/*! 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 12 An electron moves at \(2.50 \tim... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 12

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 force on the electron due to the magnetic field? (b) If the force on the electron is half of its maximum value, what is the angle between the electron's velocity and the magnetic field?

Short Answer

Expert verified
(a) Max force at 90掳: \( 2.96 \times 10^{-14} \mathrm{~N} \); Min force: 0 at 0掳 or 180掳. (b) Angle for half max force: 30掳 or 150掳.

Step by step solution

01

Understanding the Magnetic Force Formula

The magnetic force on a charged particle like an electron moving through a magnetic field is given by \( F = qvB \sin(\theta) \), where \( q \) is the charge of the electron, \( v \) is the speed, \( B \) is the magnetic field's magnitude, and \( \theta \) is the angle between the velocity and the magnetic field. The charge of an electron is \( q = -1.60 \times 10^{-19} \mathrm{~C} \).
02

Calculating Maximum Force

The maximum force occurs when \( \sin(\theta) = 1 \), i.e., when \( \theta = 90^{\circ} \). Thus, \( F_{\text{max}} = |q|vB = 1.60 \times 10^{-19} \mathrm{~C} \times 2.50 \times 10^{6} \mathrm{~m/s} \times 7.40 \times 10^{-2} \mathrm{~T} \). Calculate \( F_{\text{max}} \).
03

Calculating Minimum Force

The minimum force occurs when \( \sin(\theta) = 0 \), i.e., when \( \theta = 0^{\circ} \) or \( \theta = 180^{\circ} \). In these cases, \( F_{\text{min}} = 0 \) because the magnetic field has no component perpendicular to the velocity.
04

Calculate Force for Half Maximum

To find the angle when the force is half of the maximum, set \( F = \frac{1}{2}F_{\text{max}} \). This gives \( \frac{1}{2}qvB = qvB \sin(\theta) \), simplifying to \( \sin(\theta) = \frac{1}{2} \).
05

Find the Angle for Half Maximum Force

Solve \( \sin(\theta) = \frac{1}{2} \) for \( \theta \). The angle \( \theta \) that satisfies this is \( \theta = 30^{\circ} \) or \( 150^{\circ} \).
06

Verify and Summary

Verify each calculation and ensure that all steps show the path from question to answer clearly. The maximum force was calculated, and it occurs at \( \theta = 90^{\circ} \), while \( \theta = 0^{\circ} \) gives the minimum force. An angle of \( 30^{\circ} \) or \( 150^{\circ} \) provides half the maximum force.

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.

Electron Velocity
In physics, velocity is a vector quantity, meaning it has both magnitude and direction. When we talk about "electron velocity," we are considering how fast and in which direction the electron is moving. In the context of moving through a magnetic field, this velocity becomes a key factor.

For an electron moving through a magnetic field, its velocity is one of the variables used to determine the magnetic force it experiences. The force is described by the equation:
  • \( F = qvB \sin(\theta) \), where:
    • \( F \) is the magnetic force.
    • \( q \) is the charge of the electron.
    • \( v \) is the velocity of the electron.
    • \( B \) is the magnetic field's magnitude.
    • \( \theta \) is the angle between the velocity and the magnetic field.
An electron with a higher velocity will interact more strongly with the magnetic field, resulting in a larger force acting upon it.

In the context of this problem, the electron's velocity is given as \( 2.50 \times 10^{6} \mathrm{~m/s} \). This speed is crucial for calculating the potential forces involved.
Magnetic Field
Magnetic fields are regions where magnetic forces operate. They are created by moving electric charges or by magnetic materials. The strength of a magnetic field is measured in teslas (T).

In physics problems involving charged particles, like electrons, the magnetic field plays a vital role. It affects how the particle moves if it is within the field. The direction of the magnetic field can be specified in different ways, but its impact is considered in terms of its perpendicular component to the movement of a charged particle.

For the particular problem at hand, the magnetic field has a magnitude of \( 7.40 \times 10^{-2} \mathrm{~T} \). While the direction isn't specified, the force depends on both its magnitude and the angle with the electron's path.

The larger the magnetic field, the greater the potential force exerted on the electron, assuming all other factors, such as velocity and angle, stay constant.
Angle of Force
The angle of force in the context of magnetic force is crucial. It's the angle between the electron's velocity and the direction of the magnetic field. This angle determines the magnitude of the magnetic force acting on the electron.

The equation \( F = qvB \sin(\theta) \) shows us that the force is maximized when the angle \( \theta \) is \( 90^{\circ} \), because the sine of \( 90^{\circ} \) is 1. Conversely, when \( \theta \) is \( 0^{\circ} \) or \( 180^{\circ} \), the force is minimized to zero because the sine of these angles is 0.

In practical terms, understanding this concept allows us to predict the force exerted on electrons in magnetic fields given different orientations. If the force experienced is half the maximum possible, using \( \sin(\theta) = \frac{1}{2} \), we find that \( \theta \) can be \( 30^{\circ} \) or \( 150^{\circ} \). This demonstrates how the orientation can drastically change the force's strength, providing insight into its dynamics in magnetic environments.

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

A \(3.25 \mathrm{~g}\) bullet picks up an electric charge of \(1.65 \mu \mathrm{C}\) as it travels down the barrel of a rifle. It leaves the barrel at a speed of \(425 \mathrm{~m} / \mathrm{s}\), traveling perpendicular to the earth's magnetic field, which has a magnitude of \(5.50 \times 10^{-4} \mathrm{~T}\). Calculate (a) the magnitude of the magnetic force on the bullet and (b) the magnitude of the bullet's acceleration due to the magnetic force at the instant it leaves the rifle barrel.

An electric bus operates by drawing current from two parallel overhead cables, at a potential difference of \(600 \mathrm{~V},\) and spaced \(55 \mathrm{~cm}\) apart. When the power input to the bus's motor is at its maximum power of \(65 \mathrm{hp},\) (a) what current does it draw and (b) what is the attractive force per unit length between the cables?

If a proton is exposed to an external magnetic field of \(2 \mathrm{~T}\) that has a direction perpendicular to the axis of the proton's spin, what is the torque on the proton? A. 0 B. \(1.4 \times 10^{-26} \mathrm{~N} \cdot \mathrm{m}\) C. \(2.8 \times 10^{-26} \mathrm{~N} \cdot \mathrm{m}\) D. \(0.7 \times 10^{-26} \mathrm{~N} \cdot \mathrm{m}\)

Two long current-carrying wires run parallel to each other. Show that if the currents run in the same direction, these wires attract each other, whereas if they run in opposite directions, the wires repel.

A plastic circular loop has radius \(R,\) and a positive charge \(q\) is distributed uniformly around the circumference of the loop. The loop is now rotated around its central axis, perpendicular to the plane of the loop, with angular speed \(\omega\). If the loop is in a region where there is a uniform magnetic field \(\vec{B}\) directed parallel to the plane of the loop, calculate the magnitude of the magnetic torque on the loop.
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Wave Optics

Read Explanation

Turning Points in Physics

Read Explanation

Classical Mechanics

Read Explanation

Engineering Physics

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.