/*! 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 5 A group of particles is travelin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 5

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at \(1.50 \mathrm{~km} / \mathrm{s}\) in the \(+x\)-direction experiences a force of \(2.25 \times 10^{-16} \mathrm{~N}\) in the \(+y\)-direction, and an electron moving at \(4.75 \mathrm{~km} / \mathrm{s}\) in the \(-z\)-direction experiences a force of \(8.50 \times\) \(10^{-16} \mathrm{~N}\) in the \(+y\)-direction. (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the \(-y\)-direction at \(3.20 \mathrm{~km} / \mathrm{s} ?\)

Short Answer

Expert verified
(a) Magnetic field: 0.9375 T in the +z-direction. (b) Force on electron: 4.80 x 10^{-16} N in the +x-direction.

Step by step solution

01

Use the Formula for Magnetic Force

The force on a charged particle moving through a magnetic field is given by the formula \( F = qvB \sin \theta \), where \( F \) is the force, \( q \) is the charge of the particle, \( v \) is the velocity, \( B \) is the magnetic field, and \( \theta \) is the angle between the velocity and the magnetic field. For both the proton and the electron, the force is perpendicular to their velocity, making \( \theta = 90^\circ \). Thus, \( \sin \theta = 1 \).
02

Calculate the Magnetic Field's Magnitude Using the Proton's Data

For the proton moving in the \(+x\)-direction, \( q = 1.6 \times 10^{-19} \mathrm{C} \), \( v = 1.50 \times 10^3 \mathrm{m/s} \), and \( F = 2.25 \times 10^{-16} \mathrm{N} \). Using \( F = qvB \) gives: \[ 2.25 \times 10^{-16} = (1.6 \times 10^{-19})(1.50 \times 10^3)B \] Solve for \( B \): \[ B = \frac{2.25 \times 10^{-16}}{(1.6 \times 10^{-19})(1.50 \times 10^3)} \approx 0.9375 \mathrm{T} \]
03

Determine the Magnetic Field's Direction from Proton's Data

The magnetic force on the proton is in the \(+y\)-direction while it moves in the \(+x\)-direction. The right-hand rule indicates that the magnetic field must be in the \(+z\)-direction to yield this result. Thus, \( \vec{B} = 0.9375 \mathrm{T} \, \hat{z} \).
04

Validate Using Electron's Data

For the electron moving in the \(-z\)-direction with \( F = 8.50 \times 10^{-16} \mathrm{N} \), \( q = -1.6 \times 10^{-19} \mathrm{C} \), and \( v = 4.75 \times 10^3 \mathrm{m/s} \). Using \( F = |q|vB \), \[ 8.50 \times 10^{-16} = (1.6 \times 10^{-19})(4.75 \times 10^3)(0.9375) \] Checks out; the calculation is consistent.
05

Calculate Force on an Electron in the -y-direction

For an electron moving in the \(-y\)-direction with \( v = 3.20 \times 10^3 \mathrm{m/s} \) in a \(+z\)-directed magnetic field \(0.9375 \mathrm{T}\), \[ F = qvB \] \[ F = (1.6 \times 10^{-19} \mathrm{C})(3.20 \times 10^3 \mathrm{m/s})(0.9375) \approx -4.80 \times 10^{-16} \mathrm{N} \] The force, by the right-hand rule, acts in the \(+x\)-direction.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Magnetic Force
Magnetic force is the force that a charged particle experiences when it moves within a magnetic field. It is crucial to understand because it impacts the motion and path of particles such as electrons and protons. The magnetic force can be calculated using the formula \( F = qvB \sin \theta \), where \( F \) is the force, \( q \) is the charge, \( v \) is the velocity of the particle, \( B \) is the magnitude of the magnetic field, and \( \theta \) is the angle between the velocity and the magnetic field.

Importantly, when the particle's velocity is perpendicular to the magnetic field, as it often is in textbook problems, the sine of the angle \( \theta \) is 1. This simplifies the formula to \( F = qvB \). In these conditions, the force is maximized and always perpendicular to both the velocity of the charged particle and the magnetic field itself.
Right-Hand Rule
The right-hand rule is a helpful tool used to determine the direction of the magnetic force on a moving charged particle. Here's how it works:
  • Point your thumb in the direction of the particle's velocity.
  • Point your fingers in the direction of the magnetic field.
  • Your palm will face in the direction of the force exerted on a positive charge.

For negative charges like electrons, the force direction is opposite to the way your palm faces.

In our exercise, when the proton鈥攎oving in the \(+x\)-direction鈥攅xperiences a force in the \(+y\)-direction, and since it's positive, the magnetic field must be in the \(+z\)-direction because that's where your fingers would align when you apply the right-hand rule. Knowing this helps us deduce both the magnitude and direction of forces acting on particles in magnetic fields.
Charged Particle Motion
A charged particle's motion becomes complex as it travels through a magnetic field because of the force it experiences. This force does not increase the speed of the particle but rather changes its direction, often causing the particle to move in a circular or helical path.

In the scenario given, both a proton and an electron were observed. The proton moved in the \(+x\)-direction while the electron in the \(-z\)-direction, each experiencing a force perpendicular to their motion. Such observations allow us to confirm the consistent direction of the magnetic field using different charged particles.
  • For a proton, the magnetic force acts perpendicular to its velocity, giving us key insights into the field's direction and magnitude.
  • For an electron, this same principle helps validate those findings with additional data points.

Understanding these principles is vital for interpreting and predicting the paths of charged particles in magnetic fields.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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?

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?

A \(150-g\) ball containing \(4.00 \times 10^{8}\) excess electrons is dropped into a \(125-\mathrm{m}\) vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude \(0.250 \mathrm{~T}\) and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field.

Magnetic Moment of the Hydrogen Atom. In the Bohr model of the hydrogen atom (see Section \(8.5\) ), in the lowest energy state the electron orbits the proton at a speed of \(2.2 \times\) \(10^{6} \mathrm{~m} / \mathrm{s}\) in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{~m}\). (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current \(I ?\) (c) What is the magnetic moment of the atom due to the motion of the electron?

An Electromagnetic Rail Gun. A conducting bar with mass \(m\) and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\overrightarrow{\boldsymbol{B}}\) fills the region between the rails (Fig. P7.41). (a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m\), find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v .\) (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth \((11.2 \mathrm{~km} / \mathrm{s}) .\) Let \(B=0.80 \mathrm{~T}, I=2.0 \times 10^{3} \mathrm{~A}, m=25 \mathrm{~kg}\), and \(L=50 \mathrm{~cm}\). For simplicity asssume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Fields in Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Turning Points in Physics

Read Explanation

Dynamics

Read Explanation

Kinematics Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.