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, the dielectric is strontium titanate 34. What is the capacitance of a metal sphere of radius? a. b. c. d. Performance Task 18.5 Capacitors and Dielectrics 35. Newton’s law of universal gravitation is where. This describes the gravitational force between two point masses m1 and m2. Coulomb’s law is 18.46 18.47 596 Chapter 18 • Test Prep. This describes the where electric force between two point charges q1 and q2. (a) Describe how the force in each case depends on the distance rbetween the objects. How do the forces change if the distance is reduced by half? If the distance is doubled? (b) Describe the similarities and differences between the two laws. Consider the signs of the quantities that create the interaction (i.e., mass and charge), the constants G and k, and their dependence on separation r. (c) Given that the electric force is much stronger than TEST PREP Multiple Choice 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 36. A neutral hydrogen atom has one proton and one electron. If you remove the electron, what will be the leftover sign of the charge? a. negative b. positive c. zero d. neutral 37. What is the charge on a proton? a. +8.99 × 10–9 C b. −8.99 × 10–9 C c. + 1.60 × 10–19 C d. −1.60 × 10–19 C 38. True or false—Carbon is more conductive than pure water. a. b. true false 39. True or false—Two insulating objects are polarized. To cancel the polarization, it suffices to touch them together. true a. false b. 40. How is the charge of the proton related to the charge of the electron? a. The magnitudes of charge of the proton and the electron are equal, but the charge of the proton is positive, whereas the charge of the electron is negative. b. The magnitudes of charge of the proton and the electron are unequal, but the charge of the proton is positive, whereas the charge of the electron is negative. c. The magnitudes of charge of the proton and the electron are equal, but the charge of the proton is Access for free at openstax.org. the gravitational force, discuss why the
law for gravitational force was discovered much earlier than the law for electric force. (d) Consider a hydrogen atom, which is a single proton orbited by a single electron. The electric force holds the electron and proton together so that the hydrogen atom has a radius of about force between electron and proton does not change, what would be the approximate radius of the hydrogen atom if. Assuming the? negative, whereas the charge of the electron is positive. d. The magnitudes of charge of the proton and the electron are unequal, but the charge of the proton is negative, whereas the charge of the electron is positive. 18.2 Coulomb's law 41. If you double the distance between two point charges, by which factor does the force between the particles change? 1/2 a. b. 2 c. 4 d. 1/4 42. The combined charge of all the electrons in a dime is hundreds of thousands of coulombs. Because like charges repel, what keeps the dime from exploding? a. The dime has an equal number of protons, with positive charge. b. The dime has more protons than electrons, with positive charge. c. The dime has fewer protons than electrons, with positive charge. d. The dime is polarized, with electrons on one side and protons on the other side. 43. How can you modify the charges on two particles to quadruple the force between them without moving them? a. Increase the distance between the charges by a factor of two. Increase the distance between the charges by a factor of four. Increase the product of the charges by a factor of two Increase the product of the charges by a factor of four. b. c. d. 18.3 Electric Field 44. What is the magnitude of the electric field 12 cm from a charge of 1.5 nC? a. 9.4 × 107 N/ C 1.1 × 102 N/C b. c. 9.4 × 102 N/C d. 9.4 × 10–2 N/C 45. A charge distribution has electric field lines pointing into it. What sign is the net charge? a. positive b. neutral c. final d. negative 46. If five electric field lines come out of point charge q1 and 10 electric-field lines go into point charge q2, what is the ratio q1/q2? a. –2 b. –1 c. –1/2 d. 0 47. True or false
—The electric-field lines from a positive point charge spread out radially and point outward. a. b. false true 18.4 Electric Potential 48. What is the potential at 1.0 m from a point charge Q= − 25 nC? a. 6.6 × 102 V b. −2.3 × 102 V c. −6.6 × 102 V d. 2.3 × 102 V 49. Increasing the distance by a factor of two from a point charge will change the potential by a factor of how Short Answer 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 54. Compare the mass of the electron with the mass of the proton. a. The mass of the electron is about 1,000 times that of the proton. b. The mass of the proton is about 1,000 times that of the electron. c. The mass of the electron is about 1,836 times that of the proton. d. The mass of the proton is about 1,836 times that of the electron. Chapter 18 • Test Prep 597 much? a. 2 b. 4 c. d. 1/2 1/4 50. True or false—Voltageis the common word for potential difference, because this term is more descriptive than potential difference. a. b. false true 18.5 Capacitors and Dielectrics 51. Which magnitude of charge is stored on each plate of a 12 µF capacitor with 12 V applied across it? a. –1.0 × 10–6 C 1.0 × 10–6 C b. c. –1.4 × 10–4 C 1.4 × 10–4 C d. 52. What is the capacitance of a parallel-plate capacitor with an area of 200 cm2, a distance of 0.20 mm between the plates, and polystyrene as a dielectric? a. 2.3 nC b. 0.89 nC c. 23 nC d. 8.9 nC 53. Which factors determine the capacitance of a device? a. Capacitance depends only on the materials that make up the device. b. Capacitance depends on the electric field surrounding the device. c. Capacitance depends on the geometric and material parameters of the device. d. Capacitance depends only on the mass of the capacitor 55. The positive terminal of a battery is connected to one connection of a lightbulb,
and the other connection of the lightbulb is connected to the negative terminal of the battery. The battery pushes charge through the circuit but does not become charged itself. Does this violate the law of conservation of charge? Explain. a. No, because this is a closed circuit. b. No, because this is an open circuit. c. Yes, because this is a closed circuit. d. Yes, because this is an open circuit. 56. Two flat pieces of aluminum foil lay one on top of the other. What happens if you add charge to the top piece of aluminum foil? a. The charge will distribute over the top of the top 598 Chapter 18 • Test Prep piece. b. The charge will distribute to the bottom of the bottom piece. c. The inner surfaces will have excess charge of the opposite sign. d. The inner surfaces will have excess charge of the same sign. 57. The students in your class count off consecutively so each student has a number. The odd-numbered students are told to act as negative charge, and the evennumbered students are told to act as positive charge. How would you organize them to represent a polarized material? a. The even-numbered and odd-numbered students will be arranged one after the other. b. Two even-numbered will be followed by two odd- numbered, and so on. c. Even-numbered students will be asked to come to the front, whereas odd-numbered students will be asked to go to the back of the class. d. Half even-numbered and odd-numbered will come to the front, whereas half even-numbered and oddnumbered will go to the back. 58. An ion of iron contains 56 protons. How many electrons must it contain if its net charge is +5e? a. five electrons 51 electrons b. c. 56 electrons d. 61 electrons 59. An insulating rod carries of charge. After rubbing it with a material, you find it carries charge. How much charge was transferred to it? a. b. c. d. of 60. A solid cube carries a charge of +8e. You measure the charge on each face of the cube and find that each face carries +0.5eof charge. Is the cube made of conducting or insulating material? Explain. a. The cube is made of insulating material, because all the charges are on the surface of the cube. b. The cube is made of conducting material, because some of the charges are inside the cube
. c. The cube is made of insulating material, because all the charges are on the surface of the cube. d. The cube is made of insulating material, because some of the charges are inside the cube. 61. You have four neutral conducting spheres and a charging device that allows you to place charge qon any neutral object. You want to charge one sphere with a Access for free at openstax.org. charge q/2 and the other three with a charge q6. How do you proceed? a. Charge one sphere with charge q. Touch it simultaneously to the three remaining neutral spheres. b. Charge one sphere with charge q. Touch it to one other sphere to produce two spheres with charge. Touch one of these spheres to one other neutral sphere. c. Charge one sphere with charge q. Touch it to one other sphere to produce two spheres with charge. Touch one of these spheres simultaneously to the two remaining neutral spheres. d. Charge one sphere with charge q. Touch it simultaneously to two other neutral spheres to produce three spheres with charge q/3. Touch one of these spheres to one other neutral sphere. 18.2 Coulomb's law 62. Why does dust stick to the computer screen? a. The dust is neutral. b. The dust is polarized. c. The dust is positively charged. d. The dust is negatively charged. 63. The force between two charges is 4 × 10–9 N. If the magnitude of one charge is reduced by a factor of two and the distance between the charges is reduced by a factor of two, what is the new force between the charges? a. 2 × 10–9 N b. 4 × 10–9 N c. 6 × 10–9 N d. 8 × 10–9 N 64. True or false—Coulomb’s constant is k= 8.99 × 109 N·m2/C2. Newton’s gravitational constant is G= 6.67 × 10−11 m3/kg⋅s2. This tells you about the relative strength of the electrostatic force versus that of gravity. a. b. true false 65. An atomic nucleus contains 56 protons, for iron. Which force would this nucleus apply on an electron at a distance of 10×10–12 m? a. 0.65 × 10–4 N b. 0.02 × 10–4 N 1.3 × 10–4 N c. 72.8 × 10–4 N
d. 18.3 Electric Field 66. The electric field a distance of 10 km from a storm cloud is 1,000 N/C. What is the approximate charge in the cloud? a. 0.0011 C 11 C b. 110 C c. 1,100 C d. Chapter 18 • Test Prep 599 d. +400 mC 72. Given the potential difference between two points and the distance between the points, explain how to obtain the electric field between the points. a. Add the electric potential to the distance to obtain 67. Which electric field would produce a 10 N force in the the electric field. +x- direction on a charge of – 10 nC? a. − 1.0 × 109 N/C 1.0 × 109 N/C b. 1.0 × 1010 N/C c. 1.0 × 1011 N/C d. 68. A positive charge is located at x= 0. When a negative charge is placed at x= 10 cm, what happens to the electric field lines between the charges? a. The electric field lines become denser between the charges. b. The electric field lines become denser between the charges. c. The electric field lines remains same between the charges. d. The electric field lines will be zero between the charges. 18.4 Electric Potential 69. The energy required to bring a charge q= − 8.8 nC from far away to 5.5 cm from a point charge Q is 13 mJ. What is the potential at the final position of q? a. −112 MV b. −1.5 MV c. −0.66 MV d. +1.5 MV b. Divide the electric potential by the distance to obtain the electric field. c. Multiply the electric potential and the distance to obtain the electric field. d. Subtract the electric potential from the distance to obtain the electric field. 18.5 Capacitors and Dielectrics 73. If you double the voltage across the plates of a capacitor, how is the stored energy affected? a. Stored energy will decrease two times. b. Stored energy will decrease four times. c. Stored energy will increase two times. d. Stored energy will increase four times. 74. A capacitor with neoprene rubber as the dielectric stores 0.185 mJ of energy with a voltage of 50 V across the plates. If the area of the plates is 500 cm2, what is
the plate separation? a. 20 µm b. 20 m c. 80 µm d. 80 m 75. Explain why a storm cloud before a lightning strike is like a giant capacitor. a. The storm cloud acts as a giant charged capacitor, 70. How is electric potential related to electric potential as it can store a large amount of charge. energy? a. Electric potential is the electric potential energy per unit mass at a given position in space. b. Electric potential is the electric potential energy per unit length at a given position in space. This relation is not dimensionally correct. c. Electric potential is the electric potential energy per unit area in space. d. Electric potential is the electric potential energy per unit charge at a given position in space. 71. If it takes 10 mJ to move a charge qfrom xi = 25 cm to xf = − 25 cm in an electric field of charge q? a. −1.0 mC b. +0.25 mC c. + 1.0 mC what is the b. The storm cloud acts as a giant charged capacitor, as it contains a high amount of excess charges. c. The storm cloud acts as a giant charged capacitor, as it splits in two capacitor plates with equal and opposite charge. d. The storm cloud acts as a giant charged capacitor, as it splits in two capacitor plates with unequal and opposite charges. 76. A storm cloud is 2 km above the surface of Earth. The lower surface of the cloud is approximately 2 km2 in area. What is the approximate capacitance of this storm cloud-Earth system? a. 9 × 10–15 F b. 9 × 10-9 F c. d. 17.7 × 10-15 F 17.7 × 10-9 F 600 Chapter 18 • Test Prep Extended Response 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 77. Imagine that the magnitude of the charge on the electron differed very slightly from that of the proton. How would this affect life on Earth and physics in general? a. Many macroscopic objects would be charged, so we would experience the enormous force of electricity on a daily basis. b. Many macroscopic objects would be charged, so we would experience the small force of electricity on a daily basis. c. Many macroscopic objects would be charged, but it would not affect life on Earth and physics in general. d. Macroscopic objects would remain neutral, so it would not affect life on
Earth and physics in general. 78. True or false—Conservation of charge is like balancing a budget. a. b. true false 79. True or false—Although wood is an insulator, lightning can travel through a tree to reach Earth. a. b. true false touched to a second small metal sphere that is initially neutral. The spheres are then placed 20 cm apart. What is the force between the spheres? 1.02 × 10−7 N a. b. 2.55 × 10−7 N 5.1 × 10−7 N c. d. 20.4 × 10−7 N 18.3 Electric Field 83. Point charges are located at each corner of a square with sides of 5.0 cm. The top-left charge is q1 = 8.0 nC The top right charge is q2 = 4.0 nC. The bottom-right charge is q3 = 4.0 nC. The bottom-left charge is q4 = 8.0 nC. What is the electric field at the point midway between charges q2 and q3? a. b. c. d. 84. A long straight wire carries a uniform positive charge distribution. Draw the electric field lines in a plane containing the wire at a location far from the ends of the wire. Do not worry about the magnitude of the charge on the wire. a. Take the wire on the x-axis, and draw electric-field lines perpendicular to it. b. Take the wire on the x-axis, and draw electric-field lines parallel to it. 80. True or false—An eccentric inventor attempts to levitate c. Take the wire on the y-axis, and draw electric-field by first placing a large negative charge on himself and then putting a large positive charge on the ceiling of his workshop. Instead, while he attempts to place a large negative charge on himself, his clothes fly off. a. b. true false 18.2 Coulomb's law 81. Electrostatic forces are enormous compared to gravitational force. Why do you not notice electrostatic forces in everyday life, whereas you do notice the force due to gravity? a. Because there are two types of charge, but only one type of mass exists. b. Because there is only one type of charge, but two types of mass exist. c. Because opposite charges cancel each other, while gravity does not cancel out. d. Because opposite charges do not cancel each other, while gravity cancels out.
82. A small metal sphere with a net charge of 3.0 nC is lines along it. d. Take the wire on the z-axis, and draw electric-field lines along it. 18.4 Electric Potential 85. A square grid has charges of Q= 10 nC are each corner. The sides of the square at 10 cm. How much energy does it require to bring a q= 1.0 nC charge from very far away to the point at the center of this square? 1.3 × 10−6 J a. b. 2.5 × 10−6 J 3.8 × 10−6 J c. 5.1 × 10−6 J d. 86. How are potential difference and electric-field strength related for a constant electric field? a. The magnitude of electric-field strength is equivalent to the potential divided by the distance. b. The magnitude of electric-field strength is equivalent to the product of the electric potential and the distance. c. The magnitude of electric-field strength is Access for free at openstax.org. Chapter 18 • Test Prep 601 equivalent to the difference between magnitude of the electric potential and the distance. d. The magnitude of electric-field strength is equivalent to the sum of the magnitude of the electric potential and the distance. 88. Explain why capacitance should be inversely proportional to the separation between the plates of a capacitor. a. Capacitance is directly proportional to the electric field, which is inversely proportional to the distance between the capacitor plates. 18.5 Capacitors and Dielectrics b. Capacitance is inversely proportional to the electric 87. A 12 μF air-filled capacitor has 12 V across it. If the surface charge on each capacitor plate is σ= 7.2 mC / m2, what is the attractive force of one capacitor plate toward the other? a. 0.81 × 105 N b. 0.81 × 106 N 1.2 × 105 N c. 1.2 × 106 N d. field, which is inversely proportional to the distance between the capacitor plates. c. Capacitance is inversely proportional to the electric field, which is directly proportional to the distance between the capacitor plates. d. Capacitance is directly proportional to the electric field, which is directly proportional to the distance between the capacitor plates. 602 Chapter 18 • Test Prep Access for free at openstax.org. CHAPTER 19 Electrical Circuits Figure 19.1
Electric energy in massive quantities is transmitted from this hydroelectric facility, the Srisailam power station located along the Krishna River in India, by the movement of charge—that is, by electric current. (credit: Chintohere, Wikimedia Commons) Chapter Outline 19.1 Ohm's law 19.2 Series Circuits 19.3 Parallel Circuits 19.4 Electric Power The flicker of numbers on a handheld calculator, nerve impulses carrying signals of vision to the brain, an INTRODUCTION ultrasound device sending a signal to a computer screen, the brain sending a message for a baby to twitch its toes, an electric train pulling into a station, a hydroelectric plant sending energy to metropolitan and rural users—these and many other examples of electricity involve electric current, which is the movement of charge. Humanity has harnessed electricity, the basis of this technology, to improve our quality of life. Whereas the previous chapter concentrated on static electricity and the fundamental force underlying its behavior, the next two chapters will be devoted to electric and magnetic phenomena involving current. In addition to exploring applications of electricity, we shall gain new insights into the workings of nature. 604 Chapter 19 • Electrical Circuits 19.1 Ohm's law Section Learning Objectives By the end of this section, you will be able to do the following: • Describe how current is related to charge and time, and distinguish between direct current and alternating current • Define resistance and verbally describe Ohm’s law • Calculate current and solve problems involving Ohm’s law Section Key Terms alternating current ampere conventional current direct current electric current nonohmic ohmic Ohm’s law resistance Direct and Alternating Current Just as water flows from high to low elevation, electrons that are free to move will travel from a place with low potential to a place with high potential. A battery has two terminals that are at different potentials. If the terminals are connected by a conducting wire, an electric current (charges) will flow, as shown in Figure 19.2. Electrons will then move from the low-potential terminal of the battery (the negativeend) through the wire and enter the high-potential terminal of the battery (the positiveend). Figure 19.2 A battery has a wire connecting the positive and negative terminals, which allows electrons to move from the negative terminal to the positive terminal. Electric current is the rate at which electric charge moves. A large current, such as that used to start a truck engine,
moves a large amount very quickly, whereas a small current, such as that used to operate a hand-held calculator, moves a small amount of charge more slowly. In equation form, electric current Iis defined as is the amount of charge that flows past a given area and where The SI unit for electric current is the ampere (A), which is named in honor of the French physicist André-Marie Ampère (1775–1836). One ampere is one coulomb per second, or is the time it takes for the charge to move past the area. Electric current moving through a wire is in many ways similar to water current moving through a pipe. To define the flow of water through a pipe, we can count the water molecules that flow past a given section of the pipe. As shown in Figure 19.3, electric current is very similar. We count the number of electrical charges that flow past a section of a conductor; in this case, a wire. Access for free at openstax.org. 19.1 • Ohm's law 605 Figure 19.3 The electric current moving through this wire is the charge that moves past the cross-section A divided by the time it takes for this charge to move past the section A. Assume each particle qin Figure 19.3 carries a charge, in which case the total charge shown would be. If these charges move past the area Ain a time, then the current would be 19.1 Note that we assigned a positive charge to the charges in Figure 19.3. Normally, negative charges—electrons—are the mobile charge in wires, as indicated in Figure 19.2. Positive charges are normally stuck in place in solids and cannot move freely. However, because a positive current moving to the right is the same as a negative current of equal magnitude moving to the left, as shown in Figure 19.4, we define conventional current to flow in the direction that a positive charge would flow if it could move. Thus, unless otherwise specified, an electric current is assumed to be composed of positive charges. Also note that one Coulomb is a significant amount of electric charge, so 5 A is a very large current. Most often you will see current on the order of milliamperes (mA). Figure 19.4 (a) The electric field points to the right, the current moves to the right, and positive charges move to the right. (b) The equivalent situation but with negative charges moving to the left
. The electric field and the current are still to the right. Snap Lab Vegetable Current This lab helps students understand how current works. Given that particles confined in a pipe cannot occupy the same space, pushing more particles into one end of the pipe will force the same number of particles out of the opposite end. This creates a current of particles. Find a straw and dried peas that can move freely in the straw. Place the straw flat on a table and fill the straw with peas. When you push one pea in at one end, a different pea should come out of the other end. This demonstration is a model for 606 Chapter 19 • Electrical Circuits an electric current. Identify the part of the model that represents electrons and the part of the model that represents the supply of electrical energy. For a period of 30 s, count the number of peas you can push through the straw. When finished, calculate the pea currentby dividing the number of peas by the time in seconds. Note that the flow of peas is based on the peas physically bumping into each other; electrons push each other along due to mutually repulsive electrostatic forces. GRASP CHECK Suppose four peas per second pass through a straw. If each pea carried a charge of be through the straw? a. The electric current would be the pea charge multiplied by b. The electric current would be the pea current calculated in the lab multiplied by c. The electric current would be the pea current calculated in the lab. d. The electric current would be the pea charge divided by time.., what would the electric current. The direction of conventional current is the direction that positive charge would flow. Depending on the situation, positive charges, negative charges, or both may move. In metal wires, as we have seen, current is carried by electrons, so the negative charges move. In ionic solutions, such as salt water, both positively charged and negatively charged ions move. This is also true in nerve cells. Pure positive currents are relatively rare but do occur. History credits American politician and scientist Benjamin Franklin with describing current as the direction that positive charges flow through a wire. He named the type of charge associated with electrons negative long before they were known to carry current in so many situations. As electrons move through a metal wire, they encounter obstacles such as other electrons, atoms, impurities, etc. The electrons scatter from these obstacles, as depicted in Figure 19.5. Normally, the electrons lose energy with each interaction.
