32_InstructorSolutions

# 32_InstructorSolutions - ELECTROMAGNETIC WAVES 32 32.1....

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32-1 E LECTROMAGNETIC W AVES 32.1. IDENTIFY: Since the speed is constant, distance . x ct = SET UP: The speed of light is 8 3.00 10 m/s c . 7 1 yr 3.156 10 s. EXECUTE: (a) 8 8 3.84 10 m 1.28 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 xc t × × = × EVALUATE: The speed of light is very great. The distance between stars is very large compared to terrestrial distances. 32.2. IDENTIFY: Since the speed is constant the difference in distance is . ct Δ SET UP: The speed of electromagnetic waves in air is 8 3.00 10 m/s. c EXECUTE: A total time difference of 0.60 s μ corresponds to a difference in distance of 86 ( 3 . 0 01 0 m / s ) ( 0 . 6 0 s ) 1 8 0 m . Δ= × × = EVALUATE: 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. IDENTIFY: Apply . cf λ = SET UP: 8 c EXECUTE: (a) 8 4 3.0 10 m s 6.0 10 Hz. 5000 m c f × (b) 8 7 3.0 10 m s 5.0 m c f × (c) 8 13 6 3.0 10 m s 5.0 10 m c f × × (d) 8 16 9 3.0 10 m s 6.0 10 Hz. 5.0 10 m c f × × EVALUATE: f increases when decreases. 32.4. IDENTIFY: = and 2 . k π = SET UP: 8 c . EXECUTE: (a) c f = . UVA: 14 7.50 10 Hz × to 14 9.38 10 Hz × . UVB: 14 × to 15 1.07 10 Hz × . (b) 2 k = . UVA: 7 1.57 10 rad/m × to 7 1.96 10 rad/m × . UVB: 7 × to 7 2.24 10 rad/m × . EVALUATE: Larger corresponds to smaller f and k . 32.5. IDENTIFY: = . max max E cB = . 2 / k πλ = . 2 . f ω = SET UP: Since the wave is traveling in empty space, its wave speed is 8 c . EXECUTE: (a) 8 14 9 6.94 10 Hz 432 10 m c f × × (b) max max (3.00 10 m/s)(1.25 10 T) 375 V/m Ec B ==× × = 32

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