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Chapter 21: Problem 6

A particle that has an \(8.2 \mu \mathrm{C}\) charge moves with a velocity of magnitude \(5.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) along the \(+x\) axis. It experiences no magnetic force, although there is a magnetic field present. The maximum possible magnetic force that the charge could experience has a magnitude of \(0.48 \mathrm{N}\). Find the magnitude and direction of the magnetic field. Note that there are two possible answers for the direction of the field.

Short Answer

Expert verified
The magnetic field has a magnitude of 1.17 T and can be either in the +x or -x direction.

Step by step solution

01

Understanding the Magnetic Force Formula

The force experienced by a charged particle in a magnetic field is given by the formula: \( F = qvB \sin \theta \), where \( F \) is the magnetic force, \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field, and \( \theta \) is the angle between the velocity and the magnetic field.
02

Analyzing Given Conditions

It is given that the particle experiences no magnetic force. This implies \( \sin \theta = 0 \), meaning \( \theta = 0^\circ \) or \( \theta = 180^\circ \). The velocity and magnetic field must be parallel or anti-parallel.
03

Considering Maximum Magnetic Force

The maximum possible magnetic force \( F_{max} = 0.48 \ \mathrm{N} \) occurs when \( \sin \theta = 1 \), i.e., \( \theta = 90^\circ \). Therefore, the particle must be moving perpendicular to the magnetic field in this scenario to achieve this force.
04

Solving for Magnetic Field Magnitude

Rearrange the magnetic force formula for the magnetic field when maximum force occurs: \( B = \frac{F_{max}}{qv} \). Substitute the given values: \( q = 8.2 \times 10^{-6} \ \mathrm{C} \), \( v = 5.0 \times 10^{5} \ \mathrm{m/s} \), and \( F_{max} = 0.48 \ \mathrm{N} \).
05

Calculating Magnetic Field Magnitude

Plug the values into the equation: \[ B = \frac{0.48}{8.2 \times 10^{-6} \times 5.0 \times 10^{5}} \approx \frac{0.48}{4.1} \times 10^{-1} \approx 1.17 \ \mathrm{T} \]
06

Identifying Possible Directions

Since the force experience is zero, the direction of the magnetic field must be along the \(+x\) or \(-x\) axis. This means that the magnetic field can either be parallel (\(+x\)) or anti-parallel (\(-x\)) to the velocity. Therefore, the two possible directions for the magnetic field are \(+x\) or \(-x\) axis.

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Key Concepts

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

Magnetic Force
The magnetic force is an essential concept in understanding how particles behave in a magnetic field. It is calculated using the equation: \[ F = qvB \sin \theta \] where: - \( F \) is the magnetic force experienced by the particle. - \( q \) is the charge of the particle. - \( v \) is the velocity of the particle. - \( B \) is the strength of the magnetic field. - \( \theta \) is the angle between the direction of the velocity and the magnetic field. These components work together to define the magnetic force. When the angle \( \theta \) is 90 degrees, the force is maximized because \( \sin \theta \) reaches its peak value of 1. Conversely, if \( \theta \) is 0 or 180 degrees, the force will be zero as \( \sin \theta \) is zero, indicating no interaction.
This concept explains why a charged particle may not experience a force despite the presence of a magnetic field.
Charged Particle
A charged particle, an entity with an electric charge, interacts with magnetic fields through magnetic forces. In this context, the charge is expressed in microcoulombs (\( \mu C \)), which is one-millionth of a coulomb. The particle in the problem has a charge of \( 8.2 \times 10^{-6} \) coulombs.
  • Positive or negative charges can affect how the particle experiences force in a magnetic field.
  • The interaction depends on the orientation of the charge relative to the magnetic field.
The charged nature of the particle is fundamental to determining if there will be an interaction with the magnetic field, and how strong that interaction could be depending on the values of other contributing factors.
Velocity of a Particle
The velocity of a particle is a crucial vector quantity. It not only includes the speed of the particle but also the direction in which the particle is moving. In our example, the particle has a velocity of \( 5.0 \times 10^{5} \, \mathrm{m/s} \) along the \(+x\) axis.
  • Velocity is a key component in the magnetic force formula \( F = qvB \sin \theta \).
  • The direction of velocity significantly impacts whether the magnetic force is felt.
  • A change in speed or direction can alter the force experienced.
Since velocity plays together with other factors like charge and field strength, it determines the motion and behavior of particles within a magnetic field.
Magnetic Field Direction
The direction of a magnetic field is as important as its magnitude when determining the effects on a charged particle. In this problem, since the magnetic force is zero, the magnetic field's direction must be along the same axis as the velocity.
When the velocity and magnetic field are parallel or anti-parallel (\( \theta = 0 \) or \( \theta = 180^{\circ} \)), the particle experiences no force.
  • Magnetic fields run from the north to the south pole.
  • When dealing with two possible directions, as the problem suggests, the magnetic field can be in the direction of the \(+x\) or \(-x\) axis.
  • This gives rise to two scenarios: parallel or anti-parallel orientation with respect to the particle's velocity.
The directionality informs us of the orientation a magnetic field must take to result in no exerted force.

