9 - kavo (ak22862) HW09 Janow (12121) 1 This print-out...

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Unformatted text preview: kavo (ak22862) HW09 Janow (12121) 1 This print-out should have 19 questions. Multiple-choice 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.

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9 - kavo (ak22862) HW09 Janow (12121) 1 This print-out...

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