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A cosmic-ray proton in interstellar space has an energy of 21.8 MeV and executes a circular orbit having a radius equal to that of Mercury's orbit around the Sun (5.80 x 1010 m). What is the magnetic field in that region of space?

A cosmic-ray proton in interstellar space has an energy of 21.8 MeV and executes a circular orbit having a radius equal to that of Mercury's orbit around the Sun (5.80 x 1010 m). What is the magnetic field in that region of space?

University Physics Volume 1
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter1: Units And Measurement
Section: Chapter Questions
Problem 15P: Use the orders of magnitude you found in the previous problem to answer the following questions to...
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This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise A cosmic-ray proton in interstellar space has an energy of 21.8 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 = mv? = eAV, so solving for the speed, we have the following. 2eAV V = V By Newton's second law, for the circular motion of the proton in the magnetic field, we have mv2 = evB sin 0, R where R is the radius of the circular orbit and 0 is 90°. Substituting the expression for v that we found above from the kinetic energy and solving for B, we have the following. mv m 2eAV 1 2mAV В 3 eR %3D eR m e 2(1.67 x 10-27 kg x 107 v 5.80 x 1010 m) \| (1.6 x 10-19 c) x 10-12 T
Transcribed Image Text:This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise A cosmic-ray proton in interstellar space has an energy of 21.8 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 = mv? = eAV, so solving for the speed, we have the following. 2eAV V = V By Newton's second law, for the circular motion of the proton in the magnetic field, we have mv2 = evB sin 0, R where R is the radius of the circular orbit and 0 is 90°. Substituting the expression for v that we found above from the kinetic energy and solving for B, we have the following. mv m 2eAV 1 2mAV В 3 eR %3D eR m e 2(1.67 x 10-27 kg x 107 v 5.80 x 1010 m) \| (1.6 x 10-19 c) x 10-12 T
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