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Transcript
W15D2 Poynting Vector and EM Waves Radiation Pressure Final Exam Review P36 - 1 Announcements Final Exam Monday Morning May 20 from 9 am-12 noon Johnson Athletic Center Track 2nd floor P36 - 2 Before Starting… All of your grades should now be posted (with possible exception of last problem set). If this is not the case contact your grad TA immediately. We have given you enough information about how we grade and what the cut lines are that you can estimate what you need on the final to get the letter grade of your heart’s desire. P36 - 3 Poynting Vector and EM Waves P36 - 4 Energy in EM Waves Energy densities: Consider cylinder: 1 1 2 2 u E = e 0 E , uB = B 2 2 m0 2 æ 1 B ö 2 dU = (uE + uB ) Adz = ç e 0 E + ÷ Acdt 2è m0 ø What is the energy flow per unit area? 2 ö cæ B EB ö 1 dU c æ 2 S = ç e 0 E + ÷ = ç e 0 cEB + A dt 2 è m0 ø 2 è cm0 ÷ø ( ) EB EB 2 = e 0 m0 c + 1 0 2 m0 P36 - 5 Poynting Vector and Intensity Direction of energy flow = direction of wave propagation S= E´B m0 : Poynting vector units: Joules per square meter per sec Intensity I: 2 0 2 0 E0 B0 E cB 1 2 I S c 0 E0 2 0 2 0 c 2 0 2 P36 - 6 Momentum & Radiation Pressure EM waves transport energy: They also transport momentum: They exert a pressure: S= E´B m0 p =U / c F 1 dp 1 dU S P= = = = A A dt cA dt c This is only for hitting an absorbing surface. For hitting a perfectly reflecting surface the values are doubled: 2U Momentum transfer : p = ; c 2S Radiation pressure : P = c P36 - 7 Group Problem: Radiation A light bulb puts out 100 W of electromagnetic radiation. What is the time-average intensity of radiation from this light bulb at a distance of one meter from the bulb? What is the maximum value of the electric field, E , at this same distance from the bulb in V/m? What is the pressure this radiation will exert on a very small perfectly conducting plate at 1 meter. For simplicity, you may assume the radiation is a plane wave of wavelength λ. o 1.25664 E 06 o 8.84194 E 12 c 3 x108 m / s P36 - 8 Electromagnetism Review 1 òò E ××n̂ da = e òòò r dV (Gauss's Law) òò B × ×n̂ da = 0 (Magnetic Gauss's Law) d òC E × d s = - dt òòS B × n̂ da (Faraday's Law) S 0 V S d ò B × d s = m0 òò J × n̂ da + m0e 0 dt òò E × n̂ da (Maxwell - Ampere's Law) C S S F = q(E + v ´ B) òò closed surface (Lorentz Force Law) d J × dA = r dV òòò dt volume enclosed (Charge Conservation) P36 - 9 Circuits P36 - 10 Sign Conventions - Battery Moving from the negative to positive terminal of a battery increases your potential DV = Vb - Va = +e Moving from the positive to negative terminal of a battery decreases your potential DV = Vb - Va = -e P36 - 11 Sign Conventions - Resistor Moving across a resistor in the direction of current decreases your potential DV = Vb - Va = -IR Moving across a resistor opposite the direction of current increases your potential DV = Vb - Va = +IR P36 - 12 Sign Conventions - Capacitor Moving across a capacitor from the negatively to positively charged plate increases the electric potential DV = Vb - Va = +Q / C Moving across a capacitor from the positively to negatively charged plate decreases the electric potential DV = Vb - Va = -Q / C P36 - 13 (Dis)Charging a Capacitor 1. When the direction of current flow is toward the positive plate of a capacitor, then dQ I=+ dt 2. When the direction of current flow is away from the positive plate of a capacitor, then dQ I=dt P36 - 14 Sign Conventions - Inductor Moving across an inductor in the direction of current contributes dI L dt Moving across an inductor opposite the direction of current contributes e dI = +L dt P36 - 15 Kirchhoff’s Modified 2nd Rule dF B å D Vi = - ò E × d s = + d t i dF B Þ å D Vi =0 dt i If all inductance is ‘localized’ in inductors then our problems go away – we just have: dI å D Vi - L d t = 0 i P36 - 16 Steps For Setting Up Circuit Equations for Circuit with N loops and M junctions 1. 