Chapter 32. Maxwell s Equations and Electromagnetic Waves
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1 Chapter 32 Maxwell s Equatins and Electrmagnetic Waves
2 Maxwell s Equatins and EM Waves Maxwell s Displacement Current Maxwell s Equatins The EM Wave Equatin Electrmagnetic Radiatin MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/2012 2
3 Smething is Missing Frm Ampere s Law Bidl = µ JidA = µ I C C S The surface S in the integral abve can be any surface whse bundary is C. If the surface S 2 is chsen fr use in the abve integral the result will be that the magnetic field arund C is zer. But there is current flwing thrugh the wire s we knw there is a magnetic field present. MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/2012 3
4 Smething is Missing Frm Ampere s Law Bidl = µ JidA = µ I C C S The surface S 2 has the same bundary as S 1 but there is n current passing thrugh S 2. The charge is accumulating n the capacitr. Maxwell nticed this deficiency in Ampere s law and fixed it by defining the Displacement Current I d. He began by taking surfaces S 1 and S 2, putting them tgether and treating them like ne clsed surface S MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/2012 4
5 Bidl = µ JidA = µ I C C S The Displacement Current Charge is building up n the disk within the clsed surface S. Therefre there are electric field lines, E, that are crssing the surface S. We can use Gauss s Law here. φ = E i da = e S φe 1 Q 1 = = I t ε t ε I =ε d φ e t Q enclsed ε d MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/2012 5
6 Maxwell fixed the prblem with Ampere s law by adding anther current t the right hand side f the equatin belw Bidl = µ JidA = µ I C C The Displacement Current S I =ε d φ t e C Bidl = µ JidA = µ I + µ I = µ I + µε C d C 0 S φ e t MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/2012 6
7 Displacement Current Example In calculating the displacement current we will be making the apprximatin that the electric field is everywhere unifrm. This requires that the plate separatin be much smaller than R, the radius f the plate. The surface S must nt extend past the edge f the capacitr plates. S r must be less than R. MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/2012 7
8 Displacement Current Example φ = e In calculating the displacement current we will need t cmpute the electric flux acrss the surface S. EiˆndA = EA S Q σ A Q E = = = ε ε ε A ( ) I =ε d EA de d Q dq I = ε = ε A = ε A d = dt dt dt ε A dt d dφ dt e MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/2012 8
9 B-Field frm the Displacement Current In calculating the B-Field frm the displacement current we will be making the same apprximatins that were made in the last example: C B i dl = µ I + µε C 0 φ t e the electric field is everywhere unifrm. C B i dl = B 2πr ( ) There is n current thrugh S s I C is zer dφ B i dl = B ( 2πr ) = 0 +µε dt C e MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/2012 9
10 B-Field frm the Displacement Current C B i dl = B 2πr = 0 +µε ( ) dφ dt The size f S will vary s φ e will depend n r 2 2 φ e = AE =πr E =πr ε 2 2 σ 2 Q Qr φ e =πr =πr = 2 2 ε επr εr 2 2 d Qr r dq B( 2πr ) = µε = µ 2 2 dt ε R R dt µ r dq µ r B = = I 2 2 2π R dt 2π R σ e MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
11 Maxwell s Equatins MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
12 Maxwell s Equatins C B i dl = µ I + I = µ I + µε ( ) C d C 0 φ t e Ampere s Law ε φ i m B [ ] [ ] n = E dl = 0 - = 0 - da t t C S Q Faraday s Law inside EndA = Gauss s Law ε 0 BndA = 0 S N name - there are n magnetic mnples MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
13 Maxwell s Equatins C E i ( ) n B dl = µ I + I = µ I + µε da C d C 0 t ε φ i m B n = E dl = - = - da t t C n S 1 E da = ρdv = ε Q S inside ε 0 V 0 BndA = 0 S Ampere s Law Faraday s Law Gauss s Law N name - there are n magnetic mnples MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
14 EM Wave Equatin MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
15 Cnservative Frces and Ptentials frm Vectr Analysis W = F dl = 0 C S C F dl = F da ( ) S ( ) F da = 0 F = 0 F = - V since V = 0 Wrk arund a clsed lp = 0 Stkes Therem ( ) Therefre a ptential functin V exists fr a cnservative frce. MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
16 Vectr Analysis φ and ψ are scalar functins F and G are vectr functins φ = grad φ = gradient f φ F = div F = divergence f F F = curl F = curl f F MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
17 Vectr Analysis Gradient Divergence Curl φ ˆ φ ˆ φ φ = i + j + k ˆ x y z F F x y F F = + + x y z ˆi ˆj kˆ F = x y z F F F x y z z MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
