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Lecture 21. Energy Transfer in Electromagnetism Energy and Intensity of Electromagnetic Waves Outline: Transfer of EM Energy in Space: Poynting Formalism. Energy Transfer by EM Waves. Radiation Pressure Final exam, Tuesday, December 20, 4:00 to 7:00 pm on CAC. If you have 3 exams on December 20 or have 3 exams in a row that includes Physics 227, ASAP but not later than 5:00 pm on December 5 send your ENTIRE exam schedule to Prof. Cizewski [email protected]. Iclicker Question Consider the magnetic field π΅ = π₯ sin ππ₯ + ππ‘ . Can this field represent the plane electromagnetic wave in free space? A. sure, why not B. no, because itβs a sin and not cos function of the phase. C. no, because π΅ is directed along the direction of phase variation. D. no, because of a wrong dependence of the phase on π‘. E. no, because of B - D. 2 Iclicker Question Consider the magnetic field π΅ = π₯ sin ππ₯ + ππ‘ . Can this field represent the plane electromagnetic wave in free space? A. sure, why not B. no, because itβs a sin and not cos function of the phase. C. no, because π΅ is directed along the direction of phase variation. D. no, because of a wrong dependence of the phase on π‘. E. no, because of B - D. 3 Energy Density in E.-M. Waves EM waves: πΈ = ππ΅ This doesnβt mean that B is βweakerβ than E (βapplesβ vs. βorangesβ). The meaningful comparison - the energy densities in the πΈ and π΅ fields that form the wave: Energy density in any πΈ field: Energy density in any π΅ field: π’πΈ = π’π΅ = π0 2 πΈ π, π‘ 2 1 2 π΅ π, π‘ 2π0 (lecture 23) π0 2 π0 2 2 1 1 2 2 π’πΈ = πΈ = π π΅ = π = = π΅ = π’π΅ 2 2 π0 π0 2π0 Energy densities of the πΈ and π΅ fields in e.-m. waves are equal. 4 Energy Density in E.-M. Waves (contβd) Energy density in e.-m. waves: π’πΈπ΅ 2 1 π΅ 1 2 1 2 2 = π πΈ + = π0 πΈ = π΅ = πΈπ΅ 2 0 π0 π0 ππ0 The time-averaged energy density in a monochromatic e.-m. wave: πππ 2 ππ₯ β ππ‘ π‘ = 1 1 + cos 2 ππ₯ β ππ‘ 2 2 πΈ0 2 πππ 2 ππ₯ β ππ‘ π’πΈπ΅ π‘ π‘ = π‘ 1 2 1 = πΈ0 2 2 π0 2 1 1 2 = πΈ0 = π΅ = πΈ π΅ 2 2π0 0 2ππ0 0 0 πΈ0 , π΅0 - the amplitudes 5 Intensity of Electromagnetic Waves Intensity: the average energy transported by the e.-m. wave per unit area per 1s (the flux of energy per 1m per 1s, scalar) π’πΈπ΅ π‘ 1 = πΈπ΅ 2π0 π πΌ = π π’πΈπ΅ π‘ ππ¬ , ππ© 1m2 π π β 1π 1 1 = πΈ0 π΅0 = ππ0 πΈ0 2 2π0 2 Units: W/m2 Example: The intensity of sunlight hitting the Earth is ~ 1,400 W/m2. Find the amplitudes of the electric and magnetic fields in the e.-m. wave. 1 1 πΌ= πΈ0 π΅0 = πΈ0 2 2π0 2π0 π πΈ0 = 2πΌπ0 π = 2800 × 3 β 108 × 4π β 10β7 πΈ0 103 π/π β6 π π΅0 = = β 3 β 10 π 3 β 108 π/π π π β 1,000 π π (~ 1% of the Earthβs magnetic field) 6 Example In a fairly brightly-lit room, the intensity of light is about 100 W/m2. What is the amplitude of the electric field of the light waves in the room? If the roomβ volume is 50π3 and the light intensity is fairly uniform throughout the room, about how much energy is stored in the room in the form of light waves? πΌ = π π’πΈπ΅ π’πΈπ΅ Total energy: π‘ π‘ π’πΈπ΅ π‘ πΌ = π 100 π/π2 β7 π½/π3 = β 3.3 β 10 3 β 108 π/π π’πΈπ΅ π‘ β π£πππ’ππ = π½ β7 3.3 β 10 π3 β 50π3 β 1.7 β 10β5 π½ 7 Iclicker Question Imagine that when you switch on your lamp, you increase the intensity of light shinning on your textbook by a factor of 16. By what factor does the average electric field strength in this light increase? A. 256 1 πΌ = ππ0 πΈ0 2 2 B. 16 C. 4 D. 2 E. 1 8 Iclicker Question Imagine that when you switch on your lamp, you increase the intensity of light shinning on your textbook by a factor of 16. By what factor does the average electric field strength in this light increase? A. 256 1 πΌ = ππ0 πΈ0 2 2 B. 16 C. 4 D. 2 E. 1 9 Poynting Vector How does the electromagnetic energy get from one place to another? 