

∣∣∣∣d≈ V0 −V4

h(5.41)

Knowing that ∂2V∂x2 = ∂( ∂V

∂x )/∂x we can approximate the second derivative of V at thepoint 0 by

∂2V∂x2 ≈

∂V∂x

∣∣∣b− ∂V

∂x

∣∣∣a

h=

V2 −V0 −V0 + V1

h2 (5.42)

Similarly for y

∂2V∂y2 ≈

∂V∂y

∣∣∣c− ∂V

∂y

∣∣∣d

h=

V3 −V0 −V0 + V4

h2 (5.43)


Prof. S. Peik

Substituting these equations into Laplace’s equation yields

∂2V∂x2 +

∂2V∂y2 ≈

V1 + V2 + V3 + V4 − 4V0

h2 ≈ 0 (5.44)

Solving for V0 we find that

V0 ≈14(V1 + V2 + V3 + V4) (5.45)

Including a space charge we can extend the equation using Poisson’s equation

V0 ≈14(V1 + V2 + V3 + V4 + h2 ρ

ε) (5.46)

5.14.1 Relaxation Method

The Potential at a given point is always the average of its surrounding points. Wecan find the Potential field by iteratively checking this condition. Starting with allnodes zero inside the region and the boundary nodes with the given potential, wego through all nodes iteratively and compute the average of the surrounding nodes.We repeat this procedure until the solution converges.

This method is known as the iterative finite difference method or relaxation method.


Prof. S. Peik

Example 21:We can use Excel or OOCalc in the iteration mode, to find the potential field inside a box withgiven boundary values (here green). The calculated cells are simple the average value fromthe surrounding cells.


Prof. S. Peik

5.15 Summary

• The electric flux density ~D is defined as the density of flux lines caused bycharges Q on a surface A.

• The electric flux density ~D has the unit Asm2 .

• The d-field is proportional to the E-field by the factor ε

• Gauss’s law describes the equality of charges inside a volume to the flux leav-ing the volume

• Influence is the compensation movement of free charges inside a conductorwhen an external field is applied

• Polarization is the rotation of dipoles inside a dielectric when an external fieldis applied

• The tangential E-field on boundaries is continuous

• The normal D-field on boundaries is continuous

• The method of images allows to solve problems involving grounded conductingplates.

• The Laplace Equation and Poisson equation describe the potentials throughdifferential equations


Moving Charges 6In the previous chapters we assumed that all charges are at rest. The charge distri-bution was static. In this chapter we investigate the effects arising due to movingcharges.

6.1 Current I

The movement of charges is known as electrical current. .

The current is definition is:

The current is defined as the electric charges passing a given area in aconductor per unit time

For a conductor with the charge quantity ∆Q per time interval ∆t we get

I =∆Q∆t

(6.1)

Figure 6.1: Definition of current

For the derivation of an instantaneous current in time varying arrangements we usethe differential

I =dQdt

(6.2)

74

Prof. S. Peik

Figure 6.2: Andre Marie Ampere (1775-1836), French physicist and professor ofmathematics. About one week after Oersted he discovered the magnetic fieldgenerated by an electric current in 1820. Ampere explained the observed phe-nomenon and then developed the force relations between current-carrying con-ductors. By doing so, he laid the foundation of the electromagnetic theory. Histheory and the work of Michael Faraday are the main pillars of Maxwell’s the-ory on electromagnetism. The unit of current and the law bearing his name (seeSection 7.7 ) attest the importance of his contribution.

The unit of the current is the Ampere Cs = A.

Following this definition, the current direction is defined as the flow direction of pos-itive charges in a conductor. We call this the technical current direction. However, inmost conductors the moving charges are electrons carrying a negative charge. Thetrue movement of the charges is hence in the opposite direction of the current I. Wecall this the true current direction.

6.2 U-I-Relation at Capacitors in AC Circuits

We can determine the current through a capacitor using I = dQdt and the capacitance

definition C = QU as follows

I = CdUdt

(6.3)

Using a harmonic sinusoidal signal with the voltage phasor U = Uejωt and the cur-rent phasor I = Iejωt we can rewrite eqn 6.3 as

I = CdUdt

= Cddt

(Uejωt

)= C jω Uejωt

︸︷︷︸U

(6.4)

U =1

jωCI (6.5)

following an “AC-Resistance”, the so called impedance of

ZC =1

jωC(6.6)

6.3 Current Density ~J

The current is a macroscopic unit. The current specifies the total flux of chargesthrough a conductor. It does not indicate a field unit in a particular point in space.


Prof. S. Peik

We can define a current density ~J that fills that gap. The current density is definedas the current per area element. The current density is a vector pointing always inthe direction of the current flow.

Figure 6.3: Definition of Current Density~J

In differential notation we get:

~J =dIdA

n (6.7)

where ndenotes the direction of the current density.

The unit of the current density is

[J] =A

m2 (6.8)

The current density~J inside a conductor can vary even though the current I remainsconstant. Figure 6.4 shows such an arrangement with a non uniform cross sectionwire.

Figure 6.4: Current Density Field in a Conductor of Non-Uniform Cross Section

The total current I through a cross section A can be derived from integrating thecurrent density over the area A.

I =∫

crosssection

~J d~A (6.9)

In case of a homogeneous current density we get

I = ~J · ~A (6.10)


Prof. S. Peik

6.4 Current Density Fields

The current density describes a property of space with magnitude and direction, andis, hence, a vector field. The current density fields behave similar to the flow fieldsof a non turbulent flow of liquids and gases. As the charges follow the force lines (i.e.E-field) the current density is always directed in the direction of the ~E field drivingthe charges. Consequently, E- and J-field are aligned in space in isotropic media.

For current density fields the following applies:

• J-fields are always limited to the boundaries of the conductor

• On boundaries to dielectrics (J = 0 in dielectric!) the current density flowsparallel to that border.

• The flow is perpendicular to the surface of perfect conductors.

Example 22:Below an example for a current density field is given

non perfect conductor

perfect conductor

DielectricCurrent Density J

6.5 Relation Between E and J

As mentioned before, J- and E-field align in space, following

~J ∝ ~E (6.11)

This proportionality can be rewritten as equality using the constant σ and we get

~J = σ ~E (6.12)

The factor σ denotes the conductivity of of the conducting material. This equationis known as the point form of Ohm’s law. Note that in a perfect dielectric ~J = 0regardless of the electrical field ~E and in a perfect conductor ~E = 0 regardless of thecurrent density ~J.

The reciprocal value of σ is called the resistivity ρc. We have

ρc =1σ

(6.13)

Typical values for σ are given in Table


Prof. S. Peik

Material Conductivity σ Sm

Silver 6.1 · 107

Copper 5.8 · 107

Aluminum 3.5 · 107

Tungsten 1.8 · 107

Zinc 1.7 · 107

Brass 1.7 · 107

Sea Water 4dest. Water 10−4

Glass 10−12

Paper 10−11

Quartz (fused) 10−17

Table 6.1: Conductivity of some materials

6.6 Resistance R

We can now define a unit similar to the capacitance, that relates the current to thevoltage across conducting bodies. Figure 6.5 shows a cylindrical conductor withuniform cross section A and length l with two terminals left and right. .

Figure 6.5: Resistor

The current flowing through this cylinder is

I =∫

A~Jd~S =

∫

Aσ~Ed~S (6.14)

The Voltage across the cylinder is

V =∫

l~Ed~l (6.15)

We can now define the ratio of voltage across the cylinder and current though thecylinder by

R =VI=

∫l~Ed~l∫

A σ~Ed~S(6.16)

known as the resistance of the cylinder. Equation 6.16 is well known as Ohm’s law.


Prof. S. Peik

For homogeneous fields we get

R =UI=|~E|l|~J|A

=|~E|l

σ|~E|A=

lσ A

(6.17)

The resistance R depends from geometrical parameters only and uses the unit VA

abbreviated as Ω (say “Ohm” in honour of the Physicist Georg Simon Ohm. Thereciprocal value of the resistance is called the conductance G = 1

R with the unit Sfor Siemens. In North America the unit symbol f (say:”mho”) is also often used.

When the resistor has a non-uniform cross section or non-uniform conductivity inone direction only, e.g. the x-direction, we can determine the total resistance byintegrating infinite small slices of thickness dx

R =∫ dx

σ(x)A(x)(6.18)

Figure 6.6: Georg Simon Ohm (1789-1854) physicist and professor at the Univer-sity of Nuremberg. Inspired by his work on galvanic chains he developed the thelaw of the proportionality between current and voltage in a conductor in 1826.The unit of this proportionality constant is named in his honor “Ohm” with thesymbol Ω.

Circuits with Resistors

Obviously from equation 6.16, resistors in series have a total resistance of

RGes = R1 + R2 + R3 + · · · (6.19)

Resistors in parallel have a total resistance of

1RGes

=1

R1+

1R2

+ · · · (6.20)


Prof. S. Peik

Example 23:Determine the resistance, the Fields ~E and ~J, the voltage drops V1 and V2 and the current Iof the arrangement below. The voltage V0 is given.

A

σ1

I

x

V0

2σ1

l ll

V1 V2

The Resistance is calculated from two resistors in series with

R = R1 + R2 =l

σ1 A+

2l2σ1 A

=2l

σ1 A(6.21)

The partial resistances are equal.Following

I =VR

=σ1 A2l

V0 (6.22)

The J-Field is constant, as the cross-section is constant, and x-directed

~J =IA

=σ12l

V0 (6.23)

The E-field is not constant, as σ changes end derived by ~E =~Jσ

~E1 =V02l

(6.24)

~E2 =V04l

(6.25)

The voltage drops are from V = E · d

V1 =V02

(6.26)

V2 =V02

(6.27)

This can also be found by Ohm’s law V1 = I · R1and V2 = I · R2

Example 24:Determine the resistance between inner conductor (radius a) and outer conductor (radius b)of a coaxial cable of length l. The filling of the cable has a conductance of σ.The E-field is ~E = E0

ρ ρ between the conductors. Hence the current density is ~J = σ E0ρ ρ.

The voltage between inner and outer conductor is

V =∫ b

a

E0ρ

ρ d~ρ = [E0 ln ρ]ba = E0 lnba

(6.28)

The current across the cable is

I =∫ 2π

0

∫ l

0~J ρ dφ dz =

∫∫σ

E0ρ

ρ dφ dz = σ2πE0l (6.29)

Using the definition of the Resistance

R =VI=

ln ba

2πσl[Ω] (6.30)


Prof. S. Peik

6.7 Drift Velocity

Alternatively to the description by ~J the flow of current inside a conductor can beexpressed by the drift speed of the charges in the conductor.

Inside the conductor we have a density of free charges given by ρ. This density is aconstant of the conductor material.

In order to have a current I flowing through the cross section A a total number ofdQ charges must cross the cross section A within the time dt. This total charge isdistributed over the volume V = A · dl, where dl is the distance traveled in the timeinterval dt. Hence

dQ = ρ dl A (6.31)

This volume moves by dl within dt. In other words the charges inside the volumemove by the

~v =dldt

(6.32)

Using the current definition

I =dQdt

=ρ∆l · A

∆t= ρ~A ·~v (6.33)

and ~J = IA we get

~J = ρ~v (6.34)

The velocity ~v is known as the drift velocity inside the conductor. It represents theactual speed of the charges inside the conductor.

Example 25:The investigate the drift velocity in a typical domestic wiring system. A current of 0.5 Aflows through a copper wire with the cross section A = 1.5mm2. Copper has a free electrondensity of 8.49 ·1028Electrons per m3.Solution:Current density is J = I

A = 3.33 · 105 Am2

The charge density is

ρ = 8.49 1028 · 1.6 10−19 = 13.58 109 C/m3 (6.35)

following:

v =Jρ= 2.45 10−5 m

s= 88

mmh

(6.36)

The electrons move with 88 mm per hour.

6.8 Converted Power

The power in watts is defined as the rate of change of energy (in Joules) or forcetimes velocity, hence


Prof. S. Peik

∫~F~v =

∫ρ dv ~E~v =

∫~E ρ~v dV (6.37)

or

P =∫

V~E ·~J dV (6.38)

which is known as Joule’s law. The power density wp in Watts/m3is given by theintegrand

wp =dPdV

= ~E ·~J = σ~E2 (6.39)

for a uniform cross section resistance we get for eqn. 6.38

P =∫

L~E∫

A~J dS dl = V · I (6.40)

or

P = V · I = I2R =V2

R(6.41)

Example 26:Calculate the dissipated power inside the coaxial cable from the previous example usingJoule’s law

P =∫∫∫

~E~J ρ dz dφ dρ (6.42)

=∫∫∫ E0

ρρ σ

E0ρ

ρ ρ dz dφ dρ (6.43)

=∫ b

a

∫ 2π

0

∫ 1

0

E20

ρσ dz dφ dρ (6.44)

= 2π E20 σ l ln

ba

(6.45)

using V = E0 ln ba follows

P = 2πσlV2

ln ba

(6.46)

using P = V2

R yields the same result.

6.9 Continuity Equation

Due to the conservation of charges, the total outflow of current from a volume isequal to the negative net change of charge dQ

dt inside the volume, or

Iout =∮

A~Jd~S = −dQ

dt(6.47)

Invoking the divergence theorem to the left side we get


Prof. S. Peik

∫

V∇~J dv = −dQ

dt(6.48)

The right side can be written as dQdt = d

dt

∫V ρdv, and hence

∫

V∇~J dv = −

∫

V

dρ

dtdv (6.49)

or

∇ ·~J = −dρ

dt(6.50)

This is known as the continuity equation. With constant charge density inside theconductor, we get ∇~J = 0. This is known as Kirchhoff’s current law.


Magnetostatics 7The second major phenomenon of electromagnetics is the magnetism. Magnetismis a phenomenon by which materials exert an attractive or repulsive force on othermaterials. Magnetism arises whenever electrically charged particles are in motion.Some well known materials that exhibit magnetic properties are iron, some steels,and the mineral lodestone. All materials are influenced to one degree or another bythe presence of a magnetic field, although in some cases the influence is too smallto detect without special equipment.

7.1 History of Magnetism

Around 600 B.C. the magnetic properties of natural ferrite (Fe3O4) stones (lode-stone) were described by Greek philosophers. The effect was not further investi-gated and did not have any practical application for a long time. Around the 10thcentury it was discovered by Chinese sailors, that a magnet always orients itself innorth-south direction. The first Reference to a compass in Europe was given 1175by Alexander Neckem an English monk of St. Albans.

In 1600, William Gilbert, later physician to Queen Elizabeth I of England, pub-lished his great study of magnetism, "De Magnete"–"On the Magnet". It gave thefirst rational explanation to the mysterious ability of the compass needle to pointnorth-south: the Earth itself was magnetic. "De Magnete" opened the era of modernphysics and astronomy and started a century marked by the great achievements ofGalileo, Kepler, Newton and others.

Figure 7.1: William Gilbert (1544-1603) and his book De Magnete

84

Prof. S. Peik

In 1820, a physicist Hans Christian Oersted, learned that a current flowing througha wire would move a compass needle placed beside it. This showed that an electriccurrent produced a magnetic field.

In 1830, Joseph Henry (1797-1878), discovered that a change in magnetism canmake currents flow, but he failed to publish this. In 1832 he described self-inductance- the basic property of inductor. In recognition of his work, inductance is measuredin Henries. The stage was then set for the encompassing electromagnetic theory ofJames Clerk Maxwell.

Michael Faraday (1791-1867) an Englishman, made one of the most significantdiscoveries in the history of electricity: Electromagnetic induction. He thought ifelectricity could produce magnetism, why couldn’t magnetism produce electricity. In1831, Faraday found the solution. Electricity could be produced through magnetismby motion. He discovered that when a magnet was moved inside a coil of copperwire, a tiny electric current flows through the wire.

7.2 Permanent Magnets

Permanent magnets exert a magnetic field permanently. The following applies topermanent magnets

• they are always dipoles.

• The poles are referred to as north pole (pointing north) and south pole (pointingsouth)

• Unlike pole attract each other, like poles repel each other.

7.3 Magnetic Field

Magnetic fields are produced by electric currents, which can be macroscopic cur-rents in wires, or microscopic currents associated with electrons in atomic orbits,as in permanent magnets. Magnetic field sources are essentially bipolar in nature,having a north and south magnetic pole.

Magnetic fields are usually denoted by the symbol ~B. Historically, this field is de-scribing the effect of the field. The B-field is known as magnetic flux density ormagnetic induction, The field describing the cause of the field is called the mag-netic field (strength) ~H. Even though using opposite definition to the electric fieldquantities, this terminology is still often used to distinguish the two in the context ofmagnetic materials. Otherwise, however, this distinction is often ignored, and bothsymbols are frequently referred to as the magnetic field. Some authors call H theauxiliary field, instead. We might refer just to the H-field and B-field.

7.4 Magnetic Field Lines

When placing a test magnet (usually a compass needle) into a magnetic field, weobserve, that the test magnet aligns itself in space. That means, that we can find,


Prof. S. Peik

similar to the electric field, force lines filling the space. These force lines are thedirection of the magnetic field. The direction of the field is the equilibrium directionof a compass needle placed in the field.

NS

Figure 7.2: Force Lines (Field) of a Bar Magnet

Therefore, a magnetic field is a vector field. It associates with every point in spacea vector that may vary in time.