1 To keep the electrons moving thus requires a force, which is supplied by an electric field. The electric field in a wire points from the end of the wire at the higher potential to the end of the wire at the lower potential. Electrons, carrying a negative charge, move on average (or drift) in the direction opposite the electric field, as shown in Figure 19.5. Figure 19.5 Free electrons moving in a conductor make many collisions with other electrons and atoms. The path of one electron is shown. The average velocity of free electrons is in the direction opposite to the electric field. The collisions normally transfer energy to the conductor, so a constant supply of energy is required to maintain a steady current. So far, we have discussed current that moves constantly in a single direction. This is called direct current, because the electric charge flows in only one direction. Direct current is often called DCcurrent. Many sources of electrical power, such as the hydroelectric dam shown at the beginning of this chapter, produce alternating current, in which the current direction alternates back and forth. Alternating current is often called AC current. Alternating current moves back and forth at regular time intervals, as shown in Figure 19.6. The alternating current that comes from a normal wall socket does not suddenly switch directions. Rather, it increases smoothly up to a maximum current and then smoothly decreases back to zero. It then grows again, but in the opposite direction until it has reached the same maximum value. After that, it decreases smoothly back to zero, and the cycle starts over again. 1This energy is transferred to the wire and becomes thermal energy, which is what makes wires hot when they carry a lot of current. Access for free at openstax.org. 19.1 • Ohm's law 607 Figure 19.6 With alternating current, the direction of the current reverses at regular time intervals. The graph on the top shows the current versus time. The negative maxima correspond to the current moving to the left. The positive maxima correspond to current moving to the right. The current alternates regularly and smoothly between these two maxima. Devices that use AC include vacuum cleaners, fans, power tools, hair dryers, and countless others. These devices obtain the power they require when you plug them into a wall socket. The wall socket is connected to the power grid that provides an alternating potential (AC potential). When your device is plugged in, the AC potential pushes charges back and forth in the circuit of the device, creating
an alternating current. Many devices, however, use DC, such as computers, cell phones, flashlights, and cars. One source of DC is a battery, which provides a constant potential (DC potential) between its terminals. With your device connected to a battery, the DC potential pushes charge in one direction through the circuit of your device, creating a DC current. Another way to produce DC current is by using a transformer, which converts AC potential to DC potential. Small transformers that you can plug into a wall socket are used to charge up your laptop, cell phone, or other electronic device. People generally call this a chargeror a battery, but it is a transformer that transforms AC voltage into DC voltage. The next time someone asks to borrow your laptop charger, tell them that you don’t have a laptop charger, but that they may borrow your converter. WORKED EXAMPLE Current in a Lightning Strike A lightning strike can transfer as many as average electric current in the lightning? STRATEGY electrons from the cloud to the ground. If the strike lasts 2 ms, what is the Use the definition of current,. The charge from electrons is, where is the number of electrons and is the charge on the electron. This gives The time is the duration of the lightning strike. Solution The current in the lightning strike is Discussion 19.2 19.3 608 Chapter 19 • Electrical Circuits The negative sign reflects the fact that electrons carry the negative charge. Thus, although the electrons flow from the cloud to the ground, the positive current is defined to flow from the ground to the cloud. WORKED EXAMPLE Average Current to Charge a Capacitor In a circuit containing a capacitor and a resistor, it takes 1 min to charge a 16 μF capacitor by using a 9-V battery. What is the average current during this time? STRATEGY We can determine the charge on the capacitor by using the definition of capacitance:. This gives a charge of a 9-V battery, the voltage across the capacitor will be. When the capacitor is charged by By inserting this expression for charge into the equation for current,, we can find the average current. Solution The average current is 19.4 19.5 Discussion This small current is typical of the current encountered in circuits such as this. Practice Problems 1. 10 nC of charge flows through a circuit in 3.0 × 10−6 s. What is the current during this time? a. The current passes through the circuit is 3.3 ×
10−3 A. b. The current passes through the circuit is 30 A. c. The current passes through the circuit is 33 A. d. The current passes through the circuit is 0.3 A. 2. How long would it take a current to charge a capacitor with? a. b. c. d. Resistance and Ohm’s Law As mentioned previously, electrical current in a wire is in many ways similar to water flowing through a pipe. The water current that can flow through a pipe is affected by obstacles in the pipe, such as clogs and narrow sections in the pipe. These obstacles slow down the flow of current through the pipe. Similarly, electrical current in a wire can be slowed down by many factors, including impurities in the metal of the wire or collisions between the charges in the material. These factors create a resistance to the electrical current. Resistance is a description of how much a wire or other electrical component opposes the flow of charge through it. In the 19th century, the German physicist Georg Simon Ohm (1787–1854) found experimentally that current through a conductor is proportional to the voltage drop across a current-carrying conductor. Access for free at openstax.org. 19.1 • Ohm's law 609 The constant of proportionality is the resistance Rof the material, which leads to This relationship is called Ohm’s law. It can be viewed as a cause-and-effect relationship, with voltage being the cause and the current being the effect. Ohm’s law is an empirical law like that for friction, which means that it is an experimentally observed phenomenon. The units of resistance are volts per ampere, or V/A. We call a V/A an ohm, which is represented by the uppercase Greek letter omega ( ). Thus, Ohm’s law holds for most materials and at common temperatures. At very low temperatures, resistance may drop to zero (superconductivity). At very high temperatures, the thermal motion of atoms in the material inhibits the flow of electrons, increasing the resistance. The many substances for which Ohm’s law holds are called ohmic. Ohmic materials include good conductors like copper, aluminum, and silver, and some poor conductors under certain circumstances. The resistance of ohmic materials remains essentially the same for a wide range of voltage and current. WATCH PHYSICS Introduction to Electricity, Circuits, Current, and Resistance This video presents
Ohm’s law and shows a simple electrical circuit. The speaker uses the analogy of pressure to describe how electric potential makes charge move. He refers to electric potential as electric pressure. Another way of thinking about electric potential is to imagine that lots of particles of the same sign are crowded in a small, confined space. Because these charges have the same sign (they are all positive or all negative), each charge repels the others around it. This means that lots of charges are constantly being pushed towards the outside of the space. A complete electric circuit is like opening a door in the small space: Whichever particles are pushed towards the door now have a way to escape. The higher the electric potential, the harder each particle pushes against the others. GRASP CHECK, two resistors each with resistance If, instead of a single resistor what can you say about the current through the circuit? a. The amount of current through the circuit must decrease by half. b. The amount of current through the circuit must increase by half. c. The current must remain the same through the circuit. d. The amount of current through the circuit would be doubled. are drawn in the circuit diagram shown in the video, Virtual Physics Ohm’s Law Click to view content (http://www.openstax.org/l/28ohms_law) This simulation mimics a simple circuit with batteries providing the voltage source and a resistor connected across the batteries. See how the current is affected by modifying the resistance and/or the voltage. Note that the resistance is modeled as an element containing small scattering centers. These represent impurities or other obstacles that impede the passage of the current. GRASP CHECK In a circuit, if the resistance is left constant and the voltage is doubled (for example, from current change? Does this conform to Ohm’s law? a. The current will get doubled. This conforms to Ohm’s law as the current is proportional to the voltage. b. The current will double. This does not conform to Ohm’s law as the current is proportional to the voltage. to ), how does the 610 Chapter 19 • Electrical Circuits c. The current will increase by half. This conforms to Ohm’s law as the current is proportional to the voltage. d. The current will decrease by half. This does not conform to Ohm’s law as the current is proportional to the voltage. WORKED EXAMPLE Resistance of
a Headlight What is the resistance of an automobile headlight through which 2.50 A flows when 12.0 V is applied to it? STRATEGY Ohm’s law tells us the battery, headlight.. The voltage drop in going through the headlight is just the voltage rise supplied by. We can use this equation and rearrange Ohm’s law to find the resistance of the Solution Solving Ohm’s law for the resistance of the headlight gives 19.6 Discussion This is a relatively small resistance. As we will see below, resistances in circuits are commonly measured in kW or MW. WORKED EXAMPLE Determine Resistance from Current-Voltage Graph Suppose you apply several different voltages across a circuit and measure the current that runs through the circuit. A plot of your results is shown in Figure 19.7. What is the resistance of the circuit? Access for free at openstax.org. 19.1 • Ohm's law 611 Figure 19.7 The line shows the current as a function of voltage. Notice that the current is given in milliamperes. For example, at 3 V, the current is 0.003 A, or 3 mA. STRATEGY The plot shows that current is proportional to voltage, which is Ohm’s law. In Ohm’s law ( proportionality is the resistance R. Because the graph shows current as a function of voltage, we have to rearrange Ohm’s law in that form: Figure 19.7, we can calculate the resistance R.. This shows that the slope of the line of Iversus Vis. Thus, if we find the slope of the line in ), the constant of Solution The slope of the line is the risedivided by the run. Looking at the lower-left square of the grid, we see that the line rises by 1 mA (0.001 A) and runs over a voltage of 1 V. Thus, the slope of the line is Equating the slope with and solving for Rgives or 1 k-ohm. Discussion 19.7 19.8 This resistance is greater than what we found in the previous example. Resistances such as this are common in electric circuits, as we will discover in the next section. Note that if the line in Figure 19.7 were not straight, then the material would not be ohmic and we would not be able to use Ohm’s law
. Materials that do not follow Ohm’s law are called nonohmic. Practice Problems 3. If you double the voltage across an ohmic resistor, how does the current through the resistor change? a. The current will double. b. The current will increase by half. c. The current will decrease by half. d. The current will decrease by a factor of two. 4. The current through a resistor is. What is the voltage drop across the resistor? a. b. c. d. 612 Chapter 19 • Electrical Circuits Check Your Understanding 5. What is electric current? a. Electric current is the electric charge that is at rest. b. Electric current is the electric charge that is moving. c. Electric current is the electric charge that moves only from the positive terminal of a battery to the negative terminal. d. Electric current is the electric charge that moves only from a region of lower potential to higher potential. 6. What is an ohmic material? a. An ohmic material is a material that obeys Ohm’s law. b. An ohmic material is a material that does not obey Ohm’s law. c. An ohmic material is a material that has high resistance. d. An ohmic material is a material that has low resistance. 7. What is the difference between direct current and alternating current? a. Direct current flows continuously in every direction whereas alternating current flows in one direction. b. Direct current flows continuously in one direction whereas alternating current reverses its direction at regular time intervals. c. Both direct and alternating current flow in one direction but the magnitude of direct current is fixed whereas the magnitude of alternating current changes at regular intervals of time. d. Both direct and alternating current changes its direction of flow but the magnitude of direct current is fixed whereas the magnitude of alternating current changes at regular intervals of time. 19.2 Series Circuits Section Learning Objectives By the end of this section, you will be able to do the following: • Interpret circuit diagrams and diagram basic circuit elements • Calculate equivalent resistance of resistors in series and apply Ohm’s law to resistors in series and apply Ohm’s law to resistors in series Section Key Terms circuit diagram electric circuit equivalent resistance in series resistor steady state Electric Circuits and Resistors Now that we understand the concept of electric current, let’s see what we can do with it. As you are no doubt aware, the modern lifestyle relies heavily on electrical
devices. These devices contain ingenious electric circuits, which are complete, closed pathways through which electric current flows. Returning to our water analogy, an electric circuit is to electric charge like a network of pipes is to water: The electric circuit guides electric charge from one point to the next, running the charge through various devices along the way to extract work or information. Electric circuits are made from many materials and cover a huge range of sizes, as shown in Figure 19.8. Computers and cell phones contain electric circuits whose features can be as small as roughly a billionth of a meter (a nanometer, or pathways that guide the current in these devices are made by ultraprecise chemical treatments of silicon or other semiconductors. Large power systems, on the other hand, contain electric circuits whose features are on the scale of meters. These systems carry such large electric currents that their physical dimensions must be relatively large. ). The Access for free at openstax.org. 19.2 • Series Circuits 613 Figure 19.8 The photo on the left shows a chipthat contains complex integrated electric circuitry. Chips such as this are at the heart of devices such as computers and cell phones. The photograph on the right shows some typical electric circuitry required for high-power electric power transmission. The pathways that form electric circuits are made from a conducting material, normally a metal in macroscopic circuits. For example, copper wires inside your school building form the electrical circuits that power lighting, projectors, screens, speakers, etc. To represent an electric circuit, we draw circuit diagrams. We use lines and symbols to represent the elements in the circuit. A simple electric circuit diagram is shown on the left side of Figure 19.9. On the right side is an analogous water circuit, which we discuss below. Figure 19.9 On the left is a circuit diagram showing a battery (in red), a resistor (black zigzag element), and the current I. On the right is the analogous water circuit. The pump is like the battery, the sand filter is like the resistor, the water current is like the electrical current, and the reservoir is like the ground. There are many different symbols that scientists and engineers use in circuit diagrams, but we will focus on four main symbols: the wire, the battery or voltage source, resistors, and the ground. The thin black lines in the electric circuit diagram represent the pathway that the electric charge must follow. These pathways are assumed to be perfect conductors, so electric charge can
move along these pathways without losing any energy. In reality, the wires in circuits are not perfect, but they come close enough for our purposes. The zigzag element labeled Ris a resistor, which is a circuit element that provides a known resistance. Macroscopic resistors are often color coded to indicate their resistance, as shown in Figure 19.10. The red element in Figure 19.9 is a battery, with its positive and negative terminals indicated; the longer line represents the positive terminal of the battery, and the shorter line represents the negative terminal. Note that the battery icon is not always colored red; this is done in Figure 19.9 just to make it easy to identify. Finally, the element labeled groundon the lower left of the circuit indicates that the circuit is connected to Earth, which is a large, essentially neutral object containing an infinite amount of charge. Among other things, the ground determines the potential of the negative terminal of the battery. Normally, the potential of the ground is defined to be zero:. This 614 Chapter 19 • Electrical Circuits means that the entire lower wire in Figure 19.10 is at a voltage of zero volts. Figure 19.10 Some typical resistors. The color bands indicate the value of the resistance of each resistor. The electric current in Figure 19.9 is indicted by the blue line labeled I. The arrow indicates the direction in which positive charge would flow in this circuit. Recall that, in metals, electrons are mobile charge carriers, so negative charges actually flow in the opposite direction around this circuit (i.e., counterclockwise). However, we draw the current to show the direction in which positive charge would move. On the right side of Figure 19.9 is an analogous water circuit. Water at a higher pressure leaves the top of the pump, which is like charges leaving the positive terminal of the battery. The water travels through the pipe, like the charges traveling through the wire. Next, the water goes through a sand filter, which heats up as the water squeezes through. This step is like the charges going through the resistor. When charges flow through a resistor, they do work to heat up the resistor. After flowing through the sand filter, the water has converted its potential energy into heat, so it is at a lower pressure. Likewise, the charges exiting the resistor have converted their potential energy into heat, so they are at a lower voltage. Recall that voltage is just potential energy per charge. Thus, water pressure is analogous to electric potential energy (i
.e., voltage). Coming back to the water circuit again, we see that the water returns to the bottom of the pump, which is like the charge returning to the negative terminal of the battery. The water pump uses a source of energy to pump the water back up to a high pressure again, giving it the pressure required to go through the circuit once more. The water pump is like the battery, which uses chemical energy to increase the voltage of the charge up to the level of the positive terminal. The potential energy per charge at the positive terminal of the battery is the voltage rating of the battery. This voltage is like water pressure in the upper pipe. Just like a higher pressure forces water to move toward a lower pressure, a higher voltage forces electric charge to flow toward a lower voltage. The pump takes water at low pressure and does work on it, ejecting water at a higher pressure. Likewise, a battery takes charge at a low voltage, does work on it, and ejects charge at a higher voltage. Note that the current in the water circuit of Figure 19.9 is the same throughout the circuit. In other words, if we measured the number of water molecules passing a cross-section of the pipe per unit time at any point in the circuit, we would get the same answer no matter where in the circuit we measured. The same is true of the electrical circuit in the same figure. The electric current is the same at all points in this circuit, including inside the battery and in the resistor. The electric current neither speeds up in the wires nor slows down in the resistor. This would create points where too much or too little charge would be bunched up. Thus, the current is the same at all points in the circuit shown in Figure 19.9. Although the current is the same everywhere in both the electric and water circuits, the voltage or water pressure changes as you move through the circuits. In the water circuit, the water pressure at the pump outlet stays the same until the water goes through the sand filter, assuming no energy loss in the pipe. Likewise, the voltage in the electrical circuit is the same at all points in a given wire, because we have assumed that the wires are perfect conductors. Thus, as indicated by the constant red color of the upper wire in Figure 19.11, the voltage throughout this wire is constant at through the resistor, but once you reach the blue wire, the voltage stays at its new level of terminal of the battery (i.e., the blue terminal of the battery
).. The voltage then drops as you go all the way to the negative Access for free at openstax.org. 19.2 • Series Circuits 615 Figure 19.11 The voltage in the red wire is constant at from the positive terminal of the battery to the top of the resistor. The voltage in the blue wire is constant at from the bottom of the resistor to the negative terminal of the battery. If we go from the blue wire through the battery to the red wire, the voltage increases from we go from the blue wire up through the resistor to the red wire, the voltage also goes from Ohm’s law, we can write to to. Likewise, if. Thus, using is measured from the bottom of the resistor to the top, meaning that the top of the resistor is at a higher Note that voltage than the bottom of the resistor. Thus, current flows from the top of the resistor or higher voltage to the bottom of the resistor or lower voltage. Virtual Physics Battery-Resistor Circuit Click to view content (http://www.openstax.org/l/21batteryresist) Use this simulation to better understand how resistance, voltage, and current are related. The simulation shows a battery with a resistor connected between the terminals of the battery, as in the previous figure. You can modify the battery voltage and the resistance. The simulation shows how electrons react to these changes. It also shows the atomic cores in the resistor and how they are excited and heat up as more current goes through the resistor. Draw the circuit diagram for the circuit, being sure to draw an arrow indicating the direction of the current. Now pick three spots along the wire. Without changing the settings, allow the simulation to run for 20 s while you count the number of electrons passing through that spot. Record the number on the circuit diagram. Now do the same thing at each of the other two spots in the circuit. What do you notice about the number of charges passing through each spot in 20 s? Remember that that current is defined as the rate that charges flow through the circuit. What does this mean about the current through the entire circuit? GRASP CHECK With the voltage slider, give the battery a positive voltage. Notice that the electrons are spaced farther apart in the left wire than they are in the right wire. How does this reflect the voltage in the two wires? a. The voltage between static charges is directly proportional to the distance between them. b. The voltage between static charges is directly proportional to square of
the distance between them. c. The voltage between static charges is inversely proportional to the distance between them. d. The voltage between static charges is inversely proportional to square of the distance between them. Other possible circuit elements include capacitors and switches. These are drawn as shown on the left side of Figure 19.12. A switch is a device that opens and closes the circuit, like a light switch. It is analogous to a valve in a water circuit, as shown on the right side of Figure 19.12. With the switch open, no current passes through the circuit. With the switch closed, it becomes part of the wire, so the current passes through it with no loss of voltage. The capacitor is labeled C on the left of Figure 19.12. A capacitor in an electrical circuit is analogous to a flexible membrane in a 616 Chapter 19 • Electrical Circuits water circuit. When the switch is closed in the circuit of Figure 19.12, the battery forces electrical current to flow toward the capacitor, charging the upper capacitor plate with positive charge. As this happens, the voltage across the capacitor plates increases. This is like the membrane in the water circuit: When the valve is opened, the pump forces water to flow toward the membrane, making it stretch to store the excess water. As this happens, the pressure behind the membrane increases. Now if we open the switch, the capacitor holds the voltage between its plates because the charges have nowhere to go. Likewise, if we close the valve, the water has nowhere to go and the membrane maintains the water pressure in the pipe between itself and the valve. If the switch is closed for a long time in the electric circuit or if the valve is open for a long time in the water circuit, the current will eventually stop flowing because the capacitor or the membrane will have become completely charged. Each circuit is now in the steady state, which means that its characteristics do not change over time. In this case, the steady state is characterized by zero current, and this does not change as long as the switch or valve remains in the same position. In the steady state, no electrical current passes through the capacitor, and no water current passes through the membrane. The voltage difference between the capacitor plates will be the same as the battery voltage. In the water circuit, the pressure behind the membrane will be the same as the pressure created by the pump. Although the circuit in Figure 19.12 may seem a bit pointless because all that happens when the switch is closed is that the capacitor
charges up, it does show the capacitor’s ability to store charge. Thus, the capacitor serves as a reservoir for charge. This property of capacitors is used in circuits in many ways. For example, capacitors are used to power circuits while batteries are being charged. In addition, capacitors can serve as filters. To understand this, let’s go back to the water analogy. Suppose you have a water hose and are watering your garden. Your friend thinks he’s funny, and kinksthe hose. While the hose is kinked, you experience no water flow. When he lets go, the water starts flowing again. If he does this really fast, you experience water-no water-water-no water, and that’s really no way to water your garden. Now imagine that the hose is filling up a big bucket, and you are watering from the bottom of the bucket. As long as you had water in your bucket to begin with and your friend doesn’t kink the water hose for too long, you would be able to water your garden without the interruptions. Your friend kinking the water hose is filteredby the big bucket’s supply of water, so it does not impact your ability to water the garden. We can think of the interruptions in the current (be it water or electrical current) as noise. Capacitors act in an analogous way as the water bucket to help filter out the noise. Capacitors have so many uses that it is very rare to find an electronic circuit that does not include some capacitors. Figure 19.12 On the left is an electrical circuit containing a battery, a switch, and a capacitor. On the left is the analogous water circuit with a pump, a valve, and a stretchable membrane. The pump is like the battery, the valve is like the switch, and the stretchable membrane is like the capacitor. When the switch is closed, electrical current flows as the capacitor charges and its voltage increases. Likewise in the water circuit, when the valve is open, water current flows as the stretchable membrane stretches and the water pressure behind it increases. Access for free at openstax.org. 19.2 • Series Circuits 617 WORK IN PHYSICS What It Takes to be an Electrical Engineer Physics is used in a wide variety of fields. One field that requires a very thorough knowledge of physics is electrical engineering. An electrical engineer can work on anything from the large-scale power systems
