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Electromagnetic Waves Chapte r 1

Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves Faraday’s Law Transformer and Motional EMFs Displacement Current

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Page 1: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Electromagnetic Waves

Chapter 1

Chapter 1

Page 2: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

2

Chapter OutlinesChapter OutlinesChapter 1 Electromagnetic Waves

Faraday’s Law Transformer and Motional EMFs Displacement Current Maxwell’s Equations Lossless TEM Waves EM Wave Fundamental and Equations EM Wave Propagation in Different Media EM Wave Reflection and Transmission at Normal or

Oblique Incidence

Page 3: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

3

IntroductionIntroduction

In your previous experience in studying electromagnetic, you have learned about and experimented with electrostatics and magnetostatics … concentrating on static, or time invariant electromagnetic fields (EM Fields).

Henceforth, we shall examine situations where electric and magnetic fields are dynamic or time varying !!

Page 4: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

4

Where :

• In static EM Fields, electric and magnetic fields

are independent each other, but in dynamic field

both are interdependent.

• Time varying EM Fields, represented by E(x,y,z,t)

and H(x,y,z,t) are of more practical value than

static EM Fields.

• In time varying fields, it usually due to

accelerated charges or time varying currents.

Introduction (Cont’d..)Introduction (Cont’d..)

Page 5: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

5

In summary:

Stationary charges electrostatic fields

Steady currents magnetostatic fields

Time varying currents electromagnetic fields or waves

Introduction (Cont’d..)Introduction (Cont’d..)

Page 6: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

1.1 Faraday’s Law1.1 Faraday’s Law

According to faraday’s experiment, a static magnetic field produces no current flow, but a time varying field produces an induced voltage called electromotive force or emf in a closed circuit, which causes a flow of current.

Faraday’s Law – the induced emf, Vemf in volts, in

any closed circuit is equal to the time rate of change of the magnetic flux linkage by the circuit.

Page 7: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Faraday’s Law (Cont’d..)Faraday’s Law (Cont’d..)

tVemf

Where,

The negative sign is a consequence of Len’z Law. If we consider a single loop, Faraday’s Law can be written as:

SB dtt

Vemf

Page 8: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

An increasing magnetic field out of the page induces a current in (a) or an emf in (b). (c) The distributed resistance in a continuous conductive loop can be modeled as lumped resistor Rdist in series

with a perfectly conductive loop.

Faraday’s Law (Cont’d..)Faraday’s Law (Cont’d..)

Page 9: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Faraday’s Law (Cont’d..)Faraday’s Law (Cont’d..)

Generating emf requires a time varying magnetic flux linking the circuit. This occurs if the magnetic field changes with time ‘transformer emf’’ or if the surface containing the flux changes with time ‘motional emf’.The emf is measured around the closed path enclosing the area through which the flux is passing, can be written as:

SBLE dt

dVemf

It is clear that in time varying situation, both E and B are present and interrelated.

Page 10: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

10

Example 1Example 1

Consider the rectangular loop moving with velocity

u=uyay in the field from an infinite length line current

on the z axis. Assume the loop has a distributed

resistance Rdist. Find an expression for the current in

the loop including its direction.

Page 11: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 1Solution to Example 1

First calculate the flux through the loop at an instant time,

aB2

0I aaa l

Where, la unit vector along the line current

a unit vector perpendicular from the line current to the field point

Remember ?

So,xyzl aaaaaa

xy

IIaaB

2200

Page 12: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 1 (Cont’d..)Solution to Example 1 (Cont’d..)

Arbitrarily choose dS in the +ax direction,

xdydzd aS So the flux can be easily calculated as:

yayIb

dzy

dyI

dydzy

Id

ay

y

b

lnln2

2

2

0

0

0

0

SB

Page 13: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 1 (Cont’d..)Solution to Example 1 (Cont’d..)

Then, we want to find how this flux changes with time,

yaydt

dIb

dt

dlnln

20

dt

dy

yay

Ib

dt

d

11

20

By chain rule,

By considering uy=dy/dt,

ayy

Iabu

dt

d y

20

Page 14: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Our emf is negative of this, where:

Solution to Example 1 (Cont’d..)Solution to Example 1 (Cont’d..)

ayy

IabuV

yemf

20

Since we considered dS in the +ax direction, our emf is

taken counterclockwise circulation. But since the emf is negative, our induced current is apparently going in the clockwise direction with value of:

dist

yind Rayy

IabuI

20

Page 15: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

1.2 Transformer and motional emf1.2 Transformer and motional emf

The variation of flux with time as in previous equation maybe caused in three ways:

By having a stationary loop in a time varying B field.

(transformer emf)

By having a time varying loop area in a static B field.

