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Chapter 27: Problem 56

A cycloron is to accelerate protons to an energy of 5.4 MeV. The superconducting electromagnet of the cyclotron produces a \(3.5-\) T magnetic field perpendicular to the proton orbits. (a) When the protons have achieved a kinetic energy of 2.7 \(\mathrm{MeV}\) , what is the radius of their circular orbit and what is their angular speed? (b) Repeat part (a) when the protons have achieved their final kinetic energy of 5.4 \(\mathrm{MeV}\) .

Short Answer

Expert verified
At 2.7 MeV, radius 鈮� 0.095 m, angular speed 鈮� 2.1 脳 10鈦� rad/s. At 5.4 MeV, radius 鈮� 0.135 m, angular speed 鈮� 1.5 脳 10鈦� rad/s.

Step by step solution

01

Understanding the Problem

We need to find the radius of the circular orbit and the angular speed of protons when they have kinetic energies of 2.7 MeV and 5.4 MeV in a cyclotron with a magnetic field of 3.5 T.
02

Recall Relevant Formulas

Use the formula for the radius of a charged particle moving in a magnetic field, given by \( r = \frac{mv}{qB} \), where \( m \) is the mass, \( v \) is the velocity, \( q \) is the charge, and \( B \) is the magnetic field. The angular speed \( \omega \) is given by \( \omega = \frac{v}{r} \). The kinetic energy can be related to velocity by \( KE = \frac{1}{2}mv^2 \).
03

Calculate Velocity from Kinetic Energy for 2.7 MeV

The kinetic energy \( KE = 2.7 \, \text{MeV} = 2.7 \times 10^6 \times 1.6 \times 10^{-19} \, \text{J} \). Well-known constants are the proton mass \( m = 1.67 \times 10^{-27} \, \text{kg} \) and charge \( q = 1.6 \times 10^{-19} \, \text{C} \). Calculate the velocity using \( v = \sqrt{\frac{2KE}{m}} \).
04

Compute Radius for 2.7 MeV

Calculate the radius \( r \) of the orbit using \( r = \frac{mv}{qB} \), substituting the values found for \( m, v, q, \) and \( B = 3.5 \, \text{T} \).
05

Find Angular Speed for 2.7 MeV

Calculate the angular speed \( \omega \) using \( \omega = \frac{v}{r} \) with the previously computed \( v \) and \( r \).
06

Repeat Calculations for 5.4 MeV

Use the same process in Steps 3 to 5 but substitute \( KE = 5.4 \, \text{MeV} \) to find the radius and angular speed for protons at this energy.

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Key Concepts

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

Proton Acceleration
A cyclotron is a type of particle accelerator that is used to speed up charged particles, like protons, to high energies. When working with a cyclotron, one of the main goals is to increase the kinetic energy of these protons so that they move faster and with more force. This acceleration process is achieved using both electric and magnetic fields.

Electric fields provide the push that protons need to get moving. Every time the proton crosses the gap between the D-shaped electrodes (commonly known as dees), it gains energy from the electric field. This energy boosts the speed of the proton, incrementally increasing its kinetic energy. Over several cycles within the cyclotron chamber, this continuous increase in speed helps the proton achieve a desired energy level, such as 2.7 MeV or 5.4 MeV.

In this cyclotron exercise, the role of acceleration is crucial, as we track how kinetic energy changes the motion characteristics of protons. To achieve such high energy, careful engineering and precise control of the electric fields are necessary to ensure the protons follow the intended path and gain energy efficiently.
Magnetic Field
The magnetic field in a cyclotron plays a crucial role in directing and controlling the path of protons. As the protons gain energy and speed through the cyclotron's electric field, the magnetic field ensures that they move in circular paths inside the chamber.

In this scenario, the cyclotron uses a magnetic field of 3.5 Tesla, which is a powerful force. This magnetic field acts perpendicular to the path of the protons. According to the principles of electromagnetism, when a charged particle, such as a proton, enters a magnetic field perpendicularly, it experiences a force that causes it to move in a circular trajectory. This force is known as the Lorentz force.

The strength of the magnetic field determines how tight or wide the proton's circular path will be. A stronger magnetic field will make the circle smaller, while a weaker field allows for a larger orbit. This characteristic of the magnetic field is essential for calculating the radius of the orbit and understanding how it affects the proton's motion as it gains energy.
Circular Orbit Radius
The radius of the circular orbit in a cyclotron is a key metric that describes the path of a proton as it travels within the magnetic field. This radius depends on various parameters, including the speed of the proton, the magnetic field strength, and the specific properties of the proton itself, such as its mass and charge.

In the formula for calculating the radius of the circular orbit, expressed as \( r = \frac{mv}{qB} \), each element plays a vital role:
  • \( m \) is the mass of the proton.
  • \( v \) is the velocity of the proton.
  • \( q \) is the charge of the proton.
  • \( B \) is the magnetic field's strength.
As the kinetic energy of the proton increases, its velocity \( v \) does too, causing changes in the orbit radius. For example, at 2.7 MeV, the orbit radius will have a certain value. At a higher energy level of 5.4 MeV, this radius will increase, reflecting the increase in the proton's speed. Understanding this dynamic relationship helps in designing cyclotrons and predicting proton behavior within the magnetic field.
Angular Speed
Angular speed \( \omega \) is a measure of how fast an object is rotating or moving along a circular path. In the context of a cyclotron, understanding the angular speed of protons gives insight into their dynamic behavior as they move within the magnetic field.