that provide power to big cities to the nanoscale electronic circuits that are found in computers and cell phones (Figure 19.13). In working with power companies, you can be responsible for maintaining the power grid that supplies electrical power to large areas. Although much of this work is done from an office, it is common to be called in for overtime duty after storms or other natural events. Many electrical engineers enjoy this part of the job, which requires them to race around the countryside repairing high-voltage transformers and other equipment. However, one of the more unpleasant aspects of this work is to remove the carcasses of unfortunate squirrels or other animals that have wandered into the transformers. Other careers in electrical engineering can involve designing circuits for cell phones, which requires cramming some 10 billion transistors into an electronic chip the size of your thumbnail. These jobs can involve much work with computer simulations and can also involve fields other than electronics. For example, the 1-m-diameter lenses that are used to make these circuits (as of 2015) are so precise that they are shipped from the manufacture to the chip fabrication plant in temperature-controlled trucks to ensure that they are held within a certain temperature range. If they heat up or cool down too much, they deform ever so slightly, rendering them useless for the ultrahigh precision photolithography required to manufacture these chips. In addition to a solid knowledge of physics, electrical engineers must above all be practical. Consider, for example, how one corporation managed to launch some anti-ballistic missiles at the White Sands Missile Test Range in New Mexico in the 1960s. Before launch, the skin of the missile had to be at the same voltage as the rail from which it was launched. The rail was connected to the ground by a large copper wire connected to a stake driven into the sandy earth. The missile, however, was connected by an umbilical cord to the equipment in the control shed a few meters away, which was grounded via a different grounding circuit. Before launching the missile, the voltage difference between the missile skin and the rail had to be less than 2.5 V. After an especially dry spell of weather, the missile could not be launched because the voltage difference stood at 5 V. A group of electrical engineers, including the father of your author, stood around pondering how to reduce the voltage difference. The situation was resolved when one of the engineers realized that urine contains electrolytes and conducts electricity quite well. With that, the four engineers quickly resolved the problem by ur
inating on the rail spike. The voltage difference immediately dropped to below 2.5 V and the missile was launched on schedule. Figure 19.13 The systems that electrical engineers work on range from microprocessor circuits (left)] to missile systems (right). Virtual Physics Click to view content (http://www.openstax.org/l/21phetcirconstr) Amuse yourself by building circuits of all different shapes and sizes. This simulation provides you with various standard circuit elements, such as batteries, AC voltage sources, resistors, capacitors, light bulbs, switches, etc. You can connect these in any configuration you like and then see the result. Build a circuit that starts with a resistor connected to a capacitor. Connect the free side of the resistor to the positive terminal of a battery and the free side of the capacitor to the negative terminal of the battery. Click the reset dynamics button to see how the current flows starting with no charge on the capacitor. Now right click on the resistor to change its 618 Chapter 19 • Electrical Circuits value. When you increase the resistance, does the circuit reach the steady state more rapidly or more slowly? GRASP CHECK When the circuit has reached the steady state, how does the voltage across the capacitor compare to the voltage of the battery? What is the voltage across the resistor? a. The voltage across the capacitor is greater than the voltage of the battery. In the steady state, no current flows through this circuit, so the voltage across the resistor is zero. b. The voltage across the capacitor is smaller than the voltage of the battery. In the steady state, finite current flows through this circuit, so the voltage across the resistor is finite. c. The voltage across the capacitor is the same as the voltage of the battery. In the steady state, no current flows through this circuit, so the voltage across the resistor is zero. d. The voltage across the capacitor is the same as the voltage of the battery. In the steady state, finite current flows through this circuit, so the voltage across the resistor is finite. Resistors in Series and Equivalent Resistance Now that we have a basic idea of how electrical circuits work, let’s see what happens in circuits with more than one circuit element. In this section, we look at resistors in series. Components connected in series are connected one after the other in the same branch of a circuit, such as the resistors connected in series on the left side of Figure 19.14. Figure 19.14
On the left is an electric circuit with three resistors R1, R2, and R3 connected in series. On the right is an electric circuit with one resistor Requiv that is equivalent to the combination of the three resistors R1, R2, and R3. We will now try to find a single resistance that is equivalent to the three resistors in series on the left side of Figure 19.14. An equivalent resistor is a resistor that has the same resistance as the combined resistance of a set of other resistors. In other words, the same current will flow through the left and right circuits in Figure 19.14 if we use the equivalent resistor in the right circuit. where Iis the current in According to Ohm’s law, the voltage drop Vacross a resistor when a current flows through it is amperes (A) and Ris the resistance in ohms ( ). Another way to think of this is that Vis the voltage necessary to make a current Iflow through a resistance R. Applying Ohm’s law to each resistor on the left circuit of Figure 19.14, we find that the voltage drop across, that across voltage output of the battery, that is. The sum of these voltages equals the, and that across is is is You may wonder why voltages must add up like this. One way to understand this is to go once around the circuit and add up the successive changes in voltage. If you do this around a loop and get back to the starting point, the total change in voltage should be zero, because you end up at the same place that you started. To better understand this, consider the analogy of going for a stroll through some hilly countryside. If you leave your car and walk around, then come back to your car, the total height you gained in your stroll must be the same as the total height you lost, because you end up at the same place as you started. Thus, the gravitational potential energy you gain must be the same as the gravitational potential energy you lose. The same reasoning holds for voltage in going around an electric circuit. Let’s apply this reasoning to the left circuit in Figure 19.14. We start just below the battery and move up through the battery, which contributes a voltage gainof resistors. The voltage dropsby. Next, we got through the in going through resistor in going through resistor in going through, and by, by 19.9 Access for free at openstax.org
. 19.2 • Series Circuits 619 resistor. After going through resistor and set the sum equal to zero. This gives, we arrive back at the starting point, so we add up these four changes in voltage which is the same as the previous equation. Note that the minus signs in front of drops, whereas is a voltage rise. 19.10 are because these are voltage Ohm’s law tells us that gives,, and. Inserting these values into equation Applying this same logic to the right circuit in Figure 19.14 gives Dividing the equation by, we get 19.11 19.12 19.13 This shows that the equivalent resistance for a series of resistors is simply the sum of the resistances of each resistor. In general, Nresistors connected in series can be replaced by an equivalent resistor with a resistance of WATCH PHYSICS Resistors in Series This video discusses the basic concepts behind interpreting circuit diagrams and then shows how to calculate the equivalent resistance for resistors in series. Click to view content (https://www.openstax.org/l/02resistseries) GRASP CHECK True or false—In a circuit diagram, we can assume that the voltage is the same at every point in a given wire. a. b. false true WORKED EXAMPLE Calculation of Equivalent Resistance In the left circuit of the previous figure, suppose the voltage rating of the battery is 12 V, and the resistances are STRATEGY FOR (A) Use the equation for the equivalent resistance of resistors connected in series. Because the circuit has three resistances, we only need to keep three terms, so it takes the form. (a) What is the equivalent resistance? (b) What is the current through the circuit? 19.14 Solution for (a) Inserting the given resistances into the equation above gives 620 Chapter 19 • Electrical Circuits Discussion for (a) We can thus replace the three resistors with a single 20- resistor. STRATEGY FOR (B) Apply Ohm’s law to the circuit on the right side of the previous figure with the equivalent resistor of 20. Solution for (b) The voltage drop across the equivalent resistor must be the same as the voltage rise in the battery. Thus, Ohm’s law gives 19.15 19.16 Discussion for (b) To check that this result is reasonable, we calculate the voltage drop across each resistor and verify that they add up to the
voltage rating of the battery. The voltage drop across each resistor is Adding these voltages together gives which is the voltage rating of the battery. WORKED EXAMPLE 19.17 19.18 Determine the Unknown Resistance The circuit shown in figure below contains three resistors of known value and a third element whose resistance Given that the equivalent resistance for the entire circuit is 150, what is the resistance? is unknown. STRATEGY The four resistances in this circuit are connected in series, so we know that they must add up to give the equivalent resistance. We can use this to find the unknown resistance. Solution For four resistances in series, the equation for the equivalent resistance of resistors in series takes the form 19.19 Access for free at openstax.org. Solving for R3 and inserting the known values gives 19.3 • Parallel Circuits 621 19.20 Discussion The equivalent resistance of a circuit can be measured with an ohmmeter. This is sometimes useful for determining the effective resistance of elements whose resistance is not marked on the element. Check your Understanding 8. Figure 19.15 What circuit element is represented in the figure below? a. a battery b. a resistor c. a capacitor d. an inductor 9. How would a diagram of two resistors connected in series appear? a. b. c. d. 19.3 Parallel Circuits Section Learning Objectives By the end of this section, you will be able to do the following: • Interpret circuit diagrams with parallel resistors • Calculate equivalent resistance of resistor combinations containing series and parallel resistors Section Key Terms in parallel Resistors in Parallel In the previous section, we learned that resistors in series are resistors that are connected one after the other. If we instead combine resistors by connecting them next to each other, as shown in Figure 19.16, then the resistors are said to be connected in parallel. Resistors are in parallel when both ends of each resistor are connected directly together. Note that the tops of the resistors are all connected to the same wire, so the voltage at the top of the each resistor is the same. Likewise, the bottoms of the resistors are all connected to the same wire, so the voltage at the bottom of each resistor is the same. This means that the voltage drop across each resistor is the same. In this case, the voltage drop is the voltage rating Vof the battery, because the top and bottom wires connect to the positive and negative terminals of the battery,
respectively. Although the voltage drop across each resistor is the same, we cannot say the same for the current running through each resistor. Thus, are not necessarily the same, because the resistors do not necessarily have the same 622 Chapter 19 • Electrical Circuits resistance. Note that the three resistors in Figure 19.16 provide three different paths through which the current can flow. This means that the equivalent resistance for these three resistors must be less than the smallest of the three resistors. To understand this, imagine that the smallest resistor is the only path through which the current can flow. Now add on the alternate paths by connecting other resistors in parallel. Because the current has more paths to go through, the overall resistance (i.e., the equivalent resistance) will decrease. Therefore, the equivalent resistance must be less than the smallest resistance of the parallel resistors. Figure 19.16 The left circuit diagram shows three resistors in parallel. The voltage Vof the battery is applied across all three resistors. The currents that flow through each branch are not necessarily equal. The right circuit diagram shows an equivalent resistance that replaces the three parallel resistors. To find the equivalent resistance voltage drop across each resistor is V, we obtain of the three resistors, we apply Ohm’s law to each resistor. Because the or 19.21 19.22 We also know from conservation of charge that the three currents through the battery. If this were not true, current would have to be mysteriously created or destroyed somewhere in the circuit, which is physically impossible. Thus, we have must add up to give the current Ithat goes Inserting the expressions for into this equation gives or This formula is just Ohm’s law, with the factor in parentheses being the equivalent resistance. Thus, the equivalent resistance for three resistors in parallel is 19.23 19.24 19.25 19.26 19.27 The same logic works for any number of resistors in parallel, so the general form of the equation that gives the equivalent resistance of Nresistors connected in parallel is 19.28 Access for free at openstax.org. 19.3 • Parallel Circuits 623 WORKED EXAMPLE Find the Current through Parallel Resistors The three circuits below are equivalent. If the voltage rating of the battery is the circuit and what current runs through the circuit?, what is the equivalent resistance of STRATEGY The three resistors are connected in parallel and the voltage drop across them is Vbattery. Thus, we
can apply the equation for the equivalent resistance of resistors in parallel, which takes the form 19.29 The circuit with the equivalent resistance is shown below. Once we know the equivalent resistance, we can use Ohm’s law to find the current in the circuit. Solution Inserting the given values for the resistance into the equation for equivalent resistance gives The current through the circuit is thus 19.30 19.31 Discussion Although 0.62 A flows through the entire circuit, note that this current does not flow through each resistor. However, because 624 Chapter 19 • Electrical Circuits electric charge must be conserved in a circuit, the sum of the currents going through each branch of the circuit must add up to the current going through the battery. In other words, we cannot magically create charge somewhere in the circuit and add this new charge to the current. Let’s check this reasoning by using Ohm’s law to find the current through each resistor. 19.32 As expected, these currents add up to give 0.62 A, which is the total current found going through the equivalent resistor. Also, note that the smallest resistor has the largest current flowing through it, and vice versa. WORKED EXAMPLE Reasoning with Parallel Resistors Without doing any calculation, what is the equivalent resistance of three identical resistors Rin parallel? STRATEGY Three identical resistors Rin parallel make three identical paths through which the current can flow. Thus, it is three times easier for the current to flow through these resistors than to flow through a single one of them. Solution If it is three times easier to flow through three identical resistors Rthan to flow through a single one of them, the equivalent resistance must be three times less: R/3. Discussion Let’s check our reasoning by calculating the equivalent resistance of three identical resistors Rin parallel. The equation for the equivalent resistance of resistors in parallel gives 19.33 Thus, our reasoning was correct. In general, when more paths are available through which the current can flow, the equivalent resistance decreases. For example, if we have identical resistors Rin parallel, the equivalent resistance would be R/10. Practice Problems 10. Three resistors, 10, 20, and 30 Ω, are connected in parallel. What is the equivalent resistance? a. The equivalent resistance is 5.5 Ω b. The equivalent resistance is 60 Ω c. The equivalent resistance is 6 × 103 Ω d. The equivalent resistance is 6 × 104 Ω
11. If a drop occurs across a. Voltage drop across is b. Voltage drop across is c. Voltage drop across is d. Voltage drop across is, and.... is connected in parallel to, what is the voltage drop across? Resistors in Parallel and in Series More complex connections of resistors are sometimes just combinations of series and parallel. Combinations of series and parallel resistors can be reduced to a single equivalent resistance by using the technique illustrated in Figure 19.17. Various parts are identified as either series or parallel, reduced to their equivalents, and further reduced until a single resistance is left. The Access for free at openstax.org. process is more time consuming than difficult. 19.3 • Parallel Circuits 625 Figure 19.17 This combination of seven resistors has both series and parallel parts. Each is identified and reduced to an equivalent resistance, and these are further reduced until a single equivalent resistance is reached. Let’s work through the four steps in Figure 19.17 to reduce the seven resistors to a single equivalent resistor. To avoid distracting. In step 1, we reduce the two sets of parallel resistors circled by the blue dashed loop. algebra, we’ll assume each resistor is 10 The upper set has three resistors in parallel and will be reduced to a single equivalent resistor resistors in parallel and will be reduced to a single equivalent resistor resistors in parallel, we obtain. Using the equation for the equivalent resistance of. The lower set has two These two equivalent resistances are encircled by the red dashed loop following step 1. They are in series, so we can use the 19.34 626 Chapter 19 • Electrical Circuits equation for the equivalent resistance of resistors in series to reduce them to a single equivalent resistance step 2, with the result being. This is done in The equivalent resistor pair can be replaced by the equivalent resistor, which is given by appears in the green dashed loop following step 2. This resistor is in parallel with resistor 19.35, so the 19.36 This is done in step 3. The resistor These two resistors are combined in the final step to form the final equivalent resistor, as shown in the purple dashed loop following step 3., which is is in series with the resistor 19.37 Thus, the entire combination of seven resistors may be replaced by a single resistor with a resistance of about 14.5. That was a lot of work, and you might be asking why we do it. It’s
important for us to know the equivalent resistance of the entire circuit so that we can calculate the current flowing through the circuit. Ohm’s law tells us that the current flowing through a circuit depends on the resistance of the circuit and the voltage across the circuit. But to know the current, we must first know the equivalent resistance. Here is a general approach to find the equivalent resistor for any arbitrary combination of resistors: Identify a group of resistors that are only in parallel or only in series. 1. 2. For resistors in series, use the equation for the equivalent resistance of resistors in series to reduce them to a single equivalent resistance. For resistors in parallel, use the equation for the equivalent resistance of resistors in parallel to reduce them to a single equivalent resistance. 3. Draw a new circuit diagram with the resistors from step 1 replaced by their equivalent resistor. 4. If more than one resistor remains in the circuit, return to step 1 and repeat. Otherwise, you are finished. FUN IN PHYSICS Robot Robots have captured our collective imagination for over a century. Now, this dream of creating clever machines to do our dirty work, or sometimes just to keep us company, is becoming a reality. Roboticshas become a huge field of research and development, with some technology already being commercialized. Think of the small autonomous vacuum cleaners, for example. Figure 19.18 shows just a few of the multitude of different forms robots can take. The most advanced humanoid robots can walk, pour drinks, even dance (albeit not very gracefully). Other robots are bio-inspired, such as the dogbotshown in the middle photograph of Figure 19.18. This robot can carry hundreds of pounds of load over rough terrain. The photograph on the right in Figure 19.18 shows the inner workings of an M-block,developed by the Massachusetts Institute of Technology. These simplelooking blocks contain inertial wheels and electromagnets that allow them to spin and flip into the air and snap together in a variety of shapes. By communicating wirelessly between themselves, they self-assemble into a variety of shapes, such as desks, chairs, and someday maybe even buildings. All robots involve an immense amount of physics and engineering. The simple act of pouring a drink has only recently been mastered by robots, after over 30 years of research and development! The balance and timing that we humans take for granted is in fact a very tricky act to follow, requiring excellent balance, dexterity, and feedback. To master this requires sensors
to detect balance, computing power to analyze the data and communicate the appropriate compensating actions, and joints and actuators to implement the required actions. In addition to sensing gravity or acceleration, robots can contain multiple different sensors to detect light, sound, temperature, smell, taste, etc. These devices are all based on the physical principles that you are studying in this text. For example, the optics used for robotic vision are similar to those used in your digital cameras: pixelated semiconducting detectors in which light is Access for free at openstax.org. converted into electrical signals. To detect temperature, simple thermistors may be used, which are resistors whose resistance changes depending on temperature. Building a robot today is much less arduous than it was a few years ago. Numerous companies now offer kits for building robots. These range in complexity something suitable for elementary school children to something that would challenge the best professional engineers. If interested, you may find these easily on the Internet and start making your own robot today. 19.3 • Parallel Circuits 627 Figure 19.18 Robots come in many shapes and sizes, from the classic humanoidtype to dogbotsto small cubes that self-assemble to perform a variety of tasks. WATCH PHYSICS Resistors in Parallel This video shows a lecturer discussing a simple circuit with a battery and a pair of resistors in parallel. He emphasizes that electrons flow in the direction opposite to that of the positive current and also makes use of the fact that the voltage is the same at all points on an ideal wire. The derivation is quite similar to what is done in this text, but the lecturer goes through it well, explaining each step. Click to view content (https://www.openstax.org/l/28resistors) GRASP CHECK True or false—In a circuit diagram, we can assume that the voltage is the same at every point in a given wire. a. b. false true WATCH PHYSICS Resistors in Series and in Parallel This video shows how to calculate the equivalent resistance of a circuit containing resistors in parallel and in series. The lecturer uses the same approach as outlined above for finding the equivalent resistance. Click to view content (https://www.openstax.org/l/28resistorssp) GRASP CHECK Imagine connected Nidentical resistors in parallel. Each resistor has a resistance of R. What is the equivalent resistance for this group of parallel resistors? a. The equivalent resistance is (
R)N. 628 Chapter 19 • Electrical Circuits b. The equivalent resistance is NR. c. The equivalent resistance is d. The equivalent resistance is WORKED EXAMPLE Find the Current through a Complex Resistor Circuit The battery in the circuit below has a voltage rating of 10 V. What current flows through the circuit and in what direction? STRATEGY Apply the strategy for finding equivalent resistance to replace all the resistors with a single equivalent resistance, then use Ohm’s law to find the current through the equivalent resistor. Solution The resistor combination and can be reduced to an equivalent resistance of 19.38 by 19.39 Replacing and with this equivalent resistance gives the circuit below. We now replace the two upper resistors their equivalent resistor and. These resistors are in series, so we add them together to find the equivalent resistance. and the two lower resistors by the equivalent resistor and Replacing the relevant resistors with their equivalent resistor gives the circuit below. Access for free at openstax.org. 19.3 • Parallel Circuits 629 Now replace the two resistors, which are in parallel, with their equivalent resistor. The resistance of is Updating the circuit diagram by replacing with this equivalent resistance gives the circuit below. Finally, we combine resistors, which are in series. The equivalent resistance is The final circuit is shown below. We now use Ohm’s law to find the current through the circuit. 19.40 19.41 The current goes from the positive terminal of the battery to the negative terminal of the battery, so it flows clockwise in this circuit. Discussion This calculation may seem rather long, but with a little practice, you can combine some steps. Note also that extra significant digits were carried through the calculation. Only at the end was the final result rounded to two significant digits. 630 Chapter 19 • Electrical Circuits WORKED EXAMPLE Strange-Looking Circuit Diagrams Occasionally, you may encounter circuit diagrams that are not drawn very neatly, such as the diagram shown below. This circuit diagram looks more like how a real circuit might appear on the lab bench. What is the equivalent resistance for the resistors in this diagram, assuming each resistor is 10 and the voltage rating of the battery is 12 V. STRATEGY Let’s redraw this circuit diagram to make it clearer. Then we’ll apply the strategy outlined above to calculate the equivalent resistance. Solution To redraw the diagram, consider the figure below. In the upper circuit
