( ) Vrms = = = 23. A. Choice (a) is the correct answer.

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1 5 hapter 4. At a frequency of f 5., the inductive reactance, capacitive reactance, and impedance are X π f, X, and Z R + ( X X). This yields π f and Z (. ) + π ( 5. )(. H). A. hoice (a) is the correct answer. Z 5 π ( 5. )(. 75 F) 5, ANSWERS TO EEN NUMBERED ONEPTUA QUESTONS. At resonance, X X. This means that the impedance Z R + X X reduces to Z R. 4. The fundamental source of an electromagnetic wave is a moving charge. For example, in a transmitting antenna of a radio station, charges are caused to move up and down at the frequency of the radio station. These moving charges set up electric and magnetic fields, the electromagnetic wave, in the space around the antenna. 6. Energy moves. No matter moves. You could say that electric and magnetic fields move, but it is nicer to say that the fields stay at that point and oscillate. The fields vary in time, like sports fans in the grandstand when the crowd does the wave. The fields constitute the medium for the wave, and energy moves.. The average value of an alternating current is zero because its direction is positive as often as it is negative, and its time average is zero. The average value of the square of the current is not zero, however, since the squares of positive and negative values are always positive and cannot cancel.. The brightest portion of your face shows where you radiate the most. Your nostrils and the openings of your ear canals are particularly bright. Brighter still are the pupils of your eyes.. The changing magnetic field of the solenoid induces eddy currents in the conducting core. This is accompanied by Rconversion of electrically transmitted energy into internal energy in the conductor. 4. The voltages are not added in a scalar form, but in a vector form, as shown in the phasor diagrams throughout the chapter. Kirchhoff s loop rule is true at any instant, but the voltages across different circuit elements are not simultaneously at their imum values. Do not forget that an inductor can induce an emf in itself and that the voltage across it is 9 ahead of the current in the circuit in phase. PROBEM SOUTONS. (a) R, R (. A )(. ) 96. R, ( R, ) ( 96. ) 6 (. A ). A (d) P av R (. A ) (. ) 76W

2 Alternating urrent ircuits and Electromagnetic Waves 59. (a) R, ( R, ).. 7 P av. R R R P 6. W Because R P av av 4. (see above), if the bulbs are designed to operate at the same voltage, the. W will have the lower resistance.. The meters measure the values of potential difference and current. These are. 7 7, and 7. 7 R A.4 All lamps are connected in parallel with the voltage source, so for each lamp. Also, the current is P and the resistance is R. av 5 W,, 5. A and R R.5 A 96. W,. A and R. A 44.5 The total resistance (series connection) is R eq R + R , so the current in the circuit is A.6 R eq The power to the speaker is then P av.6 The general form of the generator voltage is R speaker. 6 A W v ( t) sin ω, so by inspection ω 6π rad s (a) R, 7 and f π π rad R, R, 7. (d) R, R. 6. A (e) ( 6. A ). 49 A (f) P av R( 6. A ) (. ) 7W (g) The argument of the sine function has units of ωt At t. 5 s, the instantaneous current is [ ] rad s s radians. i v 7 sin 6π rad s. 5 s R. 7 sin 94. rad 69. A.

3 6 hapter.7 X, so its units are π f Sec Farad Sec oulomb olt. ( ) ( ) π f X.9 (a) 6 olt olt Ohm oulomb Sec Amp π 6.. /.4 A 4 ma 6 4 π 5.. /.5 A 5 ma π f( ), so X f π. A π ( 4. 6 F )( ) 4.. (a) X π f π 6.. F, 6. X. 6 A (. 6 A). A (d) No. The current reaches its imum value one-quarter cycle before the voltage reaches its imum value. From the definition of capacitance, the capacitor reaches its imum charge when the voltage across it is also a imum. onsequently, the imum charge and the imum current do not occur at the same time.. π f X π f so π f 75. A π F 7 µ F. (a) By inspection,, 9., so,, ω π rad s Also by inspection, ω π rad s, so f 4. π π rad. 5 A. 54 A (d) X, 9..5 A 96 (e) X f π ω, so ωx π rad s 96. F. µf 5

