Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 22 - Magnetism - Problems and Conceptual Exercises - Page 797: 90

Answer

a) $B=1.285\times 10^{-3}T$ b) $\frac{F}{L}=290\frac{N}{m}$

Work Step by Step

(a) We know that $B=\frac{\mu_{\circ}I}{2\pi r}$ We plug in the known values to obtain: $B=\frac{(4\pi \times 10^{-7})(255\times 10^3)}{2\pi(35)}$ $B=1.285\times 10^{-3}T$ (b) As $\frac{F}{L}=\frac{\mu_{\circ}I_1I_2}{2\pi d}$ We plug in the known values to obtain: $\frac{F}{L}=\frac{4\pi \times 10^{-7}(255\times 10^3)^2}{2\pi(35)}$ $\frac{F}{L}=290\frac{N}{m}$
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