/*! 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 25 An electron in the beam of a TV ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 25

An electron in the beam of a TV picture tube is accelerated by a potential difference of 2.00 \(\mathrm{kV}\) . Then it passes through a region of transverse magnetic field, where it moves in a circular arc with radius 0.180 \(\mathrm{m} .\) What is the magnitude of the field?

Short Answer

Expert verified
The magnitude of the magnetic field is approximately 5.33 mT.

Step by step solution

01

Understand the Problem

We are given an electron that is accelerated by a potential difference of 2.00 kV and then moves through a magnetic field in which it follows a circular path with a radius of 0.180 m. We need to find the magnitude of the magnetic field.
02

Calculate Electron's Velocity

The kinetic energy gained by the electron is equal to the work done by the electric field, given by the potential difference. Use the equation \( KE = eV \), where \( e \) is the charge of the electron \( 1.60 \times 10^{-19} \, C \) and \( V \) is the voltage \( 2.00 \times 10^{3} \, V \). Thus, the kinetic energy \( KE = 1.60 \times 10^{-19} \, C \times 2.00 \times 10^{3} \, V = 3.20 \times 10^{-16} \, J \). The kinetic energy is also given by \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the electron \( 9.11 \times 10^{-31} \, kg \). Solving for \( v \), we obtain:\[ v = \sqrt{\frac{2 \times 3.20 \times 10^{-16} \, J}{9.11 \times 10^{-31} \, kg}} \].
03

Apply the Circular Motion in a Magnetic Field Formula

When an electron moves in a magnetic field, it follows a circular path due to the Lorentz force. The centripetal force required for circular motion is provided by the magnetic force \( F = evB \), where \( B \) is the magnetic field. The centripetal force is also given by \( \frac{mv^2}{r} \), where \( r = 0.180 \, m \). Equating the two gives us:\[ evB = \frac{mv^2}{r} \].
04

Rearrange and Solve for Magnetic Field

Rearranging the equation from Step 3 to solve for \( B \), we get:\[ B = \frac{mv}{er} \].Substitute the values of \( m = 9.11 \times 10^{-31} \, kg \), \( v \) from Step 2, \( e = 1.60 \times 10^{-19} \, C \), and \( r = 0.180 \, m \), to find \( B \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Electron Acceleration
Electron acceleration in the context of a TV picture tube involves increasing the speed of electrons by applying a potential difference. This process is crucial in various electronic applications where electron beams are directed precisely. When an electron is accelerated across a potential difference, it gains kinetic energy. This can be described with the term work done by the electric field, calculated as the potential energy difference converted into kinetic energy of the electron.
The equation that describes this energy conversion is:
  • Potential Difference (V) provides energy to the electron, expressed as \( KE = eV \), where \( KE \) is kinetic energy, \( e \) is the elementary charge, and \( V \) is potential difference.
Understanding this concept helps clarify how potential differences can set electrons into motion, directly affecting their velocity and subsequent behavior in magnetic fields.
Potential Difference
The potential difference is a measure of the energy difference per charge between two points in an electric field. This concept is especially important in electromagnetism and electronics, including devices like cathode ray tubes found in old television sets.
When a charged particle such as an electron moves through a potential difference, it experiences a change in energy, which manifests as kinetic energy if the electron is accelerated.
  • This means that a potential difference of 2.00 kV can significantly increase an electron's speed, converting electrical energy into kinetic energy.
Such transformations are essential in many devices, allowing us to control electron behavior and trajectories by electric fields, using the potential difference as a driving force. This is crucial in analyzing the behavior of electrons before entering a magnetic field.
Circular Motion in Magnetic Field
When a charged particle, like an electron, enters a magnetic field, it experiences a force perpendicular to its velocity and the field direction, causing it to move in a circular path. This circular movement is due to centripetal force, which in magnetic fields comes from the magnetic force (a result of the Lorentz force).
  • The radius of circular motion depends on the particle's speed, the charge, mass, and magnetic field strength, described by the formula \( F_c = \frac{mv^2}{r} \), where \( r \) is the radius of the circular path.
Understanding how charged particles move in magnetic fields helps in various applications. It is fundamental for designing instruments that guide and control electron beams in technology and scientific research, ensuring precise manipulations of particles.
Lorentz Force
The Lorentz force is the force experienced by a charged particle moving through electric and magnetic fields. When electrons enter a magnetic field, the magnetic component of the Lorentz force influences their trajectory by making them move in circular paths.
  • This magnetic part of the Lorentz force is described by \( F = evB \), where \( B \) is the magnetic field strength, \( v \) is the velocity of the moving charge, and \( e \) is the charge of the electron.
By understanding and applying the Lorentz force, one can determine trajectories and required magnetic field strengths for charged particles. It is integral to various technological applications, such as in particle accelerators and electronic devices, where precise control of particles' paths is crucial.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determining Diet. One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) technique. As corn photosynthesizes, it concentrates the isotope carbon-13, whereas most other plants concentrate carbon-12. Overreliance on corn consumption can then be correlated with certain diseases, because corn lacks the essential amino acid lysine. Archaeologists use a mass spectrometer to separate the 12\(\mathrm { C }\) and \(^ { 13 } \mathrm { C }\) isotopes in samples of human remains. Suppose you use a velocity selector to obtain singly ionized (missing one electron) atoms of speed \(8.50 \mathrm { km } / \mathrm { s } ,\) and you want to bend them within a uniform magnetic field in a semicircle of diameter 25.0\(\mathrm { cm }\) for the 12\(\mathrm { C }\) . The measured masses of these isotopes are \(1.99 \times 10 ^ { - 26 } \mathrm { kg } \left( ^ { 12 } \mathrm { C } \right)\) and \(2.16 \times 10 ^ { - 26 } \mathrm { kg } \left( ^ { 13 } \mathrm { C } \right) .\) (a) What strength of magnetic field is required? (b) What is the diameter of the 13 C semicircle? (c) What is the separation of the \(^ { 12 } \mathrm { C }\) and \(^ { 13 } \mathrm { C }\) ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?

