A cosmic-ray electron moves at $5.8 \times 10^6$ m/s perpendicular to Earth's magnetic field at an altitude where the field strength is $2 \times 10^{-5}$ T. What is the radius of the circular path the electron follows? r = m
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The formula is given by: r = (m*v) / (q*B) where: r is the radius of the circular path m is the mass of the charged particle v is the velocity of the charged particle q is the charge of the charged particle B is the magnetic field strength Show more…
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