1. Chapter 29 Electromagnetic Induction and Faraday’s Law
Chapter 29 opener. One of the great laws of physics is Faraday’s law of induction, which says that a changing magnetic flux produces an induced emf. This photo shows a bar magnet moving inside a coil of wire, and the galvanometer registers an induced current. This phenomenon of electromagnetic induction is the basis for many practical devices, including generators, alternators, transformers, tape recording, and computer memory.

2. 29-6 Transformers and Transmission of Power
Energy must be conserved; therefore, in the absence of losses, the ratio of the currents must be the inverse of the ratio of turns:

3. 29-6 Transformers and Transmission of Power
Example 29-12: Cell phone charger.
The charger for a cell phone contains a transformer that reduces 120-V ac to 5.0-V ac to charge the 3.7-V battery. (It also contains diodes to change the 5.0-V ac to 5.0-V dc.) Suppose the secondary coil contains 30 turns and the charger supplies 700 mA. Calculate (a) the number of turns in the primary coil, (b) the current in the primary, and (c) the power transformed.
Solution: a. NP = NS VP/VS = 720 turns.
b. IP = IS NS/NP = 29 mA.
c. P = ISVS = 3.5 W.

4. 29-6 Transformers and Transmission of Power
Transformers work only if the current is changing; this is one reason why electricity is transmitted as ac.
Figure The transmission of electric power from power plants to homes makes use of transformers at various stages.

5. 29-6 Transformers and Transmission of Power
Example: Transmission lines.
An average of 120 kW of electric power is sent to a small town from a power plant 10 km away. The transmission lines have a total resistance of 0.40 Ω. Calculate the power loss if the power is transmitted at (a) 120 V and (b) 24,000 V.
Solution: a. The total current is 120 kW/240 V = 500 A. Then the power loss is I2R = 100 kW.
b. Same reasoning, different numbers: the current is 5.0 A, and the power loss is 10 W. This is why electricity is transmitted at very high voltages.

6. ConcepTest 29.12a Transformers I
1) 30 V
2) 60 V
3) V
4) V
5) V
What is the voltage across the lightbulb?
120 V

7. ConcepTest 29.12a Transformers I
1) 30 V
2) 60 V
3) V
4) V
5) V
What is the voltage across the lightbulb?
The first transformer has a 2:1 ratio of turns, so the voltage doubles. But the second transformer has a 1:2 ratio, so the voltage is halved again. Therefore, the end result is the same as the original voltage.
120 V
240 V

8. 29-7 A Changing Magnetic Flux Produces an Electric Field
A changing magnetic flux induces an electric field; this is a generalization of Faraday’s law. The electric field will exist regardless of whether there are any conductors around: .

9. 29-7 A Changing Magnetic Flux Produces an Electric Field
Example 29-14: E produced by changing B; the betatron.
A magnetic field B between the pole faces of an electromagnet is nearly uniform at any instant over a circular area of radius r0. The current in the windings of the electromagnet is increasing in time so that B changes in time at a constant rate dB/dt at each point. Beyond the circular region (r > r0), we assume B = 0 at all times. Determine the electric field E at any point P a distance r from the center of the circular area due to the changing B.
Figure (a) Side view of nearly constant B. (b) Top view, for determining the electric field E at point P. (c) Lines of E produced by increasing B (pointing outward). (d) Graph of E vs. r.
Example 29–14.
Solution: Because of symmetry, E will be perpendicular to B and constant at radius r. Integrate around a circle of radius r as shown. For r < r0, the enclosed flux is Bπr2, and E = r/2 dB/dt. For r > r0, the enclosed flux is Bπr02, and E = r02/2r dB/dt.

10. 29-8 Applications of Induction: Sound Systems, Computer Memory, Seismograph, GFCI
A seismograph has a fixed coil and a magnet hung on a spring (or vice versa), and records the current induced when the Earth shakes.
Figure One type of seismograph, in which the coil is fixed to the case and moves with the Earth. The magnet, suspended by springs, has inertia and does not move instantaneously with the coil (and case), so there is relative motion between magnet and coil.