We define, that field lines start on a north pole and end on a south pole on permanentmagnets. When taking a closer look, magnetic field lines are actually always closedlines, i.e. the magnetic field is a rotational field and not conservative.

Magnetic field lines can be made visible by iron chips as demonstrated in Figure 7.3.

Figure 7.3: Field of a Bar Magnet Made Visible by Iron Chips, from PracticalPhysics, publ. 1914 by Macmillan and Company

7.5 Magnetic Field of the Earth

The earth’s magnetic field is similar to that of a bar magnet tilted 11 degrees fromthe spin axis of the earth. Magnetic fields surround electric currents, so we surmisethat circulating electric currents in the Earth’s molten metallic core are the origin ofthe magnetic field. A current loop gives a field similar to that of the earth.

The earth’s magnetic field is attributed to a dynamo effect of circulating electriccurrent, but it is not constant in direction. Rock specimens of different age in similarlocations have different directions of permanent magnetization. Evidence for 171magnetic field reversals during the past 71 million years has been reported.

Although the details of the dynamo effect are not known in detail, the rotation of theEarth plays a part in generating the currents which are presumed to be the source ofthe magnetic field. Mariner 2 found that Venus does not have such a magnetic fieldalthough its core iron content must be similar to that of the Earth. Venus’s rotationperiod of 243 Earth days is just too slow to produce the dynamo effect.

Interaction of the terrestrial magnetic field with particles from the solar wind sets upthe conditions for the aurora phenomena near the poles.


Prof. S. Peik

Figure 7.4: Magnetic Field of the Earth, from [email protected]

7.6 The Magnetic Field H

The magnetic field strength ~H describes the direction and magnitude of the mag-netic field in space. The H-field is defined by the cause of the field which is a movingcharge or current.

7.6.1 Magnetic Field of a Current Through a Long Wire

Permanent magnets are not the only source of magnetic fields. In 1820, Hans Chris-tian Oersted discovered that an electric current flowing through a wire caused anearby compass to deflect. This indicated that the current in the wire was generat-ing a magnetic field.

Oersted studied the nature of the magnetic field around the long straight wire. Hefound that the magnetic field existed in circular form around the wire and that theintensity of the field was directly proportional to the amount of current carried bythe wire, as shown in figure 7.5. He also found that the strength of the field wasstrongest close to the wire and diminished with distance from the conductor until itcould no longer be detected.

H

I

Figure 7.5: Magnetic Field around a Current Flowing Through a Wire, Experi-ment using iron chips, from [?]

Oersted also noticed that the direction of the magnetic field was dependent on thedirection of the electrical current in the wire. There is a simple rule for rememberingthe direction of the magnetic field around a conductor. It is called the right-handrule. If a person grasps a conductor in ones right hand with the thumb pointing in


Prof. S. Peik

the direction of the current, the fingers will circle the conductor in the direction ofthe magnetic field.

We say the current and the H-field form a right-handed system.

Figure 7.6: Right-Hand-Rule also found in the IEEE Logo

7.7 Ampere’s Law, Definition of H

There is a relation between current in the space and the H-field in the space: Themagnetic field in space around an electric current is proportional to the electric cur-rent which serves as its source, just as the electric field in space is proportional to thecharge which serves as its source (Gauss’s law!). This is summarized in Ampere’slaw.

Ampere’s Law states that for any closed loop path, the sum of the length elementstimes the magnetic field in the direction of the length element is equal to the electriccurrent enclosed in the loop. Ampere’s law is the magnetic version of Gauss’s law inelectrostatics.

∮~H d~l =

∫

A

~J d~A =n

∑i=1

Ii = current enclosed (7.1)

The integral of the magnetic field in a closed loop is equal to the totalcurrent passing through the loop

Consequently the unit of ~H is A/m.

I4

I3

I2

I1

H

Fläche A

Figure 7.7: Ampere’s Law

If there are no currents enclosed, the integral∮~H dl becomes zero. This means,

H-field lines must always circle currents.


Prof. S. Peik

We can use Ampere’s law to calculate H-fields when the current distribution is knownin space. However, we must know the qualitative H-field distribution in order toperform an integration.

James Clerk Maxwell noticed a logical inconsistency when applying Ampere’s law oncharging capacitors, and thus concluded that this law had to be incomplete. To re-solve the problem, he came up with the concept of displacement current and madea generalized version of Ampere’s law which was incorporated into Maxwell’s equa-tions. The displacement current describes the change of electric flux in the electricfield. The generalized formula is as follows:

∮~H d~l =

∫

A

~J d~A +dΨdt

(7.2)

A detailed discussion of displacement currents is given in Chapter 11.'

&

$

%

Clip-On Ammeter

A clip-on ampere meter can measure currents through wires without opening thecurrent loop. By measuring the magnetic fields surrounding the wire, we can deter-mine the current by Ampere’s law.

7.8 Magnetic Fields Due to Current in a Long Straight Wire

A long straight wire is carrying the current I as shown in Figure 7.8. With Ampere’slaw we get

∮~H d~l = I (7.3)

Since we have cylindrical symmetry we can assume that |~H| is constant a for con-stant distance r from the wire. Integrating on one of these circles with radius rbecomes a multiplication of ~H · 2πr.

Hence

|~H| · (2 π r) = I (7.4)


Prof. S. Peik

H

I

r

Figure 7.8: Magnetic Field around a Current Carrying Wire

Solved for H

|~H| = I2π r

(7.5)

7.8.1 The Curl Operator

We completed our study of Gauss’s law by applying it to a differential volume ele-ment and were led to the concept of divergence. We now apply Ampere’s circuitallaw to the perimeter of a differential surface element and discuss the third and lastof the special derivatives of vector analysis, the curl. Our immediate objective is toobtain the point form of Ampere’s circuital law.

For that purpose we create a square shaped loop lying in the xy-plane.

xΔx

ΔyH1x

H2x

H1y H2y

zy

Jz

Figure 7.9: z-directed Current Related to a Square H-Loop in the xy-Plane

The∮~H · ~dl is then equal to the total current in z-direction through the loop, i.e

I = Jz∆A, where ∆A = ∆x∆y∮

~H · ~dl = H1x∆x + H2y∆y− H2x∆x− H1y∆y = Jz∆x∆y (7.6)

Note, that at the H2x∆x and H1y∆y are subtracted as the integration direction (greendashes) is against the direction of the coordinates.

The H-Field strength in x-direction may differ over the length ∆y in y-direction by∆Hx. For Hy respectively. Hence H2x = H1x + ∆Hx and H2y = H1y + ∆Hy

H1x∆x + (H1y + ∆Hy)− (H1x + ∆Hx)∆x− H1y∆y = Jz∆x∆y (7.7)


Prof. S. Peik

∆Hy∆y− ∆Hx∆x = Jz∆x∆y (7.8)

∆Hy

∆x− ∆Hx

∆y= Jz (7.9)

For finite lengths ∆x and ∆y we create an error, as we integrated and assume theH to be constant along each section. When shrinking the ∆x and ∆y to an infinitesmall dx and dy we get the exact current density Jz for a single point

dHy

dx− dHx

dy= Jz (7.10)

x

z

y

Jz

Jz

Jx

Jy

Jx

Jy

H

Figure 7.10: Operation Applied to All Three Direction

We can repeat the same calculation for y- and x-directed currents, as seen in Figureand get

dHz

dy− dHy

dz= Jx (7.11)

dHx

dy− dHz

dx= Jy (7.12)

dHy

dx− dHx

dy= Jz (7.13)

This operation on the ~H-vector is known as the curl operation. Applying the curl toa vector yields a new vector. Hence the curl is defined in Cartesian coordinates as:

curl~H =

(dHz

dy− dHy

dz

)x +

(dHx

dy− dHz

dx

)y +

(dHy

dx− dHx

dy

)z (7.14)


Prof. S. Peik

We can formulate the curl operator from our well known nabla operator ∇ applyingthe cross product

∇× ~H = ~J (7.15)

The curl can also be written in cylindrical and spherical coordinates. See the formulasheet for the exact definition.

Figure 7.11: Paddle Wheel Visualization: The curl of a vector field is related tothe rate at which a rigid paddle wheel would rotate if placed in the flow. (a)The paddle wheel rotates because the velocity of the fluid varies in a directionperpendicular to the flow. (b) Rotation is generated by the curvature of the flow.

Example 27:Take the vector field, which depends on x and y linearly:~F(x, y, z) = yx− xy.Simply by visual inspection, we can see that the field is rotating. If we place a paddle wheelanywhere, we see immediately its tendency to rotate clockwise. Using the right-hand rule,we expect the curl to be into the page. If we are to keep a right-handed coordinate system,into the page will be in the negative z direction. The lack of x and y directions is analogousto the cross product operation.

The curl of the field isIf we calculate the curl:∇× ~F = 0x + 0y +

[∂

∂x (−x)− ∂∂y y]

z = −2z

7.9 Ampere’s Law in Point Form

The equation

∇× ~H = ~J (7.16)

is known as Ampere’s law in point form, known as the third Maxwell’s equation.Since ∇× ~H 6= 0 magnetostatic fields are not conservative!


Prof. S. Peik

Example 28:A cylindrical wire with radius R carries a current of I. Assume a homogeneous current densityinside the wire.Find ~H in cylindrical coordinates.

x

x=5cm

H-Field outside:

∮~H d~l = I (7.17)

2πρ|~H| = I (7.18)

|~H| =1

2πρ(7.19)

The field shows rotational symmetry around the wire. Hence, H has a φ-component, only.

~H =I

2πρφ (7.20)

H inside wire:The current enclosed inside the Amperian loop is Θ = |~J| · A with J = I

πR2 and A = πρ2

(7.21)

Using Ampere’s law∮

~H d~l = |~J| · A (7.22)

2πr|~H| =I · πρ2

2πR2 (7.23)

|~H| =I

2πR2 ρ (7.24)

The field shows rotational symmetry around the wire. Hence, H has a φ-component, only.

~H =I

2πR2 ρφ (7.25)


Prof. S. Peik

Example 29:Find the field Hy along the x-axis of the last example:In Cartesian coordinates using eqn A.38 to A.39 we get for H outside

~Hout =I

2π√

x2 + y2(− sin φx + cos φy) (7.26)

=I

2π√

x2 + y2

(− y√

x2 + y2x +

x√x2 + y2

y

)(7.27)

=I

2π(x2 + y2)(−y x + x y) (7.28)

On the x-axis ~H has a y-component only, namely

Hy,out(y = 0) =I

2π(x2 + y2)x∣∣∣∣y=0

=I

2πx(7.29)

Similarly for the inside Field

Hy,in(y = 0) =I

2πR2 x (7.30)

-5

0

5

-0.4 -0.2 0 0.2 0.4

Hy

x


Prof. S. Peik

Example 30:Lets double-check the above result using Ampere’s law in point form ∇× ~H = ~JOutside:

∇× ~H = ∇×(

I2πρ

φ

)(7.31)

=

(1ρ

∂Hz

∂φ− ∂Hφ

∂z

)r +

(∂Hρ

∂z− ∂Hz

∂ρ

)φ +

1ρ

(∂(ρHφ)

∂ρ− ∂Hρ

∂φ

)z (7.32)

= 0 r + 0 φ +1ρ

(∂(ρHφ)

∂ρ

)z (7.33)

=1ρ

I2π

(∂(ρ 1

ρ )

∂ρ

)z (7.34)

= 0 = ~Joutside (7.35)

Inside:

∇× ~H = ∇×(

I2πR2 ρ φ

)(7.36)

=

(1ρ

∂Hz

∂φ− ∂Hφ

∂z

)r +

(∂Hρ

∂z− ∂Hz

∂ρ

)φ +

1ρ

(∂(ρHφ)

∂ρ− ∂Hρ

∂φ

)z (7.37)

=1ρ

I2πR2

(∂(ρρ)

∂ρ

)z (7.38)

=1ρ

I2πR2 (2ρ) z (7.39)

=I

πR2 z = ~Jinside (7.40)

7.9.1 H-Field on a Sheet of Current

An infinite sheet in the z = 0 plane carries the y-directed surface current Ky~y (unit ishere A/m). Applying

3

2

K

Amperian Path

1

4

Ampere’s law yields

∮~H~dl = Kyy (7.41)

The H-field is composed of an infinite number of parallel currents in y-direction. Us-ing superposition we can deduct, that the H-field is x-directed with

~H =

H0 x for z > 0−H0 x for z < 0

(7.42)


Prof. S. Peik

Now integrating along the sketched Amperian Path of width b

∮~H =

∫ 2

1+∫ 3

2+∫ 4

3+∫ 1

4~H~dl (7.43)

= 0 + (−H0)(−b) + o + H0b (7.44)

= 2H0b (7.45)

with the total current of I = bKy through the loop we have for Ampere’s law

bKy = 2Hob (7.46)

H0 =Ky

2(7.47)

following for the field

~H =

Ky2 x for z > 0−Ky

2 x for z < 0(7.48)

or in general

H =12~K×~an (7.49)

where~an is the unit normal vector of the sheet.

7.10 Solenoids

A long straight coil of wire can be used to generate a nearly uniform magnetic fieldsimilar to that of a bar magnet. Such coils, called solenoids, have an enormous num-ber of practical applications. The field can be greatly strengthened by the additionof an iron core. Such cores are typical in electromagnets.

Figure 7.12: Solenoid with Field, from [?]

Shown in Figure 7.13 is a cross section of such a solenoid of length l with N turns.The superposition of the circular fields inside the solenoid gives rise to an almosthomogeneous field inside. The field outside is typically very weak.


Prof. S. Peik

Cross Section:Solenoid:

N Turns

l

I

Fields from Wires

Superposition

homogeous Field Inside,

Neclectible Filed Outside

Figure 7.13: Magnetic Field of a Solenoid

A

Path of Integration

Figure 7.14: Path of Integration for H-field Derivation


Prof. S. Peik

Taking a path about which to evaluate Ampere’s Law as shown in Figure 7.14 givesonly a contribution inside the coil. As the field outside is very weak we can neglectcontributions from outside.

Applying Ampere’s law

∑ I = N I =∮

H dl (7.50)

Now assuming the field to be zero outside we can split our integration into two sec-tions ∮

H ds =∫ 2

1H dl

︸︷︷︸H l

+∫ 1

2Hdl

︸︷︷︸0

= H l (7.51)

following

|~H| = N Il

(7.52)

The field inside a solenoid is constant and depends only from the length l and thenumber of turns N. Interestingly it is independent from the cross section of thesolenoid.

7.11 Biot-Savart’s Law

A more general form of Ampere’s law in differential form is Biot-Savart’s law. TheBiot-Savart’s Law relates magnetic fields to the currents which are their sources. Ina similar manner, Coulomb’s law relates electric fields to the point charges whichare their sources. Finding the magnetic field resulting from a current distributioninvolves the vector product, and is inherently a calculus problem when the distancefrom the current to the field point is continuously changing.

An infinitesimal current element makes a contribution to the magnetic field at pointP which is perpendicular to the current element, and perpendicular to the radiusvector from the current element to the field point P. The direction of the magneticfield contribution follows the right hand rule illustrated for a straight wire. This direc-tion arises from the vector product nature of the dependence upon electric current.

I ds

dH

ϕ

r

Figure 7.15: Biot-Savart’s Law

The Field contribution of a current element with current I and length dl is hence

dH =I dl

4πr2 sin ϕ (7.53)


Prof. S. Peik

where r is the distance to the field point P and ϕ the angle between current and thevector leading to the point P. Figure 7.15 shows the definitions. The H-field vector isalways perpendicular to the current direction. Hence we can write in vector notation

d~H =I d~l × r4πr2 = d~H =

I d~l ×~r4π|~r|3 (7.54)

where ~R is the vector leading from the current element (source point)~r′ to the fieldpoint~r, e.g.~r′ −~r.

Example 31:The H-field in the center of a circular loop is :With ϕ = 90, we get

dH =I

4πr2 ds (7.55)

dH

I ds

r

The total field is the integration of dH

H =∫

dH =∮ I

4πr2 ds =I

4πr2 2πr =I

2r(7.56)

We cannot calculate this result using Ampere’s law, as the H-field distribution is not known.

Just as for different charge configurations, we can now derive H-fields for differentcurrent distributions.

H =∫

L

I d~l × r4πr2 (line current) (7.57)

H =∫

L

~K dS× r4πr2 (surface current) (7.58)

H =∫

L

~J dV × r4πr2 (volume current) (7.59)

7.12 Magnetic Dipoles and Current Loop Analogy

There are two different ways of defining sources of magnetic fields as Figure 7.16illustrates.

Gilbert Model: Using the analogy to electrostatics, we can define as sources andsinks of magnetic field lines two “magnetic charges” usually referred to as


Prof. S. Peik

south and north pole. These “charges” arranged as dipoles, create exactly theobserved field. We must know though, that the dipoles must not be split, asthere are no magnetic monopoles. The model is kind of bad physics, as weexplain an observation based on our knowledge from electrostatics and ignorethe problem of the single charges.

Ampere’s Model: Every magnetic field is created by a moving charge, a.k.a cur-rent. It is hence a purely electrical phenomenon. Every moving charge givesrise to a magnetic field, that is circling this current. This can be nicely seen inBiot-Savart’s law above. A magnetic dipole field is created by a current flow-ing in a loop. A very small loop is an elementary magnet. Note that electronsspin and create a magnetic dipole field. This is the origin of any permanentmagnetic field.

