Physics 11bLecture #14InductanceS&J Chapter 32Midterm #2 Midterm #2 will be on April 7th by popular vote Covers lectures #8 through #14 inclusive Up to the end of today’s lecture Textbook chapters from 27 up to 32.4 There will be 5 problems again One is multiple-choice One is from the homework Other rules are same as #1What We Did Last Time Changing magnetic field induces emf Faraday’s law: where Lenz’s law: induced current reduces the change in ΦBMoving-rod example – connection with Lorentz force Faradays’ law in terms of E field One of the four basic laws of E&M Others: Gauss’s law for E/B, Ampère’s law AC generator Eddy currentBdΦ= ⋅∫BAvBddtΦ=−EBdddtΦ⋅=−∫EsvsinNBA tωω=EToday’s Goals Introduce Inductance Mutual inductance M between two circuits Self inductance L of one circuit Calculate for simple example: solenoid What does an inductor do in a circuit? Define the rule that works with Kirchhoff Example: RL circuits Time constant Consider energy stored in an inductor Which in turn is stored in the magnetic fieldMutual Inductance Two wire loops are close to each other Turn on the switchÎ I increasesÎ B increasesÎ emf on the right loop Define mutual inductance between loop 1 and loop 2: ObviouslyEIBBddIdt dtΦ=− ∝−EGenerally true for any pair of circuits 1212dIMdt=−Eemf on loop 2current in loop 12121MIΦ=flux through loop 2Mutual Inductance Currents I1, I2and emfs E1, E2are related by It can be shown (mathematically) that M12= M21 We can just call it M and write Let’s look at an example1212dIMdt=−E2121dIMdt=−Eand12dIMdt=−E21dIMdt=−EandSolenoid and Loop A loop of wire is wrapped around a solenoid Solenoid has N turns inlength ℓ, current is I We’ve done the B field insidesolenoid in Lecture #12 It’s uniform inside ΦBthrough the loop is If we vary I, the loop will get BA0NIBµ=A220BrNIrBπµπΦ= =A20rNMπµ=A20rNdI dIMdt dtπµ=− =−EASelf Inductance We don’t really need two circuits to find inductance Any circuit produces B field B field changes when the currentI changes emf appears on the circuit itself Define self inductance L as Minus sign form Lenz’s law Î whenever the current changes, the circuit tends to resist the changeIBdILdt=−EBLIΦ=orInductor (a.k.a. Coil) Wire wound in a coil has a large self inductance Core can be empty (air-core) Ferromagnetic (e.g. iron) coreenhances the inductance Unit of inductance is Henry (H) Let’s calculate the inductanceof an air-core coilBI increasesemf against IBI decreasesemf along Iemf V V sHrate of change of current A s A⋅===dILdt=−EAir-Core Inductor We know the B field of solenoid Flux ΦBmust be B × area A Caution: the coil wraps around the same B field N times If the core is not air, use µmof the material instead of µ0 Permeabilityµmof ferrite is ~10000 times larger than µ00NIBµ=AB20BNIANBAµΦ= =A20BNALIµΦ==AHaving many turns helpInductors in Circuits What does an inductor do in a circuit? It resists the current change Compare this with R & C These rules + Kirchhoff determines currentIVaVbbadIVVV Ldt∆= − =−IVaVbbaVVV RI∆= − =−+Q −QIVaVbbaQVVVC∆= − =−dQIdt=RL Circuit Let’s consider a simple case: R and L Switch is closed at t = 0 What follows? Step 1: define the direction of I Step 2: apply the loop rule Differential equation with 3 terms! Note I is the only variable – everything else is constant Switch variable from I to +−REIL0dILRIdt−−=ExRI=−Edx dIRdt dt=−0LdxxRdt−=We’ve seen this beforeRL Circuit Solution for is exponential  We defined Fix x0with the initial condition  At t = 0, current I = 00LdxxRdt−=0expRxx tL⎛⎞=−⎜⎟⎝⎠+−REILxRI=−E0() expxRIt tRRL⎛⎞=− −⎜⎟⎝⎠E0(0) 0xxt R===−⋅=EE() 1 expRIt tRL⎧⎫⎛⎞=−−⎨⎬⎜⎟⎝⎠⎩⎭ERL Circuit Current in a simple RL circuit approaches its final value exponentially Final value = what I would be if L were not there Time constantLRτ=+−REIL() 1tIteRτ−⎧⎫=−⎨⎬⎩⎭EtREτAnother RL Circuit What happens when an RL circuit is switched off? Inductor resists change – but current cannot flow once the switch is open Let’s make it a little more interesting Consider this circuit Î Switch is initially closed andopens at t = 0 Before t = 0, current I0is constant L has ∆V = 0 No current flows through R2 That means+−R1EI0LR201IR=EIs this obvious?Another RL Circuit Once the switch is open, the lower half does nothing At t = 0, the current is still I0– L is trying to keep it Loop rule:  Solution is exponential+−R1EILR220dILRIdt−−=201() exptRItI t eLRτ−⎛⎞=−=⎜⎟⎝⎠E2LRτ=0()tItIeτ−=t1REτEnergy Conservation In the second RC circuit, the current kept flowing after the switch was opened R2dissipates power Integrating from t = 0 to infinity This much energy was spent by R2 Who supplied it?+−R1EILR22222220RtLPIV RI RIe−=∆ = =222202002022RtLLILUPdtRI eR∞∞−⎡⎤==− =⎢⎥⎣⎦∫Energy in Inductor Inductor L, with current I flowing, contains energy  When current is increased, it requires power The energy is recovered when current is decreased cf. a capacitor stores energy22LIU =dIPIV ILdt=− ∆ =I22dI LIU Pdt LI dt LIdIdt== = =∫∫ ∫22CVU =Energy in Magnetic Field Inductor is nothing but a wire in an empty space Energy is stored in the empty space itself Consider the air-core coil again We know and  The energy is Energy density (per unit volume) is20NALµ=AB0NIBµ=A222200222NAILI AUBµµ== =AA202BUBuAµ==Acf. We saw in Lecture #6202EEuε=Summary Inductance Mutual inductance Self inductance or For air-core solenoid Inductors in a circuit: RL circuit has a time constant Energy in an inductor , density12dIMdt=−E21dIMdt=−EdILdt=−EBLIΦ=20NALµ=AIVaVbbadIVVV Ldt∆= − =−LRτ=22LIU