Chapter 34 Solutions

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1 Chapter 34 Solutions 34.1 Sine the light from this star travels at m/s, the last bit of light will hit the Earth in m m/s s 680 years. Therefore, it will disappear from the sky in the year A.D. 34. v 1 kµ 0 e m/s 34.3 E B or 0 B ; so B T 733 nt 34.4 E max B max v is the generalized version of Equation B max E max v V/m N m (/3)( m/s) V C T C m N s T 38.0 pt 34.5 (a) f λ or f (50.0 m) m/s so f Hz 6.00 MHz E B or.0 B so B max (73.3 nt)( k) max () k π λ π m 1 and ω π f π ( s 1 ) rad/s B B max os(kx ω t) (73.3 nt) os (0.16x t)( k) 34.6 ω π f 6.00π 10 9 s s -1 k π λ ω 6.00π 109 s m/s 0.0π 6.8 m 1 B max E 300 V/m m/s 1.00 µt E 300 V m os ( 6.8x t) B (1.00 µt) os (6.8x t) 000 by Harourt, In. All rights reserved.

2 Chapter 34 Solutions (a) B E 100 V/m m/s T µt λ π k π µm m () f λ m/s m Hz 34.8 E E max os(kx ω t) E x E max sin(kx ω t)(k) E t E max sin(kx ω t)( ω) E x E max os(kx ω t)(k ) E t E max os(kx ω t)( ω) We must show: E x µ 0 e E 0 t That is, ( k )E max os( kx ω t) µ 0 e 0 ( ω) E max os( kx ω t) But this is true, beause k ω 1 fλ 1 µ 0 e 0 The proof for the wave of magneti field is preisely similar. *34.9 In the fundamental mode, there is a single loop in the standing wave between the plates. Therefore, the distane between the plates is equal to half a wavelength. λ L (.00 m) 4.00 m Thus, f λ m/s 4.00 m Hz 75.0 MHz *34.10 d A to A 6 m ± 5% λ λ 1 m ± 5% 000 by Harourt, In. All rights reserved.

3 300 Chapter 34 Solutions v λ f ( 0.1 m ± 5% )( s 1 ) ms ± 5% S I U At U V u Energy Unit Volume u I 1000 W/m m/s 3.33 µ J/m S av P 4πr W 7.68 µw/m 4π ( m) E max µ 0 S av V / m V max E max L (76.1 mv/m)(0.650 m) 49.5 mv (amplitude) or 35.0 mv (rms) r ( 5.00 mi) ( 1609 m / mi) m S P 4πr W 4π ( µw/m m)

4 Chapter 34 Solutions 301 Goal Solution What is the average magnitude of the Poynting vetor 5.00 miles from a radio transmitter broadasting isotropially with an average power of 50 kw? G : O : As the distane from the soure is inreased, the power per unit area will derease, so at a distane of 5 miles from the soure, the power per unit area will be a small fration of the Poynting vetor near the soure. The Poynting vetor is the power per unit area, where A is the surfae area of a sphere with a 5-mile radius. A : The Poynting vetor is In meters, r ( 5.00 mi) ( 1609 m / mi) 8045 m and the magnitude is S W (4π)(8045) W/m L : The magnitude of the Poynting vetor ten meters from the soure is 199 W/m, on the order of a million times larger than it is 5 miles away! It is surprising to realize how little power is atually reeived by a radio (at the 5-mile distane, the signal would only be about 30 nw, assuming a reeiving area of about 1 m ). 100 W I 4π (1.00 m) 7.96 W/m u I J/m n J/m 3 (a) u E 1 u 13.3 n J/m3 u B 1 u 13.3 n J/m3 () I 7.96 W/m Power output (power input)(effiieny) Thus, Power input power out eff W W and A P I W W/m m 000 by Harourt, In. All rights reserved.

