1P28-Class 28: Outline Hour 1: Displacement Current Maxwell’s Equations Hour 2: Electromagnetic waves2P28-Finally: Bringing it All Together3P28-Displacement Current4P28-Ampere’s Law: Capacitor Consider a charging capacitor: I Use Ampere’s Law to calculate the magnetic field just above the top plate 1) Red Amperian Area, Ienc= I 2) Green Amperian Area, I = 0 What’s Going On? 0Ampere's law: enc d Iµ⋅ =∫ Bs 5P28-Displacement Current 0 E d ddQ Idt dtε Φ = ≡ We don’t have current between the capacitor plates but we do have a changing E field. Can we “make” a current out of that? 0 0 0 E QE Q EA A ε εε = ⇒ = = Φ This is called (for historic reasons) the Displacement Current6P28-Maxwell-Ampere’s Law 0 0 0 0 ( )encl d C E encl d I I dI dt µ µ µε ⋅ = + Φ = + ∫ Bs 7P28-PRS Questions: Capacitor8P28-Maxwell’s Equations9P28-Electromagnetism Review • E fields are created by: (1) electric charges (2) time changing B fields • B fields are created by (1) moving electric charges (NOT magnetic charges) (2) time changing E fields • E (B) fields exert forces on (moving) electric charges Gauss’s Law Faraday’s Law Ampere’s Law Maxwell’s Addition Lorentz Force10P28-Maxwell’s Equations 0 0 0 0 (Gauss's Law) (Faraday's Law) 0 (Magnetic Gauss's Law) (Ampere-Maxwell Law) ( (Lorentz force Law) in S B C S E enc C Qd dd dt d dd I dt q ε µ µε ⋅ = Φ ⋅ = − ⋅ = Φ ⋅ = + = + × ∫∫ ∫ ∫∫ ∫ EA E s BA B s F EvB) 11P28-Electromagnetic Radiation12P28-A Question of Time… http://ocw.mit.edu/ans7870/8/ 8.02T/f04/visualizations/light/ 05-CreatingRadiation/05-pith_f220_320.html13P28-14P28-Electromagnetic Radiation: Plane Waves http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/light/07-EBlight/07-EB_Light_320.html15P28-Traveling Waves Consider f(x) = x=0 What is g(x,t) = f(x-vt)? x=0 t=0 x=vt0 t=t0 x=2vt0 t=2t0 f(x-vt) is traveling wave moving to the right!16P28-Traveling Sine Wave Now consider f(x) = y = y0sin(kx): x Amplitude (y0) 2Wavelength ( ) wavenumber ( )k πλ = What is g(x,t) = f(x+vt)? Travels to left at velocity v y = y0sin(k(x+vt)) = y0sin(kx+kvt)17P28-Traveling Sine Wave Amplitude (y0) 1Period ( ) frequency ( ) 2 angular frequency ( ) T f π ω = = ( )0 siny y kx kvt= + 0 0sin( ) sin( )yy kvt y tω= ≡At x=0, just a function of time:18P28-Traveling Sine Wave 0 sin( )yy kx tω= −Wavelength: Frequency : 2Wave Number: Angular Frequency: 2 1 2Period: Speed of Propagation: Direction of Propagation: f k f T f v fk x λ π λ ω π π ω ω λ = = = = = = + i i i i i i i19P28-Electromagnetic Waves Remember: f cλ = Hz20P28-Electromagnetic Radiation: Plane Waves Watch 2 Ways: 1) Sine wave traveling to right (+x) 2) Collection of out of phase oscillators (watch one position) Don’t confuse vectors with heights – they are magnitudes of E (gold) and B (blue) http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/light/07-EBlight/07-EB_Light_320.html21P28-PRS Question: Wave22P28-Group Work: Do Problem 123P28-Properties of EM Waves 8 00 1 310 m vc sµε == = × 0 0 EE cB B = = Travel (through vacuum) with speed of light At every point in the wave and any instant of time, E and B are in phase with one another, with E and B fields perpendicular to one another, and to the direction of propagation (they are transverse): Direction of propagation = Direction of ×EB Direction of Propagation EE= E sin( k pˆ ⋅r −ωt); BB = B sin( k pˆ ⋅r −ωt) ˆ0 () ˆ0 () P28-24 ( ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆˆ z x y z x y ⋅ − − − − − − E B p p r i j k j k i k i j j i k k j i i k j ˆˆ ˆ×=EB p25P28-PRS Question: Direction of Propagation26P28-In Class Problem: Plane EM Waves27P28-Energy & the Poynting Vector28P28-Energy in EM Waves 2 2 0 0 1 1 ,2 2E Bu E u Bε µ = =Energy densities: Consider cylinder: 2 2 0 0 1( ) 2E B BdU u u Adz E Acdtε µ ⎛ ⎞ = + = +⎜ ⎟ ⎝ ⎠ What is rate of energy flow per unit area? 0 02 c EB cEB c ε µ ⎛ ⎞ = +⎜ ⎟ ⎝ ⎠ ( )2 00 0 1 2 EB cεµµ = + 0 EB µ = 1 dUS Adt = 2 2 0 02 c BEεµ ⎛ ⎞ = +⎜ ⎟ ⎝ ⎠29P28-Poynting Vector and Intensity 0 : Poynting vector µ × = EBS units: Joules per square meter per sec Direction of energy flow = direction of wave propagation Intensity I: 2 2 00 0 0 0 0 02 2 2 E B E cB I S cµ µ µ ≡< >= = =30P28-Energy Flow: Resistor 0µ × = EBS On surface of resistor is INWARD31P28-PRS Questions: Poynting Vector32P28-Energy Flow: Inductor On surface of inductor with increasing current is INWARD0µ × = EBS 33P28-Energy Flow: Inductor On surface of inductor with decreasing current is OUTWARD0µ × = EBS
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