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Chapter 21: Problem 78

A particle that has an \(8.2-\mu C\) charge moves with a velocity of magnitude \(5.0 \times 10^{5} \mathrm{~m} / \mathrm{s}\). When the velocity points along the \(+x\) axis, the particle 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. Make sure that your answers are consistent with your answers to the Concept Questions.

Short Answer

Expert verified
The magnitude of the magnetic field is approximately 1.17 T, with possible directions along the +x or -x axis.

Step by step solution

01

Understanding the problem

We are given a charged particle experiencing zero magnetic force along the +x-axis, which indicates that the magnetic field must be parallel (or anti-parallel) to the velocity for no force to occur. We are also given the maximum force the particle can experience when the velocity and magnetic field are perpendicular. We need to find the magnitude and the possible directions of the magnetic field.
02

Using the Lorentz force equation

The magnetic force experienced by a charged particle is given by the equation: \[ F = qvB \sin(\theta) \], where \( F \) is the magnetic force, \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the field. When \( \theta = 0\), the force is zero; when \( \theta = 90^\circ\), the force is maximum.
03

Solve for the magnitude of the magnetic field

Given the maximum force \( F = 0.48 \, \mathrm{N} \), the charge \( q = 8.2 \, \mu \mathrm{C} = 8.2 \times 10^{-6} \, \mathrm{C} \), and velocity \( v = 5.0 \times 10^{5} \, \mathrm{m/s} \), you can solve for the magnetic field \( B \) when \( \sin(\theta) = 1 \): \[ B = \frac{F}{qv} = \frac{0.48}{8.2 \times 10^{-6} \times 5.0 \times 10^{5}} \approx 1.17 \, \mathrm{T} \].So, the magnitude of the magnetic field is approximately \( 1.17 \, \mathrm{T} \).
04

Determine direction possibilities for the magnetic field

Since the velocity is along the +x-axis and the magnetic force is zero when field and velocity are parallel or anti-parallel, two possible directions exist for the magnetic field when it causes no force: along the +x direction or the -x direction. For maximum force, the field must be perpendicular, implying it could also be in the y or z direction relative to the initial setup to satisfy vector cross-product rules.

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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 central to understanding how charged particles interact with magnetic and electric fields. It can be described by the equation: \[ F = q(E + v \times B) \] where:
  • \( F \) is the Lorentz force acting on the charged particle.
  • \( q \) is the electric charge of the particle.
  • \( E \) is the electric field.
  • \( v \) is the particle's velocity.
  • \( B \) is the magnetic field.
In scenarios dealing purely with magnetic fields, the electric field component \( E \) is zero. Thus, the force depends on the velocity of the particle and the magnetic field, as well as the angle \( \theta \) between \( v \) and \( B \).
The force reaches a maximum when the velocity is perpendicular to the magnetic field, causing a maximum effect on the charged particle's trajectory.
Charged particle
Charged particles, such as electrons or protons, are fundamental units carrying an electric charge. A charge can be either positive or negative, and it enables particles to interact with electromagnetic fields.
In this exercise, we examine a particle with a charge of \(8.2\,\mu C\), which means it carries an electric charge that is \(8.2 \times 10^{-6}\, C\). The interaction of this charge with a magnetic field is governed by the Lorentz force.
Charged particles move in straight lines only if unperturbed. However, within a magnetic field, they can experience a force that alters their trajectory depending on the orientation and strength of the magnetic field relative to the particle's path.
Velocity of particle
The velocity of a particle is a vector quantity that includes both the speed and direction of movement. In the context of a magnetic field, this value is crucial in determining the magnetic force experienced by a charged particle.
The particle in the given exercise moves at a velocity of \(5.0 \times 10^{5}\, \text{m/s}\) along the \(+x\) axis.
This specific direction becomes relevant because if the magnetic field is aligned with the velocity (i.e., they are parallel), the particle will experience no magnetic force. Conversely, when the field is perpendicular to velocity, maximum force is experienced.
Magnetic force
Magnetic force is a component of the Lorentz force, and it only acts on moving charged particles within a magnetic field. The magnitude of the force is calculated using: \[ F = qvB \sin(\theta) \] where \( \sin(\theta) \) reflects the angle between the velocity vector and the magnetic field vector.
Without a perpendicular component, the force becomes zero. In this exercise, the maximum magnetic force is \(0.48 \, \text{N}\), occurring when the velocity is perpendicular to the magnetic field. The magnetic force does not alter the speed of the particle but instead changes its direction, bending its path until no force is present when the vectors align.
Directions of magnetic field
The direction of a magnetic field determines the nature and direction of the force exerted on a charged particle. Two potential scenarios exist in this exercise:
  • When the particle's velocity is parallel or anti-parallel to the magnetic field, no magnetic force acts on it.
  • In contrast, if the magnetic field is perpendicular, this results in the maximum magnetic force acting on the particle.
For the given problem, the magnetic field could either be along the \(+x\) or \(-x\) direction to cause zero force when aligned with the particle's velocity. To achieve maximum force, the field could also be oriented in directions perpendicular to the \(+x\) axis, such as along the \(y\)- or \(z\)-axis. Understanding these possible directions is critical in predicting and observing the behavior of charged particles in magnetic fields.

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Most popular questions from this chapter

The length of the wire is \(L=1.00 \mathrm{~m}\). The current in the coil is \(I=1.7 \mathrm{~A}\), and the magnetic field of the motor is \(0.34 \mathrm{~T}\). Find the maximum torque when the wire is used to make a single- turn square coil and a two-turn square coil. Verify that your answers are consistent with your answer to the Concept Question.

In the operating room, anesthesiologists use mass spectrometers to monitor the respiratory gases of patients undergoing surgery. One gas that is often monitored is the anesthetic isoflurane (molecular mass \(\left.=3.06 \times 10^{-25} \mathrm{~kg}\right)\). In a spectrometer, a singly ionized molecule of isoflurane (charge \(=+e\) ) moves at a speed of \(7.2 \times 10^{3} \mathrm{~m} / \mathrm{s}\) on a circular path that has a radius of \(0.10 \mathrm{~m}\). What is the magnitude of the magnetic field that the spectrometer uses?

Review Interactive Solution \(\underline{21.55} 21.55\) at for one approach to this problem. Two circular coils are concentric and lie in the same plane. The inner coil contains 140 turns of wire, has a radius of \(0.015 \mathrm{~m}\), and carries a current of \(7.2 \mathrm{~A}\). The outer coil contains 180 turns and has a radius of \(0.023 \mathrm{~m}\). What must be the magnitude and direction (relative to the current in the inner coil) of the current in the outer coil, such that the net magnetic field at the common center of the two coils is zero?

A wire has a length of \(7.00 \times 10^{-2} \mathrm{~m}\) and is used to make a circular coil of one turn. There is a current of \(4.30 \mathrm{~A}\) in the wire. In the presence of a 2.50-T magnetic field, what is the maximum torque that this coil can experience?

A proton is projected perpendicularly into a magnetic field that has a magnitude of \(0.50 \mathrm{~T}\). The field is then adjusted so that an electron will follow the exact same circular path when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron? Verify that your answer is consistent with your answers to the Concept Questions.
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