f02x2aan - V 15 m 80 2 1 2 1 = Φ = dt d r E B mV/m 30 V/m...

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NAME PHYS-1200 PHYSICS II QUIZ 2 OCTOBER 30, 2002 SOLUTIONS PART A. 1. B 7. A 2. D 8. C 3. C 4. D 5. D 6. D PART B . 1. a) The displacement current flows in the same direction as the conventional current. b) Since the current is increasing, the capacitor is discharging. That means that the upper plate in positive. c) The emf will oppose the increase in current, as Lenz’s law predicts. 2. a) s t x v 05 . 0 m 0 . 1 = = v = 20 m/s b) The wavelength can be read directly from the graph. λ = 4.0 m c) m 0 . 4 m/s 20 = = λ v f f = 5.0 Hz
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NAME PART C . 1. a) BA B = Φ , so A B B Φ = , and V 15 . 0 m) 30 . 0 ( 1 1 1 2 2 π π = Φ = Φ = dt d R dt d A dt dB B B T/s 53 . 0 = dt dB (or V/m²) b) TO THE TOP OF THE PAGE According to Lenz’s law, the direction of the induced field will oppose the change that produced it. c) dt d d B Φ - = s E , and from symmetry, dt d r E B Φ - = π 2 , so dt d r E B Φ - = π 2 1 . Since the
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Unformatted text preview: V 15 . m) 80 . ( 2 1 2 1 = Φ = dt d r E B mV/m 30 V/m 10 . 3 2 = × =-E 2. a) f beat = ∆ f = 440 Hz – 400 Hz f beat = 40 Hz b) At the fundamental frequency, Hz) 2(400 m/s 600 2 2 = = = f v L string λ L string = 0.75 m c) At the fundamental frequency, Hz) 2(440 m/s 343 2 2 = = = f v L pipe L pipe = 0.39 m d) TOWARD THE MICROPHONE The source must approach the detector to result in a higher frequency. e) For a moving source string v v v f f-= ' . Then, v f f v v string ' =-, and m/s) 343 ( Hz 440 Hz 400 1 ' 1 -= -= v f f v string v string = 31 m/s...
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