Chapter S34

8/7/2019 Chapter S34

1/40

General form of Faradays Lawb

b aba b a

a

U UV V V E ds

q

! | !

r rg

So the electromotive force around a closed path is:

E ds!I r rgAnd Faradays Law becomes:

BdE ds dt*! ! I r

rgA changing magnetic flux produces an electric field.

This electric field is necessarily non-conservative.


2/40

Eproduced by changing B

dd

dt

*! ! I

rrg l

2d B

E 2 r r dt

T ! Trr

d BrE2 dt

!

r

r

How about outside ro ?


3/40

Problems with Amperes Law

o enclCd I ! Q

rrl

o2 r I! Qr

oI2 rQ!

Tr


4/40

But what if..

o enclCB d I ! Q

rrl

oB 2 r 0T ! Qr

B 0!r


5/40

Maxwells correction to Amperes Law

Q V!

oAd

I! V Ed!

o

o o E

A

Q Ed AEd

I ! ! I ! I *

Eo

dQ dI

dt dt

*! ! I alled displacement current, Id


6/40

Maxwells Equations

Gauss's law electric

0 Gauss's law in magnetism

Faraday's law

Ampere-Maxwell lawI

oS

S

B

Eo o o

qd

d

dd

dt

dd

dt

!

!

* !

* !

E A

B A

E s

B s

The two Gausss laws are symmetrical, apart from the absence of the term formagnetic monopoles in Gausss law for magnetismFaradays law and the Ampere-Maxwell law are symmetrical in that the lineintegrals ofE and B around a closed path are related to the rate of change of

the respective fluxes


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Gausss law (electrical): The total electric flux through any

closed surface equals the net chargeinside that surface divided by Io

This relates an electric field to thecharge distribution that creates it

Gausss law (magnetism): The total magnetic flux through

any closed surface is zero This says the number of field lines

that enter a closed volume mustequal the number that leave that

volume This implies the magnetic fieldlines cannot begin or end at anypoint

Isolated magnetic monopoles havenot been observed in nature

oS

qd

! E A

0S

d !

B A


8/40

Faradays law of Induction: This describes the creation of an electric field by a

changing magnetic flux The law states that the emf, which is the line

integral of the electric field around any closedpath, equals the rate of change of the magnetic fluxthrough any surface bounded by that path

One consequence is the current induced in aconducting loop placed in a time-varying B

The Ampere-Maxwell law is a generalization ofAmperes law

It describes the creation of a magnetic field by anelectric field and electric currents

The line integral of the magnetic field around anyclosed path is the given sum

Bdddt

* ! E s

I Eo o o dd dt* ! B s


9/40

The Lorentz Force Law Once the electric and magnetic fields are known at

some point in space, the force acting on a particle

of charge q can be calculated F =qE +qv x B This relationship is called the Lorentz force law Maxwells equations, together with this force law,

completely describe all classical electromagneticinteractions


10/40

Maxwells Equations in integral form

A Vo o

Q 1E dA dV ! ! VI I

rr

A dA 0

!

rr

Gausss Law

Gausss Law forMagnetism

A

d dE d dA

dt dt

* ! !

r rr rl

Eo encl o o o oA

d dEd I J dA

dt dt

* ! Q Q I ! Q I

rr rr rl

Faradays Law

Amperes Law


11/40

Maxwells Equations in free space(no charge or current)

AE dA 0 !

rr

AdA 0 ! rr

Gausss Law


A

d d

E d dAdt dt

*

! !

r rr r

lE

o o o o A

d dd E dA

dt dt

* ! Q I ! Q I

r rr rl

Faradays Law

Amperes Law


12/40

Hertzs Experiment An induction coil is connected to a

transmitter The transmitter consists of two spherical

electrodes separated by a narrow gap The discharge between the electrodes

exhibits an oscillatory behavior at a veryhigh frequency

Sparks were induced across the gap of thereceiving electrodes when the frequency ofthe receiver was adjusted to match that ofthe transmitter

In a series of other experiments, Hertz alsoshowed that the radiation generated by thisequipment exhibited wave properties Interference, diffraction, reflection, refraction

and polarization He also measured the speed of the radiation


13/40

Implication A magnetic field will be produced in empty space if thereis a changing electric field. (correction to Ampere)

This magnetic field will be changing. (originally therewas none!)