5 30 Chapter 34 Solutions *34.16 I B max P µ 0 4π r B max P 4π r µ π ( )( ) ( 4π 10 7 ) ( ) ( ) T Sine the magneti field of the Earth is approximately T, the Earth's field is some 100,000 times stronger (a) P I R 150 W; A π rl π ( m)( m) m S P A B µ 0 33 kw/m (points I π r µ 0 (1.00) π ( ) radially inward) µt E V x IR L 150 V m 1.88 kv/m Note: S EB µ 0 33 kw/m

6 Chapter 34 Solutions (a) E B (80.0 i j 64.0 k)(n/c) (0.00 i j k)µt E B ( )N s/c m 0 S 1 (80.0 i j 64.0 k)(n/c) (0.00 i j k)µt µ E B 0 4π 10 7 T m/a S (6.40 k 3. j 6.40 k i 1.8 j i)10 6 W/m 4π 10 7 S (11.5 i 8.6 j) W/m 30.9 W/m at 68. from the +x axis We all the urrent I rms and the intensity I. The power radiated at this frequeny is P (0.0100)( V rms )I rms ( V rms) R 1.31 W If it is isotropi, the intensity one meter away is I P A B max 1.31 W 4π (1.00 m) W/m S av µ 0 I 4π 10 7 T m/a m/s µ 0 B max ( )( W / m ) 9.5 nτ *34.0 (a) effiieny useful power output total power input S av P A 700 W m ( )( m) Wm 100% 700 W 100% 50.0% 1400 W S av 69 kw m toward the oven hamber () S av E max µ 0 E max 4π 10 7 T m m A s W m V 14. kv m m 000 by Harourt, In. All rights reserved.

7 304 Chapter 34 Solutions 34.1 (a) B max E max N/C m/s.33 mt I E max µ 0 ( ) (4π 10 7 )( ) 650 MW/m () I P A : P I A ( W/m ) π 4 ( m) 510 W 34. Power SA E max µ 0 (4πr ); solving for r, r Pµ 0 π E max (100 W)µ 0 π(15.0 V / m) 5.16 m 3 ( ) W 34.3 (a) I 3 π( m) 4.97 kw/m u av I J/m s m/s 16.6 µ J/m (a) E B ( m/s)( T) 540 V/m u av B µ 0 ( ) 4π µ J/m 3 () S av u av ( )( ) 773 W/m (d) This is 77.3% of the flux in Example It may be loudy, or the Sun may be setting For omplete absorption, P S npa *34.6 (a) P ( S av )( A) ( 6.00 W / m )( m ) J/s In one seond, the total energy U impinging on the mirror is J. The momentum p transferred eah seond for total refletion is p U ( J) m/s 10 kg m s (eah seond) F dp dt kg m/s 1 s N

8 Chapter 34 Solutions (a) The radiation pressure is ()(1340 W / m ) m/s N/m Multiplying by the total area, A m gives: F 5.36 N The aeleration is: a F m 5.36 N 6000 kg m/s () It will take a time t where: d 1 at or t d a m m/s ( ) ( ) s 10.7 days 34.8 The pressure P upon the mirror is P S av where A is the ross-setional area of the beam and S av P A The fore on the mirror is then F PA P A P A Therefore, F ( ) ( ) N 34.9 I P π r E max µ 0 (a) E max P( µ 0 ) π r 1.90 kn/c J/s m/s (1.00 m) 50.0 pj () p U kg m/s 000 by Harourt, In. All rights reserved.

9 306 Chapter 34 Solutions (a) If P S is the total power radiated by the Sun, and r E and r M are the radii of the orbits of the planets Earth and Mars, then the intensities of the solar radiation at these planets are: I E r Thus, I M I E E r M P S 4πr E and I M P S 4πr M ( ) m 1340 W m m 577 W/m Mars interepts the power falling on its irular fae: P M I M ( π R M ) ( 577 W m )π( m) W () If Mars behaves as a perfet absorber, it feels pressure P S M I M and fore F PA I M ( π R M ) P M W ms N (d) The attrative gravitational fore exerted on Mars by the Sun is F g GM S M M r M ( N m kg ) kg ( )( kg) ( m) N whih is ~10 13 times stronger than the repulsive fore of () (a) The total energy absorbed by the surfae is ( ) 60.0 s U ( 1 I )At W m m ( ) 11.3 kj The total energy inident on the surfae in this time is U.5 kj, with U 11.3 kj being absorbed and U 11.3 kj being refleted. The total momentum transferred to the surfae is p ( momentum from absorption )+ ( momentum from refletion ) p U + U 3U ( 103 J) ms kg ms 34.3 S av µ 0 J max 8 or 570 (4π 10 7 )J max ( ) 8 so J max 3.48 A/m

Chapter 34 Solutions 307 34.33 (a) P S av A µ 0 J max 8 A P 4π 10 7 (10.0) (3.00 10 8 ) 8 (1.0 0.400).

