34 Electromagnetic Waves CHAPTER OUTLINE 34.1 Displacement Current and the General Form of Ampère’s Law 34.2 Maxwell’s Equations and Hertz’s Discoveries 34.3 Plane Electromagnetic Waves 34.4 Energy Carried by Electromagnetic Waves 34.5 Momentum and Radiation Pressure 34.6 Production of Electromagnetic Waves by an Antenna 34.7 The Spectrum of Electromagnetic Waves * An asterisk indicates a question or problem new to this edition ANSWERS TO OBJECTIVE QUESTIONS OQ34.1 (i) Answer (c) Both the light intensity and the gravitational force follow inverse-square laws (ii) Answer (a) The smaller grain presents less face area and feels a smaller force due to light pressure OQ34.2 (i) Answer (c) (ii) Answer (c) (iii) Answer (c) (iv) Answer (b) (v) Answer (b) The same amount of energy passes through concentric spheres of increasing area as the wave travels outward from its source, so the amplitude and the intensity, which is proportional to the square of the amplitude, decrease OQ34.3 Answer (b) Frequency, wavelength, and the speed of light are related: fλ = c → λ= c 3.00 × 108 m/s = = 0.122 m = 12.2 cm f 2.45 ì 109 Hz 572 â 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 34 OQ34.4 (i) 573 Answer (a) According to f = 2π LC , to make f half as large, the capacitance should be made four times larger (ii) Answer (b) According to f λ = c, if frequency is halved, wavelength is doubled OQ24.5 Answer (e) Accelerating charge, changing electric field, or changing magnetic field can be the source of a radiated electromagnetic wave OQ34.6 Answers (c) and (d) The relationship between frequency, wavelength, and the speed of a wave is f λ = v In a vacuum, all electromagnetic waves travel at the same speed: v = c Electromagnetic waves, consisting of oscillating electric and magnetic fields, are transverse waves OQ34.7 (i) through (v) have the same answer (c) The same amount of energy passes through equal areas parallel to the yz plane as the wave travels in the +x direction, so the amplitude and the intensity, which is proportional to the square of the amplitude, not change OQ34.8 (i) Answer (b) Electric and magnetic fields both carry the same energy, so their amplitudes are proportional to each other (ii) Answer (a) The intensity is proportional to the square of the amplitude OQ34.9 Answer (d) The peak values of the electric and magnetic field components of an electromagnetic wave are related by Emax Bmax = c, where c is the speed of light in vacuum Thus, Emax = cBmax = ( 3.00 × 108 m/s ) ( 1.50 × 10−7 T ) = 45.0 N/C OQ34.10 (i) The ranking is c > b > d > e > a Gamma rays have the shortest wavelength (ii) The ranking is a > e > d > b > c According to f λ = c, as wavelength decreases, frequency increases (iii) The ranking is a = b = c = d = e All electromagnetic waves travel at the speed of light c in vacuum, which is assumed here OQ34.11 Answer (d) An electromagnetic wave travels in the direction of the Poynting vector: S = E ì B à0 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 574 Electromagnetic Waves ANSWERS TO CONCEPTUAL QUESTIONS CQ34.1 The entire room and its contents have a soft glow Incandescent light bulbs shine brightly in the infrared, but fluorescent lights not The top of a computer monitor glows brighter than the screen, which glows faintly Windowpanes appear dark if they are cool, and a patch of wall where sunlight falls glows brighter than where the sunlight does not fall Heating resistors or warm air outlets shine, and the air near to them has a faint glow, but cold air outlets are dark, and the nearby air has no glow CQ34.2 Electromagnetic waves carry momentum Recalling what we learned in Chapter 9, the impulse imparted by a particle that bounces elastically off a wall is twice that imparted by an object that sticks to a wall Similarly, the impulse, and hence the pressure exerted by a wave reflecting from a surface, must be twice that exerted by a wave that is absorbed CQ34.3 