, the blue resistors constitute a path from the positive terminal of the battery to the negative terminal. In parallel with this circuit are the red resistors, which constitute another path from the positive to negative terminal of the battery. The blue and red paths are shown more cleanly drawn in the lower circuit diagram. Note that, in both the upper and lower circuit diagrams, the blue and red paths connect the positive terminal of the battery to the negative terminal of the battery. Now it is easier to see that turn is in parallel with the series combination of. The equivalent resistance is are in parallel, and the parallel combination is in series with. This combination in. First, we calculate the blue branch, which contains 19.42 where we show the contribution from the parallel combination of resistors and from the series combination of resistors. We now calculate the equivalent resistance of the red branch, which is 19.43 Inserting these equivalent resistors into the circuit gives the circuit below. Access for free at openstax.org. 19.3 • Parallel Circuits 631 These two resistors are in parallel, so they can be replaced by a single equivalent resistor with a resistance of 19.44 The final equivalent circuit is show below. Discussion Finding the equivalent resistance was easier with a clear circuit diagram. This is why we try to make clear circuit diagrams, where the resistors in parallel are lined up parallel to each other and at the same horizontal position on the diagram. We can now use Ohm’s law to find the current going through each branch to this circuit. Consider the circuit diagram with and is. The voltage across each of these branches is 12 V (i.e., the voltage rating of the battery). The current in the blue branch 19.45 19.46 The current across the red branch is The current going through the battery must be the sum of these two currents (can you see why?), or 1.4 A. Practice Problems 12. What is the formula for the equivalent resistance of two parallel resistors with resistance R1 and R2? a. Equivalent resistance of two parallel resistors b. Equivalent resistance of two parallel resistors c. Equivalent resistance of two parallel resistors d. Equivalent resistance of two parallel resistors 632 Chapter 19 • Electrical Circuits 13. Figure 19.19 What is the equivalent resistance for the two resistors shown below? a. The equivalent resistance is 20 Ω b. The equivalent resistance is 21 Ω c. The equivalent resistance is 90 Ω
d. The equivalent resistance is 1,925 Ω Check Your Understanding 14. The voltage drop across parallel resistors is ________. a. the same for all resistors b. greater for the larger resistors less for the larger resistors c. d. greater for the smaller resistors 15. Consider a circuit of parallel resistors. The smallest resistor is 25 Ω. What is the upper limit of the equivalent resistance? a. The upper limit of the equivalent resistance is 2.5 Ω. b. The upper limit of the equivalent resistance is 25 Ω. c. The upper limit of the equivalent resistance is 100 Ω. d. There is no upper limit. 19.4 Electric Power Section Learning Objectives By the end of this section, you will be able to do the following: • Define electric power and describe the electric power equation • Calculate electric power in circuits of resistors in series, parallel, and complex arrangements Section Key Terms electric power Power is associated by many people with electricity. Every day, we use electric power to run our modern appliances. Electric power transmission lines are visible examples of electricity providing power. We also use electric power to start our cars, to run our computers, or to light our homes. Power is the rate at which energy of any type is transferred; electric power is the rate at which electric energy is transferred in a circuit. In this section, we’ll learn not only what this means, but also what factors determine electric power. To get started, let’s think of light bulbs, which are often characterized in terms of their power ratings in watts. Let us compare a 25-W bulb with a 60-W bulb (see Figure 19.20). Although both operate at the same voltage, the 60-W bulb emits more light intensity than the 25-W bulb. This tells us that something other than voltage determines the power output of an electric circuit. Access for free at openstax.org. Incandescent light bulbs, such as the two shown in Figure 19.20, are essentially resistors that heat up when current flows through them and they get so hot that they emit visible and invisible light. Thus the two light bulbs in the photo can be considered as two different resistors. In a simple circuit such as a light bulb with a voltage applied to it, the resistance determines the current by Ohm’s law, so we can see that current as well as voltage must determine the power. 19.4 • Electric Power
633 Figure 19.20 On the left is a 25-W light bulb, and on the right is a 60-W light bulb. Why are their power outputs different despite their operating on the same voltage? The formula for power may be found by dimensional analysis. Consider the units of power. In the SI system, power is given in watts (W), which is energy per unit time, or J/s Recall now that a voltage is the potential energy per unit charge, which means that voltage has units of J/C We can rewrite this equation as and substitute this into the equation for watts to get 19.47 19.48 But a Coulomb per second (C/s) is an electric current, which we can see from the definition of electric current, Qis the charge in coulombs and tis time in seconds. Thus, equation above tells us that electric power is voltage times current, or, where This equation gives the electric power consumed by a circuit with a voltage drop of Vand a current of I. For example, consider the circuit in Figure 19.21. From Ohm’s law, the current running through the circuit is Thus, the power consumed by the circuit is Where does this power go? In this circuit, the power goes primarily into heating the resistor in this circuit. 19.49 19.50 Figure 19.21 A simple circuit that consumes electric power. In calculating the power in the circuit of Figure 19.21, we used the resistance and Ohm’s law to find the current. Ohm’s law gives the current:, which we can insert into the equation for electric power to obtain 634 Chapter 19 • Electrical Circuits This gives the power in terms of only the voltage and the resistance. We can also use Ohm’s law to eliminate the voltage in the equation for electric power and obtain an expression for power in terms of just the current and the resistance. If we write Ohm’s law as and use this to eliminate Vin the equation, we obtain This gives the power in terms of only the current and the resistance. Thus, by combining Ohm’s law with the equation terms of voltage and resistance and one in terms of current and resistance. Note that only resistance (not capacitance or anything else), current, and voltage enter into the expressions for electric power. This means that the physical characteristic of a circuit that determines how much power it dissipates is its resistance. Any capacitors in the circuit do not dissip
ate electric power—on the contrary, capacitors either store electric energy or release electric energy back to the circuit. for electric power, we obtain two more expressions for power: one in To clarify how voltage, resistance, current, and power are all related, consider Figure 19.22, which shows the formula wheel. The quantities in the center quarter circle are equal to the quantities in the corresponding outer quarter circle. For example, to express a potential V in terms of power and current, we see from the formula wheel that. Figure 19.22 The formula wheel shows how volts, resistance, current, and power are related. The quantities in the inner quarter circles equal the quantities in the corresponding outer quarter circles. WORKED EXAMPLE Find the Resistance of a Lightbulb A typical older incandescent lightbulb was 60 W. Assuming that 120 V is applied across the lightbulb, what is the current through the lightbulb? STRATEGY We are given the voltage and the power output of a simple circuit containing a lightbulb, so we can use the equation find the current Ithat flows through the lightbulb. to Solution Solving for the current and inserting the given values for voltage and power gives 19.51 Thus, a half ampere flows through the lightbulb when 120 V is applied across it. Access for free at openstax.org. 19.4 • Electric Power 635 Discussion This is a significant current. Recall that household power is AC and not DC, so the 120 V supplied by household sockets is an alternating power, not a constant power. The 120 V is actually the time-averaged power provided by such sockets. Thus, the average current going through the light bulb over a period of time longer than a few seconds is 0.50 A. WORKED EXAMPLE Boot Warmers To warm your boots on cold days, you decide to sew a circuit with some resistors into the insole of your boots. You want 10 W of heat output from the resistors in each insole, and you want to run them from two 9-V batteries (connected in series). What total resistance should you put in each insole? STRATEGY We know the desired power and the voltage (18 V, because we have two 9-V batteries connected in series), so we can use the equation to find the requisite resistance. Solution Solving for the resistance and inserting the given voltage and power, we obtain 19.52 Thus, the total resistance in
each insole should be 32 Discussion Let’s see how much current would run through this circuit. We have 18 V applied across a resistance of 32, so Ohm’s law gives 19.53 All batteries have labels that say how much charge they can deliver (in terms of a current multiplied by a time). A typical 9-V alkaline battery can deliver a charge of 565 ), so this heating system would function for a time of (so two 9 V batteries deliver 1,130 19.54 WORKED EXAMPLE Power through a Branch of a Circuit Each resistor in the circuit below is 30. What power is dissipated by the middle branch of the circuit? STRATEGY The middle branch of the circuit contains resistors the equivalent resistance in this branch, and then use in series. The voltage across this branch is 12 V. We will first find to find the power dissipated in the branch. Solution The equivalent resistance is circuit is. The power dissipated by the middle branch of the 636 Chapter 19 • Electrical Circuits Discussion Let’s see if energy is conserved in this circuit by comparing the power dissipated in the circuit to the power supplied by the battery. First, the equivalent resistance of the left branch is 19.55 The power through the left branch is 19.56 19.57 The right branch contains only, so the equivalent resistance is. The power through the right branch is The total power dissipated by the circuit is the sum of the powers dissipated in each branch. The power provided by the battery is 19.58 19.59 19.60 where Iis the total current flowing through the battery. We must therefore add up the currents going through each branch to obtain I. The branches contributes currents of The total current is and the power provided by the battery is 19.61 19.62 19.63 This is the same power as is dissipated in the resistors of the circuit, which shows that energy is conserved in this circuit. Practice Problems 16. What is the formula for the power dissipated in a resistor? a. The formula for the power dissipated in a resistor is b. The formula for the power dissipated in a resistor is c. The formula for the power dissipated in a resistor is P= IV. d. The formula for the power dissipated in a resistor is P= I2V. 17. What is the formula for power dissipated by a resistor given its resistance and the voltage across it? a. The formula for
the power dissipated in a resistor is b. The formula for the power dissipated in a resistor is c. The formula for the power dissipated in a resistor is d. The formula for the power dissipated in a resistor is Access for free at openstax.org. 19.4 • Electric Power 637 Check your Understanding 18. Which circuit elements dissipate power? a. b. c. d. capacitors inductors ideal switches resistors 19. Explain in words the equation for power dissipated by a given resistance. a. Electric power is proportional to current through the resistor multiplied by the square of the voltage across the resistor. b. Electric power is proportional to square of current through the resistor multiplied by the voltage across the resistor. c. Electric power is proportional to current through the resistor divided by the voltage across the resistor. d. Electric power is proportional to current through the resistor multiplied by the voltage across the resistor. 638 Chapter 19 • Key Terms KEY TERMS alternating current electric current whose direction alternates back and forth at regular intervals ampere unit for electric current; one ampere is one coulomb per second ( ) circuit diagram schematic drawing of an electrical circuit including all circuit elements, such as resistors, capacitors, batteries, and so on conventional current flows in the direction that a positive charge would flow if it could move same as the combined resistance of a group of resistors in parallel when a group of resistors are connected side by side, with the top ends of the resistors connected together by a wire and the bottom ends connected together by a different wire in series when elements in a circuit are connected one after the other in the same branch of the circuit nonohmic material that does not follow Ohm’s law Ohm’s law electric current is proportional to the voltage direct current electric current that flows in a single applied across a circuit or other path direction electric circuit physical network of paths through which electric current can flow electric current electric charge that is moving electric power in a circuit equivalent resistor rate at which electric energy is transferred resistance of a single resistor that is the ohmic material that obeys Ohm’s law resistance how much a circuit element opposes the passage of electric current; it appears as the constant of proportionality in Ohm’s law circuit element that provides a known resistance resistor steady state when the characteristics of a system do not change over time SECTION SUMMARY 19.1 Ohm's law 19.3 Parallel Circuits • Direct current
is constant over time; alternating current • The equivalent resistance of a group of Nidentical alternates smoothly back and forth over time. resistors Rconnected in parallel is R/N. • Electrical resistance causes materials to extract work • from the current that flows through them. In ohmic materials, voltage drop along a path is proportional to the current that runs through the path. 19.2 Series Circuits • Circuit diagrams are schematic representations of electric circuits. • Resistors in series are resistors that are connected head to tail. • The same current runs through all resistors in series; however, the voltage drop across each resistor can be different. • The voltage is the same at every point in a given wire. • Connecting resistors in parallel provides more paths for the current to go through, so the equivalent resistance is always less than the smallest resistance of the parallel resistors. • The same voltage drop occurs across all resistors in parallel; however, the current through each resistor can differ. 19.4 Electric Power • Electric power is dissipated in the resistances of a circuit. Capacitors do not dissipate electric power. • Electric power is proportional to the voltage and the current in a circuit. • Ohm’s law provides two extra expressions for electric power: one that does not involve current and one that does not involve voltage. KEY EQUATIONS 19.1 Ohm's law electric current Iis the charge passes a plane per unit time that an ampere is the coulombs per unit time that pass a plane Access for free at openstax.org. Ohm’s law: the current Iis proportional to the voltage V, with the resistance Rbeing the constant of proportionality 19.2 Series Circuits 19.4 Electric Power Chapter 19 • Chapter Review 639 the equivalent resistance of N resistors connected in series 19.3 Parallel Circuits the equivalent resistance of Nresistors connected in parallel CHAPTER REVIEW Concept Items 19.1 Ohm's law 1. You connect a resistor across a battery. In which direction do the electrons flow? a. The electrons flow from the negative terminal of the battery to the positive terminal of the battery. b. The electrons flow from the positive terminal of the battery to the negative terminal of the battery. 2. How does current depend on resistance in Ohm’s law? a. Current is directly proportional to the resistance. b. Current is inversely proportional to the resistance. c. Current is proportional to the square
of the resistance. d. Current is inversely proportional to the square of the resistance. 3. In the context of electricity, what is resistance? a. Resistance is the property of materials to resist the passage of voltage. b. Resistance is the property of materials to resist the passage of electric current. c. Resistance is the property of materials to increase the passage of voltage. d. Resistance is the property of materials to increase the passage of electric current. 4. What is the mathematical formula for Ohm’s law? a. b. c. d. 19.2 Series Circuits 5. In which circuit are all the resistors connected in series? for a given current Iflowing through a potential difference V, the electric power dissipated for a given current Iflowing through a resistance R, the electric power dissipated for a given voltage difference Vacross a resistor R, the electric power dissipated a. b. c. d. 6. What is the voltage and current through the capacitor in the circuit below a long time after the switch is closed? a. 0 V, 0 A b. 0 V, 10 A 10 V, 0 A c. 10 V, 10 A d. decreases the overall resistance. d. Adding resistors in parallel gives the current longer path through which it can flow hence decreases the overall resistance. 19.4 Electric Power 9. To draw the most power from a battery, should you connect a small or a large resistance across its terminals? Explain. a. Small resistance, because smaller resistance will lead to the largest power b. Large resistance, because smaller resistance will lead to the largest power 10. If you double the current through a resistor, by what factor does the power dissipated by the resistor change? a. Power increases by a factor of two. b. Power increases by a factor of four. c. Power increases by a factor of eight. d. Power increases by a factor of 16. material is ohmic. 19.2 Series Circuits 13. Given three batteries (5V, 9V, 12V) and five resistors (10, 20, 30, 40, 50Ω) to choose from, what can you choose to form a circuit diagram with a current of 0.175A? You do not need to use all of the components. a. Batteries (5V, 9V) and resistors (30Ω, 50Ω) connected in series b. Batteries (5V4,
12V) and resistors (10Ω, 20Ω, 40Ω, and 50Ω) connected in series. c. Batteries (5V, 9V, and 12V) and resistors (10Ω, 20Ω, and 30Ω) connected in series. 14. What is the maximum resistance possible given a and a resistor of? resistor of a. b. c. d. 15. Rank the points A, B, C, and D in the circuit diagram from lowest voltage to highest voltage. 640 Chapter 19 • Chapter Review 19.3 Parallel Circuits 7. If you remove resistance from a circuit, does the total resistance of the circuit always decrease? Explain. a. No, because for parallel combination of resistors, the resistance through the remaining circuit increases. b. Yes, because for parallel combination of resistors, the resistance through the remaining circuit increases. 8. Explain why the equivalent resistance of a parallel combination of resistors is always less than the smallest of the parallel resistors. a. Adding resistors in parallel gives the current a shorter path through which it can flow hence decreases the overall resistance. b. Adding resistors in parallel gives the current another path through which it can flow hence decreases the overall resistance. c. Adding resistors in parallel reduce the number of paths through which the current can flow hence Critical Thinking Items 19.1 Ohm's law 11. An accelerator accelerates He nuclei (change = 2e) to a speed of v= 2 × 106 m/s. What is the current if the linear density of He nuclei is λ= 108 m–1? a. I= 9.6 × 10–5 A b. I= 3.2 × 10–5 A c. I= 12.8 × 10–5 A d. I= 6.4 × 10–5 A 12. How can you verify whether a certain material is ohmic? a. Make a resistor from this material and measure the current going through this resistor for several different voltages. If the current is proportional to the voltage, then the material is ohmic. b. Make a resistor from this material and measure the current going through this resistor for several different voltages. If the current is inversely proportional to the voltage, then the material is ohmic. c. Make a resistor from this material and measure the current going through this resistor for several different voltages. If the current is proportional to the square of the
voltage, then the material is ohmic. d. Make a resistor from this material and measure the current going through this resistor for several different voltages. If the current is inversely proportional to the square of the voltage, then the Access for free at openstax.org. Chapter 19 • Chapter Review 641 a. The equivalent resistance of the circuit 14 Ω. b. The equivalent resistance of the circuit 16.7 Ω. c. The equivalent resistance of the circuit 140 Ω. d. The equivalent resistance of the circuit 195 Ω. 19.4 Electric Power 18. Two lamps have different resistances. (a) If the lamps are connected in parallel, which one is brighter, the lamp with greater resistance or the lamp with less resistance? (b) If the lamps are connected in series, which one is brighter? Note that the brighter lamp dissipates more power. a. (a) lamp with greater resistance; (b) lamp with less resistance (a) lamp with greater resistance; (b) lamp with greater resistance (a) lamp with less resistance; (b) lamp with less resistance (a) lamp with less resistance; (b) lamp with greater resistance b. c. d. 19. To measure the power consumed by your laptop computer, you place an ammeter (a device that measures electric current) in series with its DC power supply. When the screen is off, the computer draws 0.40 A of current. When the screen is on at full brightness, it draws 0.90 A of current. Knowing the DC power supply delivers 16 V, how much power is used by the screen? a. The power used by the screen is −8.0 W. b. The power used by the screen is 0.3 W. c. The power used by the screen is 3.2 W. d. The power used by the screen is 8.0 W. which of the following functions would be best to fit the data? Assume that a, b, and care nonzero constants adjusted to fit the data. a. b. c. d. 22. A battery of unknown voltage is attached across a resistor in series with. You add a second battery with so that the voltage across is now and measure of current through resistor. You add a third battery with series with the first two batteries so that the voltage of across current through?. What is the resistance of and measure is in a. A, B, C, D b. B,
C, A, D c. C, B, A, D d. D, A, B, C 19.3 Parallel Circuits 16. Can all resistor combinations be reduced to series and parallel combinations? a. No, all practical resistor circuits cannot be reduced to series and parallel combinations. b. Yes, all practical resistor circuits can be reduced to series and parallel combinations. 17. What is the equivalent resistance of the circuit shown below? Figure 19.23 Problems 19.1 Ohm's law 20. What voltage is needed to make 6 C of charge traverse a 100-Ω resistor in 1 min? a. The required voltage is 1 × 10−3 V. b. The required voltage is 10 V. c. The required voltage is 1,000 V. d. The required voltage is 10,000 V. 21. Resistors typically obey Ohm’s law at low currents, but show deviations at higher currents because of heating. Suppose you were to conduct an experiment measuring the voltage, V, across a resistor as a function of current, I, including currents whose deviations from Ohm’s law start to become apparent. For a data plot of Vversus I, 642 Chapter 19 • Chapter Review a. b. c. d. 19.2 Series Circuits 23. What is the voltage drop across two 80-Ω resistors connected in series with 0.15 A flowing through them? a. 12 V b. 24 V c. 36 V d. 48 V 24. In this circuit, the voltage drop across the upper resistor is 4.5 V. What is the battery voltage? a. 4.5V 7.5V b. 12V c. 18V d. 19.3 Parallel Circuits 25. What is the equivalent resistance of this circuit? Performance Task 19.4 Electric Power 29. 1. An incandescent light bulb (i.e., an old-fashioned light bulb with a little wire in it) 2. A lightbulb socket to hold the light bulb 3. A variable voltage source 4. An ammeter Procedure • Screw the lightbulb into its socket. Connect the Access for free at openstax.org. a. The equivalent resistance of the circuit is 32.7 Ω. b. The equivalent resistance of the circuit is 100 Ω. c. The equivalent resistance of the circuit is 327 Ω. d. The equivalent resistance of the circuit is 450 Ω. 26. What is the
equivalent resistance of the circuit shown below? a. The equivalent resistance is 25 Ω. b. The equivalent resistance is 50 Ω. c. The equivalent resistance is 75 Ω. d. The equivalent resistance is 100 Ω. 19.4 Electric Power 27. When 12 V are applied across a resistor, it dissipates 120 W of power. What is the current through the resistor? a. The current is 1,440 A. b. The current is 10 A. c. The current is 0.1 A. d. The current is 0.01 A. 28. Warming 1 g of water requires 1 J of energy per. How long would it take to warm 1 L of water from 20 to 40 °C if you immerse in the water a 1-kW resistor connected across a 9.0-V batteries aligned in series? a. 10 min b. 20 min c. 30 min d. 40 min positive terminal of the voltage source to the input of the ammeter. Connect the output of the ammeter to one connection of the socket. Connect the other connection of the socket to the negative terminal of the voltage source. Ensure that the voltage source is set to supply DC voltage and that the ammeter is set to measure DC amperes. The desired circuit is shown below. Chapter 19 • Test Prep 643 10 rows. Label the left column voltsand the right column current.Adjust the voltage source so that it supplies from between 1 and 10 volts DC. For each voltage, write the voltage in the volts column and the corresponding amperage measured by the ammeter in the current column. Make a plot of volts versus current, that is, a plot with volts on the vertical axis and current on the horizontal axis. Use this data and the plot to answer the following questions: 1. What is the resistance of the lightbulb? 2. What is the range of possible error in your result 3. for the resistance? In a single word, how would you describe the curve formed by the data points? • On a piece of paper, make a two-column table with TEST PREP Multiple Choice 19.1 Ohm's law 30. What are the SI units for electric current? a. a battery b. a capacitor the ground c. d. a switch a. b. c. d. 31. What is the SI unit for resistance? a. b. c. d. 32. The equivalent unit for an ohm is a ________.
a. V/A b. C/m c. d. V/s 33. You put DC across resistor and measure the current through it. With the same voltage across resistor, you measure twice the current. What is the ratio 1? a. b. c. 4 d. 2 19.2 Series Circuits 34. What does the circuit element shown represent? 35. How many 10-Ω resistors must be connected in series to make an equivalent resistance of 80 Ω? a. 80 b. 8 c. 20 d. 40 36. Which two circuit elements are represented in the circuit diagram? a. a battery connected in series with an inductor b. a capacitor connected in series with a resistor c. a resistor connected in series with a battery d. an inductor connected in series with a resistor 37. How much current will flow through a 10-V battery with a 100-Ω resistor connected across its terminals? a. 0.1 A 1.0 A b. c. 0 d. 1,000 A 19.3 Parallel Circuits 38. A 10-Ω resistor is connected in parallel to another resistor R. The equivalent resistance of the pair is 8 Ω. What is the resistance R? 10 Ω a. b. 20 Ω 30 Ω c. 644 Chapter 19 • Test Prep d. 40 Ω 39. Are the resistors shown connected in parallel or in series? Explain. a. The resistors are connected in parallel because the same current flows through all three resistors. b. The resistors are connected in parallel because different current flows through all three resistors. c. The resistors are connected in series because the same current flows through all three resistors. d. The resistors are connected in series because different current flows through all three resistors. 19.4 Electric Power 40. Which equation below for electric power is incorrect? a. b. Short Answer 19.1 Ohm's law 44. True or false—it is possible to produce nonzero DC current by adding together AC currents. a. b. false true 45. What type of current is used in cars? a. alternating current indirect current b. c. direct current d. straight current 46. If current were represented by, voltage by, and resistance by, what would the mathematical expression be for Ohm’s law? a. b. c. d. 47. Give a verbal expression for Ohm’s law. a.