18 Vectr Identities ( fg ) = f g + g f i i i ( AiB ) = ( Bi ) A+ ( Ai ) B + B ( A ) + A ( B) ( fa ) = ( f ) A+ f ( A) i( A B ) = B i( A) - A i( B) ( fa ) = ( f ) A+ f ( A) ( A B ) = ( Bi ) A - ( Ai ) B + ( ib) A - ( ia) B i 2 A = A - A ( ) MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
19 Vectr Identities B B B x y z By By By + A ˆ x + A y + Az j x y z ( ) x x x Ai B = A ˆ x + A y + Az i Bz Bz Bz + ˆ A x + A y + Az k x y z MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
20 Maxwell s Equatins: Integral Frm t Differential Frm Stkes Therem Eidl = C S EindA ˆ Divergence Therem S FindA ˆ = V ifdv MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
21 Maxwell s Equatins C C B i dl = µ I + µε E i dl = - C 0 B t n da S E t n da Ampere s Law Faraday s Law n S 1 E da = ρdv ε 0 BndA = 0 S V Gauss s Law N name - there are n magnetic mnples MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
22 Maxwell s Equatins 1 E da = E i nda ˆ = ρdv n ε S S 0 V Gauss s Law Use the Divergence Therem t recast the surface integral int a vlume integral 1 EindA ˆ = iedv = ρdv ε S V V V ρ ie - dv = 0 ε ρ ie - = 0 ε MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
23 Maxwell s Equatins C B i dl = µ I + µε C Integral frm E i dl = - C 0 S B t n da E t n da Ampere s Law Faraday s Law Differential frm B - 1 E t µ ε 0 0 B E + = 0 t = µ J m n S 1 E da = ρdv ε 0 V Gauss s Law ρ ie = ε BndA = 0 S ib = 0 MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
24 Wave Eqn frm Maxwell s Eqn The differential frm f Maxwell s equatins brings ut the symmetry and nn-symmetry f the E and B fields We will use the fllwing vectr identity with the E-field ( i ) 2 A = A - A ( i ) 2 E = E - E B E + = 0 t ( ) 2 E - ie = B t ( ) 2 E E - ie = µj + εµ f t t MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
25 Wave Eqn frm Maxwell s Eqn ( ) 2 E E - ie = µj + εµ f t t 2 2 E J f ρ E -εµ =µ + 2 t t ε In free space there are n surces 2 2 E E - ε µ = 0 2 t E 1 E - = 0; where c = 2 2 c t ε µ These are the surce terms This is the frm f a wave equatin is the speed f light MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
26 Slutins f the Wave Equatin In free space the slutins f the wave equatins shw that E and B are in phase. Ex Ex = sin kx -ωt By By ( ) 2π 2π k = ; ω = = 2πf λ T These equatins describe plane waves that are unifrm thrugh ut any plane perpendicular t the x-axis. MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
27 Plane Plarized Waves Examining the E and B cmpnents shw that this represents a plane plarized wave. The E vectr is riented in the x directin and the B vectr is riented in the y directin. Ex Ex = sin kx -ωt By By ( ) 2π 2π k = ; ω = = 2πf λ T MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
28 Relatinships Between E and B Vectrs k 1 E B = E = E = ω c c E = cb The Pynting Vectr describes the prpagatin f the electrmagnetic energy E B S = µ With E in the x-directin and B in the y-directin the energy flws in the z-directin. MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
29 Relatinships Between E and B Vectrs The Pynting Vectr describes the prpagatin f the electrmagnetic energy E B S = µ E B = E ( ) ˆ ( ) ˆ sin kx -ωt i B sin kx -ωt j 2 E B = E B sin kx -ωt kˆ ( ) The energy is prprtinal t E and B and is flwing in the z-directin, perpendicular t E and B. MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
30 The Principle f Invariance Einstein s fundamental pstulate f relativity can be stated: It is physically impssible t detect the unifrm mtin f a frame f reference frm bservatins made entirely within that frame. The laws f Physics shuld be the same fr all nn-accelerated bservers. MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
31 The Principle f Invariance The laws f Physics shuld be the same fr all nn-accelerated bservers. If tw bservers watch the mtin f an bject frm tw different inertial reference systems (n acceleratin), mving at a relative velcity v, they shuld find the same laws f Physics F 1 = m 1 a 1 and F 2 =m 2 a 2 MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
32 Galilean and Lrentz Transfrmatins MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
33 Galilean and Lrentz Transfrmatins The inertial reference frames are related by a Galilean transfrmatin. Newtn s laws are invariant under these transfrmatins but nt Maxwell s Equatins Prir t Einstein s Thery f Special Relativity it was determined that a Lrentz transfrmatin kept Maxwell s equatin invariant. Hwever, n ne knew exactly what they meant. MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
34 Galilean and Lrentz Transfrmatins MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
35 Electrmagnetic Radiatin MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
36 MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