4 Alternating urrent ircuits and Electromagnetic Waves 6. X π f, and from ε t, we have ε ( t ). The units of self inductance are then t [ ] [ ε ][ ] olt sec. The units of inductive reactance are given by amp [ ] [ X ] [ f][ ] sec olt sec Amp olt Ohm Amp.4 (a) X π f π. 5. H. 6, 7. X A ( 6. 9 A ). 75 A.5 The required inductive reactance is X,, and the needed inductance is X, 4. π f π f π.. A. 75 H.6 Given: v. sin πt and. 5 H ω π rad s (a) By inspection, ω π rad s, so f π π 5. Also by inspection,,., so that,, X π f ω ( π rad s )(. 5 H ) 47. (d), 4. 9 X 47.. A (e) (. A ). 55 A (f) The phase difference between the voltage across an inductor and the current through the inductor is φ 9, so the average power delivered to the inductor is P, av, cos, φ cos( 9 ) (g) When a sinusoidal voltage with a peak value, is applied to an inductor, the current through the inductor also varies sinusoidally in time, with the same frequency as the applied voltage, and has a imum value of, X. However, the current lags behind the voltage in phase by a quarter-cycle or π radians. Thus, if the voltage is given by v, sin( ω t), the current as a function of time is i sin( ω t ) π. n the case of the given inductor, the current through it will be i ( 55. A) sin πt π. continued on next page

5 6 hapter (h) When i +. A, we have: sin πt π. A.55 A, or πt π sin (. A.55 A) sin (. 9). 4 rad π rad +. 4 rad and t 9. s 9. ms π rad s.7 From NΦ B (see Section.6 in the textbook), the total flux through the coil is ΦB,total NΦB, where Φ B is the flux through a single turn on the coil. Thus, ( Φ B, total ) X ( ) rm s. 45 T m π f π 6.. (a) The applied voltage is sin ωt. sin 5t, so we have that. and ω π f 5 rad s. The impedance for the circuit is v + ( ) + ( ) Z R + X X R π f π f R ω ω 4 5 rad s ( H) ( 5 rads )( 5 F) or Z (. ) Z A.9 X π f π F Z R + X X ( ) (a). Z.. 6 A R, R (. 6 A )( 5. )., X (. 6 A )( 66. ). 9 X X 66. (d) φ tan tan R so, the voltage lags behind the current by 5

6 Alternating urrent ircuits and Electromagnetic Waves 6. (a) X π f π 5. 4 H 6 X f π π ( 5. ) 44. F 79 Z R + X X + ( ) Z ( 5 A )( 776 ) 94. X X 6 79 φ tan tan R Thus, the current leads the voltage by (a) X π f π 4.5 H. 77 X f π π( 4 ). 5 F 65. Z R + X X ( ) + (.77 ) k 4 Z A φ tan X X R tan ( ) 9 5. (d) φ >, so the voltage leads the current. X π f π 6.. H 7. 7 X π f π 6.. F Z R + X X 65 + ( ) (a) R, R ( 75. A )( 5. ) (e), X ( 75. A )( 77. ) 4 X (. 75 A )( 65 ) 79 (d) Z (. 75 )( ) 64 A

