Tue. Nov. 11, 2008Physics 208, Lecture 211 From last time… EM waves Inductors in circuits I? + -

Tue. Nov. 11, 2008 Physics 208, Lecture 21 1

From last time…

EM waves

Inductors in circuits I?

+

-


•A Transverse wave.

•Electric/magnetic fields perpendicular to propagation direction

•Can travel in empty space

f = v/, v = c = 3 x 108 m/s (186,000 miles/second)


A microwave oven irradiates food with electromagnetic radiation that has a frequency of about 1010 Hz. The wavelengths of these microwaves are on the order of

A. kilometers

B. meters

C. centimeters

D. micrometers

Quick Quiz

€

=c / f =3 ×108 m /s

1010 /s= 3cm


Electromagnetic waves

€

Bo = Eo /c

€

rE =

r E o cos kz −ωt( )

r B =

r B o cos kz −ωt( )

€

c =1/ εoμo =1/ 8.85 ×10−12C2 /N ⋅m2( ) 4π ×10−7 N / A2

( )

= 2.9986 ×108 m /s

€

rE ⊥

r B

€

k =2π

λ, ω =

2π

f

z

x

y


Energy and EM Waves

Energy density in E-field

Energy density in B-field

€

uE = εoE 2 r, t( ) /2

€

uB = B2 r, t( ) /2μo

Total

€

uTot = εoE 2 /2 + B2 /2μo

= εoE 2 /2 + E 2 /2c 2μo = εoE 2 r, t( ) = B2 r, t( ) /μo

€

uTot = εoE 2 = εoEo2 cos2 kz −ωt( ) moves w/ EM wave

at speed c


Power and intensity in EM waves Energy density uE moves at c

Instantaneous energy transfer = energy passing plane per second. = This is power density W/m2

Time average of this is Intensity =

€

cuTot = cεoE 2 r, t( ) = cB2 r, t( ) /μo

€

cεoEmax2 /2 = cBmax

2 /2μo


Example: E-field in laser pointer

1 mW laser pointer.

Beam diameter at board ~ 2mm

Intensity =

€

10−3W

π 0.001m( )2 = 318W /m2

How big is max E-field?

€

cεoEmax2 /2 = 318W /m2

Emax =2 318W /m2

( )

3×108 m /s( ) 8.85 ×10−12C2 /N ⋅m2( )

= 489N /C = 489V /m


Spherical waves Sources often radiate EM wave in all directions

Light bulb The sun Radio/tv transmission tower

Spherical wave, looks like plane wave far away Intensity decreases with distance

Power spread over larger area

€

I =Psource

4π r2

Source power

Spread over thissurface area


QuestionA radio station transmits 50kW of power from its

antanna. What is the amplitude of the electric field at your radio, 1km away.

€

I =50,000W

4π 1000m( )2 = 4 ×10−3W / m2

€

cεoEmax2 /2 = 4 ×10−3W /m2

Emax =2 4 ×10−3W /m2( )

3 ×108 m /s( ) 8.85 ×10−12C2 /N ⋅m2( )

=1.73N /C =1.73V /m

A. 0.1 V/m

B. 0.5 V/m

C. 1 V/m

D. 1.7 V/m

E. 15 V/m


The Poynting Vector Rate at which energy flows through a unit area perpendicular

to direction of wave propagation

Instantaneous power per unit area (J/s.m2 = W/m2) is also

Its direction is the direction of propagation of the EM wave

This is time dependent Its magnitude varies in time Its magnitude reaches a maximum at the

same instant as E and B

€

rS =

1

μo

r E ×

r B ≡ Poynting Vector


Radiation Pressure Saw EM waves carry energy They also have momentum When object absorbs energy U from EM wave:

Momentum p is transferred

Result is a force

Pressure = Force/Area = €

p = U /c ( Will see this later in QM )

€

F = Δp /Δt =U /Δt

c= P /c

€

prad =P / A

c= I /c

Radiation pressure on perfectly absorbing object

Power

Intensity


Radiation pressure & force

EM wave incident on surface exerts a radiation pressure prad (force/area) proportional to intensity I.

Perfectly absorbing (black) surface:

Perfectly reflecting (mirror) surface:

Resulting force = (radiation pressure) x (area) €

prad = I /c

€

prad = 2I /c


QuestionA perfectly reflecting square solar sail is 107m X 107m. It has

a mass of 100kg. It starts from rest near the Earth’s orbit, where the sun’s EM radiation has an intensity of 1300 W/m2.

How fast is it moving after 1 hour?

€

prad = 2I /c

Frad = prad A = 2IA /c =2 1300W /m2( ) 1.145 ×104 m2

( )

3 ×108 m /s= 0.1N

a = Frad /m =10−3 m /s2

v = at = 10−3 m /s2( ) 3600s( ) = 3.6m /s

A. 100 m/s

B. 56 m/s

C. 17 m/s

D. 3.6 m/s

E. 0.7 m/s


Polarization of EM waves Usually indicate the polarization direction by

indicating only the E-field. Can then be indicated with a line:

Unpolarized Plane Polarizedx

yz

€

rE = Eo cos kz −ωt( ) ˆ x r B = Bo cos kz −ωt( ) ˆ y

Superposition of plane polarized waves


Producing polarized light Polarization by selective absorption: material that transmits

waves whose E-field vibrates in a plain parallel to a certain direction and absorbs all others

This polarizationabsorbed

This polarizationtransmitted transmission axis

Polaroid sheet

Long-chain hydrocarbon molecules Demo on MW and metal grid


Transmission at an angle

Incident wave is equivalent to superposition

x

y

transmission

€

rE inc =

r E o cos kx −ωt( )

Plane-polarizedincident wave

polarizer

€

E inc cosθ( ) ˆ x + E inc sinθ( ) ˆ y

absorbedtransmitted

Transmitted wave =

€

rE trans = Eo cosθ cos kx −ωt( ) ˆ x


Detecting polarized light Polarizer

transmits component of E-field parallel to transmission axis absorbs component of E-field perpendicular to transmission axis

Transmitted intensity: I = I0cos2 I0 = intensity of polarized beam on analyzer (Malus’ law)

Allowed componentparallel to analyzer axis

Polaroid sheets


Malus’ law

Transmitted amplitude is Eocos (component of polarization along polarizer axis)

Transmitted intensity is Iocos2( square of amplitude)

Perpendicular polarizers give zero intensity.


Polarization by reflectionUnpolarizedIncident light

Reflection polarized with E-field parallel to surface

Refractedlight

Unpolarized light reflected from a surface becomes partially polarized

Degree of polarization depends on angle of incidence n


Reducing glare

Transmission axis

Reflected sunlight partially polarized.

Horizontal reflective surface ->the E-field vector of reflected light has strong horizontal component.


Circular and elliptical polarization

Circularly polarized light is a superposition of two waves with orthogonal linear polarizations, and 90˚ out of phase.

The electric field rotates in time with constant magnitude.

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Tue. Nov. 11, 2008Physics 208, Lecture 211 From last time… EM waves Inductors in circuits I? + -