No Radio waves travel at a finite speed, the speed of light Radio waves can travel around the curved surface of the Earth, bouncing between the ground and the ionosphere, which has an altitude that is small when compared to the radius of the Earth The distance across the lower forty-eight states is approximately 000 km, requiring a × 106 m transit time of ~ 10−2 s × 10 m/s CQ34.4 Sound Light 1) Sound is a longitudinal wave 1) Light is a transverse wave 2) Sound requires a material medium 2) Light does not require a material medium 3) Sound in air moves at hundreds of meters per second 3) Light in air moves at hundreds of millions of meters per second 4) The speed of sound through a medium, depending the material of the medium, can be faster or slower than that in air 4) The speed of light through materials is less than in vacuum © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 34 575 5) Sound propagates by a chain reaction of density and pressure disturbances recreating each other 5) Light propagates by a chain reaction of electric and magnetic fields recreating each other 6) Audible sound has frequencies over a range of three decades (ten octaves) from 20 Hz to 20 kHz 6) Visible light has frequencies over a range of less than one octave, from 430 to 700 THz (THz = Terahertz = 1012 Hz) 7) Audible sound has wavelengths of ordinary size (1.7 cm to 17 m) 7) Visible light has wavelengths of very small size (400 nm to 750 nm) CQ34.5 The changing magnetic field of the solenoid induces eddy currents in the conducting core This is accompanied by I2 R conversion by heating of electrically-transmitted energy into internal energy in the conductor CQ34.6 (a) The electric and magnetic fields of the light wave oscillate in time at each point in space, like sports fans in a grandstand when the crowd does “the wave.” (b) The wave transports energy CQ34.7 An infrared photograph records the infrared light reflected, but also emitted by a person’s face When a person blushes or exercises or becomes excited, warmer areas glow brighter in the infrared A person’s nostrils and the openings of the ear canals are bright; brighter still are just the pupils of the eyes CQ34.8 No, they not Specifically, Gauss’s law in magnetism prohibits magnetic monopoles If magnetic monopoles existed, then the magnetic field lines would not have to be closed loops, but could begin or terminate on a magnetic monopole, as they can in Gauss’s law in electrostatics CQ34.9 Different stations have transmitting antennas at different locations For best reception align your rabbit ears perpendicular to the straight-line path from your TV to the transmitting antenna The transmitted signals are also polarized The polarization direction of the wave can be changed by reflection from surfaces—including the atmosphere—and through Kerr rotation—a change in polarization axis when passing through an organic substance In your home, the plane of polarization is determined by your surroundings, so antennas need to be adjusted to align with the polarization of the wave © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 576 Electromagnetic Waves CQ34.10 CQ34.11 Consider a typical metal rod antenna for a car radio Charges in the rod respond to the electric field portion of the carrier wave Variations in the amplitude of the incoming radio wave cause the electrons in the rod to vibrate with amplitudes emulating those of the carrier wave Likewise, for frequency modulation, the variations of the frequency of the carrier wave cause constant-amplitude vibrations of the electrons in the rod but at frequencies that imitate those of the carrier The Poynting vector S describes the energy flow associated with an electromagnetic wave The direction of S is along the direction of propagation and the magnitude of S is the rate at which electromagnetic energy crosses a unit surface area perpendicular to the direction of S CQ24.12 The frequency of EM