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kll sin(kr-wt); fx 2.| 4π We now want to evaluate the Poynting vector S = ReEx ReH. Taking the real part of E, ReE= woll sin(kr-wt) sine. Taking the real part of B/µg, ReH = - Therefore, S = 4π sin² (kr-wt) sin² f. (a) show that the time-averaged Poynting vector (S) = lim S dt = T-00 Ho ( sin² 0 f. 4πr

kll sin(kr-wt); fx 2.| 4π We now want to evaluate the Poynting vector S = ReEx ReH. Taking the real part of E, ReE= woll sin(kr-wt) sine. Taking the real part of B/µg, ReH = - Therefore, S = 4π sin² (kr-wt) sin² f. (a) show that the time-averaged Poynting vector (S) = lim S dt = T-00 Ho ( sin² 0 f. 4πr

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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ReE 4. We now want to evaluate the Poynting vector S = ReEx ReH. Taking the real part of E, woll sin(kr-wt) = 4π sin 0 6. Taking the real part of B/μo, ReH == kll sin(kr-wt) 4π fx 2.| r 2 kll Therefore, S = 4πr sin² (kr wt) sin f. (a) show that the time-averaged Poynting vector (S) = lim S dt = T-00 T 2 kll sin² 0 f. 4πr
Transcribed Image Text:ReE 4. We now want to evaluate the Poynting vector S = ReEx ReH. Taking the real part of E, woll sin(kr-wt) = 4π sin 0 6. Taking the real part of B/μo, ReH == kll sin(kr-wt) 4π fx 2.| r 2 kll Therefore, S = 4πr sin² (kr wt) sin f. (a) show that the time-averaged Poynting vector (S) = lim S dt = T-00 T 2 kll sin² 0 f. 4πr
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