solution4_pdf - white (rw22679) – Hw04 – Ross –...

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Unformatted text preview: white (rw22679) – Hw04 – Ross – (10219) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A proton moves perpendicularly to a uni- form magnetic field B with a speed of 9 × 10 6 m / s and experiences an acceleration of 4 × 10 13 m / s 2 in the positive x direction when its velocity is in the positive z direction. The mass of a proton is 1 . 673 × 10 − 27 kg. Find the magnitude of the field. Correct answer: 0 . 0464722 T. Explanation: Let : q = 1 . 6 × 10 − 19 C , a x = 4 × 10 13 m / s 2 , v z = 9 × 10 6 m / s , and m = 1 . 673 × 10 − 27 kg . The magnetic force provides the accelera- tion F = F magnetic ma = q v B . Thus B = ma q v = ( 1 . 673 × 10 − 27 kg )( 4 × 10 13 m / s 2 ) (1 . 6 × 10 − 19 C) (9 × 10 6 m / s) = . 0464722 T . 002 (part 2 of 2) 10.0 points What is its direction? 1. negative x direction 2. None of these 3. positive x direction 4. positive z direction 5. negative y direction correct 6. negative z direction 7. positive y direction Explanation: Apply the right-hand rule (positive charge): Force directed out of the palm of the hand, fingers in the direction of the field, thumb in the direction of the velocity. The palm faces to the positive x direction, and the thumb points in the positive z di- rection, so fingers point in the negative y direction. + y + x + z F v 003 10.0 points A positively charged particle moving parallel to the z-axis enters a magnetic field (pointing out of of the page), as shown in the figure below....
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This note was uploaded on 12/01/2010 for the course PHYSICS 21900 taught by Professor Rhoads during the Spring '10 term at IUPUI.

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solution4_pdf - white (rw22679) – Hw04 – Ross –...

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