eleg 413 spring 2017 lecture #1 - university of delawaremirotzni/eleg413/eleg413lec1.pdf · eleg...

ELEG 413Spring 2017Lecture #1

Mark Mirotznik, Ph.D.Professor

The University of DelawareTel: (302)831-4221

Email: [email protected]

ELEG 413 Engineering Electromagnetics

Instructor: M. Mirotznik, Tel (302)831-4241, [email protected]: Engineering Electromagnetics, Constantine Balanishttp://www.eecis.udel.edu/~mirotzni/ELEG413/ELEG413_2017.htmOffice Hours: Tuesday/Thursday 1:00-2:00 in Evans 106

Homework 0%In class quizzes 20%1st Exam 25%2nd Exam 25%Final Exam 30%

100%

Grading Your grade will be based on, quizzes and exams broken down as follows:

mailto:[email protected]

What was physics like at the time of Maxwell?

Newton’s laws had been around for almost 200 years and seemed to explain almost everything.

Most felt that physics was largely a solved problem with just some odds and ends to figure out.

It was felt that everything when broken down to its most fundamental level was just an application of mechanics (Newton’s laws).

What was known about electromagnetics to Maxwell?

(1) That electrical charges come in two types (negative and positive) and that they produce forces between them that proportional to the amount of charge and inversely proportional to the square of the distance between the charges. Like charges repel and unlike charges attract.


(2) That magnets have two poles (N and S) and also produce a force between magnets (like poles repel and unlike poles attract) but unlike electrical charges we cannot separate the two poles. If a break a magnet into two parts I get two magnets each having a N/S poles.


(3) That electrical currents on two wires would create a force between them (like magnets) - Ampere 1820


(3) That an electric current inside a magnetic field produces a force that was at right angles to the direction of the magnetic lines of force and the direction of the current).

(4) Mutual inductance (Faraday) –also made the first electric motor and generator



Nobody had a good explanation for why these things were happening? Some of it was also a bit spooky!

How does Q1 even know about Q2?

Faraday had an idea!Lines of force – first time someone starting thinking about electromagnetism as a field theory.

Michael Faraday(September 1791 – August 1867)

Along came Maxwell

He was crazy smart!

Published his first scientific paper at the age of 14.

Besides his historic work in electromagnetic theory he also Published the first paper on the

theory of feedback control systems Published one of the first papers on

the kinetics theory of gases. Let to modern thermodynamic theory.

Studied color vision and even printed the first color photograph. We still use the Maxwell color charts to quantify color vision.

Along came Maxwell:

"A Dynamical Theory of the Electromagnetic Field" is the third of James Clerk Maxwell's papers regarding electromagnetism, published in 1865

Along came Maxwell:

"A Dynamical Theory of the Electromagnetic Field" is the third of James Clerk Maxwell's papers regarding electromagnetism, published in 1865

In section III of "A Dynamical Theory of the Electromagnetic Field", which is entitled "General Equations of the Electromagnetic Field", Maxwell formulated twenty equations with twenty unknowns which were to become known as Maxwell's equations.

In this amazing piece of work Maxwell put forth the foundations of a field theory that could explain all known phenomena regarding electromagnetism. He also added some unknown terms that resulted in the prediction of electromagnetic waves which travel near 3x108

m/s (speed of light). Also connecting for the first time the fields of optics with electricity and magnetism.

This is how Maxwell’s equations looked in his

original notation. These are 20 coupled differential

equations (Ugh!).

Unfortunately, this work was largely ignored for nearly 20

years. The math was too hard to solve for most .

Enter the Maxwellians

George Francis FitzGerald Oliver Lodge Oliver Heaviside Heinrich Hertz

These men saw the brilliance of Maxwell’s ideas and worked to:

(1) Validate the existence of EM waves experimentally (Hertz)(2) Reformulate Maxwell’s 20 equations into a more digestible 4

vector equations (Heaviside and FitzGerald)(3) Demonstrated through various solutions to the new Maxwell’s

equations how they can be used to predict all that is know about electromagnetic phenomenon and some stuff that was yet to be shown.