The changing magnetic field will produce an electric field.(Faraday)

This changes the electric field.

This produces a new magnetic field.

This is a change in the magnetic field.


14/40

An antenna

We have changed the magneticfield near the antenna

Hook up anA source

An electric field results! This isthe start of a radiation field.


15/40

Look at the cross section

E and B are perpendicular (transverse)We say that the waves are polarized.

E and B are in phase (peaks and zeros align)

alled:Electromagnetic Waves

Accelerating

electric charges

give rise to

electromagneticwaves


16/40

Angular Dependence of Intensity

This shows the angulardependence of the radiationintensity produced by a dipole

antenna The intensity and power

radiated are a maximum in aplane that is perpendicular tothe antenna and passingthrough its midpoint

The intensity varies as(sin2 /r2


17/40

Active Figure 34.3

(SLIDESHOW MODE ONLY)


18/40

Harmonic Plane Waves

x

At t=

0

At x = 0

P

P!spatial period orwavelength

88!temporal period

phase velocity

2v

T T 2 k

P T P [! ! P ! !

T

t

r

r


19/40

Applying Faraday to radiation

BdE ddt

* !

rrl

E d E dE y E y dE y ! ( ( ! (rr

lBd dB dx y

dt dt

*! (

dBdE y dx ydt

( ! (

dE dB

dx dt!


20/40

Applying Ampere to radiation

Eo o

dB d

dt

* ! Q I

rrl

B d B z B dB z dB z ! ( ( ! ( rr lEd dE dx z

dt dt

*! (

o odEdB z dx zdt

( ! Q I (

o o

dB dE

dx dt

! Q I


21/40

Fields are functions of bothposition (x) and time (t)

o od ddx dt

! Q I

d d

dx dt!

x t

x x!

x x

o ox tx x! Q Ix x

2

2x x t

x x x! x x x

2

o o 2t x t

x x x

! Q Ix x x

Partial derivatives

are appropriate

2 2

o o2 2

x t

x x! Q I

x x

This is a waveequation!


22/40

The Trial Solution The simplest solution to the partial differential

equations is a sinusoidal wave:

E=Emax cos (kxt)B =Bmax cos (kxt)

The angular wave number is k= 2/ is the wavelength

The angular frequency is = 2 is the wave frequency


23/40

The trial solution

y oE E E sin kx t! ! [2 2

o o2 2

E E

x t

x x! Q I

x x

2

2o2

EE sin kx t

t

x! [ [

x

22

o2

Ek E sin kx t

x

x! [

x

2 2o o o ok E sin kx t E sin kx t [ ! Q I [ [2

2

o o

1

k

[!

Q I


24/40

The speed of light(or any other electromagnetic radiation)

o o

1v c

k

[! ! !

Q I

2v f

T T 2 k

P T P [! ! P ! !

T


25/40

3. The speed of an electromagnetic wave traveling in a transparentnonmagnetic substance is , where is the dielectric constant of the substance.Determine the speed of light in water, which has a dielectric constant at opticalfrequencies of 1.78.

5. Figure 34.3 shows a plane electromagnetic sinusoidal wave propagatingin thex direction. Suppose that the wavelength is 50.0 m, and the electric fieldvibrates in thexyplane with an amplitude of 22.0 V/m. alculate (a) thefrequency of the wave and (b) the magnitude and direction ofB when the electricfield has its maximum value in the negativey direction. (c) Write an expressionforB with the correct unit vector, with numerical values forBmax, k, and , andwith its magnitude in the form

6. Write down expressions for the electric and magnetic fields of asinusoidal plane electromagnetic wave having a frequency of 3.00 GHz andtraveling in the positivex direction. The amplitude of the electric field is 300V/m.


26/40

The electromagnetic spectrum

2v

T T 2 k

P T P [! ! P ! !

