Presentation is loading. Please wait.

Presentation is loading. Please wait.

Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures.

Similar presentations


Presentation on theme: "Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures."— Presentation transcript:

1 Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures by Wayne Anderson Chapter 30 Inductance

2 Copyright © 2012 Pearson Education Inc. Energy through space for free?? A puzzler! Increasing current in time Creates increasing flux INTO ring

3 Copyright © 2012 Pearson Education Inc. Energy through space for free?? A puzzler! Increasing current in time Creates increasing flux INTO ring Induce counter- clockwise current and B field OUT of ring

4 Copyright © 2012 Pearson Education Inc. Energy through space for free?? But… If wire loop has resistance R, current around it generates energy! Power = i 2 /R!! Increasing current in time Induced current i around loop of resistance R

5 Copyright © 2012 Pearson Education Inc. Energy through space for free?? Yet…. NO “potential difference”! Increasing current in time Induced current i around loop of resistance R

6 Copyright © 2012 Pearson Education Inc. Energy through space for free?? Answer? Energy in B field!! Increasing current in time Increased flux induces EMF in coil radiating power

7 Copyright © 2012 Pearson Education Inc. Goals for Chapter 30 - Inductance To learn how current in one coil can induce an emf in another unconnected coil To relate the induced emf to the rate of change of the current To calculate the energy in a magnetic field

8 Copyright © 2012 Pearson Education Inc. Goals for Chapter 30 Introduce circuit components called INDUCTORS Analyze circuits containing resistors and inductors Describe electrical oscillations in circuits and why the oscillations decay

9 Copyright © 2012 Pearson Education Inc. Introduction How does a coil induce a current in a neighboring coil. A sensor triggers the traffic light to change when a car arrives at an intersection. How does it do this? Why does a coil of metal behave very differently from a straight wire of the same metal? We’ll learn how circuits can be coupled without being connected together.

10 Copyright © 2012 Pearson Education Inc. Mutual inductance Mutual inductance: A changing current in one coil induces a current in a neighboring coil. Increase current in coil 1

11 Copyright © 2012 Pearson Education Inc. Mutual inductance Mutual inductance: A changing current in one coil induces a current in a neighboring coil. Increase current in coil 1 Increase B flux

12 Copyright © 2012 Pearson Education Inc. Mutual inductance Mutual inductance: A changing current in one coil induces a current in a neighboring coil. Increase current in coil 1 Increase B flux Induce current in loop 2 Induce flux opposing change

13 Copyright © 2012 Pearson Education Inc. Mutual inductance EMF induced in single loop 2 = - d (  B2 )/dt –Caused by change in flux through second loop of B field –Created by the current in the single loop 1 So…  B2 is proportional to i 1 What does that proportionality constant depend upon?

14 Copyright © 2012 Pearson Education Inc. Mutual inductance  B2 is proportional to i 1 and is affected by: –# of windings in loop 1 –Area of loop 1 –Area of loop 2 Define “M” as mutual inductance of coil 1 on coil 2

15 Copyright © 2012 Pearson Education Inc. Mutual inductance Define “M” as mutual inductance on coil 2 from coil 1  B2 = M 21 I But what if loop 2 also has many turns? Increase #turns in loop 2? => increase flux!

16 Copyright © 2012 Pearson Education Inc. Mutual inductance IF you have N turns in coil 2, each with flux  B, total flux is Nx larger Total Flux = N 2  B2

17 Copyright © 2012 Pearson Education Inc. Mutual inductance  B2 proportional to i 1 So N 2  B = M 21 i 1 M 21 is the “Mutual Inductance” [Units] = Henrys = Wb/Amp

18 Copyright © 2012 Pearson Education Inc. Mutual inductance EMF 2 = - N 2 d [  B2 ]/dt and N 2  B = M 21 i 1 so EMF 2 = - M 21 d [i 1 ]/dt Because geometry is “shared” M 21 = M 12 = M

19 Copyright © 2012 Pearson Education Inc. Mutual inductance Define Mutual inductance: A changing current in one coil induces a current in a neighboring coil. M = N 2  B2 /i 1 M = N 1  B1 /i 2

20 Copyright © 2012 Pearson Education Inc. Mutual inductance examples Long solenoid with length l, area A, wound with N1 turns of wire N2 turns surround at its center. What is M?

21 Copyright © 2012 Pearson Education Inc. Mutual inductance examples M = N 2  B2 /i 1 We need  B2 from the first solenoid (B 1 =  o ni 1 ) n = N 1 /l  B2 = B 1 A M = N 2  o i 1 AN 1 /li 1 M =  o AN 1 N 2 /l All geometry!

