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Unformatted text preview: Anderson, Matthew Homework 9 Due: Apr 2 2008, 11:00 pm Inst: Vitaly 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the magnitude of the magnetic force on a proton moving with velocity 4 . 47 Mm / s in the positive x direction in a magnetic field of 1 . 7 T in the positive z direction. Correct answer: 1 . 21749 pN. Explanation: Let : q = 1 . 60218 10 19 C , ~v = (4 . 47 Mm / s) = (4 . 47 10 6 m / s) , and ~ B = (1 . 7 T) k . ~ F = q~v ~ B = (1 . 60218 10 19 C) (4 . 47 10 6 m / s) / h (1 . 7 T) k i 10 12 pN 1 N = (1 . 21749 pN) , so the magnitude is 1 . 21749 pN . keywords: 002 (part 1 of 2) 10 points An electron is projected into a uniform mag netic field given by ~ B = B x + B y , where B x = 3 . 8 T and B y = 2 T. The magnitude of the charge on an electron is 1 . 60218 10 19 C . x y z v = 390000 m / s electron 3 . 8 T 2 T B Find the direction of the magnetic force when the velocity of the electron is v , where v = 390000 m / s. 1. b F = 2. b F = 3. b F = 1 2 (  ) 4. b F = 5. b F = 6. b F = 1 2 ( + ) 7. b F = 1 2 (  ) 8. b F = k correct 9. b F = k Explanation: Let : q = 1 . 60218 10 19 C , B x = 3 . 8 T , and B y = 2 T . Basic Concepts: Magnetic force on a mov ing charge is given by ~ F = q~v ~ B . Solution: ~ B = (3 . 8 T) + (2 T) v = (390000 m / s) for the electron. Find: The vector expression for the force on the electron. This solves both part 1 and part 2. We will go through two methods of doing the problem. The first is more mathematically oriented and the second uses more of a reasoning argu ment. Anderson, Matthew Homework 9 Due: Apr 2 2008, 11:00 pm Inst: Vitaly 2 Method 1: The force acting on a charge q with velocity ~v in the presence of an external magnetic field ~ B is given by ~ F = q~v ~ B Taking the cross product of ~v with ~ B we obtain ~ F = q~v ~ B = q fl fl fl fl fl fl k v B x B y fl fl fl fl fl fl = q n [( B y )(0) ( B x )( v )] k [(0)(0) ( B x )(0)] + [( v )(0) ( B x )(0)] o = q B x v k = ( 1 . 60218 10 19 C)(3 . 8 T) (390000 m / s) k = (2 . 37443 10 13 N) k , and the direction is + k . Method 2: The other method is to real ize that the only component of the magnetic field which affects the electron is the compo nent perpendicular to its velocity. Therefore, F = q  ~v ~ B  = q v B with the direction given by the right hand rule to be in the neg ative k direction; but recalling to reverse the direction because the electron has a negative instead of positive charge....
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This note was uploaded on 03/11/2010 for the course PHYS 121 taught by Professor Opyrchal during the Spring '08 term at NJIT.
 Spring '08
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