2002 Ch. 24

Ch. 24 Electromagnetic WavesTake a single positive charge and wiggle it up and down:The charge changes position as a function of time, so as the electric field E.II TimeBut since the charge is moving, it constitutes a current:The current points up when the charge moves upup, and the current pointsdown when the charge moves down.gfield!This current, like all currents, creates a magneticThe direction of the field is given by RHR-2.

zBXIxI yBBy RHR-2, we see that when the current points up, the mag. fieldpoints into the screen, and when the current points down, themag. field points out of the screen.Thus, I have a changing magnetic field and a changing electric field which areoriented at right angles to each other!The electric field is in the xz-plane, and the magnetic field is in the xy-plane.The fields move out away from the source (our accelerating charge):Propagation of Electromagnetic (EM) WavesAn EM wave is a transverse wave:The wave motion is at right anglesto the direction of propagation.

It was James Clerk Maxwell (1831-1879) who worked out the mathematics ofthe wave propagation:In words: The changing electric field induces a magnetic field (which alsochanges), and this changing magnetic field induces an electric field, etc.This is how the wave propagates!EM wavesa es dodon’tt needeed a mediumed u to ttravela e tthrough.oug Theyey cacan ppropagateopagatethrough a vacuum.How fast do EM waves travel?We can answer this question by looking at Maxwell’s wave equation:dE B μ o J μ oε odtFrom partialFti l differentialdiffti l equations,tiwe identifyid tif theth speedd2 off theth wave asone over the coefficient on dE/dt.

Only fields exist in the free space, the wave equation becomes:1 dE B 2v dtSov 21μ oε odB E dt v 1μ oε o11 v v 9 7 123.334 10(4π 10 )(8.85 10 ) v 2.999 108 m/sWow! EM waves propagate at the speed of light!!!!

Review:1 Stationary charges create electric fields.1.fields2. Moving charges (constant velocity) create magnetic fields.3. Accelerating charges create electromagnetic waves.24.2 Electromagnetic SpectrumSince Maxwell discovered that EM waves move at the speed oflight, he hypothesized that light itself must be an EM wave!Like any wave, EM waves have a frequency, a period, and an amplitude.From Ch. 16, we know that:v fλAnd since v c, we get for EM waves:Speed of lightc fλFrequency((Hz))Wavelength(m)

f cλHigher frequencies mean shorter wavelengths!The Electromagnetic Spectrum

24.3 The Speed of LightVery fast!!!!.but finite.c 3.00 10 m/s8MMoont EtoEarthth 1.31 3 secondsdSun to Earth 8 minutesDistant stars and other astronomical objects are so far away that astronomers usea unit of distance called the light year (ly).1 ly The distance light travels in 1 year 9.5 1015 m

Question 24 - 1The FM station broadcastingg at 103.3 MHz pplaysymusic for the “Diva in all of us”. What is thewavelength of these radio waves?1.22.3.44.0.344 m29m2.93.4 10-7 m2 9 106 m2.9

24.4 Energy Carried by EM WavesEM waves carryy energygy jjust like anyy other wave.An EM wave consists of both an electric and magnetic field, and energy iscontained in both fields.Energy density in electric field (in free space):Electrical Energy 1Electrical Energy Density 2 εoE2VolumeEEnergyddensityi ini magnetici fieldfi ld (in(i freefspace):)Magnetic Energy Density 12μoB2Notice that the energy goes as the square of the fields.

So the total energy density in an EM wave is the sum of these two:u εoE 21212μoB2But in an EM wave, the electric field and magnetic field carry the same energy.12Thus,εoE 212μoB2This allows me to write the total energy density in terms of just Ε or just Β:u εoE2 SinceSince,12εoE 212μoB21μoB2 E 21μ oε oB21 2 E 2 Bc2

E cBSo in an EM wave, the magnitude of the electric field is proportional to themagnitude of the magnetic field, and the proportionality constant is c, thespeed of light!The magnitude of the electric and magnetic fields in an EM wave fluctuatein time. It is useful to consider an average value of the two fields:This is called the rms or root-mean-square value of the fields:ErmsEo 2BrmsBo 2HHere,Eo andd Bo are theth peakk valuesloff ththe ttwo fields.fi ldNow we can calculate an average energy density using the rms values:u εoE122rms 12μo2rmsB

