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A cosmic-ray electron moves at 7.5×106 m/s7.5 \times 10^6 \mathrm{~m} / \mathrm{s} perpendicular to Earth's magnetic field at an altitude where the field strength is 1.0×105 T1.0 \times 10^{-5} \mathrm{~T}. What is the radius of the circular path the electron follows?

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Answered 2 years ago
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If we're given the electron's speed and the Earth's magnetic field:

v=7.5106m/sB=1.0105Tv=7.5\cdot10^6\,\text{m/s}\\ B=1.0\cdot10^{-5}\,\text{T}

we need to calculate the radius of the circular path of the electron's movement, assuming the velocity is always perpendicular to the field. We know the electron's charge value is:

q=1.6021019Cq=-1.602\cdot10^{-19}\,\text{C}

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