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 Quiz 1, October 20, 2005 6.013 Formula Sheet attached. Problem 1 (35 Points) rJ=J () i = J0r i amperes/meter 2 0 < r < R1z z zR1 z 1R 2R r φ A coaxial cable of very long length carries a z directed current density that varies with radial position on the inner cylinder as: rJ=J () i = J0r i amperes/meter2 0 < r < R1z z zR1 A perfectly conducting outer cylinder of radius R2 carries all the return current so that =H0 for r > R2 . a) Find the H field for 0r<< R1. b) What are the magnitude and direction of the surface current density (amperes/meter) on the rR2 surface?=Problem 2 (35 Points) Two flat electrodes at angle α extend from radius 12toRR and have a depth d in the z direction (out of the paper). The electrodes enclose a lossy dielectric medium with permittivity ε and conductivity σ. There is no free volume charge within the lossy dielectric. The electric potentials on the electrodes are ()()000and VφφαΦ== Φ= =. a) The scalar electric potential Φ is of the form ()ABφφΦ=+. What values of A and B satisfy the boundary conditions? b) Find the electric field E(r, )φwithin the lossy dielectric. c) What is the free surface charge density, sfσ, on the electrode at φα= ? d) What is the capacitance of this device? You may neglect fringing fields. z 1R 2R0V,εσφ0 r Depth d α r iProblem 3 (30 Points) x • • ,εµ z y iθ iH iE An electromagnetic wave is traveling at an angle θ with respect to the z axis within a medium iwith dielectric permittivity ε and magnetic permeability µ. The magnetic field is given as: ⎛ ⎞π×⎡ j⎜2108 t−π(x+ 3z)⎟⎤ Hi = H0 Re ⎢e ⎝ ⎠⎥ iy amperes/meter . ⎢ ⎥⎦⎣ a) Find the frequency in Hertz. b) Find the wavelength in meters. c) Find the numerical value of the speed of light in the medium in meters/second. d) Find the angle θ.i6.013 Quiz 1 Formula Sheet October 20, 2005 Cartesian Coordinates (x,y,z): ∂Ψ ∂Ψ∇Ψ = xˆ ∂Ψ + yˆ ∂y + zˆ∂x ∂z ∂Ax ∂Ay +∂Az∇iA = +∂x ∂y ∂z ∇×A = x ⎛∂Az −∂Ay ⎞⎟+ yˆ ⎝⎜ ∂z −∂Az ⎞+ zˆ ⎛⎜ ∂Ay −∂Ax ⎞ˆ ⎝⎜ ∂y ∂z ⎠ ⎛∂Ax ∂x ⎠⎟ ⎝∂x ∂y ⎟ ⎠ 22 ∂Ψ ∂Ψ2∇Ψ =∂Ψ+ 2 + ∂x2 ∂y2 ∂z2 Cylindrical coordinates (r,φ,z): ˆ1 ∂Ψ + zˆ ∂Ψ∇Ψ = rˆ ∂Ψ +φ∂r r ∂φ ∂z 1 ∂(rAr )+1 ∂Aφ∂Az ∇iA = + r ∂r r ∂φ ∂z rˆ rφˆ zˆ⎛1 ∂Az −∂Aφ⎞ ˆ ⎛∂Ar −∂Az ⎞ 1 ∂ ⎛ (rAφ)∇×A = rˆ ⎝⎜ r ∂φ ⎟+φ⎜ ⎟+ zˆr ⎝⎜ ∂r −∂Ar ⎟⎞ = 1det ∂ ∂ r ∂ ∂φ ∂ ∂ z∂z ⎠ ⎝∂z ∂r ⎠ ∂φ⎠ r Ar rA φ Az 2 2 ∇Ψ = 1 ∂∂ (r ∂Ψ)+ 1 ∂ Ψ ∂ Ψ 2 + r r ∂r r2 ∂φ2 ∂z2 Spherical coordinates (r,θ,φ): ˆ1 ∂Ψ∇Ψ = rˆ ∂Ψ +θˆ1 ∂Ψ +φ∂r r ∂θ r sin θ ∂φ 2 ∇iA = 1 ∂(rAr )+ ∂(sin θAθ)+ 1 ∂Aφ1 2 ∂r r sin θ ∂θ r sin θ ∂φ r 1 ⎛∂(sin θAφ) ∂Aθ⎞ ⎛ 1 ∂Ar 1 ∂(rA φ)⎞ 1 ⎛∂(rAθ) ∂Ar ⎞ˆ ˆ∇×A = rˆ ⎜ −⎟+θ⎜ − ⎟+φ⎜ −⎟r sin θ⎝ ∂θ ∂φ⎠ ⎝ r sin θ ∂φ r ∂r ⎠ r ⎝∂r ∂θ ⎠ rˆ rθˆ rsin θφˆ ∂ ∂ r ∂ ∂θ ∂ ∂φ = 1 det 2rsin θ Ar rA θ r sin θAφ 221 ∂Ψ 1 ∂ Ψ ∇Ψ = 1 ∂(r2 ∂Ψ)+ ∂(sin θ )+2 ∂r ∂r r2 sin θ ∂θ 2r ∂θ r2 sin2 θ∂φ Gauss’ Divergence Theorem: Vector Algebra: ∇=∂ ∂ + ∂ ∂ + ∂ ∂ xˆ x yˆ y zˆ z ∫V ∫• xx + A By + A Bz∇iG dv = AGinˆ da AB = A B y z (∇•∇× A)= 0 ( (Stokes’ Theorem: ∇× ∇× A)=∇ ∇• A)−∇ 2A(∇×G )inˆ da = G d i ∫A ∫CBasic Equations for Electromagnetics and Applications Fundamentals fq E + v ×µ H)[]=( No ∇×E = −∂ B ∂t ∫cEds =− dB • da• dt ∫A ∇×HJ +∂ D ∂t = ∫cHds =∫AJ •da + dD • da• dt ∫A ∇•D =ρ→ ∫AD • da = ρdv ∫V ∇•B0 →∫AB • da = 0= ∇• ρ∂=−∂J t E = electric field (Vm-1) H = magnetic field (Am-1) D = electric displacement (Cm-2) B = magnetic flux density (T) Tesla (T) = Weber m-2 = 10,000 gauss ρ = charge density (Cm-3) J = current density (Am-2) σ = conductivity (Siemens m-1) εJs = surface current density (Am-1) ρs = surface charge density (Cm-2) o ≈ 8.854 × 10-12 Fm-1 µo = 4π × 10-7 Hm-1 -0.5 ≅ 3 × 108c = (εoµo) ms-1 e = -1.60 × 10-19 C ηo ≅ 377 ohms = (µo/εo)0.5 ∇−µε∂ ∂ 2 2( t2 )E = 0 [Wave Eqn.] jtEy(z,t) = E+(z-ct) + E-(z+ct) = Re{E (z)e ω}y Hx(z,t) = ηo-1[E+(z-ct)-E-(z+ct)] [or(ωt-kz) or (t-z/c)] 2 2 )∫A (EH)• da +(d dt )∫V (ε× E 2 +µ H 2 dv =−∫VE • J dv (Poynting Theorem) Media and Boundaries D =ε E + Po ∇•D =ρ f, τ=εσ ∇•ε E =ρ +ρ o f p ∇•P =−ρ , J =σ Ep B =µH =µ (H + M )o 0.52 2ε=ε (1−ω ω2 ), ω=(Ne mεo )(Plasma)o p p ε=ε(1j− σ ωε)eff skin depth δ = (2/ωµσ)0.5 [m] = 0E1// − =×nˆH1// − Js − = 02 nˆ1 D B1⊥ 1⊥− 2 //E2 //H2B⊥ 2D⊥ =ρs 0 = if σ= ∞ Electromagnetic Quasistatics E =−∇Φ (r), Φ(r) =∫V' (ρ()r 4πε r '− r )dv ' ∇2Φ = -ρf ε C = Q/V = Aε/d [F] L = Λ/I i(t) = C dv(t)/dt v(t) = L di(t)/dt = dΛ/dt we = Cv2(t)/2; wm = Li2(t)/2 Lsolenoid = N2µA/W τ = RC, τ = L/R • Λ=∫A Bda (per turn) FI -1 ] =×µ H[ Nm o Electromagnetic Waves ∇−µε∂ ∂ 2 2( t2 )E = 0 [Wave Eqn.] 2 2 ) jk ri(∇+k E = 0, E= E e− k = ω(µε)0.5 = ω/c = 2π/λ o 2kx2 + ky2 + kz2 = ko = ω2µε vp = ω/k, vg = (∂k/∂ω)-1 θr = θi sin θt sin = iθ=kki tnni t θ=sin−1 (n ni )c t 0.5θ=tan−1 ( i ) for TM B ε ε t xjkzz θ>θ ⇒ Et = E Te+α − ci kk ' =−jk '' T1 Γ=− TTE = (2 1 +[η cos cos θi ])i θηt t TTM = (2 1 +ηt cos [ θ η t i cos
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