(motional emf)

By having a time varying loop area in a time varying

B field.

Page 16: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Transformer and motional emf (Cont’d..)Transformer and motional emf (Cont’d..)

• Stationary Loop in Time Varying B Field

This is the case where a stationary conducting loop is in a time varying magnetic B field. The equation becomes:

SB

LE dt

dVemf

By applying Stokes Theorem in the middle term, we obtain:

SB

SE dt

dVemf

Page 17: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Transformer and motional emf (Cont’d..)Transformer and motional emf (Cont’d..)

This leads us to the point or differential form of Faraday’s Law,

t

B

E

Based on this equation, the time varying electric field is not conservative, or not equal to zero. The work done in taking a charge about a closed path in a time varying electric field, for example, is due to the energy from the time varying magnetic field.

Page 18: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Transformer and motional emf (Cont’d..)Transformer and motional emf (Cont’d..)

• Moving Loop in static B Field

When a conducting loop is moving in a static B field, an emf is induced in the loop. Recall that the force on a charge moving with uniform velocity in magnetic field,

BuQFm

BuQ

FE m

m

So then, we define the

motional electric field Em,

Page 19: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Transformer and motional emf (Cont’d..)Transformer and motional emf (Cont’d..)

If we consider a conducting loop moving with uniform velocity u as consisting of a large number of free electrons, the emf induced in the loop is:

L

memf ddV LBuLE

• Moving Loop in Time Varying Field

The total emf would be:

LS

memf ddt

dV LBuSB

LE

Page 20: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

20

Example 2Example 2

The loop shown is inside a uniform magnetic field

B = 50 ax mWb/m2 . If side DC of the loop cuts

the flux lines at the frequency of 50Hz and the loop

lies in the yz plane at time t = 0, find the induced

emf at t = 1 ms.

Page 21: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 2Solution to Example 2

Since the B field is time invariant, the induced emf is motional, that is:

L

memf ddV LBuLE

Where,zDC dzdd aLL

aa

Lu

dt

d

dt

d loopmoving

1002,4 fcm

Page 22: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

As u and dL is in cylindrical coordinates, transform B field into cylindrical coordinate (Chapter 1 in Electromagnetic Theory !! ):

Solution to Example 2 (Cont’d..)Solution to Example 2 (Cont’d..)

a-aaB sincos00 BB x

Where B0 = 0.05 , therefore:

z

z

B

BB

a

aaa

Bu

cos

0sincos

00 0

00

Page 23: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

And

dz

dzdzBd

cos2.0

cos05.010004.0cos0

LBu

Solution to Example 2 (Cont’d..)Solution to Example 2 (Cont’d..)

mVdzVz

emf cos6cos2.003.0

0

To determine recall that,

Ctdtd

C is constant

at because the loop is in the yz plane! 2,0 t

Page 24: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 2 (Cont’d..)Solution to Example 2 (Cont’d..)

Hence,

2

t

Therefore,

mVt

tVemf

100sin6

2cos6cos6

So that at t = 1 ms,

mV.Vemf 8255)001.0(100sin6

Page 25: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

1.3 Displacement Current1.3 Displacement Current

We recall from Ampere’s Circuital Law for static field,

cJH

EJ c

‘c’ subscript is used to identify it as a conduction current density, which related to electric field Ohm’s Law by:

But divergence of curl of a vector is identically zero,

JH 0

Page 26: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Displacement Current (Cont’d..)Displacement Current (Cont’d..)

The current continuity equation,

tc

vJ

dc JJH

We see that the static form of Ampere’s Law is clearly invalid for time varying fields since it violates the law of current continuity, and it was resolved by Maxwell introduction which what we called displacement current density,

Where Jd is the rate of

change of the electric flux density,

td

D

J

Page 27: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Displacement Current (Cont’d..)Displacement Current (Cont’d..)

The insertion of Jd was one of the major contribution

of Maxwell. Without Jd term, electromagnetic wave

propagation (e.g. radio or TV waves) would be

impossible. At low frequencies, Jd is usually

neglected compared with Jc. But at radio

frequencies, the two terms are comparable.

tc

D

JH

Therefore,

Page 28: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Displacement Current (Cont’d..)Displacement Current (Cont’d..)

By applying the divergence of curl, rearrange, integrate and apply Stoke’s Theorem, we can get the integral form of Ampere’s circuital Law:

dcc iidt

dd

SDSJLH

Do you really understand this displacement current??

Only formula and formula…????

Page 29: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Displacement Current (Cont’d..)Displacement Current (Cont’d..)

To have clear understanding of displacement current, consider the simple capacitor circuit of figure below.

