Intermediate Physics PHYS102

Transcription

1 Intermediate Physics PHYS102

2 Dr Richard H. Cyburt Assistant Professor of Physics My office: 402c in the Science Building My phone: (304) My My webpage: In person or is the best way to get a hold of me. PHYS102

3 My Office Hours TWR 9:30-11:00am W 4:00-5:00pm Meetings may also be arranged at other times, by appointment PHYS102

4 Problem Solving Sections I would like to have hour-long sections for working through problems. This would be an extra component to the course and count towards extra credit TR 1-2 pm WF am S308 If you can t make these, you can still pick up the problem worksheet. PHYS102

5 Midterm 2 Thursday, March 2 during class 8-9:15 Allowed one half sheet (8.5x11) piece of paper w/ notes/formuli Calculator pencil or blue/black pen Review, Wednesday, March 1 7-9pm Bring questions!!! PHYS102

6 Intermediate Physics PHYS102

7 Douglas Adams Hitchhiker s Guide to the Galaxy PHYS102

8 In class!! PHYS102

9 This lecture will help you understand: Faraday s Law Alternating Current AC Electricity and Transformers PHYS102

10 Lenz s Law Text: p. 813

11 Lenz s Law Text: p. 813

12 Lenz s Law Text: p. 813

13 QuickCheck 25.8 The bar magnet is pushed toward the center of a wire loop. Which is true? There is a clockwise induced current in the loop. There is a counterclockwise induced current in the loop. There is no induced current in the loop.

14 QuickCheck 25.8 The bar magnet is pushed toward the center of a wire loop. Which is true? There is a clockwise induced current in the loop. There is a counterclockwise induced current in the loop. There is no induced current in the loop. 1. Upward flux from magnet is increasing. 2. To oppose the increase, the field of the induced current points down. 3. From the right-hand rule, a downward field needs a cw current.

15 QuickCheck 25.9 The bar magnet is pushed toward the center of a wire loop. Which is true? There is a clockwise induced current in the loop. There is a counterclockwise induced current in the loop. There is no induced current in the loop.

16 QuickCheck 25.9 The bar magnet is pushed toward the center of a wire loop. Which is true? There is a clockwise induced current in the loop. There is a counterclockwise induced current in the loop. There is no induced current in the loop. Magnetic flux is zero, so there s no change of flux.

17 QuickCheck A bar magnet sits inside a coil of wire that is connected to a meter. For each of the following circumstances 1.The bar magnet is at rest in the coil, 2.The bar magnet is pulled out of the coil, 3.The bar magnet is completely out of the coil and at rest, 4.The bar magnet is reinserted into the coil, What can we say about the current in the meter? A. The current goes from right to left. B. The current goes from left to right. C. There is no current in the meter.

18 QuickCheck A bar magnet sits inside a coil of wire that is connected to a meter. For each of the following circumstances 1.The bar magnet is at rest in the coil, C 2.The bar magnet is pulled out of the coil, A 3.The bar magnet is completely out of the coil and at rest, C 4.The bar magnet is reinserted into the coil, B What can we say about the current in the meter? A. The current goes from right to left. B. The current goes from left to right. C. There is no current in the meter.

19 QuickCheck A magnetic field goes through a loop of wire, as at right. If the magnitude of the magnetic field is 1.Increasing, 2.Decreasing, 3.Constant, What can we say about the current in the loop? Answer for each of the stated conditions. A. The loop has a clockwise current. B. The loop has a counterclockwise current. C. The loop has no current.

20 QuickCheck A magnetic field goes through a loop of wire, as at right. If the magnitude of the magnetic field is 1.Increasing, B 2.Decreasing, A 3.Constant, C What can we say about the current in the loop? Answer for each of the stated conditions. A. The loop has a clockwise current. B. The loop has a counterclockwise current. C. The loop has no current.

21 QuickCheck The magnetic field is confined to the region inside the dashed lines; it is zero outside. The metal loop is being pulled out of the magnetic field. Which is true? A. There is a clockwise induced current in the loop. B. There is a counterclockwise induced current in the loop. C. There is no induced current in the loop.

22 QuickCheck The magnetic field is confined to the region inside the dashed lines; it is zero outside. The metal loop is being pulled out of the magnetic field. Which is true? A. There is a clockwise induced current in the loop. B. There is a counterclockwise induced current in the loop. C. There is no induced current in the loop. 1. The flux through the loop is into the screen and decreasing. 2. To oppose the decrease, the field of the induced current must point into the screen. 3. From the right-hand rule, an inward field needs a cw current.

