/*! 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 72 A beam of electrons whose kineti... [FREE SOLUTION] | 91影视

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Chapter 28: Problem 72

A beam of electrons whose kinetic energy is \(K\) emerges from a thin-foil "window" at the end of an accelerator tube. A metal plate at distance \(d\) from this window is perpendicular to the direction of the emerging beam (Fig. 28-53). (a) Show that we can prevent the beam from hitting the plate if we apply a uniform magnetic field such that $$ B \geq \sqrt{\frac{2 m K}{e^{2} d^{2}}} $$ in which \(m\) and \(e\) are the electron mass and charge. (b) How should \(\vec{B}\) be oriented?

Short Answer

Expert verified
(a) Apply \( B \geq \sqrt{\frac{2mK}{e^2d^2}} \). (b) \( \vec{B} \) should be perpendicular to the electron beam's velocity.

Step by step solution

01

Understanding the Electron's Motion

When an electron moves through a magnetic field, it experiences a force that causes it to move in a circular path. This is due to the Lorentz force, given by the equation \( F = qvB \), where \( q \) is the charge, \( v \) is the velocity, and \( B \) is the magnetic field strength.
02

Relating Kinetic Energy to Velocity

The kinetic energy \( K \) of the electron is given by \( K = \frac{1}{2}mv^2 \). Solving for velocity \( v \), we get \( v = \sqrt{\frac{2K}{m}} \). This relationship will help in finding how the electron moves under the influence of the magnetic field.
03

Circular Motion in Magnetic Fields

In a magnetic field, the electron follows a circular path with a radius \( r \) given by \( r = \frac{mv}{eB} \). This is derived from equating the centripetal force \( \frac{mv^2}{r} \) to the magnetic force \( evB \).
04

Ensuring Electron Path Avoids the Plate

To prevent the beam from hitting the plate, the circle's radius must be such that the electron completes a semicircle at or before hitting the distance \( d \). This requires that \( r = \frac{mv}{eB} \leq \frac{d}{2} \), meaning the electron will bend away before reaching the plate.
05

Solving the Inequality for Magnetic Field

Substituting \( v = \sqrt{\frac{2K}{m}} \) into \( r = \frac{mv}{eB} \leq \frac{d}{2} \), we get \( B \geq \sqrt{\frac{2mK}{e^2d^2}} \). This inequality provides the minimum magnetic field needed to ensure the beam does not hit the plate.
06

Determine Magnetic Field Orientation

The magnetic field \( \vec{B} \) should be oriented perpendicular to the velocity of the electrons. This means \( \vec{B} \) should be applied either into or out of the page (using the right-hand rule) for when the beam moves along the horizontal plane.

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Key Concepts

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

Lorentz Force
When charged particles such as electrons move through a magnetic field, they experience what is known as the Lorentz Force. This force is described by the equation \( F = qvB \), where:
  • \( F \) is the force acting on the particle,
  • \( q \) is the charge of the particle (for an electron, \( q = -e \)),
  • \( v \) is the velocity of the particle,
  • \( B \) is the magnetic field strength.
This force is perpendicular to both the velocity of the particle and the magnetic field. As a result, it causes the particle to travel in a circular path.
Understanding the behavior of electrons in a magnetic field is crucial, as it helps manipulate their paths in devices like cathode ray tubes and particle accelerators.
For practical applications, knowing how to orient the magnetic field to guide electrons ensures they reach their target or avoid obstacles, such as a metal plate in the given exercise.
Kinetic Energy
Kinetic energy represents the energy a particle possesses due to its motion. For an electron with mass \( m \) and velocity \( v \), its kinetic energy \( K \) can be expressed as:
  • \( K = \frac{1}{2}mv^2 \).
From this equation, we can derive the velocity of the electron when the kinetic energy is known by rearranging the formula to find \( v \):
  • \( v = \sqrt{\frac{2K}{m}} \).

