22 MAGNETISM - AP Physicsapphysicsisps.weebly.com/uploads/5/1/8/0/51801459/ap2... · 2018-08-31 · 22 MAGNETISM Figure 22.1The magnificent spectacle of the Aurora Borealis, or northern

22 MAGNETISM

Figure 22.1 The magnificent spectacle of the Aurora Borealis, or northern lights, glows in the northern sky above Bear Lake near Eielson Air ForceBase, Alaska. Shaped by the Earth’s magnetic field, this light is produced by radiation spewed from solar storms. (credit: Senior Airman Joshua Strang,via Flickr)

Chapter Outline22.1. Magnets

22.2. Ferromagnets and Electromagnets

22.3. Magnetic Fields and Magnetic Field Lines

22.4. Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field

22.5. Force on a Moving Charge in a Magnetic Field: Examples and Applications

22.6. The Hall Effect

22.7. Magnetic Force on a Current-Carrying Conductor

22.8. Torque on a Current Loop: Motors and Meters

22.9. Magnetic Fields Produced by Currents: Ampere’s Law

22.10. Magnetic Force between Two Parallel Conductors

22.11. More Applications of Magnetism

Connection for AP® CoursesMagnetism plays a major role in your everyday life. All electric motors, with uses as diverse as powering refrigerators, startingcars, and moving elevators, contain magnets. Magnetic resonance imaging (MRI) has become an important diagnostic tool in thefield of medicine, and the use of magnetism to explore brain activity is a subject of contemporary research and development.Other applications of magnetism include computer memory, levitation of high-speed trains, the aurora borealis, and, of course,the first important historical use of magnetism: navigation. You will find all of these applications of magnetism linked by a smallnumber of underlying principles.

In this chapter, you will learn that both the internal properties of an object and the movement of charged particles can generate amagnetic field, and you will learn why all magnetic fields have a north and south pole. You will also learn how magnetic fieldsexert forces on objects, resulting in the magnetic alignment that makes a compass work. You will learn how we use this principleto weigh the smallest of subatomic particles with precision and contain superheated plasma to facilitate nuclear fusion.

Big Idea 1 Objects and systems have properties such as mass and charge. Systems may have internal structure.

Enduring Understanding 1.E Materials have many macroscopic properties that result from the arrangement and interactions ofthe atoms and molecules that make up the material.

Essential Knowledge 1.E.5 Matter has a property called magnetic permeability.

Chapter 22 | Magnetism 965

Essential Knowledge 1.E.6 Matter has a property called magnetic dipole moment.

Big Idea 2 Fields existing in space can be used to explain interactions.

Enduring Understanding 2.D A magnetic field is caused by a magnet or a moving electrically charged object. Magnetic fieldsobserved in nature always seem to be produced either by moving charged objects or by magnetic dipoles or combinations ofdipoles and never by single poles.

Essential Knowledge 2.D.1 The magnetic field exerts a force on a moving electrically charged object. That magnetic force isperpendicular to the direction of the velocity of the object and to the magnetic field and is proportional to the magnitude of thecharge, the magnitude of the velocity, and the magnitude of the magnetic field. It also depends on the angle between the velocityand the magnetic field vectors. Treatment is quantitative for angles of 0°, 90°, or 180° and qualitative for other angles.

Essential Knowledge 2.D.2 The magnetic field vectors around a straight wire that carries electric current are tangent toconcentric circles centered on that wire. The field has no component toward the current-carrying wire.

Essential Knowledge 2.D.3 A magnetic dipole placed in a magnetic field, such as the ones created by a magnet or the Earth, willtend to align with the magnetic field vector.

Essential Knowledge 2.D.4 Ferromagnetic materials contain magnetic domains that are themselves magnets.

Big Idea 3 The interactions of an object with other objects can be described by forces.

Enduring Understanding 3.C At the macroscopic level, forces can be categorized as either long-range (action-at-a-distance)forces or contact forces.

Essential Knowledge 3.C.3 A magnetic force results from the interaction of a moving charged object or a magnet with othermoving charged objects or another magnet.

Big Idea 4 Interactions between systems can result in changes in those systems.

Enduring Understanding 4.E The electric and magnetic properties of a system can change in response to the presence of, orchanges in, other objects or systems.

Essential Knowledge 4.E.1 The magnetic properties of some materials can be affected by magnetic fields at the system.Students should focus on the underlying concepts and not the use of the vocabulary.

Figure 22.2 Engineering of technology like iPods would not be possible without a deep understanding of magnetism. (credit: Jesse! S?, Flickr)

22.1 Magnets

Learning ObjectivesBy the end of this section, you will be able to:

• Describe the difference between the north and south poles of a magnet.• Describe how magnetic poles interact with each other.

966 Chapter 22 | Magnetism

This content is available for free at http://cnx.org/content/col11844/1.13

Figure 22.3 Magnets come in various shapes, sizes, and strengths. All have both a north pole and a south pole. There is never an isolated pole (amonopole).

All magnets attract iron, such as that in a refrigerator door. However, magnets may attract or repel other magnets.Experimentation shows that all magnets have two poles. If freely suspended, one pole will point toward the north. The two polesare thus named the north magnetic pole and the south magnetic pole (or more properly, north-seeking and south-seekingpoles, for the attractions in those directions).

Universal Characteristics of Magnets and Magnetic Poles

It is a universal characteristic of all magnets that like poles repel and unlike poles attract. (Note the similarity withelectrostatics: unlike charges attract and like charges repel.)

Further experimentation shows that it is impossible to separate north and south poles in the manner that + and − chargescan be separated.

Figure 22.4 One end of a bar magnet is suspended from a thread that points toward north. The magnet’s two poles are labeled N and S for north-seeking and south-seeking poles, respectively.

Misconception Alert: Earth’s Geographic North Pole Hides an S

The Earth acts like a very large bar magnet with its south-seeking pole near the geographic North Pole. That is why thenorth pole of your compass is attracted toward the geographic north pole of the Earth—because the magnetic pole that isnear the geographic North Pole is actually a south magnetic pole! Confusion arises because the geographic term “NorthPole” has come to be used (incorrectly) for the magnetic pole that is near the North Pole. Thus, “North magnetic pole” isactually a misnomer—it should be called the South magnetic pole.


Figure 22.5 Unlike poles attract, whereas like poles repel.

Figure 22.6 North and south poles always occur in pairs. Attempts to separate them result in more pairs of poles. If we continue to split the magnet, wewill eventually get down to an iron atom with a north pole and a south pole—these, too, cannot be separated.

Real World Connections: Dipoles and Monopoles

Figure 22.6 shows that no matter how many times you divide a magnet the resulting objects are always magnetic dipoles.Formally, a magnetic dipole is an object (usually very small) with a north and south magnetic pole. Magnetic dipoles have avector property called magnetic momentum. The magnitude of this vector is equal to the strength of its poles and thedistance between the poles, and the direction points from the south pole to the north pole.

A magnetic dipole can also be thought of as a very small closed current loop. There is no way to isolate north and southmagnetic poles like you can isolate positive and negative charges. Another way of saying this is that magnetic fields of amagnetic object always make closed loops, starting at a north pole and ending at a south pole.

With a positive charge, you might imagine drawing a spherical surface enclosing that charge, and there would be a net fluxof electric field lines flowing outward through that surface. In fact, Gauss’s law states that the electric flux through a surfaceis proportional to the amount of charge enclosed.

With a magnetic object, every surface you can imagine that encloses all or part of the magnet ultimately has zero net flux ofmagnetic field lines flowing through the surface. Just as many outward-flowing lines from the north pole of the magnet passthrough the surface as inward-flowing lines from the south pole of the magnet.