7 64 hapter. X π f π 6..4 H 5 X π f π 6.. F Z R + X X 4 + ( ) R and Z R (a) Z + X X X X 7., Z Z Z 9 75 R 7 9. Z R + X R +. R, ZR ZR Z 9 75 R 6.4 X π f π 6. H. Z R + X X + ( )...6 (a) 7 Z.6. A R, R (. A ).., X (. A ).. When the instantaneous current is a imum i, the instantaneous voltage across the resistor is v R ir R R,.. The instantaneous voltage across an inductor is always 9 or a quarter cycle out of phase with the instantaneous current. Thus, when i, v. Kirchhoff s loop rule always applies to the instantaneous voltages around a closed path. Thus, for this series circuit, v v + source R v and at this instant when i we have v source R+.. (d) When the instantaneous current i is zero, the instantaneous voltage across the resistor is v R ir. Again, the instantaneous voltage across an inductor is a quarter cycle out of phase with the current. Thus, when i, we must have v,.. Then, applying Kirchhoff s loop rule to the instantaneous voltages around the series circuit at the instant when i gives v v + v +,.. source R

8 Alternating urrent ircuits and Electromagnetic Waves 65.5 X f π π( 6. ). F. Z R + X 5... R + ( ) and 5 Z R. secondary.76 5 A ( ) 5 Therefore, R b, b A.6 (a) X π f π 6.. F Z R + X R X (d). Z 7. A The phase angle in this R circuit is X X. 4 φ tan tan R Since φ <, the voltage lags behind the current by 55.. Adding an inductor will change the impedance and hence the current in the circuit..7 (a) X π f π H 7. 5 X f 6 π π( 5. ). F k Z R + X X + ( ) k (d). Z 5.. A ma (e) X X φ tan tan R 5 4. (f) R, R (. A )( 5 ). 7, X (. A )( 7. 5 )., X (. A ). 59 9

9 66 hapter. The voltage across the resistor is a imum when the current is a imum, and the imum value of the current occurs when the argument of the sine function is π. The voltage across the capacitor lags the current by 9 or π radians, which corresponds to an argument of in the sine function and a voltage of v. Similarly, the voltage across the inductor leads the current by 9 or π radians, corresponding to an argument in the sine function of π radians or, giving a voltage of v..9 X π f π 5..5 H 5. and X f π π( 5. ) 65. F Z R + X X ( ) ad Z ad Z ad A. 4. (a) Z ab R 4., so Z ab 59. A 4. 4 ab Zbc X 5., and Zbc. 59 A 5. 5 bc Zcd X 49., and Zcd. 59 A cd (d) Zbd X X 9., so Zbd. 59 A X f π π ( 6 ). 5 F Z R + X X bd. + ( )...6 (a) 7 Z.6. A R, R (. A ).., X (. A ).. When the instantaneous current i is zero, the instantaneous voltage across the resistor is v R ir. The instantaneous voltage across a capacitor is always 9 or a quarter cycle out of phase with the instantaneous current. Thus, when i, v,. and q (. 6 F )(. ). v 5 4 µ Kirchhoff s loop rule always applies to the instantaneous voltages around a closed path. Thus, for this series circuit, vsource + vr + v, and at this instant when i, we have v + v,.. source continued on next page

10 Alternating urrent ircuits and Electromagnetic Waves 67 (d) When the instantaneous current is a imum ( i ), the instantaneous voltage across the resistor is v R ir R R,.. Again, the instantaneous voltage across a capacitor is a quarter cycle out of phase with the current. Thus, when i, we must have v and q v. Then, applying Kirchhoff s loop rule to the instantaneous voltages around the series circuit at the instant when i and v gives v + v + v v v,. source R source R R 4. (a) Z.5 A P av R gives R P av. W.5 A 4. Z R + X, so X Z R. 4 4 and X 4 π f π H. Given v sin ( ωt) ( 9. ) sin( 5t), observe that 9. and ω 5 rad s. Also, the net reactance is X X π f π f ω ω. (a) X X ω ( 5 rad s )(. H ) ω ( 5 so the impedance is Z R X X rad s) 5. F + ( ) Z Z A The phase difference between the applied voltage and the current is X X 44. φ tan tan R so the average power delivered to the circuit is P av cosφ cos φ. 95 A 9. cos ( 4. 5 ) W. Please see the textbook for the statement of Problem. and the answers for that problem. There, you should find that 4,. 97 A, and φ 5.. The average power delivered to the circuit is then P av cosφ.4 The current in the circuit is 6 Z.. A cosφ and the average power delivered to the circuit is. 97 A 4 cos ( 5. ) 4. 6 W P av ( φ) cos R R, R. A.. W