A sinusoidal voltage source is applied to the capacitor, and from circuit theory we know the voltage is related to the current by the capacitance. i(t) here is the conduction current.

Page 30: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Displacement Current (Cont’d..)Displacement Current (Cont’d..)

Consider the loop surrounding the plane surface S1. By

static form of Ampere’s Law, the circulation of H must be equal to the current that cuts through the surface. But,

the same current must pass through S2 that passes

between the plates of capacitor.

Page 31: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Displacement Current (Cont’d..)Displacement Current (Cont’d..)

But, there is no conduction current passes through an ideal capacitor, (where J=0, due to σ=0 for an ideal

dielectric ) flows through S2. This is contradictory in view

of the fact that the same closed path as S1 is used.

But to resolve this conflict, the current passing through S2

must be entirely a displacement current, where it needs to be included in Ampere’s Circuital Law.

122 SSS

d dIt

Qd

tdd SJSDSJLH

So we obtain the same current for either surface though it is conduction current in S1 and displacement current in S2.

Page 32: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Displacement Current (Cont’d..)Displacement Current (Cont’d..)

The ratio of conduction current magnitude to the displacement current magnitude is called loss tangent, where it is used to measure the quality of the dielectric good dielectric will have very low loss tangent.

Other example for physical meaning:

Ji = current sourceJc= conducted current through resistorJd=displacement current through dielectric material

mi= magnetic current sourcemd=displacement magnetic current

Page 33: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

1.4 Maxwell Equations1.4 Maxwell Equations

Below is the generalized forms of Maxwell Equations:

Maxwell Equations

Point or Differential

Form

Integral Form

Gauss’s Law

Gauss’s Law for Magnetic Field

Faraday’s Law

Ampere’s CircuitalLaw

v D

0 B

t

B

E

tc

D

JH

encQd SD

0 SB d

SBLE dt

d

SDSJLH dt

dd c

Page 34: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Maxwell Equations (Cont’d..)Maxwell Equations (Cont’d..)

It is worthwhile to mention other equations that go hand in hand with Maxwell’s equations.

BuEF q

EJ

HB

ED

tv

J

Lorent’z Force Equation

Constitutive Relations

Current Continuity Equation

Page 35: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Circuit - Field relations:Circuit - Field relations:

EJc GVVR

i 1

HB iL

ED

t

iLVL

dt

Hmd

VCqe

t

VCi c

c

dt

Ee

Field Relation Circuit Relation

Maxwell Equations (Cont’d..)Maxwell Equations (Cont’d..)

Page 36: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

1.5 Lossless TEM Waves1.5 Lossless TEM Waves

Let’s use Maxwell’s equations to study the relationship between the electric and magnetic field components of an electromagnetic wave.

Consider an x-polarized wave propagating in the +z direction in some ideal medium characterized by µ and ε, with σ = 0.

Electromagnetic wave as x-polarized means that the E field vector is always pointing in the x or –x direction. Choose σ = 0 to make medium lossless for simplicity, as given by:

xztEtz aE cos, 0

Page 37: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Lossless TEM Waves (Cont’d..)Lossless TEM Waves (Cont’d..)

A plot of the equation E(z,0) = E0cos(z)ax at 10 MHz

in free space with E0 = 1 V/m.

Page 38: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Lossless TEM Waves (Cont’d..)Lossless TEM Waves (Cont’d..)

Upon application of Maxwell equations, we would also find the magnetic field propagates in the +z direction, but the field is always normal or perpendicular to the electric field vector the wave is said to propagate in a transverse electromagnetic wave mode, or TEM.

TEM Waves has no E field or H field components along the direction of propagation.We can apply Faraday’s Law, to

the propagating electric field equation previously.tt

HB

E

Solve the right hand side and the left hand side to get H !!

Page 39: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Lossless TEM Waves (Cont’d..)Lossless TEM Waves (Cont’d..)

yy

zyx

ztEztEz

ztEzyx

aa

aaa

E

sincos

00cos

00

0

dtztE

d

ztEt

y

y

aH

aH

sin

sin

0

0

Where,

CztE

y aH

cos0

Page 40: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

We could see that the time varying E is the only source of H, if no conduction current given that can also generate H Thus C must be zero.

Plot of the equation

H(z,0) = (E0/) cos(–

z)ay at 10 MHz in free

space with

E0 = 1 V/m along with

the lighter plot of

E(z,0).

The amplitudes of E

and H are related by

Maxwell equations.

Lossless TEM Waves (Cont’d..)Lossless TEM Waves (Cont’d..)

Page 41: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Lossless TEM Waves (Cont’d..)Lossless TEM Waves (Cont’d..)

