Question

A cosmic ray electron moves at 7.50×106m/s7.50 \times 10^{6} \mathrm{m} / \mathrm{s} perpendicular to the Earth’s magnetic field at an altitude where field strength is 1.00×105T.1.00 \times 10^{-5} \mathrm{T}. What is the radius of the circular path the electron follows?

Solution

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Answered 10 months ago
Answered 10 months ago
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Strategy\textit{\textbf{Strategy}}:

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

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