37 Electrmagnetic Radiatin These are all different frms f electrmagnetic radiatin. Anytime yu accelerate r decelerate a charged particle it gives ff electrmagnetic radiatin. Electrns circulating abut their nuclei dn't give ff radiatin unless they change energy levels. Thermal mtin gives ff cntinuus EM radiatin. Example Infrared radiatin which peaks belw the visible spectrum. MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
38 Electric Diple Radiatin MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
39 Electric Diple Radiatin MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
40 Diple Antenna - Radiatin Distributin I(θ) 2 sinθ r 2 Nte the different rientatin f the angle measurement MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
41 Diple Antenna - Radiatin Distributin I(θ) = I 2 sin θ 2 r In these prblems yu will need t determine the value f I r else take a rati s that the I factr will cancel ut. MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
42 Diple Antenna - Radiatin Distributin (a.) Find I 1 at r 1 = 10m and θ = 90 (b.) Find I 2 at r 2 = 30m and θ = 90 I(r,θ) = I 2 sin θ 2 r (a.) (b.) (c.) (c.) Rati f I 2 / I 1 I = I(r = 10,θ = 90 ) = I 1 2 I = I(r = 30,θ = 90 ) = I 2 2 I I 2 1 = 900 = I I I = I = MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
43 MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
44 Electric - Diple Antenna MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
45 Oscillating Ring Antenna MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
46 Oscillating Ring Pair Antenna MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
47 Electric - Diple Antenna Plane wave Far frm surce antenna - Far Field MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
48 Magnetic - Lp Antenna Plane wave Far frm surce antenna - Far Field MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
49 ε ε Magnetic - Lp Antenna ( ) dφ d BA m 2 B = - = - = - πr dt dt t 2 B = πr t rms rms ( ) B = B sin kx -ωt B = -ωb cs ( kx -ωt ) t B ωb ( ) = ωb -cs kx -ωt = = ωb rms t 2 rms ε ε rms rms Find ε rms? rms 2 B 2 = πr = πr ωbrms t rms Erms 2π 2 = πr ωb = πr ω = r fe rms c c MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/ rms rms
50 Magnetic - Lp Antenna r = 10.0cm N = 1 E rms = V/m f : (a.) 600 khz; (b.) 60.0 MHz (a.) ε ε rms ε rms 2π c 2 2 = r f Erms 2π ( ) ( ) = c 2 2π 2 6 = x mV c = x µV (b.) ( ) ( ) = rms MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
51 Energy and Mmentum in an Electrmagnetic Wave MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
52 Energy and Mmentum in an Electrmagnetic Wave MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
53 E-M Energy and Mmentum The Pynting Vectr describes the prpagatin f the electrmagnetic energy E B S = µ U u LA avg avg P avg = = = uavg Ac t L c Pavg I = = uavgc A U avg is the ttal energy and u avg is the energy density. I is the intensity, the average pwer per unit area. MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
54 E-M Energy and Mmentum These are the electric and magnetic energy densities 1 2 B u = ε E and u = e m 2 2µ Since E = cb ( E c) 2 B 2 E m 2 e 2µ 2µ 2µ c 2 u = = = = ε E = u Therefre the energy density can be expressed in different ways. 2 2 B EB u = u + u = ε E = = e m µ µ c 2 MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
55 E-M Energy and Mmentum The energy density 2 2 B EB u = u + u = ε E = = e m µ µ c E rmsbrms 1 EB I = uavgc = = = S µ 2 µ The intensity I is the energy/(m 2 sec) = pwer/m 2 ; E B S = µ This is the Pynting vectr, its magnitude is the intensity. avg MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
56 Radiatin Pressure p r Mmentum I p r = = = u Unit AreaiUnit Time c avg I E B E B E B p = = = = = 2 2 rms rms r 2 c 2µ c µ c 2µ c 2µ MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
57 Radiatin Pressure p r - Example A lightbulb emits spherically symmetric electrmagnetic waves in al directins. Assume 50W f electrmagnetic radiatin is emitted. Find (a) the intensity, (b) the radiatin pressure, (c) the electric and magnetic field magnitudes at 3.0m frm the bulb. The energy spreads ut unifrmly ver a sphere f radius r. The surface area f the sphere is 4πr 2. (a.) (b.) Pwer 50 W I = Intensity = = = Area 4πr m I r c 8 3.0x10-9 p = = = 1.47x10 Pa 2 2 MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
58 Radiatin Pressure p r - Example A lightbulb emits spherically symmetric electrmagnetic waves in al directins. Assume 50W f electrmagnetic radiatin is emitted. Find (a) the intensity, (b) the radiatin pressure, (c) the electric and magnetic field magnitudes at 3.0m frm the bulb. (c.) Remember I E B p = = = and E = cb 2 2 r 2 c 2µ c 2µ r -8 B = 6.08x10 T ( -9 )( )( -7 ) B = 2µ p = 1.47x10 2 4πx10 0 E 0 = 18.2 m ( ) E = cb = 3.0x x10 V 8-8 MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
59 Extra Slides MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
60 MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/
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