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Question

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 $1.00 \times 10^{-5} \mathrm{T}.$ What is the radius of the circular path the electron follows?

Solution

VerifiedAnswered 10 months ago

Answered 10 months ago

Step 1

1 of 3$\textit{\textbf{Strategy}}$:

To find the circular path the electron follows, we will use the equation $r = \frac{mv}{qB}$ and solve it for $r$ where $m$ is the mass of the electron.

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