lect25 - Physics 227: Lecture 25 Light II Lecture 24...

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Physics 227: Lecture 25 Light II Lecture 24 review: E 0 = B 0 c c = 1/ ( μ 0 ε 0 ) Generated For “low” Frequencies / energies by oscillating / accelerated charges - e.g., dipole antenna or circular motion Generated For high Frequencies / energies by atomic electron transitions between energy levels (visible light, UV, X-rays) Thursday, December 8, 2011
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EM Wave iclicker A. has only an x component B. has only a y component C. has only a z component D. has components in multiple directions E. Not enough information is given to know. In a sinusoidal electromagnetic wave in a vacuum, the electric Feld has only an x-component. This component is given by E x = E max cos (ky + ω t) The magnetic Feld of this wave. .. Thursday, December 8, 2011
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Energy density of light Previously we have learned that the energy density of an electric Feld is ε 0 E 2 /2, and the energy density of a magnetic Feld is B 2 /2 μ 0 . ±or light, B = E/c = ( ε 0 μ 0 )E, so. .. u = ε 0 E 2 /2 + B 2 /2 μ 0 = ε 0 E 2 /2 + ε 0 μ 0 E 2 /2 μ 0 = ε 0 E 2 . The energy densities of the electric and magnetic Felds are the same. The formulas are instantaneous - for same point in space and time. They are not averaged over time or space. (Average cos 2 θ = 1/2.) Thursday, December 8, 2011
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For a light plane wave, what power per unit area passes through a plane? The energy density u = ε 0 E 2 . In a time dt the light wave moves a distance cdt. Thus Poynting’s vector S = du/dt = ε 0 E 2 cdt / dt = ε 0 cE 2 . Units of S: power / unit area. There are other ways to write this. E.g.: Since E = cB, S = ε 0 c 2 EB. Or: S = ε 0 c 2 EB = ε 0 EB/ ε 0 μ 0 = EB/ μ 0 . Since the energy travels in the direction given by E x B , S = E x B / μ 0 . The total power through a surface is P = S . d
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This note was uploaded on 01/03/2012 for the course PHYSICS 750:227 taught by Professor Ronaldgilman during the Fall '11 term at Rutgers.

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lect25 - Physics 227: Lecture 25 Light II Lecture 24...

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