hw04b - aljabr (faa335) Hw04 Ross (89251) 1 This print-out...

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Unformatted text preview: aljabr (faa335) Hw04 Ross (89251) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A 120 cm diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 8 . 18 10 5 N m 2 / C. What is the electric field strength? Correct answer: 7 . 23271 10 5 N / C. Explanation: Let : r = 60 cm = 0 . 6 m and = 8 . 18 10 5 N m 2 / C . By Gauss law, = contintegraldisplay vector E d vector A The position of maximum electric flux will be that position in which the plane of the loop is perpendicular to the electric field; i.e. , when vector E d vector A = E dA . Since the field is constant, = E A = E r 2 E = r 2 = 8 . 18 10 5 N m 2 / C (0 . 6 m) 2 = 7 . 23271 10 5 N / C . 002 (part 1 of 3) 10.0 points An electric field of magnitude 4330 N / C is applied along the x axis. Calculate the electric flux through a rect- angular plane 0 . 377 m wide and 0 . 74 m long if the plane is parallel to the yz plane. Correct answer: 1207 . 98 N m 2 / C. Explanation: Let : E = 4330 N / C , = 0 , L = 0 . 74 m , and W = 0 . 377 m . The area of the rectangle is A = L W = (0 . 74 m) (0 . 377 m) = 0 . 27898 m 2 . Since the field is uniform,the flux is = E A = E A cos = (4330 N / C) (0 . 27898 m 2 ) (cos0 ) = 1207 . 98 N m 2 / C ....
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hw04b - aljabr (faa335) Hw04 Ross (89251) 1 This print-out...

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