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Transcript
21 Alternating Current Circuits and Electromagnetic Waves Clicker Questions Question O1.01 Description: Relating object color to the wave nature of light. Question A particular object appears to be red. The reason it appears red is: 1. 2. 3. 4. 5. 6. It absorbs only red frequencies as they hit. It reflects only red frequencies as they hit. It absorbs red frequencies more than other frequencies. It reflects red frequencies more than other frequencies. It emits red frequencies more than other frequencies. Not enough information. Commentary Purpose: To hone the concept of “color” as a property of an object. Discussion: If an object appears red when you look at it, the light emanating from it and reaching your eye must be predominantly red in frequency. Physically, answers (2), (4), and (5) are all possible. Case (5) would describe a glowing object such as a neon sign or red LED. Most common objects, however, do not emit light, but rather reflect some of the light striking them. A real red object will reflect predominantly red frequencies of light, but will not perfectly absorb all other frequencies. Thus, (4) is a better answer than (2). If you interpret “a particular object” as a typical non-luminous object, (4) would be the best answer. However, (6) — not enough information — is defensible, since you were not told whether the object is luminous, which would make (5) accurate. Key Points: • The apparent color of an object is determined by the mix of frequencies in the light traveling from it to the observer; the light from a red object will be dominated by light in the red portion of the spectrum. • An object may emit light if it is luminous (glowing), but most objects merely reflect some of the light incident upon them. For Instructors Only Questions that may support productive discussion: “What will you see if you look at a red object through a piece of red-tinted glass? How about through blue-tinted glass? If you shine red light on it? Or blue light? How does an RGB monitor display millions of different colors?” 252 56157_21_ch21_p252-280.indd 252 3/19/08 5:49:33 PM Alternating Current Circuits and Electromagnetic Waves 253 Question O1.02 Description: Linking optics to real-world experience of color. Question The Sun appears to be yellow when overhead because: 1. 2. 3. 4. 5. 6. The human eye is most sensitive to yellow. Yellow frequencies are transmitted better through the atmosphere. Air does not absorb yellow frequencies. The temperature of the Sun is 6000 K. More than one of the above. None of the above. Commentary Purpose: To probe your understanding of the colors of glowing objects, and to connect physics processes to your everyday experience. Discussion: Hot matter emits photons of many frequencies, but more of some frequencies than others. The higher the temperature of the matter, the more high-frequency photons are emitted. If the temperature high enough, the emitted light will be visible (have enough photons in the visible part of the spectrum), the color will be determined by the mix of photon frequencies. The higher the temperature, the closer to the blue end of the spectrum the radiated light will appear. At a temperature of about 6000 K, the mix of radiated photon frequencies is such that the object will appear to glow yellow. This accounts for the sun’s color when overhead. Other stars, having different surface temperatures, have different colors: there are red stars, blue stars, etc. Key Points: • Objects glow when hot enough. The mix of photon frequencies emitted, and thus the color, depends on the temperature of the object. • The sun’s surface temperature is approximately 6000 K, resulting in its yellow color. For Instructors Only This question is most effectively used before covering the relevant material on light and black-body radiation, to elicit student’s preconceptions and motivate learning of the material. Question O1.03 Description: Linking optics to real-world experience of color. Question The sky appears to be blue during the day because: 1. 2. 56157_21_ch21_p252-280.indd 253 Air absorbs blue light less than other frequencies (i.e., acts like a blue filter). Air molecules emit blue light after being struck by sunlight. 3/19/08 5:49:34 PM 254 Chapter 21 3. 4. 5. The sky reflects blue light from the oceans. The temperature high in the Earth’s upper atmosphere is 1000 K. None of the above. Commentary Purpose: To connect physics processes to your everyday experience. Discussion: When light travels through the atmosphere, some of the photons collide with air molecules and change directions (a process called “scattering”), and most travels through unscattered. Some photons are more likely to be scattered than others, however; photons toward the blue end of the visible spectrum are far, far more likely to scatter than those near the red end. Thus, when sunlight travels through the atmosphere, much of the blue light scatters off in various directions. An observer looking at the sun sees photons of all colors traveling directly without scattering; an observer looking elsewhere in the sky sees photons that have reached her after scattering off an air molecule, and these are predominantly blue. Thus, the proper answer to “Why is the sky blue?” is that “Blue photons are much more likely to scatter from air molecules than photons of other colors.” This is not on the list, so “none of the above” (5) is the best choice. Key Points: • When light travels through air, some of the photons collide with air molecules and scatter off in different directions. • The frequency (color) of a photon strongly affects how likely it is to scatter off of an air molecule. The closer to the blue end of the visible spectrum, the more likely scattering is. • The sky’s color is due to photons that reach the observer after scattering from an air molecule. For Instructors Only This also is a good question to ask before students should “know” the right answer, to elicit preconceptions, stimulate discussion, and motivate subsequent learning. Depending on your students’ mathematical sophistication, they may appreciate knowing that the scattering probability of a photon from an air molecule goes as the fourth power of the frequency. This is not because blue molecules are “bigger,” but rather because of the quantum-mechanical nature of the scattering interaction. Discussion of the sky’s color near sunrise and sunset has been omitted here, as that subject is taken up in Question O1.04 (which makes a good follow-up to this one). We do want students to take “none of the above” answers seriously. Question O1.04 Description: Linking optics to real-world experience of color. Question The Sun appears to be red as it sets because: 1. 