11. Summary of Chapter 29 Magnetic flux:
Changing magnetic flux induces emf:
Induced emf produces current that opposes original flux change.

12. Summary of Chapter 29
Changing magnetic field produces an electric field.
General form of Faraday’s law:
Electric generator changes mechanical energy to electrical energy; electric motor does the opposite. .

13. Summary of Chapter 29
Transformer changes magnitude of voltage in ac circuit; ratio of currents is inverse of ratio of voltages:
and

14. Chapter 30 Inductance, Electromagnetic Oscillations, and AC Circuits
Chapter 30 Opener. A spark plug in a car receives a high voltage, which produces a high enough electric field in the air across its gap to pull electrons off the atoms in the air–gasoline mixture and form a spark. The high voltage is produced, from the basic 12 V of the car battery, by an induction coil which is basically a transformer or mutual inductance. Any coil of wire has a self-inductance, and a changing current in it causes an emf to be induced. Such inductors are useful in many circuits.

15. Units of Chapter 30 Mutual Inductance Self-Inductance
Energy Stored in a Magnetic Field
LR Circuits
LC Circuits and Electromagnetic Oscillations
LC Circuits with Resistance (LRC Circuits)
AC Circuits with AC Source

16. Units of Chapter 30 LRC Series AC Circuit Resonance in AC Circuits
Impedance Matching
Three-Phase AC

17. 30-1 Mutual Inductance
Mutual inductance: a changing current in one coil will induce a current in a second coil:
And vice versa; note that the constant M, known as the mutual inductance, is the same:

18. 30-1 Mutual Inductance Unit of inductance: the henry, H:
1 H = 1 V·s/A = 1 Ω·s.
A transformer is an example of mutual inductance.
Figure A changing current in one coil will induce a current in the second coil.

19. 30-1 Mutual Inductance Example 30-1: Solenoid and coil.
A long thin solenoid of length l and cross-sectional area A contains N1 closely packed turns of wire. Wrapped around it is an insulated coil of N2 turns. Assume all the flux from coil 1 (the solenoid) passes through coil 2, and calculate the mutual inductance.
Solution: Assuming all the flux from the solenoid stays within the secondary coil, the flux is BA = μ0 (N1/l) I1A, and M = μ0 (N1N2/l) A.

20. 30-2 Self-Inductance
A changing current in a coil will also induce an emf in itself:
Here, L is called the self-inductance:

21. 30-2 Self-Inductance Example 30-3: Solenoid inductance.
(a) Determine a formula for the self-inductance L of a tightly wrapped and long solenoid containing N turns of wire in its length l and whose cross-sectional area is A.
(b) Calculate the value of L if N = 100, l = 5.0 cm, A = 0.30 cm2, and the solenoid is air filled.
Solution: a. B = μ0nI and is constant; the flux is μ0NIA/l and the self-inductance is μ0N2A/l.
b. Plugging in the numbers gives L = 7.5 μH.

22. 30-2 Self-Inductance
Conceptual Example 30-4: Direction of emf in inductor.
Current passes through a coil from left to right as shown. (a) If the current is increasing with time, in which direction is the induced emf? (b) If the current is decreasing in time, what then is the direction of the induced emf?
Figure Example 30–4. The + and - signs refer to the induced emf due to the changing current, as if points A and B were the terminals of a battery (and the coiled loops were the inside of the battery).
Solution: a. The induced emf opposes the change that caused it, so the induced emf acts to oppose the current.
b. Now the induced emf acts to reinforce the current (as it is decreasing).

23. 30-2 Self-Inductance Example 30-5: Coaxial cable inductance.
Determine the inductance per unit length of a coaxial cable whose inner conductor has a radius r1 and the outer conductor has a radius r2. Assume the conductors are thin hollow tubes so there is no magnetic field within the inner conductor, and the magnetic field inside both thin conductors can be ignored. The conductors carry equal currents I in opposite directions.
Solution: The flux through a rectangle of width dr and length l is dΦB = (μ0I/2πr) l dr. Integrating over r to find the total flux gives ΦB = (μ0Il/2π) ln (r2/r1). Therefore, L = (μ0l/2π) ln (r2/r1).

