Lecture1-4_NG

1/19/20091/19/2009 BITS, PILANI Dr. Navneet BITS, PILANI Dr. Navneet GuptaGupta

11

EEE C 433EEE C 433Lecture 1Lecture 1--44

Dr. Dr. NavneetNavneet GuptaGuptaAssistant Professor, EEE GroupAssistant Professor, EEE Group

Electromagnetic Fields and Waves

1/19/2009 BITS, PILANI Dr. Navneet Gupta 2

Course DescriptionCourse No. : EEE C433Course Title : Electromagnetic Fields

and WavesInstructor-in-charge : Dr. Navneet Gupta

(email: [email protected])(Chamber: 2210-H, FD-II)

Instructors : Dr. Niladri Sarkar,Mr. G. Meenakshi Sundran, Mr. Mahesh Angira


Scope and objective of the course:

Most fundamental topics in Electrical and Electronics Engineering. Thorough understanding of many areas such as VLSI, PCBs operating at GHz clocks, rotating machines, microwaves and antennas depends upon electromagnetics.To provide the students with the basic understanding of electromagnetic fields and waves.


Text Book

Text Book :John D. Kraus and Daniel A. Fleisch, “Electromagnetics”, 5th ed., McGraw-Hill, New York, 1999.Total: 11 Chapters


Reference BooksReference Books :

David K.Cheng, “Field and Wave Electromagnetics” 2nd ed. Addison Wesley Longman, Singapore 1999. (R1)Fawwaz T. Ulaby, “Electromagnetics for Engineers”, Pearson Education India, 2005.(R2)EDD Notes: “Smith Chart and its Applications”, BITS, Pilani, 2008 (R3)R.K.Shevgaonkar, “Electromagnetic Waves”, TataMcGraw-Hill Publishing Company Ltd., New Delhi, 2006.(R4)Constantine A. Balanis “Antenna Theory: Analysis and Design”, 2nd. Ed. John Wiley & Sons (ASIA) Pte Ltd, Singapore 2005. (R5)


Course PlanTotal Lectures: 40Broad Topics:

Foundation of ElectromagneticsEM Wave PropagationTransmission linesSmith ChartWaveguidesAntennas Radio Wave Propagation


Evaluation Scheme

No make-up will be given for Assignment and Quizzes however for Tests and Comprehensive Examination make-up examination will be given only in genuine cases for which prior permission of theinstructor-in-charge is required.


Introduction to Electromagnetics

What is electromagnetics?

Electromagnetics is the study of the mutual interactions between electric charges. Charges may be stationary, they may move with constant velocity (v=const) or they may be in accelerated motion .

What is electrostatics?

Electrostatics deals with the interactions between stationary charges.


Introduction to Electromagnetics

Electromagnetics is important because it provides a real-world, three-dimensional understanding of electricity and magnetism.


Center for Electromagnetic Theory and Applications (CETA)

Research Laboratory of ElectronicsMassachusetts Institute of Technology


Foundation of Electromagnetics:M. Faraday (1831):

Faraday showed that a magnetic field which varied in time--like the one produced by an alternating current (AC)--could drive electric currents, if (say) copper wires were placed in it in the appropriate way. That was "magnetic induction," the phenomenon on which electric transformers are based.

Changing magnetic fields could produce electric currents We already know that electric currents produce magnetic fields. Would it perhaps be possible for space to support a wave motion alternating between the two?


Foundation of Electromagnetics:magnetic field ---> electric current ---> magnetic field ---> electric current ---> ...

There was one stumbling block. Such a wave could not exist in empty space, because empty space contained no copper wires and could not carry the currents needed to complete the above cycle. A brilliant young Scotsman, James Clerk Maxwell, solved the riddle in 1861 by proposing the equations that relate electricity and magnetism


Foundation of Electromagnetics:Electricity and magnetism were originally thought to be unrelatedin 1861, James Clerk Maxwell provided a mathematical theory that showed a close relationship between all electric and magnetic phenomena

FOUNDER OF EMT


Foundation of Electromagnetics:Then Heinrich Hertz in Germany showed that an electric current bouncing back and forth in a wire (nowadays it would be called an "antenna") could be the source of such waves. Electric sparks create such back-and-forth currents when they jump across a gap--hence the crackling caused by lightning on AM radio--and Hertz in 1886 used such sparks to send a radio signal across his lab. Later the Italian Marconi, with more sensitive detectors, extended the range of radio reception, and in 1903 detected signals from Europe as far as Cape Cod, Massachussets.


