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 E)0N

Question

Quizplus · Accepted Answer

The magnitude of the magnetic force on a charged particle moving perpendicular to a magnetic field is given by the equation F = qvB, where F is the force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field strength. In this case, the particle is an electron with a charge of -1.60 × 10^-19 C, a velocity of 3.0 × 10^14 m/s, and a magnetic field strength of 0.40 T. Plugging these values into the formula, we get:
F = (-1.60 × 10^-19 C)(3.0 × 10^14 m/s)(0.40 T) = 1.92 × 10^-15 N
Therefore, the magnitude of the magnetic force on the electron is 1.9 × 10^-15 N, or choice B.

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