waves in a microwave oven, typically 2.45 GHz, is chosen to be in a band of frequencies absorbed by water molecules The plastic and the glass contain no water molecules Plastic and glass have very different absorption frequencies from water, so they may not absorb any significant microwave energy and remain cool to the touch CQ34.13 Maxwell included a term in Ampère’s law to account for the contributions to the magnetic field by changing electric fields, by treating those changing electric fields as “displacement currents.” SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 34.1 P34.1 Displacement Current and the Generalized Form of Ampère’s Law (a) We use the right-hand rule for both real and displacement currents Thus, the direction of B is counterclockwise, and the direction of B at P is upwards (b) We use the extended form of Ampère’s law, Equation 34.7 Since no moving charges are present, I = and we have dΦE B ∫ ⋅ d = µ0 ∈0 d t In order to evaluate the integral, we make use of the symmetry of the ANS FIG P34.1 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 34 577 situation Symmetry requires that no particular direction from the center can be any different from any other direction Therefore, there must be circular symmetry about the central axis We know the magnetic field lines are circles about the axis Therefore, as we travel around such a magnetic field circle, the magnetic field remains constant in magnitude Setting aside until later the determination of the direction of B, we integrate ∫ B ⋅ d around the circle at R = 0.150 m to obtain π RB Differentiating the expression ΦE = AE, we have dΦE ⎛ π d ⎞ dE =⎜ dt ⎝ ⎟⎠ dt Thus, 2⎞ ⎛ µ0 ∈0 d dE ⎜ π d ⎟ dE B ⋅ d = π RB = µ ∈ → B = ⎟ 0 ⎜ ∫ ⎜ ⎟ dt R dt ⎝ ⎠ Substituting numerical values, ( 4π × 10 B= −7 T ⋅ m/A ) ( 8.85 × 10−12 C2 /N ⋅ m ) ( 0.100 m ) ( 0.150 m ) × ( 20.0 V/m ⋅ s ) = 1.85 × 10−18 T P34.2 For the capacitor, dQ dt dΦE d I = ( EA ) = = dt dt ∈0 ∈0 (a) dE I 0.200 A = = −12 dt ∈0 A ( 8.85 × 10 C /N ⋅ m ) ⎡π ( 10.0 × 10−2 m ) ⎤ ⎣ ⎦ = 7.19 × 1011 V m ⋅ s (b) ∫ B ⋅ ds = ∈0 µ0 dΦE : dt 2π rB = ∈0 µ0 ⎤ d⎡ Q ⋅ π r2 ⎥ ⎢ dt ⎢⎣ ∈ A ⎥⎦ −2 µ0 Ir µ0 ( 0.200 A ) ( 5.00 × 10 m ) B= = = 2.00 × 10−7 T −2 2A ⎡π ( 10.0 × 10 m ) ⎤ ⎣ ⎦ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 578 P34.3 Electromagnetic Waves The electric field in the space between the plates is E = The flux of this field is (a) σ Q = ∈0 ∈0 A ⎛ Q ⎞ Q ΦE = E ⋅ A = ⎜ A cos 0 = ⎟ ∈0 ⎝ ∈0 A ⎠ The rate of change of flux is dΦE dQ dt I 0.100 A = = = dt ∈0 ∈0 8.85 × 10−12 C2 /N ⋅ m = 11.3 × 109 V ⋅ m/s (b) The displacement current is defined as dΦE dt = (8.85 × 10 -12 C2 / N ⋅ m )(1.13 × 1010 N ⋅ m / C ⋅ s) I d = ∈0 = 0.100 A Section 34.2 P34.4 Maxwell’s Equations and Hertz’s Discoveries F = ma = qE + qv × B so −e ⎡E + v × B ⎤⎦ where a= m⎣ ˆi ˆj kˆ v × B = 10.0 = − ( 4.00 T ⋅ m/s ) ˆj 0 0.400 Then ⎛ −1.60 × 10−19 C ⎞ a=⎜ −31 ⎝ 9.11 × 10 kg ⎟⎠ × ⎡⎣( 2.50 V/m ) ˆi + ( 5.00 V/m ) ˆj − ( 4.00 T ⋅ m/s ) ˆj ⎤⎦ = ( −1.76 × 1011 ) ⎡⎣ 2.50ˆi + 1.00ˆj ⎤⎦ m/s a= ( −4.39ˆi − 1.76j) ì 10 11 m/s â 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 34 P34.5 579 The net force on the proton is the Lorentz force, as described by ∑ F = ma = qE + qv × B so that a = Taking the cross product of v and B, ˆi v × B = 200 0.200 Then, ˆj e ⎡E + v × B ⎤⎦ m⎣ kˆ = –200(0.400)ˆj + 200(0.300)kˆ 0 0.300 0.400 ⎛ 1.60 × 10 –19 ⎞ e ⎡ 50.0 ˆj – 80.0 ˆj + 60.0 kˆ ⎤ m/s a = ⎡⎣E + v × B ⎤⎦ = ⎜ –27 ⎟ ⎣ ⎦ ⎝ 1.67 × 10 ⎠ m ( ) = –2.87 × 109 ˆj + 5.75 × 109 kˆ m/s P34.6 (a) The very long rod creates the same electric field that it would if stationary We apply Gauss’s law to a cylinder, concentric with the rod, of radius r = 20.0 cm and length : qinside E ∫ ⋅ dA = ∈ E ( 2π r ) cos 0° = E= = ANS FIG P34.6 λ ∈0 λ radially outward 2π ∈0 r 35.0 × 10−9 C/m ˆj −12 2 2π ( 8.85 × 10 C /N ⋅ m ) ( 0.200 m ) = 3.15 × 103 ˆj N/C (b) The charge in motion constitutes a current of (35.0 × 10–9 C/m) × (15.0 × 106 