Maxwell’s Equations in Differential Form(as formulated by Heaviside)

m

ic

BD

JJtDH

MtBE

ρ

ρ

=⋅∇

=⋅∇

++∂∂

=×∇

−∂∂

−=×∇

Faraday’s Law

Ampere’s Law

Gauss’s Law

Gauss’s Magnetic Law

Maxwell’s mechanical vortex model

Did Maxwell and his Disciples Have a Good Feeling for What Electric and

Magnetic Fields Are? FitzGerald’s Wheel and

Band Model (1885)

Lodge’s string and beads model(1876)

Vector Analysis Review:

mB

D

JtDH

MtBE

ρ

ρ

=⋅∇

=⋅∇

+∂∂

=×∇

−∂∂

−=×∇

The electric and magnetic

fields in Maxwell’s equations are vector fields that vary in

both time and space.

zzyyxx

zzyyxx

zzyyxx

zzyyxx

atzyxHatzyxHatzyxHtzyxH

atzyxDatzyxDatzyxDtzyxD

atzyxBatzyxBatzyxBtzyxB

atzyxEatzyxEatzyxEtzyxE

),,,(),,,(),,,(),,,(

),,,(),,,(),,,(),,,(

),,,(),,,(),,,(),,,(

),,,(),,,(),,,(),,,(

++=

++=

++=

++=

sourcescalartzyxsourcescalartzyx

atzyxMatzyxMatzyxMtzyxM

atzyxJatzyxJatzyxJtzyxJ

m

zzyyxx

zzyyxx

==

++=

++=

),,,(),,,(

),,,(),,,(),,,(),,,(

),,,(),,,(),,,(),,,(

ρρ

The Vector Electromagnetic Fields

The Electromagnetic Sources

The main mathematics that is needed to understand these

equations and solve real problems is vector

calculus.

Vector Analysis Review:

AAa

AaA A

=

=

a = unit vector

AaA A

=

1. Dot Product (projection)

)cos( ABBABA θ=⋅

2. Cross Product

)sin( ABn BAaBA θ=×

AaA A

=

A

BaB B

=

ABθna

Orthogonal Coordinate Systems:

23

22

21

213

312

321

332211

uuu

uuu

uuu

uuu

Auuuuuu

AAAA

aaa

aaa

aaa

aAAaAaAaA

++=

×=

×=

×=

=++=

332211 uuuuuu BABABABA ++=⋅

)()()(

12213

3113223321

uuuuu

uuuuuuuuuu

BABAaBABAaBABAaBA

−+−+−=×

321

321

321

uuu

uuu

uuu

BBBAAAaaa

BA

=×

Orthogonal Coordinate Systems:

dl332211 dladladlald uuu

++=

Sd

na

dSaSd n

=

321 dldldldv =

dl1dl2

dl3

Cartesian Coordinate Systems:

yxz

zxy

zyx

Azzyyxx

aaa

aaa

aaa

aAAaAaAaA

×=

×=

×=

=++=

zzyyxx BABABABA ++=⋅

zyx

zyx

zyx

BBBAAAaaa

BA

=×

x

yz

Cartesian Coordinate Systems (cont):

dzdydxdvdydxads

dzdxadsdzdyads

dzdydxdl

dzadyadxald

zz

yy

xx

zyx

==

==

++=

++=

222

Cylindrical Coordinate Systems:

Azzrr aAAaAaAaA =++= φφ

dzdrdrdvdzrdads

dzdradsdzrdads

dzardadrald

zz

rr

zr

φφ

φ

φ

φφ

φ

==

==

++=

x

y

z

φ r

z

(r,φ,z)

Spherical Coordinate Systems:

ARR aAAaAaAaA =++= φφθθ

φθθ

θϕθφθθ

φθθ

φϕ

θθ

φθ

dddRRdv

dRdRadsddRRads

ddRads

dRaRdadRald

RR

R

)sin(

)sin()sin(

)sin(

2

2

=

===

++=

x

y

z

φ

R

(R,θ,φ)

θ

Vector Coordinate Transformation:

z

r

z

y

x

AAA

AAA

φφφφφ

−=

1000)cos()sin(0)sin()cos(

ϕ

θ

θθφφθφθφφθφθ

AAA

AAA R

z

y

x

−

−=

0)sin()cos()cos()sin()cos()sin()sin()sin()cos()cos()cos()sin(

Gradient of a Scalar Field:Assume f(x,y,z) is a scalar fieldThe maximum spatial rate of change of f at some locationis a vector given by the gradient of f denoted byGrad(f) or f∇

φθθ

φ

φθ

φ

∂∂

+∂∂

+∂∂

=∇

∂∂

+∂∂

+∂∂

=∇

∂∂

+∂∂

+∂∂

=∇

)sin(Rfa

rfa

Rfaf

zfa

rfa

rfaf

zfa

yfa

xfaf

R

zr

zyx

Divergence of a Vector Field:Assume E(x,y,z) is a vector field. The divergence of E is defined as the net outward flux of E in some volume as thevolume goes to zero. It is denoted by E

⋅∇

φθ

θθθ

φ

φ

θ

φ

∂∂

+

∂∂

+∂∂

=⋅∇

∂∂

+∂∂

+∂∂

=⋅∇

∂∂

+∂∂

+∂∂

=⋅∇

ER

ER

ERRR

E

zEE

rrE

rrE

zE

yE

xEE

R

zr

zyx

)sin(1

))(sin()sin(

1)(1

1)(1

22

Curl of a Vector Field:Assume E(x,y,z) is a vector field. The curl of E is measureof the circulation of E also called a “vortex” source. Itis denoted by E

×∇

∂∂

∂∂

∂∂

=×∇

∂∂

∂∂

∂∂

=×∇

∂∂

∂∂

∂∂

=×∇

φθ

φθ

φ

φ

θφθ

θ

θ

φ

ERREER

aRaRa

RE

ErEEzr

aara

rE

EEEzyx

aaa

E

R

R

zr

zr

zyx

zyx

)sin(

)sin(

)sin(1

1

2

Laplacian of a Scalar Field:

)( V∇⋅∇

Assume f(x,y,z) is a scalar field. The Laplacian is defined as and denoted by V2∇

2

2

22

22

2

2

2

2

2

22

2

2

2

2

2

22

)(sin1

))(sin()sin(

1)(1

1)(1

φθ

θθ

θθ

φ

∂∂

+

∂∂

∂∂

+∂∂

∂∂

=∇

∂∂

+∂∂

+∂∂

∂∂

=∇

∂∂

+∂∂

+∂∂

=∇

VR

VR

VR

RRR

V

zVV

rV

rr

rrV

zV

yV

xVV

Basic Theorems:1. Divergence Theorem or Gauss’s Law

sdEdvEsv

⋅=⋅∇ ∫∫∫∫∫

2. Stokes Theorem

∫∫∫ ⋅=⋅×∇cs

ldEsdE

)(

Examples:1. Verify the Divergence Theorem for

zarazrE zr 2),( 2 +=

on a cylindrical region enclosed by r=5, z=0 and z=4

r = 5

z = 0

z = 4



r = 5

z = 0

z = 4

( ) 23231

)2(01)(1

1)(1

2

2

+=+=⋅∇

∂∂

+∂∂

+⋅∂∂

=⋅∇

∂∂

+∂∂

+∂∂

=⋅∇

rrr

E

zz

rrr

rrE

zEE

rrE

rrE z

r

φ

φφ


sdEdvEsv

⋅=⋅∇ ∫∫∫∫∫



r = 5

z = 0

z = 4

( ) 23231 2 +=+=⋅∇ rrr

E


sdEdvEsv

⋅=⋅∇ ∫∫∫∫∫

( )

