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: 16


(a) $\mu = 0.025~A~m^2$ (b) $B = 1.5~\mu T$

Work Step by Step

(a) We can find the magnetic moment: $B = \frac{\mu_0~\mu}{2\pi~r^3}$ $\mu = \frac{2\pi~r^3~B}{\mu_0}$ $\mu = \frac{(2\pi)~(0.10~m)^3~(5.0\times 10^{-6}~T)}{4\pi\times 10^{-7}~T~m/A}$ $\mu = 0.025~A~m^2$ (b) We can find the on-axis field strength: $B = \frac{\mu_0~\mu}{2\pi~r^3}$ $B = \frac{(4\pi\times 10^{-7}~T~m/A)(0.025~A~m^2)}{(2\pi)~(0.15~m)^3}$ $B = 1.5\times 10^{-6}~T$ $B = 1.5~\mu T$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.