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Most popular questions from this chapter

The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole human bodies has a field strength of \(7.0 \mathrm{T},\) and the current in the solenoid is \(2.0 \times 10^{2} \mathrm{A}\). What is the number of turns per meter of length of the solenoid? Note that the solenoid used to produce the magnetic field in this type of system has a length that is not very long compared to its diameter. Because of this and other design considerations, your answer will be only an approximation.

A particle has a charge of \(q=+5.60 \mu \mathrm{C}\) and is located at the coordinate origin. As the drawing shows, an electric field of \(E_{x}=+245 \mathrm{N} / \mathrm{C}\) exists along the \(+x\) axis. A magnetic field also exists, and its \(x\) and \(y\) components are \(B_{x}=+1.80 \mathrm{T}\) and \(B_{y}=+1.40 \mathrm{T} .\) Calculate the force (magnitude and direction) exerted on the particle by each of the three fields when it is (a) stationary, (b) moving along the \(+x\) axis at a speed of \(375 \mathrm{m} / \mathrm{s},\) and \((\mathrm{c})\) moving along the \(+z\) axis at a speed of \(375 \mathrm{m} / \mathrm{s}\)

An electron is moving through a magnetic field whose magnitude is \(8.70 \times 10^{-4} \mathrm{T}\). The electron experiences only a magnetic force and has an acceleration of magnitude \(3.50 \times 10^{14} \mathrm{m} / \mathrm{s}^{2} .\) At a certain instant, it has a speed of \(6.80 \times 10^{6} \mathrm{m} / \mathrm{s}\). Determine the angle \(\theta\) (less than \(90^{\circ}\) ) between the electron's velocity and the magnetic field.

An \(\alpha\) -particle has a charge of \(+2 e\) and a mass of \(6.64 \times 10^{-27} \mathrm{kg} .\) It is accelerated from rest through a potential difference that has a value of \(1.20 \times 10^{6} \mathrm{V}\) and then enters a uniform magnetic field whose magnitude is \(2.20 \mathrm{T}\). The \(\alpha\) -particle moves perpendicular to the magnetic field at all times. What is (a) the speed of the \(\alpha\) -particle, (b) the magnitude of the magnetic force on it, and (c) the radius of its circular path?

Two parallel rods are each \(0.50 \mathrm{m}\) in length. They are attached at their centers to either end of a spring (spring constant \(=150 \mathrm{N} / \mathrm{m}\) ) that is initially neither stretched nor compressed. When 950 A of current is in each rod in the same direction, the spring is observed to be compressed by \(2.0 \mathrm{cm} .\) Treat the rods as long, straight wires and find the separation between them when the current is present.
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