Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves Faraday’s Law...

Electromagnetic Waves

Chapter 1

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

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 !!

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..)

In summary:

Stationary charges electrostatic fields

Steady currents magnetostatic fields

Time varying currents electromagnetic fields or waves

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

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.

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

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

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..)

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

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

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.

Solution to Example 1Solution to Example 1

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

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

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:

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

By chain rule,

By considering uy=dy/dt,

Our emf is negative of this, where:

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:

yind Rayy

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.

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:

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

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

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.

• 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,

So then, we define the

motional electric field Em,

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:

memf ddV LBuLE

• Moving Loop in Time Varying Field

The total emf would be:

memf ddt

dV LBuSB

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.

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

memf ddV LBuLE

Where,zDC dzdd aLL

d loopmoving

1002,4 fcm

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

a-aaB sincos00 BB x

Where B0 = 0.05 , therefore:

0sincos

dzdzBd

cos2.0

cos05.010004.0cos0

mVdzVz

emf cos6cos2.003.0

To determine recall that,

C is constant

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

Hence,

Therefore,

100sin6

2cos6cos6

So that at t = 1 ms,

mV.Vemf 8255)001.0(100sin6

1.3 Displacement Current1.3 Displacement Current

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

‘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,

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

The current continuity equation,

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,

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.

Therefore,

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

SDSJLH

Do you really understand this displacement current??

Only formula and formula…????

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.

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.

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

tdd SJSDSJLH

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

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

1.4 Maxwell Equations1.4 Maxwell Equations

Below is the generalized forms of Maxwell Equations:

Maxwell Equations

Point or Differential

Integral Form

Gauss’s Law

Gauss’s Law for Magnetic Field

Faraday’s Law

Ampere’s CircuitalLaw

encQd SD

0 SB d

SBLE dt

SDSJLH dt

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

Lorent’z Force Equation

Constitutive Relations

Current Continuity Equation

Circuit - Field relations:Circuit - Field relations:

EJc GVVR

Field Relation Circuit Relation

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

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

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.

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

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

ztEztEz

ztEzyx

sincos

Where,

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.

We can apply Ampere’s Circuital Law, to

the propagating magnetic field equation previously.tc

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

Try this!!!

And propagation velocity relation:

puto get

-Solution-

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).

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

To find H,

zt zto y y

e tex y z zte

a a aH

2 210 10, =zt zt

y yo o

td e dt te dt

H a H a

Therefore,

udv uv vdu 2 and .ztu t dv e dt

This integral is solved by parts

where we let

We arrive at:

2 4zt zt

t Ae e

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 :

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.

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 ).

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.

Where,

Initial amplitude at z = 0

and exponential terms attenuation

Angular frequency

Phase constant

Phase shift

And important relations:

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.

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 :

0H , 0E

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,

So now we have :

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)

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-

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

ss jj EE 2s

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 )(

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

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 )(

0)()()()( 2

xsxsxs

022 zxsE-

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

Hence,

With first and second term is zero :

0)(0)()( 2

zExsxs

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

0,022 or

So, this equation, becomes:

Which has two solutions,

eEEAeE

,,,0)2(

,,,0)1(

zzxs eEeEzE 00)(

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

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

xz zteEtz aE cos, 0

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

s eEeE aE 00or generally as :

xz zteEzteEtz aaE coscos, 00

Evaluating the curl of E to find:

s eEeE a-E 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

We would have been led to the expression,

s eHeH aH 00

0EBy making comparison, we can find a relationship between

and , or and

0intrinsic impedance (in ohms)

E 2tan

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

Jloss tangent

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

• Good (lossless or perfect) dielectric if

• Good conductor if

, 1tan

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)

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 :

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

SSP HE

PIn addition to properties of cross product,

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.

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).

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

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

7 212 cos 2 10

30 4 y

Vx t x

Since free space is stated,

and then

Example 5 Example 5

Suppose:

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

Find :

H (x,y,t).

The direction of propagation, ap

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

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,

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

The direction of propagation :

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

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.

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

jjj )(

jjjjj 22)0(

Evaluate the propagation constant,

Where,

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

The propagation velocity,

Evaluate the intrinsic impedance,

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

r 1200

intrinsic impedance of free space

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

883 10

0.75 1016

c x mu x

The propagation velocity:

The angular frequency:

8 80.75 10 10 7.5 10p

radu x x

From given H field :

8( , ) 100cos 7.5 10 10 z

mAy t x t y

So, the time harmonic H field is:

0.100 ,

1200.100 3

j ys z

j y j ys P s y z x

E a H a a aWhere,

8( , ) 9.4cos 7.5 10 10 x

Vy t x t y

Finally, the instantaneous expression for E field is:

• 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.

• Conductor

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

Therefore,

where interior

brackets becomes:

The intrinsic impedance approximated by:

By considering leading to:

and also..

E leads H by 450

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.

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)

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

1 we have:

200, or 10.13

1 10 4 10o

f mx x

45 452 28j je e

We could also find the intrinsic impedance,

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

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)

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:

We have the following sets of equations:

xzjztt

xzjzrr

xzjzii

Incident Fields

Reflected Fields

Transmitted Fields

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

The boundary conditions:

212121

HHKHHa

Applying these boundary conditions at z=0 to get:

Reflection CoefficientTransmissio

n Coefficient

Try this!!