22 Copyright © 2012 Pearson Education Inc. Mutual inductance examples M =  o AN 1 N 2 /l If N 1 = 1000 turns, N 2 = 10 turns, A = 10 cm 2, l = 0.50 m –M = 25 x 10 -6 Wb/A –M = 25  H

23 Copyright © 2012 Pearson Education Inc. Mutual inductance examples Using same system ( M = 25  H) Suppose i 2 varies with time as = (2.0 x 10 6 A/s)t At t = 3.0  s, what is average flux through each turn of coil 1? What is induced EMF in solenoid?

24 Copyright © 2012 Pearson Education Inc. Mutual inductance examples Suppose i 2 varies with time as = (2.0 x 10 6 A/s)t At t = 3.0  s, i 2 = 6.0 Amps M = N 1  B1 /i 2 = 25  H  B1 = Mi 2 /N 1 = 1.5x10 -7 Wb Induced EMF in solenoid? –EMF 1 = -M(di 2 /dt) –-50Volts

25 Copyright © 2012 Pearson Education Inc. Self-inductance Self-inductance: A varying current in a circuit induces an emf in that same circuit. Always opposes the change! Define L = N  B /i Li = N  B If i changes in time: d(Li)/dt = Nd  B /dt = -EMF or EMF = -Ldi/dt

26 Copyright © 2012 Pearson Education Inc. Inductors as circuit elements! Inductors ALWAYS oppose change: In DC circuits: –Inductors maintain steady current flow even if supply varies In AC circuits: –Inductors suppress (filter) frequencies that are too fast.

27 Copyright © 2012 Pearson Education Inc. Potential across an inductor The potential across an inductor depends on the rate of change of the current through it. The self-induced emf does not oppose current, but opposes a change in the current. V ab = -Ldi/dt

28 Copyright © 2012 Pearson Education Inc. Magnetic field energy Inductors store energy in the magnetic field: U = 1/2 LI 2 Units: L = Henrys (from L = N  B /i ) N  B /i = B-field Flux/current through inductor that creates that flux Wb/Amp = Tesla-m 2 /Amp [U] = [Henrys] x [Amps] 2 [U] = [Tesla-m 2 /Amp] x [Amps] 2 = Tm 2 Amp But F = qv x B gives us definition of Tesla [B] = Teslas= Force/Coulomb-m/s = Force/Amp-m

29 Copyright © 2012 Pearson Education Inc. Magnetic field energy Inductors store energy in the magnetic field: U = 1/2 LI 2 [U] = [Tesla-m 2 /Amp] x [Amps] 2 = Tm 2 Amp [U] = [Newtons/Amp-m] m 2 Amp = Newton-meters = Joules = Energy!

30 Copyright © 2012 Pearson Education Inc. Magnetic field energy The energy stored in an inductor is U = 1/2 LI 2. The energy density in a magnetic field (Joule/m 3 ) is u = B 2 /2  0 (in vacuum) u = B 2 /2  (in a magnetic material) Recall definition of  0 (magnetic permeability) B =  0 i/2  r (for the field of a long wire)  0 = Tesla-m/Amp [u] = [B 2 /2  0 ] = T 2 /(Tm/Amp) = T-Amp/meter

31 Copyright © 2012 Pearson Education Inc. Magnetic field energy The energy stored in an inductor is U = 1/2 LI 2. The energy density in a magnetic field (Joule/m 3 ) is u = B 2 /2  0 (in vacuum) [u] = [B 2 /2  0 ] = T 2 /(Tm/Amp) = T-Amp/meter [U] = Tm 2 Amp = Joules So… energy density [u] = Joules/m 3

32 Copyright © 2012 Pearson Education Inc. Calculating self-inductance and self-induced emf Toroidal solenoid with area A, average radius r, N turns. Assume B is uniform across cross section. What is L?

33 Copyright © 2012 Pearson Education Inc. Calculating self-inductance and self-induced emf Toroidal solenoid with area A, average radius r, N turns. L = N  B /i  B = BA = (  o Ni/2  r)A L =  o N 2 A/2  r (self inductance of toroidal solenoid) Why N 2 ?? If N =200 Turns, A = 5.0 cm2, r = 0.10 m L = 40  H

34 Copyright © 2012 Pearson Education Inc. Potential across an inductor The potential across a resistor drops in the direction of current flow V ab = V a -V b > 0 The potential across an inductor depends on the rate of change of the current through it.

