Oscillations and Electromagnetic Waves. March 30, 2014 Chapter 31 1

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1 Oscillations and Electromagnetic Waves March 30, 2014 Chapter 31 1

2 Three Polarizers! Consider the case of unpolarized light with intensity I 0 incident on three polarizers! The first polarizer has a polarizing direction that is vertical, θ 1 = 0! The second polarizer has a polarizing angle of θ 2 = 45 with respect to the vertical! The third polarizer has a polarizing angle of θ 3 = 90 with respect to the vertical PROBLEM! What is the intensity of the light passing through all the polarizers in terms of the initial intensity? March 30, 2014 Chapter 31 3

3 Three Polarizers SOLUTION March 30, 2014 Chapter 31 4

4 Three Polarizers! The intensity of the unpolarized light is I 0! The intensity of the light passing the first polarizer is I 1 = 1 2 I 0! The intensity of the light passing the second polarizer is I 2 = I 1 cos 2 ( 45 0 ) = I 1 cos 2 ( 45 ) = 1 2 I 0 ( cos2 45 )! The intensity of the light passing the third polarizer is I 3 = I 2 cos 2 ( ) = I 2 cos 2 ( 45 ) = 1 2 I 0 ( cos4 45 ) = I 0 / 8! The fact that 1/8 th of the intensity of the light is transmitted is somewhat surprising because polarizers 1 and 3 have polarizing angles that are perpendicular to each other March 30, 2014 Chapter 31 5

5 LC circuit! Energy: U =U E +U B = 1 2 q 2 C Li2! Sinusoidal oscillation q = q max cos ω 0 t φ i = i max sin(ω 0 t φ) ( ) ( Note the sign)! With i max = ω 0 q max and ω 0 = 1 LC March 30, 2014 Chapter 30 6

6 LC Clicker! The capacitor in a LC circuit is initially uncharged, but there is a current I 0 flowing through the inductor. Will the circuit oscillate? A. No, because the initial voltage is zero. B. No, because the current is only in the inductor C. Yes, because the initial voltage is not zero. D. Yes, because the initial current will decrease which results in an induced voltage. E. None of the above. March 30, 2014 Chapter 31 7

7 LC circuit! An oscillating LC circuit consists of a 75mH inductor and a 3.60 μf capacitor. The maximum charge on the capacitor is 2.9 μc. What is the maximum current?! Key ideas: calculate the total energy U=U E contained in the circuit from what you know about the capacitor. Then use energy conservation and U B =0.5i 2 L Physics for Scientists & Engineers 2, Lecture 55 8

8 LC circuit! An oscillating LC circuit consists of a 75mH inductor and a 3.60 μf capacitor. The maximum charge on the capacitor is 2.9 μc. What is the maximum current?! Key ideas: calculate the total energy U=U E contained in the circuit from what you know about the capacitor. Then use energy conservation and U B =0.5i 2 L 1. All the energy in the circuit resides in the capacitor when it has its maximum charge. The current is then When the capacitor is fully discharged, the current is maximum and the total energy resides in the inductor: Physics for Scientists & Engineers 2, Lecture 55 9

9 ! Amplitude q = q max e Rt 2L cos ωt ( )! Frequency RLC circuit ω = ω 2 0 R 2L 2 ω 0 = 1 LC! Energy U E = q 2 Rt max L 2C e cos 2 ωt ( ) March 30, 2014 Chapter 30 10

10 Driven RLC Circuit! The driving voltage in a time-varying RLC circuit is given by V emf = V max sinωt! We describe the time-varying current in driven RLC circuit elements using a phasor I m i = I m sin( ωt φ)! The current is in phase with the voltage drop across the resistor! We describe the voltage in terms of a phasor V R! The time-varying currents and voltages in the circuit can have different phases March 30, 2014 Chapter 30 11

11 Resistance and Reactance! Time-varying emf V emf = V max sinωt! Time-varying emf V R with resistor i R = V R R = I R sinωt Resistance R! Time-varying emf V C with capacitor X C = 1 i ωc C = V C sin(ωt + 90 ) X C Capacitive Reactance X C! Time-varying emf V L with inductor i L = V L X L = ωl sin ωt 90 X L ( ) Inductive Reactance X L March 30, 2014 Chapter 30 12

12 Driven RLC Circuit! The sum of the two phasors V L - V C and V R must equal V m V 2 m = V 2 R + ( V L V C ) 2! The impedance Z is given by Z = R 2 + ( X L X C ) 2! The current is given by I m = V m Z = V m R 2 + ωl 1 2 ωc! The phase is given by φ = tan 1 V L V C V R = tan 1 X L X C R March 30, 2014 Chapter 30 13

13 Driven RLC Circuit X L > X C Inductive X C > X L Capacitive X L = X C Φ = 0 Resonance frequency: ω 0 = 1 LC March 30, 2014 Chapter 30 14

14 Power I rms = I 2 V rms = V m 2 I rms = V rms Z! Average power dissipated: 2 P = I rms R! Phase: cosφ = V R V m = = V rms R 2 + ωl 1 2 ωc = V rms Z IR IZ = I rmsr R Z = I rms V rms R Z March 30, 2014 Chapter 30 16

15 Power! We can then express the power in the circuit as P = I rms V rms cosφ! We can see that the maximum power is dissipated when ϕ = 0, i.e. at the resonance frequency! We call cosϕ the power factor! Quality factor of a series RLC circuit: Q = ω 0 L R = 1 R L C March 30, 2014 Chapter 30 17

