Four equations (integral form) :

Gauss’s law

Gauss’s law for magnetism

Faraday’s law

Ampere-Maxwell law

0

ˆ

insideqdAnE

dAnBdt

dldE ˆ

dt

dIldB elec

pathinside 0_0

+ Lorentz force BvqEqF

Maxwell’s Equations

0ˆ AnB

fT

f

/12

Acceleration:

tydt

yda sin2

max2

2

rc

aqEradiative 2

04

1

jsin4

12

2max

0

trc

qyEradiative

Sinusoidal E/M field

Sinusoidal Electromagnetic Radiation

According to particle theory of light: photons have energy and momentum

Classical E/M model of light: E/M radiation must carry energy and momentum

Energy and Momentum of E/M Radiation

A particle will experience electric force during a short time d/c:

qEFelec

p p 0 Felect qE d

c

What will happen to the ball?

It will oscillate

Energy was transferred from E/M field to the ball

mc

qEd

m

pKK

2

1

20

22Amount of energy in the pulse is ~ E2

Energy of E/M Radiation

mc

qEdK

2

12

Ball gained energy:

Pulse energy must decrease

)(J/m 32

0

20

1

2

1

2

1BE

Volume

Energy

E/M radiation: E=cB

200

20

2

0

20

11

2

11

2

1

2

1

cE

c

EE

Volume

Energy

2

0E

Energy of E/M Radiation

There is E/M energy stored in the pulse: )(J/m 320E

Volume

Energy

Pulse moves in space: there is energy flux

Units: J/(m2s) = W/m2

During t:

Energy 0E2 Act

2

0EctA

Energyflux

EBflux0

1

used: E=cB, 00=1/c2

Energy Flux

EBflux0

1

The direction of the E/M radiation was given by BE

Energy flux, the “Poynting vector”:

)1

0

2W/m (in BES

• S is the rate of energy flux in E/M radiation• It points in the direction of the E/M radiation

John HenryPoynting(1852-1914)

Energy Flux: The Poynting Vector

Intensity,

In the vicinity of the Earth, the energy density of radiation emitted by the sun is ~1400 W/m2. What is the approximate magnitude of the electric field in the sunlight?

Solution:EBflux

0

1

2

0Ec

N/C 7250

c

fluxE

Note: this is an average (rms) value

Exercise

A laser pointer emits ~5 mW of light power. What is the approximate magnitude of the electric field?

Solution:

1. Spot size: ~2 mm2. flux = (5.10-3 W)/(3.14.0.0012 m2)=1592 W/m2

3. Electric field:

N/C 7730

c

fluxE (rms value)

What if we focus it into 2 a micron spot?

Flux will increase 106 times, E will increase 103 times:

N/C 000,773E

Exercise

• E field starts motion• Moving charge in magnetic field:

BvqFmag

Fmag

What if there is negative charge?

BvqFmag

Fmag

‘Radiation pressure’: What is its magnitude?

Average speed: v/2

c

EvqB

vqFmag 22

12

)/(2

c

vqE

c

Evq

F

F

elec

mag

Momentum of E/M Radiation

Net momentum: in transverse direction: 0 in longitudinal direction: >0

Relativistic energy:

E2 pc 2 mc2 2

Quantum view: light consists of photons with zero mass: 22 pcE

Classical (Maxwell): it is also valid, i.e. momentum = energy/speed

BES

0

1

)N/m (in 2BEcc

S

0

1

Momentum flux:

Momentum Flux

Units of Pressure

What is the magnitude of the electric field due to sunlight near the Earth (1400 W/m2)? What is the force due to sunlight on a sail with the area 1 km2 at this location?Solution:

If reflective surface?2N/m 6103.9

Total force on the sail: N 3.9F

Exercise: Solar Sail

Atmospheric pressure is ~ 105 N/m2

|𝑆|= 1𝜇0

𝐸𝐵= 1𝜇0

𝐸 𝐸𝑐= 𝐸2

𝜇0𝑐725 N/C

If absorbed, pressure =N/m2

Electric fields are not blocked by matter: how can E decrease?

Re-radiation: Scattering

Positive charge

Why there is no light going through a cardboard?

Electric fields are not blocked by matterElectrons and nucleus in cardboard reradiate lightBehind the cardboard reradiated E/M field cancels original field

Cardboard

In which of these situations will the bulb light?

A) AB) BC) CD) NoneE) B and C

In transparent media, the superposition can result in change of wavelength and speed of wavefront

Index of refraction of medium,

Depends upon wavelengthand properties of medium

Refraction: Bending of Light

Rays perpendicular to wavefront bend at surface

A ray bends as it goes from one transparent media to another

Refraction: Snell’s Law

𝜃1

𝜃2

𝜃𝑎𝑖𝑟=45 °

𝜃𝑤𝑎𝑡𝑒𝑟 ≈ ?33 °

Reflection and transmission

Total Internal Reflection

𝜃𝑔𝑙𝑎𝑠𝑠

𝑛𝑔𝑙𝑎𝑠𝑠≈ 1.5

=.75

𝜃𝑎𝑖𝑟

𝜃𝑔𝑙𝑎𝑠𝑠For small

W?

𝜃𝑎𝑖𝑟 ≈ si n−1 [𝑛𝑔𝑙𝑎𝑠𝑠 sin (𝜃𝑔𝑙𝑎𝑠𝑠 ) ]

=.96

=1.15

𝜃𝑎𝑖𝑟 ≈ 49 °

𝜃𝑎𝑖𝑟 𝑑𝑜𝑒𝑠𝑛′ 𝑡 𝑒𝑥𝑖𝑠𝑡…𝑛𝑜𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑜𝑛

𝜃𝑎𝑖𝑟 ≈ 75 °