We can apply Ampere’s Circuital Law, to

the propagating magnetic field equation previously.tc

DJH

By considering lossless characteristic, taking the curl of H, equating the equation and then integrate it, we could get :

Try this!!!

pu

And propagation velocity relation:

1

puto get

-Solution-

Page 42: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

42

Example 3Example 3

Suppose in free space that:

E(z,t) = 5.0 e-2zt ax V/m.

Is the wave lossless?

Find H(z,t).

Page 43: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 3Solution to Example 3

Since the wave has an attenuation term (e-2zt) it is clearly not lossless.

To find H,

2 2

2

5 10

5 0 0

x y z

zt zto y y

zt

e tex y z zte

a a aH

E a a

2 210 10, =zt zt

y yo o

td e dt te dt

H a H a

Therefore,

Page 44: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 3 (Cont’d..)Solution to Example 3 (Cont’d..)

udv uv vdu 2 and .ztu t dv e dt

This integral is solved by parts

where we let

We arrive at:

2 22

10 10

2 4zt zt

yo o

t Ae e

z z m

H a

Page 45: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

1.6 EM Wave Fundamental and equations1.6 EM Wave Fundamental and equations

In free space, the constitutive parameters are σ = 0, µr = 1, εr = 1, so the Ampere’s Law and Faraday’s

Law equations become :

tt

H

E B

E 0

ttc

E

H D

JH 0

If there is some point in space a source of time varying E field, a H field is induced in the surrounding region. As this H field also changing with time, it in turn induces an E field. Energy is pass back and forth between E and H fields as they radiate away from the source at the speed of light.

Page 46: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

The EM waves radiates spherically, but at a remote distance away from the source they resemble uniform plane wave. In a uniform plane wave, the E and H fields are orthogonal, or transverse to the direction of propagation ( to propagate in TEM mode ).

Page 47: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

We will briefly review some of the fundamental features of waves before employing them in the study of electromagnetic. Consider time harmonic waves, represented by sine waves, rather than transient waves (pulses or step functions), generally as:

xz zteEtz aE cos, 0

The E field is a function of position (z) and time (t). It is always pointing in +x or –x direction x-polarized wave.

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

Page 48: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

Where,

0

0

E

eE z

fT

12

f 2

Initial amplitude at z = 0

and exponential terms attenuation

Angular frequency

Phase constant

Phase shift

And important relations:

2

fdt

dzu p

Page 49: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

E(0,t) = Exax = E0cos(t)ax. E(0,t) =Exax = E0cos(t + )ax.

E(z, 0) = E0cos(–z)ax. E(z, 0) = E0e–zcos (–z)ax.

Page 50: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

Use Maxwell’s equations to derive formulas governing EM wave propagation.

Consider that the medium is free of any charge, where:

And linear, isotropic, homogeneous, and time invariant (simple media), whereby the Maxwell’s equation can be rewritten as :

0D

0H , 0E

HE ,

EEH

tt

Page 51: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

Take curl of both sides of Faraday’s Law,

Consider position derivative acting on a time derivative in a homogeneous material,

Exchange the Faraday’s Law for the curl of H

Invoking a vector identity,

t

HE

HE

t

2

2

tttt

EEE

EE

AAA 2

Page 52: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

So now we have :

2

22

tt

EE

EE

2

22

tt

EE

E

But, medium is charge free, so divergence of E is zero,

Helmholtz Wave Equation for E.

This can be broken up into three vector equations ( Ex, Ey and Ez)

Page 53: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

53

Task 1 Task 1

By starting with Maxwell’s equations for

simple and charge free media, derive the

Helmholtz Wave Equation for magnetic

field, H

Solve this! Please submit the solution after class!

-Solution-

Page 54: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

But, with interest on Helmholtz equations for time harmonic fields, the previous Helmholtz wave equation becomes :

ss jj EE 2s

s jt

EE

022 ss EE

because

Generally written in the form:

Where the propagation constant, γ, with real part is attenuation and an imaginary part is phase constant,

jjj )(

Page 55: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

The value of and β in terms of the material’s constitutive parameters:

112

112

2

2

Page 56: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

The Helmholtz wave equation for time harmonic magnetic fields, H :

022 ss HH

With loss of generality, we assume that x polarized traveling in the +az direction, where Es has only an x

component with function of z, since for plane wave that the fields do not vary in transverse direction, where in this case is xy plane. Then :

xxss zE aE )(

Page 57: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

0)()()()( 2

2

2

2

2

2

2

zE

z

zE

y

zE

x

zExs

xsxsxs

022 zxsE-

By substituting this equation into the Helmholtz wave equation for E field, it becomes:

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

Hence,

With first and second term is zero :

0)(0)()( 2

2

22

2

2

zE

zzE

z

zExsxs

xs

Page 58: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

This is a scalar wave equation, a linear homogeneous differential equation. A possible solution for this equation is :