11 6 hapter.5 (a) P av R R R,, so P 4 W. A, av R 5 Thus, R, 5 R.. A Z R + X, which yields X Z R R 9. A (. ) 7. and X 7. π f π 6 7. H.6 X π f π 6 6. H X f π π ( 6 ) 5 F Z R + X X + ( ) 5 (a) R, R R Z. 5 9, X X Z ( )., X X Z ( ). 9 No, R,,, The power delivered to the resistor is the greatest. No power losses occur in an ideal capacitor or inductor. P av R R Z 5. W.7 (a) The frequency of the station should match the resonance frequency of the tuning circuit. At resonance, X X or π f π f, which gives f 4π or f π π. H. 5 F ( ) M Yes, the resistance is not needed. The resonance frequency is found by simply equating the inductive reactance to the capacitive reactance, which leads to Equation (.9) as shown above.

12 Alternating urrent ircuits and Electromagnetic Waves 69. (a) The resonance frequency of an R circuit is f π. Thus, the inductance is.9 f 4 f π π.. F. 56 H 56 ph ( ) ( ) X f 9 π π H. For f f, so π π f k 5. min 5 4π 5.. H For f f ( ) 6 6 k. 6 min 6 4π. 6. H ( ).4 (a) At resonance, X X so the impedance will be + Z R + X X R R 5 When X X, we have π f π f, which yields f π π. H 75 F 5. F 5nF F 4. 9 nf 4 The current is a imum at resonance where the impedance has its minimum value of Z R. (d) At f 6, X π ( )( H ) and Z + ( ) 6. 75, X π 6 75 F , Thus, Z Z 5 ( 4 ) 5. A

13 7 hapter.4 ω π f. H F ( ) rad s Thus, ω ω rad s X ω rad s. H. X ω rad s F Z R + X X 5. + ( ) A Z. The average power is P av one period is 5. R 7. A. 77. W and the energy converted in E T π P P J π av av 77. J ω s rad s. 4.4 The resonance frequency is ω π f Also, X ω and X ω (a) At resonance, X X. H ω. F Thus, Z R + R, and P av At ω ω ; X X Z. R 4. A. 4W 4. 5, X X ω ( ) A ω Z R + X X + ( ). 5 5 and. A 5 so P av R. A. 9. W At ω ω 4 ; X X 4 continued on next page 5, X X ω 4 ( ) 4 Z 75, and. A 75 ω so P av R (. A ) (. ) 7. W. 7 mw

14 Alternating urrent ircuits and Electromagnetic Waves 7 (d) At ω ω ; X ( X ), X X ω so P av Z 5, and. A 5 R. A.. 9 W (e) At ω 4ω ; X 4 ( X ) 4, X X ω 4 Z 75, and. A 75 5 ω 5 ω so P av R (. A ) (. ) 7. W. 7 mw The power delivered to the circuit is a imum when the current is a imum. This occurs when the frequency of the source is equal to the resonance frequency of the circuit. is related to the imum input voltage ( ) by the.4 The imum output voltage N expression ( ) ( ) N, where N and N are the number of turns on the primary coil and the secondary coil respectively. Thus, for the given transformer, 5 ( ) ( 7 ). 5 and the voltage across the secondary is.44 (a) The output voltage of the transformer is N N,, Assuming an ideal transformer, P P, and the power delivered to the D player is output ( Pav ) ( Pav ) ( ) ( A ),,. 5 ( ). W input.45 The power input to the transformer is ( P av ) input (,), ( 6 )( 5 A). 5 W For an ideal transformer, ( Pav ) P ouput (, ), ( av ) so the current in the long-distance input power line is ( P ) 5 av input. W,. A (, ) The power dissipated as heat in the line is then P lost line A R,.. W The percentage of the power delivered by the generator that is lost in the line is P lost. % ost % P. input 5 W W %. %

