/*! 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 25 \(\mathrm{A} 12.5-\mu \mathrm{C}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 25

\(\mathrm{A} 12.5-\mu \mathrm{C}\) particle with a mass of \(2.80 \times 10^{-5} \mathrm{kg}\) moves perpendicular to a \(1.01-\) T magnetic field in a circular path of radius \(21.8 \mathrm{m}\). (a) How fast is the particle moving? (b) How long will it take the particle to complete one orbit?

Short Answer

Expert verified
(a) The particle moves at 9.8 m/s. (b) It takes approximately 14.0 s to complete one orbit.

Step by step solution

01

Understanding Lorentz Force

In a magnetic field, a charged particle moving with velocity \( v \) experiences a force called the Lorentz force, which causes it to move in a circular path. The force is perpendicular to the velocity and magnetic field, and its magnitude is given by \( F = qvB \), where \( q \) is the charge, \( v \) is the velocity, and \( B \) is the magnetic field strength.
02

Equate Centripetal and Magnetic Forces

The particle is moving in a circle due to the magnetic force providing the necessary centripetal force. Thus, we equate the magnetic force \( qvB \) to the centripetal force \( \frac{mv^2}{r} \), where \( m \) is mass, \( v \) is velocity, and \( r \) is the radius: \( qvB = \frac{mv^2}{r} \).
03

Solve for Velocity

Rearrange the equation \( qvB = \frac{mv^2}{r} \) to solve for velocity \( v \):\[v = \frac{qBr}{m}\]Substitute the known values: \( q = 12.5 \times 10^{-6} \) C, \( B = 1.01 \) T, \( r = 21.8 \) m, and \( m = 2.80 \times 10^{-5} \) kg. Calculate \( v \).
04

Calculate Orbital Time Period

The time \( T \) to complete one orbit is the circumference of the circle \( 2\pi r \) divided by the velocity \( v \):\[T = \frac{2\pi r}{v}\]Substitute the radius \( r \) and the velocity \( v \) from the previous step to find the time period \( T \).
05

Final Calculations and Solutions

Substitute the values into the equations to calculate: 1. Velocity \( v = \frac{12.5 \times 10^{-6} \times 1.01 \times 21.8}{2.80 \times 10^{-5}} = 9.8 \) m/s.2. Time period \( T = \frac{2 \pi \times 21.8}{9.8} \approx 14.0 \) s.

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.

Lorentz Force
The Lorentz force is a fundamental concept of electromagnetism, describing the force experienced by a charged particle when it moves through a magnetic field.
This force plays a critical role in determining the path of the particle.
  • Direction: The force acts perpendicularly to both the velocity of the particle and the magnetic field, orienting it into a circular motion.
  • Magnitude: The magnitude of this force can be calculated using the equation: \( F = qvB \).
    Here, \( q \) represents the charge of the particle, \( v \) is its velocity, and \( B \) is the strength of the magnetic field.
Because the Lorentz force is perpendicular to the motion at each point, it doesn't work in terms of energy but alters the direction of the velocity, leading to circular movement.
Centripetal Force
Centripetal force is necessary for any object to move in a circular path.
It acts towards the center of the circle in which the object is moving.
This force helps balance the body's inertia, pulling it inward and keeping it on its circular trajectory.
  • The need for centripetal force is essential because objects naturally move in straight lines, a consequence of Newton's first law of motion.
  • The expression for centripetal force \( F = \frac{mv^2}{r} \) demonstrates the relationship between mass \( m \), velocity \( v \), and radius \( r \) of the circle.
In the case of a charged particle in a magnetic field, the Lorentz force provides the required centripetal force for circular motion.
Circular Motion
Circular motion arises when a force consistently changes an object's direction, yet not its speed. This change must always be perpendicular to the velocity of the object.
  • Characteristics: It is uniform if the speed is constant, which is often assumed when discussing motion in magnetic fields.
  • Forces Involved: In the context of a magnetic field, the Lorentz force becomes the source of the centripetal force that keeps the object in circular motion.
The particle's velocity and the magnetic field's direction determine the curvature of the path, with the circular path's radius being affected by the balance of the centripetal force and the magnetic force.
Magnetic Field
A magnetic field is a region where magnetic forces are exerted, typically represented by magnetic field lines. It affects charged particles moving within the field.
  • Direction: Given by a vector quantity, its direction aligns with the direction a north pole of a compass needle points.
  • Magnetic Influence: Moving charges experience a force due to the magnetic field, which can be harnessed to manipulate paths into circular arcs.
The field's strength, measured in teslas (T), affects the Lorentz force acting on a charged particle.
In this problem, the uniform magnetic field exerts a force on the moving charged particle, steering its path into circular motion due to the continuous application of the Lorentz force.

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

Find the angle between the plane of a loop and the magnetic field for which the magnetic torque acting on the loop is equal to \(x\) times its maximum value, where \(0 \leq x \leq 1\).

The magnetic force exerted on a \(1.2-\mathrm{m}\) segment of straight wire is \(1.6 \mathrm{N}\). The wire carries a current of \(3.0 \mathrm{A}\) in a region with a constant magnetic field of 0.50 T. What is the angle between the wire and the magnetic field?

In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius \(5.29 \times 10^{-11} \mathrm{m}\) about the nucleus. Given that the charge on the electron is \(-1.60 \times 10^{-19} \mathrm{C}\), and that its speed is \(2.2 \times 10^{6} \mathrm{m} / \mathrm{s},\) find the magnitude of the magnetic field the electron produces at the nucleus of the atom.

An alpha particle (the nucleus of a helium atom) consists of two protons and two neutrons, and has a mass of \(6.64 \times 10^{-27} \mathrm{kg} .\) A horizontal beam of alpha particles is injected with a speed of \(1.3 \times 10^{5} \mathrm{m} / \mathrm{s}\) into a region with a vertical magnetic field of magnitude \(0.155 \mathrm{T}\). (a) How long does it take for an alpha particle to move halfway through a complete circle? (b) If the speed of the alpha particle is doubled, does the time found in part (a) increase, decrease, or stay the same? Explain. (c) Repeat part (a) for alpha particles with a speed of \(2.6 \times 10^{5} \mathrm{m} / \mathrm{s}\).

Charged particles pass through a velocity selector with electric and magnetic fields at right angles to each other, as shown in Figure \(22-32 .\) If the electric field has a magnitude of \(450 \mathrm{N} / \mathrm{C}\) and the magnetic field has a magnitude of \(0.18 \mathrm{T},\) what speed must the particles have to pass through the selector undeflected?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Circular Motion and Gravitation

Read Explanation

Medical Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Physical Quantities And Units

Read Explanation

Mechanics and Materials

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.