2. 56157_21_ch21_p252-280.indd 254 Air absorbs red light less than other frequencies (i.e., acts like a red filter). The sunlight has a red-shift when you’re moving fastest away from it. 3/19/08 5:49:34 PM Alternating Current Circuits and Electromagnetic Waves 3. 4. 5. 6. 255 The Sun cools down to 5000 K each evening. Light is refracted as it enters the atmosphere. The Sun dies in a glorious fireball each evening and is reborn each morning. None of the above. Commentary Purpose: To connect physics processes to your everyday experience. Discussion: When light travels through the atmosphere, some of the photons collide with air molecules and change directions (a process called “scattering”), and most travels through unscattered. Some photons are more likely to be scattered than others, however; photons toward the blue end of the visible spectrum are far, far more likely to scatter than those near the red end. Thus, when sunlight travels through the atmosphere, much of the blue light scatters off in various directions. At sunset, light from the sun skims along the Earth’s surface before reaching an observer, traveling through much more air than it does at midday. This means each photon has many more chances to scatter off in another direction, and by the time the sunlight has reached the observer, enough of the blue light has been scattered away that the remaining light looks reddish. This explanation does not appear among the listed answers, so “none of the above” is the best choice. Light does refract upon entering the atmosphere, but only slightly, about half a degree. Although refraction will cause separation of colors because of frequency dependence of the index of refraction, refraction is not the cause of the red light at sunset. At midday, more blue light scatters away than for other colors, but enough remains that the sun’s color is not significantly altered from its original yellowish-white color. Key Points: • When light travels through air, some of the photons collide with air molecules and scatter off in different directions. • The frequency (color) of a photon strongly affects how likely it is to scatter off of an air molecule. The closer to the blue end of the visible spectrum, the more likely scattering is. • As blue photons are disproportionately scattered out of the sun’s light, the remaining light is increasingly reddish in appearance. For Instructors Only This question makes a good follow-up to Question O1.03: it requires the same principles as discussed there, but asks students to reason further with them. This checks how well they grasp those principles (as opposed to just memorizing the answer) and helps them solidify the ideas. QUICK QUIZZES 1. 56157_21_ch21_p252-280.indd 255 (c). The average power is proportional to the rms current, which is nonzero even though the average current is zero. (a) is only valid for an open circuit, for which R → ∞. (b) and (d) can never be true because iav = 0 for AC currents. 3/19/08 5:49:35 PM 256 Chapter 21 2. (b). Choices (a) and (c) are incorrect because the unaligned sine curves in Figure 21.9 mean the voltages are out of phase, and so we cannot simply add the maximum (or rms) voltages across the elements. (In other words, ∆V ≠ ∆VR + ∆VL + ∆VC even though ∆v = ∆v R + ∆v L + ∆vC.) 3. (b). Closing switch A replaces a single resistor with a parallel combination of two resistors. Since the equivalent resistance of a parallel combination is always less than the lowest resistance in the combination, the total resistance of the circuit decreases, which causes the 2 impedance Z = Rtotal + ( X L − XC ) to decrease. 2 4. (a). Closing switch A replaces a single resistor with a parallel combination of two resistors. Since the equivalent resistance of a parallel combination is always less than the lowest resistance in the combination, the total resistance of the circuit decreases, which causes the phase angle, φ = tan −1 ⎡⎣( X L − XC ) R ⎤⎦ , to increase. 5. (a). Closing switch B replaces a single capacitor with a parallel combination of two capacitors. Since the equivalent capacitance of a parallel combination is greater than that of either of the individual capacitors, the total capacitance increases, which causes the capacitive reactance XC = 1 2π fC to decrease. Thus, the net reactance, X L − XC , increases causing the phase angle, φ = tan −1 ⎡⎣( X L − XC ) R ⎤⎦ , to increase. 6. (b). Inserting an iron core in the inductor increases both the self-inductance and the inductive reactance, X L = 2π fL. This means the net reactance, X L − XC, and hence the impedance, 2 Z = Rtotal + ( X L − XC ) , increases, causing the current (and therefore, the bulb’s brightness) to decrease. 2 7. (b), (c). Since pressure is force per unit area, changing the size of the area without altering the intensity of the radiation striking that area will not cause a change in radiation pressure. In (b), the smaller disk absorbs less radiation, resulting in a smaller force. For the same reason, the momentum in (c) is reduced. 8. (b), (d). The frequency and wavelength of light waves are related by the equation λ f = v or f = v λ , where the speed of light v is a constant within a given medium. Thus, the frequency and wavelength are inversely proportional to each other, when one increases the other must decrease. ANSWERS TO MULTIPLE CHOICE QUESTIONS ∆Vmax 170 V = = 120 V , so (c) is the correct choice. 2 2 1. ∆Vrms = 2. When the frequency doubles, the rms current I rms = ∆VL ,rms X L = ∆VL ,rms 2π fL is cut in half. Thus, the new current is I rms = 3.0 A 2 = 1.5 A and (e) is the correct answer. 3. ∆VC ,rms = I rms XC = 4. ∆VL ,rms = I rms X L = I rms ( 2π fL ) = ( 2.0 A ) 2π ( 60.0 Hz ) ⎡⎣(1.0 2π ) H ⎤⎦ = 120 V, and the correct answer is choice (c). 5. 56157_21_ch21_p252-280.indd 256 I rms 1.00 × 10 −3 A = 16.7 V, so choice (a) is correct. = 2π fC 2π ( 60.0 Hz ) ⎡⎣(1.00 2π ) × 10 −6 F ⎤⎦ At the resonance frequency, X L = XC and the impedance is Z = R 2 + ( X L − XC ) = R . Thus, the rms current is I rms = ∆Vrms Z = (120 V) ( 20 Ω ) = 6.0 A, and (b) is the correct choice. 2 3/19/08 5:49:35 PM Alternating Current Circuits and Electromagnetic Waves 257 6. The battery produces a constant current in the primary coil, which generates a constant flux through the secondary coil. With no change in flux through the secondary coil, there is no induced voltage across the secondary coil, and choice (e) is the correct answer. 7. 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 = ( 3.0 × 10 8 m s (1.5 × 10 −7 T = 45 N C which is choice (d). 