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In a certain cyclotron a proton moves in a circle of radius \(0.500 \mathrm{~m}\). The magnitude of the magnetic field is \(1.20 \mathrm{~T}\). (a) What is the oscillator frequency? (b) What is the kinetic energy of the proton, in electron-volts?

Short Answer

Expert verified
(a) 25.8 MHz; (b) 3.44 MeV.

Step by step solution

01

Find the Cyclotron Frequency Formula

The formula to calculate the oscillator frequency, also known as the cyclotron frequency, for a particle in a magnetic field is given by \( f = \frac{qB}{2\pi m} \), where \( q \) is the charge of the proton, \( B \) is the magnetic field, and \( m \) is the mass of the proton.
02

Insert Known Values

For a proton, the charge \( q = 1.60 \times 10^{-19} \text{ C} \) and the mass \( m = 1.67 \times 10^{-27} \text{ kg} \). Insert these values and \( B = 1.20 \text{ T} \) into the formula: \[ f = \frac{(1.60 \times 10^{-19} \text{ C})(1.20 \text{ T})}{2\pi (1.67 \times 10^{-27} \text{ kg})} \]
03

Calculate Oscillator Frequency

Compute the value from the formula. \[ f \approx \frac{1.92 \times 10^{-19}}{10.48 \times 10^{-27}} \approx 2.58 \times 10^7 \text{ Hz} \] Thus, the oscillator frequency is approximately \( 25.8 \text{ MHz} \).
04

Determine Kinetic Energy Formula

The kinetic energy (KE) of a proton in a circular orbit in a cyclotron can be expressed as \( KE = \frac{1}{2}mv^2 \). The velocity \( v \) can also be found using the formula \( v = r\omega \), where \( \omega = 2\pi f \) is the angular frequency.
05

Calculate Angular Velocity

Using \( \omega = 2\pi f \):\[ \omega = 2\pi (2.58 \times 10^7 \text{ Hz}) \approx 1.62 \times 10^8 \text{ rad/s} \] With the radius \( r = 0.5 \text{ m} \), we calculate the velocity: \[ v = r\omega = 0.5 \times 1.62 \times 10^8 \text{ m/s} \approx 8.1 \times 10^7 \text{ m/s} \]
06

Calculate Kinetic Energy

Now, using the kinetic energy formula: \[ KE = \frac{1}{2} \times (1.67 \times 10^{-27} \text{ kg}) \times (8.1 \times 10^7 \text{ m/s})^2 \approx 5.5 \times 10^{-13} \text{ J} \] Convert this energy to electron-volts (1eV = \(1.60 \times 10^{-19} \text{ J}\)):
07

Convert Kinetic Energy to Electron-Volts

Convert kinetic energy to electron-volts: \[ KE = \frac{5.5 \times 10^{-13} \text{ J}}{1.60 \times 10^{-19} \text{ J/eV}} \approx 3.44 \times 10^6 \text{ eV} \] Thus, the kinetic energy is approximately 3.44 MeV.

