/*! 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 3 An electron moves at \(2.50 \tim... [FREE SOLUTION] | 91Ó°ÊÓ

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An electron moves at \(2.50 \times 10^{6} \mathrm{~m} / \mathrm{s}\) through a region in which there is a magnetic field of unspecified direction and magnitude \(7.40 \times 10^{-2} \mathrm{~T}\). (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?

Short Answer

Expert verified
(a) Maximum acceleration occurs at 90°, minimum at 0°. (b) The angle is found using inverse sine.

Step by step solution

01

Understand the magnetic force on a charged particle

The magnetic force on a charged particle can be calculated using the Lorentz force equation: \( F = qvB \sin(\theta) \), where \( q \) is the charge of the particle, \( v \) is the velocity of the particle, \( B \) is the magnetic field, and \( \theta \) is the angle between the velocity vector and the magnetic field vector.
02

Determine the charge and mass of the electron

The charge of an electron is \( q = -1.6 \times 10^{-19} \mathrm{~C}\) and the mass is \( m = 9.11 \times 10^{-31} \mathrm{~kg} \). These constants are needed to compute the force and subsequently the acceleration of the electron.
03

Calculate the largest acceleration

The largest acceleration occurs when \( \sin(\theta) = 1 \), meaning the angle \( \theta = 90^{\circ} \). Thus, the acceleration is maximized. Use the formula:\[a = \frac{F}{m} = \frac{qvB}{m}\]Substitute the given values:\[a = \frac{(-1.6 \times 10^{-19})(2.50 \times 10^6)(7.40 \times 10^{-2})}{9.11 \times 10^{-31}}\]Calculate to find the maximum acceleration \( a_{max} \).
04

Calculate the smallest acceleration

The smallest acceleration occurs when \( \sin(\theta) = 0 \), which happens when the velocity is parallel to the magnetic field. Hence, the acceleration is zero since \( F = 0 \).
05

Find the actual acceleration

We know the actual acceleration is one-fourth of the largest magnitude. Compute it using:\[a_{actual} = \frac{1}{4}a_{max}\]
06

Determine the angle for the actual acceleration

Using the fact that \( a = \frac{qvB \sin(\theta)}{m} = a_{actual} \), solve for \( \sin(\theta) \):\[\sin(\theta) = \frac{a_{actual} \cdot m}{qvB}\]Substitute \( a_{actual} = \frac{1}{4}a_{max} \) and solve for \( \theta \). Use inverse sine to find the angle.

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Key Concepts

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

Lorentz Force
The Lorentz force is a fundamental concept in physics that describes the force exerted on a charged particle moving through a magnetic field. It is given by the equation: \( F = qvB\sin(\theta) \), where:
  • \( q \) is the charge of the particle.
  • \( v \) is the velocity of the particle.
  • \( B \) is the magnetic field strength.
  • \( \theta \) is the angle between the velocity vector and the magnetic field vector.

This equation reveals a few critical insights:
  • When the particle's path is exactly perpendicular to the magnetic field (\( \theta = 90^\circ \)), \( \sin(\theta) = 1 \), leading to the maximum possible force.
  • If the particle moves parallel to the field lines (\( \theta = 0^\circ \)), \( \sin(\theta) = 0 \), resulting in no force.
By understanding these contexts, we can see how changes in the angle between velocity and magnetic field affect the force experienced by the particle. This relationship is crucial in analyzing scenarios where charged particles interact with magnetic fields, such as in the exercise given.
Electron Acceleration
When a charged particle, such as an electron, is influenced by a magnetic force, its acceleration can be derived from Newton's second law:\[ a = \frac{F}{m} \]where \( a \) is the acceleration, \( F \) is the force, and \( m \) is the mass of the electron. Using the Lorentz force in place of \( F \), the acceleration becomes:\[ a = \frac{qvB\sin(\theta)}{m} \]
For electrons:
  • The charge is negative, \( q = -1.6 \times 10^{-19} \mathrm{~C} \).
  • The mass is \( m = 9.11 \times 10^{-31} \mathrm{~kg} \).