m/s) = 0.525 A This current creates a magnetic field àI B= r ( ì 10 = T ⋅ m/A ) ( 0.525 A ) ˆ k = 5.25kˆ × 10−7 T 2π ( 0.200 m ) −7 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 580 Electromagnetic Waves (c) The Lorentz force on the electron is F = qE + qv × B F = ( −1.60 × 10−19 C ) 3.15 × 103 ˆj N/C ( ( ) + ( −1.60 × 10−19 C ) 240 × 106 ˆi m/s ( ) × 5.25 × 10−7 kˆ T ) F = 5.04 × 10−16 − ˆj N + 2.02 × 10−17 + ˆj N ( ) ( ) ( ) = 4.83 − ˆj × 10−16 N Section 34.3 *P34.7 (a) Plane Electromagnetic Waves From Equation 34.20, λ= c 3.00 × 108 m s = = 261 m f 1 150 × 103 s −1 so 180 m = 0.690 wavelengths 261 m (b) From Equation 34.20, λ= c 3.00 × 108 m s = = 3.06 m f 98.1 × 106 s −1 so 180 m = 58.9 wavelengths 3.06 m *P34.8 From Equation 34.20, c 3.00 × 108 m s λ= = = 11.0 m f 27.33 × 106 Hz P34.9 (a) Since the light from this star travels at 3.00 × 108 m/s, the last bit of light will hit the Earth in t= d 6.44 × 1018 m = = 2.15 × 1010 s = 681 years c 2.998 ì 108 m s â 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 34 (b) From Table C.4 (in Appendix C of the textbook), the average Earth-Sun distance is d = 1.496 × 1011 m, giving the transit time as t= (c) 581 d ⎛ 1.496 × 1011 m ⎞ ⎛ ⎞ = ⎜ ⎟ = 8.32 c ⎜⎝ 2.998 × 108 m/s ⎟⎠ ⎝ 60 s ⎠ Also from Table C.4, the average Earth-Moon distance is d = 3.84 × 108 m, giving the time for the round trip as 2d ( 3.84 × 10 m ) t= = = 2.56 s c 2.998 × 108 m/s P34.10 From f λ = c, we have f= P34.11 c 2.998 × 108 m/s = = 4.738 × 1014 Hz −9 λ 632.8 × 10 m In the fundamental mode, there is a single loop in the standing wave between the plates Therefore, the distance between the plates is equal to half a wavelength λ = 2L = ( 2.00 m ) = 4.00 m f = Thus, P34.12 c 3.00 × 108 m/s = = 7.50 × 107 Hz = 75.0 MHz λ 4.00 m E 220 V/m = c or = 3.00 × 108 m/s, so B B B = 7.33 × 10−7 T = 733 nT P34.13 From Equation 34.17, v= *P34.14 1 = c = 0.750c = 2.25 ì 108 m/s à0 1.78 Time to reach object = 1 ( total time of flight ) = ( 4.00 × 10−4 s ) = 2.00 × 10−4 s 2 Thus, d = vt = ( 3.00 × 108 m s ) ( 2.00 × 10−4 s ) = 6.00 × 10 m = 60.0 km P34.15 (a) c = f λ gives the frequency as f= c 3.00 × 108 m/s = = 6.00 ì 106 Hz 50.0 m â 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 600 P34.63 Electromagnetic Waves (a) θ = 15.0°: P = Pmax cos ( 15.0° ) = 0.933Pmax = 93.3% (b) θ = 45.0°: P = Pmax cos ( 45.0° ) = 0.500Pmax = 50.0% (c) θ = 90.0°: P = Pmax cos ( 90.0° ) = The gravitational force exerted by the Sun on the particle is given by Fgrav = GMSm ⎛ GMS ⎞ ⎡ ⎛ ⎞ ⎤ = ⎜ ⎟ ⎢ρ ⎜ π r ⎟ ⎥ ⎝ R ⎠⎣ ⎝3 ⎠⎦ R2 where MS = mass of Sun, r = radius of particle, and R = distance from Sun to particle The force exerted by solar radiation on the particle is given by Frad = PA, and since the particle absorbs all the radiation, by Equation 34.28, we have S Frad = PA = π r c When the particle is in equilibrium, the gravitational force toward the Sun is balanced by the force of radiation away from the Sun, Frad = Fgrav , so S ⎛ GMS ⎞ πr =⎜ ⎟ ⎝ R ⎠ c ⎡ ⎛ 3⎞ ⎤ ⎢⎣ ρ ⎜⎝ π r ⎟⎠ ⎥⎦ Solving for r, the radius of the particle, then gives r= 3SR 4cGMS ρ Suppressing units, r= ( 214 ) ( 3.75 × 1011 ) ( 3.00 × 108 ) ( 6.67 × 10−11 ) ( 1.991 × 1030 ) ( 1 500 ) = 3.78 × 10−7 m = 378 nm P34.64 The gravitational force exerted by the Sun on the particle is given by Fgrav = GMSm ⎛ GMS ⎞ ⎡ ⎛ ⎞ ⎤ = ⎜ ⎟ ⎢ρ ⎜ π r ⎟ ⎥ ⎝ R ⎠⎣ ⎝3 ⎠⎦ R2 where MS = mass of Sun, r = radius of particle, and R = distance from Sun to particle The force exerted by solar radiation on the particle is given by Frad = PA, and since the particle absorbs all the radiation, by Equation 34.28, we have S Frad = PA = π r c © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 34 601 When the particle is in equilibrium, the gravitational force toward the Sun is balanced by the force of radiation away from the Sun, Frad = Fgrav , so S ⎛ GMS ⎞ πr =⎜ ⎟ ⎝ R ⎠ c ⎡ ⎛ 3⎞ ⎤ ⎢⎣ ρ ⎜⎝ π r ⎟⎠ ⎥⎦ Solving for r, the radius of the particle, then gives r= P34.65 (a) The