Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
EM Waves Reminders on waves This Lecture !More on EM waves !EM spectrum !Polarization ! Traveling waves on a string along x obey the wave equation: " 2 y(x,t) 1 " 2 y(x,t) = 2 "x 2 v "t 2 y=wave function General solution : y(x,t) = f1(x-vt) + f2(x+vt) y = displacement ! From previous Lecture Displacement currents !Maxwell’s equations !EM Waves ! 1 Maxwell equations in vacuum ! 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-"t) y(x,t) = sin(kx+ "t) A = amplitude k = 2#/! = wave number ! = wavelength v f = frequency T = 1/f = period " = 2#f=2#/T angular frequency 2 Solutions of these equations: sinusoidal traveling transverse waves propagating along x ! # E" ds = $ L # SB" dA = 0 d%B (Faraday - Henry) dt ! # B" ds = µ & L 0 0 pulse traveling along -x EM waves from Maxwell equations in the absence of charges (q=0) and conduction currents (I=0) #SE" dA = 0 (Gauss' Law) ! d%E (Ampere - Maxwell law) dt E = Emax cos (kx – "t) B = Bmax cos (kx – "t) E x B direction of c ! From these equations we get EM wave equations traveling in vacuum! •E and B are perpendicular oscillating vectors ! E!B=0 4 Quick Quiz on EM Waves E and B are orthogonal ! An easy way to understand this: 3 2 1 $ E I Increasing B-field !B = B A cos$ 1. Max B flux $=0 => also circular E is largest! 2. Less flux 3. Null flux $=90° => circular E smallest! B parallel to area normal and E perpendicular to circuit so E % B $ E " ds = - E orthogonal to B! 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 !/2? 1) Into the page 2) out of the page E x d#B dt !/2 What is the wave length? E!B=0 ! 5 1) 10 m 2) 5 m Quick Quiz pn EM waves Important Relation between E and B Which orientation will have the largest induced emf? E y x z B loop in xz plane loop in xy plane A B C yz in p o lo ane pl ! E = Emax cos (kx – "t) ! B = Bmax cos (kx – "t) ! First derivatives: From: "E = #kE max sin(kx # $t) "x "B = $Bmax sin(kx # $t) "t "E "B =# "x "t This relation comes from Maxwell’s equations! ! EM Waves generators: Antennas Source: atoms and molecules Human eye Visible range from red (700 nm) to violet (400 nm) The EM Spectrum Sources of EM waves: oscillating charges, accelerated/decellerated charges, electron transitions between energy levels in atoms, nuclei and molecules Gamma rays: !~ 10-14- 10-10 m Source: radioactive nuclei, cause serious damage to living tissues X-rays: ~10-12 -10-8 m source: deceleration of high-energy electrons striking a metal target Diagnostic tool in medicine UV !~ 6 x 10-10 - 4 x 10-7 m Most UV light from the sun is absorbed in the stratosphere by ozone 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) ! ! ! ! Microwaves: ! ~10-4 -0.3 m sources: electronic devices radar systems, MW ovens Energy density of E and B field Poynting vector E/B=c ! True for any geometry ! ! ! Similarly for a solenoid with current: EB E = µ0 cµ0 uB = Magnitude is time dependent ! reaches a max at the same instant as E and B do ! "0 A d 1 1" A U 1 U = CV 2 = 0 E 2 d 2 # uE = = "0 E 2 2 2 d Ad 2 Its direction is the direction of propagation of the EM wave 2 Magnitude: S= ! In a parallel plate capacitor: C = This is the power per unit area (J/s.m2 = W/m2) ! ! ! Rate at which energy flows through a unit area perpendicular to direction of wave propagation ! Radio: ! ~ 10 - 0.1 m Sources: charges accelerating through conducting wires Radio and TV Infrared: ! ~ 7 x 10-7-10-3 m Sources: hot objects and molecules B2 2µ0 True for any geometry ! Energy carried by EM waves ! ! Total instantaneous energy density of EM waves u =uE + uB = 1/2 'o ! Intensity and Poynting vector E2 + B2 /(2µo) ! uE = uB Since B = E/c and Pav = 1 B2 E2 uE = "0 E 2 = uB = = 2 2 2µ0 2c µ0 ! EM waves carry energy! ! In a given volume, the energy is shared equally by the two fields ! 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 !t=L/c The average power of the EM wave in the cylinder is: U av uav AL = = uav Ac "t "t The intensity is I= ! ! I= Sav The average energy density over one or more cycles of oscillations is: 1 2 since <sin2(kx - "t)> = 1/2 uav = 2uE ,av = "0 E 2 = "0 E max E/B=c Pav 1 E B 2 = "0cE max = max max A 2 2µ0 I & E2 ! average power per unit area (units W/m2) ExB 2 ! ! Radiation pressure and momentum ! dv dp F = ma = m = dt dt time energy U p = force " time = force " distance " = = distance velocity c Radiation momentum: ! the radiation momentum is the radiation energy/velocity Radiation pressure from the Sun prad = ! ! prad ! ! ! Radiation Pressure Power F p U Power I Sav = I = = = = = = A A A"t cA"t cA c Complete absorption on a surface: total transferred momentum p = U / c and prad=Sav/c ! Solar intensity at the Earth ( 1.5 x 108 km far from the Sun): 1350 W/m2 ! ! ! ! ! ! ! Direct sunlight pressure I/c ~4.5 x 10-6 N/m2 ! If the sail is a reflecting mirror prad = 2 x 4.5 x 10-6 N/m2 Can be used by spacecrafts for propulsion as wind for sailing boats! What is the sail area needed to accelerate a 104 kg spacecraft at a = 0.01 m/s2 assuming perpendicular incidence of the radiation on the sail? A = F/prad = ma/prad = 107 m2 Perfectly reflecting surface: p = 2U/c and prad = 2Sav/c Circular and elliptical polarization Polarization of Light Waves (34.8) ! S F Power / A I = av = = A c c c 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 ! 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 DEMO with MW generator and metal grid 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 pick up antenna connected to Ammeter Metal grid MW generator This polarization absorbed This polarization transmitted transmission axis Polaroid sheet (E. Land 1928) Long-chain hydrocarbon molecules 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 This not null current since the wave reaches the antenna polarization absorbed pick-up This polarization transmitted transmission axis Relative orientation of polarizers Polarizers and MALUS’ LAW If linearly polarized light (plane of polrization indicated by red arrow) of intensity I0 passes through a polarizing filter with transmission axis at y Polaroid sheet an angle $ along y $ Einc = E0sin$ i + E0 cos$ j After the polarizer Etransm = E0cos$ j So the intensity transmitted is Itransm = E02 cos2$ = )0cos2$ ! ! ! Transmitted amplitude is Eocos$ (component of polarization along polarizer axis) Transmitted intensity is Iocos2$ ( square of amplitude) Perpendicular polarizers give zero intensity. 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 $ respect to the polarizer transmits 100% of the incident intensity when $ = 0 and zero when $ = 90° Allowed component parallel to analyzer axis Polaroid sheets ! x transmission axis E0cos$ Long-chain hydrocarbon molecules Transmitted intensity: I = I0cos2$ I0 = intensity of polarized beam on analyzer (Malus’ law) this is called Malus’ law