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http://www.labinitio.com (nz302.jpg) adapted from http://www.labinitio.com (nz302.jpg) Today’s lecture is brought to you by the letter P. http://www.labinitio.com (nz288.jpg) Announcements There is lots of nice math in chapter 10! This lecture calls your attention to those parts of the chapter that you need to know for exams. Keep this lecture in mind when you study chapter 10. Exam 3 is the two weeks from yesterday. I will need to know by the end of next Wednesday’s lecture of any students who have special needs different than for exam 2. Exam 3 will cover material through the end of today’s lecture. Material presented in lecture next week will be covered on the final exam. Note! Note: All of these mean “average:” <S> Saverage Sav Savg Today’s agenda: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave. rarely in the course of human events have so many starting equations been given in so little time We began this course by studying fields that didn’t vary with time—the electric field due to static charges, and the magnetic field due to a constant current. In case you didn’t notice—about a half dozen lectures ago things started moving! We found that changing magnetic field gives rise to an electric field. Also a changing electric field gives rise to a magnetic field. These time-varying electric and magnetic fields can propagate through space. Electromagnetic Waves Maxwell’s Equations q enclosed E dA o d B E ds dt B dA 0 dΦ E B ds=μ 0 Iencl +μ 0ε 0 dt These four equations provide a complete description of electromagnetism. E 0 B 0 dB ×E=dt 1 dE B= 2 +μ 0 J c dt Production of Electromagnetic Waves Apply a sinusoidal voltage to an antenna. Charged particles in the antenna oscillate sinusoidally. The accelerated charges produce sinusoidally varying electric and magnetic fields, which extend throughout space. The fields do not instantaneously permeate all space, but propagate at the speed of light. y x z direction of propagation y x direction of propagation z This static image doesn’t show how the wave propagates. Here are animations, available on-line: http://phet.colorado.edu/en/simulation/radio-waves (shows electric field only) http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=35 Here is a movie. Electromagnetic waves are transverse waves, but are not mechanical waves (they need no medium to vibrate in). Therefore, electromagnetic waves can propagate in free space. At any point, the magnitudes of E and B (of the wave shown) depend only upon x and t, and not on y or z. A collection of such waves is called a plane wave. y x z direction of propagation Manipulation of Maxwell’s equations leads to the following Equations on this slide are for waves plane wave equations for E and B: propagating along x-direction. 2E y x 2 = 0 0 2E y (x, t) 2B z 2B z (x,t) = 0 0 2 x t 2 t 2 These equations have solutions: Ey =Emax sin kx - t Bz =Bmax sin kx - t where 2 k= , = 2f , and Emax and Bmax are the electric and magnetic field amplitudes f = = c. k You can verify this by direct substitution. Emax and Bmax in these notes are sometimes written by others as E0 and B0. You can also show that E y B z =x t Emax sin kx - t x =- Bmax sin kx - t t Emax k cos kx - t =Bmax cos kx - t Emax E 1 = = =c= . Bmax B k 0 0 At every instant, the ratio of the magnitude of the electric field to the magnitude of the magnetic field in an electromagnetic wave equals the speed of light. y direction of propagation x z Emax (amplitude) E(x,t) Today’s agenda: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave. Energy Carried by Electromagnetic Waves Electromagnetic waves carry energy, and as they propagate through space they can transfer energy to objects in their path. The rate of flow of energy in an electromagnetic wave is described by a vector S, called the Poynting vector.* S= 1 E B 0 The magnitude S represents the rate at which energy flows through a unit surface area perpendicular to the direction of wave propagation (energy per time per area). Thus, S represents power per unit area. The direction of S is along the direction of wave propagation. The units of S are J/(s·m2) =W/m2. *J. H. Poynting, 1884. y E B z For an EM wave E B = EB EB so S = . 0 1 S = E B 0 S c x Because B = E/c we can write E2 cB 2 S= = . 