### Problem Statement: Alpha particles (charge = +2e, mass = 6.68 × 10^-27 kg) are accelerated in a cyclotron to a final orbit radius of 0.30 m. The magnetic field in the cyclotron is 0.80 T. The period of circular motion of the alpha particles is closest to: #### Options: A. 0.40 μs B. 0.25 μs C. 0.16 μs D. 0.49 μs #### Instruction: Click Save and Submit to save and submit. Click Save All Answers to save all answers. ### Explanation: This problem involves calculating the period of circular motion of alpha particles in a cyclotron. To find the period, you can use principles from cyclotron physics and electromagnetism. --- ### Solution Outline: The period \( T \) of a particle in a cyclotron can be found using the formula: \[ T = \frac{2 \pi m}{qB} \] where: - \( m \) is the mass of the alpha particle. - \( q \) is the charge of the alpha particle. - \( B \) is the magnetic field strength. Let's calculate the period using the given values: - \( m = 6.68 × 10^{-27} \) kg - \( q = +2e \) (For alpha particles, \( e = 1.6 × 10^{-19} \) C, so \( q = 2 \times 1.6 \times 10^{-19} \) C) - \( B = 0.80 \) T Substitute these values into the formula to solve for \( T \). --- This type of problem is typical of those found in advanced physics courses dealing with particle accelerators and electromagnetism. Understanding how to manipulate these formulas and use the given constants is crucial for solving such problems effectively.
### Problem Statement: Alpha particles (charge = +2e, mass = 6.68 × 10^-27 kg) are accelerated in a cyclotron to a final orbit radius of 0.30 m. The magnetic field in the cyclotron is 0.80 T. The period of circular motion of the alpha particles is closest to: #### Options: A. 0.40 μs B. 0.25 μs C. 0.16 μs D. 0.49 μs #### Instruction: Click Save and Submit to save and submit. Click Save All Answers to save all answers. ### Explanation: This problem involves calculating the period of circular motion of alpha particles in a cyclotron. To find the period, you can use principles from cyclotron physics and electromagnetism. --- ### Solution Outline: The period \( T \) of a particle in a cyclotron can be found using the formula: \[ T = \frac{2 \pi m}{qB} \] where: - \( m \) is the mass of the alpha particle. - \( q \) is the charge of the alpha particle. - \( B \) is the magnetic field strength. Let's calculate the period using the given values: - \( m = 6.68 × 10^{-27} \) kg - \( q = +2e \) (For alpha particles, \( e = 1.6 × 10^{-19} \) C, so \( q = 2 \times 1.6 \times 10^{-19} \) C) - \( B = 0.80 \) T Substitute these values into the formula to solve for \( T \). --- This type of problem is typical of those found in advanced physics courses dealing with particle accelerators and electromagnetism. Understanding how to manipulate these formulas and use the given constants is crucial for solving such problems effectively.
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:
Alpha particles (charge = +2e, mass = 6.68 × 10^-27 kg) are accelerated in a cyclotron to a final orbit radius of 0.30 m. The magnetic field in the cyclotron is 0.80 T. The period of circular motion of the alpha particles is closest to:
#### Options:
A. 0.40 μs
B. 0.25 μs
C. 0.16 μs
D. 0.49 μs
#### Instruction:
Click Save and Submit to save and submit. Click Save All Answers to save all answers.
### Explanation:
This problem involves calculating the period of circular motion of alpha particles in a cyclotron. To find the period, you can use principles from cyclotron physics and electromagnetism.
---
### Solution Outline:
The period \( T \) of a particle in a cyclotron can be found using the formula:
\[ T = \frac{2 \pi m}{qB} \]
where:
- \( m \) is the mass of the alpha particle.
- \( q \) is the charge of the alpha particle.
- \( B \) is the magnetic field strength.
Let's calculate the period using the given values:
- \( m = 6.68 × 10^{-27} \) kg
- \( q = +2e \) (For alpha particles, \( e = 1.6 × 10^{-19} \) C, so \( q = 2 \times 1.6 \times 10^{-19} \) C)
- \( B = 0.80 \) T
Substitute these values into the formula to solve for \( T \).
---
This type of problem is typical of those found in advanced physics courses dealing with particle accelerators and electromagnetism. Understanding how to manipulate these formulas and use the given constants is crucial for solving such problems effectively.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8689a897-6f19-4f8c-bce2-bc1e854b9491%2F05b0d8f6-c98e-4ede-947a-a2188913ed84%2Fmtlt8mp.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
Alpha particles (charge = +2e, mass = 6.68 × 10^-27 kg) are accelerated in a cyclotron to a final orbit radius of 0.30 m. The magnetic field in the cyclotron is 0.80 T. The period of circular motion of the alpha particles is closest to:
#### Options:
A. 0.40 μs
B. 0.25 μs
C. 0.16 μs
D. 0.49 μs
#### Instruction:
Click Save and Submit to save and submit. Click Save All Answers to save all answers.
### Explanation:
This problem involves calculating the period of circular motion of alpha particles in a cyclotron. To find the period, you can use principles from cyclotron physics and electromagnetism.
---
### Solution Outline:
The period \( T \) of a particle in a cyclotron can be found using the formula:
\[ T = \frac{2 \pi m}{qB} \]
where:
- \( m \) is the mass of the alpha particle.
- \( q \) is the charge of the alpha particle.
- \( B \) is the magnetic field strength.
Let's calculate the period using the given values:
- \( m = 6.68 × 10^{-27} \) kg
- \( q = +2e \) (For alpha particles, \( e = 1.6 × 10^{-19} \) C, so \( q = 2 \times 1.6 \times 10^{-19} \) C)
- \( B = 0.80 \) T
Substitute these values into the formula to solve for \( T \).
---
This type of problem is typical of those found in advanced physics courses dealing with particle accelerators and electromagnetism. Understanding how to manipulate these formulas and use the given constants is crucial for solving such problems effectively.
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