Materials Science and Engineering: An Introduction

Published by Wiley
ISBN 10: 1118324579
ISBN 13: 978-1-11832-457-8

Chapter 20 - Magnetic Properties - Questions and Problems - Page 834: 20.1a


$H = 24,000 T$

Work Step by Step

Given: 0.25 m long coil with 400 turns Current = 15 A Required: magnetic field strength Solution: Using Equation 20.1: $H= \frac{NI}{l} = \frac{(400 turns)(15 A)}{0.25 m} = 24,000 T$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.