0 c 0 These equations for S apply at any instant of time and represent the instantaneous rate at which energy is passing through a unit area. EB E2 cB 2 S= = = 0 0 c 0 EM waves are sinusoidal. Ey =Emax sin kx - t Bz =Bmax sin kx - t EM wave propagating along x-direction The average of S over one or more cycles is called the wave intensity I. The time average of sin2(kx - t) is ½, so 2 2 EmaxBmax Emax cBmax I = Saverage = S = = = 20 20 c 20 Notice the 2’s in this equation. This equation is the same as 10-29 in your text, using c = 1/(00)½. The magnitude of S is the rate at which energy is transported by a wave across a unit area at any instant: energy power time S= = area area instantaneous instantaneous Thus, energy power time I= S = = area area average average Note: Saverage and <S> mean the same thing! Energy Density The energy densities (energy per unit volume) associated with electric and magnetic fields are: 1 B2 uB = 2 0 1 uE = 0E2 2 Using B = E/c and c = 1/(00)½ we can write E 1 B2 1 c uB = = 2 0 2 0 2 1 0 0E2 1 = = 0E2 2 0 2 1 1 B2 2 uB = uE = 0E = 2 2 0 remember: E and B are sinusoidal functions of time 1 1 B2 2 uB = uE = 0E = 2 2 0 For an electromagnetic wave, the instantaneous energy density associated with the magnetic field equals the instantaneous energy density associated with the electric field. Hence, in a given volume the energy is equally shared by the two fields. The total energy density is equal to the sum of the energy densities associated with the electric and magnetic fields: B2 2 u = uB +uE = 0E = 0 2 B u = uB +uE = 0E2 = 0 instantaneous energy densities (E and B vary with time) When we average this instantaneous energy density over one or more cycles of an electromagnetic wave, we again get a factor of ½ from the time average of sin2(kx - t). 1 2 uE = 0Emax , 4 2 1 B max uB = , and 4 0 2 B 1 1 2 max u = 0Emax = 2 2 0 2 2 1 Emax 1 cBmax Recall Saverage = S = = so we see that S = c u . 2 0 c 2 0 The intensity of an electromagnetic wave equals the average energy density multiplied by the speed of light. Help! “These factors of ¼, ½, and 1 are making my brain hurt!” It’s really not that bad. These are the energy densities associated with E(t) and B(t) at some time t: 1 1 B2 2 uB = uE = 0E = 2 2 0 Add uB and uE to get the total energy density u(t) at time t: 2 B u = uB +uE = 0E2 = 0 Help! Again, these are the energy densities associated with E(t) and B(t) at some time t: 2 1 1 B uB = uE = 0E2 = 2 2 0 If you average uB and uE over one or more cycles, you get an additional factor of ½ from the time average of sin2(kx-t). 1 2 uE = 0Emax 4 2 1 B max uB = , 4 0 The Emax and Bmax come from writing E = Emax sin(kx-t) and B = Bmax sin(kx-t), and canceling the sine factors. Help! These are the average energy densities associated with E(t) and B(t) over one or more complete cycles. 1 2 uE = 0Emax 4 2 1 B max uB = , 4 0 Add uE and uB to get the total average energy density over one or more cycles: 2 2 B B 1 1 1 1 2 2 max max u = uE + uB = 0Emax + = 0Emax = 4 4 0 2 2 0 Help! Summary: At time t: Average: At time t: Average: 2 1 1 B (t) 2 uB (t) = uE (t) = 0E (t) = 2 2 0 1 2 uE = 0Emax 4 2 1 B max uB = , 4 0 2 B (t) 2 u(t) = 0E (t) = 0 2 B 1 1 2 max u = 0Emax = 2 2 0 If you use a starting equation that is not valid for the problem scenario, you will get incorrect results! Example: a radio station on the surface of the earth radiates a sinusoidal wave with an average total power of 50 kW.* Assuming the wave is radiated equally in all directions above the ground, find the amplitude of the electric and magnetic fields detected by a satellite 100 km from the antenna. Strategy: we want Emax, Bmax. We are given average power. From average power we can calculate intensity, frompower intensity From the and average we we cancan calculate max and B max.from calculate E intensity, and Satellite Station intensity we can calculate Emax and Bmax. *In problems like this you need to ask whether the power is radiated into all space or into just part of space. Example: a radio station on the surface of the earth radiates a sinusoidal wave with an average total power of 50 kW.* Assuming the wave is radiated equally in all directions above the ground, find the amplitude of the electric and magnetic fields detected by a satellite 100 km from the antenna. Area=4R2/2 All the radiated power passes through the hemispherical surface* so the average power per unit area (the intensity) is Satellite R Station P power I= = 2 area average 2R 5.00 10 W = 2 1.00 10 m 4 5 2 = 7.96 10-7 W m2 Today’s lecture is brought to you by the letter P. *In problems like this you need to ask whether the power is radiated into all space or into just part of space. 