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: 20.1b

Answer

$ B_{0} = 0.0302 tesla $

Work Step by Step

Given: 0.25 m long coil with 400 turns Current = 15 A H = 24,000 A-turns-m Required: flux density in vacuum Solution: Using Equation 20.3: $ B_{0} = μ_{0}H = (1.257 \times 10^{-6} H/m)(24,000 A-turns/m) = 0.0302 tesla $
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