Electromagnetic Waves (Optional Unit) A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007.

Electromagnetic Waves (Optional Unit)

A PowerPoint Presentation by

Paul E. Tippens, Professor of Physics

Southern Polytechnic State University© 2007

Maxwell’s TheoryElectromagnetic theory developed by James Electromagnetic theory developed by James

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

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

concepts: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.

++ --qq11qq11 qq22qq22

204

qE

r 2

04

qE

r

F qEF qE

Maxwell’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

sin

qB

qv

Maxwell’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 occur by a change in area or by a change in area or by a change in the B-field:the B-field: = B = B AA

= A = A BB

Maxwell’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 Wave

Consider two metal rods connected to an Consider two metal rods connected to an ac source with sinusoidal current and ac source with sinusoidal current and

voltage.voltage.+

--

--

+

+

--

Arrows show field vectors (E)

E Wave

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

--

+

An Alternating Magnetic Field

B

I

rr

Inward B

XIn

B

I

rr

Outward B

•Out

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

of paper.of paper.

rr

+

--

X••

--

+

+

--

X••

--

+

A Magnetic Wave Generation

Arrows show magnetic field vectors (B)

B - Wave

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

current.current.

Ir

+

--BB

Ir

BB

--

+

Ir

+

--BB

I+

--

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

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

+

--

Arrows show field vectors

EM wave propagation in space

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

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

A B-field Moves Past a Charge

Relativity tells us that there is no preferred Relativity tells us that there is no preferred frame of reference. Consider that a magnetic frame of reference. Consider that a magnetic field B moves at the speed of light c past a field B moves at the speed of light c past a stationary charge q: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 F

F qcB cBq

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

Substitution Substitution shows:shows:

E cBE

cB

E

cB

An E-field Moves Past a Point

A length of wire A length of wire l l moves at velocity moves at velocity cc past past point A:point A:

Ar

+ + + + + +

cc

EE

EEWire moves at

velocity c past A

A current A current I I is is simulatedsimulated..In time In time tt, a length of , a length of

wire wire l l = ct passes = ct passes point Apoint A

q ctI c

t t

Charge Charge density:density:

q q

ct

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

Thus, the current Thus, the current I I is:is:

Simulated current I:I c

Moving E-field (Cont.)A

r

+ + + + + +

cc

EE

EE

simulated currentsimulated current:: I cA B-field is created by A B-field is created by thethe

0 0

2 2

I cB

r r

Recall from Gauss’ Recall from Gauss’ law:law:

02E

r

Eliminating Eliminating from from these two equations these two equations

gives:gives:

0 0B cE 0 0B cE

The Speed of an EM WaveA

r

+ + + + + +

cc

EE

EE

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

0 0B cE 0 0B cE Ec

BE

cB

Substituting Substituting E = cBE = cB into into latter equation gives:latter equation gives:

0 0 ( )B c cB

0 0

1c

0 0

1c

EM-waves travel at EM-waves travel at the speed of light, the speed of light,

which is:which is:c = c = 3.00 x 103.00 x 1088

m/sm/s

Important Properties for All Electromagnetic Waves

• EM waves are transverse waves. Both E EM waves are transverse waves. Both E and B are perpendicular to wave and B are perpendicular to wave velocity velocity cc..

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

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

Energy Energy density udensity u for for an E-field:an E-field:

AA dd .

U UuVol Ad

2 201 12 2 ( )

AU CV Ed

d

0Recall and :A

C V Edd

21

02 AdEUu

Ad Ad

Energy density

u: 2102u E

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

R

l

A

220 1

2; ; N A

L U LI V A

0

0

NI NI B

B

2 20

22

N IUu

A

2

02

Bu

Energy

density for B-field:

Energy Density for EM Wave

The energy of an EM wave is shared The energy of an EM wave is shared equally by the electric and magnetic equally by the electric and magnetic fields, so that the total energy density of fields, so that the total energy density of the wave is given by:the wave is given by: 2

2102

02

Bu E

Total energy density:

Or, since energy is shared equally:

22

00

Bu E

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

and 2 2m m

rms rms

E BE B and

2 2m m

rms rms

E BE B

The The average energy densityaverage energy density uuavgavg is is therefore:therefore: 21

02avg mu E 20avg rmsu Eoror

Example 1: The maximum amplitude of an E-field from sunlight is 1010 V/m. What is the root-mean-square value of the B-field?