2 1 Emax I= S = 2 0 c Satellite R Emax = 20cI Station = 2 410-7 3 108 7.96 10-7 = 2.45 10-2 V Emax Bmax = = c m 2.45 10-2 V m = 8.17 10-11 T 8 3 10 m s You could get Bmax from I = c Bmax2/20, but that’s a lot more work Example: for the radio station in the example on the previous two slides, calculate the average energy densities associated with the electric and magnetic field. 1 2 uE = 0Emax 4 2 1 Bmax uB = 4 0 1 -12 -2 2 uE = 8.85 10 2.45 10 4 1 8.17 10 uB = 4 4 10-7 uE =1.33 10 -15 J m3 -11 2 uB =1.33 10 -15 J m3 If you are smart, you will write <uB> = <uE> = 1.33x10-15 J/m3 and be done with it. Today’s agenda: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave. Momentum and Radiation Pressure EM waves carry linear momentum as well as energy. The momentum density carried by an electromagnetic wave is S dp = 2 dV c This equation is not on your equation sheet, but you have permission to use it for tomorrow’s homework (if needed) (10.28) where dp is the momentum carried in the volume dV. Today’s lecture is brought to you by the letter P. When the momentum carried by an electromagnetic wave is absorbed at a surface, pressure is exerted on that surface. If we assume that EM radiation is incident on an object for a time t and that the radiation is entirely absorbed by the object, then the object gains energy U in time t. Maxwell showed that the momentum change of the object is then: U p = (total absorption) c incident The direction of the momentum change of the object is in the direction of the incident radiation. If instead of being totally absorbed the radiation is totally reflected by the object, and the reflection is along the incident path, then the magnitude of the momentum change of the object is twice that for total absorption. incident reflected 2U p = c (total reflection along incident path) The direction of the momentum change of the object is again in the direction of the incident radiation. Radiation Pressure The radiation pressure on the object is the force per unit area: P= From Newton’s 2nd F A Law (F = dp/dt) we have: P = For total absorption, p = F 1 dp = A A dt U c dU 1 dp 1 d U 1 dt S So P = = = = A dt A dt c c A c incident (Equations on this slide involve magnitudes of vector quantities.) This is the instantaneous radiation pressure in the case of total absorption: S(t) P(t) = c For the average radiation pressure, replace S by <S>=Savg=I: Prad = S average c I = c Electromagnetic waves also carry momentum through space with a momentum density of Saverage/c2=I/c2. This is not on your equation sheet but you have special permission to use it in tomorrow’s homework, if necessary. Today’s lecture is brought to you by the letter P. I Prad = (total absorption) c incident absorbed Using the arguments above it can also be shown that: 2I Prad = (total reflection) c incident reflected If an electromagnetic wave does not strike a surface, it still carries momentum away from its emitter, and exerts Prad=I/c on the emitter. Example: a satellite orbiting the earth has solar energy collection panels with a total area of 4.0 m2. If the sun’s radiation is incident perpendicular to the panels and is completely absorbed find the average solar power absorbed and the average force associated with the radiation pressure. The intensity (I or Saverage) of sunlight prior to passing through the earth’s atmosphere is 1.4 kW/m2. Power = IA = 1.4 103 W m2 2 3 4.0 m = 5.6 10 W = 5.6 kW Assuming total absorption of the radiation: Prad = S average c I = = 1.4 103 W c 3 10 8 m F = Prad A = 4.7 10-6 N m 2 m = 4.7 10 Pa 2 s -6 2 -5 4.0 m =1.9 10 N Caution! The letter P (or p) has been used in this lecture for power, pressure, and momentum! That’s because today’s lecture is brought to you by the letter P. Revolutionary Application of Electromagnetic Waves I know you are mostly engineers, and think applications are important… So I found you a revolutionary new application that uses electromagnetic waves. The UFO Detector. Only $48.54* at amazon.com (just search for ufo detector). “The UFO detector continually monitors its surrounding area for any magnetic and electromagnetic anomalies.” *2014 price New starting equations from this lecture: 1 S = E B 0 2 2 1 Emax 1 cBmax Saverage = = 2 0 c 2 0 Emax E 1 = =c= Bmax B 0 0 1 1 B2 2 uB = uE = 0E = 2 2 0 2 k= , = 2f , f = = c k 2 B 1 1 2 max u = 0Emax = 2 2 0 U 2U p = or c c I 2I Prad = or c c There are even more on your starting equation sheet; they are derived from the above!