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A cosmic ray electron moves at $7.50 \times 10^{6} \mathrm{m} / \mathrm{s}$ perpendicular to the Earth's magnetic field at an altitude where field strength is $1.00 \times 10^{-5}$ T. What is the radius of the circular path the electron follows?

The circular path the electron follow has a path of 4.27 $\mathrm{m}$

Physics 102 Electricity and Magnetism

Chapter 22

Magnetism

Magnetic Field and Magnetic Forces

Sources of Magnetic field

Electromagnetic Induction

Inductance

University of Michigan - Ann Arbor

Simon Fraser University

Hope College

McMaster University

Lectures

08:42

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A magnetic field is a math…

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A cosmic-ray electron move…

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An electron moves in a cir…

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05:27

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A proton enters a $6.0 \ti…

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A cosmic-ray proton in int…

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An electron is moving at a…

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A proton moves at $7.50 \t…

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Assume that all charged pa…

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A beam of electrons moves …

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An electron moving at $4.0…

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Magnetic Fields in the Boh…

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An electron passes through…

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here. The radius of the circular path is gonna be equaling the mass of the particle times V the velocity developed by the charge multiplied by the magnetic field. This is gonna be equaling the mass of an electron 9.11 times 10 to the negative 31st kilograms multiplied by the velocity of 7.50 times 10 to the sixth meters per second. Divided by the charge on an electron. We know to be 1.60 times 10 to the negative 19 cool arms multiplied by be the magnitude of the magnetic field 1.0 times 10 to the negative fifth Tesla's. And we can then say that here the radius of the circular path for the electron would be 4.27 meters. This would be our final answer. That is the end of the solution. Thank you for

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