Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 33 - Electromagnetic Waves - Problems - Page 1001: 13a

Answer

$E_m=1.03\frac{kV}{m}$

Work Step by Step

We know that; $I=\frac{E_m^2}{2\mu_{\circ}c}$ This can be rearranged as: $E_m=\sqrt{2I\mu_{\circ}c}$ We plug in the known values to obtain: $E_m=\sqrt{2(1400)(4\times 3.1416\times 10^{-7})(3\times 10^8)}=1.03\times 10^3=1.03\frac{kV}{m}$
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