1m2 Intensity : the average energy transported by the e.-m. wave per unit area per 1s (scalar) πΌ = π π’πΈπ΅ π‘ 1 = πΈ0 π΅0 2π0 ππ¬ , ππ© π β 1π In order to retain the direction of the energy flow (along π), letβs try πΈ × π΅: Poynting vector: (energy transferred through 1 m2 per 1s in the direction of π) π π‘ 1 = πΈ0 π΅0 πππ 2 ππ₯ β ππ‘ π0 π‘ 1 π= πΈ×π΅ π0 1 = πΈ π΅ 2π0 0 0 For a traveling monochromatic e.-m. wave: π π - the time-averaged flux of energy π‘ 1 = πΈ0 π΅0 π = πΌπ 2π0 10 Iclicker Question The drawing shows a sinusoidal electromagnetic standing wave. The average (averaged over either x or t) Poynting vector in this wave A. points along the x-axis. B. points along the y-axis. C. points along the z-axis. D. is zero. E. none of the above 1 π= πΈ×π΅ π0 11 Iclicker Question The drawing shows a sinusoidal electromagnetic standing wave. The average (averaged over either x or t) Poynting vector in this wave A. points along the x-axis. B. points along the y-axis. C. points along the z-axis. D. is zero. E. none of the above 1 π= πΈ×π΅ π0 12 Poynting Formalism (contβd) The concept of Poynting vector applies to all πΈ and π΅ fields, not just to the e.-m. waves. Poynting formalism works in all situations: 1 π= πΈ×π΅ π0 (a) static In statics πΈ , being a potential field, is normal to the equipotential surfaces, thus π is tangential to the equipotential surfaces. (b) dynamic note that in this case πΈ is non-conservative, and the equipotential surfaces do not make sense. Few consequences: (a) Whenever we have both E and B (and πΈ β¦ π΅), the e.-m. energy flows in space! (even if we deal with static fields). (b) The flow of energy is associated with the momentum of the e.m. waves (remember, the energy has βmassβ). (c) The momentum transfer (absorption or reflection of e.-m. waves) implies radiation pressure. 13 Example: Transmission Line π© We can apply the Poynting vector formalism to obtain the well-known result: π = ππΌ. However, the fields are non-uniform, and the integration is difficult. Letβs simplify the geometry. π° π¬ R Consider a transmission cable that consists of two flat metal ribbons of width W, a small distance a<<W apart. The two conductors are held at a potential difference V, and carry current I that travels uniformly down one strip and back along the other. V π π° π° Transmitted power to the load R : π π¬ π© π= ππππ π π πππ‘πππ π π πΈ= π πΌ π΅ = π 0 πΎ = π0 π both uniform, orthogonal to one another 1 1π πΌ πΈ × π΅ β ππ΄ = π0 ππ = ππΌ π0 π0 π π 14 Circuits with Batteries and Resistors π° π¬ + π° - Iπ¬ π© π° π© Since π is directed out of the battery, the battery βradiatesβ the e.-m. energy into the circuit. π points not along the wires (we assume that the wires have no resistance). In the vicinity of the resistor, π points into the resistor, and again not along the wires. 