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 - Problems - Page 585: 35

Answer

a. $B_{net}=(2 \times10^{-5} T /A)(I-25A)$ b. $B_{net}=(2 \times10^{-5} T /A)(I+25A)$

Work Step by Step

a. The currents are in the same direction. The 2 magnetic fields, halfway between the currents, will point in opposite directions. Their magnitudes are subtracted. $$B_{net}=\frac{\mu_oI_1}{2\pi r}-\frac{\mu_oI_2}{2\pi r}$$ $$B_{net}=\frac{4\pi \times10^{-7} T\cdot m/A}{2\pi (0.010m)}(I-25A)$$ $$B_{net}=(2 \times10^{-5} T /A)(I-25A)$$ b. Now the currents run in opposite directions. The 2 magnetic fields, halfway between the currents, point in the same direction. Their magnitudes are added. $$B_{net}=\frac{\mu_oI_1}{2\pi r}+\frac{\mu_oI_2}{2\pi r}$$ $$B_{net}=\frac{4\pi \times10^{-7} T\cdot m/A}{2\pi (0.010m)}(I+25A)$$ $$B_{net}=(2 \times10^{-5} T /A)(I+25A)$$
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