15 7 hapter.46 (a) Since the transformer is to step the voltage down from volts to 6. volts, the secondary must have fewer turns than the primary. (, ), (, ), so the For an ideal transformer, ( Pav ) ( P input av ) or ouput current in the primary will be, ( ),, 6. 5 ma, 5 ma The ratio of the secondary to primary voltages is the same as the ratio of the number of turns on the secondary and primary coils, N N. Thus, the number of turns needed on the secondary coil of this step down transformer is N N 6 4. turns.47 (a) At 9% efficiency, ( P ) 9. ( P ) Thus, if ( P av ) kw output av output av input Pav output kw the input power to the primary is ( Pav ). kw input 9. 9.,, ( P ) av input.,, ( P ) av output 6 kw. W. 6 6 kw. W.,, A A ( ).4 R line m m 9 (a) The power transmitted is P av transmitted so P av transmitted 5. 6 W 5. Thus, P av loss Rline. A W 9. kw A The power input to the line is Pav P P input av transmitted av 5. loss W +.9 W 5. W and the fraction of input power lost during transmission is ( P ) av fraction ( P ) av loss input 4.9 W.5 77 or.577% 6 5. W continued on next page

16 Alternating urrent ircuits and Electromagnetic Waves 7 t is impossible to deliver the needed power with an input voltage of 4.5 k. The imum line current with an input voltage of 4.5 k occurs when the line is shorted out at the customer s end, and this current is 4 5 ( ) 5. 5 A R 9 line The imum input power is then ( P input ) ( )( ) 4 ( 45. ) ( 5. 5 A ) 6. 9 W 6. 9 kw This is far short of meeting the customer s request, and all of this power is lost in the transmission line..49 From v λ f, the wavelength is v. ms 6 λ 4. m 4 km f 75 The required length of the antenna is then λ 4 km, or about 6 miles. Not very practical at all..5 (a) t d c 644. m y. m s 7.56 s 6. y From Table.4 (in Appendix of the textbook), the average Earth-Sun distance is d. 496 m, giving the transit time as t d c. 496 m min. min. m s 6 s Also from Table.4, the average Earth-Moon distance is d 4. m, giving the time for the round trip as d. 4 m t 56. s c. m s.5 The amplitudes of the electric and magnetic components of an electromagnetic wave are related by the expression E B c; thus the amplitude of the magnetic field is B E m. 6 c. m s T.5 c ( ) µ 7 4π N s. 54 N m or c. 99 m s

17 74 hapter.5 (a) The frequency of an electromagnetic wave is f c λ, where c is the speed of light and λ is the wavelength of the wave. The frequencies of the two light sources are then Red: f red c. ms λ 66 m red and nfrared: f R c. ms 9. 9 λ 94 m R 4 The intensity of an electromagnetic wave is proportional to the square of its amplitude. f 67% of the incident intensity of the red light is absorbed, then the intensity of the emerging wave is % 67% %.. Hence, we must have of the incident intensity, or f i E E, f, i f. 57. i.54 f is the incident intensity of a light beam, and is the intensity of the beam after passing through length of a fluid having concentration of absorbing molecules, the Beer-ambert law states that log ( ) ε where ε is a constant. For 66-nm light, the absorbing molecules are oxygenated hemoglobin. Thus, if % of this wavelength light is transmitted through blood, the concentration of oxygenated hemoglobin in the blood is log. [] ε The absorbing molecules for 94-nm light are deoxygenated hemoglobin, so if 76% of this light is transmitted through the blood, the concentration of these molecules in the blood is HBO log. 76 [] ε Dividing equation [] by equation [] gives the ratio of deoxygenated hemoglobin to oxygenated hemoglobin in the blood as HB log ( 76. ). 5 or HB 5. HBO log. HB HBO Since all the hemoglobin in the blood is either oxygenated or deoxygenated, it is necessary that HB + HBO., and we now have 5. HBO + HBO.. The fraction of hemoglobin that is oxygenated in this blood is then. HBO. or % Someone with only % oxygenated hemoglobin in the blood is probably in serious trouble, needing supplemental oxygen immediately. EB.55 From ntensity µ Thus, B and E B c, we find ntensity 7 µ 4π T m A ntensity c. ms cb µ. ( ) 4 Wm 5 6 T and E B c. 5 T. m s. m

