/*! 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 2 A particle of mass \(10 \mathrm{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 28: Problem 2

A particle of mass \(10 \mathrm{~g}\) and charge \(80 \mu \mathrm{C}\) moves through a uniform magnetic field, in a region where the free-fall acceleration is \(-9.8 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}^{2} .\) The velocity of the particle is a constant \(20 \mathrm{i} \mathrm{km} / \mathrm{s},\) which is perpendicular to the magnetic field. What, then, is the magnetic field?

Short Answer

Expert verified
The magnetic field is approximately 61.25 T.

Step by step solution

01

Understanding the Given Information

We are given a mass of the particle as \( 10 \, \text{g} \) or \( 0.01 \, \text{kg} \), charge \( 80 \, \mu C \) or \( 80 \times 10^{-6} \, \text{C} \), and velocity \( \mathbf{v} = 20 \hat{i} \, \text{km/s} \) or \( 20,000 \hat{i} \, \text{m/s} \). The free-fall acceleration is \( -9.8 \hat{j} \, \text{m/s}^2 \).
02

Analyzing Forces

The particle is moving in a constant velocity, implying a net force of zero on the particle. Therefore, the gravitational force \( \mathbf{F}_g = m\mathbf{g} = (0.01) \times (-9.8 \hat{j}) \) is balanced by the magnetic force \( \mathbf{F}_B \).
03

Calculating Magnetic Force

The formula for the magnetic force on a charged particle is \( \mathbf{F}_B = q \mathbf{v} \times \mathbf{B} \). Given the net force is zero, \( \mathbf{F}_B = -\mathbf{F}_g \).Thus, \( 80 \times 10^{-6} \times 20,000 \times B \hat{k} = 0.01 \times 9.8 \hat{j} \).
04

Solving for Magnetic Field

From the equality \( 1.6 \times 10^{-3} B = 0.098 \), solve for \( B \), i.e. \[ B = \frac{0.098}{1.6 \times 10^{-3}} \approx 61.25 \, \text{T} \].

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
The magnetic force is a critical concept when considering how charged particles interact with magnetic fields. It is given by the equation \( \mathbf{F}_B = q(\mathbf{v} \times \mathbf{B}) \), where \( q \) is the charge of the particle, \( \mathbf{v} \) is its velocity, and \( \mathbf{B} \) is the magnetic field vector.
The magnetic force acts perpendicular to both the velocity of the charged particle and the magnetic field. This particular arrangement is due to the nature of the cross product in the equation:
  • **Direction**: The force direction follows the right-hand rule. If you point your fingers in the direction of velocity (\( \mathbf{v} \)), and curl them towards the magnetic field (\( \mathbf{B} \)), your thumb points in the direction of the magnetic force (\( \mathbf{F}_B \)).
  • **Magnitude**: The magnitude is proportional to the sine of the angle between \( \mathbf{v} \) and \( \mathbf{B} \). Maximum force occurs when the angle is 90 degrees, meaning \( \mathbf{v} \) is perpendicular to \( \mathbf{B} \).
This concept is essential for solving problems where charged particles move through magnetic fields, like determining the motion or the magnetic field present.
Charged Particle Motion
When a charged particle moves through a magnetic field, its motion is influenced by the magnetic force exerting a perpendicular effect.
This typically results in circular or spiral trajectories because:
  • The force is always perpendicular to its velocity, never changing the speed, only the direction.
  • The motion forms a centripetal force situation, causing circular motion if the particle moves perfectly perpendicular to a uniform field.
Mathematically, the radius of this circular path can be determined by balancing the magnetic force against the centripetal force needed to keep an object in circular motion: \[ qvB = \frac{mv^2}{r} \] solving for \( r \) gives \[ r = \frac{mv}{qB} \]
This expression shows that the radius depends on mass \( m \), velocity \( v \), charge \( q \), and magnetic field \( B \). The understanding of charged particle motion is vital for applying magnetic fields in technology and sciences, such as in particle accelerators.
Uniform Magnetic Field
A uniform magnetic field is one where the magnetic field lines are parallel and equally spaced, meaning the magnetic field strength (\( \mathbf{B} \)) is constant throughout.
This kind of field is an idealized model often used for simplifying problems involving magnetic forces and particle trajectories.
  • **Characteristics**: In a uniform magnetic field, the forces exerted on a moving charged particle will have both magnitude and direction that remain constant at given points, producing a predictable motion path.
  • **Applications**: It is commonly applied in scenarios like the field inside a solenoid (a coil of wire), or in magnetic resonance imaging (MRI) where uniform magnetic fields help produce clear imaging.
In exercises analyzing motions of charged particles, assuming a uniform magnetic field helps use simple mathematical expressions, as consistent magnetic forces and predictable circular paths simplify calculations. This was exactly the case in understanding how the magnetic force balanced with gravitational force in the given problem scenario.

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

Bainbridge's mass spectrometer, shown in Fig. 28-54, separates ions having the same velocity. The ions, after entering through slits, \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\), pass through a velocity selector composed of an electric field produced by the charged plates \(P\) and \(P^{\prime},\) and a magnetic field \(\vec{B}\) perpendicular to the electric field and the ion path. The ions that then pass undeviated through the crossed \(\vec{E}\) and \(\vec{B}\) fields enter into a region where a second magnetic field \(\vec{B}^{\prime}\) exists, where they are made to follow circular paths. A photographic plate (or a modern detector) registers their arrival. Show that, for the ions, \(q / m=E / r B B^{\prime}\), where \(r\) is the radius of the circular orbit.

A cyclotron with dee radius \(53.0 \mathrm{~cm}\) is operated at an oscillator frequency of \(12.0 \mathrm{MHz}\) to accelerate protons. (a) What magnitude \(B\) of magnetic field is required to achieve resonance? (b) At that field magnitude, what is the kinetic energy of a proton emerging from the cyclotron? Suppose, instead, that \(B=1.57 \mathrm{~T}\). (c) What oscillator frequency is required to achieve resonance now? (d) At that frequency, what is the kinetic energy of an emerging proton?

An electron follows a helical path in a uniform magnetic field of magnitude \(0.300 \mathrm{~T}\). The pitch of the path is \(6.00 \mu \mathrm{m},\) and the magnitude of the magnetic force on the electron is \(2.00 \times 10^{-15} \mathrm{~N}\). What is the electron's speed?

An electron with kinetic energy \(2.5 \mathrm{keV}\) moving along the positive direction of an \(x\) axis enters a region in which a uniform electric field of magnitude \(10 \mathrm{kV} / \mathrm{m}\) is in the negative direction of the \(y\) axis. A uniform magnetic field \(\vec{B}\) is to be set up to keep the electron moving along the \(x\) axis, and the direction of \(\vec{B}\) is to be chosen to minimize the required magnitude of \(\vec{B}\). In unit-vector notation, what \(\vec{B}\) should be set up?

A stationary circular wall clock has a face with a radius of \(15 \mathrm{~cm} .\) Six turns of wire are wound around its perimeter; the wire carries a current of \(2.0 \mathrm{~A}\) in the clockwise direction. The clock is located where there is a constant, uniform external magnetic field of magnitude \(70 \mathrm{mT}\) (but the clock still keeps perfect time). At exactly 1: 00 P.M., the hour hand of the clock points in the direction of the external magnetic field. (a) After how many minutes will the minute hand point in the direction of the torque on the winding due to the magnetic field? (b) Find the torque magnitude.
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Electricity

Read Explanation

Energy Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Kinematics Physics

Read Explanation

Further Mechanics and Thermal 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.