MIT OpenCourseWare http://ocw.mit.edu 6.013/ESD.013J Electromagnetics and Applications, Fall 2005 Please use the following citation format: Markus Zahn, Erich Ippen, and David Staelin, 6.013/ESD.013J Electromagnetics and Applications, Fall 2005. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/termsMassachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.013 Electromagnetics and Applications Lecture 1, Sept. 8, 2005 I. Maxwell’s Equations in Integral Form in Free Space 1. Faraday’s Law d ∫ Eds =-dt ∫µ0H i da i C S Circulation Magnetic Flux of E µ0=4 ×10 π-7 henries/meter [magnetic permeability of free space] EQS form: ∫ Eds 0 (Kirchoff’s Voltage Law, conservative electric i = C field) diMQS circuit form: v= L (Inductor) dt2. Ampère’s Law (with displacement current) ∫ Hds = ∫ i + d ∫ε0Ei Jda i da dtC S S Circulation Conduction Displacement of H Current Current MQS form: ∫ Hds = ∫ Jda ii C S 6.013 Electromagnetics and Applications Lecture 1 Prof. Markus Zahn Page 1 of 7dvEQS circuit form: i= C (capacitor)dt3. Gauss’ Law for Electric Field ∫ε0Eda i = ∫ρ dV S V ε0 ≈10-9 ≈ 8.854×10-12 farads/meter 36π c= 1 ≈×3 108 meters/second (Speed of electromagnetic waves in εµ00 free space) 4. Gauss’ Law for Magnetic Field µ Hda =0∫0 i S In free space: B = µ0H magnetic magnetic flux field density intensity (Teslas) (amperes/meter) 5. Conservation of Charge Take Ampère’s Law with displacement current and let contour C → 0 lim ∫H i ds = 0 = ∫J i da + d ∫ε0E i da →C0C S dtS ∫ρdV V 6.013 Electromagnetics and Applications Lecture 1 Prof. Markus Zahn Page 2 of 7d ∫ J i da + ∫ρ dV = 0 dt VSTotal current Total charge leaving volume inside volume through surface 6. Lorentz Force Law f=q E+v×µ0H( ) II. Electric Field from Point Charge ε E i da = ε E 4 π r = q∫ 0 0 r2 SE= q r 4π ε0r2 2 T sin θ =f = q c24π ε0r T cos θ =Mg 2 tan θ =q =r 24π ε0rMg 2l 6.013 Electromagnetics and Applications Lecture 1 Prof. Markus Zahn Page 3 of 71 2⎡ 3 ⎤2π ε0r Mg q= ⎢ ⎥ ⎣ l ⎦ III. Faraday Cage ∫Jdai = i = -d ∫ρdV -d () dq = -q = dt dt dtS ∫idt = q IV. Edgerton’s Boomer 1. Magnetic Field, Current, and Inductance Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. 6.013 Electromagnetics and Applications Lecture 1 Prof. Markus Zahn Page 4 of 7Ni ∫Hds i ≈ H 2 1 π a=Ni 1 1 ⇒ H1 ≈ 11 π2aCb 2a2 µ Naµ21 0 1 0λ≈N (π a )µ H =N22a π 12 1i ≈ i1 0 1 π Naµ λ12 0L= ≈ i1 2 1ω = LC 1212 C 2Li ≈ 2Cv ⇒ i ≈ vLp p p p C = 25 µf, v p = 4 k V, N 1= 50, a ≈ 7 cm L1≈ 0.1 mH 3 ωip≈2000 A, ω≈ 20 x10 / s ⇒f = ≈3k Hz 2π 5Hp ≈ 2.3 x 10 A / m ⇒ Bp = µ0 Hp ≈ 0.3 Teslas ≈ 3000 Gauss 2. Electrical Breakdown in Single Turn Coil with Small Gap ∆ R Bp ⎧0 Inside Metal Coil E ≈⎨ ⎩E0 Small Gap ∆ 6.013 Electromagnetics and Applications Lecture 1 Prof. Markus Zahn Page 5 of 7∫i d2 C E ds =E0∆ = − dt (B pπR ) B = B cos ωt p m B ωπR2 E0 = m sin ωt ∆ Take: Bm≈ 0.3 Tesla, ω≈20,000 radians/second, R ≈0.07 m, ∆= 0.01 mm ωπR2 π 2Bm 0.3(20,000) (0.07) 6Em = = −5 = 9 × 10 Volts/meter∆ 10Breakdown strength of air ≈× Volts/meter. 31063. Force on Metal Disk d 2dBp 2 ∫ i ≈π2 aE φ = −∫ i ≈−πa = πa B mωsin ωtEds Bda dt dt Cb Sa σa dBp σaJ= σE = − = B ωsin ωtφ φ m2dt 2 = ∫J x µ0 HdV V 0F=Jx Hµ , ∫ V f= F dVForce per unit volume total force ≈∆=H ⇒ H =− ∆ Kφ Jφ−r rJφ =×µ H = J i ×µ H i = −µ JHi = µ 2∆F J 0 φ φ 0 r r 0 φ r z 0Jφ iz F =µJ2∆=µ∆ ⎛σaB ω⎞2 sin 2 ωtz 0 φ 0 ⎜ m ⎟⎝2 ⎠ 224µ∆σ af =π∆F a2 = π 0 B2 ω2 sin 2 ωtz z m4 6.013 Electromagnetics and Applications Lecture 1 Prof. Markus Zahn Page 6 of 7 Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.σalu minum ≈ 3.7 × 107 Siemens/meter, a=0.07 m, ∆ =2 mm, ω= 20, 000 radians/second, Bm≈ 0.3 Tesla, M=0.08 kg f =µ0 (π∆σ a2ωB )2 sin 2 ωtz m4π = 10−7 ⎡π 2 × 10 −3 3.7 × 107 .07 220, 000 0.3 ⎤2 sin 2 ωt = 4.7 × 10 6 sin 2 ωt Mg = (0.08)9.8 ≈ 0.8 Newtons f 4.7 × 10 6 ⎣( )( )( ) ( )⎦ max≈ ≈ 5.9 × 10 6 Mg 0.8 Neglecting losses: 1 21 2CV = Mv (t = 0 ) = Mgh 2 2 + v(t = 0 ) = + CV M =M.08 kg,C = 25 f µ , V = 4000 voltspv(t = 0 ) = 70.7 meters/second (Initial velocity) + 2 h = v (t = 0+)= 255 meters (Maximum height) 2g 6.013 Electromagnetics and Applications Lecture 1 Prof. Markus Zahn Page 7 of 7 Courtesy of Hermann A. Haus and James R. Melcher. Used with
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