18 Alternating urrent ircuits and Electromagnetic Waves (a) To exert an upward force on the disk, the laser beam should be aimed vertically upward, striking the lower surface of the disk. To just levitate the disk, the upward force exerted on the disk by the beam should equal the weight of the disk. The momentum that electromagnetic radiation of intensity, incident normally on a perfectly reflecting surface of area A, delivers to that surface in time t is given by Equation (.) as p U c ( A t) c. Thus, from the impulse-momentum theorem, the average force exerted on the reflecting surface is F p t A c. Then, to just levitate the surface, F A c mgand the required intensity of the incident radiation is mgc A. mgc mgc A r π ( 5. kg )( 9. m s ). π 4. m ( ms) Wm Propulsion by light pressure in a significant gravity field is impractical because of the enormous power requirements. n addition, no material is perfectly reflecting, so the absorbed energy would melt the reflecting surface..57 The distance between adjacent antinodes in a standing wave is λ. Thus, λ 6. cm. cm. m, and 9 c λ f (. m ) m s.5 At Earth s location, the wave fronts of the solar radiation are spheres whose radius is the Sun- Pav Pav Earth distance. Thus, from ntensity, the total power is A 4πr W 4π 4 m 6 4π (. 49 m ) 74 W. P av ntensity r.59 From λ f c, we find f c..6 λ f 7. 6 ms c. ms λ 5.5 m. m 4.6 (a) For the AM band, λ min c. f 6 f min ms c. ms λ 54 m 556 m For the FM band, λ min c. f c. λ f min 6 6 ms ms 7. m 4. m

19 76 hapter.6 The transit time for the radio wave is t R dr m 4. s. ms c. m s and that for the sound wave is t s ds. m 7. s 7. ms 4 m s v sound Thus, the radio listeners hear the news.4 ms before the studio audience because radio waves travel so much faster than sound waves..6 f an object of mass m is attached to a spring of spring constant k, the natural frequency of vibration of that system is f k m π. Thus, the resonance frequency of the O double bond will be f k π m π oxygen atom Nm 66. kg 5. and the light with this frequency has wavelength c. λ f 5. ms 5. m 5. µm The infrared region of the electromagnetic spectrum ranges from λ mm down to λ min 7 nm.7 µ m. Thus, the required wavelength falls within the infrared region..64 Since the space station and the ship are moving toward one another, the frequency after being Doppler shifted is f f + u c, so the change in frequency is O S f f f f O S S u c 6. 4 ( ).. 5 ms 6. ms and the frequency observed on the spaceship is f f + f O S.65 Since you and the car ahead of you are moving away from each other (getting farther apart) at a rate of u km h km h 4 km h, the Doppler shifted frequency you will detect is f f u c, and the change in frequency is O S u f fo fs f S c 4. 4 ( ) 4 km h. 7 ms 7 6. ms km h. The frequency you will detect will be f f + f O S

20 Alternating urrent ircuits and Electromagnetic Waves The driver was driving toward the warning lights, so the correct form of the Doppler shift equation is f f + u c. The frequency emitted by the yellow warning light is O S f S c. λ 5 S 9 ms m and the frequency the driver claims that she observed is f O c. λ 56 O 9 ms m The speed with which she would have to approach the light for the Doppler effect to yield this claimed shift is f O u c f. S ( ms) ms.67 (a) At f 6., the reactance of a 5. -µ F capacitor is X π f π F and the impedance of this R circuit is 77 Z R X Z A 65 ma After addition of an inductor in series with the resistor and capacitor, the impedance of the + ( ) circuit is Z R + X X 5. π f 77. To reduce the current to one-half the value found above, the impedance of the circuit must be doubled to a value of Z ( 4 ) 6. Thus, ( 5. ) + ( π f 77 ) ( 6 ) or π f 77 ± ( 6 ) ( 5. ) 77 ± 65 Since the inductance cannot be negative, the potential solution associated with the lower sign must be discarded, leaving π H