N

SI

H H

Figure 7.16: Field from a Magnetic Dipole or Current loop

7.13 Definition of Magnetic Field by the Effect

Recall the two basic vector fields of electrostatics:

The E-Field is defined by the effect, i.e. the force on a test charge, precisely ~E =~Fq

with the unit Vm

The D-Field is defined by the cause, i.e. charges producing flux lines, where ~D =~Flux

Area = QA with the unit As

m2

Similarly, we can define the magnetic field in two ways, by the cause or by the effectrespectively:

The H-Field is defined by the cause, i.e. current producing field lines. We caneasily see the relation in Biot-Savart’s law. A current element gives rise tomagnetic field line loops.

The B-Field is defined by the effect, i.e. a force turning a magnetic dipole, asshown in Figure 7.17. When placing a magnet or current loop into a magneticfield, we can observe a torque acting on the magnet. The torque is a force ona cantilever, here the magnet length. The force on a current loop (or dipolemagnet) is proportional to the surrounding field, the B-Field.


Prof. S. Peik

N

S

F

F

F

FI

Figure 7.17: Force observed from a magn. Dipole and Current Loop of Size b× b

7.14 Magnetic Flux Density

When the B-field is perpendicular to the current I in a wire of length l we can observea force F which is

F ∝ B (7.60)

F ∝ I (7.61)

F ∝ l (7.62)

henceF = I · l · |~B| (7.63)

The direction of ~F is perpendicular to I and ~B. More details are given in Chapter 10.

The unit of the flux density is

[B] =[F][I][l]

=Ws

m

1Am

=Vs

m2 = 1 T (7.64)

The SI unit for magnetic field is the Tesla. One Tesla is equal to 1 Newton/(A/m). Fromthese units it can be seen that the flux density is a measure of the force applied toa particle by the magnetic field.

The B-Field is known as the magnetic flux density or or magnetic induction. In termsof behavior its the counterpart to the electric flux density D in electrostatics,eventhough cause and effect are switched.

The magnetic flux density ~B can be similarly derived from ~H as the electric fluxdensity ~D can be derived from ~E. Similar to the relation of field strength and fluxdensity in electrostatics we can find

~B = µ ~H (7.65)

with the permeability of µ = µ0 · µr, where the absolute permeability µ0 = 4π10−7 Vs

Am = 1,257·10−6 VsAm is the permeability of free space. Compare with ε of elec-

trostatics!


Prof. S. Peik

7.14.1 Permeability

The relative permeability µr is a constant of the material filling the space. The per-meability is a material property that describes the ease with which a magnetic fluxis established in the component.

A table is given in Table .

Type Material µr

Silver 0,999921Diamagnetism Lead 0,999984

Copper 0,99999Vacuum 1

Air 1,00000035Paramagnetism Aluminum 1,000024

Tungsten 1,000067Platinum 1,000256

Ferromagnetism Iron 1000

Table 7.1: Relative Permeabilities of Some Materials

Materials may be classified by their response to externally applied magnetic fieldsas diamagnetic, paramagnetic, or ferromagnetic. These magnetic responses differgreatly in strength.

Diamagnetism is a property of all materials and opposes applied magnetic fields,but is very weak.

Paramagnetism, when present, is stronger than diamagnetism and produces mag-netization in the direction of the applied field, and proportional to the appliedfield.

Ferromagnetic effects are very large, producing magnetizations sometimes or-ders of magnitude greater than the applied field and as such are much largerthan either diamagnetic or paramagnetic effects.

For paramagnetic and diamagnetic materials the relative permeability is very closeto 1. For ferromagnetic materials, the relative permeability may be very large.

7.15 Hysteresis Loop

Ferromagnetic materials exhibit a non-linear dependence between the magnetic flux~B and the magnetic field ~H. Hence

~B 6= µ~H for ferromagnetic materials (7.66)

or µ is not a constant, rather a function of ~H and ~B.

Ferromagnetic materials can be characterized by a hysteresis loop. A hysteresisloop shows the relationship between the induced magnetic flux density B and the


Prof. S. Peik

magnetizing force H. It is often referred to as the B-H loop. An example hysteresisloop is shown in Figure 7.18.

The loop is generated by measuring the magnetic flux density ~B of a ferromagneticmaterial while the magnetic auxiliary Field ~H is changed. A ferromagnetic materialthat has never been previously magnetized or has been thoroughly demagnetizedwill follow the dashed line as |~H| is increased. As the line demonstrates, the greaterthe amount of current applied, the stronger the magnetic field in the component. Atthe far right point almost all of the magnetic domains are aligned and an additionalincrease in the magnetizing force will produce very little increase in magnetic flux.The material has reached the point of magnetic saturation. When H is reduced backdown to zero, the curve will move to point Br. At this point, it can be seen thatsome magnetic flux remains in the material even though the magnetizing force iszero. This is referred to as the point of retentivity on the graph and indicates theremanence or level of residual magnetism in the material. (Some of the magneticdomains remain aligned but some have lost there alignment.) As the magnetizingforce is reversed, the curve moves to point Hc, where the flux has been reduced tozero. This is called the point of coercivity on the curve. (The reversed magnetizingforce has flipped enough of the domains so that the net flux within the material iszero.) The force required to remove the residual magnetism from the material, iscalled the coercive force or coercivity of the material.

B

H

New CurveBr

Hk

Br:Residual Magnetism

Hk: Point of Coercivity

Figure 7.18: Hysteresis Loop for Ferromagnetica

As the magnetizing force is increased in the negative direction, the material willagain become magnetically saturated but in the opposite direction. Reducing H tozero brings the curve to B = 0 again. It will have a level of residual magnetism equalto that achieved in the other direction. Increasing H back in the positive directionwill return B to zero. Notice that the curve did not return to the origin of the graphbecause some force is required to remove the residual magnetism. The curve willtake a different path back the saturation point where it with complete the loop.

From the hysteresis loop, a number of primary magnetic properties of a material canbe determined.

Retentivity Br- A measure of the residual flux density corresponding to the saturationinduction of a magnetic material. In other words, it is a material’s ability to retaina certain amount of residual magnetic field when the magnetizing force is removedafter achieving saturation. Residual Magnetism Br or Residual Flux - the magnetic


Prof. S. Peik

flux density that remains in a material when the magnetizing force is zero. Notethat residual magnetism and retentivity are the same when the material has beenmagnetized to the saturation point. However, the level of residual magnetism maybe lower than the retentivity value when the magnetizing force did not reach thesaturation level.

Coercive Force - The amount of reverse magnetic field which must be applied to amagnetic material to make the magnetic flux return to zero.

7.16 Magnetic Flux

The number of magnetic lines of force cutting through a plane of a given area at aright angle is known as the magnetic flux density ~B.

The total number of lines of magnetic force in a material is called magnetic flux ψ.The total flux is simply the flux density applied over an area A

Φ =∫

A~Bd~S (7.67)

Flux carries the unit of a Weber or Wb, which is simply a Tesla-square meter. Themagnetic flux is a quantity of convenience in the statement of Faraday’s Law (de-scribed in the next Chapter) and in the discussion of objects like transformers andsolenoids.

Flux lines of magnetic fields are always closed. In other words there are no sourcesand drains of magnetic flux lines, as magnetic sources are always currents or mag-netic dipoles1. It is not possible to have a isolated magnetic monopole (magneticcharge). Thus the total flux through a closed surface in a magnetic field must bezero, that is

∮~Bd~S = 0 (7.68)

This is known as the law of conservation of magnetic flux or Gauss law for mag-netostatic fields. Although the magnetic field is not conservative magnetic flux isconserved. By applying the divergence theorem of eqn. A.62 we obtain

∇ · ~B = 0 (7.69)

This is known as Maxwell’s fourth equation in point form.

1Magnetic permanent dipoles have their origin also in subatomic currents, hence, magnetic fieldsare always created by currents


Prof. S. Peik

Example 32:Inside a ferromagnetic bar (µr = 1000) with two cross sections as shown a magnetic field isgenerated by a coil with N = 100 turns. Find H,B and Φ in both sections. Assume a lengthof the coil of a. The length a is 1cm. The current I=1A.

N Turns

I

a

a

2a

2a

B1B2

The H-field is created following eqn. 7.52

H1 =NIl

=100A

0.01m= 10 000

A

m(7.70)

now

B1 = µ0µr H1 = 4π · 10−7 Vs

Am1000 · 10 000

A

m(7.71)

= 12.566 T (7.72)

The flux in section one is

Φ1 = B · A = 12.566 T · 0.012m2 (7.73)

= 0.0012566 Vs (7.74)

Since the flux must remain the same in section one and section two the B and H field changesto

B2 =ΦA2

=0.0012566 Vs

4 · 0.012m2 (7.75)

= 3.1415 T (7.76)

H2 =B2

µ0µr=

3.1415 T

µ01000(7.77)

= 2499.9A

m(7.78)

7.17 Magnetic Vector Potential

As we know from Chapter 3 the electric field ~E can always be expressed as thegradient of a scalar potential function. There is no general scalar potential for themagnetic field ~B but it can be expressed as the curl of a vector function.

This function ~A is given the name magnetic vector potential but it is not directlyassociated with work the way that scalar potential is.

In order to define the vector potential, we start with defining a scalar magneticpotential Vm similar to eqn 3.38 by

~H = −∇Vm (7.79)


Prof. S. Peik

With the identity ∇× (∇ · ~A) = 0 and Ampere’s law we can write

~J = ∇× H = ∇× (−∇Vm) = 0 (7.80)

Since Vmmust satisfy the vector identity ∇× (∇Vm) = 0 the scalar potential is onlydefined for regions with ~J = 0.

∇2Vm = 0 when ~J = 0 (7.81)

In order to satisfy Gauss’s law for magneto-statics and the identity we can find outthat

~B = ∇× ~A (7.82)

Plugging it into Ampere’s law yields

∇× ~B = ∇× (∇× A) = ∇(∇ · ~A)−∇2 ~A = µ~J (7.83)

7.17.1 Gauge Transformations

Since the magnetic field ~B is defined as the curl of ~A, and the by vector identitythe curl of a gradient is identically zero, then any arbitrary function which can beexpressed as the gradient of a scalar function f may be added to ~A without changing

the value of ~B obtained from it. That is, ~A′ can be freely substituted for ~A where

~A′ = ~A + ~∇ f (7.84)

Such transformations are called gauge transformations, and there have been a num-ber of "gauges" that have been used to advantage is specific types of calculationsin electromagnetic theory.


Prof. S. Peik

Example 33:We define an arbitrary vector potential

~A =

00

x y z

(7.85)

We can now derive the B-Field of this Potential as

~B = ∇× ~A =

x z−y z

0

(7.86)

We can use a gauge transformation with any scalar function, e.g. f = x + y2 + z3, where

∇ · f =

12y3z2

(7.87)

Our new vector potential is now ~A′ = ~A +∇ · f =

12y

x y z + 3z2

The B-field derived from this potential is still

~B = ∇× ~A′ =

x z−y z

0

(7.88)

Mupad Code:export(linalg);

print(Unquoted," First we define an arbitrary Vector Potential A=");

A:=matrix(3,1,[0,0,x*y]);

print(Unquoted," The B-field is B=curl(A)=");

B:=curl(A,[x,y,z]);

print(Unquoted," now we define a differentiable function f=");

f:=x+y*y+z*z*z;

print(Unquoted," where grad(f)=");

grad(f,[x,y,z]);

print(Unquoted," Now we can define a new Aprime=A+grad(f)");

Aprime:=A+grad(f,[x,y,z]);

print(Unquoted," The B-Field is still B=curl(Aprime)=");

B:=curl(A,[x,y,z]);

quit

The simplest gauge transformation is done by choosing ~A such that ∇ · ~A = 0. Thisis known as Coulomb’s gauge. Another gauge often used in EM-wave problems isthe Lorenz gauge ∇ · ~A = jωµεΦ = 0, including a scalar potential Φ.


Prof. S. Peik

Example 34:Using Coulomb’s gauge on the vector potential of the last example ∇ · ~A = 0 must besatisfied. This is not the case

∇ · ~A =

∂∂x∂

∂y∂∂z

00

x y z

= x · y (7.89)

We can now add the gradient of an arbitrary function which fixes the problem.

∇ · ~A′ = ∇ · (~A +∇ · f ) = ∇ ·

00

x y z

+

∂ f∂x∂ f∂y∂ f∂z

= 0 (7.90)

As we see, we can use a function f = − 12 x y z2, The gradient is

∇ · f =

− 1

2 yz2

− 12 xz2

−x y z

(7.91)

Now

∇ · ~A′ = ∇ · (~A +∇ · f ) = ∇ ·

12 yz2

12 xz2

0

= 0 (7.92)

we have a Coulomb’s gauge

7.17.2 Poisson Equation for Magnetostatics

By choosing Coulombs gauge we get

∇× ~B = −∇2 ~A = µ~J (7.93)

In contrast to the usual definition the Laplacian operator ∇2 is applied to a vector.Note, that we have to use the definition

∇2 ~A = ∇ · (∇ · ~A)−∇× (∇× ~A) (7.94)

In Cartesian coordinates, we can derive three independent differentials, as

∇2 ~A = (∇2Ax)x + (∇2Ay)y + (∇2Az)z (7.95)

In other curvilinear coordinate systems this is not the case (poisonous snake), i.e.

∇2 ~A 6= (∇2Ar)r + (∇2Aφ)φ + (∇2Aθ)θ (7.96)

in spherical coordinates.

Eqn 7.93 is the Poisson equation known from electrostatics now applied to magneto-statics. We can now solve this equation and get the solution for the vector potential


Prof. S. Peik

~A =∫

µI d~l4πR

for a line current (7.97)

~A =∫

µ~K d~S4πR

for a surface current (7.98)

~A =∫

µ~J dV4πR

for a volume current (7.99)

Idl

dA

R

I

Figure 7.19: Vector Potential from Current Distributions

As seen from the equation this vector equation are actually three scalar equationfor x,y, and z-direction. This means the d~A and d~J are always in the same direction.

However, when applying ~A =∫ µ~J dV

4πR we must express ~J in Cartesian coordinates.Again, the poisonous snake is the problem. Also, note, that the vector potential canusually not be derived for problems with currents in infinity, like an infinite straightwire. Here we have to apply other strategies, as described in [?].

From the vector potential ~A we can now derive the magnetic flux by applying Stokestheorem on Φ =

∫~B d~S with eqn. 7.82 and get

Φ =∮

~A d~l (7.100)

The total flux through a closed loop can be derived by summing up the ~A-field alongthe boundary of the loop.

The validation of the above equations is omitted here and can be found in [?] andother electromagnetics textbooks.

One rationale for the vector potential is that it may be easier to calculate the vectorpotential than to calculate the magnetic field directly from a given source currentgeometry. Its most common application is to antenna theory and the description ofelectromagnetic waves.


Prof. S. Peik

Example 35:Given the magnetic potential ~A = − 1

4 ρ2 z Wbm calculate the total magnetic flux crossing the

surface φ = π2 , 1 ≤ ρ ≤ 2mand 0 ≤ z ≤ 5m

Using ~B = ∇× ~A =ρ2 φ

now the flux is

Φ =∫ 5

0

∫ ∫ 2

1~B · d~S =

∫ 5

0

∫ ∫ 2

1Bφ dz dρ =

12

5[12

ρ2]21 =54(4− 1) =

154

Wb (7.101)

Or using eqn 7.100

Φ =∮

~A d~l =∫ 2

1+∫ 3

2+∫ 4

3+∫ 1

4(7.102)

y

x

z

1 2

4

2

12

3 4

A

π/2

Φ =∫ 3

2Az(ρ = 1) dl −

∫ 1

4Az(ρ = 2) dl (7.103)

= −14

5 +44

5 =154

Wb (7.104)

7.18 Boundary Conditions

The behavior of magnetostatic fields on boundaries is analogous to the definitionsin electrostatic fields

Ht1 = Ht2Bn1 = Bn2

(7.105)

and

µ1Hn1 = µ2Hn2µ2Bt1 = µ1Bt2

(7.106)


Inductance and Mag-netic Circuits

8In Chapter 4.1 we defined a capacitance as the ratio of electric flux (i.e. the D-field)to voltage. The device storing the electrostatic energy was the capacitor. Analo-gously, we define now an inductance as the ratio of magnetic flux to current. Thedevice storing magnetic energy is the inductor.

8.1 Inductance

Let us take a closer look at a current loop again as shown in Figure 8.1 . The currentI creates a magnetic flux Φ through the loop. Following Ampere’s law the flux islinearly dependent on the current I.

Figure 8.1: Current loop with B-field

The inductance of a current loop is defined as

L =ΦI

(8.1)

The unit of the inductance is

[L] =[Φ]

[I]=

VsA

= H (8.2)

with the short form H (Henry). Note, in comparison the unit with the capacitanceunit [C]=F=As/V

111

Prof. S. Peik

8.2 Flux Linkage

For a solenoid with more than one turn the magnetic flux multiplies with the numberof turns as the flux penetrates every loop. We can now define a total flux called theflux linkage to be

λ = NΦ (8.3)

where N is the number of turns of the coil.

Figure 8.2: Explanation Flux Linkage

Hence the inductance of a solenoid is

L =NΦ

I(8.4)

8.3 Inductance of a Long Solenoid

For a long slim solenoid of length l and cross section A and N turns we can easilyfind its inductance.

We know that H = n Il , hence B = µ 0

n Il .

