College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 22 - Problems - Page 864: 32

Answer

(a) The average energy density is $4.67\times 10^{-6}~J/m^3$ (b) $E_{rms} = 726~V/m$ $B_{rms} = 2.42\times 10^{-6}~T$

Work Step by Step

(a) We can find the average energy density: $\frac{I}{c} = \frac{1400~W/m^2}{3.0\times 10^8~m/s} = 4.67\times 10^{-6}~J/m^3$ The average energy density is $4.67\times 10^{-6}~J/m^3$ (b) We can find $E_{rms}$: $E_{rms}^2 = \frac{I}{c~\epsilon_0}$ $E_{rms} = \sqrt{\frac{I}{c~\epsilon_0}}$ $E_{rms} = \sqrt{\frac{1400~W/m^2}{(3.0\times 10^8~m/s)~(8.85\times 10^{-12}~C^2/N~m^2)}}$ $E_{rms} = 726~V/m$ We can find $B_{rms}$: $B_{rms} = \frac{E_{rms}}{c}$ $B_{rms} = \frac{726~V/m}{3.0\times 10^8~m/s}$ $B_{rms} = 2.42\times 10^{-6}~T$
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