34 P ( V) R or P ( V) V ( )E y y E y losθ y θ l eeiving ntenna V osθ so P os θ (a) θ 15.0 : P P max os ( 15.

35 (a) Construtive interferene ours when d os θ nλ for some integer n. osθ n λ d n λ n n 0, ± 1, ±,.

10 Chapter 34 Solutions (a) P S av A µ 0 J max 8 A P 4π 10 7 (10.0) ( ) 8 ( ).6 kw S av µ 0 J max (4π 10 7 (10.0) ( ) 4.71 kw/m 8 8 *34.34 P ( V) R or P ( V) V ( )E y y E y losθ y θ l eeiving ntenna V osθ so P os θ (a) θ 15.0 : P P max os ( 15.0 ) 0.933P max 93.3% θ 45.0 : P P max os ( 45.0 ) 0.500P max 50.0% () θ 90.0 : P P max os ( 90.0 ) (a) Construtive interferene ours when d os θ nλ for some integer n. osθ n λ d n λ n n 0, ± 1, ±,... λ / strong θ os , 70 Destrutive interferene ours when dosθ n + 1 λ : os θ n + 1 weak θ os 1 (±1) 0, by Harourt, In. All rights reserved.

11 308 Chapter 34 Solutions Goal Solution Two radio-transmitting antennas are separated by half the broadast wavelength and are driven in phase with eah other. In whih diretions are (a) the strongest and the weakest signals radiated? G : O : The strength of the radiated signal will be a funtion of the loation around the two antennas and will depend on the interferene of the waves. A diagram helps to visualize this situation. The two antennas are driven in phase, whih means that they both reate maximum eletri field strength at the same time, as shown in the diagram. The radio EM waves travel radially outwards from the antennas, and the reeived signal will be the vetor sum of the two waves. A : (a) Along the perpendiular bisetor of the line joining the antennas, the distane is the same to both transmitting antennas. The transmitters osillate in phase, so along this line the two signals will be reeived in phase, onstrutively interfering to produe a maximum signal strength that is twie the amplitude of one transmitter. L : Along the extended line joining the soures, the wave from the more distant antenna must travel one-half wavelength farther, so the waves are reeived 180 out of phase. They interfere destrutively to produe the weakest signal with zero amplitude. Radio stations may use an antenna array to diret the radiated signal toward a highly-populated region and redue the signal strength delivered to a sparsely-populated area λ f 536 m so h λ m λ f 188 m so h λ m For the proton: ΣF ma qvb sin 90.0 mv R The period and frequeny of the proton s irular motion are therefore: T πr v ( ) ( ) T π m qb π kg C The harge will radiate at this same frequeny, with ( ) s f Hz. λ f ms Hz 56. m For the proton, ΣF ma yields qvb sin 90.0 mv R The period of the proton s irular motion is therefore: T πr v The frequeny of the proton's motion is f 1/T π m qb The harge will radiate eletromagneti waves at this frequeny, with λ T f π m qb

12 Chapter 34 Solutions 309 *34.39 From the eletromagneti spetrum hart and aompanying text disussion, the following identifiations are made: Frequeny f Wavelength, λ f Classifiation Hz 10 0 Hz 150 Mm Radio khz 10 3 Hz 150 km Radio MHz 10 6 Hz 150 m Radio GHz 10 9 Hz 15 m Mirowave THz 10 1 Hz 150 µm Infrared PHz Hz 150 nm Ultraviolet EHz Hz 150 pm x-ray ZHz 10 1 Hz 150 fm Gamma ray YHz 10 4 Hz 150 am Gamma Ray Wavelength, λ Frequeny f λ Classifiation km 10 3 m Hz Radio m 10 0 m Hz Radio mm 10 3 m Hz Mirowave µm 10 6 m Hz Infrared nm 10 9 m Hz Ultraviolet/x-ray pm 10 1 m Hz x-ray/gamma ray fm m Hz Gamma ray am m Hz Gamma ray *34.40 (a) f λ m/s 1.7 m ~ 10 8 Hz radio wave 1000 pages, 500 sheets, is about 3 m thik so one sheet is about m thik f m/s m ~ 1013 Hz infrared *34.41 f λ m/s m Hz 34.4 (a) λ f m/s /s 61 m so 180 m 61 m wavelengths λ f m/s /s 3.06 m so 180 m 3.06 m 58.9 wavelengths 000 by Harourt, In. All rights reserved.

