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Unformatted text preview: choe (dec865) – Practice HW 20 Solutions – Weathers – (22202) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The electric field between the plates of an air capacitor of plate area 0 . 09 m 2 is changing at a rate of 23500 V / m / s. The permitivity of free space is 8 . 85419 × 10 12 C 2 / N / m 2 . What is Maxwell’s displacement current? Correct answer: 1 . 87266 × 10 8 A. Explanation: Let : ǫ = 8 . 85419 × 10 12 C 2 / N / m 2 , A = 0 . 09 m 2 , and d E dt = 23500 V / m / s , where A is the area and d E dt is the changing rate of the electric field. The displacement current is I D = ǫ A d E dt = ( 8 . 85419 × 10 12 C 2 / N / m 2 ) × ( . 09 m 2 ) (23500 V / m / s) = 1 . 87266 × 10 8 A . 002 10.0 points The electric field in an electromagnetic wave (in vacuum) is described by E = E max sin( k x − ω t ) , where E max = 20 N / C and k = 9 . 3 × 10 8 m 1 . The speed of light is 2 . 99792 × 10 8 m / s. Find the amplitude of the corresponding magnetic wave. Correct answer: 6 . 67128 × 10 8 T. Explanation: Let : c = 2 . 99792 × 10 8 m / s and E max = 20 N / C . The relation between the magnitude of the electric field and that of the magnetic field for an electromagnetic wave is given by B = E c . Thus, the magnitude of the magnetic wave is B max = E max c = 20 N / C 2 . 99792 × 10 8 m / s = 6 . 67128 × 10 8 T . 003 (part 1 of 2) 10.0 points What is the wavelength corresponding to a frequency of 2 × 10 19 Hz of electromagnetic wave in free space? The speed of light is 2 . 99792 × 10 8 m / s. Correct answer: 1 . 49896 × 10 11 m. Explanation: Let : f = 2 × 10 19 Hz and c = 2 . 99792 × 10 8 m / s . The relation between frequency and wave length is given by λ = c f = 2 . 99792 × 10 8 m / s 2 × 10 19 Hz = 1 . 49896 × 10 11 m ....
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This document was uploaded on 09/23/2010.
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