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

Question

Quizplus · Accepted Answer

The force on a charged particle moving in a magnetic field is given by the equation F = qvB sin θ, where q is the charge of the particle, v is its speed, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector. In this case, the electron has a charge of -1.60 × 10^-19 C, a speed of 4.0 × 10^5 m/s, and the magnetic field has a strength of 5.0 × 10^-5 T and is directed downward at 45° below horizontal. Plugging these values into the equation, we get:F = (-1.60 × 10^-19 C)(4.0 × 10^5 m/s)(5.0 × 10^-5 T)sin 45° = 2.3 × 10^-18 NTherefore, the magnitude of the force that the Earth's magnetic field exerts on the electron is 2.3 × 10^-18 N.

An Electron Traveling Toward the North with Speed 4