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 831: 18

Answer

$L = 1.4~cm$

Work Step by Step

We can find the magnetic moment: $B = \frac{\mu_0~\mu}{2\pi~r^3}$ $\mu = \frac{2\pi~r^3~B}{\mu_0}$ $\mu = \frac{(2\pi)~(0.50~m)^3~(7.5\times 10^{-9}~T)}{4\pi\times 10^{-7}~T~m/A}$ $\mu = 0.0046875~A~m^2$ We can find the length $L$ of each side of the square: $\mu = I~A$ $\mu = I~L^2$ $L^2 = \frac{\mu}{I}$ $L = \sqrt{\frac{\mu}{I}}$ $L = \sqrt{\frac{0.0046875~A~m^2}{25~A}}$ $L = 0.014~m$ $L = 1.4~cm$
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