( )

( )

∫ ∫

∫ ∫∫∫∫

∫ ∫ ∫

∫ ∫ ∫∫∫∫

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

⋅=

+=⋅∇

⋅+=

⋅+=⋅∇

4

0

2

0

4

0

2

0

5

0

23

4

0

2

0

5

0

2

4

0

2

0

5

0

150

23

23

z

z

z

zv

z

z

r

r

z

z

r

rv

dzd

dzdrrdvE

dzdrdrr

dzrdrdrdvE

πθ

θ

πθ

θ

πθ

θ

πθ

θ

θ

θ

θ

θ



r = 5

z = 0

z = 4

( ) 23231 2 +=+=⋅∇ rrr

E


sdEdvEsv

⋅=⋅∇ ∫∫∫∫∫

ππ

θπθ

θ

120042150

1504

0

2

0

=⋅⋅=⋅∇

⋅=⋅∇

∫∫∫

∫ ∫∫∫∫=

=

=

=

dvE

dzddvE

v

z

zv



r = 5

z = 0

z = 4


sdEdvEsv

⋅=⋅∇ ∫∫∫∫∫

( )

( )( )∫ ∫

∫ ∫

∫ ∫∫∫

=

=

=

=

=

=

=

=

=

=

=

=

⋅⋅++

⋅⋅++

⋅+=⋅

5

0

2

0

2

5

0

2

0

2

4

0

2

0

2

02

42

525

r

r zzr

r

r zzr

z

z rzrs

drrdaara

drrdaara

dzdazaasdE

θ

θ

θ

πθ

θ

πθ

θ

πθ

θ



r = 5

z = 0

z = 4


sdEdvEsv

⋅=⋅∇ ∫∫∫∫∫

( )

( )( )∫ ∫

∫ ∫

∫ ∫∫∫

=

=

=

=

=

=

=

=

=

=

=

=

⋅⋅++

⋅⋅++

⋅+=⋅

5

0

2

0

2

5

0

2

0

2

4

0

2

0

2

02

42

525

r

r zzr

r

r zzr

z

z rzrs

drrdaara

drrdaara

dzdazaasdE

θ

θ

θ

πθ

θ

πθ

θ

πθ

θ

πππ

ππ

120052128425

2128425

23

5

0

23

=⋅⋅+⋅⋅=

⋅⋅+⋅⋅=⋅∫∫ rsdEs



r = 5

z = 0

z = 4


sdEdvEsv

⋅=⋅∇ ∫∫∫∫∫

π1200=⋅∫∫ sdEs

π1200=⋅∇∫∫∫ dvEv

Odds and Ends:1. Normal component of field

n

E

nEEn =⋅

2. Tangential component of field

tEEn =×

Maxwell’s Equations in Differential Form

m

ic

BD

JJtDH

MtBE

ρ

ρ

=⋅∇

=⋅∇

++∂∂

=×∇

−∂∂

−=×∇

Faraday’s Law

Ampere’s Law

Gauss’s LawGauss’s Magnetic Law

Maxwell’s Equations Field Variables

=

=

=

=

B

H

D

E

Electric Field, V/m

Electric Displacement, Q/m2

Magnetic Field, A/m

Magnetic Flux Density, T

Maxwell’s Equations Source Variables

Conductive Current Density, A/m2

Magnetic Charge Density Wb/m3

Electric Charge Density, Q/m3

===

=

=

m

i

c

M

J

J

ρρ

Impressed Current Density, A/m2

Magnetic Current Density, V/m2

Faraday’s Law

sdBt

ldE

tBE

c s

⋅∂∂

−=⋅

∂∂

−=×∇

∫ ∫∫

S

C

tB∂∂

E

Ampere’s Law

∫∫∫ ∫∫ ⋅+⋅∂∂

=⋅

∂∂

+=×∇

sc ssdJsdD

tldH

tDJH

t

D∂∂

J

J

H

H

Gauss’s Law

∫∫∫∫∫ ==⋅

=⋅∇

v totsQdvsdD

Dρ

ρ

totQ

D

Gauss’s Magnetic Law

00

∫∫ =⋅

=⋅∇

ssdB

B

B

“all the flow of B entering the volume V must leave the volume”

