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Unformatted text preview: kunkel (bk5776) – APB CHW 10 – Atman – (19312) 1 This printout should have 32 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An electron in a vacuum is first accelerated by a voltage of 71600 V and then enters a region in which there is a uniform magnetic field of 0 . 167 T at right angles to the direction of the electron’s motion. The mass of the electron is 9 . 11 × 10 − 31 kg and its charge is 1 . 60218 × 10 − 19 C. What is the magnitude of the force on the electron due to the magnetic field? Correct answer: 4 . 24614 × 10 − 12 N. Explanation: Let : V = 71600 V , B = 0 . 167 T , m = 9 . 11 × 10 − 31 kg , q e = 1 . 60218 × 10 − 19 C . The kinetic energy K gained after acceler ation is K = 1 2 mv 2 = q e V , so the velocity is v = radicalbigg 2 q e V m = radicalBigg 2 (1 . 60218 × 10 − 19 C)(71600 V) 9 . 11 × 10 − 31 kg = 1 . 58697 × 10 8 m / s . Then the force on it is f = q v B = (1 . 60218 × 10 − 19 C) × (1 . 58697 × 10 8 m / s) (0 . 167 T) = 4 . 24614 × 10 − 12 N . 002 (part 1 of 2) 10.0 points A proton travels with a speed of 3 . 07 × 10 6 m / s at an angle of 17 . 7 ◦ with a magnetic field of . 304 T pointed in the y direction. The charge of proton is 1 . 60218 × 10 − 19 C. What is the magnitude of the magnetic force on the proton? Correct answer: 4 . 54615 × 10 − 14 N. Explanation: Let : v = 3 . 07 × 10 6 m / s , θ = 17 . 7 ◦ , and B = 0 . 304 T . F = q v B sin θ = (1 . 60218 × 10 − 19 C)(3 . 07 × 10 6 m / s) × (0 . 304 T) sin17 . 7 ◦ = 4 . 54615 × 10 − 14 N. 003 (part 2 of 2) 10.0 points The mass of proton is 1 . 67262 × 10 − 27 kg What is the proton’s acceleration? Correct answer: 2 . 71797 × 10 13 m / s 2 . Explanation: Let : m = 1 . 67262 × 10 − 27 kg , F = ma. a = F m = 4 . 54615 × 10 − 14 N 1 . 67262 × 10 − 27 kg = 2 . 71797 × 10 13 m / s 2 . 004 10.0 points A metal wire of mass 18 kg slides without fric tion on two horizontal rails spaced a distance of 65 m apart. The track lies in a vertical uniform magnetic field of 15 T. A constant current 74 A from a generator flows down one rail, across the wire, down the other rail back to the generator. Find the velocity of the wire at t = 21 s, assuming it to be at rest at t = 0. Correct answer: 84175 m / s. Explanation: Let : m = 18 kg , L = 65 m , and B = 15 T . kunkel (bk5776) – APB CHW 10 – Atman – (19312) 2 The magnetic force on the wire is I LB , so the acceleration is given by F = I LB = ma. Then, a = I LB m and the velocity after 21 s is v = at = I LB t m = (74 A) (65 m) (15 T) (21 s) (18 kg) = 84175 m / s . 005 10.0 points A wire carries a current of 13 A in a direction that makes an angle of 43 . 8 ◦ with the direction of a magnetic field of strength 0 . 408 T....
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This note was uploaded on 04/03/2011 for the course PHYS 101 taught by Professor Smith during the Spring '11 term at Arkansas Tech.
 Spring '11
 Smith
 Physics

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