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Unformatted text preview: aljabr (faa335) – Hw20 – Ross – (89251) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A plane electromagnetic sinusoidal wave of frequency 48 . 5 MHz travels in free space. The speed of light is 2 . 99792 × 10 8 m / s. Determine the wavelength of the wave. Correct answer: 6 . 18129 m. Explanation: Let : c = 2 . 99792 × 10 8 m / s , and f = 48 . 5 MHz = 4 . 85 × 10 7 Hz . The speed of light is c = λf λ = c f = 2 . 99792 × 10 8 m / s 4 . 85 × 10 7 Hz = 6 . 18129 m . 002 (part 2 of 3) 10.0 points Find the period of the wave. Correct answer: 2 . 06186 × 10 8 s. Explanation: The period T of the wave is the inverse of the frequency: T = 1 f = 1 4 . 85 × 10 7 Hz = 2 . 06186 × 10 8 s . 003 (part 3 of 3) 10.0 points At some point and some instant, the electric field has has a value of 483 N / C. Calculate the magnitude of the magnetic field at this point and this instant. Correct answer: 1 . 61111 × 10 6 T. Explanation: Let : E = 483 N / C . The magnitudes of the electric and the mag netic fields are related by E B = c , B = E c = 483 N / C 2 . 99792 × 10 8 m / s = 1 . 61111 × 10 6 T . 004 (part 1 of 3) 10.0 points High power lasers in factories are used to cut through cloth and metal. One such laser has a beam diameter of 0 . 481 mm and generates an electric field at the target having an amplitude . 845 MV / m. The speed of light is 2 . 99792 × 10 8 m / s the permeability of free space is 4 π × 10 7 T · N / A. What is the amplitude of the magnetic field produced? Correct answer: 0 . 00281862 T. Explanation: Let : E = 0 . 845 MV / m = 8 . 45 × 10 5 V / m , and c = 2 . 99792 × 10 8 m / s ....
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This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue University.
 Spring '10
 Gavrine

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