Chapter 32C Chapter 32C -- Electromagnetic Electromagnetic Waves Waves

A PowerPoint Presentation by

Paul E. Tippens, Professor of Physics

Southern Polytechnic State University

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor of Physics

Southern Polytechnic State UniversitySouthern Polytechnic State University

© 2007

Much of this material is Much of this material is NOTNOT in Tippens Textbookin Tippens Textbook

Objectives: Objectives: After completing this After completing this module, you should be able to:module, you should be able to:

•• Explain and discuss with appropriate diagrams Explain and discuss with appropriate diagrams the general properties of all the general properties of all electromagnetic electromagnetic waveswaves..

•• Discuss and apply the mathematical relationship Discuss and apply the mathematical relationship between the between the electric Eelectric E and and magnetic Bmagnetic B components of an EM wave.components of an EM wave.

•• Define and apply the concepts of Define and apply the concepts of energy densityenergy density, , intensityintensity, and , and pressurepressure due to EM waves.due to EM waves.

This module is OPTIONAL: check with instructor.

MaxwellMaxwell’’s Theorys TheoryElectromagnetic theory developed by James

Maxwell (1831 – 1879) is based on four concepts: Electromagnetic theory developed by James Electromagnetic theory developed by James

Maxwell (1831 Maxwell (1831 –– 1879) is based on four concepts:1879) is based on four concepts:

1. Electric fields E begin on positive charges and end on negative charges and Coulomb’s law can be used to find the field E and the force on a given charge.

1. Electric fields E begin on positive charges and end on negative charges and Coulomb’s law can be used to find the field E and the force on a given charge.

++ --q1 qq11 q2 qq22

204

qEr

204

qEr

F qEF qE

MaxwellMaxwell’’s Theory (Cont.)s Theory (Cont.)

2. Magnetic field lines

do not begin or end, but rather consist of entirely closed loops.

2. Magnetic field lines

do not begin or end, but rather consist of entirely closed loops.

sinB

A

sinqB

qv

MaxwellMaxwell’’s Theory (Cont.)s Theory (Cont.)

3. A changing magnetic field B induces an emf and therefore an electric field E (Faraday’s Law).

3. A changing magnetic field B induces an emf and therefore an electric field E (Faraday’s Law).

Faraday’s Law:

-Nt

E =

A change in flux A change in flux can can occur by a change in area or occur by a change in area or by a change in the Bby a change in the B--field:field:

= B = B AA

= A = A BB

MaxwellMaxwell’’s Theory (Cont.)s Theory (Cont.)

4. Moving charges (or an electric current) induce a magnetic field B.

4. Moving charges (or an electric current) induce a magnetic field B.

R

Inductance L

lB

Solenoid

0NIB

Current I induces B field

B I

Lenz’s law

xxxx

xxxxxxxx

B

Production of an Electric WaveProduction of an Electric WaveConsider two metal rods connected to an ac Consider two metal rods connected to an ac source with sinusoidal current and voltage.source with sinusoidal current and voltage.

+

--

--

+

+

--

Arrows show field vectors (E)

E Wave

Vertical transverse sinusoidal EVertical transverse sinusoidal E--waves.waves.

--

+

An Alternating Magnetic FieldAn Alternating Magnetic Field

B

I

rr

Inward B

XIn

B

I

rr

Outward B

•Out

The ac sinusoidal current also generates a The ac sinusoidal current also generates a magnetic wave alternating in and out of paper.magnetic wave alternating in and out of paper.

rr

+

--

X••

--

+

+

--

X••

--

+

A Magnetic Wave GenerationA Magnetic Wave Generation

Arrows show magnetic field vectors (B)

B - Wave

The generation of a magnetic wave The generation of a magnetic wave due to an oscillating ac current.due to an oscillating ac current.

Ir

+

--BB

Ir

BB

--

+

Ir

+

--BB

I+

--Horizontal transverse sinusoidal BHorizontal transverse sinusoidal B--waves.waves.