Maxwell’s Equations in Differential Form: No Magnetic Sources

0=⋅∇

=⋅∇

++∂∂

=×∇

∂∂

−=×∇

B

D

JJtDH

tBE

ic

ρ

Faraday’s Law

Ampere’s Law


CONSTITUTIVE RELATIONS

ED

ε=ε=εr εo=permittivity (F/m)εo=8.854 x 10-12 (F/m)


HB

µ=

µ=µoµr µo=permeability of free space (H/m)µo=4π x 10-7 (H/m)


EJc

σ= σ=conductivity (S/m)

Maxwell’s Equations in Differential Form: No Magnetic Sources

0=⋅∇

=⋅∇

++∂∂

=×∇

∂∂

−=×∇

B

D

JEtEH

tHE

i

ρ

σε

µ Faraday’s Law

Ampere’s Law


Maxwell’s Equations in Differential Form: No Magnetic Sources and No Loss

0=⋅∇

=⋅∇

+∂∂

=×∇

∂∂

−=×∇

B

D

JtEH

tHE

i

ρ

ε

µ Faraday’s Law

Ampere’s Law


Maxwell added this term

Boundary Conditions

(1) Tangential Component of E?

11,µε

22 ,µε)1(

tE

)2(tE

Boundary Conditions

(1) Tangential Component of E?

11,µε

22 ,µε)1(

tE

)2(tEn̂

n̂

)1()2( ˆˆ EnEn

×=×

or

( ) 0ˆ )1()2( =−× EEn

Boundary Conditions

(2) Tangential Component of H?

11,µε

22 ,µε)1(

tH

)2(tH

Boundary Conditions

(2) Tangential Component of H?

11,µε

22 ,µε)1(

tH

)2(tH

( ) sJHHn

=−× )1()2(ˆ

n̂

Boundary Conditions

(3) Normal Component of E?

11,µε

22 ,µε)1(nE

)2(nE

Boundary Conditions

(3) Normal Component of E?

11,µε

22 ,µε)1(nE

)2(nE

( ) sDDn ρ=−⋅ )1()2(ˆ

( ) sEEn ρεε =−⋅ )1(1

)2(2ˆ

or

Boundary Conditions

(4) Normal Component of H?

11,µε

22 ,µε)1(nH

)2(nH

Boundary Conditions

(4) Normal Component of H?

11,µε

22 ,µε)1(nH

)2(nH

( ) 0ˆ )1()2( =−⋅ BBn

or

( ) 0ˆ )1(1

)2(2 =−⋅ HHn

µµ

Boundary Conditions(ALWAYS TRUE)

( ) 0ˆ )1()2( =−⋅ BBn

( ) sDDn ρ=−⋅ )1()2(ˆ

( ) sJHHn

=−× )1()2(ˆ

( ) 0ˆ )1()2( =−× EEn

Boundary Conditions(ALWAYS TRUE)

( ) 0ˆ )1()2( =−⋅ BBn

( ) sDDn ρ=−⋅ )1()2(ˆ

( ) sJHHn

=−× )1()2(ˆ

( ) 0ˆ )1()2( =−× EEn

11,µε)1(

nE

)2(nE

(PEC)

How do these simplify if one of the materials is a perfect electrical conductor (PEC)?