Some physicists have theorized that magnetic monopoles exist. These would be isolated magnetic “charges” that would onlygenerate field lines that flow outward or inward (not loops). Despite many searches, we have yet to experimentally verify theexistence of magnetic monopoles.



The fact that magnetic poles always occur in pairs of north and south is true from the very large scale—for example, sunspotsalways occur in pairs that are north and south magnetic poles—all the way down to the very small scale. Magnetic atoms haveboth a north pole and a south pole, as do many types of subatomic particles, such as electrons, protons, and neutrons.

Making Connections: Take-Home Experiment—Refrigerator Magnets

We know that like magnetic poles repel and unlike poles attract. See if you can show this for two refrigerator magnets. Willthe magnets stick if you turn them over? Why do they stick to the door anyway? What can you say about the magneticproperties of the door next to the magnet? Do refrigerator magnets stick to metal or plastic spoons? Do they stick to all typesof metal?

22.2 Ferromagnets and Electromagnets


• Define ferromagnet.• Describe the role of magnetic domains in magnetization.• Explain the significance of the Curie temperature.• Describe the relationship between electricity and magnetism.

The information presented in this section supports the following AP® learning objectives and science practices:

• 2.D.3.1 The student is able to describe the orientation of a magnetic dipole placed in a magnetic field in general and theparticular cases of a compass in the magnetic field of the Earth and iron filings surrounding a bar magnet. (S.P. 1.2)

• 2.D.4.1 The student is able to use the representation of magnetic domains to qualitatively analyze the magneticbehavior of a bar magnet composed of ferromagnetic material. (S.P. 1.4)

• 4.E.1.1 The student is able to use representations and models to qualitatively describe the magnetic properties of somematerials that can be affected by magnetic properties of other objects in the system. (S.P. 1.1, 1.4, 2.2)

FerromagnetsOnly certain materials, such as iron, cobalt, nickel, and gadolinium, exhibit strong magnetic effects. Such materials are calledferromagnetic, after the Latin word for iron, ferrum. A group of materials made from the alloys of the rare earth elements arealso used as strong and permanent magnets; a popular one is neodymium. Other materials exhibit weak magnetic effects, whichare detectable only with sensitive instruments. Not only do ferromagnetic materials respond strongly to magnets (the way iron isattracted to magnets), they can also be magnetized themselves—that is, they can be induced to be magnetic or made intopermanent magnets.

Figure 22.7 An unmagnetized piece of iron is placed between two magnets, heated, and then cooled, or simply tapped when cold. The iron becomes apermanent 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 tothe south pole of the original magnet. Note that there are attractive forces between the magnets.

When a magnet is brought near a previously unmagnetized ferromagnetic material, it causes local magnetization of the materialwith unlike poles closest, as in Figure 22.7. (This results in the attraction of the previously unmagnetized material to the magnet.)What happens on a microscopic scale is illustrated in Figure 22.8. The regions within the material called domains act like smallbar magnets. Within domains, the poles of individual atoms are aligned. Each atom acts like a tiny bar magnet. Domains aresmall and randomly oriented in an unmagnetized ferromagnetic object. In response to an external magnetic field, the domainsmay grow to millimeter size, aligning themselves as shown in Figure 22.8(b). This induced magnetization can be madepermanent if the material is heated and then cooled, or simply tapped in the presence of other magnets.


Figure 22.8 (a) An unmagnetized piece of iron (or other ferromagnetic material) has randomly oriented domains. (b) When magnetized by an externalfield, the domains show greater alignment, and some grow at the expense of others. Individual atoms are aligned within domains; each atom acts like atiny 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 the size of the domains. There isa well-defined temperature for ferromagnetic materials, which is called the Curie temperature, above which they cannot bemagnetized. The Curie temperature for iron is 1043 K (770ºC) , which is well above room temperature. There are several

elements and alloys that have Curie temperatures much lower than room temperature and are ferromagnetic only below thosetemperatures.

ElectromagnetsEarly in the 19th century, it was discovered that electrical currents cause magnetic effects. The first significant observation wasby 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 charges had any connection with magnets.Electromagnetism is the use of electric current to make magnets. These temporarily induced magnets are calledelectromagnets. Electromagnets are employed for everything from a wrecking yard crane that lifts scrapped cars to controllingthe beam of a 90-km-circumference particle accelerator to the magnets in medical imaging machines (See Figure 22.9).

Figure 22.9 Instrument for magnetic resonance imaging (MRI). The device uses a superconducting cylindrical coil for the main magnetic field. Thepatient goes into this “tunnel” on the gurney. (credit: Bill McChesney, Flickr)

Figure 22.10 shows that the response of iron filings to a current-carrying coil and to a permanent bar magnet. The patterns aresimilar. In fact, electromagnets and ferromagnets have the same basic characteristics—for example, they have north and southpoles that cannot be separated and for which like poles repel and unlike poles attract.



Figure 22.10 Iron filings near (a) a current-carrying coil and (b) a magnet act like tiny compass needles, showing the shape of their fields. Theirresponse to a current-carrying coil and a permanent magnet is seen to be very similar, especially near the ends of the coil and the magnet.

Making Connections: Compasses

Figure 22.11 Compass needles and a bar magnet.

In Figure 22.11, a series of tiny compass needles are placed in an external magnetic field. These dipoles respond exactlylike the iron filings in Figure 22.10. For needles close to one pole of the magnet, the needle is aligned so that the oppositepole of the needle points at the bar magnet. For example, close to the north pole of the bar magnet, the south pole of thecompass needle is aligned to be closest to the bar magnet.

As an experimenter moves each compass around, the needle will rotate in such a way as to orient itself with the field of thebar magnet at that location. In this way, the magnetic field lines may be mapped out precisely.

The strength of the magnetic field depends on the medium in which the magnetic field exists. Some substances (like iron)respond to external magnetic fields in a way that amplifies the external magnetic field. The magnetic permeability of asubstance is a measure of the substance’s ability to support or amplify an already existing external magnetic field.Ferromagnets, in which magnetic domains in the substance align with and amplify an external magnetic field, are examplesof objects with high permeability.

Combining a ferromagnet with an electromagnet can produce particularly strong magnetic effects. (See Figure 22.12.) Wheneverstrong magnetic effects are needed, such as lifting scrap metal, or in particle accelerators, electromagnets are enhanced byferromagnetic materials. Limits to how strong the magnets can be made are imposed by coil resistance (it will overheat and meltat sufficiently high current), and so superconducting magnets may be employed. These are still limited, because superconductingproperties are destroyed by too great a magnetic field.


Figure 22.12 An electromagnet with a ferromagnetic core can produce very strong magnetic effects. Alignment of domains in the core produces amagnet, the poles of which are aligned with the electromagnet.

Figure 22.13 shows a few uses of combinations of electromagnets and ferromagnets. Ferromagnetic materials can act asmemory devices, because the orientation of the magnetic fields of small domains can be reversed or erased. Magneticinformation storage on videotapes and computer hard drives are among the most common applications. This property is vital inour digital world.

Figure 22.13 An electromagnet induces regions of permanent magnetism on a floppy disk coated with a ferromagnetic material. The information storedhere is digital (a region is either magnetic or not); in other applications, it can be analog (with a varying strength), such as on audiotapes.

Current: The Source of All MagnetismAn electromagnet creates magnetism with an electric current. In later sections we explore this more quantitatively, finding thestrength and direction of magnetic fields created by various currents. But what about ferromagnets? Figure 22.14 shows modelsof how electric currents create magnetism at the submicroscopic level. (Note that we cannot directly observe the paths ofindividual electrons about atoms, and so a model or visual image, consistent with all direct observations, is made. We candirectly observe the electron’s orbital angular momentum, its spin momentum, and subsequent magnetic moments, all of whichare explained with electric-current-creating subatomic magnetism.) Currents, including those associated with othersubmicroscopic particles like protons, allow us to explain ferromagnetism and all other magnetic effects. Ferromagnetism, forexample, results from an internal cooperative alignment of electron spins, possible in some materials but not in others.

