Physics: Principles with Applications (7th Edition)

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

Chapter 18 - Electric Currents - Search and Learn - Page 525: 7

Answer

$1.33\times10^{-6}\;\rm m$

Work Step by Step

The electron in an ac current will move back and forth through the wire since the direction of the current changes at a constant rate. We can call that a simple harmonic motion. This means that the maximum distance traveled by the electron is twice its amplitude from its original position. And since the maximum current is related to the maximum drift velocity, $$I_{max} =ne Av_{max}\tag 1$$ And the maximum distance traveled by the electron in one direction is given by its amplitude $$v={\rm A}\omega={\rm A} \cdot 2\pi f$$ note that the italic $A$ refers to Area while ${\rm A}$ refers to amplitude. To distinguish them, we will use $\Delta x$ rather than ${\rm A}$. $$v_{max}=\Delta x \cdot 2\pi f\tag 2$$ We also know that $$\overline{ P } =I_{\rm rms}V_{\rm rms}=\dfrac{I_{max}V_{max}}{\sqrt{2}}\tag 3$$ Now we know that the maximum distance is twice the amplitude, so $$\Delta x_{max}=2\Delta x$$ Plugging $\Delta x$ from (2); $$\Delta x_{max}=2\cdot \dfrac{v_{max}}{2\pi f}=\dfrac{v_{max}}{ \pi f}$$ Plugging $v_{max}$ from (1); $$\Delta x_{max}= \dfrac{I_{max}}{ \pi f\cdot neA}$$ Plugging $I_{max}$ from (3); $$\Delta x_{max}= \dfrac{\sqrt{2}\;\; \overline{P}}{ \pi fneAV_{max}} =\dfrac{\sqrt{2}\;\; \overline{P}}{ \pi fne(\pi r^2)V_{max}}$$ $$\Delta x_{max}=\dfrac{\sqrt{2}\;\; \overline{P}}{ \pi^2 fne(\frac{D}{2})^2 V_{max}}$$ Plugging the known and recall that $n$ for copper's electrons is found in ex: 18-14 in your textbook. $$\Delta x_{max}=\dfrac{\sqrt{2}\cdot 650}{\pi^2\cdot 60\cdot 8.4\times 10^{28}\cdot 1.6\times10^{-19}\cdot(\frac{1.7\times10^{-3}}{2})^2\cdot 120 }$$ $$\Delta x_{max}=\color{red}{\bf 1.33\times10^{-6}}\;\rm m$$
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