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Unformatted text preview: kavo (ak22862) HW09 Janow (12121) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find the magnitude of the magnetic force on a proton moving with velocity 4 . 38 Mm / s in the positive x direction in a magnetic field of 1 . 8 T in the positive z direction. Correct answer: 1 . 26316 pN. Explanation: Let : q = 1 . 60218 10 19 C , vectorv = (4 . 38 Mm / s) = (4 . 38 10 6 m / s) , and vector B = (1 . 8 T) k . vector F = qvectorv vector B = (1 . 60218 10 19 C) bracketleftbig (4 . 38 10 6 m / s) bracketrightbig bracketleftBig (1 . 8 T) k bracketrightBig 10 12 pN 1 N = (1 . 26316 pN) , so the magnitude is 1 . 26316 pN . 002 (part 1 of 2) 10.0 points An electron is projected into a uniform mag netic field given by vector B = B x + B y , where B x = 4 . 7 T and B y = 1 . 9 T. The magnitude of the charge on an electron is 1 . 60218 10 19 C . x y z v = 3 . 9 10 5 m / s electron 4 . 7 T 1 . 9 T B Find the direction of the magnetic force when the velocity of the electron is v , where v = 3 . 9 10 5 m / s. 1. hatwide F = 1 2 ( + ) 2. hatwide F = 1 2 ( ) 3. hatwide F = 4. hatwide F = 5. hatwide F = 1 2 ( ) 6. hatwide F = k 7. hatwide F = k correct 8. hatwide F = 9. hatwide F = Explanation: Let : q = 1 . 60218 10 19 C , B x = 4 . 7 T , and B y = 1 . 9 T . Basic Concepts: Magnetic force on a mov ing charge is given by vector F = qvectorv vector B . Solution: vector B = (4 . 7 T) + (1 . 9 T) v = (3 . 9 10 5 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. Method 1: The force acting on a charge q with velocity vectorv in the presence of an external magnetic field vector B is given by vector F = qvectorv vector B kavo (ak22862) HW09 Janow (12121) 2 Taking the cross product of vectorv with vector B we obtain vector F = qvectorv vector B = q vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle k v B x B y vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = q braceleftBig [( B y )(0) ( B x )( v )] k [(0)(0) ( B x )(0)] + [( v )(0) ( B x )(0)] bracerightBig = q B x v k = ( 1 . 60218 10 19 C)(4 . 7 T) (3 . 9 10 5 m / s) k = (2 . 93679 10 13 N) k , and the direction is + k ....
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This note was uploaded on 09/16/2009 for the course PHYS 121 taught by Professor Opyrchal during the Fall '08 term at NJIT.
 Fall '08
 Opyrchal
 Physics, Force

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