23 Section 25.4 Faraday s Law

24 Faraday s Law An induced emf ℇ is the emf associated with a changing magnetic flux. The direction of the current is determined by Lenz s law. The size of the induced emf is determined by Faraday s law.

25 Faraday s Law Faraday s law is a basic law of electromagnetic induction. It says that the magnitude of the induced emf is the rate of change of the magnetic flux through the loop:

26 Faraday s Law A coil wire consisting of N turns acts like N batteries in series, so the induced emf in the coil is

27 Faraday s Laws There are two fundamentally different ways to change the magnetic flux through a conducting loop: 1. The loop can move or expand or rotate, creating a motional emf. 2. The magnetic field can change. The induced emf is the rate of change of the magnetic flux through the loop, regardless of what causes the flux to change.

28 Faraday s Laws Text: p. 815

29 Faraday s Laws Text: p. 815

30 Example Problem The following figure shows a 10-cm-diameter loop in three different magnetic fields. The loop s resistance is 0.1 Ohms. For each situation, determine the magnitude and direction of the induced current.

31 Eddy Currents There are two loops lying entirely in a metal sheet between two magnets. As the sheet is pulled, the loop on the right is leaving the magnetic field, and the flux is decreasing. According to Faraday s law, the flux change induces a current to flow around the loop. Lenz s law says the current flows clockwise.

32 Eddy Currents The loop on the left side of the metal enters the field and so the flux through it is increasing. Lenz s law requires the induced whirlpool current on the left loop to be counterclockwise.

33 Eddy Currents Eddy currents are the spreadout whirlpools of an induced current in a solid conductor. Both whirlpools of current are moving in the same direction as they pass through the magnet. The magnetic field exerts a force on the current, opposite the direction of pull, acting as a braking force.

34 Eddy Currents Because of the braking force exerted by the magnetic field, an external force is required to pull a metal through a magnetic field. If the pulling force ceases, the magnetic braking force quickly causes the metal to decelerate until it stops.

35 Section 26.1 Alternating Current

36 Alternating Current A battery produces a constant emf so a flashlight has a constant glow. The electricity distributed to homes is different. Rather than a constant emf, it has a sinusoidal variation. A lightbulb in your home is on when the emf is positive, but not when it is negative. The resulting flicker is too rapid to notice under normal circumstances.

37 Alternating Current In Chapter 25 we saw that an electrical generator works by rotating a coil of wire in a magnetic field. The steady rotation of the coil causes the emf and the induced current in the coil to oscillate sinusoidally, alternating positive and then negative. The oscillation forces charges to flow first in one direction and then the other a half cycle later. This is an alternating current: AC. If the emf is constant and the current is always in the same direction, the electricity is called direct current: DC.

38 Alternating Current The instantaneous emf of an AC voltage source can be written as ℇ 0 is the peak or maximum emf, T is the period of oscillation, and f = 1/T is the frequency.

39 Resistor Circuits In circuits where the current and voltage are oscillating, we use i to represent the instantaneous current and v for the instantaneous voltage. The potential difference across a resistor R is called the resistor voltage v R. It is related to the current i R through the resistor by Ohm s law: v R = i R R

40 Resistor Circuits

41 Resistor Circuits We can analyze a circuit with a resistor connected across an AC emf the same way we did with a DC resistor circuit. Kirchhoff s loop law says that the sum of all the potential differences around a closed path is zero:

42 Resistor Circuits The resistor voltage is a sinusoidal voltage at frequency f: v R = V R cos(2pft) V R is the peak or maximum voltage, the amplitude of the sinusoidally varying voltage. In the single resistor case, V R = ℇ 0, so the current through the resistor is I R = V R /R is the peak current.

43 Resistor Circuits The resistor s instantaneous current and voltage oscillate in phase. The current is at its maximum and minimum values when the voltage is at its maximum and minimum values.

44 AC Power in Resistors The instantaneous power is written as p = i R2 R = [I R cos(2pft)] 2 R = I R2 R[cos(2pft)] 2

45 AC Power in Resistors The power oscillates twice during every cycle of the emf.

46 AC Power in Resistors The current in a lightbulb reverses direction 120 times per second, so the power reaches a maximum twice per second. However the bulb glows steadily, so the average power is more useful. The average power P R is

47 AC Power in Resistors Recall that the power in a DC circuit is P R = I 2 R = V 2 /R. The root-mean-square current and the root-mean-square voltage are more useful expressions of power: Using the root-mean-square values, we write the average power dissipated in an AC circuit as: As long as you work with rms voltages and currents, all the expressions for DC power carry over to AC power.