So as the EM waves propagate thru space, they carry energy along with them.The transportpof energygy is related to the intensityy of the wave.Back in Ch. 16 we defined the intensity of a wave as the power per unit area:PWEnergyS AtAtAWhat is the relationship between the intensity, S, and the energy density, u?Sparing you the derivation, we find thatS cuSo the intensity and energy density are just related by the speed of light, c.In terms of the fields then:If we use the rms values forthe fields, then S S, theaverage intensity.S cu cε o E 12S cε o E 2S c2μoB22c2μoB2

Question 24 - 2Both the electric and magneticgfield of an EM aredoubled in magnitude. What happens to the intensityof the wave?1.2.3.4.5.NothingIt doublesIt quadruplesIt decreases by a factorof 2It decreases by a factorof 4

24.5 The Doppler EffectWhen the observer of a wave,, or source of the wave (or( both)) is moving,g,the observed wave frequency is different than that emitted by the source.EM waves also exhibit a Doppler effect. But .1. They don’t require a medium thru which to propagate, and.2. Only the relative motion of the source to the observer isp, since the speedpat which all EM waves move is theimportant,same, the speed of light.So how do we calculate the shift in frequency?If the EM wave, the source, and the observer all travel along the same line, then: vrel f o f s 1 c fo is the observed frequencyfs isi ththe ffrequency emitteditt d bby theth sourcevrel is the relative velocity between observer and sourceThe signg is used when the objectjand source move toward each other.The – sign is used when the object and source move away from each other.(*This is valid for speeds vrel c.)

Astronomers can use the Doppler effect to determine how fast distant objectsare moving relative to the earth.EExamplelA distant galaxy emits light that has a wavelength of 500.7 nm. On earth, thewavelengthg of this lightg is measured to be 503.7 nm. ((a)) Decide if the ggalaxyy is movinggaway from or toward the earth. (b) Find the speed of the galaxy relative to earth.SolutionW startWet t withith theth DopplerDl equation:ti vrel f o f s 1 c The light is shifted to longer wavelengths, which means smaller frequencies: f c/λThus, fo fs. Which means that the parenthesis 1 vrel must be 1.c Therefore, the correct sign in the parentheses is the – sign: the galaxy ismovingi away ffrom earth.th(b) From the Doppler equation:Thus,vrel c (1 )λsλo(vrel c 1 fofs). 3 10 (1 8But500.7 nm503.7 nmf c .λ) 1.8 10 m/s6

24.6 PolarizationEM waves are transverse waves and can be polarized.Consider the electric field part of an EM wave.It oscillates up and down as the wave propagates:Thus, the wave oscillations are perpendicular to thedirection of travel and occur in only one direction.We refer to this wave as linearly polarized.polarizedA vertical slit would allow the waveto ppass through,g , since the slit isparallel to the oscillations:Direction ofelectric fieldA horizontal slit would block thewave and not allow it to pass,since the slit is perpendicular tothe oscillations:

So how is polarized light produced?We could use the antenna on a radio station:AntennaThe up and down vibrations of the electrons in the antenna produce polarizedwaves whose direction is determined by the orientation of the antenna.Incandescent light is unpolarized, resulting from many atoms vibrating in allpossible orientations.Antenna-IncandescentLight

We can convert unpolarized light into polarized light by blocking all but one ofthe electric field orientations.A device that does this is called a polarizer or polaroidpolaroid.The one direction that a polarizer allows light to pass thru it is called thetransmission axisaxis.Let’s start with unpolarized light and pass it thru a polarizer:If the intensity of the unpolarized light is S before it passes thru thepolarizer, then the intensity of the polarized light coming out will be ½S.

Malus’ LawOnce the light has been polarized, it’s possible for another polarizer (called theanalyzer) to change the direction and intensity of the polarized light.We know from our earlier discussion on intensity that:Therefore:S E 2.Out of the analyzer then,S cε o E 2S E 2 cos 2θ .So both the intensity and polarization direction can be adjusted by changingthe angle of the analyzer.The average intensity of the light leaving the analyzer is:S o is the average intensity of the light entering the analyzer.S S o cos 2 θMalus’ Law

Out of the analyzer then, S E2cos2θ. So both the intensity and polarization direction can be adjusted by changing the angle of the analyzer. The average intensity of the light leaving the analyzer is:The average intensity of the light leaving the analyzer is: S S cos2 θ o Sois the average intensity of the light entering the analyzer .