magnetic-field amplitude is Bmax = (b) 3SR 4cGMS ρ Emax 0.200 × 10−6 V/m = = 6.67 × 10−16 T c 3.00 × 108 m/s The intensity is the Poynting vector averaged over one or more cycles, given by 0.200 × 10−6 V/m ) ( Emax = = µ0 c ( 4π × 10−7 T ⋅ m/A ) ( 3.00 × 108 m/s ) Savg = 5.31 × 10−17 W/m (c) The power tells how fast the antenna receives energy It is ⎛ d⎞ ⎛ 20.0 m ⎞ P = Savg A = Savgπ ⎜ ⎟ = ( 5.31 × 10−17 W/m ) π ⎜ ⎝ 2⎠ ⎝ ⎟⎠ = 1.67 × 10−14 W (d) The force tells how fast the antenna receives momentum It is ⎛ 5.31 × 10−17 W/m ⎞ ⎛ 20.0 m ⎞ ⎛ Savg ⎞ F = PA = ⎜ A=⎜ π⎜ ⎟ ⎝ c ⎟⎠ ⎝ 3.00 × 108 m/s ⎟⎠ ⎝ ⎠ = 5.56 × 10−23 N (approximately the weight of 000 hydrogen atoms!) P34.66 Of the intensity S = 370 W/m2, the 38.0% that is reflected exerts a pressure P1 = 2Sr ( 0.380 ) S = c c The absorbed light exerts pressure P2 = Sa 0.620S = c c © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 602 Electromagnetic Waves Altogether the pressure at the subsolar point on Earth is 1.38S 1.38 ( 1370 W/m ) = = 6.30 × 10−6 Pa c 3.00 × 10 m/s (a) Ptotal = P1 + P2 = (b) Compared to normal atmospheric pressure, Pa 1.01 × 105 N/m = Ptotal 6.30 × 10−6 N/m = 1.60 × 1010 times smaller than atmospheric pressure P34.67 The mirror intercepts power P = I1 A1 = ( 1.00 × 103 W/m ) ⎡⎣π ( 0.500 m ) ⎤⎦ = 785 W (a) In the image, I = I2 = (b) P , so A2 785 W π ( 0.020 m ) = 625 kW/m 2 Emax , so I2 = µ0c Emax = µ0 cI = ( 4π × 10−7 T ⋅ m/A ) ( 3.00 × 108 m/s ) ( 6.25 × 105 W/m ) = 21.7 kN C (c) Bmax = Emax = 72.4 µT c (d) We obtain the time interval from 0.400 ( PΔt ) = mcΔT solving, Δt = mcΔT ( 1.00 kg ) ( 186 J/kg ⋅°C )( 100°C − 20.0°C ) = 0.400P 0.400 ( 785 W ) = 1.07 × 103 s = 17.8 P34.68 (a) In E = q Φ 487 N ⋅ m /C ˆ ˆ r = r = rˆ , 4π ∈0 r 4π r 4π r 38.8 E = rˆ where E is in volts per meter and r is in meters r © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 34 (b) 603 The radiated intensity is I= P Emax = 4π r 2 µ0 c solving, µ0 cP µ0 cP = 4π r r 2π Emax = = r Emax = (c) ( 4π × 10 −7 T ⋅ m/A ) ( 3.00 × 108 m/s )( 25.0 W ) 2π 38.7 where E is in volts per meter and r is in meters r 38.7 = 3.00 × 106 → r = 1.29 × 10−5 = 12.9 × 10−6 , so r is r 12.9 µm, but the expression in part (b) does not apply if this point is inside the source For Emax = (d) From part (c), we see that in the radiated wave, the field amplitude is inversely proportional to distance As the distance doubles, the amplitude is cut in half (e) In the static case, the field is inversely proportional to the square of the distance As the distance doubles, the field is reduced by a factor of P34.69 (a) At steady state, Pin = Pout and the power radiated out is Pout = eσ AT Thus, P T = ⎡⎢ out ⎤⎥ ⎣ eσ A ⎦ 14 ⎡ ⎤ 900 W/m =⎢ −8 ⎥ ⎢⎣ 0.700 ( 5.67 × 10 W/m ⋅ K ) ⎥⎦ 14 = 388 K = 115°C (b) The box of horizontal area A presents projected area A sin 50.0° perpendicular to the sunlight Then by the same reasoning, 0.900 ( 000 W/m ) A sin 50.0° = 0.700 ( 5.67 × 10−8 W/m ⋅ K ) AT or ⎡ ⎤ 900 W/m ) sin 50.0° ( T=⎢ −8 ⎥ ⎢⎣ 0.700 ( 5.67 × 10 W/m ⋅ K ) ⎥⎦ 14 = 363 K = 90.0°C © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 604 P34.70 Electromagnetic Waves (a) See ANS FIG P34.70 ANS FIG P34.70 (b) uE = 1 ∈0 E = ∈0 Emax cos (kx) 2 (c) uB = 2 B = Bmax cos (kx) µ0 µ0 (d) Note that 2 Emax Emax uB = cos (kx) = cos (kx) 2 µ0 c µ0 ( µ0 ∈0 ) = ∈0 Emax cos (kx) = uE 2 cos (kx) Therefore, u = uE + uB = ∈0 Emax (e) λ Eλ = ∫0 uA dx Eλ = ∫0 ∈0 Emax cos λ 2 (kx)A dx = λ ⎡1 ⎤ ∫0 ∈0 Emax A ⎢⎣ + cos(2kx)⎥⎦ A dx 2 ∈0 Emax λ λ = ∈0 Emax A x + A sin(2kx) 4k ∈0 Emax = ∈0 Emax Aλ + A [ sin(4π ) − sin(0)] 4k = (f) (g) P= ∈0 Emax λA 2 Eλ ∈0 Emax λA 1 2 = = ∈0 Emax λ f ) A = ∈0 cEmax A ( T (1 f ) 2 P ∈0 cEmax A I= = = ∈0 cEmax A A © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 34 (h) 605 From part (g), we have µ ∈ cE cE cE E2 ∈0 cEmax = 0 max = ( µ0 ∈0 ) max = max = max µ0 2 µ0 c µ0 µ0 c Emax The result in part (g) agrees with I = in Equation 34.24 µ0 c P34.71 The bead is black, so