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

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Quizplus · Accepted Answer

The force on a charged particle moving in a magnetic field is given by the Lorentz force equation:$$F = qvBsinθ$$where:* q is the charge of the particle* v is the velocity of the particle* B is the magnetic field strength* θ is the angle between the velocity vector and the magnetic field vectorIn this case, the electron has a charge of -1.60 × 10^-19 C, a velocity 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.The angle between the velocity vector and the magnetic field vector is 45°, so:$$sinθ = sin45° = 0.707$$Plugging these values into the Lorentz force equation, we get:$$F = (-1.60 × 10^-19 C)(4.0 × 10^5 m/s)(5.0 × 10^-5 T)(0.707)$$$$F = -2.3 × 10^-18 N$$The magnitude of the force is 2.3 × 10^-18 N, so the correct answer is A.

An Electron Traveling Toward the North with Speed 4