Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 100

A strip of copper carries current in the \(+x\) -direction. There is an external magnctic field directed out of the page. What is the direction of the Hall electric field?

Short Answer

Expert verified
Answer: The direction of the Hall electric field is in the \(-y\) direction.

Step by step solution

01

Understanding the problem and setting up the coordinates

The magnetic field is given in the \(z\) direction, pointing out of the page. The current is in the \(x\) -direction, flowing through the copper strip. Our goal is to find the direction of the Hall electric field, which is perpendicular to both the current and the magnetic field.
02

Right-hand rule for the magnetic force on moving charges

We'll use the right-hand rule to determine the direction of the force experienced by the mobile charges as they move through the strip due to the external magnetic field. Place the thumb of your right hand in the direction of the current (in this case, the \(+x\) direction), and let your fingers curl in the direction of the magnetic field (coming out of the page, or in the \(+z\) direction). The direction in which your palm faces represents the direction of the force experienced by the positively charged particles (\(+q\)) in the copper strip. In this case, your palm will be facing in the \(+y\) direction.
03

Determine the direction of the Hall electric field

Now that we know the direction of the force experienced by the positive charges in the copper strip, we can conclude that the Hall electric field will be established in the opposite direction to this force. Therefore, the Hall electric field is in the \(-y\) direction. So the direction of the Hall electric field is in the \(-y\) 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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

In a certain region of space, there is a uniform electric field \(\overrightarrow{\mathbf{E}}=2.0 \times 10^{4} \mathrm{V} / \mathrm{m}\) to the east and a uniform magnetic field \(\overline{\mathbf{B}}=0.0050 \mathrm{T}\) to the west. (a) What is the electromagnetic force on an electron moving north at \(1.0 \times 10^{7} \mathrm{m} / \mathrm{s} ?(\mathrm{b})\) With the electric and magnetic fields as specified, is there some velocity such that the net electromagnetic force on the electron would be zero? If so, give the magnitude and direction of that velocity. If not, explain briefly why not.
An electron moves with speed \(2.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a uniform magnetic field of \(1.4 \mathrm{T},\) pointing south. At one instant, the electron experiences an upward magnetic force of $1.6 \times 10^{-14} \mathrm{N} .$ In what direction is the electron moving at that instant? Be specific: give the angle(s) with respect to $\mathrm{N}, \mathrm{S}, \mathrm{E}, \mathrm{W},$ up, down. (If there is more than one possible answer, find all the possibilitics.)
A long straight wire carries a \(4.70-\) A current in the positive \(x\) -direction. At a particular instant, an electron moving at $1.00 \times 10^{7} \mathrm{m} / \mathrm{s}\( in the positive \)y\( -direction is \)0.120 \mathrm{m}$ from the wire. Determine the magnetic force on the electron at this instant. Sce the figure with Problem \(65 .\)
The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ Natural carbon consists of two different isotopes (excluding $^{14} \mathrm{C},$ which is present in only trace amounts). The isotopes have different masses, which is due to different numbers of neutrons in the nucleus; however, the number of protons is the same, and subsequently the chemical properties are the same. The most abundant isotope has an atomic mass of \(12.00 \mathrm{u} .\) When natural carbon is placed in a mass spectrometer, two lines are formed on the photographic plate. The lines show that the more abundant isotope moved in a circle of radius \(15.0 \mathrm{cm},\) while the rarer isotope moved in a circle of radius \(15.6 \mathrm{cm} .\) What is the atomic mass of the rarer isotope? (The ions have the same charge and are accelerated through the same potential difference before entering the magnetic field.)
An early cyclotron at Cornell University was used from the 1930 s to the 1950 s to accelerate protons, which would then bombard various nuclei. The cyclotron used a large electromagnet with an iron yoke to produce a uniform magnetic field of \(1.3 \mathrm{T}\) over a region in the shape of a flat cylinder. Two hollow copper dees of inside radius \(16 \mathrm{cm}\) were located in a vacuum chamber in this region. (a) What is the frequency of oscillation necessary for the alternating voltage difference between the dees? (b) What is the kinetic energy of a proton by the time it reaches the outside of the dees? (c) What would be the equivalent voltage necessary to accelerate protons to this energy from rest in one step (say between parallel plates)? (d) If the potential difference between the dees has a magnitude of $10.0 \mathrm{kV}$ each time the protons cross the gap, what is the minimum number of revolutions each proton has to make in the cyclotron?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Work Energy and Power

Read Explanation

Fundamentals of Physics

Read Explanation

Physics of Motion

Read Explanation

Mechanics and Materials

Read Explanation

Circular Motion and Gravitation

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.

Sign-up for free