EM Waves. From previous Lecture. This Lecture More on EM waves EM spectrum Polarization. Displacement currents Maxwell s equations EM Waves

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1 EM Waves This Lecture More on EM waves EM spectrum Polarization From previous Lecture Displacement currents Maxwell s equations EM Waves 1

Reminders on waves Traveling waves on a string along x obey the wave equation: 2 y(x,t) x 2 = 1 v 2 2 y(x,t) t 2 y=wave function General solution : y(x,t) = f1(x-vt) + f2(x+vt) y = displacement λ

2 Reminders on waves Traveling waves on a string along x obey the wave equation: 2 y(x,t) x 2 = 1 v 2 2 y(x,t) t 2 y=wave function General solution : y(x,t) = f1(x-vt) + f2(x+vt) y = displacement λ pulse traveling along +x pulse traveling along -x Traveling wave: superposition of sinusoidal waves (produced by a source that oscillates with simple harmonic motion): y(x,t) = A sin(kx-ωt) y(x,t) = sin(kx+ ωt) A = amplitude k = 2π/λ = wave number λ = wavelength f = frequency T = 1/f = period v ω = 2πf=2π/T angular frequency 2

3 Maxwell equations in vacuum in the absence of charges (q=0) and conduction currents (I=0) E da = 0 (Gauss' Law) B da = 0 S S L E ds = dφ B dt (Faraday - Henry) B ds = L µ 0 ε 0 dφ E dt (Ampere - Maxwell law) From these equations we get EM wave equations traveling in vacuum!

4 EM waves from Maxwell equations Solutions of these equations: sinusoidal traveling transverse waves propagating along x E = E max cos (kx ωt) B = B max cos (kx ωt) λ E x B direction of c E and B are perpendicular oscillating vectors E B = 0 4

5 E and B are orthogonal E I An easy way to understand this: Φ B = B A cosθ θ Increasing B-field 1. Max B flux θ=0 => also circular E is largest! 2. Less flux 3. Null flux θ=90 => circular E smallest! B parallel to area normal and E perpendicular to circuit so E B E orthogonal to B! E ds = - dφ B dt E B = 0 5

6 Quick Quiz on EM Waves Shown below is the E-field of an EM wave broadcast at 30 MHz and traveling to the right. What is the direction of the magnetic field during the first λ/2? 1) Into the page 2) out of the page E λ/2 What is the wave length? 1) 10 m 2) 5 m x

7 Quick Quiz pn EM waves Which orientation will have the largest induced emf? E y z x B loop in xy plane loop in xz plane A B C loop in yz plane

8 Important Relation between E and B E = E max cos (kx ωt) B = B max cos (kx ωt) From: First derivatives: E x = ke max sin(kx ωt) B t = ωb max sin(kx ωt) E x = B t This relation comes from Maxwell s equations!

charges oscillate between the rods (a) As oscillations continue, the rods become less charged, the field near the

9 EM Waves generators: Antennas Sources of EM waves: oscillating charges, accelerated/decellerated charges, electron transitions between energy levels in atoms, nuclei and molecules 2 rods connected to alternate current generator; charges oscillate between the rods (a) As oscillations continue, the rods become less charged, the field near the charges decreases and the field produced at t = 0 moves away from the rod (b) The charges and field reverse (c) The oscillations continue (d)

The EM Spectrum Gamma rays: λ~ 10-14 - 10-10 m Source: radioactive nuclei, cause serious damage to living tissues X-rays: ~10-12 -10-8 m source: deceleration of

by ozone Infrared: λ ~ 7 x 10-7 -10-3 m Sources: hot objects and molecules Microwaves: λ ~10-4 -0.

10 The EM Spectrum Gamma rays: λ~ m Source: radioactive nuclei, cause serious damage to living tissues X-rays: ~ m source: deceleration of high-energy electrons striking a metal target Diagnostic tool in medicine UV λ~ 6 x x 10-7 m Most UV light from the sun is absorbed in the stratosphere by ozone Infrared: λ ~ 7 x m Sources: hot objects and molecules Microwaves: λ ~ m sources: electronic devices radar systems, MW ovens Source: atoms and molecules Human eye Visible range from red (700 nm) to violet (400 nm) Radio: λ ~ m Sources: charges accelerating through conducting wires Radio and TV

11 Poynting vector Rate at which energy flows through a unit area perpendicular to direction of wave propagation This is the power per unit area (J/s. m 2 = W/m 2 ) E/B=c Its direction is the direction of propagation of the EM wave Magnitude: S = EB = E 2 µ 0 cµ 0 Magnitude is time dependent reaches a max at the same instant as E and B do

12 Energy density of E and B field In a parallel plate capacitor: U = 1 2 CV 2 = 1 ε 0 A 2 d E 2 d 2 u E = U Ad = 1 2 ε 0E 2 C = ε 0A d True for any geometry Similarly for a solenoid with current: u B = B2 2µ 0 True for any geometry