Crucial to the statement that electric current is the source of all magnetism is the fact that it is impossible to separate north andsouth magnetic poles. (This is far different from the case of positive and negative charges, which are easily separated.) A currentloop always produces a magnetic dipole—that is, a magnetic field that acts like a north pole and south pole pair. Since isolatednorth and south magnetic poles, called magnetic monopoles, are not observed, currents are used to explain all magneticeffects. If magnetic monopoles did exist, then we would have to modify this underlying connection that all magnetism is due toelectrical current. There is no known reason that magnetic monopoles should not exist—they are simply never observed—and so



searches at the subnuclear level continue. If they do not exist, we would like to find out why not. If they do exist, we would like tosee evidence of them.

Electric Currents and Magnetism

Electric current is the source of all magnetism.

Figure 22.14 (a) In the planetary model of the atom, an electron orbits a nucleus, forming a closed-current loop and producing a magnetic field with anorth pole and a south pole. (b) Electrons have spin and can be crudely pictured as rotating charge, forming a current that produces a magnetic fieldwith a north pole and a south pole. Neither the planetary model nor the image of a spinning electron is completely consistent with modern physics.However, they do provide a useful way of understanding phenomena.

PhET Explorations: Magnets and Electromagnets

Explore the interactions between a compass and bar magnet. Discover how you can use a battery and wire to make amagnet! Can you make it a stronger magnet? Can you make the magnetic field reverse?

Figure 22.15 Magnets and Electromagnets (http://cnx.org/content/m55375/1.2/magnets-and-electromagnets_en.jar)

22.3 Magnetic Fields and Magnetic Field Lines


• Define magnetic field and describe the magnetic field lines of various magnetic fields.

Einstein is said to have been fascinated by a compass as a child, perhaps musing on how the needle felt a force without directphysical contact. His ability to think deeply and clearly about action at a distance, particularly for gravitational, electric, andmagnetic forces, later enabled him to create his revolutionary theory of relativity. Since magnetic forces act at a distance, wedefine a magnetic field to represent magnetic forces. The pictorial representation of magnetic field lines is very useful invisualizing the strength and direction of the magnetic field. As shown in Figure 22.16, the direction of magnetic field lines isdefined to be the direction in which the north end of a compass needle points. The magnetic field is traditionally called the B-field.


http://cnx.org/content/m55375/1.2/magnets-and-electromagnets_en.jar

Figure 22.16 Magnetic field lines are defined to have the direction that a small compass points when placed at a location. (a) If small compasses areused to map the magnetic field around a bar magnet, they will point in the directions shown: away from the north pole of the magnet, toward the southpole of the magnet. (Recall that the Earth’s north magnetic pole is really a south pole in terms of definitions of poles on a bar magnet.) (b) Connectingthe arrows gives continuous magnetic field lines. The strength of the field is proportional to the closeness (or density) of the lines. (c) If the interior ofthe magnet could be probed, the field lines would be found to form continuous closed loops.

Small compasses used to test a magnetic field will not disturb it. (This is analogous to the way we tested electric fields with asmall test charge. In both cases, the fields represent only the object creating them and not the probe testing them.) Figure 22.17shows how the magnetic field appears for a current loop and a long straight wire, as could be explored with small compasses. Asmall compass placed in these fields will align itself parallel to the field line at its location, with its north pole pointing in thedirection of B. Note the symbols used for field into and out of the paper.

Figure 22.17 Small compasses could be used to map the fields shown here. (a) The magnetic field of a circular current loop is similar to that of a barmagnet. (b) A long and straight wire creates a field with magnetic field lines forming circular loops. (c) When the wire is in the plane of the paper, thefield is perpendicular to the paper. Note that the symbols used for the field pointing inward (like the tail of an arrow) and the field pointing outward (likethe tip of an arrow).

Making Connections: Concept of a Field

A field is a way of mapping forces surrounding any object that can act on another object at a distance without apparentphysical connection. The field represents the object generating it. Gravitational fields map gravitational forces, electric fieldsmap electrical forces, and magnetic fields map magnetic forces.

Extensive exploration of magnetic fields has revealed a number of hard-and-fast rules. We use magnetic field lines to representthe field (the lines are a pictorial tool, not a physical entity in and of themselves). The properties of magnetic field lines can besummarized by these rules:

1. The direction of the magnetic field is tangent to the field line at any point in space. A small compass will point in thedirection of the field line.

2. The strength of the field is proportional to the closeness of the lines. It is exactly proportional to the number of lines per unitarea perpendicular to the lines (called the areal density).

3. Magnetic field lines can never cross, meaning that the field is unique at any point in space.

4. Magnetic field lines are continuous, forming closed loops without beginning or end. They go from the north pole to the southpole.

The last property is related to the fact that the north and south poles cannot be separated. It is a distinct difference from electricfield lines, which begin and end on the positive and negative charges. If magnetic monopoles existed, then magnetic field lineswould begin and end on them.



22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field


• Describe the effects of magnetic fields on moving charges.• Use the right-hand rule 1 to determine the velocity of a charge, the direction of the magnetic field, and the direction of

magnetic force on a moving charge.• Calculate the magnetic force on a moving charge.


• 2.D.1.1 The student is able to apply mathematical routines to express the force exerted on a moving charged object bya magnetic field. (S.P. 2.2)

• 3.C.3.1 The student is able to use right-hand rules to analyze a situation involving a current-carrying conductor and amoving electrically charged object to determine the direction of the magnetic force exerted on the charged object due tothe magnetic field created by the current-carrying conductor. (S.P. 1.4)

What is the mechanism by which one magnet exerts a force on another? The answer is related to the fact that all magnetism iscaused by current, the flow of charge. Magnetic fields exert forces on moving charges, and so they exert forces on othermagnets, all of which have moving charges.

Right Hand Rule 1The magnetic force on a moving charge is one of the most fundamental known. Magnetic force is as important as theelectrostatic or Coulomb force. Yet the magnetic force is more complex, in both the number of factors that affects it and in itsdirection, than the relatively simple Coulomb force. The magnitude of the magnetic force F on a charge q moving at a speed

v in a magnetic field of strength B is given by

(22.1)F = qvB sin θ,

where θ is the angle between the directions of v and B. This force is often called the Lorentz force. In fact, this is how we

define the magnetic field strength B —in terms of the force on a charged particle moving in a magnetic field. The SI unit for

magnetic field strength B is called the tesla (T) after the eccentric but brilliant inventor Nikola Tesla (1856–1943). To determine

how the tesla relates to other SI units, we solve F = qvB sin θ for B .

(22.2)B = Fqv sin θ

Because sin θ is unitless, the tesla is

(22.3)1 T = 1 NC ⋅ m/s = 1 N

A ⋅ m(note that C/s = A).

Another smaller unit, called the gauss (G), where 1 G = 10−4 T , is sometimes used. The strongest permanent magnets havefields near 2 T; superconducting electromagnets may attain 10 T or more. The Earth’s magnetic field on its surface is only about

5×10−5 T , or 0.5 G.

The direction of the magnetic force F is perpendicular to the plane formed by v and B , as determined by the right hand rule1 (or RHR-1), which is illustrated in Figure 22.18. RHR-1 states that, to determine the direction of the magnetic force on apositive moving charge, you point the thumb of the right hand in the direction of v , the fingers in the direction of B , and a

perpendicular to the palm points in the direction of F . One way to remember this is that there is one velocity, and so the thumbrepresents it. There are many field lines, and so the fingers represent them. The force is in the direction you would push withyour palm. The force on a negative charge is in exactly the opposite direction to that on a positive charge.