21 7 hapter.6 Suppose you cover a.7 m-by-. m section of beach blanket. Suppose the elevation angle of the Sun is 6. Then the target area you fill in the Sun s field of view is 7. m. m cos 4. m. The intensity the radiation at Earth s surface is surface 6. incoming and only 5% of this is E t absorbed. Since P ( ) av, the absorbed energy is A A E 5. A t 5 6 surface.. incoming A t ( 5. )( 6. )( 4 Wm )( 4. m )( 6s) 6 5 J or ~ 6 J.69 Z R X R π f π ( ) 6 5. F 5. 7 Thus, P av R R Z 57. P av and cost E rate t rate.7 X ω, so ω X ( ) kw 4h. cents kwh. 7 cents W kw Then, X, which gives ω ( X ) ( X X) ( )(. ) or ( 96 ) [] From ω π f, we obtain ( f ) π Substituting from Equation [], this becomes ( 96 ) ( π f ) or 6. ( π f ) 96 π π 96 Then, from Equation [], 5 F 6 µ F.7 R 5 ( ) F. 5 H. 5 mh D D Z R π f..6 A A 4. Thus, Z R π f ( 4. ) ( 9. ) π( 6. ) H mh

22 Alternating urrent ircuits and Electromagnetic Waves 79.7 (a) c. ms The required frequency is f.. Therefore, the resonance λ. m frequency of the circuit is f., giving π ( ) π f π 4 H 6. F. 6 pf A d l, so d l ( ) d 6. F. m 5. N m 4. m 4. mm X X ( π f ) π. 4 H 5 ( ).7 (a) E B c, so B E. m 67. c. m s EB ntensity µ ( 6. m) π T m A 6 T ( T) 5. 7 Wm P av d ntensity A ntensity π 7 π. m ( 5. Wm) W.74 (a) Z. A 6. D R. A D 4. From Z R X R π f + +, we find Z R π f ( 6. ) ( 4. ) π( 6 ). H mh

23 hapter.75 (a) From Equation (.), the momentum imparted in time t to a perfectly reflecting sail of area A by normally incident radiation of intensity is p U c A t c. From the impulse-momentum theorem, the average force exerted on the sail is then a av F av 4 ( m ) p t A t c A 4 Wm 6. t c. m s F. 56 N 9. 5 ms m 6 kg av From x t+ at t v, with v, the time is ( x). 4 m 5 a 9. ms av ( 6 s) N d 9 d 4.64 s..76 (a) The intensity of radiation at distance r from a point source, which radiates total power P, is P A P 4π r. Thus, at distance r. infrom a cell phone radiating a total power of P. W. mw, the intensity is. mw 4π. in. 54 cm in 6. mw cm This intensity is 4% higher than the imum allowed leakage from a microwave at this distance of. inches. f when using a Blue toothheadset (emitting.5 mw of power) in the ear at distance r h. in 5. cm from center of the brain, the cell phone (emitting. W of power) is located in the pocket at distance r p. m. cm from the brain, the total radiation intensity at the brain is. mw total phone + headset 4π. cm 5. mw + 4π 5. cm mw mw cm cm mw mw or total 6. cm cm 4. mw cm. 4 mw cm

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

Oscillations and Electromagnetic Waves March 30, 2014 Chapter 31 1 Three Polarizers! Consider the case of unpolarized light with intensity I 0 incident on three polarizers! The first polarizer has a polarizing