An electron moves with a speed of 3.0 × 10⁴ m/s perpendicular to a uniform magnetic field of 0.40 T.What is the magnitude of the magnetic force on the electron? (e = 1.60 × 10^-19C) A)4.8 × 10^-14 N B)1.9 × 10^-15 N C)5 × 10^-20 N D)2.2 × 10^-24 N

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Quizplus · Accepted Answer

The magnetic force on a charged particle moving perpendicular to a magnetic field is given by F = qvB, where q is the charge, v is the velocity, and B is the magnetic field. Plugging in the given values, we get F = (1.60 × 10-19 C)(3.0 × 1014 m/s)(0.40 T) = 1.9 × 10-15 N. Therefore, the correct answer is B.

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