An Electromagnetic WaveAn Electromagnetic WaveAn electromagnetic wave consists of combination An electromagnetic wave consists of combination of a transverse electric field and a transverse of a transverse electric field and a transverse magnetic field perpendicular to each other.magnetic field perpendicular to each other.

+

--

Arrows show field vectors

EM wave propagation in space

Transmitting and Receiving Transmitting and Receiving An ac current generates an EM wave which then An ac current generates an EM wave which then

generates an ac signal at receiving antenna.generates an ac signal at receiving antenna.

A BA B--field Moves Past a Chargefield Moves Past a ChargeRelativity tells us that there is no preferred frame of Relativity tells us that there is no preferred frame of reference. Consider that a magnetic field B moves at reference. Consider that a magnetic field B moves at the speed of light c past a stationary charge q:the speed of light c past a stationary charge q:

qq

NN

SScc

B

cc Stationary positive charge

Charge Charge q q experiences a experiences a magnetic force Fmagnetic force F

or FF qcB cBq

But electric field But electric field E = F/qE = F/q::

Substitution shows:Substitution shows: E cBEcB

An EAn E--field Moves Past a Pointfield Moves Past a PointA length of wire A length of wire l l moves at velocity moves at velocity cc past point past point AA::

Ar

+ + + + + +cc

EE

EEWire moves at

velocity c past A

A current A current I I is simulatedis simulated..

In time In time tt, a length of wire , a length of wire l l = ct= ct passes point passes point AA

q ctI ct t

Charge density:Charge density:q q

ct

In time In time t: t: q = q = ctct

Thus, the current Thus, the current I I is:is: Simulated current I:I c

Moving EMoving E--field (Cont.)field (Cont.)A

r

+ + + + + +cc

EE

EE

simulated currentsimulated current:: I cA A BB--field is created by thefield is created by the

0 0

2 2I cBr r

Recall from GaussRecall from Gauss’’ law:law:

02E

r

Eliminating Eliminating

from these from these two equations gives:two equations gives:

0 0B cE

The Speed of an EM WaveThe Speed of an EM WaveA

r

+ + + + + +cc

EE

EE

For EM waves, we have seen:For EM waves, we have seen:

0 0B cE EcB

Substituting Substituting E = E = cBcB into into latter equation gives:latter equation gives:

0 0 ( )B c cB

0 0

1c

EMEM--waves travel at the waves travel at the speed of light, which is:speed of light, which is:

c = c = 3.00 x 103.00 x 1088 m/sm/s

Important Properties for All Important Properties for All Electromagnetic WavesElectromagnetic Waves

•• EM waves are EM waves are transversetransverse waves. Both waves. Both EE and and BB are perpendicular to wave velocity are perpendicular to wave velocity cc..

•• The ratio of the EThe ratio of the E--field to the Bfield to the B--field is field is constant and equal to the velocity constant and equal to the velocity cc..

Energy Density for an EEnergy Density for an E--fieldfieldEnergy density Energy density uu is the energy per unit volume is the energy per unit volume ((J/mJ/m33) carried by an EM wave. Consider ) carried by an EM wave. Consider uu for the for the electric field electric field EE of a capacitor as given below:of a capacitor as given below:

Energy density Energy density u u for an Efor an E--field:field: AA dd .