Ohm’s law says that the current through a resistor equals the voltage across the resistor multiplied by the resistance of the resistor. b. Ohm’s law says that the voltage across a resistor equals the current through the resistor multiplied by the resistance of the resistor. c. Ohm’s law says that the resistance of the resistor equals the current through the resistor multiplied Access for free at openstax.org. c. d. 41. What power is dissipated in a circuit through which flows across a potential drop of? a. b. c. Voltage drop across is d.. 42. How does a resistor dissipate power? a. A resistor dissipates power in the form of heat. b. A resistor dissipates power in the form of sound. c. A resistor dissipates power in the form of light. d. A resistor dissipates power in the form of charge. 43. What power is dissipated in a circuit through which 0.12 A flows across a potential drop of 3.0 V? a. 0.36 W b. 0.011 W c. 5 V d. 2.5 W by the voltage across a resistor. d. Ohm’s law says that the voltage across a resistor equals the square of the current through the resistor multiplied by the resistance of the resistor. 48. What is the current through a 100-Ω resistor with 12 V across it? a. 0 b. 0.12 A c. 8.33 A d. 1,200 A 49. What resistance is required to produce from a battery? a. 1 b. c. 60 120 d. 19.2 Series Circuits 50. Given a circuit with one 9-V battery and with its negative terminal connected to ground. The two paths are connected to ground from the positive terminal: the right path with a 20-Ω and a 100-Ω resistor and the left path with a 50-Ω resistor. How much current will flow in the right branch? a. b. Chapter 19 • Test Prep 645 c. d. a. b. c. d. 51. Through which branch in the circuit below does the most current flow? 19.3 Parallel Circuits 54. Ten 100-Ω resistors are connected in series. How can you increase the total resistance of the circuit by about 40 percent? a. Adding two 10-Ω resistors increases the total resistance of the circuit by about 40 percent. b. Rem
oving two 10-Ω resistors increases the total resistance of the circuit by about 40 percent. c. Adding four 10-Ω resistors increases the total resistance of the circuit by about 40 percent. d. Removing four 10-Ω resistors increases the total resistance of the circuit by about 40 percent. 55. Two identical resistors are connected in parallel across the terminals of a battery. If you increase the resistance of one of the resistors, what happens to the current through and the voltage across the other resistor? a. The current and the voltage remain the same. b. The current decreases and the voltage remains the a. All of the current flows through the left branch due to the open switch. b. All of the current flows through the right branch due to the open switch in the left branch. c. All of the current flows through the middle branch due to the open switch in the left branch d. There will be no current in any branch of the circuit due to the open switch. 52. What current flows through the resistor in the same. circuit below? c. The current and the voltage increases. d. The current increases and the voltage remains the same. 56. a. b. c. d. 53. What is the equivalent resistance for the circuit below if and? In the circuit below, through which resistor(s) does the most current flow? Through which does the least flow? Explain. a. The most current flows through the 15-Ω resistor because all the current must pass through this resistor. b. The most current flows through the 20-Ω resistor because all the current must pass through this resistor. c. The most current flows through the 25-Ω resistor because it is the highest resistance. d. The same current flows through the all the resistor because all the current must pass through each of the resistors. 19.4 Electric Power 57. You want to increase the power dissipated in a circuit. 646 Chapter 19 • Test Prep You have the choice between doubling the current or doubling the resistance, with the voltage remaining constant. Which one would you choose? a. doubling the resistance b. doubling the current battery? a. The power dissipated is 2430 W. b. The power dissipated is 270 W. c. The power dissipated is 2.7 W. d. The power dissipated is 0.37 W. 58. You want to increase the power dissipated in a circuit. You have the choice between reducing the
voltage or reducing the resistance, with the current remaining constant. Which one would you choose? a. b. reduce the voltage to increase the power reduce the resistance to increase the power 59. What power is dissipated in the circuit consisting of 310-Ω resistors connected in series across a 9.0-V Extended Response 19.1 Ohm's law 61. Describe the relationship between current and charge. Include an explanation of how the direction of the current is defined. a. Electric current is the charge that passes through a conductor per unit time. The direction of the current is defined to be the direction in which positive charge would flow. b. Electric current is the charges that move in a conductor. The direction of the current is defined to be the direction in which positive charge would flow. c. Electric current is the charge that passes through a conductor per unit time. The direction of the current is defined to be the direction in which negative charge would flow. d. Electric current is the charges that move in a conductor. The direction of the current is defined to be the direction in which negative charge would flow. 62. What could cause Ohm’s law to break down? a. b. c. d. If small amount of current flows through a resistor, the resistor will heat up so much that it will change state, in violation of Ohm’s law. If excessive amount of current flows through a resistor, the resistor will heat up so much that it will change state, in violation of Ohm’s law. If small amount of current flows through a resistor, the resistor will not heat up so much and it will not change its state, in violation of Ohm’s law. If excessive amount of current flows through a resistor, the resistor will heat up so much that it will not change its state, in violation of Ohm’s law. 63. You connect a single resistor across a battery and find that flows through the circuit. You add Access for free at openstax.org. 60. What power is dissipated in a circuit consisting of three 10-Ω resistors connected in parallel across a 9.0-V battery? a. The power dissipated is 270 W. b. The power dissipated is 30 W. c. The power dissipated is 24 W. d. The power dissipated is 1/24 W. another resistor after the first resistor and find that flows through the circuit. If you have connected in a line
one after the other, what resistors would be their total resistance? a. b. c. d. 19.2 Series Circuits 64. Explain why the current is the same at all points in the circuit below. a. b. c. If the current were not constant, the mobile charges would bunch up in places, which means that the voltage would decrease at that point. A lower voltage at some point would push the current in the direction that further decreases the voltage. If the current were not constant, the mobile charges would bunch up in places, which means that the voltage would increase at that point. But a higher voltage at some point would push the current in the direction that decreases the voltage. If the current were not constant, the mobile charges would bunch up in places, which mean that the voltage would increase at that point. A higher voltage at some point would push the current in the direction that further increases the d. voltage. If the current were not constant, the mobile charges would bunch up in places, which mean that the voltage would decrease at that point. But a lower voltage at some point would push the current in the direction that increases the voltage. 65. What is the current through each resistor in the circuit? a. Current through resistors R1, R2, R3, and R4 is 0.48 A, 0.30 A, 1.2 A, and 0.24 A, respectively. b. Current through resistors R1, R2, R3, and R4 is 1200 A, 1920 A, 480 A, and 2400 A, respectively. c. Current through resistors R1, R2, R3, and is R4 2.08 A, 3.34 A, 0.833 A, and 4.17 A, respectively. d. The same amount of current, 0.096 A, flows through all of the resistors. 19.3 Parallel Circuits 66. In a house, a single incoming wire at a high potential with respect to the ground provides electric power. How are the appliances connected between this wire and the Chapter 19 • Test Prep 647 ground, in parallel or in series? Explain. a. The appliances are connected in parallel to provide different voltage differences across each appliance. b. The appliances are connected in parallel to provide the same voltage difference across each appliance. c. The appliances are connected in series to provide the same voltage difference across each appliance. d. The appliances are connected in series to provide
different voltage differences across each appliance. 19.4 Electric Power 67. A single resistor is connected across the terminals of a battery When you attach a second resistor in parallel with the first, does the power dissipated by the system change? a. No, the power dissipated remain same. b. Yes, the power dissipated increases. c. Yes, the power dissipated decreases. 68. In a flashlight, the batteries are normally connected in series. Why are they not connected in parallel? a. Batteries are connected in series for higher voltage and power output. b. Batteries are connected in series for lower voltage and power output. c. Batteries are connected in series so that power output is a much lower for the same amount of voltage. d. Batteries are connected in series to reduce the overall loss of energy from the circuit. 648 Chapter 19 • Test Prep Access for free at openstax.org. CHAPTER 20 Magnetism Figure 20.1 The magnificent spectacle of the Aurora Borealis, or northern lights, glows in the northern sky above Bear Lake near Eielson Air Force Base, Alaska. Shaped by Earth’s magnetic field, this light is produced by radiation spewed from solar storms. (credit: Senior Airman Joshua Strang, Flickr) Chapter Outline 20.1 Magnetic Fields, Field Lines, and Force 20.2 Motors, Generators, and Transformers 20.3 Electromagnetic Induction You may have encountered magnets for the first time as a small child playing with magnetic toys or INTRODUCTION refrigerator magnets. At the time, you likely noticed that two magnets that repulse each other will attract each other if you flip one of them around. The force that acts across the air gaps between magnets is the same force that creates wonders such as the Aurora Borealis. In fact, magnetic effects pervade our lives in myriad ways, from electric motors to medical imaging and computer memory. In this chapter, we introduce magnets and learn how they work and how magnetic fields and electric currents interact. 650 Chapter 20 • Magnetism 20.1 Magnetic Fields, Field Lines, and Force Section Learning Objectives By the end of this section, you will be able to do the following: • Summarize properties of magnets and describe how some nonmagnetic materials can become magnetized • Describe and interpret drawings of magnetic fields around permanent magnets and current-carrying wires • Calculate the magnitude and direction of magnetic force in a magnetic field and the force
on a current- carrying wire in a magnetic field Section Key Terms Curie temperature domain electromagnet electromagnetism ferromagnetic magnetic dipole magnetic field magnetic pole magnetized north pole permanent magnet right-hand rule solenoid south pole Magnets and Magnetization People have been aware of magnets and magnetism for thousands of years. The earliest records date back to ancient times, particularly in the region of Asia Minor called Magnesia—the name of this region is the source of words like magnet. Magnetic rocks found in Magnesia, which is now part of western Turkey, stimulated interest during ancient times. When humans first discovered magnetic rocks, they likely found that certain parts of these rocks attracted bits of iron or other magnetic rocks more strongly than other parts. These areas are called the polesof a magnet. A magnetic pole is the part of a magnet that exerts the strongest force on other magnets or magnetic material, such as iron. For example, the poles of the bar magnet shown in Figure 20.2 are where the paper clips are concentrated. Figure 20.2 A bar magnet with paper clips attracted to the two poles. If a bar magnet is suspended so that it rotates freely, one pole of the magnet will always turn toward the north, with the opposite pole facing south. This discovery led to the compass, which is simply a small, elongated magnet mounted so that it can rotate freely. An example of a compass is shown Figure 20.3. The pole of the magnet that orients northward is called the north pole, and the opposite pole of the magnet is called the south pole. Access for free at openstax.org. 20.1 • Magnetic Fields, Field Lines, and Force 651 Figure 20.3 A compass is an elongated magnet mounted in a device that allows the magnet to rotate freely. The discovery that one particular pole of a magnet orients northward, whereas the other pole orients southward allowed people to identify the north and south poles of any magnet. It was then noticed that the north poles of two different magnets repel each other, and likewise for the south poles. Conversely, the north pole of one magnet attracts the south pole of other magnets. This situation is analogous to that of electric charge, where like charges repel and unlike charges attract. In magnets, we simply replace charge with pole: Like poles repel and unlike poles attract. This is summarized in Figure 20.4, which shows how the force between magnets depends on their relative orientation. Figure 20.
4 Depending on their relative orientation, magnet poles will either attract each other or repel each other. Consider again the fact that the pole of a magnet that orients northward is called the north pole of the magnet. If unlike poles attract, then the magnetic pole of Earth that is close to the geographic North Pole must be a magnetic south pole! Likewise, the magnetic pole of Earth that is close to the geographic South Pole must be a magnetic north pole. This situation is depicted in Figure 20.5, in which Earth is represented as containing a giant internal bar magnet with its magnetic south pole at the geographic North Pole and vice versa. If we were to somehow suspend a giant bar magnet in space near Earth, then the north pole of the space magnet would be attracted to the south pole of Earth’s internal magnet. This is in essence what happens with a compass needle: Its magnetic north pole is attracted to the magnet south pole of Earth’s internal magnet. 652 Chapter 20 • Magnetism Figure 20.5 Earth can be thought of as containing a giant magnet running through its core. The magnetic south pole of Earth’s magnet is at the geographic North Pole, so the north pole of magnets is attracted to the North Pole, which is how the north pole of magnets got their name. Likewise, the south pole of magnets is attracted to the geographic South Pole of Earth. What happens if you cut a bar magnet in half? Do you obtain one magnet with two south poles and one magnet with two north poles? The answer is no: Each half of the bar magnet has a north pole and a south pole. You can even continue cutting each piece of the bar magnet in half, and you will always obtain a new, smaller magnet with two opposite poles. As shown in Figure 20.6, you can continue this process down to the atomic scale, and you will find that even the smallest particles that behave as magnets have two opposite poles. In fact, no experiment has ever found any object with a single magnetic pole, from the smallest subatomic particle such as electrons to the largest objects in the universe such as stars. Because magnets always have two poles, they are referred to as magnetic dipoles—dimeans two. Below, we will see that magnetic dipoles have properties that are analogous to electric dipoles. Figure 20.6 All magnets have two opposite poles, from the smallest, such as subatomic particles, to the largest, such as stars. WATCH PHYSICS Introduction to Magnetism This video provides
an interesting introduction to magnetism and discusses, in particular, how electrons around their atoms contribute to the magnetic effects that we observe. Click to view content (https://www.openstax.org/l/28_intro_magn) GRASP CHECK Toward which magnetic pole of Earth is the north pole of a compass needle attracted? Access for free at openstax.org. 20.1 • Magnetic Fields, Field Lines, and Force 653 a. The north pole of a compass needle is attracted to the north magnetic pole of Earth, which is located near the geographic North Pole of Earth. b. The north pole of a compass needle is attracted to the south magnetic pole of Earth, which is located near the geographic North Pole of Earth. c. The north pole of a compass needle is attracted to the north magnetic pole of Earth, which is located near the geographic South Pole of Earth. d. The north pole of a compass needle is attracted to the south magnetic pole of Earth, which is located near the geographic South Pole of Earth. Only certain materials, such as iron, cobalt, nickel, and gadolinium, exhibit strong magnetic effects. Such materials are called ferromagnetic, after the Latin word ferrumfor iron. Other materials exhibit weak magnetic effects, which are detectable only with sensitive instruments. Not only do ferromagnetic materials respond strongly to magnets—the way iron is attracted to magnets—but they can also be magnetized themselves—that is, they can be induced to be magnetic or made into permanent magnets (Figure 20.7). A permanent magnet is simply a material that retains its magnetic behavior for a long time, even when exposed to demagnetizing influences. Figure 20.7 An unmagnetized piece of iron is placed between two magnets, heated, and then cooled, or simply tapped when cold. The iron becomes a permanent magnet with the poles aligned as shown: Its south pole is adjacent to the north pole of the original magnet, and its north pole is adjacent to the south pole of the original magnet. Note that attractive forces are created between the central magnet and the outer magnets. When a magnet is brought near a previously unmagnetized ferromagnetic material, it causes local magnetization of the material with unlike poles closest, as in the right side of Figure 20.7. This causes an attractive force, which is why unmagnetized iron is attracted to a magnet. What happens on a microscopic scale is illustrated in Figure 7(a).
Regions within the material called domains act like small bar magnets. Within domains, the magnetic poles of individual atoms are aligned. Each atom acts like a tiny bar magnet. Domains are small and randomly oriented in an unmagnetized ferromagnetic object. In response to an external magnetic field, the domains may grow to millimeter size, aligning themselves, as shown in Figure 7(b). This induced magnetization can be made permanent if the material is heated and then cooled, or simply tapped in the presence of other magnets. Figure 20.8 (a) An unmagnetized piece of iron—or other ferromagnetic material—has randomly oriented domains. (b) When magnetized by an external magnet, the domains show greater alignment, and some grow at the expense of others. Individual atoms are aligned within 654 Chapter 20 • Magnetism domains; each atom acts like a tiny bar magnet. Conversely, a permanent magnet can be demagnetized by hard blows or by heating it in the absence of another magnet. Increased thermal motion at higher temperature can disrupt and randomize the orientation and size of the domains. There is a well-defined temperature for ferromagnetic materials, which is called the Curie temperature, above which they cannot be magnetized. The Curie temperature for iron is 1,043 K (770 elements and alloys that have Curie temperatures much lower than room temperature and are ferromagnetic only below those temperatures. ), which is well above room temperature. There are several Snap Lab Refrigerator Magnets We know that like magnetic poles repel and unlike poles attract. See if you can show this for two refrigerator magnets. Will the magnets stick if you turn them over? Why do they stick to the refrigerator door anyway? What can you say about the magnetic properties of the refrigerator door near the magnet? Do refrigerator magnets stick to metal or plastic spoons? Do they stick to all types of metal? GRASP CHECK You have one magnet with the north and south poles labeled. How can you use this magnet to identify the north and south poles of other magnets? a. If the north pole of a known magnet is repelled by a pole of an unknown magnet on bringing them closer, that pole of unknown magnet is its north pole; otherwise, it is its south pole. If the north pole of known magnet is attracted to a pole of an unknown magnet on bringing them closer, that pole of unknown magnet is its north pole; otherwise, it is its south pole. b.
Magnetic Fields We have thus seen that forces can be applied between magnets and between magnets and ferromagnetic materials without any contact between the objects. This is reminiscent of electric forces, which also act over distances. Electric forces are described using the concept of the electric field, which is a force field around electric charges that describes the force on any other charge placed in the field. Likewise, a magnet creates a magnetic field around it that describes the force exerted on other magnets placed in the field. As with electric fields, the pictorial representation of magnetic field lines is very useful for visualizing the strength and direction of the magnetic field. As shown in Figure 20.9, the direction of magnetic field lines is defined to be the direction in which the north pole of a compass needle points. If you place a compass near the north pole of a magnet, the north pole of the compass needle will be repelled and point away from the magnet. Thus, the magnetic field lines point away from the north pole of a magnet and toward its south pole. Figure 20.9 The black lines represent the magnetic field lines of a bar magnet. The field lines point in the direction that the north pole of a small compass would point, as shown at left. Magnetic field lines never stop, so the field lines actually penetrate the magnet to form complete loops, as shown at right. Access for free at openstax.org. 20.1 • Magnetic Fields, Field Lines, and Force 655 Magnetic field lines can be mapped out using a small compass. The compass is moved from point to point around a magnet, and at each point, a short line is drawn in the direction of the needle, as shown in Figure 20.10. Joining the lines together then reveals the path of the magnetic field line. Another way to visualize magnetic field lines is to sprinkle iron filings around a magnet. The filings will orient themselves along the magnetic field lines, forming a pattern such as that shown on the right in Figure 20.10. Virtual Physics Using a Compass to Map Out the Magnetic Field Click to view content (http://www.openstax.org/l/28magcomp) This simulation presents you with a bar magnet and a small compass. Begin by dragging the compass around the bar magnet to see in which direction the magnetic field points. Note that the strength of the magnetic field is represented by the brightness of the magnetic field icons in the grid pattern around the magnet. Use the magnetic field meter to check the field strength at several points around the bar magnet. You
can also flip the polarity of the magnet, or place Earth on the image to see how the compass orients itself. GRASP CHECK With the slider at the top right of the simulation window, set the magnetic field strength to 100 percent. Now use the magnetic field meter to answer the following question: Near the magnet, where is the magnetic field strongest and where is it weakest? Don’t forget to check inside the bar magnet. a. The magnetic field is strongest at the center and weakest between the two poles just outside the bar magnet. The magnetic field lines are densest at the center and least dense between the two poles just outside the bar magnet. b. The magnetic field is strongest at the center and weakest between the two poles just outside the bar magnet. The magnetic field lines are least dense at the center and densest between the two poles just outside the bar magnet. c. The magnetic field is weakest at the center and strongest between the two poles just outside the bar magnet. The magnetic field lines are densest at the center and least dense between the two poles just outside the bar magnet. d. The magnetic field is weakest at the center and strongest between the two poles just outside the bar magnet and the magnetic field lines are least dense at the center and densest between the two poles just outside the bar magnet. Figure 20.10 Magnetic field lines can be drawn by moving a small compass from point to point around a magnet. At each point, draw a short line in the direction of the compass needle. Joining the points together reveals the path of the magnetic field lines. Another way to visualize magnetic field lines is to sprinkle iron filings around a magnet, as shown at right. When two magnets are brought close together, the magnetic field lines are perturbed, just as happens for electric field lines when two electric charges are brought together. Bringing two north poles together—or two south poles—will cause a repulsion, and the magnetic field lines will bend away from each other. This is shown in Figure 20.11, which shows the magnetic field lines created by the two closely separated north poles of a bar magnet. When opposite poles of two magnets are brought together, the 656 Chapter 20 • Magnetism magnetic field lines join together and become denser between the poles. This situation is shown in Figure 20.11. Figure 20.11 (a) When two north poles are approached together, the magnetic field lines repel each other and the two magnets experience a repulsive force. The same occurs if
two south poles are approached together. (b) If opposite poles are approached together, the magnetic field lines become denser between the poles and the magnets experience an attractive force. Like the electric field, the magnetic field is stronger where the lines are denser. Thus, between the two north poles in Figure 20.11, the magnetic field is very weak because the density of the magnetic field is almost zero. A compass placed at that point would essentially spin freely if we ignore Earth’s magnetic field. Conversely, the magnetic field lines between the north and south poles in Figure 20.11 are very dense, indicating that the magnetic field is very strong in this region. A compass placed here would quickly align with the magnetic field and point toward the south pole on the right. Note that magnets are not the only things that make magnetic fields. Early in the nineteenth century, people discovered that electrical currents cause magnetic effects. The first significant observation was by the Danish scientist Hans Christian Oersted (1777–1851), who found that a compass needle was deflected by a current-carrying wire. This was the first significant evidence that the movement of electric charges had any connection with magnets. An electromagnet is a device that uses electric current to make a magnetic field. These temporarily induced magnets are called electromagnets. Electromagnets are employed for everything from a wrecking yard crane that lifts scrapped cars to controlling the beam of a 90-km-circumference particle accelerator to the magnets in medical-imaging machines (see Figure 20.12). Figure 20.12 Instrument for magnetic resonance imaging (MRI). The device uses a cylindrical-coil electromagnet to produce for the main magnetic field. The patient goes into the tunnelon the gurney. (credit: Bill McChesney, Flickr) The magnetic field created by an electric current in a long straight wire is shown in Figure 20.13. The magnetic field lines form concentric circles around the wire. The direction of the magnetic field can be determined using the right-hand rule. This rule shows up in several places in the study of electricity and magnetism. Applied to a straight current-carrying wire, the right-hand rule says that, with your right thumb pointed in the direction of the current, the magnetic field will be in the direction in which your right fingers curl, as shown in Figure 20.13. If the wire is very long compared to the distance rfrom the wire, the strength B of the magnetic
field is given by where Iis the current in the wire in amperes. The SI unit for magnetic field is the tesla (T). The symbol —read “mu-zero”—is a constant called the “permeability of free space” and is given by 20.2 20.1 Access for free at openstax.org. 20.1 • Magnetic Fields, Field Lines, and Force 657 Figure 20.13 This image shows how to use the right-hand rule to determine the direction of the magnetic field created by current flowing through a straight wire. Point your right thumb in the direction of the current, and the magnetic field will be in the direction in which your fingers curl. WATCH PHYSICS Magnetic Field Due to an Electric Current This video describes the magnetic field created by a straight current-carrying wire. It goes over the right-hand rule to determine the direction of the magnetic field, and presents and discusses the formula for the strength of the magnetic field due to a straight current-carrying wire. Click to view content (https://www.openstax.org/l/28magfield) GRASP CHECK A long straight wire is placed on a table top and electric current flows through the wire from right to left. If you look at the wire end-on from the left end, does the magnetic field go clockwise or counterclockwise? a. By pointing your right-hand thumb in the direction opposite of current, the right-hand fingers will curl counterclockwise, so the magnetic field will be in the counterclockwise direction. b. By pointing your right-hand thumb in the direction opposite of current, the right-hand fingers will curl clockwise, so the magnetic field will be in the clockwise direction. c. By pointing your right-hand thumb in the direction of current, the right-hand fingers will curl counterclockwise, so the magnetic field will be in the counterclockwise direction. d. By pointing your right-hand thumb in the direction of current, the right-hand fingers will curl clockwise, so the magnetic field will be in the clockwise direction. Now imagine winding a wire around a cylinder with the cylinder then removed. The result is a wire coil, as shown in Figure 20.14. This is called a solenoid. To find the direction of the magnetic field produced by a solenoid, apply the right-hand rule to several points on the coil. You
should be able to convince yourself that, inside the coil, the magnetic field points from left to right. In fact, another application of the right-hand rule is to curl your right-hand fingers around the coil in the direction in which the current flows. Your right thumb then points in the direction of the magnetic field inside the coil: left to right in this case. Figure 20.14 A wire coil with current running through as shown produces a magnetic field in the direction of the red arrow. Each loop of wire contributes to the magnetic field inside the solenoid. Because the magnetic field lines must form closed loops, the field lines close the loop outside the solenoid. The magnetic field lines are much denser inside the solenoid than outside the solenoid. The resulting magnetic field looks very much like that of a bar magnet, as shown in Figure 20.15. The magnetic field strength deep inside a solenoid is 658 Chapter 20 • Magnetism where Nis the number of wire loops in the solenoid and is the length of the solenoid. 20.3 Figure 20.15 Iron filings show the magnetic field pattern around (a) a solenoid and (b) a bar magnet. The fields patterns are very similar, especially near the ends of the solenoid and bar magnet. Virtual Physics Electromagnets Click to view content (http://www.openstax.org/l/28elec_magnet) Use this simulation to visualize the magnetic field made from a solenoid. Be sure to click on the tab that says Electromagnet. You can drive AC or DC current through the solenoid by choosing the appropriate current source. Use the field meter to measure the strength of the magnetic field and then change the number of loops in the solenoid to see how this affects the magnetic field strength. GRASP CHECK Choose the battery as current source and set the number of wire loops to four. With a nonzero current going through the solenoid, measure the magnetic field strength at a point. Now decrease the number of wire loops to two. How does the magnetic field strength change at the point you chose? a. There will be no change in magnetic field strength when number of loops reduces from four to two. b. The magnetic field strength decreases to half of its initial value when number of loops reduces from four to two. c. The magnetic field strength increases to twice of its initial value when number of loops reduces from four
