Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 33 - Electromagnetic Induction - Exercises and Problems - Page 1000: 61

Answer

a) ${\bf 3.93\times 10^{-4}}\;\rm J/m^3 $ b) ${\bf 3.14}\;\rm A $

Work Step by Step

$$\color{blue}{\bf [a]}$$ We know that the magnetic energy density is given by $$u_B=\dfrac{B^2}{2\mu_0}\tag 1$$ where the magnetic field strength at the center of a current loop is given by $$B=\dfrac{\mu_0 I}{2r}$$ Plug into (1), $$u_B=\dfrac{1}{2\mu_0}\left[\dfrac{\mu_0 I}{2r}\right]^2 $$ $$u_B=\dfrac{1}{2\mu_0}\left[\dfrac{\mu_0^2 I^2}{4r^2}\right] $$ $$u_B= \dfrac{\mu_0 I^2}{8r^2} $$ Plug the known; $$u_B= \dfrac{(4\pi \times 10^{-7})(1)^2}{8(0.02)^2} $$ $$u_B=\color{red}{\bf 3.93\times 10^{-4}}\;\rm J/m^3 $$ $$\color{blue}{\bf [b]}$$ We know that the magnetic field strength at a distance $r$ from a straight current wire is given by $$B=\dfrac{\mu_0 I}{2\pi r}$$ Plug into (1), $$u_B=\dfrac{1}{2\mu_0}\left[\dfrac{\mu_0 I}{2\pi r}\right]^2 $$ $$u_B=\dfrac{1}{2\mu_0}\left[\dfrac{\mu_0^2 I^2}{4\pi ^2r^2}\right] $$ $$u_B= \dfrac{\mu_0 I^2}{8\pi ^2r^2} $$ Solving for $I$; $$I=\sqrt{\dfrac{8\pi ^2r^2u_B}{\mu_0}}$$ Plug the known; $$I=\sqrt{\dfrac{8\pi ^2(0.02)^2 (3.93\times 10^{-4})}{(4\pi \times 10^{-7})}}$$ $$I=\color{red}{\bf 3.14}\;\rm A $$
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