Consequently the flux is Φ = B A = µ 0A n Il

With the definition of the inductance we get:

L =NΦ

I= µ0 A

n2

l(8.5)

The circuit symbol for an inductance is a group of four semi circles as shown in Figure8.3.

Figure 8.3: Circuit Symbol of an Inductance


Prof. S. Peik

8.4 Circuits with Inductances

When connecting several inductances the total inductance of the group is derivedby the following equations.

Inductances in series:

Lges = L1 + L2 + L3 + · · · (8.6)

L1 L2 L3

Inductances in parallel:

1Lges

=1L1

+1L2

+1L3

+ · · · (8.7)

L1

L2

L3

Example 36:Calculate the external inductance per length of a coaxial cable as shown. The externalinductance is defined as the inductance due to the B-field in the dielectric region

R1

R2

R3

we know for the dielectric region

~B2 =µI

2πρφ (8.8)

The flux through an area element in this region is

dΦ = ~B2dρ dz =µI

2πρdρ dz (8.9)

Hence the total flux through the region between the conductors is

Φ =∫ R2

R1

∫ l

0dΦ =

∫ R2

R1

∫ l

0

µI2πρ

dρ dz = lµI2π

[ln ρ]R2R1

= lµI2π

lnR2R1

(8.10)

The external inductance is due to the B-field in the dielectric, hence the external inductancefor a cable of length l is

L′ext =ΦI=

µl2π

lnR2R1

(8.11)


Prof. S. Peik

8.5 Energy in the Magnetic Field

Just as the energy of electrostatic fields the magnetic field energy is derived by

W =12

∫~H ~B dV =

12

∫µ|~H|2dV (8.12)

following for the stored energy in an inductor

W =12

LI2 (8.13)

We can rearrange the eqn. and get

L =2WI2 =

1I2

∫~B · ~H dV (8.14)


Prof. S. Peik

Example 37:From last example we know

~B2 =µI

2πρφ (8.15)

Hence the inductance using eqn. 8.14 is

Lext =1I2

∫ 2π

0

∫ R2

R1

∫ l

0µH2 ρ dρ dz (8.16)

=∫ 2π

0

∫ R2

R1

∫ l

0µ

122π2ρ2 ρ dρ dz (8.17)

= 2πlµ

4π2 lnR2R1

(8.18)

=µl2π

lnR2R1

(8.19)

Or the inductance per length

L′ext ==µ

2πln

R2R1

(8.20)

Compare with the last example!We know for the inner conductor region ρ < a

~B1 =µIρ

2πR21

φ (8.21)

following

Lint =1I2

∫ 2π

0

∫ R1

0

∫ l

0µB2 ρ dρ dz (8.22)

=∫ 2π

0

∫ R1

0

∫ l

0

1µ

µ2ρ2

4π2R41

ρ dρ dz (8.23)

= 2πlµ

4π2R41[14

ρ4]R10 (8.24)

Lint =µl8π

(8.25)

or as impedance per length

L′int =µ

8π(8.26)

The total inductance is L = Lint + Lext =µ

2π

(14 + ln R2

R1

)


Prof. S. Peik

8.6 Magnetic Circuits

Magnetic field lines can be guided through ferromagnetic materials, as the µrµ0 isvery high compared to the µ0 in air. Consequently, We call ferromagnetic materi-als also magnetic conductors. The magnetic flux can be considered a “magneticcurrent” driven by the H-field. The H-field is created by currents passing throughthe H-field loop. Magnetic field lines are always closed, as the current loop is in anelectric circuit.

Using this analogy, we can treat magnetized ferromagnetic rings as a circuit anal-ogous to electric circuits. We call this a magnetic circuit. When we can assumehomogeneous fields inside the material we can define

• A coil with N turns acts like a generator with magnetic voltage Um = N · I.

• The current is represented by a magnetic current which is equal to the flux Φ

• A piece of ferromagnetic material with µr → ∞ acts like a loss-less wire

• A piece of material with µr 6= ∞ acts like a resistor Rm, also known as reluc-tance Rm

N

I

µr

lFE

Um

Um,FE

Um,G

ψ

Figure 8.4: Ferromagnetic Torus with Air Gap and equivalent. Magnetic Circuit

We can relate now the magnetic quantities to the electric quantities by comparesAmpere’s law with Kirchhoff’s law.

For Ampere’s law we have for the magnetic circuit of Figure 8.4∮

H dl = ∑ I (8.27)

HFE lFE + HG lG = N · I (8.28)

where the indexes FE and G denote the Iron and Gap, respectively. In Kirchhoff’slaw we get

UM = Um,FE + Um,G (8.29)

Comparing the two we can observe


Prof. S. Peik

Um,FE︸︷︷︸+Um,G︸︷︷︸ = Um︸︷︷︸ (8.30)

HFE lFE + HG lG = N · I (8.31)

The source can be represented as a DC voltage source with Um = N · I. The voltagedrops are represented by Um,FE = HFE lFE and Um,G = HG lG. Note that the unit ofthe magnetic voltage is Amperes.

8.7 Magnetic Resistance

We defined the flux as the magnetic current

Im = Φ (8.32)

Now we can relate the magnetic voltage to the magnetic current and find a magneticresistance also called reluctance. For the section of iron we have

Rm =Um,FE

Im=

HFE lFE

Φ=

1µ B lFE

Φ=

1µ

ΦA lFE

Φ=

1µ

lFE

A(8.33)

or in general for any cylindrical section of length land cross section Aas shown inFigure 8.5

N

I

µr

l

ψ

Um

Figure 8.5: Definition of Magnetic Reluctance

Rm =1µ

lA

(8.34)

For a magnetic circuit with homogeneous fields we can use an electric equivalentcircuit with

Definition Unit

Magnetic Voltage Um = N · I AMagnetic Current Im = Φ Vs

Magnetic Resistance (Reluctance) Rm = 1µ

lA

AVs

Analogous to electric circuits we can define an Ohm’s law


Prof. S. Peik

Um = Rm ·Φ (8.35)

and also can apply Kirchhoff’s laws

∑ Um = 0 and ∑ Φ = 0 (8.36)

on any magnetic circuit.This technique is very useful in analyzing magnetic circuits

Example 38:Find the inductance of the torus-coil in Figure 8.4 with lFE = 20cm and gap width of lg =2mm

and a cross section of A =1cm2and a µr = 1000. The coil has 200 turns.The magnetic voltage source has a voltage of Um = N · I = 200IThe magnetic reluctance of the torus is Rm,FE = lFE

µ0µr A

The magnetic reluctance of the gap is Rm,g =lg

µ0 AThe magnetic current (flux Φ) is hence

Φ =Um

RFE + Rg=

NI1

µ0 A ( lFEµr

+ lg)(8.37)

The inductance is the flux linkage N ·Φ per current I. Hence

L =NΦ

I=

NI

NI1

µ0 A ( lFEµr

+ lg)(8.38)

L = µo AN2

l f eµr

+ lg(8.39)

Using the numerical values we get L = 2.28mH

8.7.1 Perfect Magnetic Conductor

In a perfect magnetic conductor the permeability reaches infinity and the magneticvoltage drop is zero, there is still flux (magn. current) and, consequently, magneticflux density

µ = ∞ (8.40)

H = 0 (8.41)

Φ = finite (8.42)

B = finite (8.43)


Time-Varying Fields 9So far we observed static fields, only. Static means that we do not have any fieldvariations over time. Now, let us extend our field theory to dynamic problems, i.e.with time variation. Time-varying magnetic fields for example generate electricity,which is called induction. The law of induction, first observed by Faraday, is nowknown as Faraday’s law.

9.1 Faraday’s Law

Any change in the magnetic environment of a coil of wire will cause a voltage (emf)to be "induced" in the coil. No matter how the change is produced, the voltagewill be generated. The change could be produced by changing the magnetic fieldstrength, moving a magnet toward or away from the coil, moving the coil into or outof the magnetic field, rotating the coil relative to the magnet, etc.

Vi Vi Vi

Flux Decreases

Figure 9.1: Time-Varying Flux through a Loop Induces Voltage

Experimentally we can demonstrate induction by two magnetically linked coils. asshown in Figure 9.2. Changing the current in coil 1 creates a change of the fluxthrough coil 2, and, hence, a voltage Vi. Also moving the coil away or rotating it willresult in flux changes and hence in an induced voltage Vi

Faraday formulated his law of induction as

Vi = −dΦdt

(9.1)

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Prof. S. Peik

B, Φ

Ui

i

Figure 9.2: Induction from One Coil to Another

for a coil with N turn we get using flux linkage

Vi = −NdΦdt

(9.2)

Here Vi is the induced voltage of the arrangement. We seen from the equation weget an induced voltage only, when the flux through the loop changes.

In case of a stationary loop as shown in Figure 9.1 the total induced voltage becomesthe integral of the E-field along the wire and Φ =

∫~B d~S, hence we can rewrite

Faraday’s law as

Vi =∮

~E d~l = −∫

∂~B∂t

d~S (9.3)

By applying Stoke’s Theorem onto the LHS of this equation we get

∫

A(∇× ~E) d~S = −

∫∂~B∂t

d~S (9.4)

or

∇× ~E = −∂~B∂t

(9.5)

This is known as the point form of Faraday’s law.

9.2 Lenz’s Law

The minus sign in eqn. 9.1 denotes, that the induced voltage is of opposite sign ofthe flux using a right-hand system. Finding the correct sign of Vi can be very tricky,as a lot of directions and coordinate systems might be involved. We can find thesign of the voltage or induced current, however, by a very simple rule:

The induced current in a conductor is always directed such that itopposes the change in the applied field

This rule is known as Lenz’s law, after the nineteenth century Russian scientist Hein-rich Lenz who first formulated it.


Prof. S. Peik

When an emf is generated by a change in magnetic flux according to Faraday’s Law,the polarity of the induced emf is such that it produces a current whose magneticfield opposes the change which produces it. The induced magnetic field inside anyloop of wire always acts to keep the magnetic flux in the loop constant. In theexamples below, if the B field is increasing, the induced field acts in opposition to it.If it is decreasing, the induced field acts in the direction of the applied field to try tokeep it constant.

Φ increasing dΦ/dt>0

Ι

(opposes increasing Φ)

++++++

- - -- - -

Φ decreasing dΦ/dt<0Ι

(opposes decreasing Φ)

++++++

- - -- - -Ui>0Ui<0

Figure 9.3: Lenz’s Law

9.3 Moving Loop in a Magnetic Field

We can create a change of magnetic flux Φ also by moving the loop into out outof the field. A loop of Area A moved into the field will experience a positive fluxchange. Hence a negative induced voltage will be observed.

Example 39:A triangular loop is moved into a homogeneous magnetic field with constant velocity v

The flux increases first by t2, stays then constant and decreases by t2 again.The induced voltage is equal to the negative derivative of . Hence we get the run of the fluxand voltage as follows

Vi

Parabola

Parabola


Prof. S. Peik

9.4 Self- and Mutual Inductance

We can induce a voltage in a coil through a time-varying magnetic field generatedby another coil. This arrangement is shown in Figure 9.4. Thew left coil is called theprimary coil. The right coil is called the secondary coil.

B, Φ

Vi

A

N1 N2

B, Φ

Vi

i

A

N1

Self InductanceMutual Inductance

i

Figure 9.4: Mutual and Self Inductance

A current I1though the primary coil creates a flux Φ21in the secondary coil. This fluxreaches partly or in whole the secondary coil and induces the voltage

Vi2 = −N2dΦ21

dt(9.6)

9.4.1 Mutual Inductance

This follows the definition of the mutual inductance as

M21 =N2Φ21

i1(9.7)

In plain English, the mutual inductance is the flux through coil 2 generated by acurrent in coil 1 divided by that current in coil 1. The mutual inductance has the unitHenry.

Now we can write for the induced voltage

Vi2 = −M21di1dt

(9.8)

Similarly we can define a reverse mutual inductance

M12 =N1Φ12

i2(9.9)

and

Vi1 = −M12di2dt

(9.10)


Prof. S. Peik

for the induced voltage in the primary coil due to a current in the secondary coil.Interestingly the two mutual inductances are always identical due to reciprocity.

M12 = M21 (9.11)

The proof is omitted here and can be found in many electromagnetics text books.

9.4.2 Self Inductance

The current i1 gives also rise to an induced voltage in it’s own coil. That is, the fluxΦ1 creates a voltage

Ui1 = −N1dΦ1

dt(9.12)

at itself. Following with the definition of inductance from eqn 8.1

L =NΦ1

i1(9.13)

for the induced voltage

Ui = −Ldidt

(9.14)

This is known as the self induced voltage.

9.5 V-I-Dependence in coils in AC-Circuits

Using the phasor U = Uejωt as the applied voltage to a coil we get

U = −LdIdt

= −LjωIejωt = −ZL Iejωt (9.15)

As the inductor is considered a passive device, the EMF-voltage direction is switched,such that voltage and current are in the same direction. This means the impedanceof a coil is just

ZL = jωL (9.16)

9.6 Transformer

A Transformer is a ferromagnetic ring with at least two coils on it. A typical trans-former is shown in Figure 9.5. The left coil, called the primary coil, has N1turn. Theright coil, the secondary coil, has N2turns.

There is a galvanic separation between the coil, this means no charges can movefrom one coil to the other. Energy is transmitted from the primary coil to the sec-ondary coil by the magnetic field only.

As shown in Figure 9.5, the left coil creates a magnetic flux Φ1, that penetrates theright coil. For time-varying currents, this induces a voltage on the right coil.


Prof. S. Peik

L1

L2

Φ1

Φ2

i1

i2

u1

u2

N1 N2

Figure 9.5: Transformer

The flux creates the voltages by Faraday’s law

V1 =N1dΦ1

dt(9.17)

V2 =N2dΦ2

dt(9.18)

Using the definition L1 = NΦ1i1

follows

V1 = L1di1dt

. (9.19)

Or similarly

M21 = N2dΦ1

i1(9.20)

as mutual induction from coil 1 to coil 2. Hence9.17 and 9.18

V1 = L1di1dt

(9.21)

V2 = M21di1dt

(9.22)

When exciting the secondary coil 2 with i2we get

V1 = M12di2dt

(9.23)

V2 = L2di2dt

(9.24)

Note, that the mutual inductance in forward direction and backward direction is al-ways the same, i.e. M12 = M21.


Prof. S. Peik

Superposition of both cases yields

V1 = L1di1dt

+ Mdi2dt

(9.25)

V2 = Mdi1dt

+ L2di2dt

(9.26)

L1 L2

M

Figure 9.6: Circuit Schematic of a Transformer

For AC-Circuits with I1 = I1ejωt and I2 = I2ejωtwe get

U1 = jωL1 I1 + jωM12 I2 (9.27)

U2 = jωM21 I1 + jωL2 I2 (9.28)

or written as a matrix

(U1U2

)= jω

(L1 MM L2

)(I1I2

)(9.29)

Equations 9.27 and 9.28 can be extended by including ohmic losses inside the coilssuch that

U1 = (jωL1 + R1)I1 + jωM12 I2 (9.30)

U2 = jωM21 I1 + (jωL2 + R2)I2 (9.31)

We can interpret eqn. 9.30 and 9.31 by an equivalent circuit as shown in Figure 9.7.

L1-M L2-M

MU1

I1 I2

U2

R1 R2

Figure 9.7: Equivalent Circuit of a Transformer


Prof. S. Peik

Example 40:Lets assume a transformer with 200 turns on the primary side and 10 turns on the secondaryside. The core of the transformer has a total length of 10 cm a cross section of 1 cm2 and amagnetic permeability of µr = 1000.

1. Find the self and mutual inductance

2. Find the voltage on the secondary side when the secondary side is open and to theprimary side a voltage of 230 V with 50 Hz is applied.

The magnetic resistance is

RFE =1

µoµr

lA

= 7.9577e + 05 AVs (9.32)

When applying current i1 only, the flux in the ring is:

Φ =N1i1RFE

= 2.5 · 10−4Vs · i1 (9.33)

M11 =N1φ

i1= 50mH (9.34)

M21 =N2φ

i1= 2.5mH (9.35)

When applying current i2 only, the flux in the ring is:

Φ =N2i1RFE

= 1.25 · 10−5Vs · i2 (9.36)

M22 =N2φ

i2= 0.125mH (9.37)

M12 =N1φ

i2= 2.5mH (9.38)

Now assume I1 = I1ejωt and I2 = 0 and V1 = 230Vejωt

following eqns. 9.27 and 9.28

U1 = jωL1 I1 (9.39)

U2 = jωM21 I1 (9.40)

I1 =U1

jωL1= −j14.64A (9.41)

U2 = jωM21 I1 = 11.5V (9.42)

9.6.1 Ideal Transformer

An ideal transformer implements a core with high permittivity µ, no copper lossesand no fringe fields. The primary side has by definition N turns whereas the sec-ondary side has one turn. Now we have the inductances

L1 = N2µlA

(9.43)

L2 = µlA

(9.44)

M12 = M21 = NµlA

(9.45)


Prof. S. Peik

where l is the length of the ring core and A its cross-section.