13 310 Chapter 34 Solutions (a) f λ gives ( Hz)λ m/s: λ m 6.00 pm f λ gives ( Hz)λ m/s: λ m 7.50 m *34.44 Time to reah objet 1 (total time of flight) 1 ( s) s Thus, d vt ( m/s)( s) m 60.0 km The time for the radio signal to travel 100 km is: t r The sound wave to travel 3.00 m aross the room in: t s m m/s s 3.00 m 343 m/s s Therefore, listeners 100 km away will reeive the news before the people in the newsroom by a total time differene of t s s s *34.46 The wavelength of an ELF wave of frequeny 75.0 Hz is λ f ms 75.0 Hz m The length of a quarter-wavelength antenna would be L m km or 0.61 mi L ( 1000 km) 61 mi 1.00 km Thus, while the projet may be theoretially possible, it is not very pratial (a) For the AM band, λ max f min m/s Hz 556 m λ min m/s f max Hz 187 m For the FM band, λ max f min m/s Hz 3.41 m λ min f max m/s Hz.78 m

14 Chapter 34 Solutions CH 4 : f min 66 MHz λ max 4.55 m f max 7 MHz λ min 4.17 m CH 6 : f min 8 MHz λ max 3.66 m f max 88 MHz λ min 3.41 m CH 8 : f min 180 MHz λ max 1.67 m f max 186 MHz λ min 1.61 m (a) P SA (1340 W/m )4π ( m) W S B max so B max µ 0 µ 0 S (4π 10 7 N/A )(1340 W / m ) m/s 3.35 µt S E max µ 0 so E max µ 0 S 4π ( 10 7 )( )( 1340) 1.01 kv/m *34.50 Suppose you over a 1.7 m-by-0.3 m setion of beah blanket. Suppose the elevation angle of the Sun is 60. Then the target area you fill in the Sun's field of view is (1.7 m )( 0.3 m )( os 30 ) 0.4 m Now I P A E At : E IAt 1340 W m (0.6)(0.5)(0.4 m ) 3600 s ~ 10 6 J (a) ε dφ B dt d dt (BA os θ) A d dt (B max os ωt os θ) AB max ω(sin ωt os θ) ε(t) π fb max A sin π ftos θ π r fb max os θ sin π ft Thus, ε max π r fb max os θ, where θ is the angle between the magneti field and the normal to the loop. If E is vertial, then B is horizontal, so the plane of the loop should be vertial and the plane should ontain the line of sight to the transmitter. 000 by Harourt, In. All rights reserved.

15 31 Chapter 34 Solutions 34.5 (a) F grav GM s m R GM s ρ ( 4/3 )π r3 R where M s mass of Sun, r radius of partile and R distane from Sun to partile. Sine F rad Sπ r, F rad 1 3SR F grav r 4GM s ρ 1 r 3SR From the result found in part (a), when F grav F rad, we have r 4GM s ρ ( )( m) ( )( kg) ( 1500 kg m 3 )( ms) m 3 14 W / m r N m kg (a) B max E max S av E max µ T W/m () P S av A W (d) F PA S av A N ( weight of 3000 H atoms!) *34.54 (a) The eletri field between the plates is E V l, direted downward in the figure. The magneti field between the plate's edges is B µ 0 i π r ounterlokwise. The Poynting vetor is: S 1 µ 0 E B ( V)i π rl (radially outward) S B E i i The lateral surfae area surrounding the eletri field volume is ( V)i A π rl, so the power output is P SA π rl ( π rl) ( V)i () As the apaitor harges, the polarity of the plates and hene the diretion of the eletri field is unhanged. Reversing the urrent reverses the diretion of the magneti field, and therefore the Poynting vetor. The Poynting vetor is now direted radially inward.

16 Chapter 34 Solutions 313 *34.55 (a) The magneti field in the enlosed volume is direted upward, db with magnitude B µ 0 ni and inreasing at the rate dt µ 0 n di dt. The hanging magneti field indues an eletri field around any irle of radius r, aording to Faraday s Law: E( π r) µ 0 n di dt π r ( ) E µ 0 nr di dt S E B or E µ 0 nr di dt (lokwise) i Then, S 1 E B 1 µ 0 nr µ 0 µ 0 di dt µ 0 ni ( ) inward, or the Poynting vetor is S µ 0 n ri di dt (radially inward) () The power flowing into the volume is P SA lat where A lat is the lateral area perpendiular to S. Therefore, P µ 0 n ri di dt ( π rl) µ 0 π n r di li dt Taking A ross to be the ross-setional area perpendiular to B, the indued voltage between the ends of the indutor, whih has N nl turns, is V ε N db dt A ross nl µ 0 n di dt ( π r ) µ 0 π n r l di dt and it is observed that P ( V)i *34.56 (a) The power inident on the mirror is: P I IA 1340 W m [ ( ] ) W π 100 m The power refleted through the atmosphere is P R 0.746( W) W () S P R A W 0.65 π W/m m ( ) Noon sunshine in Saint Petersburg produes this power-per-area on a horizontal surfae: P N 0.746( 1340 W / m )sin W / m The radiation intensity reeived from the mirror is 0.65 W m 1 W m 100% 0.513% of that from the noon Sun in January. 000 by Harourt, In. All rights reserved.