8. When a power source, AC or DC, is first connected to a RL combination, the presence of the inductor impedes the buildup of a current in the circuit. The value of the current starts at zero and increases as the back emf induced across the inductor decreases somewhat in magnitude. Thus, the correct choice is (c). 9. The voltage across the capacitor is proportional to the stored charge. This charge, and hence the voltage ∆vC , is a maximum when the current has zero value and is in the process of reversing direction after having been flowing in one direction for a half cycle. Thus, the voltage across the capacitor lags behind the current by 90°, and (a) is the correct choice. 10. In an RLC circuit, the instantaneous voltages ∆v R , ∆v L , and ∆vC (across the resistance, inductance, and capacitance, respectively) are not in phase with each other. The instantaneous voltage ∆v R is in phase with the current, ∆v L leads the current by 90°, while ∆vC lags behind the current by 90°. The instantaneous values of these three voltages do add algebraically to give the instantaneous voltages across the RLC combination, but the rms voltages across these components do not add algebraically. The rms voltages across the three components must be added as vectors (or phasors) to obtain the correct rms voltage across the RLC combination. Among the listed choices, choice (e) is the false statement. 11. In an AC circuit, both an inductor and a capacitor store energy for one-half of the cycle of the current, and return that energy to the circuit during the other half of the cycle. On the other hand, a resistor converts electrical potential energy into thermal energy during all parts of the cycle of the current. Thus, only the resistor has a nonzero average power loss. The power delivered to a circuit by a generator is given by Pav = I rms ∆Vrms cos φ , where φ is the phase difference between the generator voltage and the current. Among the listed choices, the only true statement is choice (c). 12. If the voltage is a maximum when the current is zero, the voltage is either leading or lagging the current by 90° (or a quarter cycle) in phase. Thus, the element could be either an inductor or a capacitor. It could not be a resistor since the voltage across a resistor is always in phase with the current. If the current and voltage were out of phase by 180°, one would be a maximum in one direction when the other was a maximum value in the opposite direction. The correct choice for this question is (d). 13. At resonance, X L = XC , and the phase difference between the current and the applied voltage is φ = tan −1 ⎡⎣( X L − XC ) R ⎤⎦ = tan −1 ( 0 ) = 0° The correct answer is choice (c). 56157_21_ch21_p252-280.indd 257 3/19/08 5:49:36 PM 258 14. Chapter 21 At a frequency of f = 5.0 × 10 2 Hz, the inductive reactance, capacitive reactance, and impedance are X L = 2π fL, XC = 1 2 , and Z = R 2 + ( X L − XC ) . This yields 2π f C 2 Z= ( 20.0 Ω ) and I rms = 2 ⎡ ⎤ 1 + ⎢ 2π ( 5.0 × 10 2 Hz ( 0.120 H ) − ⎥ = 51 Ω, 2π ( 5.0 × 10 2 Hz ( 0.75 × 10 −6 F ⎥⎦ ⎢⎣ ) ) ) ∆Vrms 120 V = = 2.3 A . Choice (a) is the correct answer. Z 51 Ω ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2. At resonance, X L = XC. This means that the impedance Z = R 2 + ( X L − XC ) reduces to Z = R. 2 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. 8. 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. 10. 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. 12. The changing magnetic field of the solenoid induces eddy currents in the conducting core. This is accompanied by I 2 R conversion of electrically transmitted energy into internal energy in the conductor. 14. 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 maximum values. Do not forget that an inductor can induce an emf in itself and that the voltage across it is 90° ahead of the current in the circuit in phase. PROBLEM SOLUTIONS 21.1 56157_21_ch21_p252-280.indd 258 (a) ∆VR,rms = I rms R = ( 8.00 A ) (12.0 Ω ) = 96.0 V (b) ∆VR,max = 2 ( ∆VR,rms ) = 2 ( 96.0 V) = 136 V (c) I max = 2 I rms = 2 ( 8.00 A ) = 11.3 A (d) 2 2 Pav = I rms R = ( 8.00 A ) (12.0 Ω ) = 768 W 3/19/08 5:49:37 PM Alternating Current Circuits and Electromagnetic Waves 21.2 259 ) (a) ∆VR,max = 2 ( ∆VR,rms ) = 2 (1.20 × 10 2 V = 1.70 × 10 2 V (b) 2 Pav = I rms R= (c) Because R = 2 (1.20 × 102 V ∆Vrms ∆V 2 ⇒ R = rms = Pav 60.0 W R 2 ∆Vrms Pav ) 2 = 2.40 × 10 2 Ω (see above), if the bulbs are designed to operate at the same voltage, the 1.00 × 10 2 W will have the lower resistance . 21.3 The meters measure the rms values of potential difference and current. These are ∆Vrms = 21.4 ∆Vrms 70.7 V ∆Vmax 100 V = = 2 .95 A = = 70.7 V , and I rms = 24.0 Ω R 2 2 All lamps are connected in parallel with the voltage source, so ∆Vrms = 120 V for each lamp. Also, the current is I rms = Pav ∆Vrms and the resistance is R = ∆Vrms I rms . I1, rms = I 2, rms = I 3, rms = 21.5 150 W 120 V = 1.25 A and R1 = R2 = = 96.0 Ω 120 V 1.25 A 100 W 120 V = 0.833 A and R3 = = 144 Ω 120 V 0.833 A The total resistance (series connection) is Req = R1 + R2 = 8.20 Ω + 10.4 Ω = 18.6 Ω, so the current in the circuit is I rms = ∆Vrms 15.0 V = = 0.806 A 18.6 Ω Req The power to the speaker is then 21.6 The general form of the generator voltage is ∆v = ( ∆Vmax ) sin (ω t ) , so by inspection (a) ∆VR,max = 170 V (c) ∆VR,rms = (d) I rms = (e) I max = 2 I rms = 2 ( 6.00 A ) = 8.49 A (f) 2 2 Pav = I rms R = ( 6.00 A ) ( 20.0 Ω ) = 720 W (g) ∆VR,max 2 and = (b) f = ω 60π rad s = = 30.0 Hz 2π 2π rad 170 V = 120 V 2 ∆VR,rms 120 V = = 6.00 A 20.0 Ω R The argument of the sine function has units of [ω t ] = ( rad s ) ( s ) = radians . At t = 0.005 0 s, the instantaneous current is i= 56157_21_ch21_p252-280.indd 259 2 2 Pav = I rms Rspeaker = ( 0.806 A ) (10.4 Ω ) = 6.76 W (170 V) ∆v (170 V) sin ⎡⎣( 60π rad s ) ( 0.005 0 s ) ⎤⎦ = = sin ( 0.94 rad ) = 6.9 A 20.0 Ω R 20.0 Ω 3/19/08 5:49:38 PM 260 Chapter 21 21.7 XC = 1 , so its units are 2π fC 1 1 Volt Volt = = Ohm = = (1 Sec ) Farad (1 Sec ) (Coulomb Volt ) Coulomb Sec Amp 21.8 21.9 I max = 2 I rms = ) (a) I max = 2 (120 V) 2π ( 60.0 Hz ) ( 2.2 0 × 10 − 6 C/V = 0.141 A = 141 mA (b) I max = 2 ( 240 V) 2π ( 50.0 Hz ) ( 2.2 0 × 10 − 6 C/V = 0.235 A = 235 mA I rms = ∆Vrms = 2π f C ( ∆Vrms ) , so XC ) f = 21.10 56157_21_ch21_p252-280.indd 260 ) XC = 1 1 = = 221 Ω 2π fC 2π ( 60.0 Hz ) (12.0 × 10 −6 F (b) I rms = ∆VC ,rms 36.0 V = = 0.163 A 221 Ω XC (c) I max = 2 I rms = 2 ( 0.163 A ) = 0.231 A I rms = so 21.12 0.30 A I rms = = 4.0 × 10 2 Hz 2π C ( ∆Vrms ) 2π ( 4.0 × 10 −6 F ( 30 V) (a) (d) 21.11 2 ( ∆Vrms ) = 2 ( ∆Vrms ) 2π f C XC ) No. The current reaches its maximum value one-quarter cycle before the voltage reaches its maximum value. From the definition of capacitance, the capacitor reaches its maximum charge when the