Lecture 21. Energy Transfer in Electromagnetism Energy and Intensity of Electromagnetic Waves. Transfer of EM Energy in Space: Poynting Formalism.

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1 Lecture 21. Energy Transfer in Electromagnetism Energy and ntensity of Electromagnetic Waves Outline: Transfer of EM Energy in Space: Poynting Formalism. Energy Transfer by EM Waves. Radiation Pressure

2 Energy Density in E.-M. Waves EM waves: E B = 1 c E2 k E = cc This doesn t mean that B is weaker than E ( apples vs. oranges ). The meaningful comparison - the energy densities in the E and B fields that form the wave: Energy density in any E field: u E = ε 0 2 E2 r, t Energy density in any B field: u E = 1 B 2 r, t (lecture 22) 2μ 0 u E = ε 0 2 E2 = ε 0 2 c2 B 2 = c 2 = 1 ε 0 μ 0 = 1 2μ 0 B 2 = u B For E.M. waves traveling in vacuum, the E-field energy density = the B-field energy density! 2

3 Energy Density in E.-M. Waves (cont d) Energy density in e.-m. waves: u EE = 1 2 ε 0E 2 + B2 μ 0 = ε 0 E 2 = 1 μ 0 B 2 = 1 cμ 0 EB The time-averaged energy density in a monochromatic e.-m. wave: ccc 2 kk ωω t = cos 2 kk ωω t = 1 2 E 0 2 ccc 2 kk ωω t = 1 2 E 0 2 u EE t = ε 0 2 E 0 2 = 1 2μ 0 B 0 2 = 1 2cμ 0 E 0 B 0 E 0, B 0 - the amplitudes 3

4 ntensity of Electromagnetic Waves ntensity: the average energy transported by the e.-m. wave per unit area per 1s (the flux of energy per 1m per 1s, scalar) u E, u B c 1s u EE t = 1 2μ 0 c EB = c u EE t = 1 2μ 0 E 0 B 0 = 1 2 cc 0E 0 2 1m 2 c Units: W/m 2 Example: The intensity of sunlight hitting the Earth is ~1,400 W/m 2. Find the amplitudes of the electric and magnetic fields in the e.-m. wave. = 1 2μ 0 E 0 B 0 = 1 2μ 0 c E 0 2 E 0 = 2μ 0 c = π 10 7 V m 1,000 V m B 0 = E 0 c = 103 V/m m/s T ( ~1% of the Earth s magnetic field) 4

5 Example n a fairly brightly-lit room, the intensity of light is about 100 W/m 2. f the room s volume is 50m 3 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? = c u EE t u EE t = c u EE t = 100 W/m m/s J/m 3 7 J Total energy: u EE t vooooo = m 3 50m J 5

6 Poynting Vector How does the electromagnetic energy get from one place to another? = c u EE t = 1 E 2μ 0 B 0 0 -scalar quantity 1m 2 u E, u B c c 1s n order to retain the direction of the energy flow (along k ), let s try E B: Poynting vector: (energy transferred through 1 m 2 per 1s in the direction of k ) S = k = 1 μ 0 E B S t = 1 μ 0 E 0 B 0 ccc 2 kk ωω t = 1 2μ 0 E 0 B 0 - the time-averaged flux of energy For a traveling monochromatic e.-m. wave: S t = 1 2μ 0 E 0 B 0 k 8

7 Poynting Formalism (cont d) The concept of Poynting vector applies to all E and B fields, not just to the e.-m. waves. S = 1 μ 0 E B Poynting formalism works in all situations, both static and dynamic. Few consequences: (a) Whenever we have both E and B (and E B), 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 mass ). (c) The momentum transfer (absorption or reflection of e.-m. waves) implies radiation pressure. 11

8 Example: Transmission Line E B R We can apply the Poynting vector formalism to obtain the well-known result: P = VV. However, the fields are non-uniform, and the integration is difficult. Let s simplify the geometry. 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 that travels uniformly down one strip and back along the other. V W a E B R E = V a B = μ 0 K = μ 0 W both uniform, orthogonal to one another Power transmitted to the load R : P = 1 μ 0 E B da crrrr sssssss = 1 μ 0 V a μ 0 W aa = VV 12

