
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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![**Problem Statement:**
If the magnetic field of the wire is \(2.0 \times 10^{-4}\) T and the electron moves at \(9.0 \times 10^{6}\) m/s, what is the magnitude \( F \) of the force exerted on the electron?
Express your answer in newtons to two significant figures.
---
In this problem, you are asked to determine the force on an electron moving in a magnetic field. The relevant equation is the Lorentz force law, which is given by:
\[ F = qvB \sin(\theta) \]
where:
- \( F \) is the force,
- \( q \) is the charge of the electron (\(-1.6 \times 10^{-19}\) C),
- \( v \) is the velocity of the electron,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity and the magnetic field.
For this exercise, assume that the angle \(\theta\) is 90 degrees, meaning the velocity is perpendicular to the magnetic field, thus \(\sin(\theta) = 1\). Therefore, the force can be simplified to:
\[ F = qvB \]](https://content.bartleby.com/qna-images/question/9b0e7686-0c0d-4eba-ac81-54e736b47f2f/ab206ae6-5a20-4027-a34b-9d1e7ca99bd1/upmna9_thumbnail.jpeg)
Transcribed Image Text:**Problem Statement:**
If the magnetic field of the wire is \(2.0 \times 10^{-4}\) T and the electron moves at \(9.0 \times 10^{6}\) m/s, what is the magnitude \( F \) of the force exerted on the electron?
Express your answer in newtons to two significant figures.
---
In this problem, you are asked to determine the force on an electron moving in a magnetic field. The relevant equation is the Lorentz force law, which is given by:
\[ F = qvB \sin(\theta) \]
where:
- \( F \) is the force,
- \( q \) is the charge of the electron (\(-1.6 \times 10^{-19}\) C),
- \( v \) is the velocity of the electron,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity and the magnetic field.
For this exercise, assume that the angle \(\theta\) is 90 degrees, meaning the velocity is perpendicular to the magnetic field, thus \(\sin(\theta) = 1\). Therefore, the force can be simplified to:
\[ F = qvB \]
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