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

Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter22: Magnetic Forces And Magnetic Fields
Section: Chapter Questions
Problem 17P: The picture tube in an old black-and-white television uses magnetic deflection coils rather than...
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
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
Physics
ISBN:
9781133104261
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
College Physics
College Physics
Physics
ISBN:
9781938168000
Author:
Paul Peter Urone, Roger Hinrichs
Publisher:
OpenStax College
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
College Physics
College Physics
Physics
ISBN:
9781285737027
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics Volume 2
University Physics Volume 2
Physics
ISBN:
9781938168161
Author:
OpenStax
Publisher:
OpenStax
Glencoe Physics: Principles and Problems, Student…
Glencoe Physics: Principles and Problems, Student…
Physics
ISBN:
9780078807213
Author:
Paul W. Zitzewitz
Publisher:
Glencoe/McGraw-Hill
SEE MORE TEXTBOOKS