EM EM wavwav

ee

Earth

8

1010 V/m3.37 T

3 x 10 m/sm

m

EB

c

3.37 T; 2.38 T

1.4142m

rms rms

BB B

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

2

2 -12 Nm1 102 2 C

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

-93

J4.47 x 10

mavgu -9

3

J4.47 x 10

mavgu Note that the total energy Note that the total energy density is twice this density is twice this

value.value.

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

Area A

PI

A

EM wave moves distance EM wave moves distance ctct through area A as shown through area A as shown below:below:Total energy = density x Total energy = density x

volumevolumectct

AA

Total energy = u(ctA)Total energy = u(ctA)

EP Total uctAI uc

A Time Area tA

And And Since Since u u = =

Total intensity:2

0 mI c EP

I ucA

Calculating Intensity of Wave

In calculating intensity, you In calculating intensity, you must distinguish between must distinguish between average values and total average values and total values:values:

2 210 02avg m rmsI c E c E

2 210 02avg m rmsI c E c E

2 20 02T m rmsI c E c E

2 20 02T m rmsI c E c E

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

2 2

0 02avg m rms

c cI B B

2 2

0 02avg m rms

c cI B B

2 2

0 0

2T m rms

c cI B B

2 2

0 0

2T m rms

c cI B B

Area A

PI

A

2102avg mI c E

Example 2: A signal received from a radio station has Em = 0.0180 V/m. What is the

average intensity at that point?

2102avg mI c E 21

02avg mI c E

2

2

8 -12 2Nm12 C(3 x 10 m/s)(8.85 x 10 )(0.018 V/m)avgI

The average intensity The average intensity is:is:

-7 24.30 x 10 W/mavgI -7 24.30 x 10 W/mavgI

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

Wave Intensity and Distance

24

P PI

A r 24

P PI

A r

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

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

2(4 )avg avgP AI r I

For isotropic For isotropic conditions:conditions:

For power falling For power falling on surface of area on surface of area

A:A:P = IP = Iavgavg A A

AA

Example 3: In Example 2, an average intensity of 4.30 x 10-7 W/m2 was observed at a point. If the location is 90 km (r = 90,000 m) from the isotropic radio source, what is the

average power emitted by the source? -5 2

22.39 x 10 W/m

4avg

PI

r

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 kW P = 43.8 kW

Average Average power of power of

transmitter:transmitter:This assumes isotropic propagation, which is not This assumes isotropic propagation, which is not

likely.likely.

Radiation Pressure

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

A

Force

AreaRadiation PressureRecall that Power = F vRecall that Power = F v

F or

A

P Fc II

A A c

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

Radiation Pressure (Cont.)The change in momentum for a fully The change in momentum for a fully reflected wave is twice that for an reflected wave is twice that for an

absorbed wave, so that the radiation absorbed wave, so that the radiation pressures are as follows:pressures are as follows:

A

Force

AreaRadiation Pressure

Absorbed wave:

A

Force

AreaRadiation Pressure

Reflected wave:

F I

A c

F I

A c 2F I

A c

2F I

A c

Example 4: The average intensity of direct sunlight is around 1400 W/m2. What is the

average force on a fully absorbing surface of area 2.00 m2?

A

Force

AreaRadiation Pressure

Absorbed wave:

F I

A c

F I

A cFor absorbing For absorbing

surface:surface:

IAF

c

2 2

8

(1400 W/m )(2.00 m )

3 x 10 m/sF F = 9.33 x 10-6

NF = 9.33 x 10-6

N

The 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.

Summary

EM waves are transverse waves. Both EM waves are transverse waves. Both E and B are perpendicular to wave E and B are perpendicular to wave velocity velocity cc..

The ratio of the E-field to the B-field is The ratio of the E-field to the B-field is constant and equal to the velocity constant and equal to the velocity c.c.

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

Summary (Cont.)

Ec

BE

cB

0 0

1c

0 0

1c

EM-waves travel at EM-waves travel at the speed of light, the speed of light,

which is: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 rms

E BE B and

2 2m m

rms rms

E BE B

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

2 210 02avg m rmsI c E c E

24

P PI

A r

F I

A c

2F I

A c

Intensity and Distance

Totally Absorbin

g

Totally Reflecting