35 Copyright © 2012 Pearson Education Inc. Potential across an inductor The potential across an inductor depends on the rate of change of the current through it. The self-induced emf does not oppose current, but opposes a change in the current.

36 Copyright © 2012 Pearson Education Inc. Potential across an inductor The potential across an inductor depends on the rate of change of the current through it. The self-induced emf does not oppose current, but opposes a change in the current. The inductor acts like a temporary voltage source pointing OPPOSITE to the change.

37 Copyright © 2012 Pearson Education Inc. Potential across an inductor The inductor acts like a temporary voltage source pointing OPPOSITE to the change. This implies the inductor looks like a battery pointing the other way! Note V a > V b ! Direction of current flow from a battery oriented this way

38 Copyright © 2012 Pearson Education Inc. Potential across an inductor What if current was decreasing? Same result! The inductor acts like a temporary voltage source pointing OPPOSITE to the change. Now inductor pushes current in original direction Note V a < V b !

39 Copyright © 2012 Pearson Education Inc. The R-L circuit An R-L circuit contains a resistor and inductor and possibly an emf source. Start with both switches open Close Switch S1: Current flows Inductor resists flow Actual current less than maximum E/R E – i(t)R- L(di/dt) = 0 di/dt = E /L – (R/L)i(t)

40 Copyright © 2012 Pearson Education Inc. The R-L circuit Close Switch S1: E – i(t)R- L(di/dt) = 0 di/dt = E /L – (R/L)i(t) Boundary Conditions At t=0, di/dt = E /L i(  ) = E /R Solve this 1 st order diff eq: i(t) = E /R (1-e -(R/L)t )

41 Copyright © 2012 Pearson Education Inc. Current growth in an R-L circuit i(t) = E /R (1-e -(R/L)t ) The time constant for an R-L circuit is  = L/R. [  ]= L/R = Henrys/Ohm = ( Tesla-m 2 /Amp ) /Ohm = ( Newtons/Amp-m) (m 2 /Amp)/Ohm = (Newton-meter) / (Amp 2 -Ohm) = Joule/Watt = Joule/(Joule/sec) = seconds!

42 Copyright © 2012 Pearson Education Inc. Current growth in an R-L circuit i(t) = E /R (1-e -(R/L)t ) The time constant for an R-L circuit is  = L/R. [  ]= L/R = Henrys/Ohm EMF = -Ldi/dt [L] = Henrys = Volts /Amps/sec Volts/Amps = Ohms (From V = IR) Henrys = Ohm-seconds [  ]= L/R = Henrys/Ohm = seconds! V ab = -Ldi/dt

43 Copyright © 2012 Pearson Education Inc. The R-L circuit E = i(t)R+ L(di/dt) Power in circuit = E I E i = i 2 R+ Li(di/dt) Some power radiated in resistor Some power stored in inductor

44 Copyright © 2012 Pearson Education Inc. The R-L circuit example R = 175  ; i = 36 mA; current limited to 4.9 mA in first 58  s. What is required EMF What is required inductor What is the time constant?

45 Copyright © 2012 Pearson Education Inc. The R-L circuit Example R = 175  ; i = 36 mA; current limited to 4.9 mA in first 58  s. What is required EMF What is required inductor What is the time constant? EMF = IR = (0.36 mA)x(175  ) = 6.3 V i(t) = E /R (1-e -(R/L)t ) i(58  s) = 4.9 mA 4.9mA = 6.3V(1-e -(175/L)0.000058 ) L = 69 mH  = L/R = 390  s

46 Copyright © 2012 Pearson Education Inc. Current decay in an R-L circuit Now close the second switch! Current decrease is opposed by inductor EMF is generated to keep current flowing in the same direction Current doesn’t drop to zero immediately

47 Copyright © 2012 Pearson Education Inc. Current decay in an R-L circuit Now close the second switch! –i(t)R - L(di/dt) = 0 Note di/dt is NEGATIVE! i(t) = -L/R(di/dt) i(t) = i(0)e -(R/L)t i(0) = max current before second switch is closed

48 Copyright © 2012 Pearson Education Inc. Current decay in an R-L circuit i(t) = i(0)e -(R/L)t

49 Copyright © 2012 Pearson Education Inc. Current decay in an R-L circuit Test yourself! Signs of Vab and Vbc when S1 is closed? Vab >0; Vbc >0 Vab >0, Vbc <0 Vab 0 Vab <0, Vbc <0

50 Copyright © 2012 Pearson Education Inc. Current decay in an R-L circuit Test yourself! Signs of Vab and Vbc when S1 is closed? Vab >0; Vbc >0 Vab >0, Vbc <0 Vab 0 Vab <0, Vbc <0 WHY? Current increases suddenly, so inductor resists change

51 Copyright © 2012 Pearson Education Inc. Current decay in an R-L circuit Test yourself! Signs of Vab and Vbc when S1 is closed? Vab >0; Vbc >0 WHY? Current still flows around the circuit counterclockwise through resistor EMF generated in L is from c to b So Vb> Vc!

