To determine The magnetic force on an electron moving straight up in the given magnetic field. Explanation Given that the velocity of the electron is 3.0 × 10 6 m/s up , the strength of Earth’s magnetic field is 0.050 mT , the field direction is 70.0 ° below the horizontal. Write the expression for the magnetic force on a charged particle in a field. F → = q v → × B → (I) Here, F → is the magnetic force, q is the magnitude of charge, v → is the velocity of the particle, and B → is the magnetic field. Since the electron is moving straight up, the angle between the velocity vector and the magnetic field vector is 70.0 ° + 90.0 ° = 160.0 ° . Write the expression for the magnitude of F → . F = q v B sin θ (II) Here, θ is the angle between the velocity of the particle and the magnetic field. The direction of the magnetic force on the charged particle can be determined using the right-hand rule and the sign of the charge. Conclusion: Substitute 1.602 × 10 − 19 C for q , 3.0 × 10 6 m/s for v , 0

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ISBN: 9781260487008

Author: GIAMBATTISTA, Alan

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 19.2, Problem 19.2PP

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The magnetic force on an electron moving straight up in the given magnetic field.

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