1. Demonstrate that an electromagnetic wave propagating in free space may have the following electric field amplitude an
Posted: Sat Feb 26, 2022 11:41 am
- 1 Demonstrate That An Electromagnetic Wave Propagating In Free Space May Have The Following Electric Field Amplitude An 1 (32.19 KiB) Viewed 44 times
- 1 Demonstrate That An Electromagnetic Wave Propagating In Free Space May Have The Following Electric Field Amplitude An 2 (32.19 KiB) Viewed 44 times
- q5. (the question should more accuraty ask): Using the notes below, and one or more of the equations, find the average energy flux over time, using physical constants and amplitudes found.
Supplementy notes for q5 below.
Note that not all will be necessary to answer the question.
- 1 Demonstrate That An Electromagnetic Wave Propagating In Free Space May Have The Following Electric Field Amplitude An 3 (37.69 KiB) Viewed 44 times
- 1 Demonstrate That An Electromagnetic Wave Propagating In Free Space May Have The Following Electric Field Amplitude An 4 (48.89 KiB) Viewed 44 times
- 1 Demonstrate That An Electromagnetic Wave Propagating In Free Space May Have The Following Electric Field Amplitude An 5 (35.29 KiB) Viewed 44 times
Energy in Electromagnetic Fields: Poynting's Theorem Electromagnetic waves carry energy. The energy density in an electric field is given by: 1 = € 2 Ug = 32 (34) The energy density in a magnetic field is given by: 1 Un = (35) The energy flux (energy crossing unit area per unit time) is given by the Poynting vector: š= Ē x 7. (36) These results follow from Poynting's theorem...
Energy Flux and Impedance The Poynting vector 5 is defined by: : 5 =ĒⓇĀ. (51) The Poynting vector gives the instantaneous energy flow crossing unit area normal to the direction of flow, per unit time. Since Ēo and Ã, are perpendicular to each other and to k (the direction in which the wave is travelling): (52) E 5 k = z\ñ , Z where k is a unit vector in the direction of K, and Z is the impedance of the medium: Z= z = (53) Note that the energy flux, like the energy density, depends on the square of the field strength.
The electric field varies sinusoidally with position and with time; and the energy flux is proportional to the square of the field. 1 Therefore, since: (sin2) = 2 the average energy flux (over time or position) is: (54) Eolk 2 = (55) 22 Note that this can be written in terms of the average energy density in the wave: (5) = (U)Ű, (56) where ū is the velocity (vector) of the wave.