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Chapter 7: Problem 11

Calculate the cyclotron frequency for electrons in a B-field of \(0.2 \mathrm{~T}\).

Short Answer

Expert verified
Question: Calculate the cyclotron frequency for electrons in a magnetic field of 0.2 T. Answer: The cyclotron frequency for electrons in a magnetic field of 0.2 T is approximately -3.51 x 10^11 s^-1.

Step by step solution

01

Write down the known values

We are given the following values: Magnetic field, \(B = 0.2 \mathrm{~T}\) Electron charge, \(q = -1.6 \times 10^{-19} \mathrm{C}\) Electron mass, \(m = 9.11 \times10^{-31} \mathrm{kg}\)
02

Calculate the cyclotron frequency using the formula

The cyclotron frequency \(\omega_{c}\) is given by the formula: \(\omega_{c} = \frac{qB}{m}\) Now, plug in the given values: \(\omega_{c} = \frac{-1.6 \times 10^{-19}\mathrm{C} \times 0.2 \mathrm{T}}{9.11 \times 10^{-31} \mathrm{kg}}\)
03

Solve for the cyclotron frequency

Solve the equation to find the cyclotron frequency: \(\omega_{c} = \frac{-1.6 \times 10^{-19}\mathrm{C} \times 0.2 \mathrm{T}}{9.11 \times 10^{-31} \mathrm{kg}}\) \(\omega_{c} \approx -3.51 \times 10^{11} \mathrm{s^{-1}}\)
04

Interpret the result

The cyclotron frequency for electrons in a B-field of \(0.2 \mathrm{~T}\) is approximately \(-3.51 \times 10^{11} \mathrm{s^{-1}}\). The negative sign indicates that the electron is spinning in the opposite direction of the magnetic field.

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Key Concepts

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

Understanding the Magnetic Field
When exploring the behavior of electrons under the influence of a magnetic field, it's essential to grasp what a magnetic field is. Simply put, a magnetic field is a vector field that surrounds magnets and electric currents, denoting the direction and strength of the magnetic force that can act on other currents and magnetic materials. Magnetic fields are measured in teslas (T) in the International System of Units (SI).

In the context of a cyclotron, a magnetic field is employed to bend the path of charged particles, like electrons. It ensures that the particles move in a circular trajectory within the cyclotron. When an electron with a negative charge moves through a magnetic field, it experiences a force perpendicular to both the direction of its motion and the direction of the magnetic field, leading to circular motion. This behavior is rooted in the Lorentz force, which governs the motion of charged particles in electromagnetic fields.
The Role of Electron Charge
An electron carries a fundamental property known as charge, denoted as 'e'. This intrinsic property of matter is measured in coulombs (C). The charge of a single electron is approximately \( -1.6 \times 10^{-19} \) C, and it's this charge that interacts with a magnetic field to produce force and subsequent motion.

The negative sign signifies that the electron has a negative charge, as opposed to the positive charge carried by protons. It is precisely this charge that we factor into the equation for calculating the cyclotron frequency, which is the frequency at which an electron orbits in a magnetic field. The electron's charge, when exposed to a magnetic field, causes it to accelerate perpendicular to both the velocity of the electron and the magnetic field, leading to its spiral trajectory in a device like a cyclotron.
Electron Mass and Its Significance
The mass of the electron is another crucial constant in calculating the cyclotron frequency. The mass of an electron is approximately \( 9.11 \times 10^{-31} \) kilograms. Though the mass of an electron is quite minuscule, it plays a significant role in determining its behavior in a magnetic field.

In our equation, the mass of the electron is the denominator, which means it's inversely proportional to the cyclotron frequency. In other words, the lighter the particle, the higher the frequency with which it will orbit in a magnetic field of a given strength. This is why electrons, being so light, can reach incredibly high speeds and frequencies in a cyclotron, allowing physicists and other scientists to study them in high-energy states.

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Most popular questions from this chapter

A long thin metal strip of width \(2 L\) carries a current along its length. The current is distributed uniformly across the width with a surface density \(J\), (Sec. 1.7). Find the B-ficld (a) at a point in the plane of the strip at a distance \(R(>L)\) from its centre and (b) at a point a perpendicular distance \(R\) from its centre. Check that (a) yields \((7.19)\) as \(L / R \rightarrow 0\). What does (b) become if the sheet is of infinite width?

Two flat coils each of 10 turns have mean radii of 20 and \(2 \mathrm{~cm}\). Find an approximate value for the force between them if they are coaxial, \(10 \mathrm{~cm}\) apart and if each carries \(5 \mathrm{~A}\) in the same sense.

A circular steel disc is magnetized uniformly in a direction parallel to one of its diameters. How could this diameter be identified using no other electric or magnetic apparatus?

A particle with charge \(e\) and mass \(m\) is projected with linear momentum \(p\) at right angles to a uniform magnetic field \(B\). What are the magnetic dipole moment and angular momentum possessed by the particle as a result of its subsequent motion?

Modern magnetometers for measuring B-fields are described in Comment C12.2. In the older vibration magnetometer, a small permanent dipole was allowed to perform small angular oscillations in a horizontal plane first in a standard field \(\mathbf{B}_{0}\) and then in the unknown \(\mathrm{B}\). If the periods of oscillation were, respectively, \(T_{0}\) and \(T\), show that \(B=B_{0} T_{6}^{2} T^{2}\),
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