voltage across it is also a maximum. Consequently, the maximum charge and the maximum current do not occur at the same time. ∆Vrms ⎛ ∆V ⎞ =2π f C ⎜ max ⎟ = π f C ( ∆Vmax ) 2 ⎝ 2 ⎠ XC C= 0.75 A I = = 1.7 × 10 −5 F = 17 µ F π f ( ∆Vmax ) 2 π ( 60 Hz ) (170 V) 2 ∆VC ,max 98.0 V = 69.3 V 2 (a) By inspection, ∆VC,max = 98.0 V, so ∆VC ,rms = (b) Also by inspection, ω = 80π rad s, so f = (c) I rms = I max 0.500 A = = 0.354 A 2 2 (d) XC = ∆VC ,max 98.0 V = = 196 Ω 0.500 A I max (e) XC = 1 1 1 1 = = 2.03 × 10 −5 F = 20.3 µ F = , so C = 2π fC ω C ω XC ( 80π rad s ) (196 Ω ) 2 = ω 80π rad s = = 40.0 Hz 2π 2π rad 3/19/08 5:49:39 PM Alternating Current Circuits and Electromagnetic Waves 21.13 X L = 2π f L , and from ε ⎛ ∆I ⎞ = L ⎜ ⎟ , we have L = ⎝ ∆t ⎠ ε ( ∆t ) ∆I 261 . The units of self inductance are then ε ∆t Volt ⋅ sec . The units of inductive reactance are given by [ L ] = [ ][ ] = amp [ ∆I ] 1 ⎛ Volt ⋅ sec ⎞ Voltt = = Ohm Amp ⎟⎠ Amp [ X L ] = [ f ][ L ] = ⎛⎜⎝ sec ⎞⎟⎠ ⎜⎝ 21.14 21.15 ) (a) X L = 2π f L = 2π ( 80.0 Hz ) ( 25.0 × 10 −3 H = 12.6 Ω (b) I rms = (c) I max = 2 I rms = 2 ( 6.19 A ) = 8.75 A The required inductive reactance is X L = L= 21.16 ∆VL ,rms 78.0 V = = 6.19 A 12.6 Ω XL XL = 2π f ∆VL ,max 2 ( 2π f ) I rms ( ≥ ∆VL ,max ∆VL ,max and the needed inductance is = I max 2 I rms 4.00 V = 0.750 H 2 ( 2π ) ( 300.0 Hz ) ( 2.00 × 10 −3 A ) ) Given: v L = 1.20 × 10 2 V sin ( 30π t ) and L = 0.500 H ω 30π rad s = = 15.0 Hz 2π 2π (a) By inspection, ω = 30π rad s, so f = (b) Also by inspection, ∆VL,max = 1.20 × 10 2 V, so that ∆VL ,rms = ∆VL ,max 2 = 1.20 × 10 2 V = 84.9 V 2 (c) X L = 2π fL = ω L = ( 30π rad s ) ( 0.500 H ) = 47.1 Ω (d) I rms = (e) I max = 2 I rms = 2 (1.80 A ) = 2.55 A (f) ∆VL ,rms 84.9 V = = 1.80 A 47.1 Ω XL The phase difference between the voltage across an inductor and the current through the inductor is φ L = 90° , so the average power delivered to the inductor is PL ,av = I rms ∆VL ,rms cos φL = I rms ∆VL ,rms cos ( 90°) = 0 (g) When a sinusoidal voltage with a peak value ∆VL,max 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 maximum value of I max = ∆VL ,max X L. However, the current lags behind the voltage in phase by a quarter-cycle or π 2 radians. Thus, if the voltage is given by ∆v L = ∆VL ,max sin (ω t ) , the current as a function of time is i = I max sin (ω t − π 2 ). In the case of the given inductor, the current through it will be i = ( 2.55 A ) sin ( 30π t − π 2 ) . continued on next page 56157_21_ch21_p252-280.indd 261 3/19/08 5:49:41 PM 262 Chapter 21 (h) When i = +1.00 A, we have: sin ( 30π t − π 2 ) = (1.00 A 2.55 A ) , or 30π t − π 2 = sin −1 (1.00 A 2.55 A ) = sin −1 ( 0.392 ) = 0.403 rad and 21.17 t= π 2 rad + 0.403 rad = 2.09 × 10 −2 s = 20.9 ms 30π rad s From L = N Φ B I (see Section 20.6 in the textbook), the total flux through the coil is Φ B, total = N Φ B = L ⋅ I , where Φ B is the flux through a single turn on the coil. Thus, (Φ ) B , total max ⎡ ∆V ⎤ = L ⋅ I max = L ⋅ ⎢ max ⎥ ⎣ XL ⎦ =L 21.18 (a) 2 ( ∆Vrms ) 2 (120 V) = = 0.450 T ⋅ m 2 2π fL 2π ( 60.0 Hz ) The applied voltage is ∆v = ∆Vmax sin (ω t ) = ( 80.0 V) sin (150t ), so we have that ∆Vmax = 80.0 V and ω = 2π f = 150 rad s. The impedance for the circuit is Z = R 2 + ( X L − XC ) = R 2 + ( 2π fL − 1 2π fC ) = R 2 + (ω L − 1 ω C ) 2 2 2 2 ⎡ ⎤ 1 or Z = ( 40.0 Ω ) + ⎢(150 rad s ) 80.0 × 10 −3 H − ⎥ = 57.5 Ω −6 (150 rad s ) 125 × 10 F ⎥⎦ ⎢⎣ ( 2 21.19 ) ( ) ∆Vmax 80.0 V = = 1.39 A 57.5 Ω Z (b) I max = XC = 1 1 = = 66.3 Ω 2π f C 2π ( 60.0 Hz ) ( 40.0 × 10 −6 F ) Z = R 2 + ( X L − XC ) = 2 ( 50.0 Ω )2 + ( 0 − 66.3 Ω )2 = 83.0 Ω ∆Vrms 30.0 V = = 0.361 A Z 83.0 Ω (a) I rms = (b) ∆VR, rms = I rms R = ( 0.361 A ) ( 50.0 Ω ) = 18.1 V (c) ∆VC , rms = I rms XC = ( 0.361 A ) ( 66.3 Ω ) = 23.9 V (d) ⎛ X − XC ⎞ −1 ⎛ 0 − 66.3 Ω ⎞ φ = tan −1 ⎜ L ⎟ = tan ⎜⎝ ⎟ = −53.0° ⎝ R ⎠ 50.0 Ω ⎠ so, the voltage lags behind the current by 53° 56157_21_ch21_p252-280.indd 262 3/19/08 5:49:42 PM Alternating Current Circuits and Electromagnetic Waves 21.20 (a) 263 ) X L = 2π f L = 2π ( 50.0 Hz )( 400 × 10 −3 H = 126 Ω XC = 1 1 = = 719 Ω 2π f C 2π ( 50.0 Hz )( 4.43 × 10 −6 F ) Z = R 2 + ( X L − XC ) = 2 ( 500 Ω )2 + (126 Ω − 719 Ω )2 = 776 Ω ∆Vmax = I max Z = ( 0.250 A ) ( 776 Ω ) = 194 V (b) ⎛ 126 Ω − 719 Ω ⎞ ⎛ X − XC ⎞ = tan −1 ⎜ φ = tan −1 ⎜ L ⎟ ⎟⎠ = − 49.9° ⎝ ⎝ R ⎠ 500 Ω Thus, the current leads the voltage by 49.9° . 21.21 (a) X L = 2π f L = 2π ( 240 Hz )( 2 .50 H ) = 3.77 × 10 3 Ω XC = 1 1 = = 2 .65 × 10 3 Ω 2π f C 2π ( 240 Hz )( 0.250 × 10 −6 F ) Z = R 2 + ( X L − XC ) = 2 21.22 2 = 1.44 kΩ ∆Vmax 140 V = = 0.097 2 A Z 1.44 × 10 3 Ω (b) I max = (c) ⎡ ( 3.77 − 2 .65 ) × 10 3 Ω ⎤ ⎛ X − XC ⎞ φ = tan −1 ⎜ L = tan −1 ⎢ ⎥ = 51.2° ⎟ ⎝ R ⎠ 900 Ω ⎣ ⎦ (d) φ > 0, so the voltage leads the current X L = 2π fL = 2π ( 60.0 Hz ) ( 0.100 H ) = 37.7 Ω XC = 1 1 = = 265 Ω 2π fC 2π ( 60.0 Hz ) (10.0 × 10 −6 F ) Z = R 2 + ( X L − XC ) = 2 56157_21_ch21_p252-280.indd 263 ( 900 Ω )2 + ⎡⎣( 3.77 − 2 .65 ) × 10 3 Ω ⎤⎦ ( 50.0 Ω )2 + ( 37.7 Ω − 265 Ω )2 (a) ∆VR,rms = I rms R = ( 2.75 A ) ( 50.0 Ω ) = 138 V (b) ∆VL ,rms = I rms X L = ( 2.75 A ) ( 37.7 Ω ) = 104 V (c) ∆VC = I rms XC = ( 2.75 A ) ( 265 Ω ) = 729 V (d) ∆Vrms = I rms Z = ( 2.75 A ) ( 233 Ω ) = 641 Ω = 233 Ω (e) 3/19/08 5:49:43 PM 264 Chapter 21 X L = 2π f L = 2π ( 60.0 Hz )( 0.400 H ) = 151 Ω 21.23 XC = 1 1 = = 884 Ω 2π f C 2π ( 60.0 Hz )( 3.00 × 10 −6 F ) Z RLC = R 2 + ( X L − XC ) = 2 ( 60.0 Ω )2 + (151 Ω − 884 Ω )2 = 735 Ω ∆Vrms Z RLC and I rms = (a) Z LC = 0 + ( X L − XC ) = X L − XC = 733 Ω 2 ⎛ ∆V ⎞ ⎛ 90.0 V ⎞ ∆VLC , rms = I rms ⋅ Z LC = ⎜ rms ⎟ Z LC = ⎜ ( 733 Ω ) = 89.8 V ⎝ 735 Ω ⎟⎠ ⎝ Z RLC ⎠ (b) Z RC = R 2 + ( 0 − XC ) = 2 ( 60.0 Ω )2 + ( 884 Ω )2 = 886 Ω ⎛ ∆V ⎞ ⎛ 90.0 V ⎞ ∆VRC , rms = I rms ⋅ Z RC = ⎜ rms ⎟ Z RC = ⎜ ( 886 Ω ) = 108 V ⎝ 735 Ω ⎟⎠ ⎝ Z RLC ⎠ 21.24 X L = 2π fL = 2π ( 60 Hz ) ( 2.8 H ) = 1.1 × 10 3 Ω Z = R 2 + ( X L − XC ) = 2 (1.2 × 10 Ω ) + (1.1 × 10 3 2 3 Ω−0 ) 2 = 1.6 × 10 3 Ω ∆Vmax 170 V = = 0.11 A Z 1.6 × 10 3 Ω (a) I max = (b) ∆VR, max = I max R = ( 0.11 A ) (1.2 × 10 3 Ω = 1.3 × 10 2 V ) ) ∆VL , max = I max X L = ( 0.11 A ) (1.1 × 10 3 Ω = 1.2 × 10 2 V (c) When the instantaneous current is a maximum ( i = I max ), the instantaneous voltage across the resistor is ∆v R = iR = I max R = ∆VR, max = 1.3 × 10 V . The instantaneous voltage across an inductor is always 90° or a quarter cycle out of phase with the instantaneous current. Thus, when i = I max , ∆v L = 0 . 