This relationship is important because it allows us to understand how fast the electron is moving once it exits the accelerator tube.
Once the velocity is known, it can be applied to determine other characteristics of motion within the magnetic field, such as calculating the radius of the electron's path. This information is vital for ensuring the electron beam navigates its intended course effectively, as highlighted in the problem.
Circular Motion
In the context of electricity and magnetism, circular motion describes the path of charged particles when subjected to a perpendicular magnetic field. For an electron, this path can be characterized as a circle with a radius \( r \). This radius is governed by the formula:
  • \( r = \frac{mv}{eB} \).
Here, \( m \) is the electron mass, \( v \) is its velocity, \( e \) is its charge, and \( B \) is the magnetic field strength.
In the given exercise, ensuring that electrons do not strike the plate involves controlling this circular path. By imposing the condition that \( r \leq \frac{d}{2} \), where \( d \) is the distance to the plate, we ensure the electrons complete a semicircle and miss the plate.
Combining this with the derived velocity from kinetic energy, one can solve for the minimum magnetic field needed. This involves setting up the inequality for magnetic strength, leading to the necessary condition \( B \geq \sqrt{\frac{2mK}{e^2d^2}} \).
This careful orchestration of variables allows precise control over particle paths, avoiding potential collisions and optimizing the performance of devices relying on charged particle beams.

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Most popular questions from this chapter

A \(5.0 \mu \mathrm{C}\) particle moves through a region containing the uniform magnetic field \(-20 \hat{\mathrm{i}} \mathrm{mT}\) and the uniform electric field \(300 \hat{\mathrm{J}} / \mathrm{m} .\) At a certain instant the velocity of the particle is \((17 \mathrm{i}-11 \hat{\mathrm{j}}+7.0 \hat{\mathrm{k}}) \mathrm{km} / \mathrm{s}\). At that instant and in unit-vector nota- tion, what is the net electromagnetic force (the sum of the electric and magnetic forces) on the particle?

A magnetic dipole with a dipole moment of magnitude \(0.020 \mathrm{~J} / \mathrm{T}\) is released from rest in a uniform magnetic field of magnitude \(52 \mathrm{mT}\). The rotation of the dipole due to the magnetic force on it is unimpeded. When the dipole rotates through the orientation where its dipole moment is aligned with the magnetic field, its kinetic energy is \(0.80 \mathrm{~mJ}\). (a) What is the initial angle between the dipole moment and the magnetic field? (b) What is the angle when the dipole is next (momentarily) at rest?

A mass spectrometer (Fig. 28-12) is used to separate uranium ions of mass \(3.92 \times 10^{-25} \mathrm{~kg}\) and charge \(3.20 \times 10^{-19} \mathrm{C}\) from related species. The ions are accelerated through a potential difference of \(100 \mathrm{kV}\) and then pass into a uniform magnetic field, where they are bent in a path of radius \(1.00 \mathrm{~m}\). After traveling through \(180^{\circ}\) and passing through a slit of width \(1.00 \mathrm{~mm}\) and height \(1.00 \mathrm{~cm}\), they are collected in a cup. (a) What is the magnitude of the (perpendicular) magnetic field in the separator? If the machine is used to separate out \(100 \mathrm{mg}\) of material per hour, calculate (b) the current of the desired ions in the machine and (c) the thermal energy produced in the cup in \(1.00 \mathrm{~h}\).

An electron follows a helical path in a uniform magnetic field given by \(\vec{B}=(20 \hat{i}-50 \hat{j}-30 \hat{k}) \mathrm{m} \mathrm{T}\). At time \(t=0\), the electron's velocity is given by \(\vec{v}=(20 \hat{i}-30 \hat{j}+50 \hat{k}) \mathrm{m} / \mathrm{s}\). (a) What is the angle \(\phi\) between \(\vec{v}\) and \(\vec{B}\) ? The electron's velocity changes with time. Do (b) its speed and (c) the angle \(\phi\) change with time? (d) What is the radius of the helical path?

A positron with kinetic energy \(2.00 \mathrm{keV}\) is projected into a uniform magnetic field \(\vec{B}\) of magnitude \(0.100 \mathrm{~T}\), with its velocity vector making an angle of \(89.0^{\circ}\) with \(\vec{B}\). Find (a) the period, (b) the pitch \(p\), and (c) the radius \(r\) of its helical path.
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