Answer
$H = 24,000 T$
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:
$H= \frac{NI}{l} = \frac{(400 turns)(15 A)}{0.25 m} = 24,000 T$
Materials Science and Engineering: An Introduction
by Callister, William D.; Rethwisch, David G.
Published by
Wiley
ISBN 10:
1118324579
ISBN 13:
978-1-11832-457-8
Chapter 20 - Magnetic Properties - Questions and Problems - Page 834: 20.1a
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