/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Calculate the cyclotron frequenc... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the cyclotron frequency of a proton in a magnetic field of magnitude \(5.20 \mathrm{T}\)

Short Answer

Expert verified
The cyclotron frequency of the proton in the given magnetic field is approximately \(1.50 \times 10^{8} \, \mathrm{Hz}\)

Step by step solution

01

Identify given variables

A magnetic field strength \(B\) of \(5.20 \, \mathrm{T}\) is provided. We also know standard properties of a proton from physics: Its charge \(q\) is \(1.602 \times 10^{-19} \, \mathrm{C}\) and its mass \(m\) is \(1.673 \times 10^{-27} \, \mathrm{kg}\).
02

Apply the cyclotron frequency formula

We can apply the cyclotron frequency formula \(f = \frac{qB}{2\pi m}\). Substituting the known values in, we get \(f = \frac{(1.602 \times 10^{-19} \, \mathrm{C}) \times (5.20 \, \mathrm{T})}{2\pi \times (1.673 \times 10^{-27} \, \mathrm{kg})}\).
03

Compute

Perform the calculation to find the cyclotron frequency. This will give the result in Hertz (Hz).

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Key Concepts

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

Magnetic Field
The concept of a magnetic field is fundamental in the study of electromagnetism. It is a region around a magnetic material or a moving electric charge within which the force of magnetism acts. Magnetic fields are represented by the vector field denoted as \( B \), and its unit is the Tesla (\( \mathrm{T} \)). A magnetic field can exert a force on moving charged particles, causing them to move in circular or spiral paths. This force is perpendicular to both the magnetic field and the velocity of the particle, following the right-hand rule.

In the Original Exercise, a magnetic field strength of \( 5.20 \, \mathrm{T} \) is used. This indicates a relatively strong field, as typical laboratory-generated magnetic fields range up to only a few Teslas. The strength of the field influences phenomena such as cyclotron frequency by determining how strongly it interacts with charged particles like protons.
  • Magnetic fields influence charged particles.
  • The direction and strength are represented by vector \( B \).
  • Measured in Tesla (\( \mathrm{T} \)).
Understanding magnetic fields helps in controlling particle motions in devices like cyclotrons, which are useful for medical and scientific applications.
Proton Charge
Protons are subatomic particles found within the nucleus of an atom. Each proton carries a positive electric charge of \( +1.602 \times 10^{-19} \, \mathrm{C} \). This charge is fundamental and is considered a basic unit of charge in the field of physics. The charge of the proton plays a crucial role in the electromagnetic interactions that occur in cyclotron devices.

The proton's positive charge, when placed in a magnetic field, leads to the interactions necessary for phenomena like cyclotron frequency. Due to its charge, the proton is affected by the magnetic field, experiencing a force that makes it move in a circular path at a specific frequency. This motion is characterized by the cyclotron radius and frequency, which depend on the charge and the field strength.
  • Protons have a positive charge: \( +1.602 \times 10^{-19} \, \mathrm{C} \).
  • Essential for creating electromagnetic interactions.
  • Integral to the calculations of cyclotron frequency.
By knowing the proton charge, you can better understand how charged particles are manipulated using magnetic fields.
Cyclotron Motion
Cyclotron motion describes the circular movement of charged particles like protons when they are subjected to a perpendicular magnetic field. This motion is governed by the force exerted on the charged particle, and this interaction forms the basis for calculating the cyclotron frequency. It is an essential concept in particle physics and is utilized in devices like cyclotrons and particle accelerators.

The cyclotron frequency \( f \) is defined as the rate at which a charged particle orbits in a magnetic field and is given by the formula:\[ f = \frac{qB}{2\pi m} \]where:- \( q \) is the particle's charge,- \( B \) is the magnetic field strength,- \( m \) is the particle's mass.
This formula is derived from equating the centripetal force required for circular motion to the magnetic force that acts on the particle. Cyclotron motion is crucial for accelerating particles to high speeds in a controlled manner, which is pivotal for experiments in physics and the development of medical technology.
  • Circular motion of a particle in a magnetic field.
  • Frequency depends on charge, field strength, and mass.
  • Utilized in cyclotrons and particle accelerators.
The understanding of cyclotron motion enhances insights into manipulating and experimenting with subatomic particles.

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Most popular questions from this chapter

A small bar magnet is suspended in a uniform \(0.250-\mathrm{T}\) magnetic field. The maximum torque experienced by the bar magnet is \(4.60 \times 10^{-3} \mathrm{N} \cdot \mathrm{m} .\) Calculate the magnetic moment of the bar magnet.

. A wire having a linear mass density of \(1.00 \mathrm{g} / \mathrm{cm}\) is placed on a horizontal surface that has a coefficient of kinetic friction of \(0.200 .\) The wire carries a current of 1.50 A toward the east and slides horizontally to the north. What are the magnitude and direction of the smallest magnetic field that enables the wire to move in this fashion?

A velocity selector consists of electric and magnetic fields described by the expressions \(\mathbf{E}=E \hat{\mathbf{k}}\) and \(\mathbf{B}=B \hat{\mathbf{j}},\) with \(B=15.0 \mathrm{mT}\). Find the value of \(E\) such that a \(750-\mathrm{eV}\) electron moving along the positive \(x\) axis is undeflected.

Consider an electron orbiting a proton and maintained in a fixed circular path of radius \(R=5.29 \times 10^{-11} \mathrm{m}\) by the Coulomb force. Treating the orbiting charge as a current loop, calculate the resulting torque when the system is in a magnetic field of \(0.400 \mathrm{T}\) directed perpendicular to the magnetic moment of the electron.

A current loop with magnetic dipole moment \(\mu\) is placed in a uniform magnetic field \(\mathbf{B},\) with its moment making angle \(\theta\) with the field. With the arbitrary choice of \(U=0\) for \(\theta=90^{\circ},\) prove that the potential energy of the dipole-field system is \(U=-\mu \cdot\) B. You may imitate the discussion in Chapter 26 of the potential energy of an electric dipole in an electric field.

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