Lecture 23: THU 15 APR 2010Review Session : Midterm 3Physics 2102Jonathan DowlingEXAM 03: 6PM THU 15 APR LOCKETT 6The exam will cover: Ch.28 (second half)through Ch.32.1-3 (displacement current, andMaxwell's equations).The exam will be based on: HW07 – HW10.The formula sheet for the exam can be found here:http://www.phys.lsu.edu/classes/spring2010/phys2102/formulasheet3.pdfYou can see examples of old exam IIIs here:http://www.phys.lsu.edu/classes/spring2009/phys2102/Test3.oldtests.pdf!µ !µHighest Torque: ϕ = ±90° sinϕ = ±1Lowest Torque: ϕ = 0° & 90° sinϕ = 0Bϕ = 180°–cosϕ = +1ϕ = 0°–cosϕ = –1 !!=µB sin"= !µB cos" !F= iLB sin!Right Hand Rule: Given Current i Find Magnetic Field B!Checkpoints/QuestionsCheckpoints/QuestionsMagnetic field?Force on each wire due tocurrents in the other wires?Ampere’s Law: Find Magnitude of ∫B·ds?The current in wires A,B,D is out of thepage, current in C is into the page. Each wireproduces a circular field line going throughP, and the direction of the magnetic field foreach is given by the right hand rule. So, thecircles centers in A,B,D arecounterclockwise, the circle centered at C isclockwise. When you draw the arrows at thepoint P, the fields from B and C are pointingin the same direction (up and left).Right Hand Rule: Given Current i Find Magnetic Field B!A length of wire is formed into a closed circuit with radii a and b, asshown 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.RiB!"µ40=µ=NiARight Hand Rule & Biot-Savart: Given i Find B! !B = !Lenz’s LawInduction and InductanceInduction and Inductance• Faraday’s law: or• Inductance: L=NΦ/I– For a solenoid: L=µ0n2Al=µ0N2A/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: τ = L/R Units: [s]• Magnetic energy: U=Li2/2 Units: [J]• Magnetic energy density: u=B2/2µ0Units: [J/m3] E= !d"BdtdtdsdEBC!"=#$!!iChanging B-Flux Induces EMFiupt( )=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/ LRL CircuitsRL CircuitsE/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, theinductor begins to get charged,and the current isi=(E/R)(1−e–tR/L).When the switch is opened, theinductors begins to discharge.The current inthis case is theni= (E/R) e–tR/LEq t( )=EC cos!t +"( )i t( )=#q t( )= $!EC sin!t +"( )#i t( )=##q t( )= $!2EC cos!t +"( )UBt( )=12Li2t( )=12L!EC sin!t +"( )%&'(2Units:!L!2E2C2= [s2•C•V/s2]=[V•C]=[J]VL(t) = $L#i t( )= L!2EC cos!t +"( )Units:!L!2EC = [s2/s2•V]=[V]LC CircuitsLC Circuits q t( )=EC cos!t +"( )qmax!=!EC! i t( )=!q t( )= "#EC sin#t +$( )qiimax!=!ω ECTFrequency: f!=!ω/2π [Hertz]Angular Frequency: ω = 2πf [rad/s]Period: T!=!1/f!=!2π/ω [sec]PISI22If we add a resistor in an circuit (see figure) we mustmodify the energy equation, because now energy isbeing dissipated on the resistor: .E BRLdUi RdtqU U U= != + =Damped Oscillations in an CircuitRCL222 2Li dU q dq diLi i RC dt C dt dt+ " = + = !( )222/222 210. This is the same equation as thatof the damped harmonics osc 0, which has theillator: The angular f solution re( ) c que:os .bt mmd x dxm b kxdt dtxdq di d q d q dqi L R qdt dt dt dt dtt x e tC! "#= $ = $ + + =+ + =%= +( )2/2222ncy For the damped circuit the solution is:The angular freque1 ( ) cos . .4ncy . 4Rt LRqk bRCt Qe tmLLC Lm! " !!#% %##+== =%(31-6)/ 2Rt LQe!/ 2Rt LQe!( )q tQQ!( )q t( )/ 2( ) cosRt Lq t Qe t! "#$= +2214RLC L!"= #/222The equations above describe a harmonic oscillator with an exponentially decayingamplitude . The angular frequency of the damped oscillator1 is always smaller than the angular 4Rt LQeRLC L!"#= "221frequency of the 1undamped oscillator. If the term we can use the approximation .4LCRL LC!!
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