HW17Sol - Platt, David Homework 17 Due: Oct 17 2005, 4:00...

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Unformatted text preview: Platt, David Homework 17 Due: Oct 17 2005, 4:00 am Inst: Ken Shih 1 This print-out should have 6 questions. Multiple-choice 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.

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HW17Sol - Platt, David Homework 17 Due: Oct 17 2005, 4:00...

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