YF_ISM_32

YF_ISM_32 - ELECTROMAGNETIC WAVES 32 32.1. IDENTIFY: Since...

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32-1 E LECTROMAGNETIC W AVES 32.1. IDENTIFY: Since the speed is constant, distance . xc t = 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) 87 1 6 1 3 (3.00 10 m/s)(8.61 yr)(3.156 10 s/yr) 8.15 10 m 8.15 10 km 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 (3.00 10 m/s)(0.60 10 s) 180 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 Ec B = . 2 / k πλ = . 2. f ωπ = 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 . 0 01 0 m / s ) ( 1 . 2 51 0 T ) 3 7 5 V /m B ==× × = 32
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32-2 Chapter 32 (c) 7 9 22 r a d 1.45 10 rad/m 432 10 m k ππ λ == = × × . 14 15 ( 2 r a d ) ( 6 . 9 41 0 H z ) 4 . 3 61 0 r a d / s ωπ = × . 71 5 max cos( ) (375 V/m)cos([1.45 10 rad/m] [4.36 10 rad/s] ) EE k x t x t ω =− = × × 67 1 5 max cos( ) (1.25 10 T)cos([1.45 10 rad/m] [4.36 10 rad/s] ) BB k x t = × × × EVALUATE: The cos( ) kx t factor is common to both the electric and magnetic field expressions, since these two fields are in phase. 32.6. IDENTIFY: cf = . max max Ec B = . Apply Eqs.(32.17) and (32.19). SET UP: The speed of the wave is 8 3.00 10 m/s. c EXECUTE: (a) 8 14 9 3.00 10 m/s 6.90 10 Hz 435 10 m c f × = × × (b) 3 12 max max 8 2.70 10 V/m 9.00 10 T E B c × = × × (c) 7 2 1.44 10 rad/m k π × . 15 24 . 3 4 1 0 r a d / s f × . If max ˆ (, ) c o s ( ) =+ G zt E k z t Ei , then max ˆ ) c o s ( ) + G B k z t Bj , so that × E B GG will be in the ˆ k direction. 371 5 ˆ ( , ) (2.70 10 V/m)cos([1.44 10 rad/s) [4.34 10 rad/s] ) × + × G z t and 12 7 15 ˆ ( , ) (9.00 10 T)cos([1.44 10 rad/s) × × + × G z t . EVALUATE: The directions of E G and B G and of the propagation of the wave are all mutually perpendicular. The argument of the cosine is kz t + since the wave is traveling in the -direction z . Waves for visible light have very high frequencies. 32.7. IDENTIFY and SET UP: The equations are of the form of Eqs.(32.17), with x replaced by z . B G is along the y -axis; deduce the direction of . E G EXECUTE: 14 15 2 2( 6 . 1 01 z ) 3 . 8 31 r a d / s f × = × 15 7 8 2 2 3.83 10 rad/s 1.28 10 rad/m f k cc ππ ω × = = = × × 4 max 5.80 10 T B 84 5 max max (3.00 10 m/s)(5.80 10 T) 1.74 10 V/m B ==× × = × B G is along the y -axis. E B G G × is in the direction of propagation (the + z -direction). From this we can deduce the direction of , E G as shown in Figure 32.7.
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YF_ISM_32 - ELECTROMAGNETIC WAVES 32 32.1. IDENTIFY: Since...

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