/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 97 A proton synchrotron accelerates... [FREE SOLUTION] | 91Ó°ÊÓ

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A proton synchrotron accelerates protons to a kinetic energy of \(500 \mathrm{GeV}\). At this energy, calculate (a) the Lorentz factor, (b) the speed parameter, and (c) the magnetic field for which the proton orbit has a radius of curvature of \(750 \mathrm{~m}\).

Short Answer

Expert verified
(a) \( \gamma \approx 533.4 \); (b) \( \beta \approx 0.9999991 \); (c) \( B \approx 1.23 \text{ T} \).

Step by step solution

01

Calculate the Lorentz Factor (γ)

The Lorentz factor \( \gamma \) is defined as \( \gamma = \frac{E}{mc^2} \), where \( E \) is the total energy, and \( mc^2 \) is the rest energy of the proton. The rest energy of a proton is approximately 0.938 GeV. Thus, \( \gamma = \frac{500 + 0.938}{0.938} \approx 533.4 \).
02

Calculate the Speed Parameter (β)

The speed parameter \( \beta \) is defined as \( \beta = \frac{v}{c} = \sqrt{1 - \frac{1}{\gamma^2}} \). Substituting the value of \( \gamma \) from Step 1 gives \( \beta = \sqrt{1 - \frac{1}{533.4^2}} \approx 0.9999991 \).
03

Calculate the Magnetic Field (B)

The radius \( R \) of the proton orbit is related to the magnetic field \( B \) by the equation \( R = \frac{\gamma m v}{qB} \). Solving for \( B \), we have \( B = \frac{\gamma m v}{qR} \). Here, \( m \) is the rest mass of the proton, \( v \approx c \) (since \( \beta \approx 1 \)), \( q \) is the charge of the proton \( (1.6 \times 10^{-19} \text{ C}) \), and \( R = 750 \text{ m} \). Substituting \( \gamma = 533.4 \), mass \( m = 1.67 \times 10^{-27} \text{ kg} \), we find \( B \approx 1.23 \text{ T} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lorentz Factor
The Lorentz factor, often represented by the Greek letter \( \gamma \), plays a crucial role in the theory of relativity. It describes the effects of time dilation, length contraction, and relativistic mass increase that occur when an object moves close to the speed of light. For a proton being accelerated in a synchrotron to a high kinetic energy, calculating the Lorentz factor can help us understand how these relativistic effects come into play. In the provided exercise, the calculation begins by considering the total energy of the proton, which is the sum of its kinetic and rest energy. The rest energy of a proton is given by \( 0.938 \,\text{GeV} \). The Lorentz factor is then calculated using the formula: \[ \gamma = \frac{E}{m c^2} \]where \( E \) is the total energy (\( 500 + 0.938 \,\text{GeV} \)) and \( m c^2 \) is the rest energy. By plugging these values into the equation, you find \( \gamma \approx 533.4 \), indicating significant relativistic effects at such high energies.
Kinetic Energy
Kinetic energy is the energy that a body possesses due to its motion. In particle physics, especially in accelerators like proton synchrotrons, understanding kinetic energy is essential as particles are accelerated to incredibly high speeds. Protons in synchrotrons are brought to immense kinetic energies on the scale of gigaelectronvolts (\( \text{GeV} \)). This energy allows them to reach velocities close to the speed of light. By using that the Total energy \( E \) = Kinetic Energy + Rest Energy \( mc^2 \)we can identify how much energy is needed beyond rest energy to achieve such speeds. For instance, in this exercise, the protons are accelerated to \( 500 \text{GeV} \), which is a clear indicator of how the majority of their energy comes from the kinetic aspect, especially considering their rest energy of \( 0.938 \, \text{GeV} \).
Magnetic Field Calculation
When dealing with charged particles moving in a magnetic field, such as protons in a synchrotron, knowing how to calculate the necessary magnetic field strength is vital for precise control. The magnetic field ensures that particles maintain their desired trajectory within the accelerator. The relationship between the magnetic field \( B \), the radius of curvature \( R \), and the Lorentz factor \( \gamma \) is given by:\[ B = \frac{\gamma m v}{q R} \]where \( m \) is the rest mass of the proton, \( v \) is their velocity, and \( q \) is their charge. In this exercise, the radius is provided as \( 750 \, \text{m} \), and the calculation leads us to find \( B \approx 1.23 \, \text{T} \) (tesla), which is a strong magnetic field necessary to bend the paths of high-energy protons.
Proton Synchrotron
A proton synchrotron is a type of particle accelerator designed to accelerate protons to high energies by using magnetic fields and RF cavities. It's a circular accelerator that can increase the kinetic energy of protons to levels required for experimental particle physics, like colliding them at near-light speeds. As protons are guided along a circular path, the synchrotron's magnetic field ensures they move correctly along their designated route. The synchrotron must be accurately calibrated to account for relativistic effects, as protons gain significant energy. This exercise illustrates the concepts behind such calculations, showing how adjustments in magnetic field strength and understanding particle energy can fine-tune a synchrotron's performance.
Speed Parameter
The speed parameter, represented by \( \beta \), is defined as the ratio of an object's velocity \( v \) to the speed of light \( c \). In particle physics, this parameter is crucial for understanding the proportion of the speed at which particles are moving compared to the ultimate speed limit—light. Calculating \( \beta \) reveals just how close to the speed of light the accelerated particles are traveling. This exercise uses the relationship:\[ \beta = \sqrt{1 - \frac{1}{\gamma^2}} \]Given the Lorentz factor \( \gamma \), you can find \( \beta \approx 0.9999991 \), indicating that the protons are almost reaching the speed of light at these high energy levels. Such high speeds are essential for the particles to experience the conditions necessary for experimental observations in particle physics.

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