Figure 22.18 Magnetic fields exert forces on moving charges. This force is one of the most basic known. The direction of the magnetic force on amoving charge is perpendicular to the plane formed by v and B and follows right hand rule–1 (RHR-1) as shown. The magnitude of the force is

proportional to q , v , B , and the sine of the angle between v and B .

Making Connections: Charges and Magnets

There is no magnetic force on static charges. However, there is a magnetic force on moving charges. When charges arestationary, their electric fields do not affect magnets. But, when charges move, they produce magnetic fields that exert forceson other magnets. When there is relative motion, a connection between electric and magnetic fields emerges—each affectsthe other.

Example 22.1 Calculating Magnetic Force: Earth’s Magnetic Field on a Charged Glass Rod

With the exception of compasses, you seldom see or personally experience forces due to the Earth’s small magnetic field.To illustrate this, suppose that in a physics lab you rub a glass rod with silk, placing a 20-nC positive charge on it. Calculatethe force on the rod due to the Earth’s magnetic field, if you throw it with a horizontal velocity of 10 m/s due west in a placewhere the Earth’s field is due north parallel to the ground. (The direction of the force is determined with right hand rule 1 asshown in Figure 22.19.)

Figure 22.19 A positively charged object moving due west in a region where the Earth’s magnetic field is due north experiences a force that isstraight down as shown. A negative charge moving in the same direction would feel a force straight up.

Strategy



We are given the charge, its velocity, and the magnetic field strength and direction. We can thus use the equationF = qvB sin θ to find the force.

Solution

The magnetic force is

(22.4)F = qvb sin θ.

We see that sin θ = 1 , since the angle between the velocity and the direction of the field is 90º . Entering the other givenquantities yields

(22.5)F = ⎛⎝20×10–9 C⎞

⎠(10 m/s)⎛⎝5×10–5 T⎞

⎠

= 1×10–11 (C ⋅ m/s)⎛⎝N

C ⋅ m/s⎞⎠ = 1×10–11 N.

Discussion

This force is completely negligible on any macroscopic object, consistent with experience. (It is calculated to only one digit,since the Earth’s field varies with location and is given to only one digit.) The Earth’s magnetic field, however, does producevery important effects, particularly on submicroscopic particles. Some of these are explored in Force on a Moving Chargein a Magnetic Field: Examples and Applications.

22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications


• Describe the effects of a magnetic field on a moving charge.• Calculate the radius of curvature of the path of a charge that is moving in a magnetic field.



Magnetic force can cause a charged particle to move in a circular or spiral path. Cosmic rays are energetic charged particles inouter space, some of which approach the Earth. They can be forced into spiral paths by the Earth’s magnetic field. Protons ingiant accelerators are kept in a circular path by magnetic force. The bubble chamber photograph in Figure 22.20 shows chargedparticles moving in such curved paths. The curved paths of charged particles in magnetic fields are the basis of a number ofphenomena and can even be used analytically, such as in a mass spectrometer.

Figure 22.20 Trails of bubbles are produced by high-energy charged particles moving through the superheated liquid hydrogen in this artist’s renditionof a bubble chamber. There is a strong magnetic field perpendicular to the page that causes the curved paths of the particles. The radius of the pathcan be used to find the mass, charge, and energy of the particle.

So does the magnetic force cause circular motion? Magnetic force is always perpendicular to velocity, so that it does no work onthe charged particle. The particle’s kinetic energy and speed thus remain constant. The direction of motion is affected, but not thespeed. This is typical of uniform circular motion. The simplest case occurs when a charged particle moves perpendicular to a


uniform B -field, such as shown in Figure 22.21. (If this takes place in a vacuum, the magnetic field is the dominant factor

determining the motion.) Here, the magnetic force supplies the centripetal force Fc = mv2 / r . Noting that sin θ = 1 , we see

that F = qvB .

Figure 22.21 A negatively charged particle moves in the plane of the page in a region where the magnetic field is perpendicular into the page(represented by the small circles with x’s—like the tails of arrows). The magnetic force is perpendicular to the velocity, and so velocity changes indirection but not magnitude. Uniform circular motion results.

Because the magnetic force F supplies the centripetal force Fc , we have

(22.6)qvB = mv2

r .

Solving for r yields

(22.7)r = mvqB .

Here, r is the radius of curvature of the path of a charged particle with mass m and charge q , moving at a speed vperpendicular to a magnetic field of strength B . If the velocity is not perpendicular to the magnetic field, then v is thecomponent of the velocity perpendicular to the field. The component of the velocity parallel to the field is unaffected, since themagnetic force is zero for motion parallel to the field. This produces a spiral motion rather than a circular one.

Example 22.2 Calculating the Curvature of the Path of an Electron Moving in a Magnetic Field: AMagnet on a TV Screen

A magnet brought near an old-fashioned TV screen such as in Figure 22.22 (TV sets with cathode ray tubes instead of LCDscreens) severely distorts its picture by altering the path of the electrons that make its phosphors glow. (Don’t try this athome, as it will permanently magnetize and ruin the TV.) To illustrate this, calculate the radius of curvature of the path of

an electron having a velocity of 6.00×107 m/s (corresponding to the accelerating voltage of about 10.0 kV used in some

TVs) perpendicular to a magnetic field of strength B = 0.500 T (obtainable with permanent magnets).



Figure 22.22 Side view showing what happens when a magnet comes in contact with a computer monitor or TV screen. Electrons moving towardthe screen spiral about magnetic field lines, maintaining the component of their velocity parallel to the field lines. This distorts the image on thescreen.

Strategy

We can find the radius of curvature r directly from the equation r = mvqB , since all other quantities in it are given or known.

Solution

Using known values for the mass and charge of an electron, along with the given values of v and B gives us

(22.8)r = mv

qB =⎛⎝9.11×10−31 kg⎞

⎠⎛⎝6.00×107 m/s⎞

⎠⎛⎝1.60×10−19 C⎞

⎠(0.500 T)

= 6.83×10−4 mor

(22.9)r = 0.683 mm.Discussion

The small radius indicates a large effect. The electrons in the TV picture tube are made to move in very tight circles, greatlyaltering their paths and distorting the image.

Figure 22.23 shows how electrons not moving perpendicular to magnetic field lines follow the field lines. The component ofvelocity parallel to the lines is unaffected, and so the charges spiral along the field lines. If field strength increases in the directionof motion, the field will exert a force to slow the charges, forming a kind of magnetic mirror, as shown below.

Figure 22.23 When a charged particle moves along a magnetic field line into a region where the field becomes stronger, the particle experiences aforce that reduces the component of velocity parallel to the field. This force slows the motion along the field line and here reverses it, forming a“magnetic mirror.”

The properties of charged particles in magnetic fields are related to such different things as the Aurora Australis or AuroraBorealis and particle accelerators. Charged particles approaching magnetic field lines may get trapped in spiral orbits about thelines rather than crossing them, as seen above. Some cosmic rays, for example, follow the Earth’s magnetic field lines, enteringthe atmosphere near the magnetic poles and causing the southern or northern lights through their ionization of molecules in the


atmosphere. This glow of energized atoms and molecules is seen in Figure 22.1. Those particles that approach middle latitudesmust cross magnetic field lines, and many are prevented from penetrating the atmosphere. Cosmic rays are a component ofbackground radiation; consequently, they give a higher radiation dose at the poles than at the equator.

Figure 22.24 Energetic electrons and protons, components of cosmic rays, from the Sun and deep outer space often follow the Earth’s magnetic fieldlines rather than cross them. (Recall that the Earth’s north magnetic pole is really a south pole in terms of a bar magnet.)

