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.

29-3 EMF Induced in a Moving Conductor Example 29-7: Electromagnetic blood-flow measurement. The rate of blood flow in our body’s vessels can be measured using the apparatus shown, since blood contains charged ions. Suppose that the blood vessel is 2.0 mm in diameter, the magnetic field is 0.080 T, and the measured emf is 0.10 mV. What is the flow velocity of the blood? Remind you of the Hall effect? Solution: v = V/Bl = 0.63 m/s.

29-3 EMF Induced in a Moving Conductor Example 29-8: Force on the rod. To make the rod move to the right at speed v, you need to apply an external force on the rod to the right. (a) Explain and determine the magnitude of the required force. (b) What external power is needed to move the rod? Solution: a. The external force needs to be equal and opposite to the magnetic force (IlB) if the rod is to move at a constant speed. I = Blv/R, so F = B2l2v/R. b. The external power is Fv = B2l2v2/R, which is equal to the power dissipated in the resistance of the rod (I2R).

29-4 Electric Generators A generator is the opposite of a motor – it transforms mechanical energy into electrical energy. This is an ac generator: The axle is rotated by an external force such as falling water or steam. The brushes are in constant electrical contact with the slip rings. Figure 29-15. An ac generator.

ConcepTest 29.5 Rotating Wire Loop If a coil is rotated as shown, in a magnetic field pointing to the left, in what direction is the induced current? 1) clockwise 2) counterclockwise 3) no induced current

ConcepTest 29.5 Rotating Wire Loop If a coil is rotated as shown, in a magnetic field pointing to the left, in what direction is the induced current? 1) clockwise 2) counterclockwise 3) no induced current As the coil is rotated into the B field, the magnetic flux through it increases. According to Lenz’s law, the induced B field has to oppose this increase, thus the new B field points to the right. An induced counterclockwise current produces just such a B field.

29-4 Electric Generators If the loop is rotating with constant angular velocity ω, the induced emf is sinusoidal: For a coil of N loops, Figure 29-16. An ac generator produces an alternating current. The output emf E = E0 sin ωt, where E0 = NABω.

29-4 Electric Generators Example 29-9: An ac generator. The armature of a 60-Hz ac generator rotates in a 0.15-T magnetic field. If the area of the coil is 2.0 x 10-2 m2, how many loops must the coil contain if the peak output is to be V0 = 170 V? Solution: N = E0/BAω = 150 turns. Remember to convert 60 Hz to angular units.

ConcepTest 29.10 Generators A generator has a coil of wire rotating in a magnetic field. If the rotation rate increases, how is the maximum output voltage of the generator affected? 1) increases 2) decreases 3) stays the same 4) varies sinusoidally

ConcepTest 29.10 Generators A generator has a coil of wire rotating in a magnetic field. If the rotation rate increases, how is the maximum output voltage of the generator affected? 1) increases 2) decreases 3) stays the same 4) varies sinusoidally The maximum voltage is the leading term that multiplies sin wt and is given by e0 = NBAw. Therefore, if w increases, then e0 must increase as well.

29-5 Back EMF and Counter Torque; Eddy Currents

29-5 Back EMF and Counter Torque; Eddy Currents Induced currents can flow in bulk material as well as through wires. These are called eddy currents, and can dramatically slow a conductor moving into or out of a magnetic field. Figure 29-21. Production of eddy currents in a rotating wheel. The grey lines in (b) indicate induced current.

29-6 Transformers and Transmission of Power A transformer consists of two coils, either interwoven or linked by an iron core. A changing emf in one induces an emf in the other. The ratio of the emfs is equal to the ratio of the number of turns in each coil:

29-6 Transformers and Transmission of Power This is a step-up transformer – the emf in the secondary coil is larger than the emf in the primary: Figure 29-24. Step-up transformer (NP = 4, NS = 12).

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:

ConcepTest 29.12b Transformers Given that the intermediate current is 1 A, what is the current through the lightbulb? 1 A 120 V 240 V

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

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.

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 29-26. The transmission of electric power from power plants to homes makes use of transformers at various stages.

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.

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: .

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 29-27. (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.

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

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. .

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

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.

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

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

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:

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 30-1. A changing current in one coil will induce a current in the second coil.

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.

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

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.

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 30-4. 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).

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).

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.