T


27/40


28/40

Another lookdE dB

dx dt!

y oE E E sin kx t! ! [

o od d

E sin kx t B sin kx tdx dt

[ ! [

o oE k cos kx t B cos kx t [ ! [ [

o

o o o

E 1c

B k

[! ! !

Q I

z oB B B sin kx t! ! [


29/40

Energy in Waves

2 20

0

1 1u E B

2 2! I

Q

o

o o o

E 1c

B k

[! ! !

Q I

20u E! I

2

0

1u B!

Q0

0

u EBI

!Q


30/40

Poynting Vector

Poynting vector points in the direction the wave moves Poynting vector gives the energy passing through a unit

area in 1 sec. Units are Watts/m2

0

1S E B! v

Q

r r r

S cu!r

! ! !2 2

o o o

E cBEBS c


31/40

Intensity The wave intensity, I, is the time average of

S (the Poynting vector) over one or more

cycles When the average is taken, the time average

of cos2(kx - t) = is involved

! ! ! ! !2 2

max max max max

av2 2 2

I aveo o o

E E cS cu

c


32/40

11. How much electromagnetic energy per cubic meter iscontained in sunlight, if the intensity of sunlight at the Earthssurface under a fairly clear sky is 1 000 W/m2?

16. Assuming that the antenna of a 10.0-kW radio stationradiates spherical electromagnetic waves, compute themaximum value of the magnetic field 5.00 km from the

antenna, and compare this value with the surface magneticfield of the Earth.

21. A lightbulb filament has a resistance of 110 . Thebulb is plugged into a standard 120-V (rms) outlet, and emits

1.00% of the electric power delivered to it by electromagneticradiation of frequencyf. Assuming that the bulb is coveredwith a filter that absorbs all other frequencies, find theamplitude of the magnetic field 1.00 m from the bulb.


33/40

Radiation Pressure

Up

c

(( !

F 1 dpP

A A dt! !

ave1 dUP

Ac dt c! !

r

Maxwell showed: (Absorption of radiationby an object)

What if the radiation reflects off an object?


34/40

Pressure and

Momentum

For a perfectly reflecting surface,p = 2U/c and P= 2S/c

For a surface with a reflectivity somewherebetween a perfect reflector and a perfect absorber,the momentum delivered to the surface will besomewhere in between U/c and 2U/c

For direct sunlight, the radiation pressure is about5 x 10-6N/m2


35/40

26. A 100-mW laser beam is reflected back upon itself by amirror. alculate the force on the mirror.

27. A radio wave transmits 25.0 W/m2 of power per unitarea. A flat surface of area Ais perpendicular to the direction of

propagation of the wave. alculate the radiation pressure on it,assuming the surface is a perfect absorber.

29. A 15.0-mW heliumneon laser ( = 632.8 nm) emits a

beam of circular cross section with a diameter of 2.00 mm. (a)Find the maximum electric field in the beam. (b) What totalenergy is contained in a 1.00-m length of the beam? (c) Find themomentum carried by a 1.00-m length of the beam.


36/40

Background for the superior

mathematics student!


37/40

Harmonic Plane Waves

In general, We will only be concerned with the real part ofthe complex phasor representation of a plane wave.

Using Eulers formula:

Re cosT T

E E kx todi b! [

T T TE Ee E kx t i kx t o

i kx t

o! ! [ [ [b g b g b gcos sin

k! 2TP

= propagation number

[TX

!2

= angular frequency

(kx-[t)=

phase


38/40

Phase Velocity - Another View

J [! kx t

ddt

kdxdt

kvJ [ [! ! ! 0

Since the plane waves remain plane waves,

the phase on a plane does not change with time

v

k!

[


39/40

Vector alculus Theorems

T T T T T F dl F d S

C Az zz ! v

A VF dA FdV !

rr r r

And an Important Identity

Gauss Divergence Theorem

Stokes Theorem

T T T T T T v v ! F F F e j e j 2


40/40

Maxwells Equations In Differential Form

T T !D V

T T

!B 0

Gausss Law


T TT

v ! x

x

EB

tT T T

T

v ! xx

H Jt

Faradays Law

Amperes Law

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Chapter S34

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