Problem 10 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A cyclotron with a radius of 1.0 \(\mathrm{m}\) is to accelerate deuterons \(\left(\frac{2}{1} \mathrm{H}\right)\) to an energy of 12 \(\mathrm{MeV}\) . (a) What is the required magnetic field? (b) What frequency is needed for the voltage between the dees? \((c)\) If the potential difference between the dees averages 22 \(\mathrm{kV}\) , how many revolutions will the particles make before exiting? \((d)\) How much time does it take for one deuteron to go from start to exit? (e) Estimate how far it travels during this time.

Step-by-Step Solution

Answer

(a) 0.644 T, (b) 4.837 x 10^7 Hz, (c) ~545 revolutions, (d) ~11.3 µs, (e) ~3420 m.

Step 1: Understanding Cyclotron Motion

In a cyclotron, charged particles are accelerated in a spiral motion through the use of magnetic and electric fields. The magnetic field keeps the particles in a circular path, while the oscillating electric field accelerates them during each half-circle.

Step 2: Calculating Magnetic Field (a)

The magnetic field required can be calculated using the formula for centripetal force in a cyclotron:\[ F_c = qvB = \frac{mv^2}{r} \]and energy relation:\[ \frac{1}{2}mv^2 = E \]where \(E = 12 \mathrm{MeV} = 1.92 \times 10^{-12} \mathrm{J}\), mass of deuteron \(m = 3.34 \times 10^{-27} \mathrm{kg}\), and charge \(q = e = 1.6 \times 10^{-19} \mathrm{C}\).First, we find velocity \(v\) from energy:\[ v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2 \times 1.92 \times 10^{-12}}{3.34 \times 10^{-27}}} = 3.09 \times 10^7 \mathrm{m/s} \]Then, using \(qvB = \frac{mv^2}{r}\), solve for \(B\):\[ B = \frac{mv}{qr} = \frac{3.34 \times 10^{-27} \times 3.09 \times 10^7}{1.6 \times 10^{-19} \times 1.0} \approx 0.644 \mathrm{T} \]

Step 3: Calculating Cyclotron Frequency (b)

The cyclotron frequency is given by:\[ f = \frac{qB}{2\pi m} \]Substitute the values:\[ f = \frac{1.6 \times 10^{-19} \times 0.644}{2\pi \times 3.34 \times 10^{-27}} \approx 4.837 \times 10^7 \mathrm{Hz} \]

Step 4: Determining Number of Revolutions (c)

The total energy gained by the particle after \(n\) revolutions is given by \( n \times V = 12 \mathrm{MeV} = 1.92 \times 10^{-12} \mathrm{J}\).So, solve for \(n\):\[ n \times 22 \times 10^3 e = 1.92 \times 10^{-12} \]\[ n = \frac{1.92 \times 10^{-12}}{22 \times 10^3 \times 1.6 \times 10^{-19}} \approx 5.45 \times 10^2 \]

Step 5: Calculating Time for Acceleration (d)

The time to exit the cyclotron is the number of revolutions divided by the frequency:\[ t = \frac{n}{f} = \frac{545}{4.837 \times 10^7} \approx 1.13 \times 10^{-5} \, \mathrm{s} \]

Step 6: Estimating Distance Traveled (e)

Each revolution has a circumference \(2\pi r\), and the total number of revolutions is \(n\). Thus, the distance traveled is:\[ d = n \times 2\pi r = 545 \times 2\pi \times 1.0 \approx 3.42 \times 10^3 \, \mathrm{m} \]

Key Concepts

Magnetic Field

In a cyclotron, the magnetic field is crucial for keeping the charged particles, like deuterons, in a circular path. By applying a magnetic force perpendicular to the velocity of the particles, they are guided along a spiral trajectory.
This property allows particles to be accelerated continuously without flying out of the machine. The formula for the magnetic field in a cyclotron can be derived from the concept of centripetal force, where the magnetic field strength needed is expressed as:

• Magnetic force: \( F = qvB \)
• Centripetal force: \( F = \frac{mv^2}{r} \)

Setting these two equal allows us to solve for \( B \):

• \( B = \frac{mv}{qr} \)

For deuterons, this ensures they follow a path of a specific radius without deviating. The required magnetic field strength depends on the mass \( m \), velocity \( v \), charge \( q \), and radius \( r \) of their path.Understanding the role of the magnetic field in the cyclotron helps us manipulate and control the path of the particles according to the desired outcome.

Cyclotron Frequency

Cyclotron frequency is a key parameter in the operation of a cyclotron, defining how quickly a particle like a deuteron completes one circular motion. The frequency depends on the magnetic field strength \( B \) and the properties of the particle, such as charge \( q \) and mass \( m \).
Its formula is given by:

• \( f = \frac{qB}{2\pi m} \)

This equation shows that the cyclotron frequency is directly proportional to the magnetic field strength and the charge of the particle, while inversely proportional to the mass of the particle.
The relevance of cyclotron frequency is that it determines the timing for the alternating electric field, ensuring that the particles receive an energy boost at the precise moment they pass through the accelerating gap. If the frequency matches perfectly with the particle's orbit, the cyclotron efficiently accelerates the particles to the desired energy level.Understanding cyclotron frequency is critical for the synchronization of particle acceleration.

Deuterons

Deuterons are a type of nucleus that plays a vital role in cyclotron operations. They are composed of one proton and one neutron, and are isotopes of hydrogen (heavy hydrogen).
Deuterons have some specific properties that make them interesting for use in cyclotrons:

• Mass \( m = 3.34 \times 10^{-27} \text{ kg} \)
• Charge \( q = 1.6 \times 10^{-19} \text{ C} \)

These characteristics influence how deuterons move within a magnetic field and how they can be accelerated. Having both a beneficial mass and charge makes them an ideal candidate for acceleration to high energies, such as the 12 MeV described in the exercise.
In nuclear physics, deuterons are useful for studying nuclear reactions and for possible applications like fusion research.Understanding deuterons helps clarify their behavior within the fields in a cyclotron, ensuring we can control and predict their motion accurately.

Particle Acceleration

Particle acceleration in a cyclotron involves moving particles, like deuterons, to high speeds and increasing their energy. This is achieved by making the particles pass repeatedly through the accelerating region, where an electric field is applied.
The cyclotron works by using both a magnetic field to curve the particles' paths and an oscillating electric field to accelerate them. When a charged particle crosses from one dee (a hollow D-shaped electrode) to the other, it gains energy equal to the potential difference across the gap. After many revolutions, the particles achieve the desired energy level.
Here's the process broken down:

• A magnetic field keeps them in a circular path.
• An alternating electric field provides acceleration between each half-turn.
• The particles spiral outward as they gain speed and energy.

The potential difference and frequency of the alternating current need to be perfectly synchronized to match the cyclotron frequency.
In a typical cyclotron, once particles have reached their desired energy, they are extracted from the spiral path and directed toward a target or further experiments. Understanding the principles of particle acceleration helps us appreciate how cyclotrons increase particle energy efficiently and rapidly.