U UuVol Ad

2 201 12 2 ( )AU CV Ed

d

0Recall and :AC V Edd

2102 AdEUu

Ad Ad

Energy density u:21

02u E

Energy Density for a BEnergy Density for a B--fieldfieldEarlier we defined the energy density Earlier we defined the energy density u u for a for a BB--field field using the example of a solenoid of inductance using the example of a solenoid of inductance LL::

R

l

A

220 1

2; ; N AL U LI V A

0

0

NI NI BB

2 20

22N IUu

A

2

02Bu

Energy density for B-field:

Energy Density for EM WaveEnergy Density for EM Wave

The energy of an EM wave is shared equally by The energy of an EM wave is shared equally by the electric and magnetic fields, so that the the electric and magnetic fields, so that the total energy density of the wave is given by:total energy density of the wave is given by:

221

0202

Bu E

Total energy density:

Or, since energy is shared equally:

22

00

Bu E

Average Energy DensityAverage Energy DensityThe The EE and and BB--fields fluctuate between their fields fluctuate between their maximum values maximum values EEmm and and BBmm . An . An averageaverage value value of the energy density can be found from the of the energy density can be found from the rootroot--meanmean--square values of the fields:square values of the fields:

and 2 2m m

rms rmsE BE B

The The average energy densityaverage energy density uuavgavg is therefore:is therefore:

2102avg mu E 2

0avg rmsu Eoror

Example 1: Example 1: The maximum amplitude of an The maximum amplitude of an EE--fieldfield from sunlight is from sunlight is 1010 V/m1010 V/m. What is . What is the the rootroot--meanmean--squaresquare value of the value of the BB--fieldfield??

EM EM wavewave

Earth

8

1010 V/m 3.37 T3 x 10 m/s

mm

EBc

3.37 T ; 1.4

14

2.32

8 Tmrms rmsBBB

What is the average energy density of the wave?What is the average energy density of the wave?2

22 -12 Nm1 1

02 2 C(8.85 x 10 )(1010 V/m)avg mu E

-93

J4.47 x 10mavgu Note that the total energy Note that the total energy

density is twice this value.density is twice this value.

Wave Intensity Wave Intensity IIThe intensity of an EM wave is defined as the The intensity of an EM wave is defined as the power per unit area (power per unit area (W/mW/m22).).

Area A

PIA

EM wave moves distance EM wave moves distance ctct through area through area AA as shown below:as shown below:

Total energy = density x volumeTotal energy = density x volume

ctct

AA

Total energy =Total energy = u(ctAu(ctA)) EP Total uctAI uc

A Time Area tA

And Since And Since u = u =

Total intensity:2

0 mI c EPI ucA

Calculating Intensity of WaveCalculating Intensity of WaveIn calculating intensity, you must In calculating intensity, you must distinguish between average distinguish between average values and total values:values and total values:

2 210 02avg m rmsI c E c E

2 20 02T m rmsI c E c E

Since Since E = E = cBcB, we can also express I in terms of , we can also express I in terms of BB::

2 2

0 02avg m rmsc cI B B

2 2

0 0

2T m rms

c cI B B

Area A

PIA

2102avg mI c E

Example 2:Example 2: A signal received from a radio A signal received from a radio station has station has EEmm = 0.0180 V/m. What is the = 0.0180 V/m. What is the average intensity at that point? average intensity at that point?

2102avg mI c E

2

28 -12 2Nm1

2 C(3 x 10 m/s)(8.85 x 10 )(0.018 V/m)avgI

The The average intensityaverage intensity is:is:

-7 24.30 x 10 W/mavgI

Note that intensity is Note that intensity is power per unit areapower per unit area. The . The power of the source remains constant, but the power of the source remains constant, but the intensity decreases with the square of distance.intensity decreases with the square of distance.

Wave Intensity and DistanceWave Intensity and Distance

24P PIA r

The intensity The intensity I I at a distance r at a distance r from an from an isotropicisotropic source:source:

The The average poweraverage power of the of the source can be found from the source can be found from the intensity at a distance intensity at a distance r r ::

2(4 )avg avgP AI r I

For For isotropicisotropic conditions:conditions:For power falling on For power falling on surface of area surface of area AA::

P = P = IIavgavg AA

AA

Example 3:Example 3: In Example 2, an average intensity In Example 2, an average intensity of of 4.30 x 104.30 x 10--77 W/mW/m22 was observed at a point. If was observed at a point. If the location is the location is 90 km90 km (r = 90,000 m) from the (r = 90,000 m) from the isotropic radio source, what is the average isotropic radio source, what is the average power emitted by the source? power emitted by the source?