13 Energy carried by EM waves Total instantaneous energy density of EM waves u =u E + u B = 1/2 ε o E 2 + B 2 /(2µ o ) Since B = E/c and u E = u B u E = 1 2 ε 0E 2 = u B = B2 2µ 0 = E 2 2c 2 µ 0 EM waves carry energy! In a given volume, the energy is shared equally by the two fields The average energy density over one or more cycles of oscillations is: u av = 2u E,av = ε 0 E 2 = 1 2 ε 2 0E max since <sin 2 (kx - ωt)> = 1/2

14 Intensity and Poynting vector Let s consider a cylinder with axis along x of area A and length L and the time for the wave to travel L is Δt=L/c The average power of the EM wave in the cylinder is: P av = U av Δt = u av AL Δt = u av Ac E/B=c The intensity is I = P av A = 1 2 ε 2 0cE max I= S av I E 2 = E maxb max 2µ 0 average power per unit area (units W/m 2 ) E x B

15 Radiation pressure and momentum Radiation momentum: F = ma = m dv dt = dp dt time p = force time = force distance distance = energy velocity = U c the radiation momentum is the radiation energy/velocity Radiation Pressure p rad = F A = p AΔt = U caδt = Power ca = I c S av = I = Power A Complete absorption on a surface: total transferred momentum p = U / c and prad=sav/c Perfectly reflecting surface: p = 2U/c and p rad = 2Sav/c

Radiation pressure from the Sun p rad = F A = Power / A = I = S av c c c Solar intensity at the Earth (1.5 x 10 8 km far from the Sun): 1350 W/m 2 Direct sunlight pressure I/c ~4.

16 Radiation pressure from the Sun p rad = F A = Power / A = I = S av c c c Solar intensity at the Earth (1.5 x 10 8 km far from the Sun): 1350 W/m 2 Direct sunlight pressure I/c ~4.5 x 10-6 N/m 2 If the sail is a reflecting mirror prad = 2 x 4.5 x 10-6 N/m 2 Can be used by spacecrafts for propulsion as wind for sailing boats! What is the sail area needed to accelerate a 10 4 kg spacecraft at a = 0.01 m/s 2 assuming perpendicular incidence of the radiation on the sail? A = F/p rad = ma/prad = 10 7 m 2

17 Polarization of Light Waves (34.8) Linearly polarized waves: E-field oscillates at all times in the plane of polarization Unpolarized light: E-field in random directions. Superposition of waves with E vibrating in many different directions Linearly polarized light: E-field has one spatial orientation

Circular and elliptical polarization Circularly polarized light: superposition of 2 waves of equal amplitude with orthogonal linear polarizations, and 90 out of phase.

18 Circular and elliptical polarization Circularly polarized light: superposition of 2 waves of equal amplitude with orthogonal linear polarizations, and 90 out of phase. The tip of E describes a circle (counterclockwise = RH and clockwise=lh depending on y component ahead or behind) The electric field rotates in time with constant magnitude. If amplitudes differ elliptical polarization

19 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 polarization absorbed This polarization transmitted transmission axis Long-chain hydrocarbon molecules Polaroid sheet (E. Land 1928)

20 DEMO with MW generator and metal grid pick up antenna connected to Ammeter Metal grid MW generator If the wires of the grid are parallel to the plane of polarization the grid absorbs the E-component (electrons oscillate in the wire). The same thing happens to a polaroid: the component parallel to the direction of the chains of hydrocarbons is absorbed. If the grid is horizontal the Ammeter will measure a not null current since the wave reaches the antenna pick-up This polarization absorbed This polarization transmitted transmission axis

21 Relative orientation of polarizers Transmitted amplitude is E o cosθ (component of polarization along polarizer axis) Transmitted intensity is I o cos 2 θ ( square of amplitude) Perpendicular polarizers give zero intensity.

22 Polarizers and MALUS LAW If linearly polarized light (plane of polrization indicated by red arrow) of intensity I 0 passes through a polarizing filter with transmission axis at an angle θ along y y θ Polaroid sheet E inc = E 0 sinθ i + E 0 cosθ j After the polarizer E transm = E 0 cosθ j So the intensity transmitted is I transm = E 0 2 cos 2 θ = Ι 0 cos 2 θ Long-chain hydrocarbon molecules x transmission axis E 0 cosθ Transmitted intensity: I = I 0 cos 2 θ I 0 = intensity of polarized beam on analyzer (Malus law) this is called Malus law

23 Unpolarized light on polarizers Only I0/2 is transmitted of unpolarized light by a polarizer and it is polarized along the transmission axis. An analyzer rotated at an angle θ respect to the polarizer transmits 100% of the incident intensity when θ = 0 and zero when θ = 90 Allowed component parallel to analyzer axis Polaroid sheets

Maxwell s equations and EM waves. From previous Lecture Time dependent fields and Faraday s Law

Maxwell s equations and EM waves This Lecture More on Motional EMF and Faraday s law Displacement currents Maxwell s equations EM Waves From previous Lecture Time dependent fields and Faraday s Law 1 Radar