Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 21 - Electromagnetic Induction and Faraday's Law - General Problems - Page 623: 88

Answer

a) See the detailed answer below. b) $6\times10^8 \;\rm V/m$

Work Step by Step

a) We know that the electric field energy density is given by $$u_E=\dfrac{PE}{\rm Volume}=\frac{1}{2}\epsilon E^2$$ Plugging the known; $$u_E =\frac{1}{2}\cdot8.85\times10^{-12}\cdot (10^4)^2$$ $$u_E =\color{magenta}{\bf 4.425\times10^{-4}} \;\rm J/m^3\tag 1$$ And we know that the magnetic field energy density is given by $$u_B=\frac{B^2}{2\mu_0} $$ Plugging the known; $$u_B=\frac{2^2}{2\cdot 4\pi\times 10^{-7}} $$ $$u_B =\color{magenta}{\bf 1.59\times10^{6}} \;\rm J/m^3\tag 2$$ To compare, divide (2) by (1); $$\dfrac{u_B }{u_E}=\dfrac{1.59\times10^{6}}{4.425\times10^{-4}}=3.59\times 10^9$$ Therefore, $$ u_B =\color{magenta}{\bf3.59\times 10^9}\;u_E$$ -------------------------------------------------------------------------- b) Finding the needed magnitude electric field t produce the same energy as the magnetic field of 2T. $$u_E=u_B$$ Thus, $$ \frac{1}{2}\epsilon E^2=\frac{B^2}{2\mu_0}$$ Solving for $E$; $$ E^2=\frac{B^2}{ \mu_0 \epsilon}$$ $$ E =\sqrt{\frac{B^2}{ \mu_0 \epsilon}}= \frac{B}{ \sqrt{\mu_0 \epsilon}}$$ Plugging the known; $$ E = \frac{2}{ \sqrt{ 4\pi\times10^{-7}\cdot8.85\times10^{-12} }}=5.99\times10^8$$ $$ E =\color{magenta}{\bf6\times10^8}\;\rm V/m$$
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