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Unformatted text preview: terry (ect328) – oldhomework 33 – Turner – (59130) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A 0 . 2431 A current is charging a capacitor that has circular plates, 9 . 05 cm in radius. The plate separation is 1 . 89 mm. The permitivity or free space is 8 . 85 × 10 12 and the permeability of free space is 1 . 25664 × 10 6 T · m / A. What is the time rate of increase of electric field between the plates? Correct answer: 1 . 06757 × 10 12 V / m s. Explanation: Let : I = 0 . 2431 A , r = 9 . 05 cm = 0 . 0905 m , A = π r 2 = 0 . 0257304 m 2 , and ǫ = 8 . 85 × 10 12 . Let Φ E be the flux of the electric field, defined as Φ E = integraldisplay vector E · d vector A . Thus d Φ E dt = d dt ( E A ) = I ǫ Time rate of increase of electric field between the plates is P = d E dt = I ǫ A = . 2431 A (8 . 85 × 10 12 )(0 . 0257304 m 2 ) = 1 . 06757 × 10 12 V / m s . 002 (part 2 of 2) 10.0 points What is the magnetic field between the plates . 0518 m from the center? Correct answer: 3 . 07502 × 10 7 T. Explanation: Let : r = 0 . 0518 m and μ = 1 . 25664 × 10 6 T · m / A . Since contintegraldisplay vector B · dvectors = ǫ μ d dt Φ E Then magnetic field, B ,at the distance r from the center between the plates satisfies 2 π r B = ǫ μ d dt parenleftbigg Q ǫ A π r 2 parenrightbigg Hence B = μ I r 2 A = (0 . 2431 A)(0 . 0518 m) 2(0 . 0257304 m 2 ) × (1 . 25664 × 10 6 T · m / A) = 3 . 07502 × 10 7 T . 003 10.0 points A sinusoidal voltage is applied directly across a 37 . 94 μ F capacitor. The frequency of the source is 1 . 58 kHz, and the voltage amplitude is 49 . 2 V. Find the maximum value of the displace ment current in the capacitor. Correct answer: 18 . 531 A. Explanation: Let : C = 37 . 94 μ F = 3 . 794 × 10 5 F , f = 1 . 58 kHz = 1580 Hz , and V max = 49 . 2 V ....
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 Fall '09
 Turner
 Light, Speed of light, Correct Answer, Terry

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