2 Kirchhoff’s loop rule always applies to the instantaneous voltages around a closed path. Thus, for this series circuit, ∆vsource = ∆v R + ∆v L and at this instant when i = I max we have ∆vsource = I max R + 0 = 1.3 × 10 2 V . (d) When the instantaneous current i is zero, the instantaneous voltage across the resistor is ∆v R = iR = 0 . Again, the instantaneous voltage across an inductor is a quarter cycle out of phase with the current. Thus, when i = 0, we must have ∆v L = ∆VL , max = 1.2 × 10 2 V . Then, applying Kirchhoff’s loop rule to the instantaneous voltages around the series circuit at the instant when i = 0 gives ∆vsource = ∆v R + ∆v L = 0 + ∆VL , max = 1.2 × 10 2 V . 56157_21_ch21_p252-280.indd 264 3/19/08 6:25:32 PM Alternating Current Circuits and Electromagnetic Waves 1 1 = = 1.33 × 10 8 Ω 2π f C 2π ( 60.0 Hz )( 20.0 × 10 −12 F XC = 21.25 ) ( 50.0 × 10 Ω ) + (1.33 × 10 Z RC = R 2 + XC2 = and 265 ) ( ∆V secondary rms I rms = Z RC 3 = 2 8 Ω ) 2 = 1.33 × 10 8 Ω 5 000 V = 3..76 × 10 −5 A 1.33 × 10 8 Ω ) ) Therefore, ∆Vb, rms = I rms Rb = ( 3.76 × 10 −5 A ( 50.0 × 10 3 Ω = 1.888 V 21.26 1 1 = = 88.4 Ω 2π f C 2π ( 60.0 Hz )( 30.0 × 10 −6 F (a) XC = (b) Z = R 2 + ( 0 − XC ) = R 2 + XC2 = (c) I max = (d) The phase angle in this RC circuit is ) 2 ( 60.0 Ω )2 + ( 88.4 Ω )2 = 107 Ω ∆Vmax 1.20 × 10 2 V = = 1.12 A 107 Ω Z ⎛ X − XC ⎞ −1 ⎛ 0 − 88.4 Ω ⎞ φ = tan −1 ⎜ L ⎟ = tan ⎜⎝ ⎟ = −55.8° ⎝ R ⎠ 60.0 Ω ⎠ Since φ < 0, the voltage lags behind the current by 55.8°° . Adding an inductor will change the impedance and hence the current in the circuit. 21.27 (a) X L = 2π f L = 2π ( 50.0 Hz )( 0.250 H ) = 78.5 Ω (b) XC = (c) Z = R 2 + ( X L − XC ) = (d) I max = (e) ⎛ X − XC ⎞ ⎛ 78.5 Ω − 1590 Ω ⎞ φ = tan −1 ⎜ L = tan −1 ⎜ ⎟ ⎟⎠ = −84.3° ⎝ ⎝ R ⎠ 150 Ω (f) ∆VR,max = I max R = ( 0.138 A ) (150 Ω ) = 20.7 V 1 1 = = 1.59 × 10 3 Ω = 1.59 kΩ 2π f C 2π ( 50.0 Hz )( 2.00 × 10 − 6 F ) 2 (150 Ω )2 + ( 78.5 Ω − 1590 Ω )2 = 1.52 × 10 3 Ω = 1.52 kΩ ∆Vmax 2.10 × 10 2 V = = 0.138 A = 138 mA 1.52 × 10 3 Ω Z ∆VL ,max = I max X L = ( 0.138 A ) ( 78.5 Ω ) = 10.8 V ) ∆VC ,max = I max XC = ( 0.138 A ) (1.59 × 10 3 Ω = 219 V 56157_21_ch21_p252-280.indd 265 3/19/08 5:49:45 PM 266 21.28 Chapter 21 The voltage across the resistor is a maximum when the current is a maximum, and the maximum value of the current occurs when the argument of the sine function is π 2. The voltage across the capacitor lags the current by 90° or π 2 radians, which corresponds to an argument of 0 in the sine function and a voltage of vC = 0 . Similarly, the voltage across the inductor leads the current by 90° or π 2 radians, corresponding to an argument in the sine function of π radians or 180°, giving a voltage of v L = 0 . X L = 2π f L = 2π ( 50.0 Hz )( 0.185 H ) = 58.1 Ω 21.29 XC = 1 1 = = 49.0 Ω 2π f C 2π ( 50.0 Hz )( 65.0 × 10 −6 F ) Z ad = R 2 + ( X L − XC ) = 2 21.30 ( 40.0 Ω )2 + ( 58.1 Ω − 49.0 Ω )2 = 41.0 Ω ) ( ∆Vmax 2 ∆Vrms 150 V = = 2.59 A = Z ad Z ad ( 41.0 Ω ) 2 and I rms = (a) Z ab = R = 40.0 Ω , so ( ∆Vrms )ab = I rms Z ab = ( 2.59 A ) ( 40.0 Ω ) = 104 V (b) Z bc = X L = 58.1 Ω, and ( ∆Vrms )bc = I rms Z bc = ( 2.59 A ) ( 58.1 Ω ) = 150 V (c) Z cd = XC = 49.0 Ω, and ( ∆Vrms )cd = I rms Z cd = ( 2.59 A ) ( 49.0 Ω ) = 127 V (d) Z bd = X L − XC = 9.10 Ω, so ( ∆Vrms )bd = I rms Z bd = ( 2.59 A ) ( 9.10 Ω ) = 23.6 V XC = 1 1 = = 1.1 × 10 3 Ω 2π fC 2π ( 60 Hz ) ( 2.5 × 10 −6 F Z = R 2 + ( X L − XC ) = 2 (1.2 × 10 3 ) Ω ) + ( 0 − 1.1 × 10 Ω ) 2 3 2 = 1.6 × 10 3 Ω ∆Vmax 170 V = = 0.11 A Z 1.6 × 10 3 Ω (a) I max = (b) ∆VR, max = I max R = ( 0.11 A ) (1.2 × 10 3 Ω = 1.3 × 10 2 V ) ) ∆VC , max = I max XC = ( 0.11 A ) (1.1 × 10 3 Ω = 1.2 × 10 2 V (c) When the instantaneous current i is zero, the instantaneous voltage across the resistor is ∆v R = iR = 0 . The instantaneous voltage across a capacitor is always 90° or a quarter cycle out of phase with the instantaneous current. Thus, when i = 0 , ∆vC = ∆VC , max = 1.2 × 10 2 V and ) ) qC = C ( ∆vC ) = ( 2.5 × 10 −6 F (1.2 × 10 2 V = 3.0 × 10 −4 C = 300 µC Kirchhoff’s loop rule always applies to the instantaneous voltages around a closed path. Thus, for this series circuit, ∆vsource + ∆v R + ∆vC = 0, and at this instant when i = 0, we have ∆vsource = 0 + ∆vC = ∆VC , max = 1.2 × 10 2 V . continued on next page 56157_21_ch21_p252-280.indd 266 3/19/08 5:49:46 PM Alternating Current Circuits and Electromagnetic Waves (d) 267 When the instantaneous current is a maximum (i = I max ), the instantaneous voltage across the resistor is ∆v R = iR = I max R = ∆VR, max = 1.3 × 10 2 V . Again, the instantaneous voltage across a capacitor is a quarter cycle out of phase with the current. Thus, when i = I max , we must have ∆vC = 0 and qC = C ∆vC = 0 . Then, applying Kirchhoff’s loop rule to the instantaneous voltages around the series circuit at the instant when i = I max and ∆vC = 0 gives ∆vsource + ∆v R + ∆vC = 0 21.31 ∆vsource = ∆v R = ∆VR, maax = 1.3 × 10 2 V 104 V ∆Vrms = = 208 Ω 0.500 A I rms (a) Z= (b) 2 Pav = I rms R gives R = (c) Z = R 2 + X L2 , so X L = Z 2 − R 2 = L= and 21.32 ⇒ Pav I 2 rms = 10.0 W = 40.0 Ω ( 0.500 A )2 ( 208 Ω )2 − ( 40.0 Ω )2 = 204 Ω XL 204 Ω = = 0.541 H 2π f 2π ( 60.0 Hz ) Given v = ∆Vmax sin (ω t ) = ( 90.0 V) sin ( 350t ) , observe that ∆Vmax = 90.0 V and ω = 350 rad s. Also, the net reactance is X L − XC = 2π fL − 1 2π fC = ω L − 1 ω C . (a) X L − XC = ω L − 1 1 = ( 350 rad s ) ( 0.200 H ) − = − 44.3 Ω ωC ( 350 rad s ) ( 25.0 × 10 −6 F ) so the impedance is ∆Vrms ∆Vmax = Z Z Z = R 2 ( X L − XC ) = 2 2 = ( 50.0 Ω )2 + ( − 44.3 Ω )2 = 66.8 Ω 90.0 V = 0.953 A 2 ( 66.8 Ω ) (b) I rms = (c) The phase difference between the applied voltage and the current is ⎛ X − XC ⎞ −1 ⎛ − 44.3 Ω ⎞ φ = tan −1 ⎜ L ⎟ = tan ⎜⎝ ⎟ = − 41.5° ⎝ R ⎠ 50.0 Ω ⎠ so the average power delivered to the circuit is ∆Vmax ⎞ ⎛ 90.0 V ⎞ cosφ = ( 0.953 A ) ⎜ ⎟ cos ( − 41.5°) = 45.4 W ⎟ ⎝ ⎝ 2 ⎠ 2 ⎠ Pav = I rms ∆Vrms cosφ = I rms ⎛⎜ 21.33 Please see the textbook for the statement of Problem 21.21 and the answers for that problem. There, you should find that ∆Vmax = 140 V, I max = 0.097 2 A, and φ = 51.2°. The average power delivered to the circuit is then ( 0.097 2 A ) (140 V) cos ( 51.2°) = 4.26 W I max ⎞ ⎛ ∆Vmax ⎞ cos φ = ⎟ ⎜ ⎟ ⎝ 2 ⎠⎝ 2 ⎠ 2 Pav = I rms ∆Vrms cos φ = ⎛⎜ 21.34 The rms current in the circuit is I rms = ∆Vrms 160 V = = 2.00 A Z 80.0 Ω and the average power delivered to the circuit is Pav = I rms ( ∆Vrms cos φ ) = I rms ∆VR,rms = I rms ( I rms R ) = I r2ms R = ( 2.00 A )2 ( 22.0 Ω ) = 88.0 W 56157_21_ch21_p252-280.indd 267 3/19/08 5:49:47 PM 268 21.35 Chapter 21 (a) 2 Pav = I rms R = I rms ( I rms R ) = I rms ( ∆VR, rms ) , so I rms = Thus, (b) R= ∆VR, rms I rms Pav ∆VR, rms = 14 W = 0.28 A 50 V 50 V = 1.8 × 10 2 Ω 0.28 A = Z = R 2 + X L2 , which yields 2 ⎛ ∆V ⎞ ⎛ 90 V ⎞ − (1.8 × 10 2 Ω X L = Z 2 − R 2 = ⎜ rms ⎟ − R 2 = ⎜ ⎝ 0.28 A ⎟⎠ ⎝ I rms ⎠ and 21.36 L= 2 ) 2 = 2.7 × 10 2 Ω 2.7 × 10 2 Ω XL = = 0.72 H 2π f 2π ( 60 Hz ) ) X L = 2π fL = 2π ( 600 Hz ) ( 6.0 × 10 −3 H = 23 Ω XC = 1 1 = = 11 Ω 2π fC 2π ( 600 Hz ) ( 25 × 10 −6 F ) Z = R 2 + ( X L − XC ) = 2 (a) ( 25 Ω )2 + ( 23 Ω − 11 Ω )2 = 28 Ω ⎛ 10 V ⎞ ⎛ ∆V ⎞ ∆VR, rms = I rms R = ⎜ rms ⎟ R = ⎜ ( 255 Ω ) = 8.9 V ⎝ 28 Ω ⎟⎠ ⎝ Z ⎠ ⎛ 10 V ⎞ ⎛ ∆V ⎞ ∆VL , rms = I rms X L = ⎜ rms ⎟ X L = ⎜ ( 23 Ω ) = 8.2 V ⎝ 28 Ω ⎟⎠ ⎝ Z ⎠ ⎛ 10 V ⎞ ⎛ ∆V ⎞ ∆VC , rms = I rms XC = ⎜ rms ⎟ XC = ⎜ (11 Ω ) = 3.9 V ⎝ 28 Ω ⎟⎠ ⎝ Z ⎠ No , ∆VR, rms + ∆VL , rms + ∆VC , rms = 9.0 V + 8.2 V + 3..8 V = 21 V ≠ 10 V (b) (c) 21.37 (a) The power delivered to the resistor is the greatest. No power losses occur in an ideal capacitor or inductor. 