12 - roofner(bar784 Homework 12 Weathers(22202 This...

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roofner (bar784) – Homework 12 – Weathers – (22202) 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 10.0 points What is the radius of the smallest possible cir- cular orbit that a proton with energy 4 . 8 MeV can have in a 2 T magnetic field? The mass of a proton is 1 . 67 × 10 - 27 kg and its charge is 1 . 609 × 10 - 19 C. Correct answer: 0 . 157829 m. Explanation: Let : m = 1 . 67 × 10 - 27 kg , Q = 1 . 609 × 10 - 19 C , B = 2 T , and E = 4 . 8 MeV . The energy of a proton is E = 1 2 mv 2 , so v = radicalbigg 2 E m = radicalBigg 2 (4 . 8 MeV) (1 . 609 × 10 - 19 J / eV) (1 × 10 - 6 MeV / eV) (1 . 67 × 10 - 27 kg) = 3 . 04127 × 10 7 m / s . If the proton is shot into the magnetic field with a velocity at right angles to the direction of the field, we will get the smallest radius r = mv B q = (1 . 67 × 10 - 27 kg) (3 . 04127 × 10 7 m / s) (2 T) (1 . 609 × 10 - 19 C) = 0 . 157829 m . 002 10.0 points A singly charged positive ion has a mass of 1 . 28 × 10 - 26 kg. After being accelerated through a potential difference of 227 V, the ion enters a magnetic field of 0 . 564 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field. Correct answer: 1 . 06782 cm. Explanation: Let : m = 1 . 28 × 10 - 26 kg , q = 1 . 60218 × 10 - 19 C , V = 227 V , and B = 0 . 564 T . We use conservation of energy to find the velocity of the ion upon entering the field, 1 2 mv 2 = qV , v = radicalbigg 2 q V m = radicalBigg 2 (1 . 60218 × 10 - 19 C) (227 V) 1 . 28 × 10 - 26 kg = 75383 . 8 m / s . Therefore, the radius is r = mv q B = (1 . 28 × 10 - 26 kg) (75383 . 8 m / s) (1 . 60218 × 10 - 19 C) (0 . 564 T) = 0 . 0106782 m = 1 . 06782 cm . 003 (part 1 of 2) 10.0 points A cyclotron designed to accelerate protons has a magnetic field of magnitude 0 . 15 T over a region of radius 7 . 4 m. The charge on a proton is 1 . 60218 × 10 - 19 C and its mass is 1 . 67262 × 10 - 27 kg. What is the cyclotron frequency? Correct answer: 1 . 43682 × 10 7 rad / s. Explanation: Let : q = 1 . 60218 × 10 - 19 C , m = 1 . 67262 × 10 - 27 kg , and B = 0 . 15 T .
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roofner (bar784) – Homework 12 – Weathers – (22202) 2 The centripetal force is q v B = mv 2 r , so the cyclotron frequency is ω = v r = qB m = (1 . 60218 × 10 - 19 C) (0 . 15 T) 1 . 67262 × 10 - 27 kg = 1 . 43682 × 10 7 rad / s .
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