to two. d. The magnetic field strength increases to four times of its initial value when number of loops reduces from four to two. Magnetic Force If a moving electric charge, that is electric current, produces a magnetic field that can exert a force on another magnet, then the reverse should be true by Newton’s third law. In other words, a charge moving through the magnetic field produced by another object should experience a force—and this is exactly what we find. As a concrete example, consider Figure 20.16, which shows a Access for free at openstax.org. 20.1 • Magnetic Fields, Field Lines, and Force 659 charge qmoving with velocity force experienced by this charge is through a magnetic field between the poles of a permanent magnet. The magnitude Fof the where is the angle between the velocity of the charge and the magnetic field. The direction of the force may be found by using another version of the right-hand rule: First, we join the tails of the velocity vector and a magnetic field vector, as shown in step 1 of Figure 20.16. We then curl our right fingers from to in step (2) of Figure 20.16. The direction in which the right thumb points is the direction of the force. For the charge in Figure 20.16, we find that the force is directed into the page., as indicated 20.4 Note that the factor magnetic field because the magnetic field, because in the equation and and means that zero force is applied on a charge that moves parallel to a. The maximum force a charge can experience is when it moves perpendicular to Figure 20.16 (a) An electron moves through a uniform magnetic field. (b) Using the right-hand rule, the force on the electron is found to be directed into the page. LINKS TO PHYSICS Magnetohydrodynamic Drive In Tom Clancy’s Cold War novel “The Hunt for Red October,” the Soviet Union built a submarine (see Figure 20.17) with a magnetohydrodynamic drive that was so silent it could not be detected by surface ships. The only conceivable purpose to build such a submarine was to give the Soviet Union first-strike capability, because this submarine could sneak close to the coast of the United States and fire its ballistic missiles, destroying key military and government installations to prevent an American counterattack. Figure 20.17 A Typhoon-class Russian ballistic-missile submarine on which the fictional submarine Red October was based. A magnetohydrodynamic
drive is supposed to be silent because it has no moving parts. Instead, it uses the force experienced by charged particles that move in a magnetic field. The basic idea behind such a drive is depicted in Figure 20.18. Salt water flows through a channel that runs from the front to the back of the submarine. A magnetic field is applied horizontally across the channel, and a voltage is applied across the electrodes on the top and bottom of the channel to force a downward electric current through the water. The charge carriers are the positive sodium ions and the negative chlorine ions of salt. Using the right-hand 660 Chapter 20 • Magnetism rule, the force on the charge carriers is found to be toward the rear of the vessel. The accelerated charges collide with water molecules and transfer their momentum, creating a jet of water that is propelled out the rear of the channel. By Newton’s third law, the vessel experiences a force of equal magnitude, but in the opposite direction. Figure 20.18 A schematic drawing of a magnetohydrodynamic drive showing the water channel, the current direction, the magnetic field direction, and the resulting force. Fortunately for all involved, it turns out that such a propulsion system is not very practical. Some back-of-the-envelope calculations show that, to power a submarine, either extraordinarily high magnetic fields or extraordinarily high electric currents would be required to obtain a reasonable thrust. In addition, prototypes of magnetohydrodynamic drives show that they are anything but silent. Electrolysis caused by running a current through salt water creates bubbles of hydrogen and oxygen, which makes this propulsion system quite noisy. The system also leaves a trail of chloride ions and metal chlorides that can easily be detected to locate the submarine. Finally, the chloride ions are extremely reactive and very quickly corrode metal parts, such as the electrode or the water channel itself. Thus, the Red October remains in the realm of fiction, but the physics involved is quite real. GRASP CHECK If the magnetic field is downward, in what direction must the current flow to obtain rearward-pointing force? a. The current must flow vertically from up to down when viewed from the rear of the boat. b. The current must flow vertically from down to up when viewed from the rear of the boat. c. The current must flow horizontally from left to right when viewed from the rear of the boat. d. The current must flow horizontally from right to left when viewed from the rear of the boat. Instead of a single charge moving
through a magnetic field, consider now a steady current Imoving through a straight wire. If we place this wire in a uniform magnetic field, as shown in Figure 20.19, what is the force on the wire or, more precisely, on the electrons in the wire? An electric current involves charges that move. If the charges qmove a distance speed is Inserting this into the equation in a time t, then their gives The factor q/tin this equation is nothing more than the current in the wire. Thus, using, we obtain 20.5 20.6 This equation gives the force on a straight current-carrying wire of length angle between the current vector and the magnetic field vector. Note that for which as shown in Figure 20.19. in a magnetic field of strength B. The angle is the is the length of wire that is in the magnetic field and The direction of the force is determined in the same way as for a single charge. Curl your right fingers from the vector for Ito the vector for B, and your right thumb will point in the direction of the force on the wire. For the wire shown in Figure 20.19, the force is directed into the page. Access for free at openstax.org. 20.1 • Magnetic Fields, Field Lines, and Force 661 Figure 20.19 A straight wire carrying current Iin a magnetic field B. The force exerted on the wire is directed into the page. The length is the length of the wire that is inthe magnetic field. Throughout this section, you may have noticed the symmetries between magnetic effects and electric effects. These effects all fall under the umbrella of electromagnetism, which is the study of electric and magnetic phenomena. We have seen that electric charges produce electric fields, and moving electric charges produce magnetic fields. A magnetic dipole produces a magnetic field, and, as we will see in the next section, moving magnetic dipoles produce an electric field. Thus, electricity and magnetism are two intimately related and symmetric phenomena. WORKED EXAMPLE Trajectory of Electron in Magnetic Field A proton enters a region of constant magnetic field, as shown in Figure 20.20. The magnetic field is coming out of the page. If the electron is moving at and the magnetic field strength is 2.0 T, what is the magnitude and direction of the force on the proton? Figure 20.20 A proton enters a region of uniform magnetic field. The magnetic field is coming
out of the page—the circles with dots represent vector arrow heads coming out of the page. STRATEGY Use the equation and the velocity vector of the proton is Solution The charge of the proton is the equation gives to find the magnitude of the force on the proton. The angle between the magnetic field vectors The direction of the force may be found by using the right-hand rule.. Entering this value and the given velocity and magnetic field strength into 20.7 To find the direction of the force, first join the velocity vector end to end with the magnetic field vector, as shown in Figure 20.21. Now place your right hand so that your fingers point in the direction of the velocity and curl them upward toward the magnetic field vector. The force is in the direction in which your thumb points. In this case, the force is downward in the plane of the paper in the -direction, as shown in Figure 20.21. 662 Chapter 20 • Magnetism Figure 20.21 The velocity vector and a magnetic field vector from Figure 20.20 are placed end to end. A right hand is shown with the fingers curling up from the velocity vector toward the magnetic field vector. The thumb points in the direction of the resulting force, which is the -direction in this case. Thus, combining the magnitude and the direction, we find that the force on the proton is Discussion This seems like a very small force. However, the proton has a mass of, so its acceleration is, or about ten thousand billion times the acceleration due to gravity! We found that the proton’s initial acceleration as it enters the magnetic field is downward in the plane of the page. Notice that, as the proton accelerates, its velocity remains perpendicular to the magnetic field, so the magnitude of the force does not change. In addition, because of the right-hand rule, the direction of the force remains perpendicular to the velocity. This force is nothing more than a centripetal force: It has a constant magnitude and is always perpendicular to the velocity. Thus, the magnitude of the velocity does not change, and the proton executes circular motion. The radius of this circle may be found by using the kinematics relationship. 20.8 The path of the proton in the magnetic field is shown in Figure 20.22. Figure 20.22 When traveling perpendicular to a constant magnetic field, a charged particle will execute circular motion, as shown here for a proton. WORKED EXAMPLE
Wire with Current in Magnetic Field Now suppose we run a wire through the uniform magnetic field from the previous example, as shown. If the wire carries a current of 1.0 A in the -direction, and the region with magnetic field is 4.0 cm long, what is the force on the wire? Access for free at openstax.org. 20.1 • Magnetic Fields, Field Lines, and Force 663 STRATEGY to find the magnitude of the force on the wire. The length of the wire inside the magnetic field is Use equation 4.0 cm, and the angle between the current direction and the magnetic field direction is 90°. To find the direction of the force, use the right-hand rule as explained just after the equation Solution Insert the given values into equation to find the magnitude of the force 20.9 To find the direction of the force, begin by placing the current vector end to end with a vector for the magnetic field. The result is as shown in the figure in the previous Worked Example with and your right thumb points down the page, again as shown in the figure in the previous Worked Example. Thus, the direction of the force is in the. Curl your right-hand fingers from to -direction. The complete force is thus replaced by. Discussion The direction of the force is the same as the initial direction of the force was in the previous example for a proton. However, because the current in a wire is confined to a wire, the direction in which the charges move does not change. Instead, the entire wire accelerates in the -direction. The force on a current-carrying wire in a magnetic field is the basis of all electrical motors, as we will see in the upcoming sections. Practice Problems 1. What is the magnitude of the force on an electron moving at 1.0 × 106 m/s perpendicular to a 1.0-T magnetic field? a. 0.8 × 10–13 N 1.6 × 10–14 N b. c. 0.8 × 10–14 N 1.6 × 10–13 N d. 2. A straight 10 cm wire carries 0.40 A and is oriented perpendicular to a magnetic field. If the force on the wire is 0.022 N, what is the magnitude of the magnetic field? 1.10 × 10–2 T a. b. 0.55 × 10–2 T c. 1.10 T d. 0.55 T Check Your Understanding 3. If
two magnets repel each other, what can you conclude about their relative orientation? a. Either the south pole of magnet 1 is closer to the north pole of magnet 2 or the north pole of magnet 1 is closer to the south pole of magnet 2. b. Either the south poles of both the magnet 1 and magnet 2 are closer to each other or the north poles of both the magnet 1 664 Chapter 20 • Magnetism and magnet 2 are closer to each other. 4. Describe methods to demagnetize a ferromagnet. a. by cooling, heating, or submerging in water b. by heating, hammering, and spinning it in external magnetic field c. by hammering, heating, and rubbing with cloth d. by cooling, submerging in water, or rubbing with cloth 5. What is a magnetic field? a. The directional lines present inside and outside the magnetic material that indicate the magnitude and direction of the magnetic force. b. The directional lines present inside and outside the magnetic material that indicate the magnitude of the magnetic force. c. The directional lines present inside the magnetic material that indicate the magnitude and the direction of the magnetic force. d. The directional lines present outside the magnetic material that indicate the magnitude and the direction of the magnetic force. 6. Which of the following drawings is correct? a. b. c. d. Access for free at openstax.org. 20.2 • Motors, Generators, and Transformers 665 20.2 Motors, Generators, and Transformers Section Learning Objectives By the end of this section, you will be able to do the following: • Explain how electric motors, generators, and transformers work • Explain how commercial electric power is produced, transmitted, and distributed Section Key Terms electric motor generator transformer Electric Motors, Generators, and Transformers As we learned previously, a current-carrying wire in a magnetic field experiences a force—recall motors, which convert electrical energy into mechanical energy, are the most common application of magnetic force on currentcarrying wires. Motors consist of loops of wire in a magnetic field. When current is passed through the loops, the magnetic field exerts a torque on the loops, which rotates a shaft. Electrical energy is converted to mechanical work in the process. Figure 20.23 shows a schematic drawing of an electric motor.. Electric Figure 20.23 Torque on a current loop. A vertical loop of wire in a horizontal magnetic field is attached to a vertical shaft. When current is passed through
the wire loop, torque is exerted on it, making it turn the shaft. Let us examine the force on each segment of the loop in Figure 20.23 to find the torques produced about the axis of the vertical shaft—this will lead to a useful equation for the torque on the loop. We take the magnetic field to be uniform over the rectangular loop, which has width wand height loop. To determine the direction of the force, we use the right-hand rule. The current goes from left to right into the page, and the magnetic field goes from left to right in the plane of the page. Curl your right fingers from the current vector to the magnetic field vector and your right thumb points down. Thus, the force on the top segment is downward, which produces no torque on the shaft. Repeating this analysis for the bottom segment—neglect the small gap where the lead wires go out—shows that the force on the bottom segment is upward, again producing no torque on the shaft. as shown in the figure. First, consider the force on the top segment of the Consider now the left vertical segment of the loop. Again using the right-hand rule, we find that the force exerted on this segment is perpendicular to the magnetic field, as shown in Figure 20.23. This force produces a torque on the shaft. Repeating this analysis on the right vertical segment of the loop shows that the force on this segment is in the direction opposite that of the force on the left segment, thereby producing an equal torque on the shaft. The total torque on the shaft is thus twice the toque on one of the vertical segments of the loop. To find the magnitude of the torque as the wire loop spins, consider Figure 20.24, which shows a view of the wire loop from where Fis the applied force, ris the distance from the pivot to where the above. Recall that torque is defined as force is applied, and θis the angle between rand F. Notice that, as the loop spins, the current in the vertical loop segments is always perpendicular to the magnetic field. Thus, the equation segment as The distance rfrom the shaft to where this force is applied is w/2, so the torque created by this force is gives the magnitude of the force on each vertical Because there are two vertical segments, the total torque is twice this, or 20.10 20.11 If we have a multiple loop with Nturns, we get Ntimes the torque of a single loop. Using
the fact that the area of the loop is the expression for the torque becomes 666 Chapter 20 • Magnetism This is the torque on a current-carrying loop in a uniform magnetic field. This equation can be shown to be valid for a loop of any shape. 20.12 Figure 20.24 View from above of the wire loop from Figure 20.23. The magnetic field generates a force Fon each vertical segment of the wire loop, which generates a torque on the shaft. Notice that the currents have the same magnitude because they both represent the current flowing in the wire loop, but flows into the page and flows out of the page. From the equation a maximum positive torque of to From will oscillate back and forth. we see that the torque is zero when when The torque then decreases back to zero as the wire loop rotates to the torque is negative. Thus, the torque changes sign every half turn, so the wire loop As the wire loop rotates, the torque increases to For the coil to continue rotating in the same direction, the current is reversed as the coil passes through using automatic switches called brushes, as shown in Figure 20.25. Figure 20.25 (a) As the angular momentum of the coil carries it through the brushes reverse the current and the torque remains clockwise. (b) The coil rotates continuously in the clockwise direction, with the current reversing each half revolution to maintain the clockwise torque. Consider now what happens if we run the motor in reverse; that is, we attach a handle to the shaft and mechanically force the coil to rotate within the magnetic field, as shown in Figure 20.26. As per the equation is the angle —where between the vectors and in the wires of the loop experience a magnetic force because they are moving in a magnetic field. Again using the right-hand rule, where we curl our fingers from vector the top and bottom segments feel a force perpendicular to the wire, which does not cause a current. However, charges in the vertical wires experience forces parallel to the wire, causing a current to flow through the wire and through an external circuit if one is connected. A device such as this that converts mechanical energy into electrical energy is called a generator., we find that charges in to vector Access for free at openstax.org. 20.2 • Motors, Generators, and Transformers 667 Figure 20.26 When this coil is rotated through one-fourth of a revolution, the magnetic flux Φ changes from its maximum to zero, inducing an em
f, which drives a current through an external circuit. Because current is induced only in the side wires, we can find the induced emf by only considering these wires. As explained in Induced Current in a Wire, motional emf in a straight wire moving at velocity vthrough a magnetic field Bis where the velocity is perpendicular to the magnetic field. In the generator, the velocity makes an angle with B(see Figure 20.27), so the velocity component perpendicular to Bis Thus, in this case, the emf induced on each vertical wire segment is and they are in the same direction. The total emf around the loop is then 20.13 Although this expression is valid, it does not give the emf as a function of time. To find how the emf evolves in time, we assume that the coil is rotated at a constant angular velocity The angle is related to the angular velocity by so that Recall that tangential velocity vis related to angular velocity by Here,, so that and Noting that the area of the loop is and allowing for Nwire loops, we find that 20.14 20.15 20.16 is the emf induced in a generator coil of Nturns and area Arotating at a constant angular velocity B. This can also be expressed as in a uniform magnetic field where is the maximum (peak) emf. 20.17 20.18 668 Chapter 20 • Magnetism Figure 20.27 The instantaneous velocity of the vertical wire segments makes an angle with the magnetic field. The velocity is shown in the figure by the green arrow, and the angle is indicated. to a negative maximum of Figure 20.28 shows a generator connected to a light bulb and a graph of the emf vs. time. Note that the emf oscillates from a positive maximum of current flows through the light bulb at these times. Thus, the light bulb actually flickers on and off at a frequency of 2f, because there are two zero crossings per period. Since alternating current such as this is used in homes around the world, why do we not notice the lights flickering on and off? In the United States, the frequency of alternating current is 60 Hz, so the lights flicker on and off at a frequency of 120 Hz. This is faster than the refresh rate of the human eye, so you don’t notice the flicker of the lights. Also, other factors prevent various different types of light bulbs from switching on and off so fast, so the light output is smoothed out
a bit. In between, the emf goes through zero, which means that zero Figure 20.28 The emf of a generator is sent to a light bulb with the system of rings and brushes shown. The graph gives the emf of the generator as a function of time. is the peak emf. The period is where fis the frequency at which the coil is rotated in the magnetic field. Virtual Physics Generator Click to view content (http://www.openstax.org/l/28gen) Use this simulation to discover how an electrical generator works. Control the water supply that makes a water wheel turn a magnet. This induces an emf in a nearby wire coil, which is used to light a light bulb. You can also replace the light bulb with a voltmeter, which allows you to see the polarity of the voltage, which changes from positive to negative. GRASP CHECK Set the number of wire loops to three, the bar-magnet strength to about 50 percent, and the loop area to 100 percent. Note the maximum voltage on the voltmeter. Assuming that one major division on the voltmeter is 5V, what is the maximum voltage when using only a single wire loop instead of three wire loops? a. b. 5 V 15 V Access for free at openstax.org. c. d. 125 V 53 V 20.2 • Motors, Generators, and Transformers 669 In real life, electric generators look a lot different than the figures in this section, but the principles are the same. The source of mechanical energy that turns the coil can be falling water—hydropower—steam produced by the burning of fossil fuels, or the kinetic energy of wind. Figure 20.29 shows a cutaway view of a steam turbine; steam moves over the blades connected to the shaft, which rotates the coil within the generator. Figure 20.29 Steam turbine generator. The steam produced by burning coal impacts the turbine blades, turning the shaft which is connected to the generator. (credit: Nabonaco, Wikimedia Commons) Another very useful and common device that exploits magnetic induction is called a transformer. Transformers do what their name implies—they transform voltages from one value to another; the term voltage is used rather than emf because transformers have internal resistance. For example, many cell phones, laptops, video games, power tools, and small appliances have a transformer built into their plug-in unit that changes 120 V or 240 V AC into whatever voltage the device
uses. Figure 20.30 shows two different transformers. Notice the wire coils that are visible in each device. The purpose of these coils is explained below. Figure 20.30 On the left is a common laminated-core transformer, which is widely used in electric power transmission and electrical appliances. On the right is a toroidal transformer, which is smaller than the laminated-core transformer for the same power rating but is more expensive to make because of the equipment required to wind the wires in the doughnut shape. Figure 20.31 shows a laminated-coil transformer, which is based on Faraday’s law of induction and is very similar in construction to the apparatus Faraday used to demonstrate that magnetic fields can generate electric currents. The two wire coils are called the primary and secondary coils. In normal use, the input voltage is applied across the primary coil, and the secondary produces the transformed output voltage. Not only does the iron core trap the magnetic field created by the primary coil, but also its magnetization increases the field strength, which is analogous to how a dielectric increases the electric field strength in a capacitor. Since the input voltage is AC, a time-varying magnetic flux is sent through the secondary coil, inducing an AC output voltage. 670 Chapter 20 • Magnetism Figure 20.31 A typical construction of a simple transformer has two coils wound on a ferromagnetic core. The magnetic field created by the primary coil is mostly confined to and increased by the core, which transmits it to the secondary coil. Any change in current in the primary coil induces a current in the secondary coil. LINKS TO PHYSICS Magnetic Rope Memory To send men to the moon, the Apollo program had to design an onboard computer system that would be robust, consume little power, and be small enough to fit onboard the spacecraft. In the 1960s, when the Apollo program was launched, entire buildings were regularly dedicated to housing computers whose computing power would be easily outstripped by today’s most basic handheld calculator. To address this problem, engineers at MIT and a major defense contractor turned to magnetic rope memory, which was an offshoot of a similar technology used prior to that time for creating random access memories. Unlike random access memory, magnetic rope memory was read-only memory that contained not only data but instructions as well. Thus, it was actually more than memory: It was a hard-wired computer program. The components of magnetic rope memory were wires and iron rings—which were
called cores. The iron cores served as transformers, such as that shown in the previous figure. However, instead of looping the wires multiple times around the core, individual wires passed only a single time through the cores, making these single-turn transformers. Up to 63 wordwires could pass through a single core, along with a single bitwire. If a word wire passed through a given core, a voltage pulse on this wire would induce an emf in the bit wire, which would be interpreted as a one. If the word wire did not pass through the core, no emf would be induced on the bit wire, which would be interpreted as a zero. Engineers would create programs that would be hard wired into these magnetic rope memories. The wiring process could take as long as a month to complete as workers painstakingly threaded wires through some cores and around others. If any mistakes were made either in the programming or the wiring, debugging would be extraordinarily difficult, if not impossible. These modules did their job quite well. They are credited with correcting an astronaut mistake in the lunar landing procedure, thereby allowing Apollo 11 to land on the moon. It is doubtful that Michael Faraday ever imagined such an application for magnetic induction when he discovered it. GRASP CHECK If the bit wire were looped twice around each core, how would the voltage induced in the bit wire be affected? a. b. c. d. If number of loops around the wire is doubled, the emf is halved. If number of loops around the wire is doubled, the emf is not affected. If number of loops around the wire is doubled, the emf is also doubled. If number of loops around the wire is doubled, the emf is four times the initial value. For the transformer shown in Figure 20.31, the output voltage voltage the secondary coil gives its induced output voltage to be across the primary coil and the number of loops in the primary and secondary coils. Faraday’s law of induction for from the secondary coil depends almost entirely on the input Access for free at openstax.org. 20.19 20.2 • Motors, Generators, and Transformers 671 is the number of loops in the secondary coil and where equals the induced emf sectional area of the coils is the same on each side, as is the magnetic field strength, and so input primary voltage is the rate of change of magnetic flux. The output voltage provided coil resistance is small—a reasonable assumption for transform
ers. The cross- is also related to changing flux by is the same on each side. The 20.20 20.21 Taking the ratio of these last two equations yields the useful relationship This is known as the transformer equation. It simply states that the ratio of the secondary voltage to the primary voltage in a transformer equals the ratio of the number of loops in secondary coil to the number of loops in the primary coil. Transmission of Electrical Power Transformers are widely used in the electric power industry to increase voltages—called step-uptransformers—before longdistance transmission via high-voltage wires. They are also used to decrease voltages—called step-downtransformers—to deliver power to homes and businesses. The overwhelming majority of electric power is generated by using magnetic induction, whereby a wire coil or copper disk is rotated in a magnetic field. The primary energy required to rotate the coils or disk can be provided by a variety of means. Hydroelectric power plants use the kinetic energy of water to drive electric generators. Coal or nuclear power plants create steam to drive steam turbines that turn the coils. Other sources of primary energy include wind, tides, or waves on water. Once power is generated, it must be transmitted to the consumer, which often means transmitting power over hundreds of kilometers. To do this, the voltage of the power plant is increased by a step-up transformer, that is stepped up, and the current decreases proportionally because 20.22 The lower current flow. This heating is caused by the small, but nonzero, resistance environment through this heat is in the transmission wires reduces the Joule losses, which is heating of the wire due to a current of the transmission wires. The power lost to the which is proportional to the current squaredin the transmission wire. This is why the transmitted current small as possible and, consequently, the voltage must be large to transmit the power 20.23 must be as Voltages ranging from 120 to 700 kV are used for transmitting power over long distances. The voltage is stepped up at the exit of the power station by a step-up transformer, as shown in Figure 20.32. Figure 20.32 Transformers change voltages at several points in a power distribution system. Electric power is usually generated at greater than 10 kV, and transmitted long distances at voltages ranging from 120 kV to 700 kV to limit energy losses. Local power distribution to neighborhoods or industries goes through a substation and is sent short distances at voltages ranging from 5 to 13 k