-5 22 2.39 x 10 W/m

4avgPIr

PP = (4= (4rr22)(4.30 x 10)(4.30 x 10--7 7 W/mW/m22))

90 km

PP = 4= 4(90,000 m)(90,000 m)22(4.30 x 10(4.30 x 10--7 7 W/mW/m22))

P = 43.8 kWP = 43.8 kWAverage power Average power of transmitter:of transmitter:

This assumes This assumes isotropicisotropic propagation, which is not likely.propagation, which is not likely.

Radiation PressureRadiation PressureEMEM--waves not only carry energy, but also carry waves not only carry energy, but also carry momentum and exert pressure when absorbed momentum and exert pressure when absorbed or reflected from objects.or reflected from objects.

A

ForceArea

Radiation PressureRecall that Power = F vRecall that Power = F v

F or A

P Fc IIA A c

The pressure is due to the transfer of The pressure is due to the transfer of momentummomentum. . The above relation gives the pressure for a The above relation gives the pressure for a completely absorbingcompletely absorbing surface.surface.

Radiation Pressure (Cont.)Radiation Pressure (Cont.)The change in momentum for a fully reflected The change in momentum for a fully reflected wave is twice that for an absorbed wave, so wave is twice that for an absorbed wave, so that the radiation pressures are as follows:that the radiation pressures are as follows:

A

ForceArea

Radiation Pressure

Absorbed wave:

A

ForceArea

Radiation Pressure

Reflected wave:

F IA c

2F IA c

Example 4:Example 4: The average intensity of direct The average intensity of direct sunlight is around 1400 W/msunlight is around 1400 W/m22. What is the . What is the average force on a fully absorbing surface of average force on a fully absorbing surface of area 2.00 marea 2.00 m22??

A

ForceArea

Radiation Pressure

Absorbed wave: F IA cFor absorbing For absorbing

surface:surface:

IAFc

2 2

8

(1400 W/m )(2.00 m )3 x 10 m/s

F F = 9.33 x 10-6 NF = 9.33 x 10-6 N

The RadiometerThe RadiometerA radiometer is a device which demonstrates the existence of radiation pressure: A radiometer is a device which demonstrates the existence of radiation pressure:

RadiometerRadiometer

One side of the panels is black (totally absorbing) and the other white (totally reflecting). The panels spin under light due to the pressure differences.

One side of the panels is black (totally absorbing) and the other white (totally reflecting). The panels spin under light due to the pressure differences.

SummarySummary

EM waves are EM waves are transversetransverse waves. Both waves. Both EE and and BB are perpendicular to wave velocity are perpendicular to wave velocity cc..

The ratio of the EThe ratio of the E--field to the Bfield to the B--field is field is constant and equal to the velocity constant and equal to the velocity cc..

Electromagnetic waves carry both energy Electromagnetic waves carry both energy and momentum and can exert pressure on and momentum and can exert pressure on surfaces.surfaces.

Summary (Cont.)Summary (Cont.)

EcB

0 0

1c

EMEM--waves travel at the waves travel at the speed of light, which is:speed of light, which is:

c = c = 3.00 x 103.00 x 1088 m/sm/s

221

0202

Bu E

Total Energy Density:

and 2 2m m

rms rmsE BE B

Summary (Cont.)Summary (Cont.)The The average energy densityaverage energy density::

2102avg mu E 2

0avg rmsu Eoror

2 210 02avg m rmsI c E c E

24P PIA r

F IA c

2F IA c

Intensity and Distance

Totally Absorbing

Totally Reflecting

CONCLUSION: Chapter 32CCONCLUSION: Chapter 32C Electromagnetic WavesElectromagnetic Waves