HW07-Solutions

# HW07-Solutions - Physics 201 Solution Set 07 CHAPTER 30 AC...

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Physics 201 Solution Set 07 CHAPTER 30: AC Circuits 30[ 39, 40, 62] CHAPTER 31: Electromagnetic Waves 31[ Q5, 9, 10, 14, 15, 49] 1 CHAPTER 30 39. We find the frequency from Eq. 30-23b for the reactance of an inductor. ( ) 660 2 3283Hz 3300Hz 2 2 0.0320H L L X X fL f L π π π Ω = = = = 40. The reactance of a capacitor is given by Eq. 30-25b, 1 2 C X fC π = . ( a ) ( ) ( ) 6 1 1 290 2 2 60.0 Hz 9.2 10 F C X fC π π = = = Ω × ( b ) ( )( ) 2 6 6 1 1 1.7 10 2 2 1.00 10 Hz 9.2 10 F C X fC π π = = = × Ω × × 62. The resonant frequency is found from Eq. 30-32. The resistance does not influence the resonant frequency. ( )( ) 5 0 6 12 1 1 1 1 5.1 10 Hz 2 2 26.0 10 H 3800 10 F f LC π π = = = × × × CHAPTER 31 Q5. The magnetic field vector will oscillate up and down, perpendicular to the direction of propagation and to the electric field vector. 9. Use Eq. 31-11 with . v c = ( ) ( ) 9 8 0 0 0 0 12.5 10 T 3.00 10 m s 3.75V m E c E B c B = = = × × = 10. The frequency of the two fields must be the same: 80.0kHz . The rms strength of the electric field can be found from Eq. 31-11 with . v c = ( )( ) 8 9 rms rms 3.00 10 m s 7.75 10 T 2.33V m E cB = = × × = The electric field is perpendicular to both the direction of travel and the magnetic field, so the electric field oscillates along the horizontal north-south line.

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