Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 22 - Magnetism - Problems and Conceptual Exercises - Page 798: 98

Answer

$1.42\times 10^{-3}T$

Work Step by Step

We know that $B_{solenoid}=\mu_{\circ}nI_s$ $\implies B_{solenoid}=(4\pi \times 10^{-7}T.m/A)(2200turns/m)(0.50A)=1.38\times 10^{-3}T$ and $B_{straight \space wire}=\frac{\mu_{\circ}I}{2\pi r}$ $\implies B_{straight\space wire}=\frac{(4\pi\times 10^{-7}T.m/A)(13A)}{2\pi(0.75\times 10^{-2}m)}$ $B_{straight \space wire}=0.346\times 10^{-3}T$ Now $B_{net}=\sqrt{(B_{solenoid})^2+(B_{straight \space wire})^2}$ We plug in the known values to obtain: $B_{net}=\sqrt{(1.38\times 10^{-3}T)^2+(0.346\times 10^{-3}T)^2}$ $B_{net}=1.42\times 10^{-3}T$
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