Homework 9-solutions

# Homework 9-solutions - im(gi768 Homework 9 markert(55325...

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im (gi768) – Homework 9 – markert – (55325) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of6)10.0points A particle of mass 7 . 138 × 10 26 kg and charge of magnitude 1 . 6 × 10 19 C is acceler- ated from rest in the plane of the page through a potential difference of 369 V between two parallel plates as shown. The particle is in- jected through a hole in the right-hand plate into a region of space containing a uniform magnetic field of magnitude 0 . 0528 T ori- ented perpendicular to the plane of the page. The particle curves in a semicircular path and strikes a detector. q m Region of Magnetic Field B E hole What is the sign of the charge of the parti- cle? Neglect relativistic effects. 1. The charge q is positive (+) . correct 2. The charge q is negative ( ) . 3. The charge q cannot be determined. Explanation: The charge accelerates toward the negative plate and away from the positive plate, so the charge is positive. 002(part2of6)10.0points Which way does the magnetic field point? 1. Into the page 2. Cannot be determined 3. Toward the bottom of page 4. Toward the top of page 5. Out of the page correct 6. To the left 7. To the right Explanation: B + q m E hole + + + + Because the particle curves down, the di- rection of vectorv × vector B is down. By the right-hand rule, vector B points out of the page. 003(part3of6)10.0points What is the speed of the charged particle as it enters the region of the magnetic field? Correct answer: 40672 . 4 m / s. Explanation: Let : m = 7 . 138 × 10 26 kg , V = 369 V , and | q | = 1 . 6 × 10 19 C . The change in kinetic energy of the charged particle is equal to the work done on it by the potential difference: 1 2 m v 2 = q V v = radicalbigg 2 q V m = radicalBigg 2 (1 . 6 × 10 19 C) (369 V) 7 . 138 × 10 26 kg = 40672 . 4 m / s . 004(part4of6)10.0points

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im (gi768) – Homework 9 – markert – (55325) 2 What is the magnitude of the force exerted on the charged particle as it enters the region of the magnetic field vector B ? Correct answer: 3 . 43601 × 10 16 N. Explanation: Let : B = 0 . 0528 T . The force on the particle is given by the Lorentz force law vector F = qvectorv × vector B bardbl vector F bardbl = q v B = (1 . 6 × 10 19 C) (40672 . 4 m / s) (0 . 0528 T) = 3 . 43601 × 10 16 N . 005(part5of6)10.0points What is the distance from the point of injec- tion to the detector? Correct answer: 0 . 68731. Explanation: The acceleration of the particle will be cen- tripetal: m v 2 r = q v B r = m v q B = (7 . 138 × 10 26 kg) (40672 . 4 m / s) (1 . 6 × 10 19 C) (0 . 0528 T) = 0 . 343655 m , and the distance from the point of injection to the detector is 2 r = 2 (0 . 343655 m) = 0 . 68731 m . 006(part6of6)10.0points What is the work done by the magnetic field on the charged particle during the semicircu- lar trip? 1. W = 1 . 1808 × 10 16 J 2. W = 2 . 952 × 10 17 J 3. W = 5 . 904 × 10 17 J 4. W = 0 J correct 5. W = 2 . 952 × 10 17 J 6. W = 3 . 7586 × 10 17 J 7. W = 3 . 7586 × 10 17 J 8. W = 1 . 1808 × 10 16 J 9. W = 5 . 904 × 10 17 J Explanation: W = vector F · vector d .
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