24. 30-3 Energy Stored in a Magnetic Field
Just as we saw that energy can be stored in an electric field, energy can be stored in a magnetic field as well, in an inductor, for example.
Analysis shows that the energy density of the field is given by

25. 30-4 LR Circuits
A circuit consisting of an inductor and a resistor will begin with most of the voltage drop across the inductor, as the current is changing rapidly. With time, the current will increase less and less, until all the voltage is across the resistor.
Figure (a) LR circuit; (b) growth of current when connected to battery.

26. 30-4 LR Circuits
The sum of potential differences around the loop gives
Integrating gives the current as a function of time: .
The time constant of an LR circuit is . .

27. 30-4 LR Circuits
If the circuit is then shorted across the battery, the current will gradually decay away: .
Figure The switch is flipped quickly so the battery is removed but we still have a circuit. The current at this moment (call it t = 0) is I0.
Figure Decay of the current in Fig. 30–7 in time after the battery is switched out of the circuit.

28. 30-4 LR Circuits Example 30-6: An LR circuit.
At t = 0, a 12.0-V battery is connected in series with a 220-mH inductor and a total of 30-Ω resistance, as shown. (a) What is the current at t = 0? (b) What is the time constant? (c) What is the maximum current? (d) How long will it take the current to reach half its maximum possible value? (e) At this instant, at what rate is energy being delivered by the battery, and (f) at what rate is energy being stored in the inductor’s magnetic field?
Solution: a. At t = 0 the current is zero.
b. The time constant is L/R = 7.3 ms.
c. The maximum current is V0/R = 0.40 A.
d. When I = ½ Imax, e-t/τ = ½ then t = 5.0 ms.
e. P = IV = 2.4 W.
f. U = ½ LI2, so dU/dt = LI dI/dt = I(V0-RI) = 1.2 W. The rest of the power is dissipated in the resistor.

29. 30-5 LC Circuits and Electromagnetic Oscillations
An LC circuit is a charged capacitor shorted through an inductor.
Figure An LC circuit.

30. 30-5 LC Circuits and Electromagnetic Oscillations
Summing the potential drops around the circuit gives a differential equation for Q:
This is the equation for simple harmonic motion, and has solutions . .

31. 30-5 LC Circuits and Electromagnetic Oscillations
Substituting shows that the equation can only be true for all times if the frequency is given by
The current is sinusoidal as well:

32. 30-5 LC Circuits and Electromagnetic Oscillations
The charge and current are both sinusoidal, but with different phases.
Figure Charge Q and current I in an LC circuit. The period T = 1/f = 2π/ω = 2π(L/C)1/2.

33. 30-5 LC Circuits and Electromagnetic Oscillations
The total energy in the circuit is constant; it oscillates between the capacitor and the inductor:
Figure Energy UE (red line) and UB (blue line) stored in the capacitor and the inductor as a function of time. Note how the energy oscillates between electric and magnetic. The dashed line at the top is the (constant) total energy U = UE + UB.

34. 30-5 LC Circuits and Electromagnetic Oscillations
Example 30-7: LC circuit.
A 1200-pF capacitor is fully charged by a 500-V dc power supply. It is disconnected from the power supply and is connected, at t = 0, to a 75-mH inductor. Determine: (a) the initial charge on the capacitor; (b) the maximum current; (c) the frequency f and period T of oscillation; and (d) the total energy oscillating in the system.
Solution: a. Q0 = CV = 6.0 x 10-7 C.
b. Imax = ωQ0 = 63 mA.
c. f = ω/2π = 17 kHz; T = 1/f = 6.0 x 10-5 s.
d. U = Q02/2C = 1.5 x 10-4 J.

