Physics 012 1st Edition Lecture 20 Outline of Last Lecture I. Intensitya. Savg. = Pavg./Areab. E = cBc. Savg. = EmaxBmax)/2μ0II. Electromagnetic Spectruma. c = λfb. increasing f = decreasing λ = higher energyc. visible light falls between λ = 400nm and λ = 700nmIII. Polarizationa. electric field can point in any directioni. polarized light contains an electric field only oriented in one directionb. the intensity of initially unpolarized light is reduced by a factor of ½ when it passes through a polarizerc. Malus’ Law: I = I0cos2θThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.i. for when already polarized light passes through a second filterii. θ = angle between direction of electric field and transmission axisOutline of Current Lecture IV. Problem 28 from ch. 22a. coil: N = 105, r = 4x10-2, R = 0.48 Ωb. ΔB/Δt = 0.783 T/sc. Bcenter = (Nμ0I)/2rd. ΦB = ΣBΔAcosθ = Bπr2e. εind = -Nπr2 (ΔB/Δt)f. Iind = |εind|/R = [Nπr2 (ΔB/Δt)]/Rg. B = (Nμ0I)/2r = [(N2πrμ0)/2R][ΔB/Δt] = V. Problem 53 from ch. 24a. A = 135m2, Pavg = 1.2 x 104 Wb. Savg = Pavg/A = E2max/2μ0cc. Erms = Emax/√(2)d. Pavg/A = 2E2rms/2μ0c = E2rms/μ0ce. Erms = [(Pavgμ0c)/A]1/2 f. Brms = Erms/cVI. Problem 40 from ch.21a. ΣFx = FB – 2Tsinφ = 0b. ΣFy = 2Tcosφ –mg = 0c. φ =
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