56157_21_ch21_p252-280.indd 268 2 ( 25 Ω ) = 3.2 W The frequency of the station should match the resonance frequency of the tuning circuit. At resonance, X L = XC or 2π f0 L = 1 2π f0 C , which gives f02 = 1 4π 2 LC or f0 = (b) 2 ∆Vrms ⎞ ⎛ 10 V ⎞ R=⎜ ⎝ 28 Ω ⎟⎠ ⎝ Z ⎟⎠ ⎛ 2 R=⎜ Pav = I rms 1 2π LC = 1 2π ( 3.00 × 10 −6 ) H ( 2.50 × 10 −12 F ) = 5.81 × 10 7 Hz = 58.1 MHz 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 (21.19) as shown above. 3/19/08 5:49:48 PM Alternating Current Circuits and Electromagnetic Waves 21.38 (a) The resonance frequency of an RLC circuit is f0 = 1 2π LC . Thus, the inductance is L= (b) For 1 1 = 4π 2 f02 C 4π 2 ( 9.00 × 10 9 Hz 1 2π LC (a) F ) = 1.56 × 10 −10 H = 156 pH ) , so C = 1 4π 2 f02 L f0 = ( f0 )min = 500 kHz = 5.00 × 10 5 Hz 1 4π 2 ( 5.00 × 10 5 Hz ) ( 2 .0 × 10 2 −6 H = 5.1 × 10 −8 F = 51 nF ) f0 = ( f0 )max = 1600 kHz = 1.60 × 10 6 Hz C = Cmin = 21.40 −12 ) C = Cmax = For ) ( 2.00 × 10 2 X L = 2π f0 L = 2π ( 9.00 × 10 9 Hz (1.56 × 10 −10 H = 8.82 Ω f0 = 21.39 269 1 4π (1.60 × 10 Hz 2 6 ) 2 ( 2 .0 × 10−6 H ) = 4.9 × 10 −9 F = 4.9 nF At resonance, X L = XC so the impedance will be Z = R 2 + ( X L − XC ) = R 2 + 0 = R = 15 Ω 2 (b) When X L = XC, we have 2π fL = f = (c) (d) 1 2π LC 1 2π ( 0.20 H ) ( 75 × 10 −6 F ) = 41 Hz The current is a maximum at resonance where the impedance has its minimum value of Z = R. At f = 60 Hz, X L = 2π ( 60 Hz ) ( 0.20 H ) = 75 Ω , XC = and Z = (15 Ω ) Thus, I rms = 56157_21_ch21_p252-280.indd 269 = 1 , which yields 2π fC 2 1 = 35 Ω, 2π ( 60 Hz ) ( 75 × 10 −6 F ) + ( 75 Ω − 35 Ω ) = 43 Ω 2 ( ∆Vmax ∆Vrms = Z Z 2 )= 150 V = 2.5 A 2 ( 43 Ω ) 3/19/08 5:49:49 PM 270 21.41 Chapter 21 ω 0 = 2π f0 = 1 = LC 1 (10.0 × 10 ) H (100 × 10 −6 F −3 = 1 000 rad s ) Thus, ω = 2ω 0 = 2 000 rad s ) X L = ω L = ( 2 000 rad s ) (10.0 × 10 −3 H = 20.0 Ω XC = 1 1 = = 5.00 Ω ω C ( 2 000 rad s ) (100 × 10 −6 F ) Z = R 2 + ( X L − XC ) = (10.0 Ω )2 + ( 20.0 Ω − 5.00 Ω )2 2 I rms = = 18.0 Ω ∆Vrms 50.0 V = = 2.78 A 18.0 Ω Z The average power is one period is 2 2 Pav = I rms R = ( 2.78 A ) (10.0 Ω ) = 77.3 W and the energy converted in ⎞ J⎞ ⎛ 2π ⎛ 2π ⎞ ⎛ = 0.243 J E = Pav ⋅ T = Pav ⋅ ⎜ ⎟ = ⎜ 77.3 ⎟ ⋅ ⎜ ⎝ω⎠ ⎝ s ⎠ ⎝ 2 000 rad s ⎟⎠ 21.42 The resonance frequency is ω 0 = 2π f0 = Also, X L = ω L and XC = (a) 1 ωC ⎛ 1 ⎞ L= At resonance, XC = X L = ω 0 L = ⎜ ⎝ LC ⎟⎠ Thus, Z = R 2 + 0 2 = R, I rms = and (b) At ω = ) ( and I rms = ∆Vrms 120 V = = 4.00 A Z 30.0 Ω ( ) ( 30.0 Ω )2 + ( 500 Ω − 2 000 Ω )2 = 1 500 Ω 120 V = 0.080 0 A 1 500 Ω 2 Pav = I rms R = ( 0.080 0 A ) ( 30.0 Ω ) = 0.192 W 2 At ω = ( ) ( ) 1 1 ω0 ; XL = X L ω = 250 Ω, XC = 4 XC ω = 4 000 Ω 0 0 4 4 Z = 3 750 Ω, and I rms = so 3.00 H = 1 000 Ω 3.00 × 10 −6 F 1 1 ω 0; X L = X L ω = 500 Ω , XC = 2 XC ω = 2 000 Ω 0 0 2 2 2 so L = C 2 2 Pav = I rms R = ( 4.00 A ) ( 30.0 Ω ) = 480 W Z = R 2 + ( X L − XC ) = (c) 1 LC 120 V = 0.032 0 A 3 750 Ω 2 Pav = I rms R = ( 0.032 0 A ) ( 30.0 Ω ) = 3.07 × 10 −2 W = 30.7 mW 2 continued on next page 56157_21_ch21_p252-280.indd 270 3/19/08 5:49:50 PM Alternating Current Circuits and Electromagnetic Waves (d) ( At ω = 2 ω 0; X L = 2 X L ω 0 ) = 2 000 Ω , X (e) ) ( 1 XC ω = 500 Ω 0 2 2 Pav = I rms R = ( 0.080 0 A ) ( 30.0 Ω ) = 0.192 W 2 ( At ω = 4 ω 0; X L = 4 X L ω 0 ) = 4 000 Ω, X Z = 3 750 Ω, and I rms = so = 120 V = 0.080 0 A 1 500 Ω Z = 1 500 Ω, and I rms = so C 271 C = ) ( 1 XC ω = 250 Ω 0 4 120 V = 0.032 0 A 3 750 Ω 2 Pav = I rms R = ( 0.032 0 A ) ( 30.0 Ω ) = 3.07 × 10 −2 W = 30.7 mW 2 The power delivered to the circuit is a maximum when the rms current is a maximum. This occurs when the frequency of the source is equal to the resonance frequency of the circuit. 21.43 The maximum output voltage ( ∆Vmax )2 is related to the maximum input voltage ( ∆Vmax )1 by the N expression ( ∆Vmax )2 = 2 ( ∆Vmax )1 , where N1 and N 2 are the number of turns on the primary N1 coil and the secondary coil respectively. Thus, for the given transformer, ( ∆Vmax )2 = 1 500 (170 V) = 1.02 × 10 3 V 250 and the rms voltage across the secondary is ( ∆Vrms )2 = 21.44 (a) ( ∆Vmax )2 2 = 1.02 × 10 3 V = 721 V . 2 The output voltage of the transformer is ⎛N ⎞ ⎛ 1⎞ ∆V2,rms = ⎜ 2 ⎟ ∆V1,rms = ⎜ ⎟ (120 V) = 9.23 V ⎝ 13 ⎠ ⎝ N1 ⎠ (b) Assuming an ideal transformer, Poutput = Pinput , and the power delivered to the CD player is (Pav )2 = (Pav )1 = I1,rms ( ∆V1,rms ) = ( 0.250 A ) (120 V) = 21.45 30.0 W The power input to the transformer is (Pav )input = ( ∆V1, rms ) I1, rms = ( 3 600 V) ( 50 A ) = 1..8 × 10 5 For an ideal transformer, power line is I 2, rms = (Pav )input ( ∆V ) 2 , rms = W (Pav )ouput = ( ∆V2, rms ) I 2, rms = (Pav )input so the current in the long-distance 1.8 × 10 5 W = 1.8 A 100 000 V The power dissipated as heat in the line is then Plost = I 22, rms Rline = (1.8 A )2 (100 Ω ) = 3.2 × 10 2 W The percentage of the power delivered by the generator that is lost in the line is % Lost = 56157_21_ch21_p252-280.indd 271 ⎛ 3.2 × 10 2 W ⎞ Plost × 100% = ⎜ × 100% = 0.18% Pinput ⎝ 1.8 × 10 5 W ⎟⎠ 3/19/08 5:49:51 PM 272 21.46 Chapter 21 (a) (b) Since the transformer is to step the voltage down from 120 volts to 6.0 volts, the secondary must have fewer turns than the primary. I1, rms = (c) ) ( ( ) For an ideal transformer, (Pav )input = (Pav )ouput or ∆V1, rms I1, rms = ∆V2, rms I 2, rms so the current in the primary will be ( ∆V ) I 2 , rms 2 , rms ∆V1, rms = ( 6.0 V) ( 500 mA ) 120 V = 25 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, ∆V2 ∆V1 = N 2 N1. Thus, the number of turns needed on the secondary coil of this step down transformer is ⎛ ∆V ⎞ ⎛ 6.0 V ⎞ = 20 turns N 2 = N1 ⎜ 2 ⎟ = ( 400 ) ⎜ ⎝ 120 V ⎟⎠ ⎝ ∆V1 ⎠ 21.47 (a) (Pav )output = 0.90 (Pav )input At 90% efficiency, Thus, if (Pav )output = 1 000 kW the input power to the primary is 21.48 (b) I1, rms = (c) I 2, rms = (Pav )input ∆V1, rms = (Pav )output (Pav )output 0.90 = 1 000 kW = 1.1 × 10 3 kW 0.90 1.1 × 10 3 kW 1.1 × 10 6 W = 3.1 × 10 2 A = ∆V1, rms 3 600 V = ∆V2, rms (Pav )input = 1 000 kW 1.0 × 10 6 W = 8.3 × 10 3 A = ∆V1, rms 120 V ) ) Rline = ( 4.50 × 10 −4 Ω m ( 6.44 × 10 5 m = 290 Ω (a) The power transmitted is so I rms = Thus, (b) (Pav )transmitted ∆Vrms = (Pav )transmitted = ( ∆Vrms ) I rms 5.00 × 10 6 W = 10.0 