Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 29 - The Magnetic Field - Exercises and Problems - Page 830: 11


(a) $I = 20~A$ (b) $r = 1.6~mm$

Work Step by Step

(a) We can find the current: $B = \frac{\mu_0~I}{2~r}$ $I = \frac{2~r~B}{\mu_0}$ $I = \frac{(2)~(5.0\times 10^{-3}~m)(2.5\times 10^{-3}~T)}{4\pi\times 10^{-7}~T~m/A}$ $I = 20~A$ (b) We can find the distance $r$: $B = \frac{\mu_0~I}{2~\pi~r}$ $r = \frac{\mu_0~I}{2~\pi~B}$ $r = \frac{(4\pi\times 10^{-7}~T~m/A)(20~A)}{(2~\pi)~(2.5\times 10^{-3}~T)}$ $r = 16\times 10^{-4}~m$ $r = 1.6~mm$
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