# CH-33 - Chapter 33 1 In air light travels at roughly c =...

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Chapter 33 1. In air, light travels at roughly c = 3.0 10 8 m/s. Therefore, for t = 1.0 ns, we have a distance of d ct ( . . 30 10 0 30 8 9 m / s)(1.0 10 s) m. 2. (a) From Fig. 33-2 we find the smaller wavelength in question to be about 515 nm, (b) and the larger wavelength to be approximately 610 nm. (c) From Fig. 33-2 the wavelength at which the eye is most sensitive is about 555 nm. (d) Using the result in (c), we have 8 14 3.00 10 m/s 5.41 10 Hz 555 nm c f l . (e) The period is (5.41 10 14 Hz) –1 = 1.85 10 –15 s. 3. (a) The frequency of the radiation is f c l 30 10 10 10 6 4 10 4 7 10 8 5 6 3 . ( . )( . . m / s m) Hz. (b) The period of the radiation is T f 1 1 4 7 10 212 3 32 3 . min Hz s s. 4. Since l  l , we find f is equal to l c c l l F H G I K J 2 8 9 9 9 30 10 632 8 10 7 49 10 ( . ( . . m / s)(0.0100 10 m) m) Hz. 2 5. If f is the frequency and l is the wavelength of an electromagnetic wave, then f l = c . The frequency is the same as the frequency of oscillation of the current in the LC circuit of the generator. That is, f LC 1 2 / , where C is the capacitance and L is the inductance. Thus 1311

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CHAPTER 33 l 2 LC c . The solution for L is L Cc l 2 2 2 9 2 2 12 8 2 21 4 550 10 4 17 10 2 998 10 500 10 m F m / s H. c h c hc h . . This is exceedingly small. 6. The emitted wavelength is ( ) ( ) ( ) 8 6 12 2 2 2.998 10 m/s 0.253 10 H 25.0 10 F 4.74m. c c LC f l      7. The amplitude of the magnetic field in the wave is B E c m m 320 10 2 998 10 107 10 4 8 12 . . . V / m m / s T. 8. (a) The amplitude of the magnetic field is 9 9 8 2.0V/m 6.67 10 T 6.7 10 T. 2.998 10 m/s m m E B c . (b) Since the -wave E r oscillates in the z direction and travels in the x direction, we have B x = B z = 0. So, the oscillation of the magnetic field is parallel to the y axis. (c) The direction (+ x ) of the electromagnetic wave propagation is determined by E B r r . If the electric field points in + z , then the magnetic field must point in the – y direction. With SI units understood, we may write ( ) ( ) 15 15 8 9 15 2.0cos 10 / cos 10 3.0 10 6.7 10 cos 10 y m t x c x B B t c x t c  9. If P is the power and t is the time interval of one pulse, then the energy in a pulse is 32
E P t 100 10 10 10 10 10 12 9 5 W s J. c hc h . . 10. The intensity of the signal at Proxima Centauri is I P r 4 10 10 4 4 3 9 46 10 4 8 10 2 6 15 2 29 . . . . . W ly m / ly W / m 2 b gc h 11. The intensity is the average of the Poynting vector: I S cB m avg 2 m / s T H / m W / m 2 0 8 4 2 6 2 6 2 30 10 10 10 2 126 10 12 10 . .

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