35. 30-6 LC Oscillations with Resistance (LRC Circuit)
Any real (nonsuperconducting) circuit will have resistance.
Figure An LRC circuit.

36. 30-6 LC Oscillations with Resistance (LRC Circuit)
Now the voltage drops around the circuit give
The solutions to this equation are damped harmonic oscillations. The system will be underdamped for R2 < 4L/C, and overdamped for R2 > 4L/C. Critical damping will occur when R2 = 4L/C.

37. 30-6 LC Oscillations with Resistance (LRC Circuit)
This figure shows the three cases of underdamping, overdamping, and critical damping.
Figure Charge Q on the capacitor in an LRC circuit as a function of time: curve A is for underdamped oscillation (R2 < 4L/C), curve B is for critically damped (R2 = 4L/C), and curve C is for overdamped (R2 > 4L/C).

38. 30-6 LC Oscillations with Resistance (LRC Circuit)
The angular frequency for underdamped oscillations is given by .
The charge in the circuit as a function of time is .

39. 30-6 LC Oscillations with Resistance (LRC Circuit)
Example 30-8: Damped oscillations.
At t = 0, a 40-mH inductor is placed in series with a resistance R = 3.0 Ω and a charged capacitor C = 4.8 μF. (a) Show that this circuit will oscillate. (b) Determine the frequency. (c) What is the time required for the charge amplitude to drop to half its starting value? (d) What value of R will make the circuit nonoscillating?
Solution: a. R2 is less than 4L/C, so this circuit will oscillate.
b. f = ω/2π = 360 Hz.
c. t = 2L/R ln 2 = 18 ms.
d. For R2 ≥ 4L/C, R must be ≥ 180 Ω.

40. 30-7 AC Circuits with AC Source
Resistors, capacitors, and inductors have different phase relationships between current and voltage when placed in an ac circuit.
The current through a resistor is in phase with the voltage.
Figure (a) Resistor connected to an ac source. (b) Current (blue curve) is in phase with the voltage (red) across a resistor.

41. 30-7 AC Circuits with AC Source
The voltage across the inductor is given by or .
Figure (a) Inductor connected to an ac source. (b) Current (blue curve) lags voltage (red curve) by a quarter cycle or 90°.
Therefore, the current through an inductor lags the voltage by 90°.

42. 30-7 AC Circuits with AC Source
The voltage across the inductor is related to the current through it: .
The quantity XL is called the inductive reactance, and has units of ohms:

43. 30-7 AC Circuits with AC Source
Example 30-9: Reactance of a coil.
A coil has a resistance R = 1.00 Ω and an inductance of H. Determine the current in the coil if (a) 120-V dc is applied to it, and (b) 120-V ac (rms) at 60.0 Hz is applied.
Solution: a. For dc current, there is no magnetic induction, so the current is determined by the resistance: I = V/R = 120 A.
b. The inductive reactance is XL = 2πfL = 113 Ω. The current is Irms = Vrms/XL = 1.06 A.

44. 30-7 AC Circuits with AC Source
The voltage across the capacitor is given by .
Therefore, in a capacitor, the current leads the voltage by 90°.
Figure (a) Capacitor connected to an ac source. (b) Current leads voltage by a quarter cycle, or 90°.

45. 30-7 AC Circuits with AC Source
The voltage across the capacitor is related to the current through it: .
The quantity XC is called the capacitive reactance, and (just like the inductive reactance) has units of ohms:

46. 30-7 AC Circuits with AC Source
Example 30-10: Capacitor reactance.
What is the rms current in the circuit shown if C = 1.0 μF and Vrms = 120 V? Calculate (a) for f = 60 Hz and then (b) for f = 6.0 x 105 Hz.
Solution: a. XC = 1/2πfC = 2.7 kΩ; I = V/XC = 44 mA.
b. Now XC = 0.27 Ω and I = 440 A.