If we let, then, zsx AeE zsxzsx Ae

z

EAe

z

E 22

2,

0,022 or

So, this equation, becomes:

Which has two solutions,

zsx

zsx

zsx

zsx

eEEAeE

eEEAeE

0

0

,,,0)2(

,,,0)1(

or

or

0)(22

2

zE

zxs

Page 59: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

zzxs eEeEzE 00)(

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

The general solution is linear superposition of these two:

For the represents the E field amplitude of +z

traveling wave at z=0, by reinserting the vector, multiply

by time factor apply Euler’s identity and use the

propagation constant to get :

xz zteEtz aE cos, 0

tje

0E

Page 60: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

For the represents the E field amplitude of negative z traveling wave at z=0, and similarly by previous approach to get :

0E

xz zteEtz aE cos, 0

The general instantaneous solution is superposition of these two solutions above.

xzz

s eEeE aE 00or generally as :

xz

xz zteEzteEtz aaE coscos, 00

Page 61: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

Evaluating the curl of E to find:

yzz

s eEeE a-E 00

yzz

s ej

Ee

j

EaH

00

We can solve for H ,

If we assume that the wave propagates along +az and Hs has

only an y component, as what we had assume for Es yyss zH aH )(

The magnetic field can be found by applying Faraday’s Law :

ss j HE

Page 62: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

We would have been led to the expression,

yzz

s eHeH aH 00

0H

0EBy making comparison, we can find a relationship between

and , or and

0E

0H

j

H

E

0

0intrinsic impedance (in ohms)

njn e

j

j

0

0

H

E 2tan

Page 63: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

By plugged in the equation of intrinsic impedance into

equation of magnetic field, we could get:

ynz zte

Etz aH

cos, 0

Where E and H are out of phase by θn at any instant of time

due to the complex intrinsic impedance of the medium. Thus

at any time, E leads H or H lags E by θn . The ratio of

magnitude of conduction current density to displacement

current density in a lossy medium is:

tanjj s

s

d

c

E

E

J

Jloss tangent

Page 64: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

Tan δ is used to determine how lossy a medium is, i.e.

• Good (lossless or perfect) dielectric if

• Good conductor if

,1tan

, 1tan

Page 65: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

Basically, we can use Fleming’s Left Hand Rule to

determine the E, H and propagation direction ;

Propagation direction (first finger)

H direction (second finger)

E direction (thumb)

Page 66: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

By knowing the EM wave’s direction of propagation,

given as unit vector ap, is the same as the cross

product of Es with unit vector, aE and Hs with unit

vector aH :

SPS

SPS

HaE

EaH

1

And also, a pair of simple formulas can be derived:

SS

SSP HE

HEa

HE

EH

HE

aaa

aaa

aaa

P

P

PIn addition to properties of cross product,

Page 67: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

Representation of waves; in (a), the wave travels in the ap

= +az direction and has Es = E0+ e–z ax and Hs = (E0

+/)e–z

ay. In (b), the wave travels in the ap = –az direction and has

Es = E0–ez ax along with Hs = –(E0

–/)ez ay.

Page 68: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

68

Example 4 Example 4

Suppose in free space :

H(x,t) = 100 cos(2π x 107t – βx + π/4) az

mA/m. Find E(x,t).

Page 69: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 4 Solution to Example 4

0.100 , , 4

120 0.100 12

j x js z P x

j x j j x js P s x z y

e e

e e e e

H a a a

E a H a a a

We could find:

12 cos yt x E a

So then,

2 22 30 rad m

c f

7 212 cos 2 10

30 4 y

Vx t x

m

E a

Since free space is stated,

and then

Page 70: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

70

Example 5 Example 5

Suppose:

E (x,y,t) = 5 cos(π x 106t – 3.0x + 2.0y) az

V/m.

Find :

H (x,y,t).

The direction of propagation, ap

Page 71: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 5 Solution to Example 5

We could find:

3 25 j x j ys ze eE a

Assume nonmagnetic material and therefore have:

3 2 3 210 15j x j y j x j ys s x yj j e e j e e E H a a

3 2 3 2 3 2 3 210 152.53 3.8j x j y j x j y j x j y j x j y

s x y x yo

j je e e e e e e e

j j

H a a a a

6 6 A( , , ) 2.53cos 10 3 2 3.80cos 10 3 2

mx yx y t x t x y x t x y H a a

So that,

Page 72: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 5 (Cont’d..) Solution to Example 5 (Cont’d..)

