This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 electrons 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 , = 17 . 7 , and B = 0 . 304 T . F = q v B sin = (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 protons 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....
View Full
Document
 Spring '11
 Smith
 Physics

Click to edit the document details