we assume it absorbs all light that strikes it The bead presents an effective face of area A = π r2 to the light Since we assume the bead to be perfectly absorbing, the light pressure, from Equation 34.28, is P= Sav I F = = c c A so the light force is F = (a) I A c The light force balances the weight, F = Fg , so I π r = mg c solving, ρ ⎛ π r ⎞ gc ⎝3 ⎠ mgc I= = = ρ cgr 2 πr πr ⎛ 0.200 × 10−3 kg ⎞ = ⎜ 3.00 × 108 m/s ) ( 9.80 m/s ) ( −6 ⎟ ⎠ 3⎝ 10 m × ( 0.500 × 10−3 m ) I = 3.92 × 108 W/m (b) The minimum power required is P = IA = ( 3.92 × 108 W/m ) π ( 0.500 × 10−3 m ) = 308 W P34.72 The bead is black, so we assume it absorbs all light that strikes it The bead presents an effective face of area A = π r2 to the light Since we assume the bead to be perfectly absorbing, the light pressure, from Equation 34.28, is P= Sav I F = = c c A so the light force is F = I A c © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 606 Electromagnetic Waves (a) The light force balances the weight, F = Fg , so I π r = mg c solving, mgc I= = π r2 (b) ρ ⎛ π r ⎞ gc ⎝3 ⎠ = ρ cgr πr The minimum power required is ⎛4 ⎞ P = IA = ⎜ ρ cgr ⎟ (π r ) = πρ cgr ⎝3 ⎠ P34.73 (a) A hemisphere is half a sphere: ⎡1⎛ ⎞⎤ m = ρV = ρ ⎢ ⎜ π r ⎟ ⎥ = 5.50 + 4(0.800) kg = 8.70 kg ⎠⎦ ⎣2 ⎝ ⎛ 6m ⎞ r=⎜ ⎝ ρ 4π ⎟⎠ 13 ⎛ ( 8.7 kg ) ⎞ =⎜ ⎟ ⎝ ( 990 kg/m ) 4π ⎠ 13 = 0.161 m 4π r = 2π ( 0.161 m ) = 0.163 m 2 (b) A= (c) P = eσ AT and T = 31.0 + 273.0 = 304 K: P = 0.970 ( 5.67 × 10−8 W/m ⋅ K ) ( 0.163 m ) ( 304 K ) = 76.8 W (d) P = eσ T A I = 0.970 ( 5.67 × 10−8 W m ⋅ K ) ( 304 K ) I= = 470 W m (e) I= Emax µ0c Emax = ( µ0 cI ) 12 = ⎡⎣ ( 4π × 10−7 Tm/A ) ( 3.00 × 108 m/s ) ( 470 W/m ) ⎤⎦ 12 = 595 N/C © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 34 607 595 N/C = 1.98 µT × 108 m/s (f) Emax = cBmax → Bmax = (g) ⎛ 6m ⎞ Each kitten has radius rk = ⎜ ⎝ ρ 4π ⎟⎠ 13 ⎡ ( 0.800 ) ⎤ =⎢ ⎥ ⎣ 990 × 4π ⎦ 13 = 0.072 m and radiating area 2π ( 0.072 m ) = 0.033 m The mother cat ⎡ ( 5.50 ) ⎤ has area 2π ⎢ ⎥ = 0.120 m The total glowing area is 990 × π ⎣ ⎦ 0.120 m + 0.033 m = 0.254 m and has power output 23 ( P = IA = ( 470 W/m P34.74 (a) )( 0.254 m ) = 119 W On the right side of the equation, (C (b) ) C2 ( m/s ) 2 / N ⋅ m ) ( m/s ) = N ⋅ m ⋅ C2 ⋅ m ⋅ s N ⋅ m J = = =W C2 ⋅ s ⋅ m s s F = ma = qE, or −19 qE ( 1.60 × 10 C ) ( 100 N/C ) a= = = m 9.11 × 10−31 kg (c) 1.76 × 1013 m/s The radiated power is then: (1.60 × 10−19 C) (1.76 × 1013 m/s2 ) q2 a2 P= = 6π ∈0 c 6π ( 8.85 × 10−12 C2 /N ⋅ m ) ( 3.00 × 108 m/s ) 2 = 1.75 × 10−27 W (d) ⎛ v2 ⎞ F = mac = m ⎜ ⎟ = qvB , ⎝ r ⎠ so v= qBr m The proton accelerates at −19 v q B2 r ( 1.60 × 10 C ) ( 0.350 T ) ( 0.500 m ) a= = = r m2 (1.67 × 10−27 kg ) 2 = 5.62 ì 1014 m/s â 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 608 Electromagnetic Waves The proton then radiates (1.60 × 10−19 C) ( 5.62 × 1014 m/s2 ) q2 a2 P= = 6π ∈0 c 6π ( 8.85 × 10−12 C2 /N ⋅ m ) ( 3.00 × 108 m/s ) 2 = 1.80 × 10−24 W P34.75 We take R to be the planet’s distance from its star, and r to be the radius of the planet (a) The effective area of the planet over which it absorbs light is its projection onto a plane perpendicular to the light from its sun The projected area of a planet of radius r is π r2, so the planet absorbs light over area π r2 (b) The planet radiates over its entire surface area, 4π r2 (c) At steady-state, Pin = Pout: eI in (π r ) = eσ ( 4π r ) T ⎛ 6.00 × 1023 W ⎞ e⎜ π r ) = eσ ( 4π r ) T , so that ( ⎟ 4π R ⎝ ⎠ 6.00 × 1023 W = 16πσ R 2T R= = 6.00 × 1023 W 16πσ T 6.00 × 1023 W = 4.77 × 10 m −8 16π ( 5.67 × 10 W/m ⋅ K ) ( 310 K ) Challenge Problems P34.76 We are given f = 90.0 MHz and Emax = 200 mV/m = 2.00 × 10−3 V/m c = 3.33 m f (a) The wavelength of the wave is λ = (b) Its period is T = (c) We obtain the maximum value of the magnetic field from Bmax = = 1.11 × 10−8 s = 11.1 ns f Emax = 6.67 × 10−12 T = 6.67 pT c © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 34 (d) 609 ⎛ x ⎞ E = ( 2.00 × 10−3 ) cos 2π ⎜ − 90.0 × 