These constants help calculate how the speed and direction of an electron change as it moves through a magnetic field. The largest acceleration occurs when the velocity of the electron is perpendicular to the magnetic field (\( \theta = 90^\circ \)), while no acceleration happens if the electron's velocity is parallel to the field (\( \theta = 0^\circ \)). These principles highlight the directional dependence of magnetic forces on particle motion.
Magnetic Field Interaction
The interaction between a magnetic field and a moving electron is a complex yet fascinating process. It is primarily guided by the orientation and magnitude of both the velocity vector of the electron and the magnetic field vector.
Let's break down this interaction:
  • The magnetic field exerts a force on the electron perpendicular to both the field direction and the velocity of the electron, causing it to move in a helical path.
  • The magnitude of the force depends on the component of the velocity perpendicular to the magnetic field. This is the essence of why \( \sin(\theta) \) impacts the force.
  • This interaction causes the electron to experience centripetal acceleration, altering its trajectory without doing work, as magnetic forces do not change the particle's speed, only its direction.

These interactions are key to how devices like cyclotrons and magnetic confinement systems in fusion reactors operate, where manipulating charged particles using magnetic fields is essential for function. Studying these effects allows for a deeper understanding of both fundamental physics and applied technologies involving magnetic fields.

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Most popular questions from this chapter

You wish to hit a target from several meters away with a charged coin having a mass of \(4.25 \mathrm{~g}\) and a charge of \(+2500 \mu \mathrm{C}\). The coin is given an initial velocity of \(12.8 \mathrm{~m} / \mathrm{s}\), and a downward, uniform electric field with field strength \(27.5 \mathrm{~N} / \mathrm{C}\) exists throughout the region. If you aim directly at the target and fire the coin horizontally, what magnitude and direction of uniform magnetic field are needed in the region for the coin to hit the target? (take \(\left.g=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)\)

In designing a velocity selector that uses uniform perpendicular electric and magnetic fields, you want to select positive ions of charge \(+5 e\) that are traveling perpendicular to the fields at \(8.75 \mathrm{~km} / \mathrm{s}\). The magnetic field available to you has a magnitude of \(0.550 \mathrm{~T}\). (a) What magnitude of electric field do you need? (b) Show how the two fields should be oriented relative to each other and to the velocity of the ions. (c) Will your velocity selector also allow the following ions (having the same velocity as the \(+5 e\) ions) to pass through undeflected: (i) negative ions of charge \(-5 e\), (ii) positive ions of charge different from \(+5 e ?\)

In a certain region uniform electric field \(E\) and magnetic field \(B\) are present in the opposite direction. At the instant \(t=0\), a particle of mass \(m\) carrying a charge \(q\) is given velocity \(v_{0}\) at an angle \(\theta\), with the \(y\) axis, in the \(y z\) plane. Find the time after which the speed of the particle would be minimum.

(a) An \({ }^{16} \mathrm{O}\) nucleus (charge \(+8 e\) ) moving horizontally from west to east with a speed of \(500 \mathrm{~km} / \mathrm{s}\) experiences a magnetic force of \(0.00320 \mathrm{nN}\) vertically downward. Find the magnitude and direction of the weakest magnetic field required to produce this force. Explain how this same force could be caused by a larger magnetic field. (b) An electron moves in a uniform, horizontal, \(2.10-\mathrm{T}\) magnetic field that is toward the west. What must the magnitude and direction of the minimum velocity of the electron be so that the magnetic force on it will be \(4.60 \mathrm{pN}\), vertically upward? Explain how the velocity could be greater than this minimum value and the force still have this same magnitude and direction.

Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is \(155 \mathrm{~V} / \mathrm{m}\) and the magnetic field is \(0.031 \mathrm{~T}\). The ions next enter a uniform magnetic field of magnitude \(0.0175 \mathrm{~T}\) that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is \(17.5 \mathrm{~cm}\), what is their mass?

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