V. This is reduced to 120, 240, or 480 V for safety at the individual user site. Once the power has arrived at a population or industrial center, the voltage is stepped down at a substation to between 5 and 30 672 Chapter 20 • Magnetism kV. Finally, at individual homes or businesses, the power is stepped down again to 120, 240, or 480 V. Each step-up and stepdown transformation is done with a transformer designed based on Faradays law of induction. We’ve come a long way since Queen Elizabeth asked Faraday what possible use could be made of electricity. Check Your Understanding 7. What is an electric motor? a. An electric motor transforms electrical energy into mechanical energy. b. An electric motor transforms mechanical energy into electrical energy. c. An electric motor transforms chemical energy into mechanical energy. d. An electric motor transforms mechanical energy into chemical energy. 8. What happens to the torque provided by an electric motor if you double the number of coils in the motor? a. The torque would be doubled. b. The torque would be halved. c. The torque would be quadrupled. d. The torque would be tripled. 9. What is a step-up transformer? a. A step-up transformer decreases the current to transmit power over short distance with minimum loss. b. A step-up transformer increases the current to transmit power over short distance with minimum loss. c. A step-up transformer increases voltage to transmit power over long distance with minimum loss. d. A step-up transformer decreases voltage to transmit power over short distance with minimum loss. 10. What should be the ratio of the number of output coils to the number of input coil in a step-up transformer to increase the voltage fivefold? a. The ratio is five times. b. The ratio is 10 times. c. The ratio is 15 times. d. The ratio is 20 times. 20.3 Electromagnetic Induction Section Learning Objectives By the end of this section, you will be able to do the following: • Explain how a changing magnetic field produces a current in a wire • Calculate induced electromotive force and current Section Key Terms emf induction magnetic flux Changing Magnetic Fields In the preceding section, we learned that a current creates a magnetic field. If nature is symmetrical, then perhaps a magnetic field can create a current. In 1831, some 12 years after the discovery that an electric current generates a magnetic field, English scientist Michael Faraday (17
91–1862) and American scientist Joseph Henry (1797–1878) independently demonstrated that magnetic fields can produce currents. The basic process of generating currents with magnetic fields is called induction; this process is also called magnetic induction to distinguish it from charging by induction, which uses the electrostatic Coulomb force. When Faraday discovered what is now called Faraday’s law of induction, Queen Victoria asked him what possible use was electricity. “Madam,” he replied, “What good is a baby?” Today, currents induced by magnetic fields are essential to our technological society. The electric generator—found in everything from automobiles to bicycles to nuclear power plants—uses magnetism to generate electric current. Other devices that use magnetism to induce currents include pickup coils in electric guitars, transformers of every size, certain microphones, airport security gates, and damping mechanisms on sensitive chemical balances. Access for free at openstax.org. 20.3 • Electromagnetic Induction 673 One experiment Faraday did to demonstrate magnetic induction was to move a bar magnet through a wire coil and measure the resulting electric current through the wire. A schematic of this experiment is shown in Figure 20.33. He found that current is induced only when the magnet moves with respect to the coil. When the magnet is motionless with respect to the coil, no current is induced in the coil, as in Figure 20.33. In addition, moving the magnet in the opposite direction (compare Figure 20.33 with Figure 20.33) or reversing the poles of the magnet (compare Figure 20.33 with Figure 20.33) results in a current in the opposite direction. Figure 20.33 Movement of a magnet relative to a coil produces electric currents as shown. The same currents are produced if the coil is moved relative to the magnet. The greater the speed, the greater the magnitude of the current, and the current is zero when there is no motion. The current produced by moving the magnet upward is in the opposite direction as the current produced by moving the magnet downward. Virtual Physics Faraday’s Law Click to view content (http://www.openstax.org/l/faradays-law) Try this simulation to see how moving a magnet creates a current in a circuit. A light bulb lights up to show when current is flowing, and a voltmeter shows the voltage drop across the light bulb. Try moving the magnet through a four-turn coil and through a two-
turn coil. For the same magnet speed, which coil produces a higher voltage? GRASP CHECK With the north pole to the left and moving the magnet from right to left, a positive voltage is produced as the magnet enters the coil. What sign voltage will be produced if the experiment is repeated with the south pole to the left? a. The sign of voltage will change because the direction of current flow will change by moving south pole of the magnet to the left. b. The sign of voltage will remain same because the direction of current flow will not change by moving south pole of the magnet to the left. c. The sign of voltage will change because the magnitude of current flow will change by moving south pole of the magnet to the left. d. The sign of voltage will remain same because the magnitude of current flow will not change by moving south pole of the magnet to the left. Induced Electromotive Force If a current is induced in the coil, Faraday reasoned that there must be what he called an electromotive forcepushing the charges through the coil. This interpretation turned out to be incorrect; instead, the external source doing the work of moving the magnet adds energy to the charges in the coil. The energy added per unit charge has units of volts, so the electromotive force is actually a potential. Unfortunately, the name electromotive force stuck and with it the potential for confusing it with a real force. For this reason, we avoid the term electromotive forceand just use the abbreviation emf, which has the mathematical symbol The emf may be defined as the rate at which energy is drawn from a source per unit current flowing through a circuit. Thus, emf is the energy per unit charge addedby a source, which contrasts with voltage, which is the energy per unit charge 674 Chapter 20 • Magnetism releasedas the charges flow through a circuit. To understand why an emf is generated in a coil due to a moving magnet, consider Figure 20.34, which shows a bar magnet moving downward with respect to a wire loop. Initially, seven magnetic field lines are going through the loop (see left-hand image). Because the magnet is moving away from the coil, only five magnetic field lines are going through the loop after a short time (see right-hand image). Thus, when a change occurs in the number of magnetic field lines going through the area defined by the wire loop, an emf is induced in the wire loop. Experiments such as this show that the induced emf
is proportional to the rate of changeof the magnetic field. Mathematically, we express this as 20.24 where is the change in the magnitude in the magnetic field during time and Ais the area of the loop. Figure 20.34 The bar magnet moves downward with respect to the wire loop, so that the number of magnetic field lines going through the loop decreases with time. This causes an emf to be induced in the loop, creating an electric current. Note that magnetic field lines that lie in the plane of the wire loop do not actually pass through the loop, as shown by the leftmost loop in Figure 20.35. In this figure, the arrow coming out of the loop is a vector whose magnitude is the area of the loop and whose direction is perpendicular to the plane of the loop. In Figure 20.35, as the loop is rotated from the contribution of the magnetic field lines to the emf increases. Thus, what is important in generating an emf in the wire loop is the component of the magnetic field that is perpendicularto the plane of the loop, which is to This is analogous to a sail in the wind. Think of the conducting loop as the sail and the magnetic field as the wind. To maximize the force of the wind on the sail, the sail is oriented so that its surface vector points in the same direction as the winds, as in the right-most loop in Figure 20.35. When the sail is aligned so that its surface vector is perpendicular to the wind, as in the leftmost loop in Figure 20.35, then the wind exerts no force on the sail. Thus, taking into account the angle of the magnetic field with respect to the area, the proportionality becomes 20.25 Figure 20.35 The magnetic field lies in the plane of the left-most loop, so it cannot generate an emf in this case. When the loop is rotated so that the angle of the magnetic field with the vector perpendicular to the area of the loop increases to (see right-most loop), the magnetic field contributes maximally to the emf in the loop. The dots show where the magnetic field lines intersect the plane defined by the loop. Access for free at openstax.org. 20.3 • Electromagnetic Induction 675 Another way to reduce the number of magnetic field lines that go through the conducting loop in Figure 20.35 is not to move the magnet but to make the loop smaller. Experiments show that changing the area of a conducting
loop in a stable magnetic field induces an emf in the loop. Thus, the emf produced in a conducting loop is proportional to the rate of change of the productof the perpendicular magnetic field and the loop area 20.26 is the perpendicular magnetic field and Ais the area of the loop. The product where proportional to the number of magnetic field lines that pass perpendicularly through a surface of area A. Going back to our sail analogy, it would be proportional to the force of the wind on the sail. It is called the magnetic flux and is represented by is very important. It is. The unit of magnetic flux is the weber (Wb), which is magnetic field per unit area, or T/m2. The weber is also a volt second (Vs). The induced emf is in fact proportional to the rate of change of the magnetic flux through a conducting loop. 20.27 20.28 Finally, for a coil made from Nloops, the emf is Ntimes stronger than for a single loop. Thus, the emf induced by a changing magnetic field in a coil of Nloops is The last question to answer before we can change the proportionality into an equation is “In what direction does the current flow?” The Russian scientist Heinrich Lenz (1804–1865) explained that the current flows in the direction that creates a magnetic field that tries to keep the flux constant in the loop. For example, consider again Figure 20.34. The motion of the bar magnet causes the number of upward-pointing magnetic field lines that go through the loop to decrease. Therefore, an emf is generated in the loop that drives a current in the direction that creates more upward-pointing magnetic field lines. By using the righthand rule, we see that this current must flow in the direction shown in the figure. To express the fact that the induced emf acts to counter the change in the magnetic flux through a wire loop, a minus sign is introduced into the proportionality, which gives Faraday’s law of induction. 20.29 Lenz’s law is very important. To better understand it, consider Figure 20.36, which shows a magnet moving with respect to a wire coil and the direction of the resulting current in the coil. In the top row, the north pole of the magnet approaches the coil, so the magnetic field lines from the magnet point toward the coil. Thus, the magnetic field right increases in the coil. According
to Lenz’s law, the emf produced in the coil will drive a current in the direction that creates a pointing to the magnetic field inside the coil pointing to the left. This will counter the increase in magnetic flux pointing to the right. To see which way the current must flow, point your right thumb in the desired direction of the magnetic field and the current will flow in the direction indicated by curling your right fingers. This is shown by the image of the right hand in the top row of Figure 20.36. Thus, the current must flow in the direction shown in Figure 4(a). In Figure 4(b), the direction in which the magnet moves is reversed. In the coil, the right-pointing magnetic field the moving magnet decreases. Lenz’s law says that, to counter this decrease, the emf will drive a current that creates an due to additional right-pointing magnetic field magnetic field, and the current will flow in the direction indicate by curling your right fingers (Figure 4(b)). in the coil. Again, point your right thumb in the desired direction of the Finally, in Figure 4(c), the magnet is reversed so that the south pole is nearest the coil. Now the magnetic field toward the magnet instead of toward the coil. As the magnet approaches the coil, it causes the left-pointing magnetic field in the coil to increase. Lenz’s law tells us that the emf induced in the coil will drive a current in the direction that creates a magnetic field pointing to the right. This will counter the increasing magnetic flux pointing to the left due to the magnet. Using the righthand rule again, as indicated in the figure, shows that the current must flow in the direction shown in Figure 4(c). points 676 Chapter 20 • Magnetism Figure 20.36 Lenz’s law tells us that the magnetically induced emf will drive a current that resists the change in the magnetic flux through a circuit. This is shown in panels (a)–(c) for various magnet orientations and velocities. The right hands at right show how to apply the right- hand rule to find in which direction the induced current flows around the coil. Virtual Physics Faraday’s Electromagnetic Lab Click to view content (http://www.openstax.org/l/Faraday-EM-lab) This simulation proposes several activities. For now, click on the tab Pickup Coil, which presents
a bar magnet that you can move through a coil. As you do so, you can see the electrons move in the coil and a light bulb will light up or a voltmeter will indicate the voltage across a resistor. Note that the voltmeter allows you to see the sign of the voltage as you move the magnet about. You can also leave the bar magnet at rest and move the coil, although it is more difficult to observe the results. GRASP CHECK Orient the bar magnet with the north pole facing to the right and place the pickup coil to the right of the bar magnet. Now move the bar magnet toward the coil and observe in which way the electrons move. This is the same situation as depicted below. Does the current in the simulation flow in the same direction as shown below? Explain why or why not. Access for free at openstax.org. 20.3 • Electromagnetic Induction 677 a. Yes, the current in the simulation flows as shown because the direction of current is opposite to the direction of flow of electrons. b. No, current in the simulation flows in the opposite direction because the direction of current is same to the direction of flow of electrons. WATCH PHYSICS Induced Current in a Wire This video explains how a current can be induced in a straight wire by moving it through a magnetic field. The lecturer uses the cross product, which a type of vector multiplication. Don’t worry if you are not familiar with this, it basically combines the righthand rule for determining the force on the charges in the wire with the equation Click to view content (https://www.openstax.org/l/induced-current) GRASP CHECK through a uniform magnetic field What emf is produced across a straight wire 0.50 m long moving at a velocity of (1.5 m/s) (0.30 T)ẑ? The wire lies in the ŷ-direction. Also, which end of the wire is at the higher potential—let the lower end of the wire be at y= 0 and the upper end at y= 0.5 m)? a. 0.15 V and the lower end of the wire will be at higher potential b. 0.15 V and the upper end of the wire will be at higher potential c. 0.075 V and the lower end of the wire will be at higher potential d. 0.075 V and the upper end of the wire will be at higher potential WORKED EX
AMPLE EMF Induced in Conducing Coil by Moving Magnet Imagine a magnetic field goes through a coil in the direction indicated in Figure 20.37. The coil diameter is 2.0 cm. If the magnetic field goes from 0.020 to 0.010 T in 34 s, what is the direction and magnitude of the induced current? Assume the coil has a resistance of 0.1 678 Chapter 20 • Magnetism Figure 20.37 A coil through which passes a magnetic field B. STRATEGY Use the equation solenoid, we find it has 16 loops, so to find the induced emf in the coil, where. Counting the number of loops in the Use the equation to calculate the magnetic flux 20.30 where dis the diameter of the solenoid and we have used in the magnetic of the flux through the solenoid is Because the area of the solenoid does not vary, the change 20.31 Once we find the emf, we can use Ohm’s law, to find the current. Finally, Lenz’s law tells us that the current should produce a magnetic field that acts to oppose the decrease in the applied magnetic field. Thus, the current should produce a magnetic field to the right. Solution Combining equations and gives Solving Ohm’s law for the current and using this result gives 20.32 20.33 Lenz’s law tells us that the current must produce a magnetic field to the right. Thus, we point our right thumb to the right and curl our right fingers around the solenoid. The current must flow in the direction in which our fingers are pointing, so it enters at the left end of the solenoid and exits at the right end. Discussion Let’s see if the minus sign makes sense in Faraday’s law of induction. Define the direction of the magnetic field to be the positive direction. This means the change in the magnetic field is negative, as we found above. The minus sign in Faraday’s law of induction negates the negative change in the magnetic field, leaving us with a positive current. Therefore, the current must flow in the direction of the magnetic field, which is what we found. Now try defining the positive direction to be the direction opposite that of the magnetic field, that is positive is to the left in Figure 20.37. In this case, you will find a negative current. But since the positive direction is to the left,
a negative current must flow to the right, which again agrees with what we found by using Lenz’s law. WORKED EXAMPLE Magnetic Induction due to Changing Circuit Size The circuit shown in Figure 20.38 consists of a U-shaped wire with a resistor and with the ends connected by a sliding conducting rod. The magnetic field filling the area enclosed by the circuit is constant at 0.01 T. If the rod is pulled to the right at speed what current is induced in the circuit and in what direction does the current flow? Access for free at openstax.org. 20.3 • Electromagnetic Induction 679 Figure 20.38 A slider circuit. The magnetic field is constant and the rod is pulled to the right at speed v. The changing area enclosed by the circuit induces an emf in the circuit. STRATEGY We again use Faraday’s law of induction, although this time the magnetic field is constant and the area enclosed by the circuit changes. The circuit contains a single loop, so The rate of change of the area is Thus the rate of change of the magnetic flux is 20.34 where we have used the fact that the angle between the area vector and the magnetic field is 0°. Once we know the emf, we can find the current by using Ohm’s law. To find the direction of the current, we apply Lenz’s law. Solution Faraday’s law of induction gives Solving Ohm’s law for the current and using the previous result for emf gives 20.35 20.36 As the rod slides to the right, the magnetic flux passing through the circuit increases. Lenz’s law tells us that the current induced will create a magnetic field that will counter this increase. Thus, the magnetic field created by the induced current must be into the page. Curling your right-hand fingers around the loop in the clockwise direction makes your right thumb point into the page, which is the desired direction of the magnetic field. Thus, the current must flow in the clockwise direction around the circuit. Discussion Is energy conserved in this circuit? An external agent must pull on the rod with sufficient force to just balance the force on a current-carrying wire in a magnetic field—recall that be balanced by the rate at which the circuit dissipates power. Using constant speed vis the force required to pull the wire at a The rate at which this force does work on the rod should
where we used the fact that the angle between the current and the magnetic field is the current into this equation gives Inserting our expression above for 20.37 20.38 The power contributed by the agent pulling the rod is 680 Chapter 20 • Magnetism The power dissipated by the circuit is 20.39 20.40 We thus see that agent that pulls the rod. Thus, energy is conserved in this system. which means that power is conserved in the system consisting of the circuit and the Practice Problems 11. The magnetic flux through a single wire loop changes from 3.5 Wb to 1.5 Wb in 2.0 s. What emf is induced in the loop? a. –2.0 V b. –1.0 V c. +1.0 V d. +2.0 V 12. What is the emf for a 10-turn coil through which the flux changes at 10 Wb/s? a. –100 V b. –10 V c. +10 V d. +100 V Check Your Understanding 13. Given a bar magnet, how can you induce an electric current in a wire loop? a. An electric current is induced if a bar magnet is placed near the wire loop. b. An electric current is induced if wire loop is wound around the bar magnet. c. An electric current is induced if a bar magnet is moved through the wire loop. d. An electric current is induced if a bar magnet is placed in contact with the wire loop. 14. What factors can cause an induced current in a wire loop through which a magnetic field passes? a. b. c. d. Induced current can be created by changing the size of the wire loop only. Induced current can be created by changing the orientation of the wire loop only. Induced current can be created by changing the strength of the magnetic field only. Induced current can be created by changing the strength of the magnetic field, changing the size of the wire loop, or changing the orientation of the wire loop. Access for free at openstax.org. Chapter 20 • Key Terms 681 KEY TERMS Curie temperature well-defined temperature for the magnetic force ferromagnetic materials above which they cannot be magnetized domain region within a magnetic material in which the magnetic flux component of the magnetic field perpendicular to the surface area through which it passes and multiplied by the area magnetic poles of individual atoms are aligned magnetic pole part of a magnet that exerts the strongest electric motor device that transforms
electrical energy into force on other magnets or magnetic material mechanical energy magnetized material that is induced to be magnetic or that electromagnet device that uses electric current to make a is made into a permanent magnet magnetic field north pole part of a magnet that orients itself toward the electromagnetism study of electric and magnetic geographic North Pole of Earth phenomena emf rate at which energy is drawn from a source per unit current flowing through a circuit ferromagnetic material such as iron, cobalt, nickel, or gadolinium that exhibits strong magnetic effects generator device that transforms mechanical energy into electrical energy permanent magnet material that retains its magnetic behavior for a long time, even when exposed to demagnetizing influences right-hand rule rule involving curling the right-hand fingers from one vector to another; the direction in which the right thumb points is the direction of the resulting vector induction rate at which energy is drawn from a source per unit current flowing through a circuit magnetic dipole term that describes magnets because they solenoid uniform cylindrical coil of wire through which electric current is passed to produce a magnetic field south pole part of a magnet that orients itself toward the always have two poles: north and south geographic South Pole of Earth magnetic field directional lines around a magnetic transformer device that transforms voltages from one material that indicates the direction and magnitude of value to another SECTION SUMMARY 20.1 Magnetic Fields, Field Lines, and Force • All magnets have two poles: a north pole and a south pole. If the magnet is free to move, its north pole orients itself toward the geographic North Pole of Earth, and the south pole orients itself toward the geographic South Pole of Earth. • A repulsive force occurs between the north poles of two magnets and likewise for two south poles. However, an attractive force occurs between the north pole of one magnet and the south pole of another magnet. • A charged particle moving through a magnetic field experiences a force whose direction is determined by the right-hand rule. • An electric current generates a magnetic field. • Electromagnets are magnets made by passing a current through a system of wires. 20.2 Motors, Generators, and Transformers • Electric motors contain wire loops in a magnetic field. Current is passed through the wire loops, which forces them to rotate in the magnetic field. The current is reversed every half rotation so that the torque on the loop is always in the same direction. • Electric generators contain wire loops in a magnetic field. An external agent provides mechanical energy to force the loops
to rotate in the magnetic field, which produces an AC voltage that drives an AC current through the loops. • Transformers contain a ring made of magnetic material and, on opposite sides of the ring, two windings of wire wrap around the ring. A changing current in one wire winding creates a changing magnetic field, which is trapped in the ring and thus goes through the second winding and induces an emf in the second winding. The voltage in the second winding is proportional to the ratio of the number of loops in each winding. • Transformers are used to step up and step down the voltage for power transmission. • Over long distances, electric power is transmitted at high voltage to minimize the current and thereby minimize the Joule losses due to resistive heating. 20.3 Electromagnetic Induction • Faraday’s law of induction states that a changing magnetic flux that occurs within an area enclosed by a conducting loop induces an electric current in the loop. • Lenz’ law states that an induced current flows in the direction such that it opposes the change that induced it. 682 Chapter 20 • Key Equations KEY EQUATIONS 20.1 Magnetic Fields, Field Lines, and Force the magnitude of the force on an electric charge the force on a wire carrying current the magnitude of the magnetic field created by a long, straight current-carrying wire CHAPTER REVIEW Concept Items 20.1 Magnetic Fields, Field Lines, and Force 1. If you place a small needle between the north poles of two bar magnets, will the needle become magnetized? a. Yes, the magnetic fields from the two north poles will point in the same directions. b. Yes, the magnetic fields from the two north poles will point in opposite directions. c. No, the magnetic fields from the two north poles will point in opposite directions. d. No, the magnetic fields from the two north poles will point in the same directions. 2. If you place a compass at the three points in the figure, at which point will the needle experience the greatest torque? Why? the magnitude of the magnetic field inside a solenoid 20.3 Electromagnetic Induction magnetic flux emf greatest torque at B. b. The density of the magnetic field is minimized at C, so the magnetic compass needle will experience the greatest torque at C. c. The density of the magnetic field is maximized at B, so the magnetic compass needle will experience the greatest torque at B. d. The density of the magnetic field is maximized at A, so the
magnetic compass needle will experience the greatest torque at A. 3. In which direction do the magnetic field lines point near the south pole of a magnet? a. Outside the magnet the direction of magnetic field lines is towards the south pole of the magnet. b. Outside the magnet the direction of magnetic field lines is away from the south pole of the magnet. 20.2 Motors, Generators, and Transformers 4. Consider the angle between the area vector and the and magnetic field in an electric motor. At what angles is the torque on the wire loop the greatest? a. b. c. d. and and and 5. What is a voltage transformer? a. A transformer is a device that transforms current to voltage. b. A transformer is a device that transforms voltages from one value to another. c. A transformer is a device that transforms resistance of wire to voltage. 6. Why is electric power transmitted at high voltage? a. The density of the magnetic field is minimized at B, so the magnetic compass needle will experience the Access for free at openstax.org. Chapter 20 • Chapter Review 683 a. To increase the current for the transmission b. To reduce energy loss during transmission c. To increase resistance during transmission d. To reduce resistance during transmission 20.3 Electromagnetic Induction 7. Yes or no—Is an emf induced in the coil shown when it is stretched? If so, state why and give the direction of the induced current. a. b. c. d. If induced current flows, its direction is such that it adds to the changes which induced it. If induced current flows, its direction is such that it opposes the changes which induced it. If induced current flows, its direction is always clockwise to the changes which induced it. If induced current flows, its direction is always counterclockwise to the changes which induced it. 9. Explain how magnetic flux can be zero when the a. No, because induced current does not depend upon the area of the coil. b. Yes, because area of the coil increases; the direction of the induced current is counterclockwise. c. Yes, because area of the coil increases; the direction of the induced current is clockwise. d. Yes, because the area of the coil does not change; the direction of the induced current is clockwise. 8. What is Lenz’s law? Critical Thinking Items 20.1 Magnetic Fields, Field Lines, and Force 10. True or false—It is not recommended
to place credit cards with magnetic strips near permanent magnets. a. b. false true 11. True or false—A square magnet can have sides that alternate between north and south poles. a. b. false true 12. You move a compass in a circular plane around a planar magnet. The compass makes four complete revolutions. How many poles does the magnet have? two poles a. b. four poles c. eight poles 12 poles d. 20.2 Motors, Generators, and Transformers 13. How can you maximize the peak emf from a generator? a. The peak emf from a generator can be maximized only by maximizing number of turns. b. The peak emf from a generator can be maximized only by maximizing area of the wired loop. magnetic field is not zero. a. If angle between magnetic field and area vector is 0°, then its sine is also zero, which means that there is zero flux. If angle between magnetic field and area vector is 45°, then its sine is also zero, which means that there is zero flux. If angle between magnetic field and area vector is 60°, then its cosine is also zero, which means that there is zero flux. If the angle between magnetic field and area vector is 90°, then its cosine is also zero, which means that there is zero flux. b. c. d. c. The peak emf from a generator can be maximized only by maximizing frequency. d. The peak emf from a generator can be maximized by maximizing number of turns, maximizing area of the wired loop or maximizing frequency. 14. Explain why power is transmitted over long distances at high voltages. a. Plost = Itransmitted Vtransmitted, so to maximize current, the voltage must be maximized b. Ptransmitted = Itransmitted Vtransmitted, so to maximize current, the voltage must be maximized c. Plost = Itransmitted Vtransmitted, so to minimize current, the voltage must be maximized d. Ptransmitted = Itransmitted Vtransmitted, so to minimize current, the voltage must be maximized 20.3 Electromagnetic Induction 15. To obtain power from the current in the wire of your vacuum cleaner, you place a loop of wire near it to obtain an induced emf. How do you place and orient the loop? a. A loop of wire should be placed nearest to the vacuum cleaner wire to maximize the magnetic flux through the loop.