106 t ⎟ ˆj ⎝ 3.33 ⎠ ⎛ x ⎞ B = ( 6.67 × 10−12 ) cos 2π ⎜ − 90.0 × 106 t ⎟ kˆ ⎝ 3.33 ⎠ where E is in V/m, B in tesla, x in meters, and t in seconds 2.00 × 10−3 V/m ) ( Emax I= = µ0 c ( 4π × 10−7 T ⋅ m/A ) ( 3.00 × 108 m/s ) (e) = 5.31 × 10−9 W m (f) From Equation 34.26, I = cuavg so uavg = (g) From Equation 34.30, the pressure is P= P34.77 (a) −9 2I ( )( 5.31 × 10 W m ) = = 3.54 × 10−17 Pa c 3.00 × 10 m/s The magnetic field has amplitude Bmax = (b) Emax 175 V/m = = 5.83 × 10−7 T = 583 nT c 3.00 × 10 m/s The wave number is k= (c) I = 1.77 × 10−17 J m c 2π 2π –1 = = 419 m λ 0.015 m The angular frequency is ω = kc = ( 419 m −1 ) ( 3.00 × 108 m/s ) = 1.26 × 1011 s −1 (d) S ∝ E × B, S is in the x direction, and E vibrates in the y direction (xy plane), so B must vibrate in the z direction, thus B vibrates in the xz plane (e) The magnitude of the average Poynting vector is the wave intensity Savg = Emax Bmax µ0 = (175 V/m ) ( 5.83 × 10–7 T ) ( 4π × 10 N/A –7 ) = 40.6 W/m The Poynting vector itself points in the direction of energy transport: Savg = 40.6ˆi W/m © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 610 Electromagnetic Waves (f) For perfect reflection, the pressure is 2S ( 40.6 W m ) Pr = = = 2.71 × 10−7 N m = 271 nPa c 3.00 × 108 m/s (g) From Newton’s second law, −7 2 F PA ( 2.71 × 10 N/m ) ( 0.750 m ) ∑ a= = = m m 0.500 kg = 4.07 × 10−7 m/s a = 407 ˆi nm/s P34.78 We can approximate the magnetic field as uniform over the area of the loop while it oscillates in time as B = Bmax cos ω t The induced voltage is ε=– or (a) ( dΦ B d d = – (BA cos θ ) = –A Bmax cos ω t cos θ dt dt dt ) ε = ABmax ω (sin ω t cos θ ) Since the angular frequency is ω = π f, and the area of the loop is π r , the amplitude of this emf is ε max = 2π r f Bmax cosθ (b) P34.79 (a) where θ is the angle between the magnetic field and the normal to the loop If E is vertical, B is horizontal, so the plane of the loop should be vertical and the plane should contain the line of sight of the transmitter From the particle under a net force model, the acceleration of the astronaut is a= F dp = m m dt where dp/dt is the rate of change of momentum of the astronaut From the momentum version of the isolated system model, the rate of change of momentum of the astronaut is equal in magnitude to that of the radiation from the flashlight The momentum of the radiation leaving the flashlight can be evaluated from Equation 34.27, assuming the same equation for complete absorption applies to complete emission Therefore, the acceleration of the astronaut can be written as a= d ⎛ TER ⎞ dTER P = ⎜⎝ ⎟⎠ = m dt c mc dt mc © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 34 611 where P is the power of the radiation leaving the flashlight Because all three variables on the right side of this equation are constant, the acceleration of the astronaut is constant and we can use the particle under constant acceleration model The position of the astronaut is given by, 1⎛ P ⎞ x = xi + vit + at = + + ⎜ ⎟t 2 ⎝ mc ⎠ where we have defined the initial position of the astronaut as x = and recognized that the astronaut begins from rest Solve for the time at which the astronaut is at a position x: t= 2mcx P Substituting numerical values, t= ( 110 kg ) ( 3.00 × 108 m/s ) ( 10.0 m ) 100 W = 8.12 × 10 s = 22.6 h (b) There are no external forces on the astronaut–flashlight system, so the system is isolated for momentum Apply the conservation of momentum principle along an axis parallel to the direction of travel of the astronaut and the flashlight: Δp = → pi = p f → ( ) = m − m f v − m f ( vrel − v ) Solve for the speed of the astronaut: ⎛ mf ⎞ v=⎜ v ⎝ m ⎟⎠ rel Because this speed is constant, we can use the particle under constant velocity model to find the time interval required for the astronaut to arrive back at her spacecraft: Δt = Δx m ⎛ Δx ⎞ = v m f ⎜⎝ vrel ⎟⎠ Substituting numerical values, ⎛ 110 kg ⎞ 10.0 m Δt = ⎜ = 30.6 s ⎝ 3.00 kg ⎟⎠ 12.0 