The direction of propagation :

s sP

s s

E H

aE H

6 4 6 419 12.65j x j y j x j ys s x ye e e e E H a a

Where,

And with the exponential terms canceling in the top

and bottom of the equation for ap, we have:

0.83 0.55 .P x y a a a

Page 73: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

1.7 EM Wave Propagation In Different Media

1.7 EM Wave Propagation In Different Media

We will consider time harmonic field propagating in different types of media :-

Lossless, Charge-Free

Dielectrics

Conductors

• Lossless, Charge - Free

Charge free, ρv=0, medium has zero conductivity, σ=0.

This is the case where waves traveling in vacuum or free space (free of any charges). Perfect dielectric is also considered as lossless media.

Page 74: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Propagation In Different Media (Cont’d..)EM Wave Propagation In Different Media (Cont’d..)

jjj )(

jjjjj 22)0(

, 0

Evaluate the propagation constant,

So,

Where,

Since , the signal does not attenuate as it travels lossless medium.

0

The propagation velocity,

1pu

Page 75: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Propagation In Different Media (Cont’d..)EM Wave Propagation In Different Media (Cont’d..)

Evaluate the intrinsic impedance,

j

j

0

For lossless materials, E and H are always in phase. Again,

00

0

r

r

r

r 1200

intrinsic impedance of free space

Page 76: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

76

In a lossless, nonmagnetic material with :

εr = 16, and H = 100 cos(ωt – 10y) az mA/m.

Determine : The propagation velocity The angular frequency The instantaneous expression for the

electric field intensity.

Example 6 Example 6

Page 77: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 6 Solution to Example 6

883 10

0.75 1016

p

r

c x mu x

s

The propagation velocity:

The angular frequency:

8 80.75 10 10 7.5 10p

radu x x

s

From given H field :

8( , ) 100cos 7.5 10 10 z

mAy t x t y

m H a

Page 78: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

So, the time harmonic H field is:

Solution to Example 6 (Cont’d..)Solution to Example 6 (Cont’d..)

0.100 ,

1200.100 3

j ys z

j y j ys P s y z x

r

e

e e

H a

E a H a a aWhere,

8( , ) 9.4cos 7.5 10 10 x

Vy t x t y

m E a

Finally, the instantaneous expression for E field is:

Page 79: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

• Dielectric

Treating a dielectric as lossless is often a good approximation, but all dielectrics are to some degree lossy finite conductivity, polarization loss etc. With finite conductivity, the E field gives rise to conduction current density results in power dissipation.

Thus, it will give a complex permittivity, complex propagation constant with attenuation constant greater than zero. The intrinsic impedance is also complex, resulting a phase difference between E and H fields.

EM Wave Propagation In Different Media (Cont’d..)EM Wave Propagation In Different Media (Cont’d..)

Page 80: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Propagation In Different Media (Cont’d..)EM Wave Propagation In Different Media (Cont’d..)

• Conductor

In any decent conductor, the loss tangent, σ/ωε>>1 or σ>>ωε so that σ ≈ ∞, so that:

11

2,

2

f2

Therefore,

where interior

brackets becomes:

Page 81: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Propagation In Different Media (Cont’d..)EM Wave Propagation In Different Media (Cont’d..)

The intrinsic impedance approximated by:

j

j

j

2

1 jj

045

2)1(

2jej

2

pu

f2

By considering leading to:

and also..

E leads H by 450

Page 82: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Propagation In Different Media (Cont’d..)EM Wave Propagation In Different Media (Cont’d..)

A wave from air (free space) penetrates rapidly in a good conductor, with wavelength clearly much shorter.

In a good conductor, the large attenuation means the

penetration depth can be quite small, confining the fields

near the surface or skin, of the conductor skin depth.

A large attenuation means the fields cannot penetrate far into the conductor.

f

11

Page 83: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

83

Example 7 Example 7

In a nonmagnetic material,

E(z,t) = 10 e-200z cos(2π x 109t - 200z) ax mV/m.

Find H(z,t)

Page 84: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 7 Solution to Example 7

Since = β, the media is a good metal, with µr =

1 we have:

22

9 7

200, or 10.13

1 10 4 10o

o

Sf

f mx x

45 452 28j je e

We could also find the intrinsic impedance,

Page 85: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 7 (Cont’d)Solution to Example 7 (Cont’d)

So, to calculate H,

1 1 1010 , 10z j z z j z z j z

s x s P s z x ye e e e e e

E a H a E a a a

The instantaneous expression for the magnetic field intensity.

200 9( , ) 360 cos 2 10 200 45zy

mAz t e x t z

m H a

Page 86: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

1.8 EM Wave Reflection & Transmission at Normal Incidence1.8 EM Wave Reflection & Transmission at Normal Incidence

What happens when a EM wave is incident on a different medium? E.g. Light wave incident with mirror, most of it gets reflected but a portion gets transmitted (rapidly attenuating in the silver backing of the mirror.Consider a plane wave that are normally incident which means the planar boundary separating the two media is perpendicular to the wave’s propagation direction.