Hence

U1 = jω(N2µlA)I1 + jω(Nµ

lA)I2 (9.46)

U2 = jω(NµlA)I1 + jω(µ

lA)I2 (9.47)

Multiplying eqn . 9.47 by N and dividing the equations yields

U1NU2

= 1 (9.48)

U1U2

= N (9.49)

Similarly we can findI1

I2=

1N

(9.50)

and, hence, an impedance Z = U2I2

on the secondary side is transformed into a new

impedance Z′ = U1I1

seen on the primary side as

Z′ = N2Z (9.51)

9.7 Eddy Currents

Faraday’s Law implies that a changing magnetic flux produces an induced electricfield even in empty space. If a metal plate is inserted into this empty space theinduced electric field produces an electric currents in the metal. These inducedcurrents are called eddy currents.

If the induced currents are created by a changing magnetic field then the Eddy cur-rents will be perpendicular to the magnetic field and flowing in circles if the B-fieldis uniform.

These induced electric fields are very different from electrostatic electric fields ofpoint charges. For one thing they do not begin and end on charges but circle aroundon them self in loops. Secondly, and more importantly practically, the induced elec-tric field is non conservative, i.e. the work done by generated electric field can notbe recovered except as heat.

Often eddy currents generate unwanted losses. For example, in a transformer core,eddy currents heat up the core. By constructing the metal core of alternating lay-ers of conducting and nonconducting materials, the size of the induced loops arereduced thereby reducing the energy lost.


Prof. S. Peik

dB/dt

Eddy CurrentElectrical Conductor

Figure 9.8: Eddy Currents

Figure 9.9: Lamented Transformer Core


Prof. S. Peik

'

&

$

%

Roller Coaster Break

Eddy current breaks are used for many applica-tions. For example, in underneath a roller coastercar aluminum plates are installed. when movinginside a magnetic field, eddy current are gener-ated and produce heat. The energy is taken fromthe movement of the car. The car breaks.


ElectromagneticForces

10Magnetic fields are often utilized to create mechanical forces. A short overview ofthese effects are given here.

10.1 Lorentz Force

The Lorentz force is the force exerted on a charged particle in an electromagneticfield. The particle will experience a magnetic force The force relationship is in theform of a vector product:

~F = Q(~v× ~B) (10.1)

B

Fv v F

B

Q

Figure 10.1: Lorentz Force

The force is perpendicular to both the velocity ~v of the charge Q and the magneticfield ~B. The direction of the force is given by the right hand rule, as shown in Figure10.2. The thumb indicates the direction of the cause (Movement), the index fingerpoints in the direction of the field (B-field) and the middle finger points in resultingforce direction. By switching cause and effect, we can also get the direction of theinduced current, when the wire moves through the field.

In contrast, in an electric field the force always acts on a charge Q in the samedirection as the E-field. Thus an electron qe will simply be accelerated in the same

130

Prof. S. Peik

linear orientation as the E field, but that electron will spiral when traveling throughthe B field, due to the orientation of the cross product operator, by the right-handrule.

Cause

Field

Effect

Figure 10.2: Right Hand Rule:

10.1.1 Force on a Current Carrying wire

We can also derive the force on a current through a wire perpendicular to a magneticfield, as shown in Figure 10.1.

We know from the continuity equation

I =dQdt

(10.2)

further the velocity of charges in the wire is

v =dldt

(10.3)

following

F = (I · dt)

(~ldt× ~B

)(10.4)

or

~F = I(~l × ~B

)(10.5)

where~lis the length of the wire in the field with the direction of the current.


Prof. S. Peik

Joch N

B

I

F

S

I

I

F

B

Figure 10.3: Lorenz Force

10.2 Force on two Parallel Wires

The force on two parallel wires with current I1and I2 is

Two force can be derived from the Lorentz force to

F =µ

2π

I1 I2 ld

(10.6)

The SI-Unit Ampere is defined by this arrangement: One ampere is that constantcurrent which, if maintained in two straight parallel conductors of infinite length,of negligible circular cross-section, and placed one meter apart in vacuum, wouldproduce between these conductors a force equal to 2 · 10−7N per meter of length.The definition for the ampere is equivalent to fixing a value of the permeability ofvacuum to µ0 = 4π · 10−7 As

Vm .


Prof. S. Peik

10.3 Force on an Air Gap

When a magnetic circuit has an air gap, we can observe a force on the circuit, suchthat the parts of the circuit are attracted.

Figure 10.4 shows such a circuit.

Β

i

N

F

F

Joch Anker

Figure 10.4: Force on an Air Gap in a Magnetic Circuit

The force on the air gap is

F =12

B2L

µ0AL (10.7)

The exact derivation of this formula is given in many text books.


ElectromagneticWaves

11In the last chapters we already realized that magnetism and electricity are not twoisolated realms of physics. As shown, electric currents can generate magnetic fieldsand magnetic fields can generate electric fields. In this chapter we will investigatethe effects of this interaction further. We will work out, that time-varying electricfields can also generate magnetic fields. This brings us to the postulation of theexistence of electromagnetic waves. EM-waves can propagate through the emptyspace by exchanging electric and magnetic energy periodically.

11.1 History

The postulation of electromagnetic waves is strongly connected to the name JamesClark Maxwell. In the middle of the 19th century He developed the mathematicalconcept that implies the existence of EM-waves. Being a mathematician he did notreinforced his theory by experiments. However, 1888 Heinrich Hertz did the cru-cial experiment. He designed an apparatus that generated waves with around 100MHz. He demonstrated that these waves could be reflected, focused, and diffracted.Further he could show that the waves are polarized.

Figure 11.1: James C. Maxwell (1831-1879) and Heinrich Hertz (1847-1894)

134

Prof. S. Peik

11.2 Maxwell’s Contribution

Michael Faraday revolutionized physics in 1830 by showing that electricity and mag-netism were interrelated phenomena. He achieved this breakthrough by carefulexperimentation. Between 1864 and 1873 James Clerk Maxwell achieved a similarbreakthrough by pure thought. Of course, this was only possible because he wasable to take the experimental results of Faraday, Ampere, etc., as his starting point.All these equations were known in it’s integral Form, i.e.

Faraday’s Law:∮

C~E · d~l = − ∂

∂t

∫

A~B · d~A (11.1)

Ampere’s Law:∮

C~H · d~l =

∫

A~J (11.2)

Gauss’s Law:∮

S~Dd~S =

∫

VρdV (11.3)

magn. Gauss’s Law.:∮

S~Bd~S = 0 (11.4)

Maxwell’s first great achievement was to realize that these laws could be expressedas a set of partial differential equations. Of course, he wrote his equations out incomponent form because modern vector notation did not come into vogue untilabout the time of the First World War. In modern notation, Maxwell first wrote

∇× ~E = − ddt~B (11.5)

∇× ~H = ~J (11.6)

∇ · ~D = ρ

∇ · ~B = 0

11.3 Displacement Current

Maxwell’s second great achievement was to realize that there was something wrongwith the equations.

We can see that there is something slightly unusual about eqns. 11.5 and 11.6. Theyare very unfair to electric fields! After all, time varying magnetic fields can induceelectric fields, but electric fields apparently cannot affect magnetic fields in any way.However, there is a far more serious problem associated with the above equations,which we alluded to earlier on. Consider the integral form of the second Maxwell’sequation (i.e., Ampere’s law)

∮

C~H d~l =

∫~J d~S (11.7)

This says that the line integral of the magnetic field ~H around a closed loop C isequal to the flux of the current density through the loop. The problem is that the fluxof the current density through a loop is not, in general, a well defined quantity. In


Prof. S. Peik

order for the flux to be well defined the integral of~J d~S over some surface S attachedto a loop C must depend on C but not on the details of S .

We can observe this from the point form equation

∇× ~H = ~J (11.8)

∇ · (∇× ~H) = ∇ ·~J (11.9)

0 = ∇ ·~J (11.10)

This means the divergence of the right hand side of eqn 11.7 in point form must bezero,

∇~J = 0 (11.11)

Compare this with the continuity equation 11.12.

∇~J = −dρ

dt(11.12)

Unfortunately, the above condition is only satisfied for non time varying fields. Thus,∇~J = 0 is only true in a steady state (i.e., when d

dt = 0 ).

The problem with Ampere’s law is well illustrated by the following example. Considera long straight wire interrupted by a parallel plate capacitor as shown in Figure 11.2.

I

H-Field

I

dΦ/dt=Displacement Current

H-Field

dΦ/dt

HTang. is not continous

H due to dΦ/dt

Before Maxwell: After Maxwell:

~ ~

Figure 11.2: Circuit with Capacitor Illustrating Problem with Ampere’s Law

Suppose that C is some loop which circles the wire. In the non time dependentsituation the capacitor acts like a break in the wire, so no current flows, and nomagnetic field is generated. There is clearly no problem with Ampere’s law in thiscase.

In the time dependent situation a transient current flows in the wire as the capacitorcharges up, or charges down, so a transient magnetic field is generated. Thus, theline integral of the magnetic field around C is (transiently) non-zero. According toAmpere’s law, the flux of the current through any surface attached to C should alsobe (transiently) non-zero. At the boundary on the capacitor plate, this condition isnot satisfied. The flux seems to be interrupted, which is not in line with Ampere’s lawand the continuity equation. As a corollary we can state, that the tangential H-Fieldis not continuous at the boundary between wire and capacitor.


Prof. S. Peik

Fixing Ampere’s Law

Perhaps, if we add a term involving to the right-hand side of eqn 11.7 we can some-how fix up Ampere’s law? This is, essentially, how Maxwell reasoned more than onehundred years ago.

The electric flux ψ into the capacitor is equal to the change of the charge on theplates. The charge Q on the plates change by

dQdt

= i (11.13)

The charge Q on the plates are also the sources of the D-Field. The sum of all chargesare equal to the sum of all D-field lines, e.g. Q =

∫V ρdV =

∫S~D d~S or simply Q = Ψ.

Hence we could come up with a term like

d~ψdt

= iv (11.14)

which describes the continuation of the current i in the wire by a special current iv.

This might fix Ampere’s law. Including our extra current iv Maxwell’s equation reads

∮H dl = i + iv (11.15)

∮H dl = i +

dΨdt

(11.16)

Rewriting with∫

S~D d~S = Ψ we get

∮~H d~l = i +

ddt

∫

S~D d~S (11.17)

or in point form

∇× ~H = ~J +ddt~D (11.18)

We find that the divergence of the right-hand side is

∇ · (~J + ddt~D) = ∇~J + d

dt∇~D (11.19)

= −dρ

dt+

dρ

dt= 0 (11.20)

as a consequence of charge conservation. We see, that the problems above arefixed.

The extra current is called the displacement current (this name was invented byMaxwell).


Prof. S. Peik

In summary, we have shown that although the flux of the real current through aloop is not well defined, if we form the sum of the real current and the displacementcurrent then the flux of this new quantity through a loop is well defined.

The extended Ampere’s law is hence

∮H dl =

∫∫~Jd~A +

ddt

∫∫~Dd~A (11.21)

or

∇ · ~H = ~J +ddt~D (11.22)

11.4 Maxwell’s Equations

Including the displacement current Maxwell wrote down his final set of equations as:

Differential Form Integral Form Name

∇× ~E = − ∂~B∂t

∮

C~E · d~l = − ∂

∂t

∫

S~B · d~S (11.23) Faraday’s Law

∇× ~H = ~J + ∂~D∂t

∮

C~H · d~l =

∫

S~J +

∂

∂t~D · d~S (11.24) Ampere’s Law

∇ · ~D = ρ

∮

S~DdS =

∫

VρdV (11.25) Gauss’s Law

∇ · ~B = 0∮

S~BdS = 0 (11.26) magn. Gauss’s Law

Together with the material equations

~D = ε · ~E (11.27)~B = µ · ~H (11.28)

these equations describe the complete electromagnetism.


Prof. S. Peik

11.5 Time Harmonic Fields

Using time harmonic phasors we get for the derivative ddt = jω. Therefore, Maxwell’s

equation become

∇× ~E = −jω~B (11.29)

∇× ~H = ~J + jω~D (11.30)

∇ · ~D = ρ (11.31)

∇ · ~B = 0

11.6 Electromagnetic Waves

The above equations suggest to solve the equation system of four equations for onevariable, let’s say ~E. In the following we demonstrate, that this can be done. Thelimitation is, that our region must not include currents or charges anywhere.

Using the operation ∇× on both sides of Maxwell’s first equation yields

∇×∇× ~E = −jω(∇× ~B) (11.32)

and substituting B = µ0H we get

∇×∇× ~E = −jωµ0(∇× ~H) (11.33)

Assuming a current free region, we now plug the above into Maxwell’s second equa-tion 11.29 and we get

∇×∇× ~E = −jωµ0(jω~D)) (11.34)

with D = ε0E and j2 = −1

∇×∇× ~E = ω2µ0ε0~E (11.35)

Now let us remember the following vector identity

∇×∇× ~E = ∇× (∇ · ~E)−∇2~E (11.36)

Since we do not have any charges ρin the region and revisiting eqn. 11.31 we have∇~E=0, therefore

∇×∇× ~E = −∇2~E (11.37)

Plugging this into eqn 11.35 yields

∇2~·E + ω2µ0ε0~E = 0 (11.38)

substituting k2 = ω2µ0ε0 we obtain


Prof. S. Peik

∇2~E + k2~E = 0 (11.39)

or

∇×∇× ~E− k2~E = 0 (11.40)

This is known as the Helmholtz equation. The constant k is called the wave number

of the medium. In vacuum we have k = ω√

µ0ε0.

These equations has the famous form

d2A(µ)

dµ2 − k2A(µ) = 0 (11.41)

which is known as the general wave equation, where

k2 =ω2

v2 (11.42)

with ω the angular frequency of the wave and v is the propagation velocity. For ourelectromagnetic wave we get

v =1√ε0µ0

= 2, 998 · 108 ms= Speed of light c (11.43)

About this velocity Maxwell wrote in 1864:

“This velocity is so nearly that of light, that it seems we have strong rea-son to conclude that light itself (including radiant heat, and other radia-tion if any) is an electromagnetic disturbance in the form of wave prop-agated through the electromagnetic field according to electromagneticlaws."

Maxwell assumed, that light is an electromagnetic wave with velocity c (as any EM-wave).

The same procedure can be applied to solve for the H-field and we get

∇2~H + k2~H = 0 (11.44)

In 1873 Maxwell published his work in a two volume book with the title "A Treatiseon Electricity and Magnetism". The book is a summary of all known effects of elec-tromagnetism including the displacement current. The Treatise is the foundation ofelectromagnetism and acts as the base for any work in electrical engineering.

The impact of these equations are regarded as the biggest achievement in naturalscience ever. Some quotes:


Prof. S. Peik

"From a long view of the history of mankind - seen from, say, ten thou-sand years from now - there can be little doubt that the most significantevent of the 19th century will be judged as Maxwell’s discovery of thelaws of electrodynamics", Richard Feynman

"Maxwell’s Equations have had a greater impact on human history thanany ten presidents.", Carl Sagan

It is notable, that the Treatise was written in a very obscure way. Virtually nobodyunderstood the book. The physicist Oliver Heaviside quit his job as telegrapher anddevoted his life to understand the book.He later wrote...

"I saw that it was great, greater, and greatest, with prodigious possibili-ties in its power. I was determined to master the book... It took me sev-eral years before I could understand as much as I possible could. Then Iset Maxwell aside and followed my own course. And I progressed muchmore quickly."

In 1873 he developed the operational calculus using operators such as ∇and first"popularised" Maxwell’s equations. The four equations as we know them today areactually set up by Oliver Heaviside.

Figure 11.3: Oliver Heaviside (1850 - 1925)

Figure 11.4: Maxwell’s Treatise


Prof. S. Peik

11.7 Plane-Wave Propagation in Medium

Let us now find a solution for the possible E- and H-field distributions for a plane E-field. Plane E-field means the E-field is non-varying with in the plane perpendicularto the propagation direction. For example, this plane is the x-y plane and the E-fieldvector lies in x direction. Now we revisit the Helmholtz equation

∇×∇× ~E− k2~E = 0 (11.45)

Assuming an E-field with an x-component only, we have for the first curl

∇×

ExEyEz

=

∂∂y Ez − ∂

∂z Ey∂∂z Ex − ∂

∂x Ez∂

∂x Ey − ∂∂y Ex

=

0∂∂z Ex

− ∂∂y Ex

(11.46)

for the second curl follows

∇×

0∂∂z Ex

− ∂∂y Ex

=

− ∂

∂y Ex − ∂∂z

∂∂z Ex

0− (− ∂∂y Ex)

∂∂x

∂∂z Ex − 0

(11.47)

When assuming a uniform field in x and y direction (the so called plane wave) wehave no variation in x and y direction, hence ∂/∂x = 0 and ∂/∂y = 0 following

∇×∇× ~E =

− ∂2

∂z2 Ex00

(11.48)

We have to investigate the x-component only for the differential equation

d2Ex

dz2 + k2Ex = 0 (11.49)

The general solution for this differential eqn. is

Ex = E+x0e−jkz + E−x0ejkz (11.50)

direction where E+x0 and E−x0 are constants to be determined. We see that the solution

is composed of two location-depending phasors. The E-Field has in time-domain asinusoidal variation in z-direction and is x-directed. Remember, that there is stillalso the the sinusoidal time variation sin(ωt).

The first solution refers to a forward traveling wave. The second solution to a back-ward traveling wave. Both wave superimpose linearly.

We can now determine the H-field for the forward wave ~E+(z) using Faraday’s laweqn 11.29.