Foundation of Electromagnetics:JAGADISH CHANDRA BOSEThe First Indian Scientist who marked his footprints in the world of Electromagnetics. In fact, Bose generated millimeter waves using a circuit developed in his laboratory and used these waves for communication, much earlier than the western scientists. He also developed microwave antennas (horns) which are still considered to be ideal feeds


Maxwell’s predictionsElectric field lines originate on positive charges and terminate on negative charges

Electric field is produced by charges

Magnetic field lines always form closed loops – they do not begin or end anywhere

Magnetic fleld is produced by currents (moving charges)

A varying magnetic field induces an emf and hence an electric field (Faraday’s Law)

Electric field is also produced by changing magnetic field

Magnetic field is also produced by changing electric field.

QuestionQuestion:: is there a is there a symmetry symmetry between electric and magnetic fields, between electric and magnetic fields, i.e. can i.e. can magnetic field magnetic field be produced by be produced by changing electric fieldchanging electric field??????

Maxwell: YES!!!


Note: Charges and FieldsStationary charges produce only electric fieldsCharges in uniform motion (constant velocity) produce electric and magnetic fieldsCharges that are accelerated produce electric and magnetic fields and electromagnetic wavesThese fields are in phase

At any point, both fields reach their maximum value at the same time


Electromagnetic Waves

The E and B fields are perpendicular to each otherBoth fields are perpendicular to the direction of motion

Therefore, em waves are transverse waves


Properties of EM WavesElectromagnetic waves are transverse wavesElectromagnetic waves travel at the speed of light

Because em waves travel at a speed that is precisely the speed of light, light is an electromagnetic wave

81 2.99792 10o o

c m sμ ε

= = ×


Properties of EM Waves, 2The ratio of the electric field to the magnetic field is equal to the speed of light

Electromagnetic waves carry energy as they travel through space, and this energy can be transferred to objects placed in their path

BEc =


The Spectrum of EM WavesForms of electromagnetic waves exist that are distinguished by their frequencies and wavelengths

c = ƒλ

Wavelengths for visible light range from 400 nm to 700 nmThere is no sharp division between one kind of em wave and the next


The EMSpectrum

Note the overlap between types of wavesVisible light is a small portion of the spectrumTypes are distinguished by frequency or wavelength


The EM Spectrum

Radio WavesUsed in radio and television communication systems

MicrowavesWavelengths from about 1 mm to 30 cmWell suited for radar systemsMicrowave ovens are an application


The EM Spectrum, 2

Infrared wavesIncorrectly called “heat waves”Produced by hot objects and moleculesReadily absorbed by most materials

Visible lightPart of the spectrum detected by the human eyeMost sensitive at about 560 nm (yellow-green)


The EM Spectrum, 3Ultraviolet light

Covers about 400 nm to 0.6 nmSun is an important source of uv lightMost uv light from the sun is absorbed in the stratosphere by ozone

X-raysMost common source is acceleration of high-energy electrons striking a metal targetUsed as a diagnostic tool in medicine


The EM Spectrum, final

Gamma raysEmitted by radioactive nucleiHighly penetrating and cause serious damage when absorbed by living tissue


Review problem: car radio

An An RLC RLC circuit is used to tune a radio to an FM station circuit is used to tune a radio to an FM station broadcasting at 88.9 MHz. The resistance in the circuit is broadcasting at 88.9 MHz. The resistance in the circuit is 12.0 12.0 ΩΩ and the capacitance is 1.40 and the capacitance is 1.40 pFpF. What inductance . What inductance should be present in the circuit? should be present in the circuit? Given:

RLC circuitf0=88.9 HzR = 12.0 ΩC = 1.40 pF

Find:

L=?