m/s © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 612 Electromagnetic Waves ANSWERS TO EVEN-NUMBERED PROBLEMS P34.2 (a) 7.19 × 1011 V m ⋅ s ; (b) 2.00 × 10–7 T P34.4 ( −4.39ˆi − 1.76ˆj) × 10 P34.6 (a) 3.15 × 103 ˆj N C ; (b) 5.25kˆ × 10−7 T ; (c) 4.83 − ˆj × 10−16 N P34.8 11.0 m P34.10 4.738 × 1014 Hz P34.12 733 nT P34.44 60.0 km P34.16 See P34.16 for full explanation P34.18 The ratio of ω to k is higher than the speed of light in a vacuum, so the wave as described is impossible P34.20 8.64 × 1010 m P34.22 (a) 6.75 W/m2; (b) 6.64 × 103 W/m2; (c) A powerful automobile running on sunlight would have to carry on its roof a solar panel that is huge compared to the size of the car; (d) Agriculture and forestry for food and fuels, space heating of large and small buildings, water heating, and heating for drying and many other processes are current and potential applications of solar energy 11 m/s ( ) ( ) P34.24 (a) 0; (b) 11.5ˆi − 28.6ˆj W/m P34.26 For the small container, 33.4° and for the larger container, 21.7° P34.28 (a) 88.8 nW/m2; (b) 11.3 MW P34.30 (a) 5.16 × 10–10 T; (b) Since the magnetic field of the Earth is approximately × 10−10 T, the Earth’s field is some 100 000 times stronger P34.32 5.16 m P34.34 (a) 540 V/m; (b) 2.58 µJ/m3; (c) 773 W/m2 P34 36 83.3 nPa P34.38 (a) P34.40 (a) 5.82 × 108 N; (b) 6.10 × 1013 times stronger µ0 cP P P ; (b) ; (c) 2 πr c c © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 34 613 P34.42 (a) 590 W/m2; (b) 2.10 × 1016 W; (c) 7.01 × 107 N; (d) ~1013 times stronger; (e) The values are similar for both planets because both the forces follow inverse-square laws The force ratios are not identical for the two planets because of their different radii and masses P34.44 (a) 1.00 × 103 km or 621 mi; (b) While the project may be theoretically possible, it is not very practical P34.46 (a) P34.48 2π mc qB P34.50 (a) ~108 Hz radio wave; (b) ~1013 Hz infrared P34.52 Listeners 100 km away will receive the news before the people in the newsroom P34.54 See table in P34.54 for full description P34.56 95.1 mV/m P34.58 E = 300 cos ( 62.8x − 1.88 × 1010 t ) and B = 1.00 cos ( 62.8x − 1.88 × 1010 t ) 1 2 ; cos ( kx − ω t ) ˆi; (c) µ0 cJ max µ0 cJ max cos ( kx − ω t ) ˆj; (b) µ0 cJ max (d) 3.48 A/m P34.60 (a) 1.50 cm; (b) 25.0 μJ; (c) 7.37 mJ/m3; (d) Emax = 40.8 kV/m, Bmax = 136 μT; (e) 83.3 μN P34.62 (a) 93.3%; (b) 50.0%; (c) P34.64 3SR 4cGMS ρ P34.66 (a) 6.30 × 10–6 Pa; (b) 1.60 × 1010 times smaller than atmospheric pressure P34.68 38.8 (a) E = rˆ , where E is in volts per meter and r is in meters; r 38.7 (b) Emax = where E is in volts per meter and r is in meters; r (c) 12.9 μm, but the expression in part (b) does not apply if this point is inside the source; (d) From part (c), we see that in the radiated wave, the field amplitude is inversely proportional to distance As the distance doubles, the amplitude is cut in half; (e) In the static case, the field is inversely proportional to the square of distance As the distance doubles, the field is reduced by a factor of © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 614 Electromagnetic Waves 2 Bmax cos (kx) ; ∈0 Emax cos (kx); (c) µ0 1 2 2 cos (kx); (e) ∈0 Emax (d) ∈0 Emax λ A; (f) ∈0 cEmax A; (g) ∈0 cEmax ; 2 E2 (h) The result in part (g) agrees with I = max in Equation 34.24 µ0c P34.70 (a) See ANS FIG P34.70; (b) P34.72 (a) P34.74 (a) See P34.74(a) for full proof; (b) 1.76 × 1013 m/s2; (c) 1.75 × 10−27 W; (d) 1.80 × 10−24 W P34.76 (a) 3.33 m; (b) 11.1 ns; (c) 6.67 pT; ⎛ x ⎞ (d) E = 2.00 × 10−3 cos 2π ⎜ − 90.0 × 106 t ⎟ ˆj and ⎝ 3.33 ⎠ ⎛ x ⎞ –9 B = ( 6.67 × 10−12 ) cos 2π ⎜ − 90.0 × 106 t ⎟ kˆ ; (e) 5.31 × 10 W/m ; ⎝ 3.33 ⎠ (f) 1.77 × 10–17 J/m2; (g) 3.54 × 10–17 Pa 4 ρcgr ; (b) πρcgr 3 ( P34.78 ) (a) ε max = 2π r f Bmax cos θ ; (b) The plane of the loop should be vertical and the plane should contain the line of sight of the transmitter © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part