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Chapter 3: Problem 19

Synchrotron current * In a 6 gigaelectron-volt \(\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)\) electron synchrotron, electrons travel around the machine in an approximately circular path 240 meters long. It is normal to have about \(10^{11}\) electrons circling on this path during a cycle of acceleration. The speed of the electrons is practically that of light. What is the current? We give this very simple problem to emphasize that nothing in our definition of current as rate of transport requires the velocities of the charge carriers to be nonrelativistic and that there is no rule against a given charged particle getting counted many times during a second as part of the current.

Short Answer

Expert verified
To find the current, calculate the total time for the electrons to travel around the synchrotron using the given conditions, then find the total charge of the electrons and divide by the calculated time. Using the information provided, and carrying out the calculations will yield the answer in Ampere.

Step by step solution

01

Step 1

First, calculate the time it would take for the electrons to travel around the synchrotron (240 meters) using the formula \(t = d / v\), where \(d = 240m\) is the distance traveled by the electrons and \(v\) is the speed of the electrons, practically the speed of light (\(c = 3 \times 10^8 m/s\)). Therefore, \(t = 240 / 3 \times 10^8\). This will give the time in seconds.
02

Step 2

There are \(10^{11}\) electrons circling on this path during a cycle of acceleration. Each electron has a known charge (\(e = 1.6 \times 10^{-19} C/e^-\)). So, calculate the total charge of the \(10^{11}\) electrons as \(Q = 10^{11} \times 1.6 \times 10^{-19}\). This will give the total charge in Coulombs.
03

Step 3

Finally, use the total charge (\(Q\)) and the time (\(t\)) calculated in the previous steps to find the synchrotron current. Use the formula \(I=Q/t\). Substitute the values of \(Q\) and \(t\) in the formula to get the current in Ampere.

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Key Concepts

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

Electron Synchrotron
An electron synchrotron is a type of particle accelerator in which electrons are accelerated to high speeds in a circular path. These machines are crucial for providing electrons with the necessary energy for experiments in high-energy physics. Electron synchrotrons utilize magnetic fields to keep the particles moving along their circular path successfully.

The synchrotron allows electrons to reach speeds that are very close to that of light. The high speed is essential for various research purposes, including particle physics and material science. Circular paths ensure that this process can continue continuously, boosting the electrons' energy with each cycle.
  • Electrons are accelerated in a circular path.
  • The circumference of the path in our exercise is 240 meters.
  • This setup is critical for high-energy physics research.
Speed of Light
In the context of synchrotrons, the speed of light plays a significant role since electrons are accelerated to speeds very close to this universal constant. The speed of light, denoted as \( c \), is approximately \( 3 \times 10^8 \text{ m/s} \).

Reaching speeds near the speed of light has fascinating implications. It allows physicists to study relativistic effects that wouldn't be observable at lower speeds. In a practical sense, knowing the speed helps calculate different parameters related to the acceleration process, like the time taken for one complete cycle in the synchrotron.
  • The speed of light is crucial for calculating travel time in synchrotrons.
  • Allows the study of relativistic effects.
  • A universal constant relevant in many areas of physics.
Relativistic Current
Current in the relativistic sense, or synchrotron current, represents the rate at which charge carries move along the electron synchrotron path. The current doesn't necessarily require the velocities to be nonrelativistic鈥攔ather, it embraces high-speed conditions.

In our exercise, the electrons travel at speeds almost equal to that of light, which means they can circle the synchrotron path multiple times in a single second. The current calculation considers both the high speed and the volume of charge carriers, emphasizing the role of relativistic speeds.
  • Defined as the rate of electron travel in the synchrotron.
  • Accounts for multiple passes per second due to high velocity.
  • Current is not limited by nonrelativistic conditions.
Charge Carriers
Charge carriers in this scenario are electrons, as they flow through the synchrotron path. Electrons carry a fundamental charge of \( 1.6 \times 10^{-19} \text{ C/e}^- \), a constant that is key to understanding electric charge and current.

In accelerators like the synchrotron, a large number of electrons are grouped together to ensure that significant currents can be generated. In our particular exercise, \( 10^{11} \) electrons are in motion, contributing to the high current. Knowing the charge per electron allows calculation of the total charge, essential for finding the synchrotron current.
  • Electrons act as the charge carriers.
  • Contribute significantly to the current.
  • Fundamental charge is \( 1.6 \times 10^{-19} \text{ C} \).

