Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 32 - The Magnetic Field - Exercises and Problems - Page 956: 4

Answer

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Work Step by Step

We know, from Biot-Savart law, that the magnetic field of a moving charged particle is given by $$B=\dfrac{\mu_0}{4\pi }\dfrac{qv\sin\theta}{r^2}$$ $$\color{blue}{\bf [a]}$$ At point $(1,0,0)$, $\theta=90^\circ$, so $$B=\dfrac{(4\pi\times 10^{-7})}{4\pi }\dfrac{(1.6\times 10^{-19})(2\times 10^7)\sin90^\circ}{0.01^2} $$ $$B=\color{red}{\bf 3.2\times 10^{-15}}\;\rm T$$ and its direction is in the negative $y$-direction (since the charge is negative). $$\color{blue}{\bf [b]}$$ At point $(0,1,0)$, $\theta=0^\circ$, so $$B=\dfrac{(4\pi\times 10^{-7})}{4\pi }\dfrac{(1.6\times 10^{-19})(2\times 10^7)\sin0^\circ}{0.01^2} $$ $$B=\color{red}{\bf 0}\;\rm T$$ and its direction is out of the page. $$\color{blue}{\bf [c]}$$ At point $(0,1,1)$, $\theta=45^\circ$, so $$B=\dfrac{(4\pi\times 10^{-7})}{4\pi }\dfrac{(1.6\times 10^{-19})(2\times 10^7)\sin45^\circ}{0.01^2+0.01^2} $$ $$B=\color{red}{\bf 11.3\times 10^{-16}} \;\rm T$$ and its direction is perpendicular to the $y$-$z$ plane which means in the $x$-direction and since the charge is negative, so it must be in the negative $x$-direction.
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