15 Example: Charging a Capacitor B I h a π¬ πΊ If the process is slow (a<<ο¬), we can neglect the magnetic field energy inside (π’π β« π’π ). The capacitor is receiving energy at a rate: πππ π 1 ππΈ 2 2 2 = ππ β β π0 πΈ = ππ β β π0 πΈ ππ‘ ππ‘ 2 ππ‘ The energy must flow into this volume from somewhere. However, the energy flow through the plates = 0 (πΈ = 0 outside, the wires have no resistance). Between the plates: ππππ π΅ β ππ = π0 π0 ππΈ β ππ΄ ππ‘ π π’ππ π΅ π = π β 2ππ = π0 π0 ππΈ 2 1 ππΈ ππ π΅ π = π0 π0 π ππ‘ 2 ππ‘ The energy flow from outside through the cylindrical surface of radius a and height h: 1 1 1 ππΈ ππΈ πππ 2 π β 2ππβ = πΈπ΅ β 2ππβ = πΈ β π0 π0 π β 2ππβ = ππ β β π0 πΈ = π0 π0 2 ππ‘ ππ‘ ππ‘ This energy isnβt coming down the wires, it comes from the πΈ and π΅ fields 16 surrounding the capacitor! Radiation Pressure 1m Pressure: force per unit area. For particles bombarding a unit area: v vο1s nβ particle density The net force: πΉπππ‘ Pressure: ππ πΉπ = ππ‘ ππ ππ 2 = π π£πππ’ππ/1π = π 1π β π£ ππ‘ ππ‘ ππ π = ππ£ ππ‘ In the e.-m. wave, the particles are photons (massless particles moving with π£ = π) Photon energy: ππΈπ΅ ππΈπ΅ π = π ππβ = πππβ πππβ π πππβ 1 π ππππβ π = ππ = = = ππ‘ ππ‘ π ππ‘ π - assuming that the energy of all photons bombarding the wall is absorbed. For reflection, the pressure is twice as large. Radiation Pressure Find the pressure of sunlight at the earthβs surface, using intensity π = 1,000π/π2 and assuming that the radiation is absorbed. ππΈπ΅ π 1,000π/π2 = = β 3 × 10β6 ππ 8 π 3 × 10 π/π Radiation pressure has had a major effect on the development of the cosmos, from the birth of the universe to ongoing formation of stars and shaping of clouds of dust and gasses on a wide range of scales. Next time: Lecture 20. Inductance. Magnetic Field Energy. §§ 30.1 - 3 19 Appendix I: Alternating Currents (AC circuits) Consider two cases (see the figures): (a) the ac transmission line is loaded by a resistor; (a) the ac transmission line is loaded by a capacitor. Use Poynting formalism to describe the energy flow in these two cases. (a) π° πΊ Energy always flows towards the resistor. π° π© π¬ (b) π° π° πΊ π© π¬ Energy flows back and forth, the energy flux on average is 0. Appendix II: Energy Conservation in Electromagnetism 1 π= πΈ×π΅ π0 πΈ and π΅ fields One can show (by using vector analysis) that indeed S is related to losses of electromagnetic energy in a volume: π π π’ππ π π β ππ΄ = β ππ‘ net flux of energy out of the volume π£πππ’ππ π0 πΈ 2 π΅2 + ππ + 2 2π0 net loss of the e.-m. energy in the volume πΈ β π ππ π£πππ’ππ Joule heat in the volume 21 Appendix III: The Total Energy Flow General statement: whenever there is a flow of energy (it might be field energy or any other kind of energy), the energy flowing through a unit area per unit time is equal to the momentum density ο΄ c2. Non-zero-mass particles 1m v Massless particles (photons) particle density n 1m c vο1s Total energy of a particle: π = cο1s ππ 2 1β Momentum of a particle: π = ππ£ Total energy of a photon: ππβ = πππβ π£2 π2 Momentum of a photon: π£2 1β 2 π Energy flux: ππ π 1π2 β π£ β 1π π= = 1π2 β 1π 1π2 β 1π photon density n ππβ Energy flux: ππ 2 1β π£2 π2 = πππ 2 π 1π2 β π β 1π π= πππβ = πππβ π 2 2 1π β 1π The Total Energy Flow General statement: whenever there is a flow of energy (it might be field energy or any other kind of energy), the energy flowing through a unit area per unit time is equal to the momentum density ο΄ c2. Massless particles (photons) c 1m photon density n cο1s Momentum density ππΈπ΅ β‘ πππβ : Radiation pressure ππΈπ΅ (the momentum transferred to 1 m2 per 1s) Energy of a photon: ππβ = πππβ ππβ Momentum of a photon: Energy flux: ππΈπ΅ = π 1π2 β π β 1π π= πππβ = πππβ π 2 2 1π β 1π 1 π = π0 π0 π π2 ππΈπ΅ = ππΈπ΅ 1π2 β π β 1π /1π = π π ππΈπ΅ = π π (this is for absorption, the pressure is twice as much for reflection)