b. A loop of wire should be placed farthest to the vacuum cleaner wire to maximize the magnetic flux through the loop. 684 Chapter 20 • Test Prep c. A loop of wire should be placed perpendicular to the vacuum cleaner wire to maximize the magnetic flux through the loop. d. A loop of wire should be placed at angle greater than 90° to the vacuum cleaner wire to maximize the magnetic flux through the loop. that is proportional to the rate of change of the magnetic flux. b. The magnetic field in the coil changes rapidly due to spinning of magnet which creates an emf in the coil that is proportional to the rate of change of the magnetic flux. 16. A magneto is a device that creates a spark across a gap 17. If you drop a copper tube over a bar magnet with its by creating a large voltage across the gap. To do this, the device spins a magnet very quickly in front of a wire coil, with the ends of the wires forming the gap. Explain how this creates a sufficiently large voltage to produce a spark. a. The electric field in the coil increases rapidly due to spinning of magnet which creates an emf in the coil north pole up, is a current induced in the copper tube? If so, in what direction? Consider when the copper tube is approaching the bar magnet. a. Yes, the induced current will be produced in the clockwise direction when viewed from above. b. No, the induced current will not be produced. Problems 20.1 Magnetic Fields, Field Lines, and Force 18. A straight wire segment carries 0.25 A. What length would it need to be to exert a 4.0-mN force on a magnet that produces a uniform magnetic field of 0.015 T that is perpendicular to the wire? a. 0.55 m b. 1.10 m c. 2.20 m d. 4.40 m 20.3 Electromagnetic Induction 19. What is the current in a wire loop of resistance 10 Ω through which the magnetic flux changes from zero to Performance Task 20.2 Motors, Generators, and Transformers 21. Your family takes a trip to Cuba, and rents an old car to drive into the countryside to see the sights. Unfortunately, the next morning you find yourself deep in the countryside and the car won’t start because the battery is too weak. Wanting to jump-start the car, you open the hood and find that you can’t tell which battery TEST PREP Multiple Choice
20.1 Magnetic Fields, Field Lines, and Force 22. For a magnet, a domain refers to ______. a. b. c. the region between the poles of the magnet the space around the magnet that is affected by the magnetic field the region within the magnet in which the Access for free at openstax.org. 10 Wb in 1.0 s? a. –100 A b. –2.0 A c. –1.0 A d. +1.0 A 20. An emf is induced by rotating a 1,000 turn, 20.0 cm diameter coil in Earth’s 5.00 × 10–5 T magnetic field. What average emf is induced, given the plane of the coil is originally perpendicular to Earth’s field and is rotated to be parallel to the field in 10.0 ms? a. –1.6 × 10-4 V b. +1.6 × 10-4 V c. +1.6 × 10-1 V d. –1.6 × 10-1 V terminal is positive and which is negative. However, you do have a bar magnet with the north and south poles labeled and you manage to find a short wire. How do you use these to determine which terminal is which? For starters, how do you determine the direction of a magnetic field around a current-carrying wire? And in which direction will the force be on another magnet placed in this field? Do you need to worry about the sign of the mobile charge carriers in the wire? d. magnetic poles of individual atoms are aligned the region from which the magnetic material is mined 23. In the region just outside the south pole of a magnet, the magnetic field lines ______. a. point away from the south pole b. go around the south pole c. are less concentrated than at the north pole d. point toward the south pole 24. Which equation gives the force for a charge moving through a magnetic field? a. b. c. d. 25. Can magnetic field lines cross each other? Explain why or why not. a. Yes, magnetic field lines can cross each other because that point of intersection indicates two possible directions of magnetic field, which is possible. b. No, magnetic field lines cannot cross each other because that point of intersection indicates two possible directions of magnetic field, which is not possible. 26. True or false—If a magnet shatters into many small pieces, all the pieces will have north and south poles a
. b. true false 20.2 Motors, Generators, and Transformers 27. An electrical generator ________. is a generator powered by electricity a. b. must be turned by hand c. converts other sources of power into electrical power d. uses magnetism to create electrons 28. A step-up transformer increases the a. voltage from power lines for use in homes b. c. current from the power lines for use in homes current from the electrical generator for transmission along power lines d. voltage from the electrical power plant for transmission along power lines Chapter 20 • Test Prep 685 b. Torque is doubled. c. Torque is quadrupled. d. Torque is halved. 30. Why are the coils of a transformer wrapped around a loop of ferrous material? a. The magnetic field from the source coil is trapped and also increased in strength. b. The magnetic field from the source coil is dispersed and also increased in strength. c. The magnetic field from the source coil is trapped and also decreased in strength. d. Magnetic field from the source coil is dispersed and also decreased in strength. 20.3 Electromagnetic Induction 31. What does emfstand for? a. electromotive force b. electro motion force c. electromagnetic factor d. electronic magnetic factor 32. Which formula gives magnetic flux? a. b. c. d. 33. What is the relationship between the number of coils in a solenoid and the emf induced in it by a change in the magnetic flux through the solenoid? a. The induced emf is inversely proportional to the number of coils in a solenoid. b. The induced emf is directly proportional to the number of coils in a solenoid. c. The induced emf is inversely proportional to the square of the number of coils in a solenoid. d. The induced emf is proportional to square of the number of coils in a solenoid. 29. What would be the effect on the torque of an electric 34. True or false—If you drop a bar magnet through a motor of doubling the width of the current loop in the motor? a. Torque remains the same. copper tube, it induces an electric current in the tube. a. b. false true Short Answer 20.1 Magnetic Fields, Field Lines, and Force 35. Given a bar magnet, a needle, a cork, and a bowl full of water, describe how to make a compass. a. Magnetize
the needle by holding it perpendicular to a bar magnet’s north pole and pierce the cork along its longitudinal axis by the needle and place the needle-cork combination in the water. The needle now orients itself along the magnetic field lines of Earth. b. Magnetize the needle by holding it perpendicular to a bar magnet’s north pole and pierce the cork along its longitudinal axis by the needle and place the needle-cork combination in the water. The needle now orients itself perpendicular to the Hence, it will attract the south pole of other magnet. c. The needle will magnetize and the point of a needle kept closer to the north pole will act as a north pole. Hence, it will repel the south pole of the other magnet. d. The needle will magnetize and the point of needle kept closer to the north pole will act as a north pole. Hence, it will attract the south pole of other magnet. 39. Using four solenoids of the same size, describe how to orient them and in which direction the current should flow to make a magnet with two opposite-facing north poles and two opposite-facing south poles. 686 Chapter 20 • Test Prep magnetic field lines of Earth. c. Magnetize the needle by holding its axis parallel to the axis of a bar magnet and pierce the cork along its longitudinal axis by the needle and place the needle-cork combination in the water. The needle now orients itself along the magnetic field lines of Earth. d. Magnetize the needle by holding its axis parallel to the axis of a bar magnet and pierce the cork along its longitudinal axis by the needle and place the needle-cork combination in the water. The needle now orients itself perpendicular to the magnetic field lines of Earth. 36. Give two differences between electric field lines and magnetic field lines. a. Electric field lines begin and end on opposite charges and the electric force on a charge is in the direction of field, while magnetic fields form a loop and the magnetic force on a charge is perpendicular to the field. b. Electric field lines form a loop and the electric force on a charge is in the direction of field, while magnetic fields begin and end on opposite charge and the magnetic force on a charge is perpendicular to the field. c. Electric field lines begin and end on opposite charges and the electric force on a charge is in the perpendicular direction of field, while magnetic fields form a loop and the magnetic force on a charge is in
the direction of the field. d. Electric field lines form a loop and the electric force on a charge is in the perpendicular direction of field, while magnetic fields begin and end on opposite charge and the magnetic force on a charge is in the direction of the field. 37. To produce a magnetic field of 0.0020 T, what current is required in a 500-turn solenoid that is 25 cm long? a. 0.80 A b. 1.60 A c. 80 A 160 A d. 38. You magnetize a needle by aligning it along the axis of a bar magnet and just outside the north pole of the magnet. Will the point of the needle that was closest to the bar magnet then be attracted to or repelled from the south pole of another magnet? a. The needle will magnetize and the point of needle kept closer to the north pole will act as a south pole. Hence, it will repel the south pole of other magnet. b. The needle will magnetize and the point of needle kept closer to the north pole will act as a south pole. Access for free at openstax.org. Chapter 20 • Test Prep 687 a. The output emf will be doubled. b. The output emf will be halved. c. The output emf will be quadrupled. d. The output emf will be tripled. 44. In a hydroelectric dam, what is used to power the electrical generators that provide electric power? Explain. a. The electric potential energy of stored water is used to produce emf with the help of a turbine. b. The electric potential energy of stored water is used to produce resistance with the help of a turbine. c. Gravitational potential energy of stored water is used to produce resistance with the help of a turbine. d. Gravitational potential energy of stored water is used to produce emf with the help of a turbine. 20.3 Electromagnetic Induction 45. A uniform magnetic field is perpendicular to the plane of a wire loop. If the loop accelerates in the direction of the field, will a current be induced in the loop? Explain why or why not. a. No, because magnetic flux through the loop remains constant. b. No, because magnetic flux through the loop changes continuously. c. Yes, because magnetic flux through the loop remains constant. d. Yes, because magnetic flux through the loop changes continuously. 46. The plane of a square wire circuit with side 4.0 cm
long is at an angle of 45° with respect to a uniform magnetic field of 0.25 T. The wires have a resistance per unit length of 0.2. If the field drops to zero in 2.5 s, what magnitude current is induced in the square circuit? 35 µA a. b. 87.5 µA 3.5 mA c. 35 A d. 47. Yes or no—If a bar magnet moves through a wire loop as shown in the figure, is a current induced in the loop? Explain why or why not. 40. How far from a straight wire carrying 0.45 A is the magnetic field strength 0.040 T? a. 0.23 µm b. 0.72 µm c. 2.3 µm 7.2 µm d. 20.2 Motors, Generators, and Transformers 41. A laminated-coil transformer has a wire coiled 12 times around one of its sides. How many coils should you wrap around the opposite side to get a voltage output that is one half of the input voltage? Explain. a. six output coils because the ratio of output to input voltage is the same as the ratio of number of output coils to input coils 12 output coils because the ratio of output to input voltage is the same as the ratio of number of output coils to input coils b. c. 24 output coils because the ratio of output to input voltage is half the ratio of the number of output coils to input coils 36 output coils because the ratio of output to input voltage is three times the ratio of the number of output coils to input coils d. 42. Explain why long-distance electrical power lines are designed to carry very high voltages. a. Ptransmitted = Itransmitted> 2 Rwire and Plost = Itransmitted Vtransmitted, so Vmust be low to make the current transmitted as high as possible. b. Ptransmitted = Itransmitted> 2 Rwire and Plost = Ilost Vlost, so Vmust be low to make the current transmitted as high as possible. c. Ptransmitted = Itransmitted> 2 Rwire and Plost = Itransmitted Vtransmitted, so Vmust be high to make the current transmitted as low as possible d. Plost = Itransmitted 2 Rwire and Ptransmitted = Itransmitted Vtransmitted, so Vmust be high to make the current transmitted as low as possible. a. No, because the net magnetic
field passing through the loop is zero. 43. How is the output emf of a generator affected if you b. No, because the net magnetic field passing through double the frequency of rotation of its coil? 688 Chapter 20 • Test Prep the loop is nonzero. c. Yes, because the net magnetic field passing through the loop is zero. d. Yes, because the net magnetic field line passing through the loop is nonzero. 48. What is the magnetic flux through an equilateral Extended Response 20.1 Magnetic Fields, Field Lines, and Force 49. Summarize the properties of magnets. a. A magnet can attract metals like iron, nickel, etc., but cannot attract nonmetals like piece of plastic or wood, etc. If free to rotate, an elongated magnet will orient itself so that its north pole will face the magnetic south pole of Earth. b. A magnet can attract metals like iron, nickel, etc., but cannot attract nonmetals like piece of plastic or wood, etc. If free to rotate, an elongated magnet will orient itself so that its north pole will face the magnetic north pole of Earth. c. A magnet can attract metals like iron, nickel, etc., and nonmetals like piece of plastic or wood, etc. If free to rotate, an elongated magnet will orient itself so that its north pole will face the magnetic south pole of Earth. d. A magnet can attract metals like iron, nickel, etc., and nonmetals like piece of plastic or wood, etc. If free to rotate, an elongated magnet will orient itself so that its north pole will face the magnetic north pole of Earth. 50. The magnetic field shown in the figure is formed by current flowing in two rings that intersect the page at the dots. Current flows into the page at the dots with crosses (right side) and out of the page at the dots with points (left side). Access for free at openstax.org. triangle with side 60 cm long and whose plane makes a 60° angle with a uniform magnetic field of 0.33 T? a. 0.045 Wb b. 0.09 Wb c. 0.405 Wb d. 4.5 Wb Where is the field strength the greatest and in what direction do the magnetic field lines point? a. The magnetic field strength is greatest where the magnetic field lines are less dense; magnetic field lines points up the page. b. The magnetic field strength is greatest where the magnetic field lines
are most dense; magnetic field lines points up the page. c. The magnetic field strength is greatest where the magnetic field lines are most dense; magnetic field lines points down the page. d. The magnetic field strength is greatest where the magnetic field lines are less dense; magnetic field lines points down the page. 51. The forces shown below are exerted on an electron as it moves through the magnetic field. In each case, what direction does the electron move? a. b. c. d. (a) left to right, (b) out of the page, (c) upwards (a) left to right, (b) into the page, (c) downwards (a) right to left, (b) out of the page, (c) upwards (a) right to left, (b) into the page, (c) downwards 20.2 Motors, Generators, and Transformers 52. Explain why increasing the frequency of rotation of the coils in an electrical generator increases the output emf. a. The induced emf is proportional to the rate of change of magnetic flux with respect to distance. b. The induced emf is inversely proportional to the rate of change of magnetic flux with respect to distance. c. The induced emf is inversely proportional to the rate of change of magnetic flux with respect to time. d. The induced emf is proportional to the rate of change of magnetic flux with respect to time. 53. Your friend tells you that power lines must carry a maximum current because P= I2R, where R is the resistance of the transmission line. What do you tell her? a. Ptransmitted = Itransmitted 2Rwire and Plost = Itransmitted Vtransmitted, so Imust be high to reduce power lost due to transmission. b. Plost = Itransmitted 2Rwire and Plost = Itransmitted Vtransmitted, so Imust be high to reduce power lost due to transmission. c. Ptransmitted = Itransmitted 2Rwire and Plost = Itransmitted Vtransmitted, so Imust be low to reduce power lost due to transmission. d. Plost = Itransmitted 2Rwire and Plost = Itransmitted Vtransmitted, so Imust be low to reduce power lost due to transmission. 20.3 Electromagnetic Induction 54. When you insert a copper ring between the poles of two bar magnets as shown in the figure, do the magnets exert an attractive or repulsive
force on the ring? Explain your reasoning. a. Magnets exert an attractive force, because magnetic field due to induced current is repulsed by the magnetic field of the magnets. b. Magnets exert an attractive force, because Chapter 20 • Test Prep 689 magnetic field due to induced current is attracted by the magnetic field of the magnets. c. Magnets exert a repulsive force, because magnetic field due to induced current is repulsed by the magnetic field of the magnets. d. Magnets exert a repulsive force, because magnetic field due to induced current is attracted by the magnetic field of the magnets. 55. The figure shows a uniform magnetic field passing through a closed wire circuit. The wire circuit rotates at an angular frequency of about the axis shown by the dotted line in the figure. What is an expression for the magnetic flux through the circuit as a function of time? a. expression for the magnetic flux through the circuit Φ(t) = BAcos ωt b. expression for the magnetic flux through the circuit c. expression for the magnetic flux through the circuit d. expression for the magnetic flux through the circuit Φ(t) = 2BAcos ωt 690 Chapter 20 • Test Prep Access for free at openstax.org. CHAPTER 21 The Quantum Nature of Light Figure 21.1 In Lewis Carroll’s classic text Alice’s Adventures in Wonderland, Alice follows a rabbit down a hole into a land of curiosity. While many of her interactions in Wonderland are of surprising consequence, they follow a certain inherent logic. (credit: modification of work by John Tenniel, Wikimedia Commons) Chapter Outline 21.1 Planck and Quantum Nature of Light 21.2 Einstein and the Photoelectric Effect 21.3 The Dual Nature of Light At first glance, the quantum nature of light can be a strange and bewildering concept. Between light acting as INTRODUCTION discrete chunks, massless particles providing momenta, and fundamental particles behaving like waves, it may often seem like something out of Alice in Wonderland. For many, the study of this branch of physics can be as enthralling as Lewis Carroll’s classic novel. Recalling the works of legendary characters and brilliant scientists such as Einstein, Planck, and Compton, the study of light’s quantum nature will provide you an interesting tale of how a clever interpretation of some small details led to the most important discoveries of the past 150 years. From the electronics revolution of the twentieth century to our
future progress in solar energy and space exploration, the quantum nature of light should yield a rabbit hole of curious consequence, within which lie some of the most fascinating truths of our time. 692 Chapter 21 • The Quantum Nature of Light 21.1 Planck and Quantum Nature of Light Section Learning Objectives By the end of this section, you will be able to do the following: • Describe blackbody radiation • Define quantum states and their relationship to modern physics • Calculate the quantum energy of lights • Explain how photon energies vary across divisions of the electromagnetic spectrum Section Key Terms blackbody quantized quantum ultraviolet catastrophe Blackbodies Our first story of curious significance begins with a T-shirt. You are likely aware that wearing a tight black T-shirt outside on a hot day provides a significantly less comfortable experience than wearing a white shirt. Black shirts, as well as all other black objects, will absorb and re-emit a significantly greater amount of radiation from the sun. This shirt is a good approximation of what is called a blackbody. A perfect blackbody is one that absorbs and re-emits all radiated energy that is incident upon it. Imagine wearing a tight shirt that did this! This phenomenon is often modeled with quite a different scenario. Imagine carving a small hole in an oven that can be heated to very high temperatures. As the temperature of this container gets hotter and hotter, the radiation out of this dark hole would increase as well, re-emitting all energy provided it by the increased temperature. The hole may even begin to glow in different colors as the temperature is increased. Like a burner on your stove, the hole would glow red, then orange, then blue, as the temperature is increased. In time, the hole would continue to glow but the light would be invisible to our eyes. This container is a good model of a perfect blackbody. It is the analysis of blackbodies that led to one of the most consequential discoveries of the twentieth century. Take a moment to carefully examine Figure 21.2. What relationships exist? What trends can you see? The more time you spend interpreting this figure, the closer you will be to understanding quantum physics! Figure 21.2 Graphs of blackbody radiation (from an ideal radiator) at three different radiator temperatures. The intensity or rate of radiation emission increases dramatically with temperature, and the peak of the spectrum shifts toward the visible and ultraviolet parts of the spectrum. The shape of the spectrum cannot be described with classical physics. TIPS FOR SUCCESS When
encountering a new graph, it is best to try to interpret the graph before you read about it. Doing this will make the following text more meaningful and will help to remind yourself of some of the key concepts within the section. Access for free at openstax.org. 21.1 • Planck and Quantum Nature of Light 693 Understanding Blackbody Graphs Figure 21.2 is a plot of radiation intensity against radiated wavelength. In other words, it shows how the intensity of radiated light changes when a blackbody is heated to a particular temperature. It may help to just follow the bottom-most red line labeled 3,000 K, red hot. The graph shows that when a blackbody acquires a temperature of 3,000 K, it radiates energy across the electromagnetic spectrum. However, the energy is most intensely emitted at a wavelength of approximately 1000 nm. This is in the infrared portion of the electromagnetic spectrum. While a body at this temperature would appear red-hotto our eyes, it would truly appear ‘infrared-hot’ if we were able to see the entire spectrum. A few other important notes regarding Figure 21.2: • As temperature increases, the total amount of energy radiated increases. This is shown by examining the area underneath each line. • Regardless of temperature, all red lines on the graph undergo a consistent pattern. While electromagnetic radiation is emitted throughout the spectrum, the intensity of this radiation peaks at one particular wavelength. • As the temperature changes, the wavelength of greatest radiation intensity changes. At 4,000 K, the radiation is most intense in the yellow-green portion of the spectrum. At 6,000 K, the blackbody would radiate white hot,due to intense radiation throughout the visible portion of the electromagnetic spectrum. Remember that white light is the emission of all visible colors simultaneously. • As the temperature increases, the frequency of light providing the greatest intensity increases as well. Recall the equation Because the speed of light is constant, frequency and wavelength are inversely related. This is verified by the leftward movement of the three red lines as temperature is increased. While in science it is important to categorize observations, theorizing as to why the observations exist is crucial to scientific advancement. Why doesn’t a blackbody emit radiation evenly across all wavelengths? Why does the temperature of the body change the peak wavelength that is radiated? Why does an increase in temperature cause the peak wavelength emitted to decrease? It is questions like these that drove significant research at the turn of

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