47. 30-7 AC Circuits with AC Source
This figure shows a high-pass filter (allows an ac signal to pass but blocks a dc voltage) and a low-pass filter (allows a dc voltage to be maintained but blocks higher-frequency fluctuations).

48. 30-8 LRC Series AC Circuit
Analyzing the LRC series AC circuit is complicated, as the voltages are not in phase – this means we cannot simply add them. Furthermore, the reactances depend on the frequency.
Figure An LRC circuit.

49. 30-8 LRC Series AC Circuit
We calculate the voltage (and current) using what are called phasors – these are vectors representing the individual voltages.
Here, at t = 0, the current and voltage are both at a maximum. As time goes on, the phasors will rotate counterclockwise.
Figure 30-20a. Phasor diagram for a series LRC circuit at t=0.

50. 30-8 LRC Series AC Circuit
Some time t later, the phasors have rotated.
Figure 30-20b. Phasor diagram for a series LRC circuit at a time t later.

51. 30-8 LRC Series AC Circuit
The voltages across each device are given by the x-component of each, and the current by its x-component. The current is the same throughout the circuit.
Figure 30-20c. Phasor diagram for a series LRC circuit. Projections on x axis reflect Eqs. 30–20, 30–22a and 30–24a.

52. 30-8 LRC Series AC Circuit
We find from the ratio of voltage to current that the effective resistance, called the impedance, of the circuit is given by

53. 30-8 LRC Series AC Circuit
The phase angle between the voltage and the current is given by or
The factor cos φ is called the power factor of the circuit.

54. 30-8 LRC Series AC Circuit Example 30-11: LRC circuit.
Suppose R = 25.0 Ω, L = 30.0 mH, and C = 12.0 μF, and they are connected in series to a 90.0-V ac (rms) 500-Hz source. Calculate (a) the current in the circuit, (b) the voltmeter readings (rms) across each element, (c) the phase angle , and (d) the power dissipated in the circuit.
Solution: a. XL = 2πfL = 94.2 Ω; XC = 1/(2πfC) = 26.5 Ω; so Z = 72.2 Ω and I = 1.25 A.
b. The voltages are the currents multiplied by the reactances (or resistance). VL = 118 V; VC = 33.1 V; VR = 31.2 V.
c. cos φ = 0.346, so φ = 69.7°.
d. P = IV cos φ = 39.0 W.

55. 30-9 Resonance in AC Circuits
The rms current in an ac circuit is
Clearly, Irms depends on the frequency.

56. 30-9 Resonance in AC Circuits
We see that Irms will be a maximum when XC = XL; the frequency at which this occurs is
f0 = ω0/2π is called the resonant frequency.
Figure Current in LRC circuit as a function of angular frequency, showing resonance peak at ω = ω0 = (1/LC)1/2.

57. 30-10 Impedance Matching
When one electrical circuit is connected to another, maximum power is transmitted when the output impedance of the first equals the input impedance of the second.
The power delivered to the circuit will be a minimum when dP/dt = 0; this occurs when R1 = R2.
Figure Output of the circuit on the left is input to the circuit on the right.

58. 30-11 Three-Phase AC
Transmission lines usually transmit three-phase ac power, with the phases being separated by 120°. This makes the power flow much smoother than if a single phase were used.
Figure The three voltages, out of phase by 120° (= 2/3 π radians) in a three-phase power line.

59. 30-11 Three-Phase AC Example 30-12: Three-phase circuit.
In a three-phase circuit, 266 V rms exists between line 1 and ground. What is the rms voltage between lines 2 and 3?
Solution: V0 = 376 V. To find V3 – V2, use trigonometric identities. Result: (V3 – V2)rms = √2 V0 sin π/3 = 460 V.

60. Summary of Chapter 30 Mutual inductance: Self-inductance:
Energy density stored in magnetic field:

61. Summary of Chapter 30 LR circuit: Inductive reactance:. .
Inductive reactance:
Capacitive reactance:

62. Summary of Chapter 30 LRC series circuit:.
Resonance in LRC series circuit: