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 AttributionNoncommercial-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/terms Massachusetts 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 ∫ E i ds = - dt ∫ µ C H i da 0 S Circulation of E Magnetic Flux µ0 = 4π ×10-7 henries/meter [magnetic permeability of free space] EQS form: ∫ E i ds = 0 (Kirchoff’s Voltage Law, conservative electric C field) MQS circuit form: v = L di (Inductor) dt 2. Ampère’s Law (with displacement current) ∫ H i ds = C ∫ J i da + S d dt Circulation Conduction Current of H MQS form: ∫ H i ds C 6.013 Electromagnetics and Applications Prof. Markus Zahn = ∫ε 0 E i da S Displacement Current ∫ J i da S Lecture 1 Page 1 of 7 dv (capacitor) dt 3. Gauss’ Law for Electric Field EQS circuit form: i = C ∫ ε E i da 0 = S ∫ ρ dV V -9 10 ≈ 8.854 ×10-12 farads/meter 36π 1 c= ≈ 3 × 108 meters/second (Speed of electromagnetic waves in ε0 µ 0 ε0 ≈ free space) 4. Gauss’ Law for Magnetic Field ∫ µ H i da 0 = 0 S In free space: B = µ0 H magnetic flux density (Teslas) magnetic field intensity (amperes/meter) 5. Conservation of Charge Take Ampère’s Law with displacement current and let contour C → 0 lim C →0 d ∫ H i ds = 0 = ∫ J i da + dt ∫ ε E i da 0 C S S ∫ ρ dV V 6.013 Electromagnetics and Applications Prof. Markus Zahn Lecture 1 Page 2 of 7 ∫ J i da S + d dt ∫ ρ dV = 0 V Total current Total charge leaving volume inside volume through surface 6. Lorentz Force Law ( f = q E + v × µ0 H ) II. Electric Field from Point Charge ∫ ε E i da = ε E 4π r 0 0 r 2 =q S Er = q 4π ε0r2 T sin θ = fc = q2 4π ε0r2 T cos θ = Mg tan θ = 6.013 Electromagnetics and Applications Prof. Markus Zahn q2 r = 4π ε0r2Mg 2l Lecture 1 Page 3 of 7 ⎡ 2π ε0r3Mg ⎤ q= ⎢ ⎥ l ⎣ ⎦ 1 2 III. Faraday Cage d d dq ∫ J i da = i = - dt ∫ ρ dV = - dt (-q) = dt S ∫ 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 Prof. Markus Zahn Lecture 1 Page 4 of 7 ∫ H i ds ≈ H 2 π a = N 1 i ⇒ H1 ≈ 1 1 Cb ( ) λ ≈ N1 π a2 µ0 H1 = L= N12 π a 2 µ0 2 π a N1 i1 2πa i1 ≈ N12 a µ0 i1 2 2 1 λ N a µ 0 ≈ i1 2 ω = 1 LC 1 1 L ip2 ≈ C vp2 ⇒ ip ≈ vp C L 2 2 C = 25 µ f, vp = 4 k V, N1 = 50, a ≈ 7 c m L1 ≈ 0.1 mH ip ≈ 2000 A, ω ≈ 20 x 103 / s ⇒ f = ω ≈ 3k Hz 2π Hp ≈ 2.3 x 105 A / m ⇒ Bp = µ0 Hp ≈ 0.3 Teslas ≈ 3000 Gauss 2. Electrical Breakdown in Single Turn Coil with Small Gap R Bp ∆ ⎧0 E≈ ⎨ ⎩E0 Inside Metal Coil Small Gap ∆ 6.013 Electromagnetics and Applications Prof. Markus Zahn Lecture 1 Page 5 of 7 d ∫ Eids = E ∆ = − dt ( B πR 0 2 p ) C Bp = Bm cos ωt E0 = BmωπR 2 sin ωt ∆ Take: Bm ≈ 0.3 Tesla, ω ≈ 20, 000 radians/second, R ≈ 0.07 m, ∆ = 0.01 mm Bm ωπR 2 0.3(20, 000)π(0.07)2 = = 9 × 106 Volts/meter ∆ 10−5 Breakdown strength of air ≈ 3 × 106 Volts/meter. Em = Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. 3. Force on Metal Disk ∫ E i ds ≈ 2 π aE φ =− Cb dBp d B i da ≈ −π a2 = πa2Bmω sin ωt ∫ dt Sa dt Jφ = σ Eφ = − F = J x µ0 H , σ a dBp σa Bm ω sin ωt = 2 dt 2 f = ∫ F dV = Force per unit volume ∫ Jx µ 0 H dV V V total force K φ ≈ Jφ ∆ = −Hr ⇒ Hr = − Jφ ∆ F = J × µ0 H = Jφ iφ × µ0Hr ir = −µ0 JφHr iz = µ0 Jφ2 ∆iz 2 ⎛ σa ⎞ Fz = µ0 J2φ ∆ = µ0 ∆ ⎜ Bmω ⎟ sin2 ωt 2 ⎝ ⎠ fz = Fz πa2 ∆ = π µ0 ∆2 σ2 a4 2 2 Bmω sin2 ωt 4 6.013 Electromagnetics and Applications Prof. Markus Zahn Lecture 1 Page 6 of 7 σaluminum ≈ 3.7 × 107 Siemens/meter, a=0.07 m, ∆ =2 mm, ω = 20, 000 radians/second, Bm ≈ 0.3 Tesla, M=0.08 kg fz = µ0 π∆σa2 ωBm 4π ( ) ( 2 sin2 ωt )( 2 ) 2 = 10−7 ⎡ π 2 × 10−3 3.7 × 107 (.07 ) 20, 000 ( 0.3) ⎤ sin2 ωt ⎣ ⎦ 6 2 = 4.7 × 10 sin ωt Mg = (0.08)9.8 ≈ 0.8 Newtons fmax 4.7 × 106 ≈ ≈ 5.9 × 106 Mg 0.8 Neglecting losses: 1 1 CV2 = Mv2 (t = 0+ ) = Mgh 2 2 C V M C = 25µf , M = .08 kg, Vp = 4000 volts v(t = 0+ ) = v(t = 0+ ) = 70.7 meters/second h= (Initial velocity) 2 v ( t = 0+ ) = 255 meters 2g (Maximum height) Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. 6.013 Electromagnetics and Applications Prof. Markus Zahn Lecture 1 Page 7 of 7