PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Checkpoints/QuestionsSlide 7Slide 8Slide 9Slide 10Induction and InductanceSlide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Lecture 32: WED 01 APRReview Session : Midterm 3Physics 2102Jonathan DowlingQuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.EXAM 03: 6PM THU 02 APR LOCKETT 5 EXAM 03 REVIEW: 6PM WED 01 APR NICHOLSON 130The exam will cover: Ch.28 (second half) through Ch.32.1-3 (displacement current, and Maxwell's equations). The exam will be based on: HW08 – HW11. The formula sheet for the exam can be found here:http://www.phys.lsu.edu/classes/spring2009/phys2102/formulasheet.pdfYou can see examples of old exam IIIs here:http://www.phys.lsu.edu/classes/spring2009/phys2102/Test3.oldtests.pdfQuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture. rμ rμHighest Torque: = ±90° sin = ±1Lowest Torque: = 0° & 90° sin = 0B = 180°–cos = +1 = 0°–cos = –1 rτ =μBsinϕ=−μB cosϕ rF =iLBsinϕQuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.Right Hand Rule: Given Current i Find Magnetic Field BBCheckpoints/QuestionsCheckpoints/QuestionsMagnetic field?Force on each wire due to currents in the other wires?Ampere’s Law: Find Magnitude of ∫B∙ds?The current in wires A,B,D is out of the page, current in C is into the page. Each wire produces a circular field line going through P, and the direction of the magnetic field for each is given by the right hand rule. So, the circles centers in A,B,D are counterclockwise, the circle centered at C is clockwise. When you draw the arrows at the point P, the fields from B and C are pointing in the same direction (up and left).Right Hand Rule: Given Current i Find Magnetic Field BBA length of wire is formed into a closed circuit with radii a and b, as shown in the Figure, and carries a current i. (a) What are the magnitude and direction of B at point P? (b) Find the magnetic dipole moment of the circuit. RiBπφμ40==NiARight Hand Rule & Biot-Savart: Given i Find BB rB =⊗QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.Lenz’s LawInduction and Induction and InductanceInductance•Faraday’s law: or•Inductance: L=N/I –For a solenoid: L=0n2Al=0N2A/l•Inductor EMF: EL= L di/dt•RL circuits: i(t)=(E/R)(1–e–tR/L) or i(t)=i0e–tR/L•RL Time Constant: = L/R Units: [s]•Magnetic energy: U=Li2/2 Units: [J]•Magnetic energy density: u=B2/20Units: [J/m3] E =−dΦBdtdtdsdEBCΦ−=⋅∫rriQuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.Changing B-Flux Induces EMFQuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.iupt( )=ER1−e−tR/L( )UBt( )=12Li2=12LE2R21−e−tR/L( )2iup′t( )=ELe−tR/LVL=−Li=−ELe−tR/L iupt( )=ER1−e−tR/L( )UBt( )=12Li2=12LE2R21−e−tR/L( )2iup′t( )=ELe−tR/LVL=−Li=−ELe−tR/LFlux UpFlux Down idnt( )=ERe−tR/LUBt( )=12Li2=12LE2R2e−2tR/Liup′t( )=−ELe−tR/LVL=−Li=ELe−tR/L idnt( )=ERe−tR/LUBt( )=12Li2=12LE2R2e−2tR/Liup′t( )=−ELe−tR/LVL=−Li=ELe−tR/LRL CircuitsQuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.RL CircuitsQuickTime™ and a decompressorare needed to see this picture.E/2t=?Checkpoints/QuestionsCheckpoints/QuestionsMagnitude of induced emf/current?Magnitude/direction of induced current?Magnitude/direction of magnetic field inducing current?Given |∫E∙ds| , direction of magnetic field?Current inducing EL?Current through the battery?Time for current to rise 50% of max value?Given B, dB/dt, magnitude of electric field?Largest current?Largest L?R,L or2R,L orR, 2L or2R,2L?When the switch is closed, the inductor begins to get charged, and the current is i=(E/R)(1e–tR/L).When the switch is opened, the inductors begins to discharge.The current in this case is then i= (E/R) e–tR/LQuickTime™ and a decompressorare needed to see this picture.Eq t( )=E C cos ωt +ϕ( )i t( )=′q t( )=−ωE Csin ωt +ϕ( )′i t( )=′′q t( )=−ω2E C cos ωt +ϕ( )UBt( )=12Li2t( )=12L ωE C sin ωt +ϕ( )⎡⎣⎤⎦2Units:*Lω2E2C2=[s2•C•V/s2]=[V•C]=[J]VL(t) =−L′i t( )=Lω2E C cos ωt +ϕ( )Units:*Lω2E C =[s2/s2•V]=[V] q t( )=E C cos ωt +ϕ( )i t( )=′q t( )=−ωE Csin ωt +ϕ( )′i t( )=′′q t( )=−ω2E C cos ωt +ϕ( )UBt( )=12Li2t( )=12L ωE C sin ωt +ϕ( )⎡⎣⎤⎦2Units:*Lω2E2C2=[s2•C•V/s2]=[V•C]=[J]VL(t) =−L′i t( )=Lω2E C cos ωt +ϕ( )Units:*Lω2E C =[s2/s2•V]=[V]LC CircuitsQuickTime™ and a decompressorare needed to see this picture.LC CircuitsQuickTime™ and a decompressorare needed to see this picture. q t( )=E C cos ωt +ϕ( )qmaxB=BECB i t( )=′q t( )=−ωE Csin ωt +ϕ( )qiimaxB=BECTFrequency: fB=B 2 [Hertz]Angular Frequency: = f[rad/s]Period: TB=B1/fB=B[sec]QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.PISIQuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare
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