2. Starting with drawing a Free Body Diagram of a charge moving in a uniform magnetic field, determine the radius of motion as it moves through the magnetic field. Then, determine the magnetic field of a moving electron with velocity v = 1.30 × 106 m/s with radius 0.350 m. Show all your work.
2. Starting with drawing a Free Body Diagram of a charge moving in a uniform magnetic field, determine the radius of motion as it moves through the magnetic field. Then, determine the magnetic field of a moving electron with velocity v = 1.30 × 106 m/s with radius 0.350 m. Show all your work.
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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![### Problem 2
#### Problem Statement:
Starting with drawing a Free Body Diagram of a charge moving in a uniform magnetic field, determine the radius of motion as it moves through the magnetic field. Then, determine the magnetic field of a moving electron with velocity \( v = 1.30 \times 10^6 \, \text{m/s} \) with a radius of \( 0.350 \, \text{m} \). Show all your work.
#### Solution:
1. **Free Body Diagram**:
- First, draw a diagram showing a charged particle moving in a circular path due to the force exerted by the magnetic field. The magnetic force acts perpendicular to both the velocity of the particle and the magnetic field, resulting in a centripetal force that keeps the particle in circular motion.
2. **Equations and Explanation**:
- The centripetal force needed to keep a particle with charge \( q \) and mass \( m \) moving at a velocity \( v \) in a circular path of radius \( r \) is provided by the magnetic force. The magnetic force \( F_B \) is given by:
\[
F_B = qvB
\]
Where \( B \) is the magnetic field strength.
- The centripetal force \( F_c \) is given by:
\[
F_c = \frac{mv^2}{r}
\]
- Setting the magnetic force equal to the centripetal force:
\[
qvB = \frac{mv^2}{r}
\]
- Solving for the radius \( r \) of the circular path:
\[
r = \frac{mv}{qB}
\]
3. **Given Data**:
- For an electron, the charge \( q = 1.602 \times 10^{-19} \, \text{C} \) (Coulombs), and the mass \( m = 9.109 \times 10^{-31} \, \text{kg} \). Given \( v = 1.30 \times 10^6 \, \text{m/s} \) and \( r = 0.350 \, \text{m} \).
4. **Determining the Magnetic Field**:
- Rearrange the radius equation to solve for the magnetic field \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe1fe53d7-fd6e-4de2-afe9-e4e8773c1dc3%2F7b90930e-cb29-4bac-9f3a-66d7db7355f9%2Frkyylfc_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 2
#### Problem Statement:
Starting with drawing a Free Body Diagram of a charge moving in a uniform magnetic field, determine the radius of motion as it moves through the magnetic field. Then, determine the magnetic field of a moving electron with velocity \( v = 1.30 \times 10^6 \, \text{m/s} \) with a radius of \( 0.350 \, \text{m} \). Show all your work.
#### Solution:
1. **Free Body Diagram**:
- First, draw a diagram showing a charged particle moving in a circular path due to the force exerted by the magnetic field. The magnetic force acts perpendicular to both the velocity of the particle and the magnetic field, resulting in a centripetal force that keeps the particle in circular motion.
2. **Equations and Explanation**:
- The centripetal force needed to keep a particle with charge \( q \) and mass \( m \) moving at a velocity \( v \) in a circular path of radius \( r \) is provided by the magnetic force. The magnetic force \( F_B \) is given by:
\[
F_B = qvB
\]
Where \( B \) is the magnetic field strength.
- The centripetal force \( F_c \) is given by:
\[
F_c = \frac{mv^2}{r}
\]
- Setting the magnetic force equal to the centripetal force:
\[
qvB = \frac{mv^2}{r}
\]
- Solving for the radius \( r \) of the circular path:
\[
r = \frac{mv}{qB}
\]
3. **Given Data**:
- For an electron, the charge \( q = 1.602 \times 10^{-19} \, \text{C} \) (Coulombs), and the mass \( m = 9.109 \times 10^{-31} \, \text{kg} \). Given \( v = 1.30 \times 10^6 \, \text{m/s} \) and \( r = 0.350 \, \text{m} \).
4. **Determining the Magnetic Field**:
- Rearrange the radius equation to solve for the magnetic field \(
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