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# chapt33 - Chapter 33 Electromagnetic Waves Problem 3 A...

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Chapter 33: Electromagnetic Waves Problem 3 A certain helium-neon laser emits red light in a narrow band of wavelengths centered at 632.8 nm and with a “wavelength width” (such as on the scale of Fig. 33-1) of 0.0100 nm. What is the corresponding “frequency width” for the emission? Answer 7.5 × 10 9 Hz Solution You can calculate the frequency width by calculating the difference between two frequencies centered on 632.8 nm and separated by 0.01 nm – i.e. 632.795 nm and 6.3205 nm. Alternatively, we can use our calculus to approximate the difference in frequency corresponding to a given difference in wavelength by approximating the differences as differentials. Since f = c/ λ the differential of f is df = c λ 2 d λ = 3 × 10 8 m/s 632.8 nm ( ) 2 0.0100 nm=-7.5 × 10 9 Hz The minus sign indicates that the frequency will decrease if wavelength increases. Problem 4 Project Seafarer was an ambitious program to construct an enormous antenna, buried underground on a site about 10 000 km 2 in area. Its purpose was to transmit signals to submarines while they were deeply submerged. If the effective wavelength were 1.0 × 10 4 Earth radii, what would be the (a) frequency and (b) period of radiations emitted? Ordinarily, electromagnetic radiations do not penetrate very far into conductors such as seawater. Answer (a) 4.7 × 10 -3 Hz; (b) 3 min 32 s. Solution (a) If then wavelength is 1.0 × 10 4 Earth radii then λ = 1.0 × 10 4 * 6.37 × 10 6 m = 6.37 × 10 10 m The frequency of these electromagnetic waves is f = c λ = 3 × 10 8 m s 6.37 × 10 10 m = 4.7 × 10 3 Hz

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33-2 Chapter 33: Electromagnetic Waves (b) The period of the radiation is the reciprocal of the frequency F = 1 f = 1 4.7 × 10 3 Hz = 212 s . Problem 5 What inductance must be connected to a 17-pF capacitor in an oscillator capable of generating 550-nm (i.e., visible) electromagnetic waves? Comment on your answer. Answer 5.0 × 10 -21 H. Solution To produce 550-nm electromagnetic waves would require an oscillator frequency of f = v λ = 3.0 × 10 8 m/s 550 nm = 5.45 × 10 14 Hz An LC oscillator of this frequency would require an inductance L = 1 ω 2 C = 1 4 π 2 f 2 C = 1 4 π 2 5.45 × 10 14 Hz ( ) 2 * 17 pF ( ) = 5.0 × 10 -21 H Good luck in constructing an inductor with such incredibly small inductance! Problem 8 A plane electromagnetic wave has a maximum electric field of 3.2 × 10 -4 V/m. Find the magnetic field magnitude. Answer 1.07 pT. Solution The maximum magnetic field is B m = E m c = 3.2 × 10 -4 V/m 3.0 × 10 8 m/s = 1.07 pT Problem 11 What is the intensity of a traveling plane electromagnetic wave if B m is 1.0 × 10 -4 T Answer 1.2 MW/m 2 .
Chapter 33: Electromagnetic Waves 33-3 Solution The average rate of energy transport per unit area, or intensity, is I = S av = cB m 2 2 μ o I = 3.0 × 10 8 m/s ( ) * 1.0 × 10 -4 T ( ) 2 2 * 4 π × 10 -7 H/m ( ) = 1.2 MW/m 2 Problem 12 Assume (unrealistically) that a TV station acts as a point source broadcasting isotropically at 1.0 MW. What is the intensity of the transmitted signal reaching Proxima Centauri, the star nearest our solar system, 4.3 ly away? (An alien civilization at that distance might be able to watch X Files .) A light-year (ly) is the distance light travels in one year.

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