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32_InstructorSolutions

# 32_InstructorSolutions - ELECTROMAGNETIC WAVES 32 32.1...

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32-1 E LECTROMAGNETIC W AVES 32.1. I DENTIFY : Since the speed is constant, distance . x ct = S ET U P : The speed of light is 8 3.00 10 m/s c = × . 7 1 yr 3.156 10 s. = × E XECUTE : (a) 8 8 3.84 10 m 1.28 s 3.00 10 m/s x t c × = = = × (b) 8 7 16 13 (3.00 10 m/s)(8.61 yr)(3.156 10 s/yr) 8.15 10 m 8.15 10 km x ct = = × × = × = × E VALUATE : The speed of light is very great. The distance between stars is very large compared to terrestrial distances. 32.2. I DENTIFY : Since the speed is constant the difference in distance is . c t Δ S ET U P : The speed of electromagnetic waves in air is 8 3.00 10 m/s. c = × E XECUTE : A total time difference of 0.60 s μ corresponds to a difference in distance of 8 6 (3.00 10 m/s)(0.60 10 s) 180 m. c t Δ = × × = E VALUATE : The time delay doesn°t depend on the distance from the transmitter to the receiver, it just depends on the difference in the length of the two paths. 32.3. I DENTIFY : Apply . c f λ = S ET U P : 8 3.00 10 m/s c = × E XECUTE : (a) 8 4 3.0 10 m s 6.0 10 Hz. 5000 m c f λ × = = = × (b) 8 7 3.0 10 m s 6.0 10 Hz. 5.0 m c f λ × = = = × (c) 8 13 6 3.0 10 m s 6.0 10 Hz. 5.0 10 m c f λ × = = = × × (d) 8 16 9 3.0 10 m s 6.0 10 Hz. 5.0 10 m c f λ × = = = × × E VALUATE : f increases when λ decreases. 32.4. I DENTIFY : c f λ = and 2 . k π λ = S ET U P : 8 3.00 10 m/s c = × . E XECUTE : (a) c f λ = . UVA: 14 7.50 10 Hz × to 14 9.38 10 Hz × . UVB: 14 9.38 10 Hz × to 15 1.07 10 Hz × . (b) 2 k π λ = . UVA: 7 1.57 10 rad/m × to 7 1.96 10 rad/m × . UVB: 7 1.96 10 rad/m × to 7 2.24 10 rad/m × . E VALUATE : Larger λ corresponds to smaller f and k . 32.5. I DENTIFY : c f λ = . max max E cB = . 2 / k π λ = . 2 . f ω π = S ET U P : Since the wave is traveling in empty space, its wave speed is 8 3.00 10 m/s c = × . E XECUTE : (a) 8 14 9 3.00 10 m/s 6.94 10 Hz 432 10 m c f λ × = = = × × (b) 8 6 max max (3.00 10 m/s)(1.25 10 T) 375 V/m E cB = = × × = 32

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32-2 Chapter 32 (c) 7 9 2 2 rad 1.45 10 rad/m 432 10 m k π π λ = = = × × . 14 15 (2 rad)(6.94 10 Hz) 4.36 10 rad/s ω π = × = × . 7 15 max cos( ) (375 V/m)cos([1.45 10 rad/m] [4.36 10 rad/s] ) E E kx t x t ω = = × × 6 7 15 max cos( ) (1.25 10 T)cos([1.45 10 rad/m] [4.36 10 rad/s] ) B B kx t x t ω = = × × × E VALUATE : The cos( ) kx t ω factor is common to both the electric and magnetic field expressions, since these two fields are in phase. 32.6. I DENTIFY : c f λ = . max max E cB = . Apply Eqs.(32.17) and (32.19). S ET U P : The speed of the wave is 8 3.00 10 m/s. c = × E XECUTE : (a) 8 14 9 3.00 10 m/s 6.90 10 Hz 435 10 m c f λ × = = = × × (b) 3 12 max max 8 2.70 10 V/m 9.00 10 T 3.00 10 m/s E B c × = = = × × (c) 7 2 1.44 10 rad/m k π λ = = × . 15 2 4.34 10 rad/s f ω π = = × . If max ± ( , ) cos( ) ω = + ! z t E kz t E i , then max ± ( , ) cos( ) ω = − + ! z t B kz t B j , so that × E B ! ! will be in the ± k direction. 3 7 15 ± ( , ) (2.70 10 V/m)cos([1.44 10 rad/s) [4.34 10 rad/s] ) = × × + × ! z t z t E i and 12 7 15 ± ( , ) (9.00 10 T)cos([1.44 10 rad/s) [4.34 10 rad/s] ) = − × × + × ! z t z t B j . E VALUATE : The directions of E ! and B ! and of the propagation of the wave are all mutually perpendicular. The argument of the cosine is kz t ω + since the wave is traveling in the -direction z . Waves for visible light have very high frequencies.
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32_InstructorSolutions - ELECTROMAGNETIC WAVES 32 32.1...

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