An electron traveling due 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 force acts on the electron?

Question

Quizplus · Accepted Answer

The force on the electron is given by the Lorentz force:$$F = qvBsinθ$$where q is the charge of the electron, v is its velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.In this case, q = -1.6 × 10^-19 C, v = 4.0 × 10^5 m/s, B = 5.0 × 10^-5 T, and θ = 45°.Plugging these values into the equation, we get:$$F = (-1.6 × 10^-19 C)(4.0 × 10^5 m/s)(5.0 × 10^-5 T)sin45°$$$$F = -2.3 × 10^-18 N$$The negative sign indicates that the force is in the opposite direction of the electron's velocity, which is to the south.

An Electron Traveling Due North with Speed 4