Problem 47 - Physics for Scientists and Engineers with Modern Physics

Question

Assume there are \(5.0 \times 10^{13}\) protons at \(1.0 \mathrm{TeV}\) stored in the 1.0 -km-radius ring of the Tevatron. (a) How much current (amperes) is carried by this beam? (b) How fast would a \(1500-\mathrm{kg}\) car have to move to carry the same kinetic energy as this beam?

Step-by-Step Solution

Answer

(a) The beam carries 0.383 A. (b) The car must travel at 103.3 m/s.

Step 1: Determine the Charge of Protons

Each proton has a charge of approximately \(1.6 \times 10^{-19}\) coulombs. With \(5.0 \times 10^{13}\) protons, the total charge \(Q\) is calculated as:\[ Q = (5.0 \times 10^{13}) \times (1.6 \times 10^{-19}) \text{ C} \]\[ Q = 8.0 \times 10^{-6} \text{ C} \]

Step 2: Calculate the Current in the Beam

The beam circulates in a ring with a 1.0-km radius at nearly the speed of light, \(c \approx 3 \times 10^8 \text{ m/s}\). The time \(t\) for one circulation is given by \(t = \frac{2 \pi R}{c}\), where \(R = 1000 \text{ m}\) is the radius. Substituting the values:\[ t = \frac{2 \pi \times 1000}{3 \times 10^8} = 2.09 \times 10^{-5} \text{ s} \]The current \(I\) is the charge \(Q\) divided by the time \(t\):\[ I = \frac{8.0 \times 10^{-6}}{2.09 \times 10^{-5}} = 0.383 \text{ A} \]

Step 3: Determine the Total Kinetic Energy of the Protons

Each proton has a kinetic energy of \(1.0 \text{ TeV} = 1.0 \times 10^{12} \text{ eV}\). Converting this to joules (\(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\)) gives:\[ 1.0 \time 10^{12} \text{ eV} = 1.0 \times 10^{12} \times 1.6 \times 10^{-19} \text{ J} = 1.6 \times 10^{-7}\text{ J} \text{ per proton} \]Thus, the total kinetic energy \(E\) is:\[ E = (5.0 \times 10^{13}) \times (1.6 \times 10^{-7}) = 8.0 \times 10^{6} \text{ J} \]

Step 4: Calculate the Speed Required for a Car

The kinetic energy \(E\) of a moving object is given by \(E = \frac{1}{2}mv^2\), where \(m = 1500 \text{ kg}\) is the mass of the car and \(v\) is the speed. Solving for \(v\):\[ v = \sqrt{\frac{2E}{m}} \]Substituting the known values:\[ v = \sqrt{\frac{2 \times 8.0 \times 10^{6}}{1500}} = \sqrt{10666.67} \approx 103.3 \text{ m/s} \]

Key Concepts

Kinetic Energy Calculation

To comprehend the kinetic energy of a moving object, think of it as the energy that an object possesses due to its motion. Kinetic energy is calculated using the formula \(E = \frac{1}{2}mv^2\), where \(m\) is the mass, and \(v\) is the velocity of the object. The unit of kinetic energy in the International System is the joule (J).

In the context of the problem, each proton within the Tevatron has a kinetic energy of \(1.0 \, ext{TeV} = 1.0 \times 10^{12} \, ext{eV}\). We convert this to joules since 1 electronvolt (eV) equals \(1.6 \times 10^{-19}\) joules. Hence, the kinetic energy per proton is \(1.6 \times 10^{-7}\) J.

The total kinetic energy for all the protons is calculated by multiplying the energy of one proton by the total number of protons. This results in \(8.0 \times 10^6\) J. To further relate this to everyday scenarios, a car weighing 1500 kg would need to travel at approximately 103.3 m/s to have the same kinetic energy as the proton beam in the Tevatron.

Charge of a Proton

Every proton carries a fundamental charge, which is approximately \(1.6 \times 10^{-19}\) coulombs. This value is essential in calculating the total charge when dealing with a large number of protons, such as those in a proton beam like the one in the Tevatron.

For this exercise, the total charge of the beam is determined by multiplying the charge of a single proton by the number of protons present, which is \(5.0 \times 10^{13}\). Thus, the total charge is \(8.0 \times 10^{-6}\) C. Understanding the charge of protons is crucial in fields like particle physics and electrical engineering, as it helps in calculations concerning electric fields and currents.

Tevatron

The Tevatron, once one of the world's largest particle accelerators, was located at Fermilab in Illinois, USA. It operated by accelerating protons and antiprotons in a circular path until they nearly reached the speed of light. This immense speed is close to \(c \approx 3 \times 10^8\, ext{m/s}\), the speed of light.

In this exercise, the Tevatron's proton beam is described as circulating in a 1.0-km-radius ring. Using this information, one can calculate how much current is carried by the beam. The current, defined as the rate of flow of charge, is determined by dividing the total charge by the time it takes for one complete circulation. Using these parameters, we find the current in the proton beam is about 0.383 A. The ingenuity of particle accelerators like the Tevatron aids in the exploration of fundamental particles and the forces of nature, playing an instrumental role in advancing our understanding of the universe.