48 Example 26.1 The resistance and current of a toaster The hot wire in a toaster dissipates 580 W when plugged into a 120 V outlet. a. What is the wire s resistance? b. What are the rms and peak currents through the wire? PREPARE We ve seen that the 120 V outlet voltage is an rms value. The filament has resistance R. It dissipates 580 W when there s an rms voltage of 120 V across it. We can solve Equation 26.9 for R and then use Equations 26.9 and 26.8 to find the rms current and the peak current.

49 Example 26.1 The resistance and current of a toaster (cont.) SOLVE a. We rearrange Equation 26.9 to find the resistance from the rms voltage and the average power:

50 Example 26.1 The resistance and current of a toaster (cont.) b. A second rearrangement of Equation 26.9 allows us to find the current in terms of the power and the resistance, both of which are known: From Equation 26.8, the peak current is

51 Example 26.1 The resistance and current of a toaster (cont.) ASSESS We can do a quick check on our work by calculating the power for the rated voltage and computed current: P R = I rms V rms = (4.8 A)(120 V) = 580 W This agrees with the value given in the problem statement, giving us confidence in our solution.

52 Example Problem A 120 V (rms) AC power supply is connected to a motor, which is rated at 100 W. A. What is the rms current in the circuit? B. Now, suppose that the wires used to connect the motor to the power supply have a resistance of 7.0 Ω. Assume that the rms current stays the same. What is the voltage drop across the resistance of the wires? What is the voltage at the motor now? C. Now, suppose the power supply is 1200 V, and the motor is rated at 100 W at this higher voltage. What is the current in the circuit assuming no resistance in the wires? D. If the wires have a resistance of 7.0 Ω, what is the voltage drop across the wires? The voltage at the motor?

53 Section 26.2 AC Electricity and Transformers

54 AC Electricity and Transformers A transformer is a device that takes an alternating voltage as an input and produces either a higher or lower voltage as output. The operation of a transformer is based on the emf produced by changing magnetic fields, so the input must be AC electricity.

55 Transformer Operation A simplified version of a transformer consists of two coils wrapped on a single iron core.

56 Transformer Operation In a simple transformer, the primary coil is connected to an AC voltage. The AC voltage creates an alternating current. The current in the coil creates a magnetic field that magnetizes the iron core to produce a much stronger net field. The strong flux follows the iron core and enters the secondary coil.

57 Transformer Operation The current in the primary coil of a transformer is an alternating current; it creates an oscillating magnetic field in the iron core. The changing magnetic field means there is a changing magnetic flux through the secondary coil, which induces an emf, an AC voltage V 2 in the coil. A resistor, or load, is connected to dissipate power.

58 Transformer Operation

59 Transformer Operation The purpose of a transformer is to change the voltage, so we need to understand how the voltage in the primary coil of a transformer relates to that in the secondary coil. According to Faraday s law, an instantaneous voltage across N 1 turns due to the magnetic flux F in the primary coil is

60 Transformer Operation In an ideal transformer, all the flux is guided by the iron core through the secondary coil. The changing flux induces an emf across the secondary coil given by Because ΔF/Δt is the same in each coil, we can write a ratio of instantaneous voltages:

61 Transformer Operation We can also relate the peak and rms voltages of the primary and secondary coils: A step-up transformer, with N 2 > N 1, increases the voltage, while a step-down transformer, with N 2 < N 1, lowers the voltage.

62 Transformer Operation We can relate the peak currents between the coils in a transformer with an expression that mirrors the expression of peak voltages between the coils: A step-up transformer raises voltage but lowers current; a stepdown transformer lowers voltage but raises current.

63 QuickCheck 26.1 If the primary coil of wire on a transformer is kept the same and the number of turns of wire on the secondary is increased, how will this affect the voltage observed at the secondary? A. The voltage will increase. B. The voltage will stay the same. C. The voltage will decrease.

64 QuickCheck 26.1 If the primary coil of wire on a transformer is kept the same and the number of turns of wire on the secondary is increased, how will this affect the voltage observed at the secondary? A. The voltage will increase. B. The voltage will stay the same. C. The voltage will decrease.