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Most popular questions from this chapter

Maximum field from a passing charge In a colliding beam storage ring an antiproton going east passes a proton going west, the distance of closest approach being \(10^{-10} \mathrm{~m}\). The kinetic energy of each particle in the lab frame is \(93 \mathrm{GeV}\), corresponding to \(\gamma=100\). In the rest frame of the proton, what is the maximum intensity of the electric field at the proton due to the charge on the antiproton? For about how long, approximately, does the field exceed half its maximum intensity?

Average of six points Let \(\phi(x, y, z)\) be any function that can be expanded in a power series around a point \(\left(x_{0}, y_{0}, z_{0}\right)\). Write a Taylor series expansion for the value of \(\phi\) at each of the six points \(\left(x_{0}+\delta, y_{0}, z_{0}\right),\left(x_{0}-\right.\) \(\left.\delta, y_{0}, z_{0}\right),\left(x_{0}, y_{0}+\delta, z_{0}\right),\left(x_{0}, y_{0}-\delta, z_{0}\right),\left(x_{0}, y_{0}, z_{0}+\delta\right),\left(x_{0}, y_{0},\right.\), \(z_{0}-\delta\) ), which symmetrically surround the point \(\left(x_{0}, y_{0}, z_{0}\right)\) at a distance \(\delta\). Show that, if \(\phi\) satisfies Laplace's equation, the average of these six values is equal to \(\left(x_{0}, y_{0}, z_{0}\right)\) through terms of the third order in \(\delta\).

Force, and potential squared * (a) In Gaussian units, show that the square of a potential difference \(\left(\phi_{2}-\phi_{1}\right)^{2}\) has the same dimensions as force. (In SI units, \(\epsilon_{0}\left(\phi_{2}-\phi_{1}\right)^{2}\) has the same units as force.) This tells us that the electrostatic forces between bodies will largely be determined, as to order of magnitude, by the potential differences involved. Dimensions will enter only in ratios, and there may be some constants like \(4 \pi\). What is the order of magnitude of force youexpect with 1 statvolt potential difference between something and something else? Practically achievable potential differences are rather severely limited, for reasons having to do with the structure of matter. The highest man-made difference of electric potential is about \(10^{7}\) volts, achieved by a Van de Graaff electrostatic generator operating under high pressure. (Billion-volt accelerators do not involve potential differences that large.) How many pounds force are you likely to find associated with a "square megavolt"? These considerations may suggest why electrostatic motors have not found much application.

Spherical resistor (a) The region between two concentric spherical shells is filled with a material with resistivity \(\rho .\) The inner radius is \(r_{1}\), and the outer radius \(r_{2}\) is many times larger (essentially infinite). Show that the resistance between the shells is essentially equal to \(\rho / 4 \pi r_{1}\) (b) Without doing any calculations, dimensional analysis suggests that the above resistance should be proportional to \(\rho / r_{1}\), because \(\rho\) has units of ohm-meters and \(r_{1}\) has units of meters. But is this reasoning rigorous?

This exercise is more of a math problem than a physics problem, so maybe it doesn't belong in this book. But it's a fun one. Consider a series of events that happen at independent random times, such as the collisions in Section \(4.4\) that led to the issue discussed in Footnote \(8 .\) Such a process can be completely characterized by the probability per unit time (call it \(p\) ) of an event happening. The definition of \(p\) is that the probability of an event happening in an infinitesimal \(^{16}\) time \(d t\) equals \(p d t\). (a) Show that starting at any particular time (not necessarily the time of an event), the probability that the next event happens between \(t\) and \(t+d t\) later equals \(e^{-p t} p d t\). You can do this by breaking the time interval \(t\) into a large number of tiny intervals, and demanding that the event does not happen in any of them, but that it does happen in the following \(d t\). (You will need to use the fact that \((1-x / N)^{N}=e^{-x}\) in the \(N \rightarrow \infty\) limit.) Verify that the integral of \(e^{-p t} p d t\) correctly equals 1 . (b) Show that starting at any particular time (not necessarily the time of an event), the average waiting time (also called the expectation value of the waiting time) to the next event equals \(1 / p\). Explain why this is also the average time between events. (c) Pick a random point in time, and look at the length of the time interval (between successive events) that it belongs to. Explain, using the above results, why the average length of this interval is \(2 / p\), and not \(1 / p\). (d) We have found that the average time between events is \(1 / p\), and also that the average length of the interval surrounding a randomly chosen point in time is \(2 / p .\) Someone might think that these two results should be the same. Explain intuitively why they are not. (e) Using the above probability distribution \(e^{-p t} p d t\) properly, show mathematically why \(2 / p\) is the correct result for the average length of the interval surrounding a randomly chosen point in time.
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