Boundary Conditions(PEC)

( ) 0ˆ )1()2( =−⋅ BBn

( ) sDDn ρ=−⋅ )1()2(ˆ

( ) sJHHn

=−× )1()2(ˆ

( ) 0ˆ )1()2( =−× EEn

11,µε)1(

nE

)2(nE

(PEC)

In a PEC all the fields must be zero!!!Why?

Boundary Conditions(PEC)

0ˆ )2( =⋅Bn

sDn ρ=⋅ )2(ˆ

sJHn

=× )2(ˆ

0ˆ )2( =×En

11,µε)1(

nE

)2(nE

(PEC)

In a PEC all the fields must be zero!!!Why?

Boundary Conditions

( ) 0ˆ )1()2( =−⋅ BBn

( ) 0ˆ )1()2( =−⋅ DDn

( ) 0ˆ )1()2( =−× HHn

( ) 0ˆ )1()2( =−× EEn

NOT A PEC PEC

0ˆ )2( =⋅Bn

sDn ρ=⋅ )2(ˆ

sJHn

=× )2(ˆ

0ˆ )2( =×En

Example Problem

oµε ,1 oµε ,2

x

x

aztRaztE

)21000cos()21000cos(1

ππ

++−=

xaztTE )21000cos(2 π−=

z=00=⋅∇

=⋅∇

++∂∂

=×∇

∂∂

−=×∇

B

D

JJtDH

tBE

ic

ρ

FIND R and T on both sides!

Example Problem

oµε ,1 oµε ,2

x

x

aztRaztE

)21000cos()21000cos(1

ππ

++−= xaztTE

)21000cos(2 π−=

z=0

tBE∂∂

−=×∇

FIND H on both sides!

Example Problem

oµε ,1 oµε ,2

x

x

aztRaztE

)21000cos()21000cos(1

ππ

++−= xaztTE

)21000cos(2 π−=

z=0

tBE∂∂

−=×∇

00)21000cos(

)21000cos(

ˆˆˆ1

ztRzt

zyx

aaa

tB

zyx

ππ++

−∂∂

∂∂

∂∂

−=∂∂

00)21000cos(

ˆˆˆ2

ztTzyx

aaa

tB

zyx

π−∂∂

∂∂

∂∂

−=∂∂

Example Problem

oµε ,1 oµε ,2

x

x

aztaztE

)21000cos(25.0)21000cos(1

ππ

+−−=

xaztE )21000cos(75.02 π−=

z=0

tBE∂∂

−=×∇

y

y

y

zyx

aztR

aztt

B

aztR

ztzt

B

ztRzt

zyx

aaa

tB

ˆ)21000sin(2

ˆ)21000sin(2

ˆ)21000cos(

)21000cos(

00)21000cos(

)21000cos(

ˆˆˆ

1

1

1

ππ

ππ

ππ

ππ

+⋅+

−−=∂∂

++

−∂∂

−=∂∂

++−

∂∂

∂∂

∂∂

−=∂∂

( )