Generally, consider a time harmonic x-polarized electric field incident from medium 1 (µr1, εr1, σr1) to

medium 2 (µr2, εr2, σr2)

Page 87: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Reflection & Transmission at Normal Incidence (Cont’d..)EM Wave Reflection & Transmission at Normal Incidence (Cont’d..)

xzii zteEtz aE 10 cos),( 1

With incident field:

Page 88: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Reflection & Transmission at Normal Incidence (Cont’d..)EM Wave Reflection & Transmission at Normal Incidence (Cont’d..)

We have the following sets of equations:

yzjz

tts

xzjztt

s

yzjz

rrs

xzjzrr

s

yzjz

iis

xzjzii

s

eeE

eeE

eeE

eeE

eeE

eeE

aH

aE

aH

aE

aH

aE

22

22

11

11

11

11

2

0

0

1

0

0

1

0

0

Incident Fields

Reflected Fields

Transmitted Fields

tri EEE 000 ,, The E field intensities at z=0

Page 89: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

The boundary conditions:

212121

21

tt

tt

HHKHHa

EE

i

riir

E

EEEE

0

0

12

1200

12

120 ,

Applying these boundary conditions at z=0 to get:

i

tiit

E

EEEE

0

0

12

200

12

20

2,

2

Reflection CoefficientTransmissio

n Coefficient

EM Wave Reflection & Transmission at Normal Incidence (Cont’d..)EM Wave Reflection & Transmission at Normal Incidence (Cont’d..)

Try this!!

Page 90: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Reflection & Transmission at Normal Incidence (Cont’d..)EM Wave Reflection & Transmission at Normal Incidence (Cont’d..)

By comparison, 1

Consider a special case when medium 1 is a perfect dielectric (lossless,σ1=0) and medium 2 is a perfect

conductor (σ2= ∞). For this case, showing that the wave is totally reflected fields in perfect conductor

must vanish, so there can be no transmitted wave, E2 = 0.

0 ,1 ,02

The totally reflected wave combines with the incident wave to form a standing wave it stands and does not

travel, it consists of two traveling waves Ei and Er of

equal amplitudes but in opposite directions.

Page 91: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Reflection & Transmission at Normal Incidence (Cont’d..)EM Wave Reflection & Transmission at Normal Incidence (Cont’d..)

Standing wave pattern for an incident wave in a lossless medium reflecting off a second medium at z=0 where = 0.5.

1

1

min

max

E

ESWR

Page 92: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

92

Example 8 Example 8

A uniform planar waves is normally incident

from media 1 (z < 0, σ = 0, µr = 1.0, εr = 4.0)

to media 2 (z > 0, σ = 0, µr = 8.0, εr = 2.0).

Calculate the reflection and transmission

coefficients seen by this wave.

Page 93: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 8 Solution to Example 8

The reflection coefficient ;

2 11 2

2 1

120 8; 60 , 120 240

24

This leads to:

240 60 30.60

240 60 5

and the transmission coefficient,

1 1.60

Page 94: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

94

Example 9 Example 9

Suppose media 1 (z < 0) is air and media 2 (z

> 0) has εr = 16. The transmitted magnetic

field intensity is known to be:

Ht = 12 cos (ωt - β2z) ay mA/m.

Determine the instantaneous value of the

incident electric field.

Page 95: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 9 Solution to Example 9

We know that,

2 2

2

12t

j z j zt os y y

EmA mAe e

m m

H a a

From transmitted H field, we could find the

transmitted E field,

2t t2 o s

2

30 , so 12 , E 0.36 , and 1.13t

j zox

E mA V Ve

m m m

E a

Page 96: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

2 1

2 1

3 21 ; , 1

5 5t i io o oE E E

Since we know the relation between transmitted E

field and incident E field,

12.83, so 2.83t

j zi ioo s x

EE e

E a

Thus, 1( , ) 2.83cos .x

Vz t t z

m E a

Solution to Example 9 (Cont’d..)Solution to Example 9 (Cont’d..)

Page 97: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

1.9 EM Wave Reflection & Transmission at Oblique Incidence1.9 EM Wave Reflection & Transmission at Oblique Incidence

A uniform plane waves traveling in the ai

direction is obliquely incident from medium 1 onto medium 2.

Plane of incidence plane containing both a normal to the boundary and the incident’s wave propagation.In figure, the propagation direction is ai and the normal is az,

so the plane incidence is the x z plane. The angle of incidence, reflection and transmission is the angle that makes the field a normal to the boundary.