To calculate angular speed, we use the formula \( \omega = \frac{v}{r} \), where \( v \) is the velocity of the proton, and \( r \) is the radius of its circular path. Angular speed increases as protons gain more kinetic energy. For instance, as a proton's energy changes from 2.7 MeV to 5.4 MeV, its velocity increases, which results in a change in angular speed.

Angular speed is particularly important in determining the frequency of the proton's rotation within the cyclotron. This frequency is crucial for ensuring that proton acceleration is synchronized with the oscillating electric field, allowing for efficient energy transfer and a stable increase in speed and energy.

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Most popular questions from this chapter

A circular loop of wire with area \(A\) lies in the \(x y\) -plane. As viewed along the \(z\) -axis looking in the \(-z\) -direction toward the origin, a current \(I\) is circulating clockwise around the loop. The torque produced by an extemal magnetic field \(\vec{B}\) is given by \(\vec{\tau}=D(4 \hat{z}-3 \hat{y}),\) where \(D\) is a positive constant, and for this orientation of the loop the magnetic potential energy \(U=-\vec{\mu} \cdot \vec{B}\) is negative. The magnitude of the magnetic field is \(B_{0}=13 D / L A\) (a) Determine the vector magnetic moment of the current loop. (b) Determine the components \(B_{x}, B_{y}\) , and \(B_{z}\) of \(\vec{B}\) .

A beam of protons traveling at 1.20 \(\mathrm{km} / \mathrm{s}\) enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular- to its original direction (Fig. 27.48\()\) . The beam travels a distance of 1.18 \(\mathrm{cm}\) while in the field. What is the magnitude of the magnetic field?

In the electron gun of a TV picture tube the electrons (charge \(-e,\) mass \(m )\) are accelerated by a voltage \(V .\) After leaving the electron gun, the electron beam travels a distance \(D\) to the screen; in this region there is a transverse magnetic field of magnitude \(B\) and no electric field. (a) Sketch the path of the electron beam in the tube. (b) Show that the approximate deflection of the beam due to this magnetic field is $$ d=\frac{B D^{2}}{2} \sqrt{\frac{e}{2 m V}} $$ (Hint: Place the origin at the center of the electron beam's arc and compare an undefiected beam's path to the deflected beam's path.) (c) Evaluate this expression for \(V=750 \mathrm{V}, D=50 \mathrm{cm},\) and \(B=5.0 \times 10^{-5} \mathrm{T}\) (comparable to the earth's field). Is this deflection significant?

A circular ring with area \(4.45 \mathrm{~cm}^{2}\) is carrying a current of 12.5 A. The ring is free to rotate about a diameter. The ring, initially at rest, is immersed in a region of uniform magnetic field given by \(\vec{B}=\left(1.15 \times 10^{-2} \mathrm{~T}\right)(12 \hat{\imath}+3 \hat{\jmath}-4 \hat{k}) .\) The ring is positioned initially such that its magnetic moment is given by \(\vec{\mu}_{1}=\mu(-0.800 \hat{\imath}+0.600 \hat{\jmath}),\) where \(\mu\) is the (positive) magnitude of the magnetic moment. The ring is released and turns through an angle of \(90.0^{\circ},\) at which point its magnetic moment is given by \(\vec{\mu}_{f}=-\mu \hat{k} .\) (a) Determine the decrease in potential energy. (b) If the moment of inertia of the ring about a diameter is \(8.50 \times 10^{-7} \mathrm{~kg} \cdot \mathrm{m}^{2},\) determine the angular speed of the ring as it passes through the second position.

A particle with negative charge \(q\) and mass \(m=2.58 \times\) 10 \(^{-15} \mathrm{kg}\) is traveling through a region containing a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=-(0.120 \mathrm{T}) \hat{\boldsymbol{k}} .\) At a particular instant of time the velocity of the particle is \(\overrightarrow{\boldsymbol{v}}=\left(1.05 \times 10^{6} \mathrm{m} / \mathrm{s}\right)(-3 \hat{\imath}+4 \hat{\jmath}+\) 12\(\hat{k} )\) and the force \(\vec{F}\) on the particle has a magnitude of 1.25 \(\mathrm{N}\) . (a) Determine the charge \(q\) . (b) Determine the acceleration \(\overrightarrow{\boldsymbol{d}}\) of the particle. (c) Explain why the path of the particle is a helix, and determine the radius of curvature \(R\) of the circular component of the helical path. (d) Determine the cyclotron frequency of the particle. (e) Although helical motion is not periodic in the full sense of the word, the \(x\) - and \(y\) -coordinates do vary in a periodic way. If the coordinates of the particle at \(t=0\) are \((x, y, z)=(R, 0,0),\) determine its coordinates at a time \(t=2 T,\) where \(T\) is the period of the motion in the \(x y\) -plane.
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