Prof. S. Peik

∇× ~E = −jωµ~H (11.51)

~H =j

ωµ(∇× ~E) (11.52)

~H =j

ωµ(∇×

Ex(z)00

) (11.53)

~H =j

ωµ

0∂E+

x (z)∂z

∂E+x (z)∂y

(11.54)

~H = =j

ωµ

0−jkE+

o e−jkz

0

(11.55)

hence, there is a y-component only with

H+y =

kωµ

E+xoe−jkz = H+

y0e−jkz (11.56)

hence

H+y0 =

kωµ

E+xo (11.57)

The H-field is perpendicular to the E-field and the direction of propagation. H-fieldand E-field are in phase and travel with the same velocity in z-direction. The ar-rangement of the fields is illustrated in Figure 11.5.

The proportion of Exoto Hyois called the intrinsic impedance

η =ωµ

k=

ωµ

ω√

µε=

√µ

ε(11.58)

For vacuum the intrinsic impedance is η0 =√

µ0ε0≈ 377Ω.

We can now find the time-variance of the E-field vector

~E(z, t) = Re~E = x |E+xo| cos(ωt− kz + φ+) (11.59)

We observe, that the field changes over time t and with distance z. The Time periodfor one complete oscillation is known as the time period T = 1

f . Now we can alsodefine a propagation distance period called the wave length

λ =2π

k(11.60)

We can also define a velocity, that follows the propagation of the points with thesame phase. Looking at the time-variance solution closely we see that the phasevelocity of the wave is

vp =ω

k=

ω

ω√

µε=

1√µε

(11.61)


Prof. S. Peik

Figure 11.5: Fields of a Planar Wave

Note that with k = vpω substituted in eqn 11.60

λ =vp

f(11.62)

In vacuum we get the velocity

vp = 3 · 108 ms

= speed of light = c (11.63)

and for the relation between the wavelength λ and frequency f

c = λ · f (11.64)

11.8 EM-Wave Reflections at Normal Incidence

(Note: This section should be studied with basic knowledge of transmission line the-ory, next Chapter)

Consider the z-directed plane EM-wave of Section 11.7. This wave is now normal in-cident to a planar boundary in the xy-plane at z = 0. The left medium (medium 1) hasa dielectric permittivity of ε1, a magnetic permeability µ0 and a intrinsic impedance

of η1 =√

1ε1µ0

. The wave number of medium 1 is k1.

Considering an incident wave E+o e−jkz and a reflected wave E−o ejkz we get a total

E-field x-component and H-field y-component of

E1x = E+1 e−jk1z + E−1 ejk1z (11.65)

H1y =1η1

(E+1 e−jk1z − E−1 ejk1z) (11.66)


Prof. S. Peik

E+

1

E-

1

E+

2

medium 1 medium 2

Figure 11.6: Incident and Reflected EM-Wave at Boundary of Two Media

Similarly, we can assume a wave in the right medium with ε2, impedance η1 =√

1ε1µ0

and the wave number k2. Since there is no reflected wave (no further boundaries)we get

E2x = E+2 e−jk2z (11.67)

H2y =1η2

E+2 e−jk2z (11.68)

At the boundary we have to apply our boundary conditions from Figure 5.12. Thetangential components must be continuous, hence

E1x(z = 0) = E2x(z = 0) (11.69)

H2y(z = 0) = H2y(z = 0) (11.70)

↓ (11.71)

E+1 + E−1 = E+

2 (11.72)

1η1

(E+1 − E−1 ) =

1η2

E+2 (11.73)

Simultaneous Solutions for E−1 and E+2 for a given incident wave E+

1 yield:

E−1 =η2 − η1

η2 − η1E+

1 = ΓE+1 (reflected wave) (11.74)

E+2 =

2η2

η2 − η1E+

1 = τE+1 (transmitted wave) (11.75)

where


Prof. S. Peik

Γ =η2 − η1

η2 + η1(11.76)

τ =2η2

η2 − η1(11.77)

The quantities Γ and τ are known as reflection coefficient and transmission coeffi-cient. In case of lossless medium η1 and η2 are real and Γ and τ are also real. Wecan easily see that

Γ = τ − 1 (11.78)

for nonmagnetic media (µ = µ0) we get

Γ =

√εr2 −

√εr1√

εr2 +√

εr1(11.79)

Example 41:The dielectric constant of Pyrex glass is 4.6. At a transition between air and glass we getfor Γ = 0.36402. As the power contained in the wave is equal to the square of the E-Field,a fraction of the power of 0.364022 = 0.1325 = 13% is reflected. Hence only 87% of lightintensity (=power) is transmitted into the glass.Note, that the dielectric constant of materials varies with frequency. Hence our assumptionof 13% might not be completely correct.


Waves on Transmis-sion Lines

12At low frequencies, lines act as a short connection between two devices. At higherfrequencies, this is not true anymore. Rather, the signal changes with the length ofthe line due to the finite velocity of the signal propagation. Details in [7, 8]

For a closer examination we have to analyze a two wire line in more detail. Weassume a line with an arbitrary cross-section as shown in Figure 12.1.

Figure 12.1: Line: Cross-section, symbol and Equivalent Circuit

The capacitance per line length is called the capacitance line parameter C′ withthe unit F

m . Similarly we can define an R′,L′ and G′. The line parameters can bedetermined through the geometry of the line cross section.

This line can be represented as a chain of infinite small line sections. Each sectionacts like a small two port with shunt capacitance C′ dz and conductance G′ dz andseries inductance L′ dz and resistance a R′ dz s shown in Figure 12.2.

147

Prof. S. Peik

L’dz

C’dz

R’dz

G’dzu u+du

-du

i i+di

dz

Figure 12.2: Equivalent circuit of a Line Element

Now we compare the input voltage u1 with the output voltage u2. Using the schematicof Figure 12.2 we get for the main loop u2 = u1 + du. Or using the impedances

−u + R′dz i + L′ dzdidt

+ u + du = 0 (12.1)

−du = R′ dz i + L′ dzdidt

(12.2)

As i2 = i1 + di at the node we get

−i + G′ dz u + C′ dzdudt

+ i + di = 0 (12.3)

Hence, we get for the voltage drop

−du = R′ dz i + L′ dzdidt

(12.4)

and the shunt branch current

−di = G′ dz u + C′ dzdudt

(12.5)

Dividing both equation by dz we get

−dudz

= R′ i + L′didt

(12.6)

− didz

= G′ u + C′dudt

(12.7)

These equation describe the line completely for time variations t as well as localvariations z.

For the lossless case , i.e. R = G = 0, we get

−dudz

= L′didt

(12.8)

− didz

= C′dudt

(12.9)


Prof. S. Peik

By differentiating both eqns with respect to dz and substitution we get

d2udz2 = L′C′

d2udt2 (12.10)

d2idz2 = L′C′

d2idt2 (12.11)

For the lossy case we get

d2udz2 = R′G′ u + (R′C′ + L′G′)

dudt

+ C′L′d2udt2 (12.12)

d2idz2 = R′G′ i + (R′C′ + L′G′)

didt

+ C′L′d2idt2 (12.13)

These equations are known as the telegraph equations. The equations for voltageand current are the well known wave equations. We can conclude, that the voltageand current on the line travels as waves on it. The velocity in the lossless case is

vp =1√εµ

=1√

ε0εrµ0µr=

c√εrµr

with c = 2, 998 · 108 ms

(12.14)

we note, that the velocity in free space, i.e. εr = µr = 1, the wave propagateswith the velocity of light c. This is a very strong indication, that light itself is anelectromagnetic wave. As seen from Figure 12.3 any shaped signal propagates onthe line with the velocity vp. We can prove that that kind of functions f (z, t) solvethe wave equation.

Uo ejωt

Zi

Zo

Propagation with vel. v

z

Propagation with vel. v

z

t=t1

t=t1=z1/v

z1

Figure 12.3: Signal propagates on lossless line

12.1 Time-Harmonic Signals on Lines

When assuming time-harmonic signals on the line we can use the phasor represen-tation


Prof. S. Peik

i(t) = ReI ejωt und u(t) = ReV ejωt (12.15)

In our differential equations we get

dVdz

= −(R′ + jωL′)I (12.16)

dIdz

= −(G′ + jωC′)V (12.17)

The solution of the coupled differential equations system is

d2Vdz2 = (R′ + jωL′)(G′ + jωC′)V (12.18)

or using the abbreviation

γ2 = (R′ + jωL′)(G′ + jωC′) (12.19)

we get

d2Udz2 = γ2 V (12.20)

We call γ the propagation constant.

This is the time-harmonic wave equation, that describes the voltage distribution onthe line completely. We can find a similar equation for the current

d2 Idz2 = γ2 I (12.21)

12.2 Solution of the Wave Equation

The general solution of the wave equation is

V(z) = V+0 e−γz + V−0 eγz (12.22)

I(z) = I+0 e−γz + I−0 eγz (12.23)

with U+0 and U−0 or. I+0 and I−0 being arbitrary constants fixed by the boundary con-

ditions. The constant can be either determined by fixing current and voltage at thebeginning of the line (constant current and voltage source) or at the end of the line(fixed impedance) or a combination.


Prof. S. Peik

Uo ejωt

Zi

Zo

λ

Backward Wave

Forward Wave

Superposition= ?

Figure 12.4: Phasor diagram for forward and backward traveling wave

12.3 Propagation Constant

The constant γ is known as the propagation constant. As we already know

γ =√(R′ + jωL′)(G′ + jωC′) = α + jβ (12.24)

The propagation constant has a real and imaginary part γ. The real part α is calledattenuation constant. The imaginary part β is called phase constant.

The attenuation constant α describes the power loss in the line per length (typicallymeter, foot or kilometer). A lossless line has attenuation constant of zero. An α =5 1

km means, that the voltage on the line is attenuated by e5per kilometer.

The phase constant β describes the phase rotation per length. A β = 45m means,

that the voltage phasor rotates by 45° per meter length. The phase shift is alwaysa delay in the direction of propagation, as the wave needs time to travel along theline. The propagation constant proportional to the phasor frequency.

12.4 Wave Impedance

When we use the voltage description of the wave equation from eqn. 12.75. Bypartial differentiation of eqn. 12.75 y z we get:

dUdz

= γ(−U+0 e−γz + U−0 eγz) (12.25)

And with eqn.12.16 we get

I = − 1R′ + jωL′

dUdz

=γ

R′ + jωL′(U+

0 e−γz −U−0 eγz) (12.26)

We we substitute R′+jωL′γ = Z0 and get


Prof. S. Peik

I =1

Z0

(U+

0 e−γz −U−0 eγz) (12.27)

I =U+

0Z0︸︷︷︸I+0

e−γz−U−0Z0︸︷︷︸

I−0

eγz (12.28)

We can now relate I+0 and I−0 to the voltages U+0 and U−0 and get

I+0 =U+

0Z0⇔ Z0 =

U+0

I+0(12.29)

I−0 = − U−0Z0⇔ Z0 = −U−0

I−0(12.30)

The Voltage V+and and current I+on the line are in a fixed relation. This ratio iscalled the line impedance or wave impedance. We see that

Z0 =R′ + jωL′

γ=

√(R′ + jωL′)

√(R′ + jωL′)√

(R′ + jωL′)(G′ + jωC′)(12.31)

Z0 =

√R′ + jωL′

G′ + jωC′(12.32)

In the lossless case

Z0 =

√L′

C′(12.33)

the line impedance is real. Note, that the real value does not mean, that the line islossy! The line impedance has the unit Ω (Ohm). This impedance characterizes theline behavior and is the most important line parameter. Is is also important to know,that we cannot measure the line impedance with ohm meters. The line impedancejust relates voltage and current of the forward traveling wave and backward travel-ing wave, respectively,

V+0

I+0= Z0 (12.34)

V−0−I−0

= Z0 (12.35)

12.5 Wave Length and Propagation Constant

As seen from eqn. ?? the voltage on a line depends on the time as well as on thelocation on the line. The periodicity over time is known as the Period T = 1

f . The

“speed” of phase change of time is known as the angular velocity ω = 2π f = 2πT . A


Prof. S. Peik

similar relation of the phase change over the length of the line can now be specified.This constant is known as the phase constant β. From eqn. ?? we get the locationzpl as the location with same phase as location z = 0 by

βzpl = 2π (12.36)

zpl =2π

β(12.37)

The spatial period on the line is known as the wave length λ. As seen from eqn. ??the wave length λ and the phase constant β are related by

λ =2π

β(12.38)

12.6 Phase Velocity

The nodes of constant phase propagate with the velocity v over the line.

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 π 2π 3π 4π

Ausbreitung

Figure 12.5: Propagation of a Forward Wave over Time

The phase velocity on the line is

v =ω

β(12.39)

also called the propagation velocity on the line.


Prof. S. Peik

Relating 12.39 with β = 2πλ we get

v =ω

2πλ (12.40)

or using ω2π = f

v = f · λ (12.41)

This equation is the most basic equation of wave propagation.


Prof. S. Peik

Example 42:Consider the following observed voltage distribution on a line at t = 0 and t = 1ms.

-1

-0.5

0

0.5

1

0 2 4 6 8 10 12 14 16 18 20

-1

-0.5

0

0.5

1

0 2 4 6 8 10 12 14 16 18 20

z [m] z [m]

λ

t= 1 mst=0ms

1/e

90°»1m90°»1m

The attenuation can be derived from the envelope e-function. The envelope decays in 10 mto 1/e of the original value. Hence

e−αz =1e

∣∣∣∣z=10m

(12.42)

1e10m α

=1e

(12.43)

10 m α = 1 (12.44)

α = 0.11m

(12.45)

The attenuation is 0.1 per meter or α = 0.1 Npm

The Wave length λ is 4m as seen from measurement. The phase changes by 90° within 1m. Hence, the phase constant is β = π/2

1m = π2

radm . The phase velocity is v = 1m

1ms = 1000 ms .

We can now calculate the frequency

f =vλ=

1000 ms

4m= 250Hz (12.46)

Alternatively we can calculate β and v from λ. Now

β =2π

λ=

π

2radm

(12.47)

and

v =ω

β=

2π250Hzπ2

1m

= 1000ms

(12.48)

Since the voltage at z = 0 is 1 V and there is no backward traveling wave , the voltagefunction is

V(z) = 1Ve−0.1z−j π2 z (12.49)

12.6.1 Lossless Line

When looking at lossless lines with R′ = G′ = 0 our line equations simplify to

Z0 =

√L′

C′(12.50)

γ = jβ =√

jωL′ jωC′ = jω√

L′C′ (12.51)

v =ω

β=

1√L′C′

(12.52)

λ =2π

β=

2π

ω√

L′C′=

1f√

L′C′=

vf

(12.53)


Prof. S. Peik

12.7 Determining the Constants by Boundary Conditions

When setting boundary conditions at the beginning of the line, i.e. fixing currentI(0) and voltage U(0) at the beginning of the line we have

V(0) = Va = U+0 + V−0 (12.54)

I(0) = Ia =1

Z0(U+

0 −V−0 ) (12.55)

and

V+0 =

Va + Z0 Ia

2uand V−0 =

Va − Z0 Ia

2(12.56)

Hence we have for the voltage and current distribution along the line

V(z) =12(Va + Z0 Ia) e−γz +

12(Va − Z0 Ia) eγz (12.57)

I(z) =12

(Va

Z0+ Ia

)e−γz − 1

2

(Va

Z0− Ia

)eγz (12.58)

this is known as the physical Form of the line voltage/current distribution, with thecoefficients

V+0 =

12(Va + Z0 Ia) (12.59)

V−0 =12(Va − Z0 Ia) (12.60)

I+0 =12

(Va

Z0+ Ia

)=

V+0

Z0(12.61)

I−0 = −12

(Va

Z0− Ia

)= −V−0

Z0(12.62)

Hence we can write

V(z) = V+0 e−γz + V−0 eγz (12.63)

I(z) = I+0 e−γz − I−0 eγz (12.64)

In mathematical form we get

V(z) = Vaeγz + e−γz

2− Z0 Ia

eγz − e−γz

2(12.65)


Prof. S. Peik

or using the definitions of hyperbolic functions

U(z) = Ua cosh(γz)− Z0 Ia sinh(γz) (12.66)

I(z) = Ia cosh(γz)− Ua

Z0sinh(γz) (12.67)

12.8 End Conditions

When fixing the current Ie = I(l) and voltage Ue = U(l) at the end of the line, i.e.z = l we get

V(l) = Ve = U+0 e−γl + V−0 eγl (12.68)

I(l) = Ie =1

Z0(U+

0 e−γl −V−0 eγl) (12.69)

where now

V+0 =

12(Ve + Z0 Ie)eγl and V−0 =

12(Ve − Z0 Ie)e−γl (12.70)

and we get

V(z) =12(Ve + Z0 Ie) eγ(l−z) +

12(Ve − Z0 Ie) e−γ(l−z) (12.71)

I(z) =12

(Ve

Z0+ Ie

)eγ(l−z) − 1

2

(Ve

Z0− Ie

)e−γ(l−z) (12.72)

In mathematical form we have

V(z) = Ve cosh(γ(l − z)) + Z0 Ie sinh(γ(l − z)) (12.73)

I(z) = Ie cosh(γ(l − z)) +Ve

Z0sinh(γ(l − z)) (12.74)

12.9 Terminated Lines

We can solve the differential equations with boundary values only. In our case, thismeans, that the lines must be terminated .