65 QuickCheck 26.2 Suppose that an ideal transformer has 400 turns in its primary coil and 100 turns in its secondary coil. The primary coil is connected to a 120 V (rms) electric outlet and carries an rms current of 10 ma. What are the rms values of the voltage and current for the secondary? A. 480 V, 40 ma B. 480 V, 2.5 ma C. 30 V, 40 ma D. 30 V, 2.5 ma

66 QuickCheck 26.2 Suppose that an ideal transformer has 400 turns in its primary coil and 100 turns in its secondary coil. The primary coil is connected to a 120 V (rms) electric outlet and carries an rms current of 10 ma. What are the rms values of the voltage and current for the secondary? A. 480 V, 40 ma B. 480 V, 2.5 ma C. 30 V, 40 ma D. 30 V, 2.5 ma

67 Example 26.2 Analyzing a step-down transformer A book light has a 1.4 W, 4.8 V bulb that is powered by a transformer connected to a 120 V electric outlet. The secondary coil of the transformer has 20 turns of wire. How many turns does the primary coil have? What is the current in the primary coil?

68 Example 26.2 Analyzing a step-down transformer (cont.) We know the voltages of the primary and the secondary, so we can compute the turns in the primary coil using Equation We know the voltage and the power of the bulb, so we can find the current in the bulb. This is the current in the secondary, which we can use in Equation to find the current in the primary, the current provided by the outlet.

69 Example 26.2 Analyzing a step-down transformer (cont.) SOLVE The bulb is rated at 4.8 V; this is the rms voltage at the secondary, so (V 2 ) rms = 4.8 V. The power outlet has the usual (V 1 ) rms = 120 V, so we can rearrange Equation to find

70 Example 26.2 Analyzing a step-down transformer (cont.) The bulb connected to the secondary dissipates 1.4 W at 4.8 V; this is an rms voltage, so the rms current in the secondary is

71 Example 26.2 Analyzing a step-down transformer (cont.) We can then rearrange Equation to find the rms current in the primary:

72 Example 26.2 Analyzing a step-down transformer (cont.) ASSESS We can check our results by looking at the power supplied by the wall outlet. This is P 1 = (120 V)(0.012 A) = 1.4 W, the same as the power dissipated by the bulb, as must be the case because we ve assumed the transformer is ideal.

73 Power Transmission

74 Example 26.3 Practical power transmission To provide power to a small city, a power plant generates 40 MW of AC electricity. The power plant is 50 km from the city (a typical distance), and the 100 km of wire used in the transmission line (to the city and back) has a resistance of 7.0 Ω.

75 Example 26.3 Practical power transmission (cont.) a. To provide 40 MW of power at the generator voltage of 25,000 V, what current is required? b. What is the power dissipated in the resistance of the transmission line for this current? c. To provide 40 MW of power at 500,000 V, what current is required? d. What is the power dissipated in the resistance of the transmission line for this higher voltage?

76 Example 26.3 Practical power transmission (cont.) PREPARE We can treat the city and the wires that transmit power to it as a load. All of the voltages are rms values and the power is an average power, so we can find the current to provide a given power and the power dissipated using the relationships in Equation 26.9.

77 Example 26.3 Practical power transmission (cont.) SOLVE a. To provide 40 MW at the generator voltage of 25,000 V, the current is

78 Example 26.3 Practical power transmission (cont.) b. Passing this current through the transmission lines will result in power dissipation in the 7.0 Ω resistance of the wires. We don t know the voltage drop across the wires, but we do know the current and resistance, so we can compute P dissipated in wires = (I rms ) 2 R = (1600 A) 2 (7.0 Ω) = 18 MW This is nearly half the power generated, clearly an unacceptable loss.

79 Example 26.3 Practical power transmission (cont.) c. Increasing the transmission voltage to 500 kv reduces the necessary current: This is a remarkably small current to supply a city. If you use several high-power appliances at one time, you could easily use this much current in your house. But the necessary current for the city can be so small because the voltage is so large.

80 Example 26.3 Practical power transmission (cont.) d. The relatively small current means that the power dissipated in the resistance of the wires will be small as well: P dissipated in wires = (I rms ) 2 R = (80 A) 2 (7.0 Ω) = MW This is only about 0.1% of the power generated, which is quite reasonable.

81 Example 26.3 Practical power transmission (cont.) ASSESS The final result the power dissipated in the wires is dramatically reduced for an increased transmission voltage is just what the example was designed to illustrate.

82 Power Transmission Transmitting electricity at high voltages means a smaller current is required, and therefore the resulting power loss is more manageable than for low voltages (and larger currents). This is why electrical transmission lines run at high voltages. In order to transmit electricity at high voltages, we need to use transformers to increase the voltage, which requires AC electricity. This is why we use AC power even though it is slightly more dangerous than DC power.