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. Suppose i 1 CHANGES Flux Changes Current (emf) is induced in 2 nd coil.

Mutual Inductance  i 1 creates a field that (partially) passes through the second coil.  As i 1 changes, the flux through coil 2 changes and an emf (and current i 2 ) are created.  The two coils are mutually linked by what we call an “inductance” 10/13/2015Induction 3 i2i2

Watch Out!  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. 10/13/2015Induction 4

10/13/2015Induction 5 This schedule is now in effect: PHY2054 Problem Solving/Office Hours Schedule Room MAP-318 MondayTuesdayWednesdayThursdayFriday Bindell8:30-9:15AM Bindell11:00-12:00PM 10: :15 AM*10:30-11:30AM Dubey12:00-1:00PM 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 Dr. Dubey's hours are for problem solving only.

Mutual Inductance 10/13/2015Induction 6 i2i2 mutual Inductance

Note the form: 10/13/2015Induction 7 INDUCTANCE Think of this when we define INDUCTANCE (L) of a small coil in the next section. UNIT: henry

The two coils 10/13/2015Induction 8 Remember – the magnetic field outside of the solenoid is pretty much zero. Two fluxes (fluxi?) are the same!

10/13/2015Induction 9 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 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:

10/13/2015Induction 10 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 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

10/13/2015Induction 11 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 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

Self-inductance –  Any circuit which carries a varying current self-induced from it’s own magnetic field is said to have INDUCTANCE (L).

CHANGES An inductor resists CHANGES in the current going through it. 10/13/2015Induction 13

CHANGES An inductor resists CHANGES in the current going through it. 10/13/2015Induction 14

CHANGES An inductor resists CHANGES in the current going through it. 10/13/2015Induction 15

Inductance Defined 10/13/2015Induction 16 If the FLUX changes a bit during a short time  t, then the current will change by a small amount  i. This is actually a calculus equation Faraday says this is the emf!

So … 10/13/2015Induction 17 E= The UNIT of “Inductance – L” of a coil is the henry. SYMBOL: There should be a (-) sign but we use Lenz’s Law instead!

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Consider “AC” voltage 10/13/2015Induction 19 V 1 Maximum  t Minimum  t

The transformer 10/13/2015Induction 20 FLUX is the same through both coils (windings).

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Input/Output Impedance (Resistance) 10/13/2015Induction 22

Remember that a Capacitor stored ENERGY? U=(1/2)CV 2 10/13/2015Induction 23 i i Li LI ii Li DU I Induction U=Area=(1/2)LI 2

SO … 10/13/2015Induction 24 Energy Stored in a capacitor The energy stored in a capacitor with capacitance C and a voltage V is U=(1/2)LI 2

The Energy stored is in the Magnetic Field 10/13/2015Induction 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.

Energy Storage in Inductor 10/13/2015Induction 26

Back to Circuits 10/13/ Induction

Series LR Circuit 10/13/2015Induction 28

RL or LR Series Circuit  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 10/13/2015Induction 29

RL or LR Series Circuit 10/13/2015Induction 30 i t E/R When the switch opens, current change is high and back emf from L is maximum. 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.

RL Circuit  When L=0, the current rises very rapidly (almost instantly)  As L increases, it takes longer for the current to get to its maximum. 10/13/2015Induction 31

RL Circuit - Kirchoff Stuff 10/13/2015Induction 32

The Graphic Result – Current Growth 10/13/2015Induction 33 e= … 63% of maximum }

Decay – Short out the battery  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. 10/13/2015Induction 34

Solution 10/13/2015Induction 35

Up and Down and Up and Down and ….. 10/13/2015Induction 36

Induction 37 NEXT: AC Circuits