A terminated line is given in Figure 12.6 . The line is fed to the left with the harmonicvoltage phasor Voejωt. The source has the source impedance Zi. The line is loadedat the end with an Impedance ZL. This is called the termination. The line has thelength l.

The solution of this boundary problem can be derived with the well known tech-niques. The voltage distribution on the line can be derived (see [?]) to be the pha-sors


Prof. S. Peik

Zl

Zi

ZoUo ejωt

Figure 12.6: Line with Source and Load

V(z) = V+0 e−γz + V−0 eγz (12.75)

I(z) = I+0 e−γz + I−0 eγz (12.76)

with V+0 and V−0 and. I+0 And I−0 being constants to be determined.

The first part of the right side describes a forward traveling wave. The second parta backward traveling wave.

12.9.1 Matched Load

When the line is terminated by an impedance equal to the line impedance we callthis a matched load as shown in Figure ??, i.e.

Zl = Z0 (12.77)

At the end we have the relation of voltage and current of

Ve

Ie= Zl = Z0 or Ve = Z0 Ie (12.78)

When plugging this into eqn 12.71 we get

V(z) = Veeγ(l−z) (12.79)

I(z) =Ve

Z0eγ(l−z) (12.80)

As for z = 0 (Beginning of Line) we get Va = Veeγl we have

U(z) = Vae−γz (12.81)

I(z) =Va

Z0e−γz (12.82)

Voltage and current distribution on the line with matched load are identical to thedistribution on an infinite line. There is no backward wave.

Matched termination is often a desirable goal, as for the matched case, we do notget problems from reflected waves and also maximize the power delivered to theload.


Prof. S. Peik

Uo ejωt

Zi

Zo

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Ausbreitung

λ

Zl

Figure 12.7: Wave on a line with Matched Termination

Uo ejωt

Zi

Zo

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Ausbreitung

λ

Zl

rücklaufende Welle

hinlaufende Welle

Figure 12.8: Not Matched Load with Reflected Wave


Prof. S. Peik

12.9.2 Arbitrary Load

In the arbitrary load case we get reflected waves.

The load forces the voltage Ve and the current Ie at the end of the line (z = l) to be

Ve

Ie= Zl (12.83)

With eqn. 12.71 we get

U(z) =12(Zl + Z0)Ie︸︷︷︸

U+0

eγ(l−z) +12(Zl − Z0)Ie︸︷︷︸

U−0

e−γ(l−z) (12.84)

In order to simplify the calculations we move and flip our coordinate system to theend of the line, i.e. z′ = l − z. Note, the variable z′ counts positive to the left. Weget now

V(z′) =12(Zl + Z0)Ie︸︷︷︸

U+0

eγz′ +12(Zl − Z0)Ie︸︷︷︸

U−0

e−γz′ (12.85)

Where

V−0V+

0=

Zl − Z0

Zl + Z0(12.86)

This ratio of the forward and backward wave voltage at z = l is called reflectioncoefficient

Γ =Zl − Z0

Zl + Z0=

ZLZ0− 1

ZLZ0

+ 1(12.87)

We can also determine ZlZ0

from the reflection coefficient Γ

ZL

Z0=

1 + Γ1− Γ

(12.88)

To summarize, the waves on the line have now the form

V(z′) = V+0

(ejβz′ + Γ e−jβz′

)(12.89)

with U+0 being the incident wave at z = l.

When using Zl = Z0 we observe that Γ = 0 as mentioned before.


Prof. S. Peik

12.10 Reflections

Wave propagate through the line. When the hit a termination or a discontinuity theywill be absorbed or reflected.

The reflected amount in amplitude and phase is described by the reflection coeffi-cient

Γ =Phasor of incident wave at terminationPhasor of reflected wave at termination

=V−

V+(12.90)

The reflection coefficient can be complex. For passive terminations, the reflectioncoefficient is always |Γ| ≤ 1, i.e. Γ lies always inside the unit circle of the complexplane.

The reflection coefficient can be derived from the line impedance Z0and the load(termination) impedance ZL by

Γ =ZL − Z0

ZL + Z0(12.91)

Depending on the type of load we get the following types of reflection

• Open line end withZL = ∞:Incident and reflected wave have the same amplitude and phase , hence Γ = 1

• shorted line with ZL = 0 Ω:Incident and reflected wave have the same amplitude and a 180° phase differ-ence , hence Γ = −1

• Matched load:The load impedance is equal to the line impedance ZL = Z0. There are noreflections, i.e. Γ = 0.

• General loadThe load is ZL 6= Z0. There is a reflected wave withΓ = ZL−Z0

ZL+Z0.

When the imaginary part of Γ is positive, the load is inductive. When the imaginarypart of Γ is negative, the load is capacitive.

12.11 Input Impedance of a Line

We assume a line as shown in Figure 12.9. As shown below we can determine theinput impedance of this arrangement.

Let us revisit eqn. 12.71 and 12.72

U(z) = Ue cosh(γ(l − z)) + Z0 Ie sinh(γ(l − z)) (12.92)

I(z) = Ie cosh(γ(l − z)) +Ue

Z0sinh(γ(l − z)) (12.93)

with the ratio UeIe

= Zl for the load impedance.


Prof. S. Peik

Zo Zl

Zin

l

Figure 12.9: Input impedance of Line with Load

Now, we determine the ratio of the voltage and the current at the beginning of theline by

Zin =U(0)I(0)

=Zl cosh(γl) + Z0 sinh(γl)

cosh(γl) + ZlZ0

sinh(γl)= Z0

Zl cosh(γl) + Z0 sinh(γl)Z0 cosh(γl) + Zl sinh(γl)

(12.94)

or using the hyperbolic function tanh = sinhcosh we get

Zin = Z0ZL + Z0 tanh γlZ0 + ZL tanh γl

(12.95)

In the loss less case with γ = jβ we get:

Zin = Z0ZL + jZ0 tan βlZ0 + jZL tan βl

(12.96)

12.11.1 Impedance Transformation

When we choose l = λ4 we get an interesting relation. Usingβl = π

2 we get now

Zin =Z2

0Zl

(12.97)

in other words: The impedance Zl at the end is transformed into a new impedanceZ2

0Zl

at the input. This only works for a length of exactly λ/4. Hence, the transformationis strongly frequency dependent. We can widen the bandwidth of the transformationby cascading several of this λ/4-Transformers.

For a line length of l = λ2 we get another interesting relation. Now β = π and

Zin = Zl (12.98)

This means, that inserting a line of half a wave length does now change the impedanceof the load. No matter if the load is matched or not. This is also valid for one fre-quency only.


Prof. S. Peik

12.11.2 Short and Open Line Impedance

The open or short ended line is a special case of Eqn. 12.96. The input impedance isalways purely imaginary. The open-ended or shorted line is, therefore, a reactance.

For shorted line:

Zin = Z0 tanh γl = jZ0 tan βl (12.99)

For open-ended line:

Zin = Z0 coth γl = −jZ0 cot βl (12.100)

-8

-6

-4

-2

0

2

4

6

8

π/4 π/2 3π/2 π 2π 3π

Zin/Z0

λ/8 λ/4 3λ/4 λ/2 λ 3π

-8

-6

-4

-2

0

2

4

6

8

π/4 π/2 3π/2 π 2π 3π

Zin/Z0

Kurzschluß am Leitungsende

Leerlauf am Leitungsende

Figure 12.10: Input Impedance of a Shorted and Open-Ended Line

With open or shorted lines we can realize reactance with arbitrary values.

With a closer look we can see that we can model these lines using L-C equivalentcircuits.


Prof. S. Peik

Zo

<λ/4

Zin

L

Kurze Leitung mit l<λ/4:

Zo

<λ/4

Zin

Zo

λ/4

Zin

Kurze Leitung mit l<λ/4;3λ/4;5λ/4...:

Zo

λ/4

Zin

C

LC

C

L

In microwave engineering L’s and C’s are often realized using line segments.


Prof. S. Peik

Example 43:A 50Ω-receiving-antenna is connected to a receiver with 50Ω input impedance. The receiv-ing antenna can be modeled as a voltage source with a 50Ω source resistance. The sourcevoltage is 10 mV. The operating frequency is 440MHz. With a direct connection betweenantenna and receiver the power delivered to the receiver is equal to the available power

PL = Pavail =(U0/2)2

50Ω= 0, 5µW (12.101)

assuming RMS voltages.Now, we insert a 4m long lossless 75 Ω-cable between antenna and receiver.

Uo

Zi

ZoUa UeZL

Ia Ie

with a phase delay of

λ = cf = 0.68mand β = 9, 23 rad

m . βl = 2πλ l = 2π

0.68 4 = 36, 92 rad.

at the cable-receiver transition we have a reflection of the wave of:

Γ =ZL − Z0ZL + Z0

= − 25125

= −0.2 (12.102)

A similar reflection occurs at the transition antenna-cable, where Γ = 0.2In other words, 20% of the voltage is reflected at the first as well as second transition.Since the power is proportional to U2, we have 4% of power reflection and 96% of powertransmission at each discontinuity.The input impedance is

Zin = Z0ZL + jZ0 tan βlZ0 + jZL tan βl

= (68, 9 + j28, 7)Ω (12.103)

Now we can determine the voltage at the cable input:Ua = Zin

Zin+50Ω 10mV = (6, 03 + j0.96)mV

Ia = UaZIN

= (0.079− j0.02)mAThe transmission line equation is hence

U(z) =12(Ua + Z0 Ia) e−jβz +

12(Ua − Z0 Ia) ejβz (12.104)

I(z) =12

(Ua

Z0+ Ia

)e−jβz − 1

2

(Ua

Z0− Ia

)ejβz (12.105)

At the end of the cable we have:U(l) = (−2, 57 + j2, 65)mV

I(l) = U(l)ZL

= (−0, 05 + j0, 053)mAThe power at the load is

PL = Ue · I∗e = 0, 46uW (12.106)

Alternatively, we can estimate the power at the load by the assumption of 4% power reflec-tion at each discontinuity, i.e. 0.96*0.96=0.92=92% transmitted power. Here, we neglectmultiple reflections. Hence, PL = Pver f · 0, 92 = 0, 46uW.The forward and backward traveling wave components are:

U+0 =

12(Ua + Z0 Ia) = (5, 99 + j0, 24)mV = 6ej0,04 (12.107)

U−0 =12(Ua − Z0 Ia) = (0, 0328− j1, 2)mV = 1, 2e−j1,54 (12.108)

Ua = U+0 + U−0 = (6, 02− j0, 96)mV (see above) (12.109)


Vector Algebra AA.1 Vectors and Fields

A scalar has a magnitude only. In contrast, vectors have a magnitude and direction.In this notes, scalars are written in italics, e.g. u, whereas vectors are written withan arrow, e.g. ~A. For electrodynamic problems we use almost exclusively three-dimensional vectors, that span into the ordinary space. The components of a vectorare written as

~A =

AxAyAz

(A.1)

The magnitude of a vector is written as |~A|.In Cartesian coordinates the magnitude of a vector is given as

|~A| =√

A2x + A2

y + A2z (A.2)

A unit vector a is defined as a vector whose magnitude is unity and its direction isalong ~A, this is

a =~A|~A|

(A.3)

The unit vectors in x, y and z direction in Cartesian coordinates are defined as x, y, z.

Vectors add and subtract, by adding (or subtracting, respectively) the vector com-ponents.

~A + ~B =

Ax + BxAy + ByAz + Bz

(A.4)

166

Prof. S. Peik

A.1.1 Fields

A field is a function that specifies a quantity (scalar or vector) anywhere in a region.Hence, a field describes a certain property of space. Fields can be vector fields orscalar fields, depending on the quantity described in the region. An example for ascalar field is a temperature field; whereas ocean currents are described by a vectorfield.

A.1.2 Vector Multiplication

Vectors can be multiplied in two different ways: by scalar or vector multiplication.

The scalar-product or dot-product of two vectors ~A and ~B, written ~A·~B is definedgeometrically as the product of the magnitudes of the two vectors and the cosine ofthe angle α between them.

~A·~B = ~|A| · |~B| · cos α (A.5)

In Cartesian coordinates, that is

~A·~B = AxBx + AyBy + AzBz (A.6)

The following rules apply

~A·~B = ~B·~A (A.7)~A·(~B + ~C) = ~A·~B + ~A·~C (A.8)

~A·~A = ~|A|2 (A.9)

The cross-product or vector-product of two vectors ~A and ~B, written as ~A × ~B is avector quantity whose magnitude is the area of a parallelogram formed by the twovectors and is in the direction perpendicular to both vectors, following the right-handrule.

Thus~A× ~B = ~|A|~B| sin α ·~n (A.10)

where ~n is normal to both vectors ~A and ~B.

In Cartesian coordinates we get

~A× ~B =

∣∣∣∣∣∣

x y zAx Ay AzBx By Bz

∣∣∣∣∣∣=

AyBz − AzByAzBx − AxBzAxBy − AyBx

(A.11)

Note that the cross product is not commutative, i.e.

~A× ~B 6= ~B×~A (A.12)

but is anti-commutative~A× ~B = −( ~B×~A) (A.13)


Prof. S. Peik

A.2 Coordinate Systems

A.2.1 Cartesian Coordinates

X

Y

Z

O

xy

z

(x,y,z)

• Cartesian (rectangular) coordinates are two-dimensional or three-dimensionalcoordinates, conventionally denoted the x- , y- and z-axes;

• Coordinates are mutually perpendicular and lie anywhere in the interval (−∞, ∞);

• - The line element is given by:

ds = dxx + dyy + dzz (A.14)

• - the volume element by:dV = dx · dy · dz (A.15)

• - The nabla operator is given by:

∆ = xd

dx+ y

ddy

+ zddz

(A.16)


Prof. S. Peik

A.2.2 Cylindrical Coordinates

Cylindrical coordinates are suitable for problems with cylindrical symmetry. The co-ordinates are defined by concentric cylinders

φ

dφρ dφ

dz

z

ρ

p=ρ ρ + φ φ + z z

Cylindrical coordinate system is an "extended" two-dimensional polar coordinatesystem into the three-dimensional by superimposing a height z axis;

The coordinates are (ρ, φ, z) or (r, θ, z) are given by:

ρ =√

x2 + y2 (A.17)

φ = tan−1(y

x

)(A.18)

z = z (A.19)

where ρ ∈ [0, ∞),φ ∈ [0, 2π) ,and z ∈ (−∞, ∞).

The Cartesian coordinates are then given by:

x = ρ cos φ (A.20)

y = ρ sin φ (A.21)

z = z (A.22)

The line element is:ds = dρρ + ρdφφ + dzz (A.23)

The volume element is:dV = ρ · dρ · dφ · dz (A.24)

The nabla operator is:

∆ = ρd

dρ+ φ

1ρ

ddφ

+ zddz

(A.25)


Prof. S. Peik

A.2.3 Spherical Coordinates

Spherical coordinates are suitable for problems with spherical symmetry. The coor-dinates are defined by concentric spheres

φ dφ

dθ

r dθ r dφ sinθ

rθ

p=r r +θ θ+φ φ

The coordinates (r, φ, θ) are given by:

ρ =√

x2 + y2 + z2 (A.26)

φ = tan−1(y

x

)(A.27)

θ = sin−1

(√x2 + y2

r

)= cos−1

( zr

)(A.28)

where ρ ∈ [0, ∞),φ ∈ [0, 2π),andθ ∈ [0, π].

The Cartesian coordinates are then given by:

x = r cos φ sin θ (A.29)

y = r sin φ sin θ (A.30)

z = r cos θ (A.31)

The line element is:ds = drr + rdθθ + r sin θdφφ (A.32)

the area element is:da = r2 sin θdφdθr (A.33)

The volume element is:dV = r2 sin θdθdφdr (A.34)

The nabla operator is:

∆ = rddr

+ θ1r

ddθ

+1

r sin θφ

ddφ

(A.35)


Prof. S. Peik

A.2.4 Relations of Coordinates

Coordinates can be converted with the following formulas (from [?])

Cylindrical ↔Cartesian

x = ρ cos φ (A.36)

y = ρ sin φ

z = z

x = cos φ ρ− sin φ φ

y = sin φ ρ + cos φ φ

z = z (A.37)

ρ =√

x2 + y2 (A.38)

φ = tan−1 yx

z = z

ρ = cos φ x + sin φ yφ = − sin φ x + cos φ yz = z (A.39)

Spherical ↔Cartesian

x = r sin θ cos φ (A.40)

y = r sin θ sin φ

z = r cos θ

x = sin θ cos φ r + cos θ cos φ θ − sin φ φ

y = sin θ sin φ r + cos θ sin φ θ − cos φ φ

z = cos θ r− sin θ θ (A.41)

r =√

x2 + y2 + z2 (A.42)

θ = tan−1

√x2 + y2

zφ = tan−1 y

x

r = sin θ cos φ x + sin θ sin φ y + cos θ zθ = cos θ cos φ x + cos θ sin φ y− sin θ zφ = − sin φ x + cos φ y (A.43)

A.2.5 The Poisonous Snake

There is a poisonous snake lurking in the spherical and cylindrical coordinate system,as Griffiths calls it in [?]. The vectors r, ρ, θ, φ change their direction depending onthe point p in space. For example, ρ always points outward in a cylindrical coordinatesystem, but this could be x or y-direction in Cartesian coordinates. For example, inFigure A.1 the vectors in cylindrical coordinates are

~a = 1ρ− π

2φ and ~b = 1ρ +

π

2φ (A.44)

Naively adding the two vectors yield

~a +~b = 2ρ (A.45)

The poisonous snake snapped. Of course, the real answer is~a +~b = 0 as seen fromFigure A.1. Since the φ points in different directions for the two vectors, the addition


Prof. S. Peik

x

y

z

~b

~a -1

1

Figure A.1: Example for the Poisonous Snake

does not work. To be correct, we must always define the unit vectors including thepoint of reference. This is usually not done because is it not practical.