A 500 × 10 3 V 2 2 Rline = (10.0 A ) ( 290 Ω ) = 2 .90 × 10 4 (Pav )loss = I rms W = 29.0 kW The power input to the line is (Pav )input = (Pav )transmitted + (Pav )loss = 5.00 × 10 6 W + 2.90 × 10 4 W = 5.03 × 10 6 W and the fraction of input power lost during transmission is fraction = (Pav )loss ( Pav )input = 2.90 × 10 4 W = 0.005 77 or 0.577% 5.033 × 10 6 W continued on next page 56157_21_ch21_p252-280.indd 272 3/19/08 5:49:52 PM Alternating Current Circuits and Electromagnetic Waves (c) 273 It is impossible to deliver the needed power with an input voltage of 4.50 kV. The maximum line current with an input voltage of 4.50 kV occurs when the line is shorted out at the customer’s end, and this current is ( I rms )max = ∆Vrms 4 500 V = = 15.5 A 290 Ω Rline The maximum input power is then (P ) input max = ( ∆Vrms ) ( I rms )max ) = ( 4.50 × 10 3 V (15.5 A ) = 6.98 × 10 4 W = 6.98 kW This is far short of meeting the customer’s request, and all of this power is lost in the transmission line. 21.49 From v = λ f , the wavelength is λ= v 3.00 × 10 8 m s = = 4.00 × 10 6 m = 4 000 km f 75 Hz The required length of the antenna is then L = λ 4 = 1 000 km , or about 621 miles. Not very practical at all. 21.50 d 6.44 × 1018 m ⎛ 1y ⎞ 2 = ⎜ ⎟ = 6.80 × 10 y c 3.00 × 10 8 m s ⎝ 3.156 × 10 7 s ⎠ (a) t= (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) Also from Table C.4, the average Earth-Moon distance is d = 3.84 × 10 8 m, giving the time for the round trip as t= 21.51 56157_21_ch21_p252-280.indd 273 ) 8 2 d 2 ( 3.84 × 10 m = = 2.56 s c 3.00 × 10 8 m s The amplitudes of the electric and magnetic components of an electromagnetic wave are related by the expression Emax Bmax = c; thus the amplitude of the magnetic field is Bmax = 21.52 d 1.496 × 1011 m ⎛ 1 min ⎞ = ⎟ = 8.31 min ⎜ c 3.00 × 10 8 m s ⎝ 60 s ⎠ 330 V m Emax = = 1.10 × 10 −6 T 3.00 × 10 8 m s c c= 1 = µ0 ∈0 1 or c = 2 .998 × 10 8 m s ( 4π × 10 −7 N ⋅s C 2 2 ) ( 8.854 × 10 −12 C2 N ⋅ m 2 ) 3/19/08 5:49:52 PM 274 21.53 Chapter 21 (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 fred = Red: c 3.00 × 10 8 m s = = 4.55 × 1014 Hz λred 660 × 10 −9 m and fIR = Infrared: (b) The intensity of an electromagnetic wave is proportional to the square of its amplitude. If 67% of the incident intensity of the red light is absorbed, then the intensity of the emerging wave is (100% − 67%) = 33% of the incident intensity, or I f = 0.33I i. Hence, we must have Emax, f Emax, i 21.54 c 3.00 × 10 8 m s = = 3.19 × 1014 Hz λIR 940 × 10 −9 m = If Ii = 0.33 = 0.57 If I 0 is the incident intensity of a light beam, and I is the intensity of the beam after passing through length L of a fluid having concentration C of absorbing molecules, the Beer-Lambert law states that log10 ( I I 0 ) = −ε CL where ε is a constant. For 660-nm light, the absorbing molecules are oxygenated hemoglobin. Thus, if 33% of this wavelength light is transmitted through blood, the concentration of oxygenated hemoglobin in the blood is CHBO2 = − log10 ( 0.33) εL [1] The absorbing molecules for 940-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 CHB = − log10 ( 0.76 ) εL [2] Dividing equation [2] by equation [1] gives the ratio of deoxygenated hemoglobin to oxygenated hemoglobin in the blood as log10 ( 0.76 ) CHB = = 0.25 CHBO2 log10 ( 0.33) or CHB = 0.25CHBO2 Since all the hemoglobin in the blood is either oxygenated or deoxygenated, it is necessary that CHB + CHBO2 = 1.00 , and we now have 0.25CHBO2 + CHBO2 = 1.0 . The fraction of hemoglobin that is oxygenated in this blood is then CHBO2 = 1.0 = 0.80 1.0 + 0.25 or 80% Someone with only 80% oxygenated hemoglobin in the blood is probably in serious trouble, needing supplemental oxygen immediately. 21.55 From Intensity = 2 Emax Bmax c Bmax E and max = c, we find Intensity = 2 µ0 Bmax 2 µ0 Thus, Bmax = and 56157_21_ch21_p252-280.indd 274 2 µ0 ( Intensity ) = c 2 ( 4π × 10 −7 T ⋅ m A 3.00 × 10 m s 8 ) ) (1 340 ) W m 2 = 3.35 × 10 −6 T ) Emax = Bmax c = ( 3.35 × 10 −6 T ( 3.00 × 10 8 m s = 1.01 × 10 3 V m 3/19/08 5:49:53 PM Alternating Current Circuits and Electromagnetic Waves 21.56 (a) 275 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 I, incident normally on a perfectly reflecting surface of area A, delivers to that surface in time ∆t is given by Equation (21.30) as ∆p = 2U c = 2 ( IA∆t ) c . Thus, from the impulse-momentum theorem, the average force exerted on the reflecting surface is F = ∆p ∆t = 2 IA c . Then, to just levitate the surface, F = 2 IA c = mg and the required intensity of the incident radiation is I = mgc 2 A . (b) (c) 21.57 ( )( ( )( ) 5.00 × 10 −3 kg 9.80 m s 2 3.00 × 10 8 m s mgc mgc = = = 1.46 × 10 9 W m 2 2 −2 2 A 2π r 2 2π 4.00 × 10 m Propulsion by light pressure in a significant gravity field is impractical because of the enormous power requirements. In addition, no material is perfectly reflecting, so the absorbed energy would melt the reflecting surface. I= ) The distance between adjacent antinodes in a standing wave is λ 2. Thus, λ = 2 ( 6.00 cm ) = 12 .0 cm = 0.120 m, and ) c = λ f = ( 0.120 m ) ( 2 .45 × 10 9 Hz = 2 .94 × 10 8 m s 21.58 At Earth’s location, the wave fronts of the solar radiation are spheres whose radius is the SunEarth distance. Thus, from Intensity = Pav = ( Intensity ) ( 4π r 2 ) = ⎛⎜ 1 340 ⎝ 21.59 From λ f = c , we find f = 21.60 λ= 21.61 (a) = Pav 4π r 2 , the total power is 2 W⎞⎡ 4π (1.49 × 1011 m ⎤ = 3.74 × 10 26 W 2⎟ ⎣ ⎦ m ⎠ ) c 3.00 × 10 8 m s = = 5.45 × 1014 Hz λ 5.50 × 10 −7 m For the AM band, λmax = c fmax c fmin = 3.00 × 10 8 m s = 188 m 1 600 × 10 3 Hz = 3.00 × 10 8 m s = 556 m 540 × 10 3 Hz = 3.00 × 10 8 m s = 2 .78 m 108 × 10 6 Hz = 3.00 × 10 8 m s = 3.4 m 88 × 10 6 Hz For the FM band, λmin = λmax = 56157_21_ch21_p252-280.indd 275 A c 3.00 × 10 8 m s = = 11.0 m f 27.33 × 10 6 Hz λmin = (b) Pav c fmax c fmin 3/19/08 5:49:54 PM 276 21.62 Chapter 21 The transit time for the radio wave is tR = dR 100 × 10 3 m = = 3.33 × 10 −4 s = 0.333 ms c 3.00 × 10 8 m s and that for the sound wave is ts = ds vsound = 3.0 m = 8.7 × 10 −3 s = 8.7 ms 343 m s Thus, the radio listeners hear the news 8.4 ms beffore the studio audience because radio waves travel so much faster than sound waves. 21.63 If 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 2π . Thus, the resonance frequency of the C=O double bond will be f = 1 2π k moxygen = 1 2π 2 800 N m = 5.2 × 1013 Hz 2.66 × 10 −26 kg atom and the light with this frequency has wavelength λ= c 3.00 × 10 8 m s = = 5.8 × 10 −6 m = 5.8 µ m f 5.2 × 1013 Hz The infrared region of the electromagnetic spectrum ranges from λmax ≈ 1 mm down to λmin = 700 nm = 0.7 µ m. Thus, the required wavelength falls within the infrareed region . 