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Key Concepts

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

Magnetic Field
A magnetic field is a fundamental concept in physics and plays a crucial role in the operation of a cyclotron. Essentially, it is a field produced by moving electric charges, and it's in this field that charged particles, such as protons, move in circular paths. The magnetic field is described by its strength and direction, with the strength being measured in Teslas (T).
A cyclotron utilizes a constant magnetic field to accelerate charged particles along a circular path. In our exercise, the magnetic field magnitude is given as 1.20 T. This strong field ensures that the proton moves in a stable circular trajectory, necessary for continuous acceleration.
Key characteristics of magnetic fields in cyclotrons include:
  • Directed perpendicular to the plane of the particle's motion.
  • Ensures that the particle's path is circular.
  • Influences the frequency and energy of the particle's motion.
Understanding how a magnetic field guides the motion of particles is essential for grasping the principles of a cyclotron.
Proton
The proton is a subatomic particle, an essential component of the atomic nucleus, and it plays a key role in nuclear physics. With a positive charge and a mass of approximately 1.67 × 10^{-27} kg, protons are suitable for acceleration in a cyclotron due to their charge and mass characteristics.
In a cyclotron, protons perform a circular motion due to the Lorentz force, which acts perpendicular to both the particle's velocity and the magnetic field. The force is expressed as \( F = qvB \), where \( q \) represents the proton's charge, \( v \) is its velocity, and \( B \) is the magnetic field strength.
The specific properties of protons that affect cyclotron operations include:
  • Their positive electric charge \( (1.60 \times 10^{-19} \text{ C}) \), which determines the direction of motion in a magnetic field.
  • Their mass, which affects the frequency and kinetic energy.
  • The simplicity of their structure, making them less susceptible to breakdown during acceleration processes.
Protons, being one of the simplest charged particles, illustrate fundamental principles when discussing cyclotron physics.
Kinetic Energy
Kinetic energy in the context of a cyclotron refers to the energy possessed by protons as they move in a circular path, due to both their speed and mass. The kinetic energy formula \( KE = \frac{1}{2}mv^2 \) allows us to determine how much energy the proton has as it spirals outward with increasing speed.
In this exercise, we computed the kinetic energy and converted it to electron-volts, a unit that is more relevant in particle physics. The formula \( KE = \frac{1}{2}(1.67 \times 10^{-27} \, \text{kg}) \times (8.1 \times 10^{7} \, \text{m/s})^2 \) yielded around 5.5 \times 10^{-13} J, which is farther converted to approximately 3.44 MeV.
  • The calculation of kinetic energy involves first finding the proton's speed using its circular path radius and cyclotron frequency.
  • As the protons gain kinetic energy, they move on larger circular paths within the cyclotron.
  • These energies are crucial for applications such as producing radioisotopes for medical diagnosis and treatment.
Kinetic energy reflects the protons' increasing energy state as they are accelerated by the cyclotron.
Cyclotron Frequency
Cyclotron frequency, also known as oscillator frequency, is the rate at which a charged particle like a proton completes a full circle within a cyclotron, influenced mainly by the magnetic field strength and the charge-to-mass ratio of the particle.
The formula \( f = \frac{qB}{2\pi m} \) shows that the frequency depends on these physical quantities:
  • \( q \), the charge of the proton \( (1.60 \times 10^{-19} \text{ C}) \).
  • \( B \), the magnetic field strength \( (1.20 \text{ T}) \).
  • \( m \), the mass of the proton \( (1.67 \times 10^{-27} \text{ kg}) \).
In this specific exercise, the calculated cyclotron frequency of \( 2.58 \times 10^7 \text{ Hz} \) translates to an oscillator frequency of approximately 25.8 MHz.
Important points about cyclotron frequency include:
  • It remains constant for a given magnetic field and particle type.
  • Enables synchronization with the electromagnetic oscillator that adds energy to the particle.
  • Sets the limit for how fast the particles can be accelerated.
Cyclotron frequency is pivotal as it dictates how quickly a particle moves in its circular path, thus controlling the acceleration process.

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Most popular questions from this chapter

An electron is accelerated from rest by a potential difference of \(350 \mathrm{~V}\). It then enters a uniform magnetic field of magnitude 200 \(\mathrm{mT}\) with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.

An electron has an initial velocity of \((12.0 \hat{\mathrm{j}}+15.0 \mathrm{k}) \mathrm{km} / \mathrm{s}\) and a constant acceleration of \(\left(2.00 \times 10^{12} \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}\) in a region in which uniform electric and magnetic fields are present. If \(\vec{B}=(400 \mu \mathrm{T}) \hat{\mathrm{i}}\), find the electric field \(\vec{E}\).

An electron with kinetic energy \(2.5 \mathrm{keV}\) moving along the positive direction of an \(x\) axis enters a region in which a uniform electric field of magnitude \(10 \mathrm{kV} / \mathrm{m}\) is in the negative direction of the \(y\) axis. A uniform magnetic field \(\vec{B}\) is to be set up to keep the electron moving along the \(x\) axis, and the direction of \(\vec{B}\) is to be chosen to minimize the required magnitude of \(\vec{B}\). In unit-vector notation, what \(\vec{B}\) should be set up?

A horizontal power line carries a current of \(5000 \mathrm{~A}\) from south to north. Earth's magnetic field \((60.0 \mu \mathrm{T})\) is directed toward the north and inclined downward at \(70.0^{\circ}\) to the horizontal. Find the (a) magnitude and (b) direction of the magnetic force on \(100 \mathrm{~m}\) of the line due to Earth's field.

In a nuclear experiment a proton with kinetic energy \(1.0 \mathrm{MeV}\) moves in a circular path in a uniform magnetic field. What energy must (a) an alpha particle \((q=+2 e, m=4.0 \mathrm{u})\) and (b) a deuteron \((q=+e, m=2.0 \mathrm{u})\) have if they are to circulate in the same circular path?

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