When we are aware of the problem, it should not cause any difficulties. Keep inmind:

• Do not add, subtract or multiply vectors using spherical or cylindrical unit vec-tors

• Do not integrate or differentiate with r, ρ, θ, φvectors.

If in doubt, re-express the problem using Cartesian coordinates.

A.3 Vector Calculus

A main concept of modern mathematics is the calculation of problems using infinitesmall quantities. This area of math is called calculus.

We can define an infinite small section in the Cartesian space in all three directionby defining dx, dy and dz.These length are defined as infinite short but still existent.Differential displacement is given by

d~l = dx x + dy y + dz z (A.46)

A differential normal area is given by

~dS =dy dz x (A.47)

for an area in the y-z-plane. The same applies to the other two areas.

A differential volume element is defined as

dV = dx dy dz (A.48)

For the definitions in other coordinate systems check [?, ?, ?].

A.3.1 Line, Surface and Volume Integrals

The line integral∫

C~A d~l is the integral of the tangential component of ~A along the

curve C. Given a vector field ~A as shown in Figure A.2, we can also write


Prof. S. Peik

C

A

dl

Figure A.2: Definition of the Line Integral

∫

C~A d~l =

∫ b

a|~A| cos α dl (A.49)

where α is the angle enclosed by ~Aand~l.

The surface integral describes the flux Ψ of ~A through a surface S, as shown inFigure A.3.

Ψ =∫

S~A dS (A.50)

The surface integral is also defined for closed surfaces. For a closed surface (enclos-ing a volume) the flux becomes

Ψ =∮∫

S~A dS (A.51)

which describes the net outward flux of ~A from S. A closed surface describes alwaysa volume.

SA

dS

Figure A.3: Definition of the Line Integral

The volume integral over a scalar ρ over the volume V is defined by

P =∫

Vρ dv (A.52)

A.3.2 Del Operator

The operator ∇ (say:nabla) is a simple definition, that helps us to shorten vectorcalculus expressions. This is sometimes referred to as the Nabla-operator.


Prof. S. Peik

∇ =∂

∂xx +

∂

∂yy +

∂

∂zz (A.53)


Prof. S. Peik

A.3.3 Gradient

The gradient of a scalar field V is a vector that represents both magnitude anddirection of the maximum rate of increase of V.


grad V =

∂V∂x∂V∂y∂V∂z

(A.54)

The gradient operator can also be expressed using the del-operator and the dotproduct, that is

grad V = ∇ ·V =

∂V∂x∂V∂y∂V∂z

(A.55)

for cylindrical coordinates we get

∆ · f = ρd fdρ

+ φ1ρ

d fdφ

+ zd fdz

(A.56)

and for spherical coordinates

∆ · f = ρd fdρ

+ θ1ρ

d fdθ

+1

ρ sin θφ

d fdφ

(A.57)


Prof. S. Peik

A.3.4 Divergence

The divergence of ~A at a given point P is the outward flow per unit volume as thevolume shrinks to that point.

Hence

div ~A = ∇ · ~A = lim∆v→0

∮~A · dS∆v

(A.58)

An inward flow means negative divergence. An outward flow means positive diver-gence, as illustrated in Figure A.4.

P P P

div A<0 div A>0 div A=0

Figure A.4: Definition of the sign of the divergence operator

The divergence of a vector field ~A(x, y, z) is given by

∇ · ~A =∂ax

∂x+

∂ay

∂y+

∂az

∂z(A.59)

We note, that the result is a scalar field. The divergence can only be taken from avector. The divergence from a scalar does not make sense.

In cylindrical coordinates the divergence becomes

∇ · ~A(ρ, φ, z) =1ρ

∂

∂ρ(ρAρ) +

1ρ

∂Aϕ

∂φ+

∂Az

∂z(A.60)

In spherical coordinates

∇ · ~A(r, θ, φ) =1r2

∂

∂r(r2Ar) +

1r sin θ

∂

∂θ(sin θAθ) +

1r sin θ

∂Aφ

∂φ(A.61)

We can find a relation between the surface integral of an arbitrary volume V and itsvolume integral, by

∮

S~AdS =

∫∫∫

V∇~AdV (A.62)

This is called the divergence theorem, otherwise known as Gauss’s law.

The divergence theorem states that the total outward flow of a vectorfield ~A through a closed surface S is the same as the volume integral of

the divergence ~A.


Prof. S. Peik

Closed Surface S

dS

Volume V

dV

Figure A.5: Divergence Theorem

A.3.5 Curl and Stokes Theorem

The curl is an axial (or rotational) vector operator whose magnitude is the maxi-mum circulation of ~A per unit area as the area tends to zero. It’s direction is thenormal direction of the area when the area is oriented so as to make the circulationmaximum.

That is the curl is

curl ~A = ∇× ~A = max

(lim

∆S→0

∮~A · d~l∆S

)n (A.63)

where ∆S is the area bounded by the curve L and n is the unit vector normal to thesurface S.

The curl can be expressed by a cross product

curl ~A = ∇× ~A =

∣∣∣∣∣∣

x y z∂

∂x∂

∂y∂∂z

Ax Ay Az

∣∣∣∣∣∣=

∂∂y Az − ∂

∂z Ay∂∂z Ax − ∂

∂x Az∂

∂x Ay − ∂∂y Ax

(A.64)

in cylindrical coordinates we obtain

∇× ~A(r, θ, z) =(

1r

∂Az

∂φ− ∂Aφ

∂z

)r +

(∂Ar

∂z− ∂Az

∂ρ

)φ +

1r

(∂(rAφ)

∂r− ∂Ar

∂φ

)z

(A.65)

in spherical coordinates we get

∇× ~A(r, θ, φ) =1

r sin θ

(∂(Aφ sin θ)

∂θ− ∂Aθ

∂φ

)r+

1r

(1

sin θ

∂Ar

∂φ− ∂(r Aθ)

∂r

)θ +

1r

(∂(rAθ)

∂r− ∂Ar

∂θ

)φ

(A.66)

There are some important properties of the curl


Prof. S. Peik

The curl of a vector field is another vector field (A.67)

The curl of a scalar field makes no sense (A.68)

The divergence of the curl of a vector field vanishes, that is ∇ · (∇× ~A) = 0(A.69)

The curl of the gradient of a scalar field vanishes, that is ∇×∇ ·V = 0(A.70)

Other properties of the curl can be found in [?, ?, ?, ?].

The curl provides the maximum circulation of the field at a given point. The di-rection of the curl is normal to the rotation. Homogeneous fields have a curl of 0everywhere. The curl is only non-zero in fields that rotate somehow.

From the definition of the curl we can derive the following

∮

L~A · d~l =

∫∫

S(∇× ~A)d~S (A.71)

This is called Stokes Theorem.

Stokes Theorem states, that the circulation of a vector field ~A around a closed pathL is equal to the surface integral of the curl of ~A over the open surface S boundedby L provided that ~A and ∇× ~A are continuous on S.

L

dS

dl

Surface S

Clo

sed P

ath L

Figure A.6: Stokes Theorem


Prof. S. Peik

A.3.6 Laplacian Operator

Besides the Nabla or Del-operator it is practical to define another operator the socalled Laplacian operator. The Laplacian operator is written as ∇2 and is the diver-gence of the gradient, i.e ∇ · ∇. The operator is always applied to a scalar field.Sometimes the Laplacian is abbreviated with ∆.

∇2 = ∆ = ∇ · ∇ = div grad (A.72)


∇2 f =∂2 f∂x2 +

∂2 f∂y2 +

∂2 f∂z2 (A.73)

in cylindrical coordinates we get

1ρ

∂

∂ρ

(ρ

∂ f∂ρ

)+

1ρ2

∂2 f∂φ2 +

∂2 f∂z2 (A.74)

in spherical coordinates

∇2 f =1r2

∂

∂r

(r2 ∂ f

∂r

)+

1r2 sin θ

∂

∂θ

(sin θ

∂ f∂θ

)+

1r2 sin2 θ

∂2 f∂φ2 (A.75)

A.4 Converting Operators between Coordinate Systems

We can generalize all above formulas and operators for any orthogonal coordinatesystem.

Lets define an arbitrary three-dimensional orthogonal coordinate system with thecoordinates u, v and w. The Their unit-vectors u,v and w are pointing in the coordi-nate direction and are mutually perpendicular. Cartesian, spherical and cylindricalsystems are examples of these coordinate systems.

The displacement vector if a coordinate system is (when moving from point (u, v, w)to (u + du, v + dv, w + dw))

d~l = f du u + g dv v + h dw w (A.76)

where f , g, h are positional functions of the particular coordinate system.

Example 44:Find the function f , g, h for the Cartesian, cylindrical and spherical system.

Cartesian: Does not depend on position, hence f = g = h = 1

cylindrical with u = r, v = φ, w = z : The coordinate-length dr and dz does not vary withposition, hence f = h = 1. The displacement in φ-direction is r dφ hence g = r

spherical with u = r, v = θ, w = φ : Similar with f = 1, g = r and h = r · sin θ


Prof. S. Peik

φ dφ

dθ

r dθ r dφ sinθ

rθ

p=r r +θ θ+φ φ

Figure A.7: Segment in spherical coordinates

If we want to calculate the gradient in a special orthogonal coordinate system wehave to compute the change dt of the scalar function t(u, v, w), if we move the dis-

tance d~l. We can write for

dt =∂t∂u

du +∂t∂v

dv +∂t∂w

dw (A.77)

We can rewrite this as a dot-product

dt = ∇t · d~l = (∇t)u · f · du + (∇t)v · g · dv + (∇t)w · h · dw (A.78)

provided we define

(∇t)u ≡1f

∂t∂u

and (∇t)v ≡1g

∂t∂v

and(∇t)w ≡1h

∂t∂w

(A.79)

The gradient of t , then, is

∇t ≡ 1f

∂t∂u

u +1g

∂t∂v

v +1h

∂t∂w

w (A.80)

We can now pick the appropriate expressions of f , g, h for the desired coordinatesystem and yield the gradient expressed in this coordinate system. Using the infor-mation from the example we get the table

System u v w f g hCartesian x y z 1 1 1Spherical r θ φ 1 r r sin θ

Cylindrical r φ z 1 r 1

A similar technique can be applied for the calculation of the divergence and the rota-tion. For more information check [?, App A]. The divergence in arbitrary orthogonalcoordinate systems is


Prof. S. Peik

∇ · ~A ≡ 1f gh

[∂

∂u(ghAu) +

∂

∂v( f hAv) +

∂

∂w( f gAw)

](A.81)

The rotational operator is

∇× ~A ≡ 1gh

[∂

∂v(hAw)−

∂

∂w(gAv)

]u+

1f h

[∂

∂w( f Au)−

∂

∂u(hAw)

]v+

1f g

[∂

∂u(gAv)−

∂

∂v( f Au)

]w

(A.82)

and the Laplacian Operator is

∇2t ≡ 1f gh

[∂

∂u

(ghf

∂t∂u

)+

∂

∂v

(f hg

∂t∂v

)+

∂

∂w

(f gh

∂t∂w

)](A.83)


Useful Tables BB.1 Greek Alphabet

Aα alpha Nν nuBβ beta Ξξ xiΓγ gamma Oo omicron∆δ delta Ππ piEε, ε epsilon Pρ rhoZζ zeta Σσ sigmaHη eta Tτ tauΘθ, ϑ theta Υυ upsilonIι iota Φφ, ϕ phiKκ kappa Xχ chiΛλ lambda Ψψ psiMµ mu Ωω omega

182

Prof. S. Peik

B.2 Constants

c Lichtgeschwindigkeit 2.998× 108m s−1

e Elementarladung 1.602× 10−19 Cmn Neutronenruhemasse 1.675× 10−27 kgmp Protonenruhemasse 1.673× 10−27 kgme Elektronenruhemasse 9.110× 10−31 kgh Plancksches Wirkungsquantum 6.626× 10−34 J sh Dirac’s Konstante (= h/2π) 1.055× 10−34 J sk Boltzmannkonstante 1.381× 10−23 J K−1

G Gravitationskonstante 6.673× 10−11 N m2 kg−2

σ Stefan-Boltzmann Konstante 5.670× 10−8 J m−2 K−4 s−1

c1 Erste Strahlungskonstante (= 2πhc2) 3.742× 10−16 J m2 s−1

c2 Zweite Strahlungskonstante 1.439× 10−2 m Kεo Permittivität im Vakuum 8.854× 10−12 C2 N−1 m−2

µo Permeabilität im Vakuum 4π × 10−7 H m−1

NA Avogadrokonstante 6.022 ×1023 mol−1

R Gaskonstante 8.314 J K−1 mol−1

a0 Bohrradius 5.292 ×10−11 mµB Bohr magneton 9.274 ×10−24 J T−1

α Fine structure constant (= 1/137.0) 7.297 ×10−3

M Sonnemasse 1.989 ×1030 kgR Sonnenradius 6.96 ×108 mL Sonneluminosität 3.827 ×1026 J s−1

M⊕ Erdmasse 5.976 ×1024 kgR⊕ mittlerer Erdradius 6.371 ×106 m1 light year Lichtjahr 9.461 ×1015 m1 AU Astronomische Einheit 1.496 ×1011 m1 pc Parsec 3.086 ×1016 m1 year


Bibliography

[1] Matthew N.O. Sadiku. Elements of Electromagnetics. Oxford University PressInc, USA, new ed of 2 revised ed edition, 6 1997.

[2] William Hayt and John Buck. Engineering Electromagnetics. McGraw-Hill Sci-ence/Engineering/Math, 8 edition, 1 2011.

[3] Kenneth R. Demarest. Engineering Electromagnetics. Prentice Hall, 1 edition, 101997.

[4] David J. Griffiths. Introduction to Electrodynamics (Pie). Prentice Hall Interna-tional, 3rd international ed. edition, 5 2003.

[5] Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli. Fundamentals of Ap-plied Electromagnetics (6th Edition). Prentice Hall, 6 edition, 3 2010.

[6] David Halliday, Robert Resnick, and Jearl Walker. Fundamentals of Physics Ex-tended. Wiley, 9 edition, 3 2010.

[7] David M. Pozar. Microwave Engineering. Wiley John + Sons, 3. a. edition, 1 2004.

[8] Hans-Georg Unger and Johann Hinken. Elektromagnetische Wellen auf Leitun-gen. Hüthig, 4. aufl. edition, 12 1995.

184

Index

amber, 8Ampere, 14Ampere’s law, 88

B-field, 85Biot-Savart’s law, 98

capacitance, 50capacitor, 50capacitor, AC circuit, 75Charge, 10charge density, 11charge distributions, 11clip-on ammeter, 89closed surface, 173coercivity, 103coil, 96compass, 84conductivity, 77constant potential surface, 37Coulomb, 14Coulomb’s gauge, 107Coulomb’s law, 18cross product, 167Curl, 177curl, 90, 91, 177current, 13, 74current density, 76current direction, 75

del operator, 173Diamagnetism, 102differential displacement, 172dipole, 46dipole moment, 47displacement current, 89displacement flux, 58Divergence, 176divergence, 176divergence theorem, 60, 176drift velocity, 81

electric flux density, 57

electric potential, 33electromagnets, 96elementary charge, 14

Faraday, 85Ferromagnetic, 102field description, 21field direction, 22field line, 22field lines, 22field strength, 22flow of current, 81flux, 173

gauge transformations, 106Gauss law, 176Gauss law for magnetostatic fields, 104Gauss’s law, 59Gauss’s law for magneto-statics, 106Gradient, 175gradient, 175gravitational field, 33

H-field, 85Helmholtz equation, 140Henry, 85hysteresis loop, 102

impedance, 75Inductance, 111integration, 34intrinsic impedance, 143iron core, 96

Joule’s law, 82Joules, 81

Kirchhoff’s current law, 83Kirchhoff’s voltage law, 38

Laplace’s equation, 68law of conservation of magnetic flux, 104line integral, 172lodestone, 84

185

Prof. S. Peik

magnetic dipoles, 104magnetic field (strength), 85magnetic field lines, 85magnetic field of the earth, 86magnetic flux, 104, 109magnetic flux density, 85, 101magnetic induction, 85magnetic monopole, 104magnetic potential, scalar, 105magnetism, history, 84magnetostatic fields on boundaries, 110Maxwell, 85, 89Maxwell’s equation for electrostatic fields,

38mho, 79moving charges, 74mutual inductance, 122

Nabla, 173

Oersted, 85, 87Ohm, 79Ohm’s law, 77

Paramagnetism, 102permanent magnets, 85permeability, 101, 102permeability, absolute, 101permittivity, 18phasor, 75Poisson’s equation, 68potential, 33Power, 81

residual magnetism, 103resistance, 78resistivity, 77retentivity, 103right-hand rule, 87

scalar product, 167Self Inductance, 123solenoid, 96source points, 12Stokes Theorem, 178surface integral, 173

Tesla, 101test charge, 21

unit vectors, 166

vector potential, 105Volt, 36volume integral, 173

work, 33


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