The resonance frequency of the circuit should be chosen to match that of the radio station


Fundamentals of ElectromagneticsFundamental vector field quantities in electromagnetics

Electric field intensity (E)Volts/meter

Electric flux density (Electric Displacement) (D)Coulombs / meter2

Magnetic field intensity (B)Tesla = Webers / meter2

Magnetic flux density (H)Amps per meter (A/m)


Fundamentals of Electromagnetics

Universal constants in electromagnetics:

Velocity of an electromagnetic wave (e.g., light) in free space (perfect vacuum)

Permeability of free space

Permittivity of free space

Intrinsic impedance of free space

m/s 103 8×≈cH/m 104 7

0−×= πμ

F/m 10854.8 120

−×≈ε

Ω≈ πη 1200



Constitutive Relations:Electric and magnetic flux densities (E and B) are related to the field intensities (D and H) by constitutive relations:

Vacuum :

Simple andmedia



Vector Analysis : Mathematical Shorthand

Coordinate Systems:Rectangular/Cartesian Coordinate SystemsCylindrical Coordinate SystemsSpherical Coordinate Systems



Rectangular/Cartesian Coordinate Systems



Cylindrical Coordinate Systems



Spherical Coordinate Systems




Example: Coordinate transformation

Transform the vector

in cartesian coordinates

( ) 5ˆ2ˆcos3ˆ zr araaA +−= φφr


Line, Surface and Volume Integral

An important application of the scalar (or dot ) product involves the line integralWe take surface integral to calculate the total quantity like flux or current crossing the area.If any physical quantity like charge or mass of an object is distributed in the volume of an object the total quantity can be found out by taking the volume integral.


Example:Use the cylindrical coordinate system to find the area of a curved surface on the right circular cylinder having radius = 3m and height = 6m

00 12030 ≤≤ φ


Coulomb’s Law

2120

211212 4ˆ

rQQrFεπ

=r

Q1

Q212r

12Fr

Force due to Q1acting on Q2

Unit vector indirection of r12

electrical permittivity of free space [ε0 = 8.854X10-12 F/m].


Electric Field and electric field strength (intensity) E

)/(4

ˆ 2120

112

2

12 CNr

QrQFE

επ==

rr

Line Charges:

rE

raarr

E

Lr

Lr

0

2

0

2

12

περ

πε

ρ

=

>>

+⎟⎠⎞

⎜⎝⎛

=


Example: dc transmission lineTwo long parallel conductors of a dc transmission line separated by 2 m have charges of ρL= 5μC/m of opposite sign. Both lines are 8 m above ground. What is the magnitude of the electric field 4 m directly below one of the wires? εr=1


Electric Potential V and its Gradient E

Gradient: an operation performed on a scalar function which results in a vector function.E= -grad V E = -∇ V (moving against the electric field results in +ve work done)


Divergence and Curl of a vector field

Divergence of a vector field A at a point is defined as the net outward flux of A per unit volume as the volume about the point tends to zero.Curl of a vector field A, is a vector whose magnitude is the maximum net circulation of A per unit area as the area tends to zero and whose direction is the normal direction of the area when the area is oriented to make the net circulation maximum.


Poisson’s Equation:Laplace’s Law (for free space)


Divergence Theorem: convert volume integral to surface integral using divergence of a vector field.

Stoke’s Theorem:convert surface integral to line integral using curl of a vector field


Laplacian

( ) AAArrr

×∇×∇−∇∇=∇ .2

Laplacian of a scalar:Laplacian of a vector:

NULL IDENTITIES:The curl of the gradient of any scalar field is identically zero.The divergence of the curl of any vector field is identically zero.


ELECTRIC DIPOLEAn electric dipole is a separation of positive and negative charge. The simplest example of this is a pair of electric charge of equal magnitude but opposite sign, separated by some, usually small, distance.


Electric Dipole Potential


Equipotential lines: dipoleEquipotential lines are like contour lines on a map which trace lines of equal altitude. In this case the "altitude" is electric potential or voltage. Equipotential lines are always perpendicular to the electric field. In three dimensions, the lines form equipotentialsurfaces.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c1

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefie.html#c1




Movement along an equipotential surface requires no work because such movement is always perpendicular to the electric field.


Superposition of PotentialThe total electric potential at a point is the algebraic sum of the individual potentials at the point.


