An electron traveling toward the north with speed 4.0 × 10⁵ m/s enters a region where the Earth's magnetic field has the magnitude 5.0 × 10^-5 T and is directed downward at 45° below horizontal. What is the magnitude of the force that the Earth's magnetic field exerts on the electron? (e = 1.60 × 10^-19C) A) 2.3 × 10^-18 N B) 3.2 × 10^-18 N C) 2.3 × 10^-19 N D) 3.2 × 10^-19 N E) 2.3 × 10^-20 N

Question

Quizplus · Accepted Answer

The magnetic force on a charged particle is given by F = qvBsinθ. Here, q = 1.60 × 10^-19 C, v = 4.0 × 10^5 m/s, B = 5.0 × 10^-5 T, and θ = 45°. Thus, F = (1.60 × 10^-19 C)(4.0 × 10^5 m/s)(5.0 × 10^-5 T)sin(45°) = 2.3 × 10^-18 N.

An Electron Traveling Toward the North with Speed 4