A proton \(\left(q=1.60 \times 10^{-19} \mathrm{C}, m=1.67 \times 10^{-27} \mathrm{kg}\right)\) moves in a uniform magnetic field \(\vec{\boldsymbol{B}}=(0.500 \mathrm{T}) \hat{\boldsymbol{l}}\) . At \(t=0\) the proton has velocity components \(v_{x}=1.50 \times 10^{5} \mathrm{m} / \mathrm{s}, v_{y}=0\) and \(v_{z}=2.00 \times 10^{5} \mathrm{m} / \mathrm{s}\) (see Example 27.4\() .\) (a) What are the magnitude and direction of the magnetic force acting on the proton? In addition to the magnetic field there is a uniform electric field in the \(+x\) -direction, \(\vec{E}=\left(+2.00 \times 10^{4} \mathrm{V} / \mathrm{m}\right) \hat{\imath}\) . (b) Will the proton have a component of acceleration in the direction of the electric field? (c) Describe the path of the proton. Does the electric field affect the radius of the helix? Explain. (d) At \(t=T / 2\), where \(T\) is the period of the circular motion of the proton, what is the \(x\) -component of the displacement of the proton from its position at \(t=0 ?\)

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 \(\vec { \boldsymbol { B } } = - ( 0.120 \mathrm { T } ) \hat { \boldsymbol { k } }\) . At a particular instant of time the velocity of the particle is \(\vec { \boldsymbol { v } } = \left( 1.05 \times 10 ^ { 6 } \mathrm { m } / \mathrm { s } \right) ( - 3 \hat { \imath } + 4 \hat { \jmath } + 12 \hat { \boldsymbol { k } } )\) and the force \(\vec { \boldsymbol { F } }\) on the particle has a magnitude of 2.45\(\mathrm { N }\) . (a) Determine the charge \(q .\) (b) Determine the acceleration \(\vec { a }\) 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. (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.

A particle with charge 7.80\(\mu \mathrm{C}\) is moving with velocity \(\vec{\boldsymbol{v}}=-\left(3.80 \times 10^{3} \mathrm{m} / \mathrm{s}\right) \hat{\boldsymbol{J}}\) . The magnetic force on the particle is measured to be \(\vec{\boldsymbol{F}}=+\left(7.60 \times 10^{-3} \mathrm{N}\right) \hat{\boldsymbol{\imath}}-\left(5.20 \times 10^{-3} \mathrm{N}\right) \hat{\boldsymbol{k}}\) (a) Calculate all the components of the magnetic field you can from this information. (b) Are there components of the magnetic field that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\vec{B} \cdot \vec{F} .\) What is the angle between \(\vec{B}\) and \(\vec{\boldsymbol{F}} ?\)

An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of \(6.64 \times 10^{-27}\) kg) traveling horizontally at 35.6 \(\mathrm{km} / \mathrm{s}\) enters a uniform, vertical, 1.10 -T magnetic field. (a) What is the diameter of the path followed by this alpha particle? (b) What effect does the magnetic field have on the speed of the particle? (c) What are the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field? (d) Explain why the speed of the particle does not change even though an unbalanced external force acts on it.
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Wave Optics

Read Explanation

Nuclear Physics

Read Explanation

Scientific Method Physics

Read Explanation

Magnetism

Read Explanation

Famous Physicists

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.