An electron enters a magnetic field of 0.75 T with a velocity perpendicular to the direction of the field. At what frequency does the electron traverse a circular path? (m_el = 9.11 × 10^-31 kg, e = 1.60 × 10^-19C) A) 2.1 × 10¹⁰ Hz B) 4.8 × 10^-7 Hz C) 2.1 × 10¹⁴ Hz D) 4.8 × 10^-11 Hz

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The correct answer is A because the frequency of the electron's circular path is given by the equation f = eB / 2πm, where e is the charge of the electron, B is the magnetic field strength, and m is the mass of the electron. Plugging in the given values, we get f = (1.60 × 10^-19 C) * (0.75 T) / (2π * 9.11 × 10^-31 kg) = 2.1 × 10^10 Hz.

An Electron Enters a Magnetic Field of 0