Lecture 24Chapter 31Induction and InductanceReview• Can produce an induced current and induced emf in a loop of wire when the number of magnetic field lines passing through the loop is changing• Magnetic flux• Faraday’s law• Lenz’s law – An induced emf gives rise to a current whose B field opposes the change in flux that produced it∫•=Φ AdBBrrdtdBΦ−=EReview• Induced emf of a conductor moving with velocity, v, in a ⊥⊥⊥⊥ B field is given by • Induced current in loop in a B field experiences a force• Found F1opposes your force FappBLv=EBLiFBrr×=1FFapprr−=• Work you do in pulling the loop appears as thermal energy in the loopInductance (20)• Put a copper ring in a uniform B field which is increasing in time so the magnetic flux through the copper ring is changing• By Faraday’s law an induced emfand current are produced• From Lenz’s law their direction is counterclockwise• If there is a current there must be an E field present to move the conduction electrons around ringInductance (21)• Induced E field is same as E field produced by static charges, it will exert a force, F=qE, on a charged particle• Restate Faraday’s law – A changing B field produces an E field• True even if no copper ringInductance (22)• Electric field lines produced by a changing B field are set of concentric circles• If B field increasing or decreasing field lines are present (decreasing have opposite direction)• If B field is constant with time, no E fieldInductance (23)• Calculate work done on particle by induced E field • Remember •For q0moving along closed path the work is defined as•But FEis dqdW=EE0qW =∫•= sdFWrrEqFE 0=• Equating equations for work gives∫•= sdEqrr0Eq0Inductance (24)• Canceling q0find that • But from Faraday’s law • So Faraday’s law can be written as• A changing B field induces an E field∫•= sdErrEdtdBΦ−=EdtdsdEBΦ−=•∫rrInductance (25)• Inductor is a device used to produce a desired B field (e.g. solenoid)• A current, i, in an inductor with N turns produces a magnetic flux, ΦB, in its central region • Inductance, L is defined as• SI unit is henry, H iNLBΦ=AmTH /112⋅=Inductance (26)• What is inductance per unit length near the middle of a long solenoid?• First find flux of single loop in solenoid• Remember # turns (N) per unit length (l) isBAAdBB=•=Φ∫rriNLBΦ=lNn /=inlBAL =Inductance (27)• B field from a solenoid• Inductance per unit length is • Depends only on geometry of device (like capacitance)AnlL20µ=inB0µ=AnliAinnlinlBAL200)(µµ===Inductance (28)• A changing current in a coil generates a self-induced emf, ЕЕЕЕLin the coil • Process is called self-induction• Change current in coil using a variable resistor, EEEEL, will appear in coil only while the current is changingiNLBΦ=dtdiLdtLiddtNddtdNBBL−=−=Φ−=Φ−=)()(EInductance (29)• Induced emf only depends on rate of change of current, not its magnitude• Determine direction of EEEELusing Lenz’s law• Self-induced VLacross inductor– Ideal inductor – Real inductor (like real battery) has some internal resistance dtdiLL−=ELLV E=iRVLL−= EInductance (30)• Checkpoint #4 – Have an induced emf in a coil. What can we tell about the current through the coil? Is it moving right or left and is it constant, decreasing or increasing?Decreasing and rightward (answer d) ORIncreasing and leftward (answer e)Inductance (31)• RL circuit is a resistor and inductor in series• Similar to RC circuit (resistor & capacitor in series) – Charging up a capacitor – Discharging capacitor–whereRCC=τ)1(cteCqτ−−= Ecteqτ−=0qInductance (32)• Analogous time dependence on rise (or fall) of current if introduce an emf into (or removie it from) an RL circuit• Initially close switch i is increasing through inductor so ELopposes rise • i through R will be • Long time later i is constant so EL=0 and i in circuit is Ri E<Ri E=Inductance (33)• Initially an inductor acts to oppose changes in current through it• Long time later inductor acts like ordinary conducting wire• Apply loop rule right after switch has been closed at a• Starting at x, go clockwise0=+−− EdtdiLiRInductance (34)• Differential equation similar to capacitors• Solution is • Where inductive time constant isdtdiLiR +=E()LteRτ−−= 1EiRLL=τ• Satisfies conditions• At t=0, i = 0 • At t=∞∞∞∞, i =EEEE/RInductance (35)• Now move switch to position b so battery is out of system• Current will decrease with time and loop rule gives• Solution is 0=+dtdiLiRLLtteieRiττ−−==0E• Satisfies conditions• At t=0, i =i0 =EEEE/R• At t=∞∞∞∞, i = 0RLL=τInductance (36)• Have a circuit with resistors and inductors• What is the current through the battery just after close the switch?• Inductor oppose change in current through it• Right after switch is closed, current through inductor is 0• Inductor acts like broken wireInductance (37)• Apply loop rule • Immediately after switch closed current through the battery is0=− iRERiE=Inductance (38)• What is the current through the battery long time after the switch has been closed?• Currents in circuit have reached equilibrium so inductor acts like simple wire• Circuit is 3 resistors in
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