9 Circuits with Batteries and Resistors + - E B E B Since S is directed out of the battery, the battery radiates the e.-m. energy into the circuit. S points not along the wires (we assume that the wires have no resistance). n the vicinity of the resistor, S points into the resistor, and again not along the wires. 13

10 h a E B Example: Charging a Capacitor f the process is slow (a<<λ), we can neglect the magnetic field energy inside (u e u m ). The capacitor is receiving energy at a rate: S du e = d πa 2 h 1 2 ε 0E 2 = πa 2 h ε 0 E The energy must flow into this volume from somewhere. However, the energy flow through the plates = 0 (E = 0 outside, the wires have no resistance). Between the plates: looo B dl = μ 0 ε 0 de da suuu B r = a 2ππ = μ 0 ε 0 de πa2 B a = 1 2 μ 0ε 0 a de The energy flow from outside through the cylindrical surface of radius a and height h: S 2πππ = 1 μ 0 EE 2πππ = 1 μ 0 E 1 2 μ 0ε 0 a 2πππ = πa2 h ε 0 E = du e This energy isn t coming down the wires, it comes from the E and B fields surrounding the capacitor! 14

11 Radiation Pressure n particle density v v 1s 1m Pressure: force per unit area. For particles bombarding a unit area: F i = The net force: F nnn = n vvvvvv/1s = n 1m2 v Pressure: P = nn n the e.-m. wave, the particles are photons (massless particles moving with v = c) Photon energy: T pp = cp pp P EE = nc dp pp = d nn pp = 1 c d nct pp = S c P EE = S c - assuming that the energy of all photons bombarding the wall is absorbed. For reflection, the pressure is twice as large.

P EE = S c = 1,000W/m2 3 10 8 m/s 3 10 6 PP Radiation pressure has had a major effect on the development of the cosmos, from the

12 Radiation Pressure Find the pressure of sunlight at the earth s surface, using intensity S 1,000W/m 2 and assuming that the radiation is absorbed. P EE = S c = 1,000W/m m/s PP 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. The most massive (non-degenerate) stars with very high internal temperature are stabilized against gravity primarily by radiation pressure.

13 Next time: Lecture 20. nductance. Magnetic Field Energy

14 Appendix : 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) S B E Energy always flows towards the resistor. (b) B E S Energy flows back and forth, the energy flux on average is 0.

Appendix : Energy Conservation in Electromagnetism S = 1 μ 0 E B S E and B fields One can show (by using vector analysis) that indeed S is related to losses of electromagnetic

15 Appendix : Energy Conservation in Electromagnetism S = 1 μ 0 E B S E and B fields One can show (by using vector analysis) that indeed S is related to losses of electromagnetic energy in a volume: S da suuu = ε 0E 2 vooooo 2 + B2 2μ 0 dτ + E ȷ dτ vooooo net flux of energy out of the volume net loss of the e.-m. energy in the volume Joule heat in the volume 19

16 Appendix : 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 c 2. Non-zero-mass particles Massless particles (photons) v 1m particle density n c 1m photon density n v 1s c 1s Total energy of a particle: T = Momentum of a particle: p = Energy flux: mc2 1 v2 mm c 2 1 v2 c 2 Total energy of a photon: Momentum of a photon: Energy flux: T pp = cp pp p pp S = nn 1m 2 1s = n 1m2 v 1s 1m 2 1s mc 2 1 v2 c 2 = nnc 2 S = n 1m2 c 1s 1m 2 1s cp pp = np pp c 2

17 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 c 2. Massless particles (photons) Energy of a photon: T pp = cp pp c c 1s 1m photon density n Energy flux: Momentum of a photon: S = n 1m2 c 1s 1m 2 1s p pp cp pp = np pp c 2 Momentum density p EE np pp : p EE = 1 c 2 S = ε 0μ 0 S Radiation pressure P EE (the momentum transferred to 1 m 2 per 1s) P EE = p EE 1m 2 c 1s /1s = S c P EE = S c (this is for absorption, the pressure is twice as much for reflection)

Lecture 22. Inductance. Magnetic Field Energy.

Lecture 22. Inductance. Magnetic Field Energy. Outline: Self-induction and self-inductance. Inductance of a solenoid. The energy of a magnetic field. Alternative definition of inductance. Mutual Inductance.