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
Figure Step-up transformer (NP = 4, NS = 12).

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 29-13: 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) 240 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 29.12b Transformers II
Given that the intermediate current is 1 A, what is the current through the lightbulb?
1 A
120 V
240 V

7. ConcepTest 29.12b Transformers II
Given that the intermediate current is 1 A, what is the current through the lightbulb?
Power in = Power out
240 V  1 A = V  ???
The unknown current is 2 A.
1 A
120 V
240 V

8. ConcepTest 29.12c Transformers III
A 6 V battery is connected to one side of a transformer. Compared to the voltage drop across coil A, the voltage across coil B is:
1) greater than 6 V
2) 6 V
3) less than 6 V
4) zero A B
6 V

9. ConcepTest 29.12c Transformers III
A 6 V battery is connected to one side of a transformer. Compared to the voltage drop across coil A, the voltage across coil B is:
1) greater than 6 V
2) 6 V
3) less than 6 V
4) zero
The voltage across B is zero. Only a changing magnetic flux induces an emf. Batteries can provide only dc current. A B
6 V

10. 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:
But! But! Isn’t the integral of E around a closed path ZERO? Kirchoff’s Loop Rule?

11. 29-7 A Changing Magnetic Flux Produces an Electric Field
Basic requirement of Conservative Potential:

12. 29-7 A Changing Magnetic Flux Produces an Electric Field
Example 29-14: E produced by changing B.
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.

13. 29-8 Applications of Induction: Sound Systems, Computer Memory, Seismograph, GFCI
This microphone works by induction; the vibrating membrane induces an emf in the coil.
Figure Diagram of a microphone that works by induction.

14. 29-8 Applications of Induction: Sound Systems, Computer Memory, Seismograph, GFCI
Differently magnetized areas on an audio tape or disk induce signals in the read/write heads.
Figure (a) Read/Write (playback/recording) head for tape or disk. In writing or recording, the electric input signal to the head, which acts as an electromagnet, magnetizes the passing tape or disk. In reading or playback, the changing magnetic field of the passing tape or disk induces a changing magnetic field in the head, which in turn induces in the coil an emf that is the output signal. (b) Photo of a hard drive showing several platters and read/write heads that can quickly move from the edge of the disk to the center.

15. 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.

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

17. 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. .

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

19. 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.

20. 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

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

22. 30-1 Mutual Inductance

23. 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.

24. 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.

25. 30-2 Self-Inductance
A changing current in a coil will also induce an emf in itself:
Here, L is called the self-inductance:

26. 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.

27. 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).

28. 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).

29. 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:

30. 30-3 Energy Stored in a Magnetic Field
Example: a solenoid

31. Questions?

32. Have a great Spring(?) Break!