1. An electron is moving near a long, current carrying wire with a speed of v = 1100 m/s. When the electron is at the point shown in the figure below (r = 0.15 m), what is the magnitude and direction of the magnetic force acting on the electron? The current in the wire is I = 500 A.
1. An electron is moving near a long, current carrying wire with a speed of v = 1100 m/s. When the electron is at the point shown in the figure below (r = 0.15 m), what is the magnitude and direction of the magnetic force acting on the electron? The current in the wire is I = 500 A.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Problem Statement
1. An electron is moving near a long, current-carrying wire with a speed of \(v = 1100 \, \text{m/s}\). When the electron is at the point shown in the figure below (\(r = 0.15 \, \text{m}\)), what is the magnitude and direction of the magnetic force acting on the electron? The current in the wire is \(I = 500 \, \text{A}\).
### Explanation
This problem involves determining the magnetic force acting on an electron moving near a long, straight current-carrying wire. To solve this, one can use the following steps:
1. **Magnetic Field due to a Long Straight Wire**:
The magnetic field \(B\) at a distance \(r\) from a long, straight conductor carrying a current \(I\) is given by:
\[
B = \frac{\mu_0 I}{2 \pi r}
\]
where \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \)).
2. **Magnetic Force on a Moving Charge**:
The magnetic force \(F\) acting on a charge \(q\) moving with velocity \(v\) in a magnetic field \(B\) is given by:
\[
F = q v B
\]
The direction of this force can be determined using the right-hand rule for electrons (considering the negative charge), but generally, you need to decide the specific vector form with respect to the orientation.
### Provided Data:
- Speed of electron, \( v = 1100 \, \text{m/s} \)
- Distance from wire, \( r = 0.15 \, \text{m} \)
- Current, \( I = 500 \, \text{A} \)
- Electron charge magnitude, \( q = 1.6 \times 10^{-19} \, \text{C} \)
### Solution Steps:
1. Calculate the magnetic field \(B\):
\[
B = \frac{\mu_0 I}{2 \pi r} = \frac{(4 \pi \times 10^{-7}) \times 500}{2 \pi \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F83bd02c6-d02f-4764-a848-15fcab9c3f4f%2F1ed8b599-ecc5-426a-ac32-af9081f429bd%2Fdvlu8y_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
1. An electron is moving near a long, current-carrying wire with a speed of \(v = 1100 \, \text{m/s}\). When the electron is at the point shown in the figure below (\(r = 0.15 \, \text{m}\)), what is the magnitude and direction of the magnetic force acting on the electron? The current in the wire is \(I = 500 \, \text{A}\).
### Explanation
This problem involves determining the magnetic force acting on an electron moving near a long, straight current-carrying wire. To solve this, one can use the following steps:
1. **Magnetic Field due to a Long Straight Wire**:
The magnetic field \(B\) at a distance \(r\) from a long, straight conductor carrying a current \(I\) is given by:
\[
B = \frac{\mu_0 I}{2 \pi r}
\]
where \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \)).
2. **Magnetic Force on a Moving Charge**:
The magnetic force \(F\) acting on a charge \(q\) moving with velocity \(v\) in a magnetic field \(B\) is given by:
\[
F = q v B
\]
The direction of this force can be determined using the right-hand rule for electrons (considering the negative charge), but generally, you need to decide the specific vector form with respect to the orientation.
### Provided Data:
- Speed of electron, \( v = 1100 \, \text{m/s} \)
- Distance from wire, \( r = 0.15 \, \text{m} \)
- Current, \( I = 500 \, \text{A} \)
- Electron charge magnitude, \( q = 1.6 \times 10^{-19} \, \text{C} \)
### Solution Steps:
1. Calculate the magnetic field \(B\):
\[
B = \frac{\mu_0 I}{2 \pi r} = \frac{(4 \pi \times 10^{-7}) \times 500}{2 \pi \times
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