52 Copyright © 2012 Pearson Education Inc. Current decay in an R-L circuit Test yourself! Signs of Vab and Vbc when S2 is closed, S1 open? Vab >0; Vbc >0 Vab >0, Vbc <0 Vab 0 Vab <0, Vbc <0

53 Copyright © 2012 Pearson Education Inc. Current decay in an R-L circuit Test yourself! Signs of Vab and Vbc when S2 is closed, S1 open? Vab >0; Vbc >0 Vab >0, Vbc <0 WHY? Current still flows counterclockwise di/dt <0; EMF generated in L is from b to c! So Vb < Vc!

54 Copyright © 2012 Pearson Education Inc. The L-C circuit An L-C circuit contains an inductor and a capacitor and is an oscillating circuit. Initially capacitor fully charged; close switch Charge flows FROM capacitor, but inductor resists that increased flow. Current builds in time. At maximum current, charge flow now decreases through inductor Inductor now resists decreased flow, and keeps pushing charge in the original direction i

55 Copyright © 2012 Pearson Education Inc. The L-C circuit An L-C circuit contains an inductor and a capacitor and is an oscillating circuit. Initially capacitor fully charged; close switch Charge flows FROM capacitor, but inductor resists that increased flow. Current builds in time. Capacitor slowly discharges At maximum current, no charge is left on capacitor; current now decreases through inductor Inductor now resists decreased flow, and keeps pushing charge in the original direction i

56 Copyright © 2012 Pearson Education Inc. The L-C circuit An L-C circuit contains an inductor and a capacitor and is an oscillating circuit.

57 Copyright © 2012 Pearson Education Inc. The L-C circuit An L-C circuit contains an inductor and a capacitor and is an oscillating circuit. Now capacitor fully drained; Inductor keeps pushing charge in the original direction Capacitor charge builds up on other side to a maximum value While that side charges, “back EMF” from capacitor tries to slow charge build-up Inductor keeps pushing to resist that change. i

58 Copyright © 2012 Pearson Education Inc. The L-C circuit An L-C circuit contains an inductor and a capacitor and is an oscillating circuit.

59 Copyright © 2012 Pearson Education Inc. The L-C circuit An L-C circuit contains an inductor and a capacitor and is an oscillating circuit. Now capacitor charged on opposite side; Current reverses direction! System repeats in the opposite direction i

60 Copyright © 2012 Pearson Education Inc. The L-C circuit An L-C circuit contains an inductor and a capacitor and is an oscillating circuit.

61 Copyright © 2012 Pearson Education Inc. Electrical oscillations in an L-C circuit Analyze the current and charge as a function of time. Do a Kirchoff Loop around the circuit in the direction shown. Remember i can be +/- Recall C = q/V For this loop: -Ldi/dt – qC = 0

62 Copyright © 2012 Pearson Education Inc. Electrical oscillations in an L-C circuit -Ldi/dt – qC = 0 i(t) = dq/dt Ld 2 q/dt 2 + qC = 0 Simple Harmonic Motion! Pendulums Springs Standard solution! q(t)= Qmax cos(  t+  )  where  1/(LC) ½

63 Copyright © 2012 Pearson Education Inc. Electrical oscillations in an L-C circuit q(t)= Q max cos(  t+  ) i(t) = -  Q max sin(  t+  ) (based on this ASSUMED direction!!)  1/(LC) ½ = angular frequency

64 Copyright © 2012 Pearson Education Inc. The L-C circuit An L-C circuit contains an inductor and a capacitor and is an oscillating circuit.

65 Copyright © 2012 Pearson Education Inc. Electrical and mechanical oscillations Table 30.1 summarizes the analogies between SHM and L-C circuit oscillations.

66 Copyright © 2012 Pearson Education Inc. The L-R-C series circuit An L-R-C circuit exhibits damped harmonic motion if the resistance is not too large.


Download ppt "Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures."

Similar presentations


Ads by Google