Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 20 - Magnetism - Problems - Page 586: 62

Answer

$2.233 \times10^{-5} T\cdot m/A $

Work Step by Step

Use equation 20–8 for the magnetic field inside a solenoid, but replace $\mu_o$ with the permeability of iron. $$B=\frac{\mu NI}{\mathcal{l}}$$ $$\mu=\frac{B\mathcal{l}}{NI}$$ $$\mu=\frac{(2.2T)(0.38m)}{(780)(48A)}$$ $$=2.233 \times10^{-5} T\cdot m/A \approx2.2 \times10^{-5} T\cdot m/A\approx 18\mu_o$$
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