17 314 Chapter 34 Solutions u 1 e 0 E max (Equation 34.1) E max u e mv/m *34.58 The area over whih we model the antenna as radiating is the lateral surfae of a ylinder, A π rl π( m) ( m) m (a) The intensity is then: S P A W W m m The standard is: mw mw W m m 1.00 mw m 1.00 m 5.70 W m While it is on, the telephone is over the standard by 3.9 W m 4.19 times 5.70 W m (a) B max E max 175 V/m m/s T k π λ π ( m) 419 rad/m ω k rad/s Sine S is along x, and E is along y, B must be in the z diretion. (That is S E B.) S av E maxb max µ W/m () P r S N/m (d) a ΣF m PA m ( )( m ) N/m kg m/s

18 Chapter 34 Solutions 315 *34.60 (a) At steady-state, P in P out and the power radiated out is P out eσat 4. Thus, W m A ( ) 5.67 W 10 8 m K 4 AT W m or T 0.700( Wm K 4 ) K 115 C The box of horizontal area A, presents projeted area A sin perpendiular to the sunlight. Then by the same reasoning, W m A sin ( ) W m K 4 AT 4 ( 900 W m )sin 50.0 or T 0.700( Wm K 4 ) K 90.0 C (a) P F A I F IA P 100 J / s m/s N ( 110 kg)a a m/s and x 1 at t x a s.6 h 0 (107 kg)v (3.00 kg)(1.0 m/s v) (107 kg)v 36.0 kg m/s + (3.00 kg)v v m/s t 30.6 s 000 by Harourt, In. All rights reserved.

19 316 Chapter 34 Solutions Goal Solution An astronaut, stranded in spae 10.0 m from his spaeraft and at rest relative to it, has a mass (inluding equipment) of 110 kg. Sine he has a 100-W light soure that forms a direted beam, he deides to use the beam as a photon roket to propel himself ontinuously toward the spaeraft. (a) Calulate how long it takes him to reah the spaeraft by this method. Suppose, instead, he deides to throw the light soure away in a diretion opposite the spaeraft. If the mass of the light soure has a mass of 3.00 kg and, after being thrown, moves at 1.0 m/s relative to the reoiling astronaut, how long does it take for the astronaut to reah the spaeraft? G : O : Based on our everyday experiene, the fore exerted by photons is too small to feel, so it may take a very long time (maybe days!) for the astronaut to travel 10 m with his photon roket. Using the momentum of the thrown light seems like a better solution, but it will still take a while (maybe a few minutes) for the astronaut to reah the spaeraft beause his mass is so muh larger than the mass of the light soure. In part (a), the radiation pressure an be used to find the fore that aelerates the astronaut toward the spaeraft. In part, the priniple of onservation of momentum an be applied to find the time required to travel the 10 m. A : (a) Light exerts on the astronaut a pressure P FA S, and a fore of F SA 100 J / s m/s N By Newton s nd law, a F m N 110 kg m/s This aeleration is onstant, so the distane traveled is x 1 at, and the amount of time it travels is t x a 10.0 ( m) m/s s.6 h Beause there are no external fores, the momentum of the astronaut before throwing the light is the same as afterwards when the now 107-kg astronaut is moving at speed v towards the spaeraft and the light is moving away from the spaeraft at ( 1.0 m / s v). Thus, p i p f gives 0 ( 107 kg)v ( 3.00 kg) ( 1.0 m / s v) 0 ( 107 kg)v ( 36.0 kg m/s)+ ( 3.00 kg)v v m / s t x v 10.0 m 0.37 m / s 30.6 s L : Throwing the light away is ertainly a more expedient way to reah the spaeraft, but there is not muh hane of retrieving the lamp unless it has a very long ord. How long would the ord need to be, and does its length depend on how hard the astronaut throws the lamp? (You should verify that the minimum ord length is 367 m, independent of the speed that the lamp is thrown.)

