/*! 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 50 An electron moves in a circle of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An electron moves in a circle of radius \(r=5.29 \times 10^{-11} \mathrm{~m}\) with speed \(2.19 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Treat the circular path as a current loop with a constant current equal to the ratio of the electron's charge magnitude to the period of the motion. If the circle lies in a uniform mag. netic field of magnitude \(B=\) \(7.10 \mathrm{mT}\), what is the maximum possible magnitude of the torque produced on the loop by the field?

Short Answer

Expert verified
The maximum torque is when the magnetic moment and magnetic field are perpendicular.

Step by step solution

01

Calculate the Current in the Loop

First, determine the current (\(I\) that flows through the loop. The current is given by \[I = \frac{q_e}{T}\]where \(q_e\) is the charge of the electron (\(1.60 \times 10^{-19}\,\mathrm{C}\)) and \(T\) is the period of the electron's motion. The period is calculated using \[T = \frac{2\pi r}{v}\]where \(r = 5.29 \times 10^{-11}\,\mathrm{m}\) and \(v = 2.19 \times 10^6\,\mathrm{m/s}\). Thus, \[T = \frac{2\pi \times 5.29 \times 10^{-11}}{2.19 \times 10^6} = 1.52 \times 10^{-16}\,\mathrm{s}\].Therefore, \[I = \frac{1.60 \times 10^{-19}}{1.52 \times 10^{-16}} = 1.05 \times 10^{-3}\,\mathrm{A}\].

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 91Ó°ÊÓ!

Key Concepts

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

Electron Motion
When discussing electron motion, it's important to understand what happens when an electron moves in a circular path. Electrons can rotate in circles due to the influence of forces acting upon them. In our example, the electron travels with a constant speed of \(2.19 \times 10^6\,\mathrm{m/s}\) while maintaining a circular path with a radius of \(5.29 \times 10^{-11}\,\mathrm{m}\). This type of movement is typical at atomic scales, such as in the orbit of an electron around a nucleus.
At this scale, electrons behave differently than larger bodies that might follow circular motion. In particular, an electron's charge and the magnetic forces it responds to play crucial roles in determining its path. The electron's path in the circle can be likened to current in a wire loop, where the consistent motion creates a electromagnetism characteristic useful in various applications.
Understanding the dynamics of such electron motion is fundamental to grasping more complex phenomena in physics, such as magnetic fields and electrical currents.
Current Loop
A current loop arises when charges like electrons move in a closed path or circuit. In our example, the electron moving in a circular path can be treated as a small current loop. It's as if there's a tiny wire carrying electricity in the same shape as the electron's path.
The concept of a current loop is essential because it helps us understand how magnetic fields interact with moving charges. Here, the current is calculated by considering the electron's charge and the time it takes to complete one full circle (the period). The formula for current \(I\) is given by:
  • \(I = \frac{q_e}{T}\)
where \(q_e = 1.60 \times 10^{-19}\,\mathrm{C}\) and the period \(T = 1.52 \times 10^{-16}\,\mathrm{s}\). With these values, the resultant current is \(1.05 \times 10^{-3}\,\mathrm{A}\).
This current loop acts similarly to a small magnet due to the magnetic effects of the moving electron charge, offering a basis for understanding electromagnetism.
Magnetic Field Effects
Magnetic fields exert their influence on current loops in various ways. As soon as a moving electron - acting as a current loop - is placed in a magnetic field, the field exerts a torque on the loop. This torque depends on several factors:
  • The strength of the magnetic field: Stronger fields exert greater forces.
  • The current in the loop: Higher currents influence the torque magnitude.
  • The orientation of the loop: A perpendicular alignment maximizes torque.
The equation for torque \(\tau\) is given by:\[\tau = nIAB\sin\theta\]Where \(n\) is the number of turns (here, just one), \(I\) is the current, \(A\) is the area of the loop, \(B\) is the magnetic field magnitude, and \(\theta\) is the angle between the loop's normal and the magnetic field direction.
In our specific problem, the maximum torque occurs when \(\theta = 90^\circ\), making \(\sin\theta = 1\). This relationship helps us realize how practically aligned magnetic fields and current loops are applied in technology, such as in electric motors and generators.
Electron Charge
The electron charge is a fundamental property of electrons, indicating their electric charge. This charge, \(q_e\), is equal to \(1.60 \times 10^{-19}\,\mathrm{C}\), a constant in physics that facilitates calculations involving electric phenomena. Since electrons are charged particles, this property ties into how they move and interact with electric and magnetic fields.
In the scenario described, the electron charge is integral to determining both the current through its motion in a circle and the resultant magnetic effects, like torque, experienced in a magnetic field.
Knowing the electron charge allows physicists to analyze experiments, design equipment, and better understand the forces at play in microscopic and macroscopic systems alike. For students, grasping the concept of electron charge provides a key piece in connecting charge interactions, electric currents, and magnetic effects, forming a base for more complex studies in electromagnetics and circuit designs.

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

What uniform magnetic field, applied perpendicular to a beam of electrons moving at \(1.30 \times 10^{6} \mathrm{~m} / \mathrm{s}\), is required to make the electrons travel in a circular arc of radius \(0.350 \mathrm{~m} ?\)

A wire \(50.0 \mathrm{~cm}\) long carries a \(0.500\) A current in the positive direction of an \(x\) axis through a magnetic field \(\vec{B}=\) \((3.00 \mathrm{mT}) \hat{\mathrm{j}}+(10.0 \mathrm{~m} \mathrm{~T}) \hat{\mathrm{k}}\). In unit-vector notation, what is the magnetic force on the wire?

A proton circulates in a cyclotron, beginning approximately at rest at the center. Whenever it passes through the gap between dees, the electric potential difference between the dees is \(200 \mathrm{~V},(\mathrm{a})\) By how much does its kinetic energy increase with each passage through the gap? (b) What is its kinetic energy as it completes 100 passes through the gap? Let \(r_{100}\) be the radius of the proton's circular path as it completes those 100 passes and enters a dee, and let \(r_{101}\) be its next radius, as it enters a dee the next time. (c) By what percentage does the radius increase when it changes from \(r_{100}\) to \(r_{10 \mathrm{I}}\) ? That is, what is $$ \text { percentage increase }=\frac{r_{101}-r_{100}}{r_{100}} 100 \% ? $$

A strip of copper \(150 \mu \mathrm{m}\) thick and \(4.5 \mathrm{~mm}\) wide is placed in a uniform magnetic field \(\vec{B}\) of magnitude \(0.65 \mathrm{~T}\), with \(\vec{B}\) perpendicular to the strip. A current \(i=23 \mathrm{~A}\) is then sent through the strip such that a Hall potential difference \(V\) appears across the width of the strip. Calculate \(V\). (The number of charge carriers per unit volume for copper is \(8.47 \times\) \(10^{28}\) electrons/m \(^{3}\).)

An electron moves through a uniform magnetic field given by \(\vec{B}=B_{x} \hat{\mathrm{i}}+\left(3.0 B_{x}\right) \hat{\mathrm{j}}\). At a particular instant, the electron has velocity \(\vec{v}=(2.0 \hat{\mathrm{i}}+4.0 \mathrm{j}) \mathrm{m} / \mathrm{s}\) and the magnetic force acting on it is \(\left(6.4 \times 10^{-19} \mathrm{~N}\right) \hat{\mathrm{k}}\). Find \(B_{x}\)

See all solutions

Recommended explanations on Physics Textbooks

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.