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.

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

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.

The reflection coefficient ;

2 11 2

120 8; 60 , 120 240

This leads to:

240 60 30.60

240 60 5

and the transmission coefficient,

1 1.60

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.

We know that,

j z j zt os y y

EmA mAe e

From transmitted H field, we could find the

transmitted E field,

2t t2 o s

30 , so 12 , E 0.36 , and 1.13t

E mA V Ve

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

Thus, 1( , ) 2.83cos .x

Vz t t z

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.

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.

• TE Polarization

For TE polarization, the fields are summarized as follows:

ztxtzxj

yzxjtt

zrxrzxj

yzxjrr

zixizxj

yzxjii

sincos

cossin

cossin0

cossin

cossin0

cossin

cossin0

Incident Fields

Reflected Fields

Transmitted Fields

By applying Snell’s Law,

tir EEE 0012

120 coscos

coscos

We could get:

it EEE 0021

20 coscos

TETE 1

Try to solve this!!!

• TM Polarization

By similar geometric arguments, TM fields are:

ztxtzxjtt

zrxrzxjrr

zixizxjii

cossin

cossin0

cossin

cossin0

cossin

cossin0

sincos

Incident Fields

Reflected Fields

Transmitted Fields

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

it EEE 0021

20 coscos

itr EEE 0012

120 coscos

coscos

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:

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.

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.

2 100 102.09 , 10.45 .

x rad rad

c x m c m

1 1 1; sin sin 40 ; 7.4

5 5t tr

(b) 1 2

120120 ; 24

cos cos0.732; 1 0.268

cos cosi t

TE TE TEi t

(c) For incident field:

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

Ve e e

( , ) 1cos 1.34 1.60iy

Vz t t x z

For reflected field:

0.732r io TE oE E

1.34 1.600.732r j x j zs y

Leading to:

( , ) 0.732cos 1.34 1.60ry

Vz t t x z

Finally for transmitted field:

0.268t io TE oE E

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,

Vzxttz y

r aE 4.1035.1cos268.0),(

Electromagnetic Waves

Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves Faraday’s Law...

Documents

Transcript of Electromagnetic Waves Chapter 1. 2 Chapter Outlines Chapter 1 Electromagnetic Waves Faraday’s Law...

Chapter 3. Electromagnetic Waves & Electronic … Science II Chapter 3 1 Chapter 3. Electromagnetic Waves & Electronic devices 3 Electromagnetic Waves The electromagnetic field vectors

Chapter 33 Electromagnetic waves Masatsugu Sei Suzuki …bingweb.binghamton.edu/~suzuki/GeneralPhysNote_PDF/LN33v... · 2019-08-08 · Chapter 33 Electromagnetic waves Masatsugu Sei

Chapter 8-Electromagnetic Waves

Chapter Eight ELECTROMAGNETIC WAVES · 2015. 4. 24. · present a descriptive account of electromagnetic waves. The broad spectrum of electromagnetic waves, stretching from γ rays

Electromagnetic Spectrum and Light Chapter 18. Electromagnetic Waves Transverse Waves Transverse Waves Consist of constantly changing fields Consist of.

Electromagnetic Waves Chapter 3 Electromagnetic Spectrum Lesson 3-1 .

Chapter 18 - Unatego 18 notes.doc · Web viewThe Electromagnetic Spectrum and Light Mr. Rosener Section 1: Electromagnetic Waves What Are Electromagnetic Waves? Transverse waves consisting

THE ELECTROMAGNETIC SPECTRUM & LIGHT Chapter 18. What types of waves are electromagnetic waves?

Electromagnetic Waves - Michigan State Universityschwier/courses/2014SpringPhy184/lecture37… · March 26, 2014 Chapter 31 6 The Electromagnetic Spectrum ! Electromagnetic waves

CollegePhysics StudentSolutionsManual Chapter24skyhawkphysics.weebly.com/.../osc_physics_student_sm_ch_24_em_… · chapter 24:"electromagnetic waves" 24.1 maxwell’s$equations:$electromagnetic$waves

CHAPTER 23 Electromagnetic Wavespeople.physics.tamu.edu/adair/phys202/CHAPTER 23 Electromagnetic Waves.pdf1 chapter 23 electromagnetic waves basic concepts propagation of light electromagnetic

Chapter 25 Electromagnetic Waves. Units of Chapter 25 The Production of Electromagnetic Waves The Propagation of Electromagnetic Waves The Electromagnetic.

Chapter 17: Electromagnetic waves

Chapter 32C - - Electromagnetic Waves

CHAPTER 33 ELECTROMAGNETIC WAVES

Chapter 18: The Electromagnetic Spectrum and Light 18.1 Electromagnetic Waves.

Physical Science CHAPTER 23 Waves Waves transmit energy!!! Mechanical Vs. Electromagnetic Waves.

This chapter is the second on electromagnetic waves. We ... · 1 This chapter is the second on electromagnetic waves. We begin with a discussion of electromagnetic waves traversing

Chapter 24 Electromagnetic Waves. Electromagnetic Waves, Introduction Electromagnetic (em) waves permeate our environment EM waves can propagate through.

Chapter Eight ELECTROMAGNETIC WAVES · 2020-04-01 · present a descriptive account of electromagnetic waves. The broad spectrum of electromagnetic waves, stretching from γ rays