Homework Help is Here – Start Your Trial Now!
Science
Physics
Tutorial Exercise A cosmic-ray proton in interstellar space has an energy of 16.8 MeV and executes a circular orbit having a radius equal to that of Mercury's orbit around the Sun (5.80 × 1010 m). What is the magnetic field in that region of space? Part 1 of 3 - Conceptualize A very large orbit radius indicates that the magnetic field in this interstellar space is a very small fraction of a tesla. Part 2 of 3 - Categorize When the proton, with a charge of the same magnitude as the electron charge, accelerates through a potential difference given in volts, the kinetic energy it receives is commonly expressed in electron volts. We will use the energy version of the isolated system model applied to the particle and the electric field that accelerates it from rest to allow us to find an expression for the speed of the particle. Then we will use the particle in a magnetic field model and the particle in uniform circular motion model to find the magnetic field. Part 3 of 3 - Analyze By conservation of energy for the system of proton and electric field, the kinetic energy of the proton is given by E=1m² = eAV, so solving for the speed, we have the following. v = 2eAV m By Newton's second law, for the circular motion of the proton in the magnetic field, we have mv2 R = evB sin 0, where R is the radius of the circular orbit and is 90°. Substituting the expression for v that we found above from the kinetic energy and solving for B, we have the following. B = mv eR eR m 2eAV 12mAV m R 2(1.67 x 10-27 kg) 107 V) 5.80 x 1010 m (1.6 x 10-19 c) x Your response differs significantly from the correct answer. Rework your solution from the beginning and

Tutorial Exercise A cosmic-ray proton in interstellar space has an energy of 16.8 MeV and executes a circular orbit having a radius equal to that of Mercury's orbit around the Sun (5.80 × 1010 m). What is the magnetic field in that region of space? Part 1 of 3 - Conceptualize A very large orbit radius indicates that the magnetic field in this interstellar space is a very small fraction of a tesla. Part 2 of 3 - Categorize When the proton, with a charge of the same magnitude as the electron charge, accelerates through a potential difference given in volts, the kinetic energy it receives is commonly expressed in electron volts. We will use the energy version of the isolated system model applied to the particle and the electric field that accelerates it from rest to allow us to find an expression for the speed of the particle. Then we will use the particle in a magnetic field model and the particle in uniform circular motion model to find the magnetic field. Part 3 of 3 - Analyze By conservation of energy for the system of proton and electric field, the kinetic energy of the proton is given by E=1m² = eAV, so solving for the speed, we have the following. v = 2eAV m By Newton's second law, for the circular motion of the proton in the magnetic field, we have mv2 R = evB sin 0, where R is the radius of the circular orbit and is 90°. Substituting the expression for v that we found above from the kinetic energy and solving for B, we have the following. B = mv eR eR m 2eAV 12mAV m R 2(1.67 x 10-27 kg) 107 V) 5.80 x 1010 m (1.6 x 10-19 c) x Your response differs significantly from the correct answer. Rework your solution from the beginning and

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
See similar textbooks
icon
Related questions
Question
None
Tutorial Exercise A cosmic-ray proton in interstellar space has an energy of 16.8 MeV and executes a circular orbit having a radius equal to that of Mercury's orbit around the Sun (5.80 × 1010 m). What is the magnetic field in that region of space? Part 1 of 3 - Conceptualize A very large orbit radius indicates that the magnetic field in this interstellar space is a very small fraction of a tesla. Part 2 of 3 - Categorize When the proton, with a charge of the same magnitude as the electron charge, accelerates through a potential difference given in volts, the kinetic energy it receives is commonly expressed in electron volts. We will use the energy version of the isolated system model applied to the particle and the electric field that accelerates it from rest to allow us to find an expression for the speed of the particle. Then we will use the particle in a magnetic field model and the particle in uniform circular motion model to find the magnetic field. Part 3 of 3 - Analyze By conservation of energy for the system of proton and electric field, the kinetic energy of the proton is given by E=1m² = eAV, so solving for the speed, we have the following. v = 2eAV m By Newton's second law, for the circular motion of the proton in the magnetic field, we have mv2 R = evB sin 0, where R is the radius of the circular orbit and is 90°. Substituting the expression for v that we found above from the kinetic energy and solving for B, we have the following. B = mv eR eR m 2eAV 12mAV m R 2(1.67 x 10-27 kg) 107 V) 5.80 x 1010 m (1.6 x 10-19 c) x Your response differs significantly from the correct answer. Rework your solution from the beginning and
Transcribed Image Text:Tutorial Exercise A cosmic-ray proton in interstellar space has an energy of 16.8 MeV and executes a circular orbit having a radius equal to that of Mercury's orbit around the Sun (5.80 × 1010 m). What is the magnetic field in that region of space? Part 1 of 3 - Conceptualize A very large orbit radius indicates that the magnetic field in this interstellar space is a very small fraction of a tesla. Part 2 of 3 - Categorize When the proton, with a charge of the same magnitude as the electron charge, accelerates through a potential difference given in volts, the kinetic energy it receives is commonly expressed in electron volts. We will use the energy version of the isolated system model applied to the particle and the electric field that accelerates it from rest to allow us to find an expression for the speed of the particle. Then we will use the particle in a magnetic field model and the particle in uniform circular motion model to find the magnetic field. Part 3 of 3 - Analyze By conservation of energy for the system of proton and electric field, the kinetic energy of the proton is given by E=1m² = eAV, so solving for the speed, we have the following. v = 2eAV m By Newton's second law, for the circular motion of the proton in the magnetic field, we have mv2 R = evB sin 0, where R is the radius of the circular orbit and is 90°. Substituting the expression for v that we found above from the kinetic energy and solving for B, we have the following. B = mv eR eR m 2eAV 12mAV m R 2(1.67 x 10-27 kg) 107 V) 5.80 x 1010 m (1.6 x 10-19 c) x Your response differs significantly from the correct answer. Rework your solution from the beginning and
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON
SEE MORE TEXTBOOKS