Some incoming charged particles become trapped in the Earth’s magnetic field, forming two belts above the atmosphere knownas the Van Allen radiation belts after the discoverer James A. Van Allen, an American astrophysicist. (See Figure 22.25.)Particles trapped in these belts form radiation fields (similar to nuclear radiation) so intense that manned space flights avoid themand satellites with sensitive electronics are kept out of them. In the few minutes it took lunar missions to cross the Van Allenradiation belts, astronauts received radiation doses more than twice the allowed annual exposure for radiation workers. Otherplanets have similar belts, especially those having strong magnetic fields like Jupiter.

Figure 22.25 The Van Allen radiation belts are two regions in which energetic charged particles are trapped in the Earth’s magnetic field. One belt liesabout 300 km above the Earth’s surface, the other about 16,000 km. Charged particles in these belts migrate along magnetic field lines and arepartially reflected away from the poles by the stronger fields there. The charged particles that enter the atmosphere are replenished by the Sun andsources in deep outer space.

Back on Earth, we have devices that employ magnetic fields to contain charged particles. Among them are the giant particleaccelerators that have been used to explore the substructure of matter. (See Figure 22.26.) Magnetic fields not only control thedirection of the charged particles, they also are used to focus particles into beams and overcome the repulsion of like charges inthese beams.



Figure 22.26 The Fermilab facility in Illinois has a large particle accelerator (the most powerful in the world until 2008) that employs magnetic fields(magnets seen here in orange) to contain and direct its beam. This and other accelerators have been in use for several decades and have allowed usto discover some of the laws underlying all matter. (credit: ammcrim, Flickr)

Thermonuclear fusion (like that occurring in the Sun) is a hope for a future clean energy source. One of the most promisingdevices is the tokamak, which uses magnetic fields to contain (or trap) and direct the reactive charged particles. (See Figure22.27.) Less exotic, but more immediately practical, amplifiers in microwave ovens use a magnetic field to contain oscillatingelectrons. These oscillating electrons generate the microwaves sent into the oven.

Figure 22.27 Tokamaks such as the one shown in the figure are being studied with the goal of economical production of energy by nuclear fusion.Magnetic fields in the doughnut-shaped device contain and direct the reactive charged particles. (credit: David Mellis, Flickr)

Mass spectrometers have a variety of designs, and many use magnetic fields to measure mass. The curvature of a chargedparticle’s path in the field is related to its mass and is measured to obtain mass information. (See More Applications ofMagnetism.) Historically, such techniques were employed in the first direct observations of electron charge and mass. Today,mass spectrometers (sometimes coupled with gas chromatographs) are used to determine the make-up and sequencing of largebiological molecules.

22.6 The Hall Effect


• Describe the Hall effect.• Calculate the Hall emf across a current-carrying conductor.




We have seen effects of a magnetic field on free-moving charges. The magnetic field also affects charges moving in a conductor.One result is the Hall effect, which has important implications and applications.

Figure 22.28 shows what happens to charges moving through a conductor in a magnetic field. The field is perpendicular to theelectron drift velocity and to the width of the conductor. Note that conventional current is to the right in both parts of the figure. Inpart (a), electrons carry the current and move to the left. In part (b), positive charges carry the current and move to the right.Moving electrons feel a magnetic force toward one side of the conductor, leaving a net positive charge on the other side. Thisseparation of charge creates a voltage ε , known as the Hall emf, across the conductor. The creation of a voltage across acurrent-carrying conductor by a magnetic field is known as the Hall effect, after Edwin Hall, the American physicist whodiscovered it in 1879.

Figure 22.28 The Hall effect. (a) Electrons move to the left in this flat conductor (conventional current to the right). The magnetic field is directly out ofthe page, represented by circled dots; it exerts a force on the moving charges, causing a voltage ε , the Hall emf, across the conductor. (b) Positivecharges moving to the right (conventional current also to the right) are moved to the side, producing a Hall emf of the opposite sign, –ε . Thus, if thedirection of the field and current are known, the sign of the charge carriers can be determined from the Hall effect.

One very important use of the Hall effect is to determine whether positive or negative charges carries the current. Note that inFigure 22.28(b), where positive charges carry the current, the Hall emf has the sign opposite to when negative charges carry thecurrent. Historically, the Hall effect was used to show that electrons carry current in metals and it also shows that positivecharges carry current in some semiconductors. The Hall effect is used today as a research tool to probe the movement ofcharges, their drift velocities and densities, and so on, in materials. In 1980, it was discovered that the Hall effect is quantized, anexample of quantum behavior in a macroscopic object.

The Hall effect has other uses that range from the determination of blood flow rate to precision measurement of magnetic fieldstrength. To examine these quantitatively, we need an expression for the Hall emf, ε , across a conductor. Consider the balance

of forces on a moving charge in a situation where B , v , and l are mutually perpendicular, such as shown in Figure 22.29.Although the magnetic force moves negative charges to one side, they cannot build up without limit. The electric field caused bytheir separation opposes the magnetic force, F = qvB , and the electric force, Fe = qE , eventually grows to equal it. That is,

(22.10)qE = qvB

or

(22.11)E = vB.

Note that the electric field E is uniform across the conductor because the magnetic field B is uniform, as is the conductor. For

a uniform electric field, the relationship between electric field and voltage is E = ε / l , where l is the width of the conductor andε is the Hall emf. Entering this into the last expression gives

(22.12)εl = vB.

Solving this for the Hall emf yields

(22.13)ε = Blv (B, v, and l, mutually perpendicular),

where ε is the Hall effect voltage across a conductor of width l through which charges move at a speed v .



Figure 22.29 The Hall emf ε produces an electric force that balances the magnetic force on the moving charges. The magnetic force producescharge separation, which builds up until it is balanced by the electric force, an equilibrium that is quickly reached.

One of the most common uses of the Hall effect is in the measurement of magnetic field strength B . Such devices, called Hallprobes, can be made very small, allowing fine position mapping. Hall probes can also be made very accurate, usuallyaccomplished by careful calibration. Another application of the Hall effect is to measure fluid flow in any fluid that has freecharges (most do). (See Figure 22.30.) A magnetic field applied perpendicular to the flow direction produces a Hall emf ε as

shown. Note that the sign of ε depends not on the sign of the charges, but only on the directions of B and v . The magnitude

of the Hall emf is ε = Blv , where l is the pipe diameter, so that the average velocity v can be determined from ε providingthe other factors are known.

Figure 22.30 The Hall effect can be used to measure fluid flow in any fluid having free charges, such as blood. The Hall emf ε is measured acrossthe tube perpendicular to the applied magnetic field and is proportional to the average velocity v .

Example 22.3 Calculating the Hall emf: Hall Effect for Blood Flow

A Hall effect flow probe is placed on an artery, applying a 0.100-T magnetic field across it, in a setup similar to that in Figure22.30. What is the Hall emf, given the vessel’s inside diameter is 4.00 mm and the average blood velocity is 20.0 cm/s?

Strategy

Because B , v , and l are mutually perpendicular, the equation ε = Blv can be used to find ε .

Solution

Entering the given values for B , v , and l gives

(22.14)ε = Blv = (0.100 T)⎛⎝4.00×10−3 m⎞

⎠(0.200 m/s)= 80.0 μV


Discussion

This is the average voltage output. Instantaneous voltage varies with pulsating blood flow. The voltage is small in this type ofmeasurement. ε is particularly difficult to measure, because there are voltages associated with heart action (ECG voltages)that are on the order of millivolts. In practice, this difficulty is overcome by applying an AC magnetic field, so that the Hall emfis AC with the same frequency. An amplifier can be very selective in picking out only the appropriate frequency, eliminatingsignals and noise at other frequencies.

22.7 Magnetic Force on a Current-Carrying Conductor


• Describe the effects of a magnetic force on a current-carrying conductor.• Calculate the magnetic force on a current-carrying conductor.



