Electromagnetic WavesAmpere’s Law UpdatedHertz’s DiscoveriesPlane Electromagnetic WavesEnergy in an EM WaveMomentum and Radiation PressureThe Electromagnetic Spectrum

Problem with Ampère’s

LawId 0. μ=∫ sB rr

Ampère’s Law

For S1 , Id 0. μ=∫ sB rr

For S2 , 0. 0 ==∫ Id μsB rr

(No current in the gap)

But there is a problem if I varies with time!

Consider a parallel plate capacitor

Solution -

Displacement CurrentCall I as the conduction current.

Specify a new current that exists when a change in the electric field occurs. Call it the displacement current.

dtdI E

dΦ

= 0ε

Electrical flux through S2 :

00 εεQA

AQEAE =⎟⎟

⎠

⎞⎜⎜⎝

⎛==Φ

IdtdQ

dtdI E

d ==Φ

= 0ε

Ampère-Maxwell Law

Magnetic fields are produced by both conduction currents and by

time-varying electrical fields.

( )dt

dIIId Ed

Φ+=+=∫ 0000. εμμμsB rr

Now the generalized form of Ampère’s Law, or Ampère-Maxwell Law becomes:

Maxwell’s Equations

BvEFrrrr

×+= qq Lorentz Force Law

Faraday’s Lawdtdd BΦ

−=∫ sE rr.

Ampère-Maxwell LawdtdId EΦ

+=∫ 000. εμμsBrr

Gauss’ Law for Magnetism – no magnetic monopoles

0. =∫ ABrr

d

Gauss’ Law0

.εQd =∫ AE

rr

Hertz’s Radio WavesBy supplying short voltage bursts from the coil to the transmitter electrode we can ionize the air between the electrodes.

In effect, the circuit can be modeled as a LC circuit, with the coil as the inductor and the electrodes as the capacitor.

A receiver loop placed nearby is able to receive these oscillations and creates sparks as well.

Waves

Circular (or Spherical in 3D) Waves

Plane Waves

Electric and Magnetic Fields in Free Space

dtdd BΦ

−=∫ sE rr.

dtdId EΦ

+=∫ 000. εμμsBrr

Free Space: I = 0, Q = 0

dtdd EΦ

=∫ 00. εμsBrr

tE

xB

tB

xE

∂∂

−=∂∂

∂∂

−=∂∂

00εμ

2

2

002

2

002

2

tE

xE

tE

txB

ttB

xxE

∂∂

=∂∂

⎟⎠⎞

⎜⎝⎛

∂∂

−∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

∂∂

−=∂∂

εμ

εμ

2

2

002

2

tB

xB

∂∂

=∂∂ εμ

( )( )tkxBB

tkxEEωω

−=−=

coscos

max

max

EM Wave Oscillation

Some Important Quantities

λπ2

=k Wavenumber

00

1εμ

=c Speed of Light

fπω 2= Angular Frequency

fc

=λ Wavelength

ck=

ω

( )( )tkxBB

tkxEEωω

−=−=

coscos

max

max

( )( )tkxBB

tkxEEωω

−=−=

coscos

max

max

tE

xB

tB

xE

∂∂

−=∂∂

∂∂

−=∂∂

00εμ

ckB

EBE

BkE

===

=ω

ω

max

max

maxmax

Properties of EM WavesThe solutions to Maxwell’s equations in free space are wavelikeElectromagnetic waves travel through free space at the speed of light.The electric and magnetic fields of a plane wave are perpendicular to each other and the direction of propagation (they are transverse).The ratio of the magnitudes of the electric and magnetic fields is c.EM waves obey the superposition principle.

An EM Wave

MHzf 40=

a) λ=?, T=?

sf

T

mfc

86

6

8

105.21040

11

5.71040103

−×=×

==

=××

==λ

b) If Emax =750N/C, then Bmax =?

Tc

EB 68

maxmax 105.2

103750 −×=×

==

Directed towards the z-direction

c) E(t)=? and B(t)=?

( ) ( )( ) ( ) )1051.2838.0cos(105.2cos

)1051.2838.0cos(/750cos86

max

8max

txTtkxBB

txCNtkxEE

×−×=−=

×−=−=−ω

ωsradf

mradk

/1051.2104022

/838.05.7

22

86 ×=×==

===

ππω

πλπ

Energy Carried by EM WavesBESrrr

×=0

1μ Poynting Vector (W/m2)

For a plane EM Wave: 2

00

2

0

Bcc

EEBSμμμ

===

2max

00

2max

0

maxmax

222Bc

cEBESI av μμμ

====The average value of S is called the intensity:

S is equal to the rate of EM energy flow per unit area (power per unit area)

0

2

2μBuB =

2

20EuEε

=

( ) 20

2

0

00

0

2

21

22EEcEuB ε

μεμ

μ===

0

22

0 221

με BEuu BE ===

For an EM wave, the instantaneous electric and magnetic energies are equal.

0

22

0 με BEuuu BE ==+=

( )0

2max2

max02

0 221

μεε BEEu avav ===

avav cuSI ==

Total Energy Density of an EM Wave

The intensity is c times the total average energy density

The Energy Density

Which gives the largest average energy densityat the distance specified and thus, at least qualitatively, the best illumination

1. a 50-W source at a distance R.2. a 100-W source at a distance 2R.3. a 200-W source at a distance 4R.

Concept Question

Fields on the Screen

cE

AI av

0

2max

2μ==

P

P lamp = 150 W, 3% efficiencyA = 15 m2

mVAcE av /42.182 0

max ==Pμ

Wlampav 5.403.0 == PP

Tc

EB 8maxmax 1014.6 −×==

A

Momentum and Radiation Pressure

cTp ER= Momentum for complete absorption

dtdp

AAFP 1==

( )A

dtdTcc

Tdtd

AP ERER 11

=⎟⎠⎞

⎜⎝⎛=

cSP =

Pressure on surface

For complete reflection:cTp ER2

=cSP 2

=

Pressure From a Laser PointerP laser = 3 mW, 70% reflection, d = 2 mm

( )2

2 /955001.0003.0 mW

AS ===

πP

268 /104.5

1039557.1

7.17.0

mNP

cS

cS

cSP

−×=×

=

=+=

AntennasThe fundamental mechanism for electromagnetic radiation is

an accelerating charge

Half-Wave Antenna

Electromagnetic Spectrum

For Next ClassReading Assignment

Chapter 35: Nature of Light and Laws of Geometric Optics

WebAssign: Assignment 12