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 - General Problems - Page 588: 82

Answer

$a=2.03\times 10^{-5}\frac{m}{s^2}$

Work Step by Step

We know that: $F=qvB$ We also know that: $B=\frac{\mu_{\circ}I}{2\pi r}$ $\implies F=qv\frac{\mu_{\circ}I}{2\pi r}$ As $a=\frac{F}{m}$; $\implies a=\frac{qv\frac{\mu_{\circ}I}{2\pi r}}{m}=\frac{qv}{m}\frac{\mu_{\circ}I}{2\pi r}$ We plug in the known values to obtain: $a=\frac{18.0\times 10^{-3}\times 3.4\times 4\pi \times 10^{-7}\times 25}{2\pi \times 0.175\times 0.086}=2.03\times 10^{-5}\frac{m}{s^2}$
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