Because charges ordinarily cannot escape a conductor, the magnetic force on charges moving in a conductor is transmitted tothe conductor itself.

Figure 22.31 The magnetic field exerts a force on a current-carrying wire in a direction given by the right hand rule 1 (the same direction as that on theindividual moving charges). This force can easily be large enough to move the wire, since typical currents consist of very large numbers of movingcharges.

We can derive an expression for the magnetic force on a current by taking a sum of the magnetic forces on individual charges.(The forces add because they are in the same direction.) The force on an individual charge moving at the drift velocity vd is

given by F = qvdB sin θ . Taking B to be uniform over a length of wire l and zero elsewhere, the total magnetic force on the

wire is then F = (qvdB sin θ)(N) , where N is the number of charge carriers in the section of wire of length l . Now,

N = nV , where n is the number of charge carriers per unit volume and V is the volume of wire in the field. Noting that

V = Al , where A is the cross-sectional area of the wire, then the force on the wire is F = (qvdB sin θ)(nAl) . Gathering

terms,

(22.15)F = (nqAvd)lB sin θ.

Because nqAvd = I (see Current),

(22.16)F = IlB sin θ

is the equation for magnetic force on a length l of wire carrying a current I in a uniform magnetic field B , as shown in Figure22.32. If we divide both sides of this expression by l , we find that the magnetic force per unit length of wire in a uniform field is

Fl = IB sin θ . The direction of this force is given by RHR-1, with the thumb in the direction of the current I . Then, with the

fingers in the direction of B , a perpendicular to the palm points in the direction of F , as in Figure 22.32.



Figure 22.32 The force on a current-carrying wire in a magnetic field is F = IlB sin θ . Its direction is given by RHR-1.

Example 22.4 Calculating Magnetic Force on a Current-Carrying Wire: A Strong Magnetic Field

Calculate the force on the wire shown in Figure 22.31, given B = 1.50 T , l = 5.00 cm , and I = 20.0 A .

Strategy

The force can be found with the given information by using F = IlB sin θ and noting that the angle θ between I and Bis 90º , so that sin θ = 1 .

Solution

Entering the given values into F = IlB sin θ yields

(22.17)F = IlB sin θ = (20.0 A)(0.0500 m)(1.50 T)(1).

The units for tesla are 1 T = NA ⋅ m ; thus,

(22.18)F = 1.50 N.Discussion

This large magnetic field creates a significant force on a small length of wire.

Magnetic force on current-carrying conductors is used to convert electric energy to work. (Motors are a prime example—theyemploy loops of wire and are considered in the next section.) Magnetohydrodynamics (MHD) is the technical name given to aclever application where magnetic force pumps fluids without moving mechanical parts. (See Figure 22.33.)

Figure 22.33 Magnetohydrodynamics. The magnetic force on the current passed through this fluid can be used as a nonmechanical pump.

A strong magnetic field is applied across a tube and a current is passed through the fluid at right angles to the field, resulting in aforce on the fluid parallel to the tube axis as shown. The absence of moving parts makes this attractive for moving a hot,chemically active substance, such as the liquid sodium employed in some nuclear reactors. Experimental artificial hearts are


testing with this technique for pumping blood, perhaps circumventing the adverse effects of mechanical pumps. (Cellmembranes, however, are affected by the large fields needed in MHD, delaying its practical application in humans.) MHDpropulsion for nuclear submarines has been proposed, because it could be considerably quieter than conventional propellerdrives. The deterrent value of nuclear submarines is based on their ability to hide and survive a first or second nuclear strike. Aswe slowly disassemble our nuclear weapons arsenals, the submarine branch will be the last to be decommissioned because ofthis ability (See Figure 22.34.) Existing MHD drives are heavy and inefficient—much development work is needed.

Figure 22.34 An MHD propulsion system in a nuclear submarine could produce significantly less turbulence than propellers and allow it to run moresilently. The development of a silent drive submarine was dramatized in the book and the film The Hunt for Red October.

22.8 Torque on a Current Loop: Motors and Meters


• Describe how motors and meters work in terms of torque on a current loop.• Calculate the torque on a current-carrying loop in a magnetic field.

Motors are the most common application of magnetic force on current-carrying wires. Motors have loops of wire in a magneticfield. When current is passed through the loops, the magnetic field exerts torque on the loops, which rotates a shaft. Electricalenergy is converted to mechanical work in the process. (See Figure 22.35.)

Figure 22.35 Torque on a current loop. A current-carrying loop of wire attached to a vertically rotating shaft feels magnetic forces that produce aclockwise torque as viewed from above.

Let us examine the force on each segment of the loop in Figure 22.35 to find the torques produced about the axis of the verticalshaft. (This will lead to a useful equation for the torque on the loop.) We take the magnetic field to be uniform over therectangular loop, which has width w and height l . First, we note that the forces on the top and bottom segments are verticaland, therefore, parallel to the shaft, producing no torque. Those vertical forces are equal in magnitude and opposite in direction,so that they also produce no net force on the loop. Figure 22.36 shows views of the loop from above. Torque is defined asτ = rF sin θ , where F is the force, r is the distance from the pivot that the force is applied, and θ is the angle between rand F . As seen in Figure 22.36(a), right hand rule 1 gives the forces on the sides to be equal in magnitude and opposite in



direction, so that the net force is again zero. However, each force produces a clockwise torque. Since r = w / 2 , the torque on

each vertical segment is (w / 2)F sin θ , and the two add to give a total torque.

(22.19)τ = w2 F sin θ + w

2 F sin θ = wF sin θ

Figure 22.36 Top views of a current-carrying loop in a magnetic field. (a) The equation for torque is derived using this view. Note that the perpendicularto the loop makes an angle θ with the field that is the same as the angle between w / 2 and F . (b) The maximum torque occurs when θ is a right

angle and sin θ = 1 . (c) Zero (minimum) torque occurs when θ is zero and sin θ = 0 . (d) The torque reverses once the loop rotates past

θ = 0 .

Now, each vertical segment has a length l that is perpendicular to B , so that the force on each is F = IlB . Entering F intothe expression for torque yields

(22.20)τ = wIlB sin θ.

If we have a multiple loop of N turns, we get N times the torque of one loop. Finally, note that the area of the loop is A = wl ;the expression for the torque becomes

(22.21)τ = NIAB sin θ.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 anyshape. The loop carries a current I , has N turns, each of area A , and the perpendicular to the loop makes an angle θ with

the field B . The net force on the loop is zero.

Example 22.5 Calculating Torque on a Current-Carrying Loop in a Strong Magnetic Field

Find the maximum torque on a 100-turn square loop of a wire of 10.0 cm on a side that carries 15.0 A of current in a 2.00-Tfield.


Strategy

Torque on the loop can be found using τ = NIAB sin θ . Maximum torque occurs when θ = 90º and sin θ = 1 .

Solution

For sin θ = 1 , the maximum torque is

(22.22)τmax = NIAB.

Entering known values yields

(22.23)τmax = (100)(15.0 A)⎛⎝0.100 m2⎞

⎠(2.00 T)= 30.0 N ⋅ m.

Discussion

This torque is large enough to be useful in a motor.

The torque found in the preceding example is the maximum. As the coil rotates, the torque decreases to zero at θ = 0 . The

torque then reverses its direction once the coil rotates past θ = 0 . (See Figure 22.36(d).) This means that, unless we do

something, the coil will oscillate back and forth about equilibrium at θ = 0 . To get the coil to continue rotating in the same

direction, we can reverse the current as it passes through θ = 0 with automatic switches called brushes. (See Figure 22.37.)