16 Series RLC circuit frequency dependence March 30, 2014 Chapter 31 18

17 Transformer! Primary voltage V P varies sinusoidally.! Secondary voltage V P is given by V P N P = V S N S V S = V P N S N P! Power and current: P P = I P V P = P S = I S V S I S = I P V P V S = I P N P N S! Effective primary resistance for a load resistance R: R P = V P N = V P P I P N S 2 R V P = N P N S 2 R March 30, 2014 Chapter 31 19

18 Electromagnetic Waves! Plane-wave ansatz to solve the Maxwell equations E( r!,t) = E max sin( kx ωt) B( r!,t) = B max sin( kx ωt) ω k = 2π f 2π λ E max B max = ω k = c = f λ = c ( c is the speed of light) Faraday E B = c c = 1 µ 0 ε 0 Ampere Maxwell Physics for Scientists & Engineers 2, Lecture 54 20

19 Energy Transport! The rate of energy transported by an electromagnetic wave is usually defined as! S = 1 µ 0! E! B! The instantaneous power per unit area S =! S = power area! Instantaneous power per unit area of an electromagnetic wave! Intensity I of the wave given by Poynting vector instantaneous S = 1 µ 0 EB S = 1 cµ 0 E 2 I = 1 2 E cµ rms 0 ( E rms = E max / 2) I = S ave = W/m 2 power area ave Physics for Scientists & Engineers 2, Lecture 54 21

20 Energy Density and Radiation Pressure! The energy density of an electric and a magnetic field is given by u E = 1 2 ε 0 E 2 u B = 1 2µ 0 B 2! The radiation pressure, p r, due to a totally absorbed electromagnetic wave is p r = I c! The radiation pressure, p r, due to a totally reflected electromagnetic wave is p r = 2I c Physics for Scientists & Engineers 2, Lecture 54 22

21 Polarization! Unpolarized light polarized light! We can change unpolarized light into polarized light by passing it through a polarizer. A polarizer is characterized by its polarizing direction! The intensity I of the light - initially unpolarized - passing through the polarizer is given by I = 1 2 I 0! Polarized light passing through a polarizer:! I 0 is the initial intensity and θ is the angle between the polarization of the incident light and the polarizing angle of the polarizer Physics for Scientists & Engineers 2, Lecture 54 23

22 ! Poynting vector S Radiation Pressure! Electromagnetic wave incident on an object! Force is F = Δp Δt = IA c with! Intensity A, speed of light c! If wave is absorbed, pressure is p r = I c S = 1 E 2 = Power cµ 0 area I = 1 2 E rms cµ 0 ( E rms = E max / 2)! If wave is reflected, pressure is p r = 2I c March 30, 2014 Chapter 31 24

23 Solar Stationary Satellite! Suppose researchers want to place a satellite above the north pole of the Sun and stationary with respect to the Sun in order to study its long-term rotational characteristics! The satellite will have a totally reflecting solar sail and be located at a distance of m from the center of the Sun! The intensity of sunlight at that distance is 1400 W/m 2! The plane of the solar sail is perpendicular to a line connecting the satellite and the center of the Sun! The mass of the satellite and sail is kg PROBLEM! What is the required area of the solar sail? March 30, 2014 Physics for Scientists & Engineers, Chapter 22 25

24 Solar Stationary Satellite SOLUTION THINK! In an equilibrium position for the satellite, the area of the solar sail times the radiation pressure from the Sun produces a force that is balanced by the gravitational force between the satellite and the Sun! We can equate these two forces and solve for the area of the solar sail SKETCH March 30, 2014 Physics for Scientists & Engineers, Chapter 22 26

25 Solar Stationary Satellite RESEARCH! The satellite will be stationary if the force of gravity is balanced by the force from the radiation pressure F g = F rp! The force corresponding to the radiation pressure from sunlight is equal to the radiation pressure times the area of the solar sail F rp = p r A! The radiation pressure can be expressed in terms of the intensity of the sunlight incident on the totally reflecting solar sail p r = 2I c March 30, 2014 Physics for Scientists & Engineers, Chapter 22 27

26 Solar Stationary Satellite! The force of gravity between the satellite and the Sun is SIMPLIFY! Combining our equations gives us F g = G mm Sun d 2 2I c A = G mm Sun d 2! Solving for the area of the solar gives us A = G cmm Sun 2Id 2 March 30, 2014 Physics for Scientists & Engineers, Chapter 22 28

27 Solar Stationary Satellite CALCULATE! Putting in our numerical values we get ( ( ) m/s) kg ( ) 2( 1400 W/m 2 )( m) 2 A = m 3 kg 1 s 2 A = 63,206.2 m 2 ROUND ( ) kg! We round our result to three significant figures A = m 2 DOUBLE-CHECK! If the solar sail were circular, the radius of the sail would be r = A π = m 2 π = 142 m March 30, 2014 Physics for Scientists & Engineers, Chapter 22 29

28 Solar Stationary Satellite! We can relate the thickness t of the sail times the density ρ of the material from which the sail is constructed to the mass per unit area of the sail tρ = m A! If the sail were composed of kapton (ρ = 1420 kg/m 3 ) and had a mass of 75 kg, the thickness of the sail would be t = m ρa = 75 kg ( 1420 km/m 3 ) m 2 ( ) = µm! Kapton is developed to remain stable in the wide range of temperatures found in space; from near absolute zero to over 600 K! Current production techniques cannot produce kapton this thin, but it may be possible in the near future March 30, 2014 Physics for Scientists & Engineers, Chapter 22 30

29 Polarizer! Unpolarized light hits a polarizer: Intensity I = 1 2 I 0! Polarized light hits a polarizer: Intensity I = I 0 cos 2 θ! θ is the angle between the light polarization and the direction of the polarizer! Law of Malus March 30, 2014 Chapter 31 31

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