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Unformatted text preview: Platt, David Homework 17 Due: Oct 17 2005, 4:00 am Inst: Ken Shih 1 This printout should have 6 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. Please learn chapter 30, section 1 by your self to solve some of these problems. This section will be included in Quiz2. 001 (part 1 of 2) 10 points Assume: The mobile charge carriers are ei ther electrons or holes. The holes have the same magnitude of charge as the elec trons. The number of mobile charge car riers for this particular material is n = 8 . 49 10 28 electrons / m 3 . Note: In the figure, the point at the upper edge P 1 and at the lower edge P 2 have the same x coordinate. A constant magnetic field of magnitude points out of the paper. There is a steady flow of a horizontal current flowing from left to right in the x direction. 2.2 m 4 . 9cm 6 c m 7 . 5 A P 1 P 2 V ~ B B = 1 . 2 T y x The charge on the electron is 1 . 6021 10 19 C. What is the magnitude of the electric field between the upper and lower surfaces? Correct answer: 2 . 2506 10 7 N / C. Explanation: Let : a = 6 cm = 0 . 06 m , b = 4 . 9 cm = 0 . 049 m , B = 1 . 2 T , n = 8 . 49 10 28 electrons / m 3 , q = 1 . 6021 10 19 C , I = 7 . 5 A , and L = 2 . 2 m . L b a I P 1 P 2 V ~ B ~ B y x For the Hall effect the magnetic force bal ances the electric force which means q v d B = q E , or E = v d B . Also we know I = nq v d A or v d = I nq A , so that the magnitude of the electric field is E = I B nq A = (7 . 5 A)(1 . 2 T) n (1 . 6021 10 19 C)(0 . 00294 m 2 ) = 2 . 2506 10 7 N / C , where the area A = a b = (0 . 06 m)(0 . 049 m) = 0 . 00294 m 2 ....
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This note was uploaded on 07/26/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Physics, Work

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