y

y

zyx

aztTt

B

aztTzt

B

ztTzyx

aaa

tB

ˆ)21000sin(2

ˆ)21000cos(

00)21000cos(

ˆˆˆ

2

2

2

ππ

π

π

−⋅−=∂∂

−∂∂

−=∂∂

−∂∂

∂∂

∂∂

−=∂∂

Example Problem

oµε ,1 oµε ,2

x

x

aztaztE

)21000cos(25.0)21000cos(1

ππ

+−−=

xaztE )21000cos(75.02 π−=

z=0

tBE∂∂

−=×∇

y

y

y

y

y

y

aztR

aztB

dtaztR

aztB

aztR

aztt

B

ˆ)21000cos(1000

12

ˆ)21000cos(1000

12

ˆ)21000sin(2

ˆ)21000sin(2

ˆ)21000sin(2

ˆ)21000sin(2

1

1

1

ππ

ππ

ππ

ππ

ππ

ππ

+⋅−

−⋅+=

+⋅+

−−=

+⋅+

−−=∂∂

∫

y

y

y

aztTB

dtaztTB

aztTt

B

ˆ)21000cos(1000

12

ˆ)21000sin(2

ˆ)21000sin(2

2

2

2

ππ

ππ

ππ

−⋅=

−⋅−=

−⋅−=∂∂

∫

Example Problemoµε ,1 oµε ,2

x

x

aztaztE

)21000cos(25.0)21000cos(1

ππ

+−−=

xaztE )21000cos(75.02 π−=

z=0

tBE∂∂

−=×∇

y

y

aztR

aztB

ˆ)21000cos(1000

12

ˆ)21000cos(1000

121

ππ

ππ

+⋅−

−⋅+=

yaztTB ˆ)21000cos(1000

122 ππ −⋅=


x

x

aztaztE

)21000cos(25.0)21000cos(1

ππ

+−−=

xaztE )21000cos(75.02 π−=

z=0

tBE∂∂

−=×∇

yo

yo

aztR

aztH

ˆ)21000cos(1000

12

ˆ)21000cos(1000

121

πµπ

πµπ

+⋅−

−⋅=

yo

aztTH ˆ)21000cos(1000

122 π

µπ

−⋅=


x

x

aztaztE

)21000cos(25.0)21000cos(1

ππ

+−−=

xaztE )21000cos(75.02 π−=

z=0

tBE∂∂

−=×∇

yo

yo

aztR

aztH

ˆ)21000cos(1000

12

ˆ)21000cos(1000

121

πµπ

πµπ

+⋅−

−⋅=

yo


122 π

µπ

−⋅=

APPLY BC at Z=0

yo

yo

yo

atTatRat ˆ)1000cos(1000

12ˆ)1000cos(1000

12ˆ)1000cos(1000

12⋅=⋅−⋅

µπ

µπ

µπ


x

x

aztaztE

)21000cos(25.0)21000cos(1

ππ

+−−=

xaztE )21000cos(75.02 π−=

z=0

tBE∂∂

−=×∇

yo

yo

aztR

aztH

ˆ)21000cos(1000

12

ˆ)21000cos(1000

121

πµπ

πµπ

+⋅−

−⋅=

yo


122 π

µπ

−⋅=

APPLY BC at Z=0

yo

yo

yo


12ˆ)1000cos(1000

12ˆ)1000cos(1000

12⋅=⋅−⋅

µπ

µπ

µπ

TR =−1


x

x

aztRaztE

)21000cos()21000cos(1

ππ

++−=


z=0

yo

yo

aztR

aztH

ˆ)21000cos(1000

12

ˆ)21000cos(1000

121

πµπ

πµπ

+⋅−

−⋅= y

o


122 π

µπ

−⋅=

tDH∂∂

=×∇

[ ] xy

y

zyx

azHzt

D

zHzyx

aaa

tD

ˆ)(

0)(0

ˆˆˆ

1

1

∂∂

−=∂∂

∂∂

∂∂

∂∂

=∂∂

[ ] xy

y

zyx

azHzt

D

zHzyx

aaa

tD

ˆ)(

0)(0

ˆˆˆ

2

2

∂∂

−=∂∂

∂∂

∂∂

∂∂

=∂∂


x

x

aztRaztE

)21000cos()21000cos(1

ππ

++−=


z=0

yo

yo

aztR

aztH

ˆ)21000cos(1000

12

ˆ)21000cos(1000

121

πµπ

πµπ

+⋅−

−⋅= y

o


122 π

µπ

−⋅=

tDH∂∂

=×∇

( )

( )y

o

yo

aztR

aztDt

ˆ)21000sin(1000

12

ˆ)21000sin(1000

12

2

2

1

πµπ

πµπ

+⋅−

−⋅−=∂∂

( )y

o

aztTDt

ˆ)21000sin(1000

12 2

2 πµπ

−⋅−=∂∂


x

x

aztRaztE

)21000cos()21000cos(1

ππ

++−=


z=0

yo

yo

aztR

aztH

ˆ)21000cos(1000

12

ˆ)21000cos(1000

121

πµπ

πµπ

+⋅−

−⋅= y

o


122 π

µπ

−⋅=

tDH∂∂

=×∇

( )( )

