Copyright © 2009 Pearson Education, Inc. Chapter 29 Electromagnetic Induction and Faraday’s Law

Copyright © 2009 Pearson Education, Inc. Energy must be conserved; therefore, in the absence of losses, the ratio of the currents must be the inverse of the ratio of turns: 29-6 Transformers and Transmission of Power

Copyright © 2009 Pearson Education, Inc 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.

Copyright © 2009 Pearson Education, Inc. Transformers work only if the current is changing; this is one reason why electricity is transmitted as ac Transformers and Transmission of Power

Copyright © 2009 Pearson Education, Inc 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.

120 V What is the voltage across the lightbulb? 1) 30 V 2) 60 V 3) 120 V 4) 240 V 5) 480 V ConcepTest 29.12a Transformers I

first transformer2:1 ratio voltage doubles second transformer 1:2 ratiovoltage is halved same as the original voltage 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 120 V What is the voltage across the lightbulb? 1) 30 V 2) 60 V 3) 120 V 4) 240 V 5) 480 V ConcepTest 29.12a Transformers I

Copyright © 2009 Pearson Education, Inc. Magnetic flux: Changing magnetic flux induces emf: Induced emf produces current that opposes original flux change. Summary of Chapter 29

Copyright © 2009 Pearson Education, Inc. 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.

Copyright © 2009 Pearson Education, Inc. Summary of Chapter 29 Transformer changes magnitude of voltage in ac circuit; ratio of currents is inverse of ratio of voltages: and

Copyright © 2009 Pearson Education, Inc. Chapter 30 Inductance, Electromagnetic Oscillations, and AC Circuits

Copyright © 2009 Pearson Education, Inc. 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

Copyright © 2009 Pearson Education, Inc. LRC Series AC Circuit Resonance in AC Circuits Impedance Matching Three-Phase AC Units of Chapter 30

Copyright © 2009 Pearson Education, Inc. 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

Copyright © 2009 Pearson Education, Inc. Unit of inductance: the henry, H: 1 H = 1 V·s/A = 1 Ω·s. A transformer is an example of mutual inductance Mutual Inductance

Copyright © 2009 Pearson Education, Inc Mutual Inductance Example 30-1: Solenoid and coil. A long thin solenoid of length l and cross-sectional area A contains N 1 closely packed turns of wire. Wrapped around it is an insulated coil of N 2 turns. Assume all the flux from coil 1 (the solenoid) passes through coil 2, and calculate the mutual inductance.

Copyright © 2009 Pearson Education, Inc. A changing current in a coil will also induce an emf in itself: Here, L is called the self-inductance: 30-2 Self-Inductance

Copyright © 2009 Pearson Education, Inc 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 cm 2, and the solenoid is air filled.

Copyright © 2009 Pearson Education, Inc 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?

Copyright © 2009 Pearson Education, Inc Self-Inductance Example 30-5: Coaxial cable inductance. Determine the inductance per unit length of a coaxial cable whose inner conductor has a radius r 1 and the outer conductor has a radius r 2. 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.

Copyright © 2009 Pearson Education, Inc. 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 30-3 Energy Stored in a Magnetic Field

Copyright © 2009 Pearson Education, Inc. 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 LR Circuits

Copyright © 2009 Pearson Education, Inc 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...

Copyright © 2009 Pearson Education, Inc. If the circuit is then shorted across the battery, the current will gradually decay away: 30-4 LR Circuits.

Copyright © 2009 Pearson Education, Inc 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?

Copyright © 2009 Pearson Education, Inc LC Circuits and Electromagnetic Oscillations An LC circuit is a charged capacitor shorted through an inductor.

Copyright © 2009 Pearson Education, Inc 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..

Copyright © 2009 Pearson Education, Inc 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:

Copyright © 2009 Pearson Education, Inc LC Circuits and Electromagnetic Oscillations The charge and current are both sinusoidal, but with different phases.

Copyright © 2009 Pearson Education, Inc LC Circuits and Electromagnetic Oscillations The total energy in the circuit is constant; it oscillates between the capacitor and the inductor:

Copyright © 2009 Pearson Education, Inc 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.

Copyright © 2009 Pearson Education, Inc LC Oscillations with Resistance ( LRC Circuit) Any real (nonsuperconducting) circuit will have resistance.

Copyright © 2009 Pearson Education, Inc 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 R 2 4L/C. Critical damping will occur when R 2 = 4L/C.

Copyright © 2009 Pearson Education, Inc LC Oscillations with Resistance ( LRC Circuit) This figure shows the three cases of underdamping, overdamping, and critical damping.

Copyright © 2009 Pearson Education, Inc 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..

Copyright © 2009 Pearson Education, Inc 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?

Copyright © 2009 Pearson Education, Inc. 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 AC Circuits with AC Source

Copyright © 2009 Pearson Education, Inc. Therefore, the current through an inductor lags the voltage by 90° AC Circuits with AC Source The voltage across the inductor is given by or.

Copyright © 2009 Pearson Education, Inc AC Circuits with AC Source The voltage across the inductor is related to the current through it: The quantity X L is called the inductive reactance, and has units of ohms:.

Copyright © 2009 Pearson Education, Inc 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.

Copyright © 2009 Pearson Education, Inc. Therefore, in a capacitor, the current leads the voltage by 90° AC Circuits with AC Source The voltage across the capacitor is given by.

Copyright © 2009 Pearson Education, Inc AC Circuits with AC Source The voltage across the capacitor is related to the current through it: The quantity X C is called the capacitive reactance, and (just like the inductive reactance) has units of ohms:.

Copyright © 2009 Pearson Education, Inc. 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 LRC Series AC Circuit

Copyright © 2009 Pearson Education, Inc. 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 LRC Series AC Circuit

Copyright © 2009 Pearson Education, Inc. Some time t later, the phasors have rotated LRC Series AC Circuit

Copyright © 2009 Pearson Education, Inc. 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 LRC Series AC Circuit

Copyright © 2009 Pearson Education, Inc. We find from the ratio of voltage to current that the effective resistance, called the impedance, of the circuit is given by 30-8 LRC Series AC Circuit

Copyright © 2009 Pearson Education, Inc. Homework # 11 Chapter 29 – 54 Chapter , 12, 20, 34, 43