Figure 22.37 (a) As the angular momentum of the coil carries it through θ = 0 , the brushes reverse the current to keep the torque clockwise. (b) Thecoil will rotate continuously in the clockwise direction, with the current reversing each half revolution to maintain the clockwise torque.

Meters, such as those in analog fuel gauges on a car, are another common application of magnetic torque on a current-carryingloop. Figure 22.38 shows that a meter is very similar in construction to a motor. The meter in the figure has its magnets shapedto limit the effect of θ by making B perpendicular to the loop over a large angular range. Thus the torque is proportional to Iand not θ . A linear spring exerts a counter-torque that balances the current-produced torque. This makes the needle deflection

proportional to I . If an exact proportionality cannot be achieved, the gauge reading can be calibrated. To produce agalvanometer for use in analog voltmeters and ammeters that have a low resistance and respond to small currents, we use alarge loop area A , high magnetic field B , and low-resistance coils.



Figure 22.38 Meters are very similar to motors but only rotate through a part of a revolution. The magnetic poles of this meter are shaped to keep thecomponent of B perpendicular to the loop constant, so that the torque does not depend on θ and the deflection against the return spring is

proportional only to the current I .

22.9 Magnetic Fields Produced by Currents: Ampere’s Law


• Calculate current that produces a magnetic field.• Use the right-hand rule 2 to determine the direction of current or the direction of magnetic field loops.


• 2.D.2.1 The student is able to create a verbal or visual representation of a magnetic field around a long straight wire ora pair of parallel wires. (S.P. 1.1)


• 3.C.3.2 The student is able to plan a data collection strategy appropriate to an investigation of the direction of the forceon a moving electrically charged object caused by a current in a wire in the context of a specific set of equipment andinstruments and analyze the resulting data to arrive at a conclusion. (S.P. 4.2, 5.1)

How much current is needed to produce a significant magnetic field, perhaps as strong as the Earth’s field? Surveyors will tellyou that overhead electric power lines create magnetic fields that interfere with their compass readings. Indeed, when Oersteddiscovered in 1820 that a current in a wire affected a compass needle, he was not dealing with extremely large currents. Howdoes the shape of wires carrying current affect the shape of the magnetic field created? We noted earlier that a current loopcreated a magnetic field similar to that of a bar magnet, but what about a straight wire or a toroid (doughnut)? How is thedirection of a current-created field related to the direction of the current? Answers to these questions are explored in this section,together with a brief discussion of the law governing the fields created by currents.

Magnetic Field Created by a Long Straight Current-Carrying Wire: Right Hand Rule 2Magnetic fields have both direction and magnitude. As noted before, one way to explore the direction of a magnetic field is withcompasses, as shown for a long straight current-carrying wire in Figure 22.39. Hall probes can determine the magnitude of thefield. The field around a long straight wire is found to be in circular loops. The right hand rule 2 (RHR-2) emerges from thisexploration and is valid for any current segment—point the thumb in the direction of the current, and the fingers curl in thedirection of the magnetic field loops created by it.


Figure 22.39 (a) Compasses placed near a long straight current-carrying wire indicate that field lines form circular loops centered on the wire. (b) Righthand rule 2 states that, if the right hand thumb points in the direction of the current, the fingers curl in the direction of the field. This rule is consistentwith the field mapped for the long straight wire and is valid for any current segment.

Making Connections: Notation

For a wire oriented perpendicular to the page, if the current in the wire is directed out of the page, the right-hand rule tells usthat the magnetic field lines will be oriented in a counterclockwise direction around the wire. If the current in the wire isdirected into the page, the magnetic field lines will be oriented in a clockwise direction around the wire. We use ⊙ to

indicate that the direction of the current in the wire is out of the page, and ⊗ for the direction into the page.

Figure 22.40 Two parallel wires have currents pointing into or out of the page as shown. The direction of the magnetic field in the vicinity of thetwo wires is shown.

The magnetic field strength (magnitude) produced by a long straight current-carrying wire is found by experiment to be

(22.24)B = µ0 I

2πr (long straight wire),

where I is the current, r is the shortest distance to the wire, and the constant µ0 = 4π × 10−7 T ⋅ m/A is the permeability

of free space. (µ0 is one of the basic constants in nature. We will see later that µ0 is related to the speed of light.) Since the

wire is very long, the magnitude of the field depends only on distance from the wire r , not on position along the wire.



Example 22.6 Calculating Current that Produces a Magnetic Field

Find the current in a long straight wire that would produce a magnetic field twice the strength of the Earth’s at a distance of5.0 cm from the wire.

Strategy

The Earth’s field is about 5.0×10−5 T , and so here B due to the wire is taken to be 1.0×10−4 T . The equation

B = µ0 I2πr can be used to find I , since all other quantities are known.

Solution

Solving for I and entering known values gives

(22.25)I = 2πrB

µ0=

2π⎛⎝5.0×10−2 m⎞

⎠⎛⎝1.0×10−4 T⎞

⎠

4π×10−7 T ⋅ m/A= 25 A.

Discussion

So a moderately large current produces a significant magnetic field at a distance of 5.0 cm from a long straight wire. Notethat the answer is stated to only two digits, since the Earth’s field is specified to only two digits in this example.

Ampere’s Law and OthersThe magnetic field of a long straight wire has more implications than you might at first suspect. Each segment of currentproduces a magnetic field like that of a long straight wire, and the total field of any shape current is the vector sum of the fieldsdue to each segment. The formal statement of the direction and magnitude of the field due to each segment is called the Biot-Savart law. Integral calculus is needed to sum the field for an arbitrary shape current. This results in a more complete law, calledAmpere’s law, which relates magnetic field and current in a general way. Ampere’s law in turn is a part of Maxwell’s equations,which give a complete theory of all electromagnetic phenomena. Considerations of how Maxwell’s equations appear to differentobservers led to the modern theory of relativity, and the realization that electric and magnetic fields are different manifestations ofthe same thing. Most of this is beyond the scope of this text in both mathematical level, requiring calculus, and in the amount ofspace that can be devoted to it. But for the interested student, and particularly for those who continue in physics, engineering, orsimilar pursuits, delving into these matters further will reveal descriptions of nature that are elegant as well as profound. In thistext, we shall keep the general features in mind, such as RHR-2 and the rules for magnetic field lines listed in Magnetic Fieldsand Magnetic Field Lines, while concentrating on the fields created in certain important situations.

Making Connections: Relativity

Hearing all we do about Einstein, we sometimes get the impression that he invented relativity out of nothing. On the contrary,one of Einstein’s motivations was to solve difficulties in knowing how different observers see magnetic and electric fields.

Magnetic Field Produced by a Current-Carrying Circular LoopThe magnetic field near a current-carrying loop of wire is shown in Figure 22.41. Both the direction and the magnitude of themagnetic field produced by a current-carrying loop are complex. RHR-2 can be used to give the direction of the field near theloop, but mapping with compasses and the rules about field lines given in Magnetic Fields and Magnetic Field Lines areneeded for more detail. There is a simple formula for the magnetic field strength at the center of a circular loop. It is

(22.26)B = µ0 I

2R (at center of loop),

where R is the radius of the loop. This equation is very similar to that for a straight wire, but it is valid only at the center of acircular loop of wire. The similarity of the equations does indicate that similar field strength can be obtained at the center of aloop. One way to get a larger field is to have N loops; then, the field is B = Nµ0I / (2R) . Note that the larger the loop, the

smaller the field at its center, because the current is farther away.


Figure 22.41 (a) RHR-2 gives the direction of the magnetic field inside and outside a current-carrying loop. (b) More detailed mapping with compassesor with a Hall probe completes the picture. The field is similar to that of a bar magnet.

