Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Let Fig. E27.49 represent a strip of an unknown metal of the same dimensions as those of the silver ribbon in Exercise 27.49. When the magnetic field is 2.29 T and the current is 78.0 A, the Hall emf is found to be 131 µV. What does the simplified model of the Hall effect presented in Section 27.9 give for the density of free electrons in the unknown metal?

Short Answer

Expert verified

Free electrons per cubic metern=3.7×1028m3

Step by step solution

01

Drift velocity

The drift velocity is given by

vd=Jxn|q|

02

Determine the drift velocity

The drift velocity is

vd=Jxn|q|

Jx=IA=Iy1z1vd=In|q|y1z1

Magnetic force is equal to the electric force

|q|Ez=|q|vdByvd=EzBy

Form these equations we find that

n=IByy1|q|E=(78.0A)(2.29T)(2.3×104m)(1.6×1019C)(1.31×104V)=3.7×1028m3

Therefore, simplified model of hall effect electrons per cubic meter is

n=3.7×1028m3

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 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

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.

Sign-up for free