2. 3. 4. 5. Simplify resistors in series/parallel Assign current in every branch Choose circulation direction for N-1 loops Assign charges to each side of capacitor. Determine relation between current and charge for each branch containing a capacitor 6. Write M-1 current conservation equations for junctions 7. Write N-1 loop equations. 8. Solve system of equations. P36 - 17 Displacement Current P36 - 18 Displacement Current Q E= Þ Q = e 0 EA = e 0 F E e0 A dF E dQ = e0 º I dis dt dt So we had to modify Ampere’s Law: d òC B × d s = m0 òòS J × n̂ da + m0e 0 dt òò E × n̂ da S = m0 (I con + I dis ) P36 - 19 EM Waves P36 - 20 Traveling Sinusoidal Wave: Summary y(x,t) y0 sin(k(x vt)) Two periodicities: Spatial period : Wavelength ; Temporal period T. Wave Number : k = 2p / l Dispersion Relation : T = l / v Direction of Propagation : + x - direction P36 - 21 Traveling Sinusoidal Wave Alternative form: y(x,t) y0 sin(k(x vt)) y0 sin(kx t) Wave Number : k 2 / Angular Frequency : 2 / T Dispersion Relation : vT kv Frequency : f 1/ T v f P36 - 22 Plane Electromagnetic Waves http://youtu.be/3IvZF_LXzcc P36 - 23 Electromagnetic Waves: Plane Sinusoidal Waves Watch 2 Ways: 1) Sine wave traveling to right (+x) 2) Collection of out of phase oscillators (watch one position) Don’t confuse vectors with heights – they are magnitudes of electric field (gold) and magnetic field (blue) http://youtu.be/3IvZF_LXzcc P36 - 24 Traveling Plane Sinusoidal Electromagnetic Waves E = E0 sin(kx - w t) ĵ B = B0 sin(kx - w t) k̂ are special solutions to the 1-dim wave equations Ey 1 Ey 2 2 2 x c t 2 2 Bz 1 Bz 2 2 2 x c t 2 2 where k º 2p / l , w º 2p / T, c=l/T P36 - 25 1 Dim’l Sinusoidal EM Waves In order for the fields E = E0 sin(kx - w t) ĵ, B = B0 sin(kx - w t) k̂ to satisfy either condition below ¶E y ¶Bz =¶t ¶x ¶Bz 1 ¶E y =- 2 ¶x c ¶t then B0 E0 / c P36 - 26 Properties of 1 Dim’l EM Waves 1. Travel (through vacuum) with speed of light c 1 m 3.0 10 s 0 0 8 2. At every point in the wave and any instant of time, electric and magnetic fields are in phase with one another, amplitudes obey E0 / B0 c 3. Electric and magnetic fields are perpendicular to one another, and to the direction of propagation (they are transverse): 4. Direction of propagation = Direction of E ´ B. P36 - 27 Concept Questions: EM Waves P36 - 28 Concept Question: Direction of Propagation The figure shows the E (yellow) and B (blue) fields of a plane wave. This wave is propagating in the 1. +x direction 2. –x direction 3. +z direction 4. –z direction P36 - 29 Concept Question Answer: Propagation Answer: 4. The wave is moving in the –z direction The propagation direction is given by the dir E ´ B (Yellow x Blue) P36 - 30 Concept Question: Traveling Wave The B field of a plane EM wave is B( y,t) = B0 sin(ky - w t)k̂ The electric field of this wave is given by 1. E( y,t) = E0 sin(ky - w t) ĵ 2. E( y,t) = E0 sin(ky - w t)(- ĵ) 3. E( y,t) = E0 sin(ky - w t)î 4. E( y,t) = E0 sin(ky - w t)(- î) P36 - 31 Concept Q. Ans.: Traveling Wave Answer: 4. E( y,t) = E0 sin(ky - w t)(- î) From the argument of the sin(ky t) , we know the wave propagates in the positive y-direction. So we have Ê ´ B̂ = ?