20 Chapter 34 Solutions The 38.0% of the intensity S 1340 W m that is refleted exerts a pressure P 1 S r ( 0.380)S The absorbed light exerts pressure P S a ( 0.60)S Altogether the pressure at the subsolar point on Earth is (a) P tot P 1 + P 1.38S 1.38(1340 W/m ) m/s Pa P a N/m P tot N/m times smaller than atmospheri pressure Think of light going up and being absorbed by the bead whih presents a fae area π r b. The light pressure is P S I. (a) F l Iπ r b mg ρ 4 3 π r 3 b g and I 4ρ g 3 3m 4πρ 1/ W/m P IA ( W/m )π ( m) 1.05 kw Think of light going up and being absorbed by the bead whih presents fae area π r b. If we take the bead to be perfetly absorbing, the light pressure is P S av I F l A (a) F l F g so I F l A F g A mg π r b From the definition of density, ρ m V m 4 3 π r b 3 so Substituting for r b, I mg π 4πρ 3m /3 g 4ρ 3 /3 m π 1/3 1 4 r πρ m 3 b ( ) 1/3 4ρg 3 3m 4πρ 1/3 P IA π r 4ρg 3 3m 4π ρ 1/3 000 by Harourt, In. All rights reserved.

21 318 Chapter 34 Solutions The mirror interepts power P I 1 A 1 ( W/m )π(0.500 m) 785 W In the image, (a) I P 785 W 65 kw/m A π m ( ) I E max µ 0 so E max (µ 0 I ) 1/ [(4π 10 7 )( )( )] 1/ 1.7 kn/c B max E max 7.4 µt () P t m T 0.400(785 W)t (1.00 kg) J 4186 kg C (100 C 0.0 C) t J 314 W s 17.8 min (a) λ f ms m 1 s U P( t) J s s ( ) J 5.0 µj () u av U V U π r ( )l U ( π r ) t u av J m mj m 3 ( ) J π( m) ms ( )( s) (d) E max u av e ( 10 3 J m 3 ) C N m Vm 40.8 kv/m B max E max Vm ms T 136 µt (e) F PA S A u av A u J ava m 3 π( m) N 83.3 µn

22 Chapter 34 Solutions (a) On the right side of the equation, C ( ms ) ( C N m ) ms ( ) 3 N m C m s 3 C s 4 m 3 N m s F ma qe or a qe ( m C) ( 100 N C) ms kg () The radiated power is then: P q a F ma r m v r The proton aelerates at 6π e 0 qvb so v qbr m a v r q B r m The proton then radiates P q a 6π e 0 ( ) ( ) ( )( ) π ( ) π ( ) ( 0.500) J s W W ( ) ms ( ) ( ) ( )( ) W P S Power A P π rl 60.0 W π( m)(1.00 m)( m\s) Pa F PA SA ( P / A)A P, τ F l Pl, and τ κ θ Therefore, ( )( ) ( )( ) deg θ Pl 3 κ *34.70 We take R to be the planet s distane from its star. The planet, of radius r, presents a projeted area π r perpendiular to the starlight. It radiates over area 4π r. At steady-state, P in P out : ei in ( π r ) eσ ( 4π r )T 4 e W 4π R ( π r ) eσ ( 4π r )T 4 so that W 16πσR T 4 R W 16πσT W 16π ( Wm K 4 )( 310 K) m 4.77 Gm 000 by Harourt, In. All rights reserved.

23 30 Chapter 34 Solutions The light intensity is I S av E µ 0 The light pressure is P S E µ 0 1 e 0 E For the asteroid, PA ma and a e 0 E A m 34.7 f 90.0 MHz, E max V / m 00 mv/m (a) λ f 3.33 m T 1 f s 11.1 ns B max E max T 6.67 pt x E (.00 mv / m) os π 3.33 m t 11.1 ns j B ( 6.67 pt )k osπ x 3.33 m t 11.1 ns () I E max µ 0 ( ) (4π 10 7 )( ) W/m (d) I u av so u av J/m 3 (e) P I ()( ) Pa

Electromagnetic waves

Electromagnetic waves Eletromagneti waves He predited eletromagneti wave propagation James Clerk Maxwell (1831-1879) Eletromagneti waves He predited eletromagneti wave propagation A singular theoretial ahievement of the 19