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 - General Problems - Page 587: 64

Answer

a) $2.67\times 10^{-6}\;\rm T$ b) $5.33\times 10^{-6}\;\rm T$ c) Not equal but opposite. d) equal and opposite.

Work Step by Step

a) We know that magnitude of the magnetic field due to a wire at some points given by $$B=\dfrac{\mu_0I}{2\pi r}$$ Thus, the magnetic field magnitude due to wire A at wire B is given by $$B_{\text{Due to A at B}}=\dfrac{\mu_0I_A}{2\pi r}=\dfrac{4\pi \times10^{-7}\cdot 2}{2\pi \cdot 0.15}=\color{red}{\bf2.67\times 10^{-6} }\;\rm T$$ b) $$B_{\text{Due to B at A}}=\dfrac{\mu_0I_B}{2\pi r}=\dfrac{4\pi \times10^{-7}\cdot 4}{2\pi \cdot 0.15}=\color{red}{\bf 5.33\times 10^{-6} }\;\rm T$$ c) They are not equal, as we see from the magnitudes above, but yes they are opposite. According to the right-hand rule, if wire A is to the East of wire B, then the direction of the magnetic field magnitude due to A at B is toward the south, while the direction of the magnetic field magnitude due to B at A is toward the north. d) According to Newton's third law, the force exerted by wire A on wire B must be equal to the force exerted by wire B on wire A and the two forces are opposing each other. There is no to do the math but to make sure. We can assume that the two wires are having the same length. $$F_{\text{due to A on B}}=B_{\text{Due to A at B}}I_B l_B=\dfrac{\mu_0I_A}{2\pi r}I_Bl_B$$ $$F_{\text{due to B on A}}=B_{\text{Due to B at A}}I_B l_B=\dfrac{\mu_0I_B}{2\pi r}I_Al_A$$ We can assume that the two wires are having the same length which means that the two forces must be equal.
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