´ k̂ = ĵ Þ Ê = - î P36 - 32 Concept Question EM Wave The electric field of a plane wave is: E(z,t) = E0 sin(kz + w t) ĵ The magnetic field of this wave is given by: 1. B(z,t) = B0 sin(kz + w t)î 2. B(z,t) = B0 sin(kz + w t)(- î) 3. B(z,t) = B0 sin(kz + w t)k̂ 4. B(z,t) = B0 sin(kz + w t)(-k̂) P36 - 33 Concept Q. Ans.: EM Wave Answer: 1. B(z,t) = B0 sin(kz + w t)î From the argument of the sin(kz t) , we know the wave propagates in the negative z-direction. So we have Ê ´ B̂ = ĵ ´ ? = -k̂ Þ B̂ = î P36 - 34 Energy Flow Poynting vector : Intensity : Radiation pressure : S= E´B m0 E0 B0 E cB I º<S >= = = 2 m0 2 m 0 c 2 m0 2 0 2 0 Pabsorb = S / c; Preflect = 2S / c P36 - 35 Also in Circuit Elements… S= E´B m0 On surface of resistor is INWARD P36 - 36 Concept Questions: Poynting Vector P36 - 37 Concept Question: Capacitor The figures above show a side and top view of a capacitor with charge Q and electric and magnetic fields E and B at time t. At this time the charge Q is: 1. 2. 3. 4. Increasing in time Constant in time. Decreasing in time. Not enough information given to determine how Q is changing. P36 - 38 Concept Q. Answer: Capacitor Answer: 3. The charge Q is decreasing in time Use the Ampere-Maxwell Law. Choose positive unit normal out of plane. Because the magnetic field points clockwise line integral is negative hence positive electric flux (out of the plane of the figure on the right) must be decreasing. Hence E is decreasing. Thus Q must be decreasing, since E is proportional to Q. P36 - 39 Concept Question: Capacitor The figures above show a side and top view of a capacitor with charge Q and electric and magnetic fields E and B at time t. At this time the energy stored in the electric field is: 1. Increasing in 2. Constant in time. 3. Decreasing in time. P36 - 40 Concept Q. Answer: Capacitor Answer: 1. The the energy stored in the electric field is increasing in time The direction of the Poynting Flux S (= E x B) inside the capacitor is inward. Therefore electromagnetic energy is flowing inward, and the energy in the electric field inside is increasing. P36 - 41 Concept Question: Inductor The figures above show a side and top view of a solenoid carrying current I with electric and magnetic fields E and B at time t. The current I is 1. increasing in time. 2. constant in time. 3. decreasing in time. P36 - 42 Concept Question Answer: Inductor Answer: 3. The current I is decreasing in time Use Faraday’s law. Choose positive unit normal out of plane. Because the electric field points counterclockwise line integral is positive, therefore the positive magnetic flux must be decreasing (out of the plane of the figure on the right). Hence B is decreasing. Thus I must be decreasing, since B is proportional to I. P36 - 43 Concept Question: Inductor The figures above show a side and top view of a solenoid carrying current I with electric and magnetic fields E and B at time t. The energy stored in the magnetic field is 1. Increasing in time 2. Constant in time. 3. Decreasing in time. P36 - 44 Concept Question Answer: Inductor Answer: 3. The energy stored in the magnetic field is decreasing in time. The Poynting Flux S (= E x B) inside the solenoid is directed outward from the center of the solenoid. Therefore EM energy is flowing outward, and the energy stored in the magnetic field inside is decreasing. P36 - 45