Page 98: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)

When EM Wave in plane wave form obliquely incident on the boundary, it can be decomposed into:

Perpendicular Polarization, or transverse electric (TE) polarization The E Field is perpendicular or transverse to the plane of incidence.

Parallel Polarization, or transverse magnetic (TM) polarization The E Field is parallel to the plane of incidence, but the H Field is transverse.

We need to decompose into its TE and TM components separately, and once the reflected an the transmitted fields for each polarization determined, it can be recombined for final answer.

Page 99: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)

• TE Polarization

Page 100: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)

For TE polarization, the fields are summarized as follows:

ztxtzxj

tts

yzxjtt

s

zrxrzxj

rrs

yzxjrr

s

zixizxj

iis

yzxjii

s

tt

tt

rr

rr

ii

ii

eE

eE

eE

eE

eE

eE

aaH

aE

aaH

aE

aaH

aE

sincos

sincos

sincos

cossin

2

0

cossin0

cossin

1

0

cossin0

cossin

1

0

cossin0

2

2

1

1

1

1

Incident Fields

Reflected Fields

Transmitted Fields

Page 101: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)

By applying Snell’s Law,

ri i

t

sin

sin

2

1

iTE

i

ti

tir EEE 0012

120 coscos

coscos

We could get:

iTE

i

it

it EEE 0021

20 coscos

cos2

TETE 1

Try to solve this!!!

Page 102: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

• TM Polarization

EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)

Page 103: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)

By similar geometric arguments, TM fields are:

yzxj

tts

ztxtzxjtt

s

yzxj

rrs

zrxrzxjrr

s

yzxj

iis

zixizxjii

s

tt

tt

rr

rr

ii

ii

eE

eE

eE

eE

eE

eE

aH

aaE

aH

aaE

aH

aaE

cossin

2

0

cossin0

cossin

1

0

cossin0

cossin

1

0

cossin0

2

2

1

1

1

1

sincos

sincos

sincos

Incident Fields

Reflected Fields

Transmitted Fields

Page 104: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)

By applying Snell’s Law and employing the boundary conditions, we could get the following expressions relating the field amplitudes:

iTM

i

ti

it EEE 0021

20 coscos

cos2

iTM

i

it

itr EEE 0012

120 coscos

coscos

t

iTMTM

cos

cos1

Page 105: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)

For TM polarizations, there exists an incidence angle

at which all of the wave is transmitted into the second

medium Brewster Angle, θi = θBA , where:

22

21

21

22

21

22

22 )(

sin

BA

2

11

1sin

r

rBA

When a randomly polarized wave such as light is incident on a material at the Brewster angle, the TM polarized portion is totally transmitted but a TE component is partially reflected.

Page 106: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

106

Example 10 Example 10

A 100 MHz TE polarized wave with amplitude 1.0 V/m is obliquely incident from air (z < 0) onto a slab of lossless,

nonmagnetic material with εr = 25 (z > 0).

The angle of incidence is 40. Calculate:(a) the angle of transmission, (b) the reflection and transmission

coefficients,(c) the incident, reflected and transmitted for

E fields.

Page 107: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 10 Solution to Example 10

(a) 6

1 28

2 100 102.09 , 10.45 .

3 10r

x rad rad

c x m c m

1

2 2

1 1 1; sin sin 40 ; 7.4

5 5t tr

(b) 1 2

120120 ; 24

25

2 1

2 1

cos cos0.732; 1 0.268

cos cosi t

TE TE TEi t

Page 108: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Solution to Example 10 (Cont’d)Solution to Example 10 (Cont’d)

(c) For incident field:

2.09 sin 40 cos40 1.34 1.601 1j x zi j x j z

s y y

Ve e e

m

E a a

( , ) 1cos 1.34 1.60iy

Vz t t x z

m E a

Thus,

For reflected field:

0.732r io TE oE E

Page 109: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

1.34 1.600.732r j x j zs y

Ve e

m E a

Leading to:

Thus,

( , ) 0.732cos 1.34 1.60ry

Vz t t x z

m E a

Solution to Example 10 (Cont’d)Solution to Example 10 (Cont’d)

Finally for transmitted field:

0.268t io TE oE E

Page 110: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

2 sin cos 1.35 10.40.268 0.268t tj x zt j x j zs y y

Ve e e

m E a a

To get:

Therefore,

m

Vzxttz y

r aE 4.1035.1cos268.0),(

Solution to Example 10 (Cont’d)Solution to Example 10 (Cont’d)

Page 111: Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves  Faraday’s Law  Transformer and Motional EMFs  Displacement Current

Electromagnetic Waves

Electromagnetic Waves

End