Physics: Principles with Applications (7th Edition)

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

Chapter 22 - Electromagnetic Waves - General Problems - Page 642: 51

Answer

13 V/m, $4.4\times10^{-8}T$.

Work Step by Step

The intensity is the power per unit area. The light’s power is distributed uniformly over the surface area of a 2.50-radius sphere. The energy per unit area per unit time is the magnitude of the average intensity, equation 22–8. Find the strength of the electric field. $$\overline{I}=\frac{P}{A}=\frac{P}{4\pi r^2}=\frac{1}{2}\epsilon_ocE_o^2$$ $$E_o=\sqrt{\frac{P}{2\pi r^2\epsilon_oc }}$$ $$E_o=\sqrt{\frac{18W }{(2\pi (2.50m)^2) (8.85\times10^{-12} C^2/(N\cdot m^2))(3.00\times10^8m/s) }}$$ $$\approx 13V/m$$ Now find the strength of the magnetic field. $$B_o=\frac{E_o}{c}=\frac{13.14V/m}{3.00\times10^8m/s }=4.4\times10^{-8}T$$
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