The total electric potential at a point:V=Vp+VL+Vs+Vv


Electric Flux Density

Uniform Case:The electric flux ψ through a surface area is the integral of the normal component of electric field (times ε) over the area.


Electric Flux DensityFlux density D = ε E (C/m2)

Isotropic medium: D and E are in the same directionAnisotropic medium: D and E may be in different direction.


Electric Flux Density

Nonuniform case:Total electric flux over the sphere is equal to the charge Q enclosed by the sphere.

QdSDs

== ∫ .r

ψ



Gauss’s LawTotal electric flux out of a closed surface is equal to the net charge within the surface.


Boundary Conditions:

Two Dielectric media:The tangential component of E is continuous across the boundaryEt1=Et2

Dn1- Dn2=ρA dielectric and a conductor:

The tangential component of the E field on a conductor surface is zero.Et=0 while En= ρ/ε


Example: Fields at charge-free boundary at oblique incidence

Find the magnitude of EF in medium 2. assume ε1 > ε2

ε1

ε2

α2

α1

D1 or E1

D2 or E2

En2

En1

Et1

Et2

Boundary between two media


CapacitanceWhen a conductor acquires charge, it is distributed on the surface, such as to nullify E inside and its tangential components on the surface.

In general, potential ∝ charge acquired,


Parallel-plate Capacitor

A

C = Q / V= DA/Ed=εEA/Ed= εA/dA

w

Capacitor cell:When w = d the capacitor becomes a capacitor cell


Capacitor Energy and Energy Density

Work done to move a charge through a potential difference of V is, dW = V dQ

W = ∫ dW = ∫ Q dQ/C

Energy density: w = εE2/2 (J/m3)


Example: Sandwich capacitor

V1

V2

Capacitor plate

d E1 εr = 2 D1

d E2 εr=3 D2

10 V


Electric CurrentElectric current is the flow of electric charge.

Conductors: motion of electronsGaseous conductors: charge is carried by electrons & +veions.Liquid conductors: charge is carried by ions (+ve and –ve).

semiconductors: charge is carried by electrons and holes.Electric current can be represented as the time rate of change of charge, or I = dQ/dtCurrent density is a measure of the density of an electric current. It is defined as a vector whose magnitude is the electric current per cross-sectional area.


Electric Current

Test charge +e

Field E Force F

Field E

vd

vd = μe E


Electric Current

I = vdρA

J = I/A = vdρ

If current density is not uniform


Electric CurrentOhm’s Law

V = IR

Joule’s Law: Power = V.I = I2 R


Divergence of J

0. =∇ Jr

Kirchhoff’s current law


Boundary Conditions For Current Density: Conducting Media

2

1

2

1

21

σσ

=

=

t

t

nn

JJ

JJ


Steady Magnetic Field

Biot-Savart Law: This law gives the magnetic field intensity dH due to a current element Idl

Motion of free charges gives current and the current produces magnetic field around it.


Ampere’s circuital lawAmpere’s Circuital Law says that the integration of H around any closed path is equal to the net current enclosed by that path.

The line integral of H around a closed path is termed the circulation of H.

JH

IldHc

rr

rr

=×∇

=∫ . Integral Form

Differential Form


Gauss’s Law for magnetic fields

For a closed surface

Total magnetic flux through any surface bound by a common closed contour is the same.

0.

0.

=∇

=∫B

SdBc

r

rr


Boundary conditions for magnetostatic field

Normal component:The normal component of B is continuous across an interfaceThe normal component of H is discontinuous by the ratio μ1/μ2.

Tangential component:Ht1 – Ht2 = KIf the surface current density is zero. Ht1 = Ht2


Inductance and Magnetic energy


Inductance and Magnetic energy


Magnetic Energy DensityImagine a current generator connected to a coil, which increases current from 0 to I.The inductor coil will stores an incremental energy

dW = V I dtwhere,

V = £dI/dtso, the total energy

W = £ ∫I’ dI’ = £ I2/2 = = BANI/2 = BHAL/2where AL is the volume of the coil (m3)

Hence, energy density w = μH2/2 (J m-3)


Inductor cell

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