Chapter 32 lecture

# Chapter 32 lecture - Chapter 32 Electro-Magnetic Waves...

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Chapter 32 Electro-Magnetic Waves

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Maxwell’s Eqs. in free space away from charge & currents dt d I l d B A d B dt d l d E q A d E E c B exclosed Φ + = = Φ - = = 0 0 0 0 0 0 ε μ μ ε dt d I l d B A d B dt d l d E A d E E c B Φ + = = Φ - = = 0 0 0 0 0 0 ε μ μ dt d l d B A d B dt d l d E A d E E B Φ = = Φ - = = 0 0 0 0 ε μ If then the equations are interchangeable in E and B SURPRISE! = C , the velocity of light…… s m / 10 3 1 8 0 0 × = ε μ 0 0 ε μ B E = We will see that light waves consist of oscillating, perpendicular E and B fields that perpetually induce each other, propagate at speed c, with E = cB.
The half-space of uniform, crossed E and B vB E thus dt dx Ba dt dA B Ea dt dA B A dt dB Ea l d E = = = + - = - = ) ( 0 a “curtain” How does Φ B change if B=constant?? Φ - = - = dt d Ea l d E Law s Faraday B 0 : ' x E=0 B=0 The “curtain” is a wavefront that propagates to right in direction of E X B. What’s velocity v??? V

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The half-space of uniform, crossed E and B c v hence B v B = = = 0 0 2 0 0 1 : ε μ ε μ a But E = vB by Faraday’s Law, substituting this in above Eav Ba dt d Ba l d B Law s Ampere E 0 0 0 0 0 : ' ε μ ε μ = Φ = - = x E=0 B=0 Note: Neither E nor B can have a component in direction of v (violates Gauss’s laws) EM waves are transverse and propagate with v=c in EXB direction v
Remembering wave mechanics…..(15.3) λ π π ϖ ϖ 2 2 ) cos( ) , ( = = - = k and f with t kx y t x y nt displaceme m transverse traveling wave on string f λ Question 1: What is the velocity, v, of the wave? 1. fλ only 2. ω/k only 3. f/λ only 4. ωk only 5. fλ and ω/k so v = fλ = ω/k The x-motion/dependence is related to t-dependence by the WAVE EQUATION 2 2 2 2 2 1 t y v x y = We can show that E and B obey the wave equation using Faraday’s and Ampere’s laws…………. y x

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Question 2: Faraday’s loop dx x E E dx x E + = + ) ( adx x E dx x E a x E Ea Ea - . 5 . 4 . 3 2 . 2 2 . 1 Consider an infinitesimally thin loop of width “dx” and height “a” in a plane where E is changing with x. Integrating CCW around the loop, the path integral of Edl is: dx E(x) adx x B l d B and adx x E l d E = = . 5
First Faraday…. hence t B adx dt d dxa x E l d E B - = Φ - = = hence t E adx dt d dxa x B l d B E = Φ = - = 0 0 0 0 ε μ ε μ 2 2 0 0 2 2 t E x B t t B x x E = - = - = ε μ dx E(x) B o ut You can’t have one without the other….now Ampere dx 2 2 0 0 2 2 t E x E =

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