Physics Solution Manual for 1100 and 2101

Solution the fraction of the suns power that is

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Unformatted text preview: the circuit must have a resonant frequency f0 that matches the frequency of the radio waves. The resonant frequency depends upon the 1 capacitance C and inductance L of the circuit via f 0 = (Equation 23.10). In order 2π LC to pick up the entire range of FM waves, the circuit must be able to attain the lowest (flow = 88 MHz) and highest (fhigh = 108 MHz) necessary resonant frequency. We will use Equation 23.10 to determine the corresponding minimum and maximum capacitance values. SOLUTION Squaring both sides of f 0 = 1 2π LC (Equation 23.10) and solving for C, we obtain = ( f0 ) 2 1 or C = ( 2π )2 LC 1 ( 2π f0 ) L 2 (1) As we see from Equation (1), the greater the frequency, the smaller the value of the capacitance. So the highest frequency fhigh = 108 MHz corresponds to the minimum value of the capacitance Cmin. From Equation (1), we obtain Cmin = 1 1 = = 3.62 ×10−12 F 2 2 −7 6 ( 2π fhigh ) L ( 2π ) (108 ×10 Hz ) ( 6.00 ×10 H ) 2 On the other end of the FM frequency range, matching the lowest frequency of flow = 88.0 MHz requires a maximum capacitance value Cmax of Cmax = 1 1 = = 5.45 ×10−12 F 2 2 6 −7 ( 2π flow ) L ( 2π ) (88.0 ×10 Hz ) ( 6.00 ×10 H ) 2 Therefore, the capacitance values should range from 3.62×10−12 F to 5.45×10−12 F . Chapter 24 Problems 5. SSM REASONING 1281 The equation that represents the wave mathematically is y = A sin ( 2π f t – 2π x / λ ) . In this expression the amplitude is A = 156 N/C. The wavelength λ can be calculated using Equation 16.1, and we obtain c 3.00 × 108 m/s = = 2.00 m λ= f 1.50 × 108 Hz SOLUTION a. For t = 0 s, the wave expression becomes 2π x 2π x y π f ( 0) – = A sin ( 2π f t – 2π x / λ ) 156 sin 2= –156 sin = = –156 sin (π x ) 2.00 2.00 In this result, the units are suppressed for convenience. The following table gives the values of the electric field obtained using this version of the wave expression with the given values of the position x. The term π x is in radians when x is in meters, and conversion from radians to degrees is accomplished using the fact that 2π rad = 360°. x y = –156 sin (π x) 0m –156 sin (0) = –156 sin (0°) = 0 0.50 m –156 sin (0.50 π ) = –156 sin (90°) = –156 1.00 m –156 sin (1.00 π ) = –156 sin (180°) = 0 1.50 m –156 sin (1.50 π ) = –156 sin (270°) = +156 2.00 m –156 sin (2.00 π ) = –156 sin (360°) = 0 y (N/C) +156 These values for the electric field are plotted in the graph shown at the right. x (m) 1.00 2.00 –156 b. For t = T/4, we use the fact that f = 1/T, and the wave expression becomes 1 T 2π x π = A sin (= 156 sin 2π = 156 sin – π x 156 cos (π x ) y 2π f t – 2π x / λ ) – = 2 T 4 2.00 In this result, the units are suppressed for convenience. The following table gives the values of the electric field obtained using this version of the wave expression with the given values 1282 ELECTROMAGNETIC WAVES of the position x. The term π x is in radians when x is in meters, and conversion from radians to degrees is accomplished using the fact that 2π rad = 360°. x y = 156 cos (π x) 0m 156 cos (0) = 156 cos (0°) = +156 0.50 m 156 cos (0.50 π ) = 156 cos (90°) = 0 1.00 m 156 cos (1.00 π ) = 156 cos (180°) = –156 1.50 m 156 cos (1.50 π ) = 156 cos (270°) = 0 2.00 m 156 cos (2.00 π ) = 156 cos (360°) = +156 These values for the electric field are plotted in the graph shown at the right. ______________________________________________________________________________ 6. REASONING AND SOLUTION The average flux change through the coil in one fourth of the wave period is, according to Faraday's law, ∆Φ = NAB = NB0A. The magnitude of the average emf is then emf = ∆Φ/∆t = NB0 A /∆t. Now ∆t = T/4 = 1/(4 f ) , so Emf = 4N f B0 A = 4(450)(1.2 × 106 Hz)(2.0 × 10–13 T)(π )(0.25 m)2 = 8.5 × 10 –5 V ______________________________________________________________________________ 7. REASONING The wavelength λ of a wave is related to its speed v and frequency f by λ = v / f (Equation 16.1). Since blue light and orange light are electromagnetic waves, they travel through a vacuum at the speed of light c; thus, v = c. SOLUTION a. The wavelength of the blue light is = λ c 3.00 ×108 m/s = = f 6.34 ×1014 Hz 4.73 ×10−7 m Since 1 nm = 10−9 m, 1 nm −9 10 m ( 4.73 ×10−7 λ =m ) 473 nm = Chapter 24 Problems 1283 b. In a similar manner, we find that the wavelength of the orange light is c 3.00 ×108 m/s = ×10−7 m = nm 6.06 606 f 4.95 ×1014 Hz ______________________________________________________________________________ = λ= 8. REASONING The frequency f of the UHF wave is related to its wavelength by c = f λ (Equation 16.1), where c is the speed of light in a vacuum and λ is the wavelength. The electric and magnetic fields are both zero at the same positions, which are separated by a distance d equal to half a wavelength (see Figure 24.3). Therefore, we can express the wavelength in terms of the distance between adjacent position...
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