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In New England, the horizontal component of the earth's magnetic field has a magnitude of \(1.6 \times 10^{-5} \mathrm{T} .\) An electron is shot vertically straight up from the ground with a speed of \(2.1 \times 10^{6} \mathrm{m} / \mathrm{s} .\) What is the magnitude of the acceleration caused by the magnetic force? Ignore the gravitational force acting on the electron.

Short Answer

Expert verified
The acceleration is approximately \(5.90 \times 10^{12} \mathrm{m/s}^2\).

Step by step solution

01

Identify Relevant Formulas

The magnetic force on a moving charge is given by the formula: \( F = qvB \sin(\theta) \). Since the electron is moving perpendicular to the magnetic field, \( \theta = 90^{\circ} \) and \( \sin(90^{\circ}) = 1 \). Therefore, \( F = qvB \).
02

Calculate the Magnetic Force

Substitute the values: charge of an electron \( q = -1.6 \times 10^{-19} \mathrm{C} \), velocity \( v = 2.1 \times 10^{6} \mathrm{m/s} \), and magnetic field \( B = 1.6 \times 10^{-5} \mathrm{T} \) into the formula. Calculate \( F = (1.6 \times 10^{-19} \mathrm{C})(2.1 \times 10^{6} \mathrm{m/s})(1.6 \times 10^{-5} \mathrm{T}) = 5.376 \times 10^{-18} \mathrm{N} \).
03

Calculate the Acceleration

Use Newton's second law: \( F = ma \), where \( m \) is the mass of an electron \( 9.11 \times 10^{-31} \mathrm{kg} \). Solve for \( a \): \( a = \frac{F}{m} = \frac{5.376 \times 10^{-18} \mathrm{N}}{9.11 \times 10^{-31} \mathrm{kg}} \).
04

Simplify to Find Acceleration

Calculate \( a = \frac{5.376 \times 10^{-18} \mathrm{N}}{9.11 \times 10^{-31} \mathrm{kg}} \approx 5.90 \times 10^{12} \mathrm{m/s}^2 \). This is the magnitude of the acceleration caused by the magnetic force.

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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 electromagnetism that explains how charged particles move in electromagnetic fields. When a charged particle, like an electron, travels through a magnetic field, it experiences a force that can change its motion.
This force is given by the equation:
  • \( F = qvB \sin(\theta) \)
In this formula:
  • \( F \) is the magnetic force.
  • \( q \) is the charge of the particle.
  • \( v \) represents the velocity of the particle.
  • \( B \) is the magnetic field.
  • \( \theta \) is the angle between the velocity and the magnetic field.
The unique property of the Lorentz force is its direction, which is always perpendicular to both the velocity of the particle and the magnetic field. This characteristic causes the particle to move in a circular path when the velocity is perpendicular to the field. For the problem involving an electron in New England's magnetic field, the angle \( \theta \) is \( 90^\circ \), making the \( \sin(90^\circ) = 1 \). Hence, the equation simplifies to \( F = qvB \).
Acceleration of Charged Particles
Acceleration refers to the change in velocity of a particle due to an applied force. When dealing with charged particles in a magnetic field, this force is the Lorentz force. To find the acceleration caused by this force, we use Newton’s second law:
  • \( F = ma \)
Here, \( F \) is the force from the Lorentz equation, \( m \) is the mass of the charged particle, and \( a \) is the acceleration. Rearranging this formula, we can find:
  • \( a = \frac{F}{m} \)
Substituting the magnetic force calculated from the Lorentz force equation yields the particle's acceleration.
For the given exercise, using the magnetic force \( 5.376 \times 10^{-18} \; N \) and the known mass of an electron \( 9.11 \times 10^{-31} \; kg \), the acceleration can be calculated as \( 5.90 \times 10^{12} \; m/s^2 \). This high acceleration showcases the significant effect of even small magnetic forces on tiny particles like electrons due to their relatively low mass.
Magnetic Field Components
Magnetic fields are vector fields characterized by both a magnitude and a direction. Understanding the components of a magnetic field helps us analyze how charged particles interact with these fields.
There are two main components concerning the interaction of particles:
  • Magnitude: The strength of the magnetic field, often measured in Tesla (T), defines how much force the field can exert on a charged particle. In the exercise, the magnetic field has a magnitude of \( 1.6 \times 10^{-5} \; T \).
  • Direction: It indicates the orientation of the magnetic field line. For example, horizontal in this scenario means it runs parallel to the ground.
By knowing these components, we can predict the trajectory and experience of charged particles in fields. The orientation greatly impacts the force, as seen with the electron traveling vertically, optimizing the enforcement of the magnetic force when combined with the horizontal component of New England's magnetic field.

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Most popular questions from this chapter

Review Conceptual Example 2 as an aid in understanding this problem. A velocity selector has an electric field of magnitude \(2470 \mathrm{N} / \mathrm{C},\) directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of \(6.50 \times 10^{3} \mathrm{m} / \mathrm{s},\) enter the velocity selector and are able to pass completely through without being deflected. When a different particle with an electric charge of \(+4.00 \times 10^{-12} \mathrm{C}\) enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is \(1.90 \times 10^{-9} \mathrm{N},\) pointing directly upward. What is the speed of this particle?

A square coil and a rectangular coil are each made from the same length of wire. Each contains a single turn. The long sides of the rectangle are twice as long as the short sides. Find the ratio \(\tau_{\text {xquar }} / \tau_{\text {rectangle }}\) of the maximum torques that these coils experience in the same magnetic field when they contain the same current.

Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The length of each rod is \(0.85 \mathrm{m},\) and the mass of each is \(0.073 \mathrm{kg}\). One rod is held in place above the ground, while the other floats beneath it at a distance of \(8.2 \times 10^{-3} \mathrm{m} .\) Determine the current in the rods.

Suppose that a uniform magnetic field is everywhere perpendicular to this page. The field points directly upward toward you. A circular path is drawn on the page. Use Ampère's law to show that there can be no net current passing through the circular surface.

A charge is moving perpendicular to a magnetic field and experiences a force whose magnitude is \(2.7 \times 10^{-3} \mathrm{N}\). If this same charge were to move at the same speed and the angle between its velocity and the same magnetic field were \(38^{\circ},\) what would be the magnitude of the magnetic force that the charge would experience?

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