An electron moves horizontally across the Earth’sequator at a speed of 2.50 x 106 m/s and in a direction 35.0° N of E. At this point, the Earth’s magnetic field has a direction due north, is parallel to the surface, and has a value of 3.00 x10-5 What is the force acting on the electron due to its interaction with the Earth’s magnetic field? (a) 6.88 x 10-18 N due west (b) 6.88 x 10-18 N toward the Earth’s surface (c) 9.83 x 10-18N toward the Earth’s surface (d) 9.83 x 10-18N away from the Earth’s surface (e) 4.00 x 10-18N away from the Earth’s surface.

speed of electron (v) = 2.50×106 m/sDirection of speed (θ) = 35o N of Emagnetic field B→ = 3 ×10-5 N…

Answered: An electron moves horizontally across…

College Physics

10th Edition

ISBN:9781285737027

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter19: Magnetism

Section: Chapter Questions

Problem 52P

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An electron moves horizontally across the Earth’sequator at a speed of 2.50 x 10⁶m/s and in a direction 35.0° N of E. At this point, the Earth’s magnetic field has a direction due north, is parallel to the surface, and has a value of 3.00 x10^-5 What is the force acting on the electron due to its interaction with the Earth’s magnetic field? (a) 6.88 x 10^-18 N due west (b) 6.88 x 10^-18 N toward the Earth’s surface (c) 9.83 x 10^-18N toward the Earth’s surface (d) 9.83 x 10^-18N away from the Earth’s surface (e) 4.00 x 10^-18N away from the Earth’s surface.

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