21.64 Since the space station and the ship are moving toward one another, the frequency after being Doppler shifted is fO = fS (1 + u c ), so the change in frequency is ⎛ 1.8 × 10 5 m s ⎞ ⎛ u⎞ ∆f = fO − fS = fS ⎜ ⎟ = ( 6.0 × 1014 Hz ⎜ = 3.6 × 1011 Hz ⎝ c⎠ ⎝ 3.0 × 10 8 m s ⎟⎠ ) and the frequency observed on the spaceship is fO = fS + ∆f = 6.0 × 1014 Hz + 3.6 × 1011 Hz = 6.003 6 × 1014 Hz 21.65 Since you and the car ahead of you are moving away from each other (getting farther apart) at a rate of u = 120 km h − 80 km h = 40 km h, the Doppler shifted frequency you will detect is fO = fS (1 − u c ), and the change in frequency is ⎛ 40 km h ⎞ ⎛ 0.278 m s ⎞ ⎛ u⎞ = − 1.6 × 10 7 Hz ∆f = fO − fS = − fS ⎜ ⎟ = − ( 4.3 × 1014 Hz ⎜ ⎝ c⎠ ⎝ 3.0 × 10 8 m s ⎟⎠ ⎜⎝ 1 km h ⎟⎠ ) The frequency you will detect will be fO = fS + ∆f = 4.3 × 1014 Hz − 1.6 × 10 7 Hz = 4.299 999 84 × 1014 Hz 56157_21_ch21_p252-280.indd 276 3/19/08 5:49:54 PM Alternating Current Circuits and Electromagnetic Waves 21.66 277 The driver was driving toward the warning lights, so the correct form of the Doppler shift equation is fO = fS (1 + u c ). The frequency emitted by the yellow warning light is fS = 3.00 × 10 8 m s c = = 5.17 × 1014 Hz λS 580 × 10 −9 m and the frequency the driver claims that she observed is fO = 3.00 × 10 8 m s c = = 5.36 × 1014 Hz λO 560 × 10 −9 m The speed with which she would have to approach the light for the Doppler effect to yield this claimed shift is ⎛f ⎛ 5.36 × 1014 Hz ⎞ ⎞ − 1 = 1.1 × 10 7 m s u = c ⎜ O − 1⎟ = ( 3.00 × 10 8 m s ⎜ ⎝ 5.17 × 1014 Hz ⎟⎠ ⎠ ⎝ fS ) 21.67 (a) At f = 60.0 Hz, the reactance of a 15.0-µ F capacitor is XC = 1 1 = = 177 Ω 2π fC 2π ( 60.0 Hz ) (15.0 × 10 −6 F ) and the impedance of this RC circuit is Z = R 2 + XC2 = ( 50.0 Ω )2 + (177 Ω )2 = 184 Ω ∆Vrms 1.20 × 10 2 V = = 0.652 A = 652 mA 184 Ω Z (b) I rms = (c) After addition of an inductor in series with the resistor and capacitor, the impedance of the circuit is Z ′ = R 2 + ( X L − XC ) = ( 50.0 Ω ) + ( 2π fL − 177 Ω ) . To reduce the current to one-half the value found above, the impedance of the circuit must be doubled to a value of Z ′ = 2 (184 Ω ) = 368 Ω . Thus, 2 2 2 ( 50.0 Ω )2 + ( 2π fL − 177 Ω )2 = ( 368 Ω )2 or 2π fL = 177 Ω ± ( 368 Ω )2 − ( 50.0 Ω )2 = 177 Ω ± 365 Ω Since the inductance cannot be negative, the potential solution associated with the lower sign must be discarded, leaving L= 56157_21_ch21_p252-280.indd 277 177 Ω + 365 Ω = 1.44 H 2π ( 60.0 Hz ) 3/19/08 5:49:55 PM 278 21.68 Chapter 21 Suppose you cover a 1.7 m-by-0.3 m section of beach 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 ) cos 30° = 0.4 m 2. The intensity the radiation at Earth’s surface is I surface = 0.6 I incoming and only 50% of this is ( ∆E ∆t ) , the absorbed energy is P absorbed. Since I = av = A A ) ( ∆E = ( 0.5 I surface ) A ( ∆t ) = ⎡⎣ 0.5 0.6 I incoming ⎤⎦ A ( ∆t ) ) ) = ( 0.5 ) ( 0.6 ) (1 340 W m 2 ( 0.4 m 2 ( 3 600 s ) = 6 × 10 5 J or Z = R 2 + ( XC ) = R 2 + ( 2π f C ) 2 21.69 −2 ( 200 Ω )2 + ⎡⎣ 2π ( 60 Hz ) ( 5.00 × 10 −6 = 2 ⎛ 120 V ⎞ ⎛ ∆V ⎞ 2 R = ⎜ rms ⎟ R = ⎜ Thus, Pav = I rms ⎝ 5.7 × 10 2 Ω ⎟⎠ ⎝ Z ⎠ and ~ 10 6 J ) F ⎤⎦ 2 −2 = 5.7 × 10 2 Ω ( 200 Ω ) = 8.9 W = 8.9 × 10 −3 kW cost = ∆E ⋅ ( rate) = Pav ⋅ ∆t ⋅ ( rate ) ) = ( 8.9 × 10 −3 kW ( 24 h ) ( 8.0 cents kWh ) = 1.7 cents X L = ω L , so ω = X L L 21.70 Then, XC = 1 1 = , which gives ω C C ( XL L ) ) L = ( X L ⋅ XC ) C = ⎡⎣(12 Ω ) ( 8.0 Ω ) ⎤⎦ C or L = ( 96 Ω 2 C 1 1 , we obtain LC = LC ( 2π f0 )2 From ω 0 = 2π f0 = ) Substituting from Equation [1], this becomes ( 96 Ω 2 C 2 = or C= 1 ( 2π f0 ) [1] 96 Ω 2 = 1 ⎡⎣ 2π ( 2 000 π Hz ) ⎤⎦ 96 Ω 2 1 ( 2π f0 ) 2 = 2 .6 × 10 −5 F = 26 µ F Then, from Equation [1], ) ) L = ( 96 Ω 2 ( 2.6 × 10 −5 F = 2 .5 × 10 −3 H = 2 .5 mH 21.71 R= ( ∆V )DC I DC = 12 .0 V = 19.0 Ω 0.630 A Z = R 2 + ( 2π f L ) = 2 Thus, L = 56157_21_ch21_p252-280.indd 278 Z 2 − R2 = 2π f ∆Vrms 24.0 V = = 42.1 Ω I rms 0.570 A ( 42.1 Ω )2 − (19.0 Ω )2 2π ( 60.0 Hz ) = 9.97 × 10 −2 H = 99.7 mH 3/19/08 5:49:56 PM Alternating Current Circuits and Electromagnetic Waves 21.72 (a) c 3.00 × 10 8 m s = = 1.0 × 1010 Hz. Therefore, the resoλ 3.00 × 10 −2 m 1 nance frequency of the circuit is f0 = = 1.0 × 1010 Hz, giving 2π LC The required frequency is f = C= (b) C= 1 ( 2π f0 ) 2 L = ( 2π × 10 Hz 10 ) 1 2 ( 400 × 10−12 H ) = 6.3 × 10 −13 F = 0.63 pF ∈0 A ∈0 2 = , so d d = 21.73 279 C ⋅d = ∈0 ( 6.3 × 10 −13 ) F (1.0 × 10 −3 m 8.85 × 10 −12 C2 N ⋅ m ) = 8.4 × 10 ) −3 m = 8.4 mm ) (c) XC = X L = ( 2π f0 ) L = 2π (1.0 × 1010 Hz ( 400 × 10 −12 H = 25 Ω (a) Emax = c, so Bmax Bmax = Emax 0.20 × 10 −6 V m = = 6.7 × 10 −16 T 3.00 × 10 8 m s c ) ) (b) −6 −16 Emax Bmax ( 0.20 × 10 V m ( 6.7 × 10 T Intensity = = = 5.3 × 10 −17 W m 2 −7 2 µ0 2 ( 4π × 10 T ⋅ m A (c) Pav = ( Intensity ) ⋅ A = ( Intensity ) ⎢ ) ⎡πd2 ⎤ ⎥ ⎣ 4 ⎦ ⎡ π ( 20.0 m )2 ⎤ −14 = ( 5.3 × 10 −17 W m 2 ⎢ ⎥ = 1.7 × 10 W 4 ⎢⎣ ⎥⎦ ) 21.74 (a) Z= ∆Vrms 12 V = = 6.0 Ω 2 .0 A I rms (b) R= ∆VDC 12 V = = 4.0 Ω 3.0 A I DC From Z = R 2 + X L2 = R 2 + ( 2π f L ) , we find 2 L= 56157_21_ch21_p252-280.indd 279 Z 2 − R2 = 2π f ( 6.0 Ω )2 − ( 4.0 Ω )2 2π ( 60 Hz ) = 1.2 × 10 −2 H = 12 mH 3/19/08 5:49:56 PM 280 21.75 Chapter 21 (a) From Equation (21.30), the momentum imparted in time ∆t to a perfectly reflecting sail of area A by normally incident radiation of intensity I is ∆p = 2U c = 2 ( IA∆t ) c. From the impulse-momentum theorem, the average force exerted on the sail is then Fav = aav = (c) 1 From ∆x = v0 t + at 2 , with v0 = 0 , the time is 2 (a) ) Fav 0.536 N = = 8.93 × 10 −5 m s 2 m 6 000 kg (b) t= 21.76 ) 4 2 2 ∆p 2 ( IA∆t ) c 2 IA 2 (1 340 W m ( 6.00 × 10 m = = = = 0.536 N ∆t ∆t c 3.00 × 10 8 m s 2 ( ∆x ) = aav 2 ( 3.84 × 10 8 m ) = ( 2 .93 × 10 s) ⎛ 1 d ⎜⎝ 8.64 × 10 6 8.93 × 10 −5 m s 2 4 ⎞ ⎟ = 33.9 d s⎠ The intensity of radiation at distance r from a point source, which radiates total power P , is I = P A = P 4π r 2 . Thus, at distance r = 2.0 in from a cell phone radiating a total power of P = 2.0 W = 2.0 × 10 3 mW, the intensity is I= 2.0 × 10 3 mW 4π ⎡⎣( 2.0 in ) ( 2.54 cm 1 in ) ⎤⎦ 2 = 6.2 mW cm 2 This intensity is 24% higher than the maximuum allowed leakage from a microwave at this distance of 2.0 inches. (b) If when using a Blue toothheadset (emitting 2.5 mW of power) in the ear at distance rh = 2.0 in = 5.1 cm from center of the brain, the cell phone (emitting 2.0 W of power) is located in the pocket at distance rp = 1.0 m = 1.0 × 10 2 cm from the brain, the total radiation intensity at the brain is I total = I phone + I headset = or 56157_21_ch21_p252-280.indd 280 I total = 1.6 × 10 −2 2.0 × 10 3 mW 4π (1.0 × 10 cm 2 ) 2 + 2.5 mW mW −2 mW + 7.6 × 10 −3 2 = 1.6 × 10 2 cm cm 2 4π ( 5.1 cm ) mW mW mW + 7.6 × 10 −3 = 2.4 × 10 −2 = 0.024 mW cm 2 cm 2 cm 2 cm 2 3/19/08 5:49:57 PM