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ADDENDUM – CHAPTER 21 Mutual inductance – Circulation of currents in one coil can generate a field in the coil that will extend to a second, close by device. Flux Changes Suppose i1 CHANGES Current (emf) is induced in 2nd coil. Mutual Inductance 3 i2 i1 creates a field that (partially) passes through the second coil. As i1 changes, the flux through coil 2 changes and an emf (and current i2) are created. The two coils are mutually linked by what we call an “inductance” Induction 5/23/2017 Watch Out! 4 Exam #2 one week from today. Chapters 20 & 21 Same format but possibly one set of multiple choice questions that you hate. You should already be studying. QUIZ on Friday – Chapter #21 Today we continue with the chapter. We should finish it on Friday. Maybe. No study session on Monday next week We will have a study session on Tuesday morning like last time. Details to follow. Induction 5/23/2017 This schedule is now in effect: PHY2054 Problem Solving/Office Hours Schedule Room MAP-318 Bindell Monday 8:30-9:15AM Tuesday Bindell Dubey 11:00-12:00PM 12:00-1:00PM Wednesday 8:30-9:15AM 10:30 - 11:15 AM* Thursday Friday 8:30-9:15AM 10:30-11:30AM 1:30-2:45PM These sessions will be used both for office hours and problem solving. Students from any section of 2054 are invited to stop by for assistance in course materials (problems, etc.) Note: There will be times when the room may not be available. In that case we will use our individual offices. * In Office 5 Dr. Dubey's hours are for problem solving only. Induction 5/23/2017 Mutual Inductance B 2 The magnetic flux in coil #2 due to the current in coil #1. 6 B 2 ~ i1 B 2 ki1 i2 N 2 B 2 M 21 i1 21 i1 emf2 N 2 M 21 t t mutual Inductance Induction 5/23/2017 Note the form: 7 M N something UNIT: henry i Think of this when we define INDUCTANCE (L) of a small coil in the next section. Induction 5/23/2017 The two coils 8 Remember – the magnetic field outside of the solenoid is pretty much zero. Two fluxes (fluxi?) are the same! Induction 5/23/2017 One solenoid is centered inside another. The outer one has a length of 50.0 cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and 0.120 cm in diameter and contains 15 coils. The current in the outer solenoid is changing at 37.5 A/s. (a) What is the mutual inductance of these solenoids? (b) Find the emf induced in the inner solenoid Length = 0.5 meters N=6750 coils n=6750/.5=1.35E04 turn/meter Magnetic field INSIDE the smaller coil is the same as in the larger coil and is given by: B1 0 ni 4 10 1.35 10 i 1.7i 7 9 4 Induction 5/23/2017 One solenoid is centered inside another. The outer one has a length of 50.0 cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and 0.120 cm in diameter and contains 15 coils. The current in the outer solenoid is changing at 37.5 A/s. (a) What is the mutual inductance of these solenoids? (b) Find the emf induced in the inner solenoid B1 0 ni 1.7i Areasmaller r 2 (0.06cm 1m / 100cm) 2 coil Area 1.13 104 m 2 B 2 BA 1.92 10 4 i B 2 i 1.92 10 4 1.92 10 4 37.5V 7.2 10 3V t t B 2 emf 2 N 2 15 7.2 10 3V 11mV t 10 Induction 5/23/2017 One solenoid is centered inside another. The outer one has a length of 50.0 cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and 0.120 cm in diameter and contains 15 coils. The current in the outer solenoid is changing at 37.5 A/s. (a) What is the mutual inductance of these solenoids? (b) Find the emf induced in the inner solenoid B1 0 ni 1.7i emf2 1110 M i 37 t 11 Induction 3 0.297mH 5/23/2017 Self-inductance – Any circuit which carries a varying current self-induced from it’s own magnetic field is said to have INDUCTANCE (L). 13 An inductor resists CHANGES in the current going through it. Induction 5/23/2017 14 An inductor resists CHANGES in the current going through it. Induction 5/23/2017 15 An inductor resists CHANGES in the current going through it. Induction 5/23/2017 Inductance Defined 16 N B L i If the FLUX changes a bit during a short time t, then the current will change by a small amount i. Li N B Faraday says this is the emf! B i N L t t This is actually a calculus equation Induction 5/23/2017 So … 17 i E= emf L t There should be a (-) sign but we use Lenz’s Law instead! The UNIT of “Inductance – L” of a coil is the henry. SYMBOL: Induction 5/23/2017 18 Induction 5/23/2017 Consider “AC” voltage 19 Minimum Change@t V1 Maximum Change@t Induction 5/23/2017 The transformer 20 emf1 V1 N1 t FLUX is the same through both coils (windings). emf 2 V2 N 2 t V1 V2 N1 N 2 Induction 5/23/2017 21 Induction 5/23/2017 Input/Output Impedance (Resistance) 22 V2 N 2 V1 N1 Powerin Powerout (Lossless) I1V1 I 2V2 So V1 R I1 ( N 2 / N1 ) 2 Induction looks like an input resistance ! 5/23/2017 Remember that a Capacitor stored ENERGY? U=(1/2)CV2 23 U=Area=(1/2)LI2 Power Vi i V L t i P Li t U Lii U Li LI i Li DU Lii interval i I Induction 5/23/2017 i SO … 24 Energy Stored in a capacitor The energy stored in a capacitor with capacitance C and a voltage V is U=(1/2)LI2 Induction 5/23/2017 The Energy stored is in the Magnetic Field 25 Consider a solenoid with N turns that is very long. We assume that the field is uniform throughout its length, ignoring any “end effects”. For a long enough solenoid, we can get away with it for the following argument. Maybe. N B 0 ni 0 i l Induction 5/23/2017 Energy Storage in Inductor 26 N B l 1 2 1 N B 2 U Li i 2 2 l B BA L 1 NiBA 2 N B 0 i l Bl Ni U 1 Bl 1 B 2lA 1 B 2V U BA 2 0 2 0 2 0 U 1 B2 Energy Density V 2 0 1 Capacitor ED 0 E 2 2 0 Induction 5/23/2017 27 Back to Circuits Induction 5/23/2017 Series LR Circuit 28 Induction 5/23/2017 RL or LR Series Circuit 29 Switch is open .. no current flows for obvious reasons. Switch closed for a long time: Steady current, voltage across the inductor is zero. All voltage (E) is across the resistor. i=E/R Induction 5/23/2017 RL or LR Series Circuit 30 When the switch opens, current change is high and back emf from L is maximum. i E/R t As the current increases, more voltage is across R, the rate of change of I decreases and as the current increases, it increases more slowly. Induction 5/23/2017 RL Circuit 31 When L=0, the current rises very rapidly (almost instantly) As L increases, it takes longer for the current to get to its maximum. Induction 5/23/2017 RL Circuit - Kirchoff Stuff 32 i emf iR L 0 i Solution E i (1 e ( R / L ) t ) R E i (1 e t / ) R L (time constant) R Induction 5/23/2017 The Graphic Result – Current Growth 33 1 1 0.63 e } 63% of maximum e= 2.71828… Induction 5/23/2017 Decay – Short out the battery 34 Magnetic field begins to collapse, sending its energy into driving the current. The energy is dissipated in the resistor. i begins at maximum (E/R) and decays. Induction 5/23/2017 Solution 35 E t / i (e ) R Induction 5/23/2017 Up and Down and Up and Down and ….. 36 Induction 5/23/2017 NEXT: AC Circuits 37 Induction