( )( ) y

o

yo

aztR

aztD

ˆ)21000cos(1000

12

ˆ)21000cos(1000

12

2

2

2

2

1

πµπ

πµπ

+⋅+

−⋅=

( )( ) y

o

aztTD ˆ)21000cos(1000

22

2

2 πµπ

−⋅=


z=0

tDH∂∂

=×∇

( )( )

( )( ) y

oo

yoo

aztR

aztE

ˆ)21000cos(1000

12

ˆ)21000cos(1000

12

21

2

21

2

1

πµεε

π

πµεε

π

+⋅+

−⋅=

( )( ) y

oo

aztTE ˆ)21000cos(1000

22

2

2

2 πµεε

π−⋅=

x

x

aztRaztE

)21000cos()21000cos(1

ππ

++−=


yo

yo

aztR

aztH

ˆ)21000cos(1000

12

ˆ)21000cos(1000

121

πµπ

πµπ

+⋅−

−⋅= y

o


122 π

µπ

−⋅=


z=0

tDH∂∂

=×∇

( )( )

( )( ) y

oo

yoo

aztR

aztE

ˆ)21000cos(1000

12

ˆ)21000cos(1000

12

21

2

21

2

1

πµεε

π

πµεε

π

+⋅+

−⋅=

( )( ) y

oo


22

2

2

2 πµεε

π−⋅=

x

x

aztRaztE

)21000cos()21000cos(1

ππ

++−=


yo

yo

aztR

aztH

ˆ)21000cos(1000

12

ˆ)21000cos(1000

121

πµπ

πµπ

+⋅−

−⋅= y

o


122 π

µπ

−⋅=

APPLY BC at Z=0( )

( )( )

( )( )

( ) yoo

yoo

yoo


2ˆ)1000cos(1000

12ˆ)1000cos(1000

122

2

2

21

2

21

2

⋅=⋅+⋅µεε

πµεε

πµεε

π


z=0

tDH∂∂

=×∇

( )( )

( )( ) y

oo

yoo

aztR

aztE

ˆ)21000cos(1000

12

ˆ)21000cos(1000

12

21

2

21

2

1

πµεε

π

πµεε

π

+⋅+

−⋅=

( )( ) y

oo


22

2

2

2 πµεε

π−⋅=

y

o

yo

aztR

aztH

ˆ)21000cos(1000

12

ˆ)21000cos(1000

121

πµπ

πµπ

+⋅−

−⋅=

yo


122 π

µπ

−⋅=

APPLY BC at Z=0( )

( )( )

( )( )

( ) yoo

yoo

yoo


2ˆ)1000cos(1000

12ˆ)1000cos(1000

122

2

2

21

2

21

2

⋅=⋅+⋅µεε

πµεε

πµεε

π

211

11εεεTR =+


APPLY BC at Z=0

211

11εεεTR =+ TR =−1


APPLY BC at Z=0

211


( )21

11εεTR =+ TR

2

11εε

=+

Eq1

Eq2

Eq1+Eq2: T

+=

2

112εε

+

=

2

11

2

εε

T


APPLY BC at Z=0

211


+

=

2

11

2

εε

T

+

−=

2

11

21

εε

R

21

21

εεεε

+−

=R

Example Problem

oµε ,1 oµε ,2

x

x

aztRaztE

)21000cos()21000cos(1

ππ

++−=


z=0

FIND R and T on both sides!

21

21

εεεε

+−

=R

+

=

2

11

2

εε

T

eleg 413 spring 2017 lecture #1 - university of delawaremirotzni/eleg413/eleg413lec1.pdf · eleg...

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