Magnetic Field Produced by a Current-Carrying SolenoidA solenoid is a long coil of wire (with many turns or loops, as opposed to a flat loop). Because of its shape, the field inside asolenoid can be very uniform, and also very strong. The field just outside the coils is nearly zero. Figure 22.42 shows how thefield looks and how its direction is given by RHR-2.

Figure 22.42 (a) Because of its shape, the field inside a solenoid of length l is remarkably uniform in magnitude and direction, as indicated by thestraight and uniformly spaced field lines. The field outside the coils is nearly zero. (b) This cutaway shows the magnetic field generated by the currentin the solenoid.

The magnetic field inside of a current-carrying solenoid is very uniform in direction and magnitude. Only near the ends does itbegin to weaken and change direction. The field outside has similar complexities to flat loops and bar magnets, but the magneticfield strength inside a solenoid is simply

(22.27)B = µ0nI (inside a solenoid),

where n is the number of loops per unit length of the solenoid (n = N / l , with N being the number of loops and l the

length). Note that B is the field strength anywhere in the uniform region of the interior and not just at the center. Large uniformfields spread over a large volume are possible with solenoids, as Example 22.7 implies.

Example 22.7 Calculating Field Strength inside a Solenoid

What is the field inside a 2.00-m-long solenoid that has 2000 loops and carries a 1600-A current?

Strategy



To find the field strength inside a solenoid, we use B = µ0nI . First, we note the number of loops per unit length is

(22.28)n = Nl = 2000

2.00 m = 1000 m−1 = 10 cm−1 .

Solution

Substituting known values gives

(22.29)B = µ0nI = ⎛⎝4π×10−7 T ⋅ m/A⎞

⎠⎛⎝1000 m−1⎞

⎠(1600 A)= 2.01 T.

Discussion

This is a large field strength that could be established over a large-diameter solenoid, such as in medical uses of magneticresonance imaging (MRI). The very large current is an indication that the fields of this strength are not easily achieved,however. Such a large current through 1000 loops squeezed into a meter’s length would produce significant heating. Highercurrents can be achieved by using superconducting wires, although this is expensive. There is an upper limit to the current,since the superconducting state is disrupted by very large magnetic fields.

Applying the Science Practices: Charged Particle in a Magnetic Field

Visit here (http://openstaxcollege.org/l/31particlemagnetic) and start the simulation applet “Particle in a Magnetic Field(2D)” in order to explore the magnetic force that acts on a charged particle in a magnetic field. Experiment with thesimulation to see how it works and what parameters you can change; then construct a plan to methodically investigate howmagnetic fields affect charged particles. Some questions you may want to answer as part of your experiment are:

• Are the paths of charged particles in magnetic fields always similar in two dimensions? Why or why not?• How would the path of a neutral particle in the magnetic field compare to the path of a charged particle?• How would the path of a positive particle differ from the path of a negative particle in a magnetic field?• What quantities dictate the properties of the particle’s path?• If you were attempting to measure the mass of a charged particle moving through a magnetic field, what would you

need to measure about its path? Would you need to see it moving at many different velocities or through different fieldstrengths, or would one trial be sufficient if your measurements were correct?

• Would doubling the charge change the path through the field? Predict an answer to this question, and then test yourhypothesis.

• Would doubling the velocity change the path through the field? Predict an answer to this question, and then test yourhypothesis.

• Would doubling the magnetic field strength change the path through the field? Predict an answer to this question, andthen test your hypothesis.

• Would increasing the mass change the path? Predict an answer to this question, and then test your hypothesis.

There are interesting variations of the flat coil and solenoid. For example, the toroidal coil used to confine the reactive particles intokamaks is much like a solenoid bent into a circle. The field inside a toroid is very strong but circular. Charged particles travel incircles, following the field lines, and collide with one another, perhaps inducing fusion. But the charged particles do not cross fieldlines and escape the toroid. A whole range of coil shapes are used to produce all sorts of magnetic field shapes. Addingferromagnetic materials produces greater field strengths and can have a significant effect on the shape of the field.Ferromagnetic materials tend to trap magnetic fields (the field lines bend into the ferromagnetic material, leaving weaker fieldsoutside it) and are used as shields for devices that are adversely affected by magnetic fields, including the Earth’s magnetic field.

PhET Explorations: Generator

Generate electricity with a bar magnet! Discover the physics behind the phenomena by exploring magnets and how you canuse them to make a bulb light.

Figure 22.43 Generator (http://cnx.org/content/m55399/1.4/generator_en.jar)


http://openstaxcollege.org/l/31particlemagnetic

http://cnx.org/content/m55399/1.4/generator_en.jar

22.10 Magnetic Force between Two Parallel Conductors


• Describe the effects of the magnetic force between two conductors.• Calculate the force between two parallel conductors.


• 2.D.2.1 The student is able to create a verbal or visual representation of a magnetic field around a long straight wire ora pair of parallel wires. (S.P. 1.1)


You might expect that there are significant forces between current-carrying wires, since ordinary currents produce significantmagnetic fields and these fields exert significant forces on ordinary currents. But you might not expect that the force betweenwires is used to define the ampere. It might also surprise you to learn that this force has something to do with why large circuitbreakers burn up when they attempt to interrupt large currents.

The force between two long straight and parallel conductors separated by a distance r can be found by applying what we havedeveloped in preceding sections. Figure 22.44 shows the wires, their currents, the fields they create, and the subsequent forcesthey exert on one another. Let us consider the field produced by wire 1 and the force it exerts on wire 2 (call the force F2 ). The

field due to I1 at a distance r is given to be

(22.30)B1 = µ0 I1

2πr .

Figure 22.44 (a) The magnetic field produced by a long straight conductor is perpendicular to a parallel conductor, as indicated by RHR-2. (b) A viewfrom above of the two wires shown in (a), with one magnetic field line shown for each wire. RHR-1 shows that the force between the parallelconductors is attractive when the currents are in the same direction. A similar analysis shows that the force is repulsive between currents in oppositedirections.

This field is uniform along wire 2 and perpendicular to it, and so the force F2 it exerts on wire 2 is given by F = IlB sin θ with

sin θ = 1 :

(22.31)F2 = I2lB1.

By Newton’s third law, the forces on the wires are equal in magnitude, and so we just write F for the magnitude of F2 . (Note

that F1 = −F2 .) Since the wires are very long, it is convenient to think in terms of F / l , the force per unit length. Substituting

the expression for B1 into the last equation and rearranging terms gives

(22.32)Fl = µ0 I1 I2

2πr .

F / l is the force per unit length between two parallel currents I1 and I2 separated by a distance r . The force is attractive if

the currents are in the same direction and repulsive if they are in opposite directions.



Making Connections: Field Canceling

For two parallel wires, the fields will tend to cancel out in the area between the wires.

Figure 22.45 Two parallel wires have currents pointing in the same direction, out of the page. The direction of the magnetic fields induced byeach wire is shown.

Note that the magnetic influence of the wire on the left-hand side extends beyond the wire on the right-hand side. To theright of both wires, the total magnetic field is directed toward the top of the page and is the result of the sum of the fields ofboth wires. Obviously, the closer wire has a greater effect on the overall magnetic field, but the more distant wire alsocontributes. One wire cannot block the magnetic field of another wire any more than a massive stone floor beneath you canblock the gravitational field of the Earth.

Parallel wires with currents in the same direction attract, as you can see if we isolate the magnetic field lines of wire 2influencing the current in wire 1. Right-hand rule 1 tells us the direction of the resulting magnetic force.

Figure 22.46 The same two wires are shown, but now only a part of the magnetic field due to wire 2 is shown in order to demonstrate how themagnetic force from wire 2 affects wire 1.

When the currents point in opposite directions as shown, the magnetic field in between the two wires is augmented. In theregion outside of the two wires, along the horizontal line connecting the wires, the magnetic fields partially cancel.

Figure 22.47 Two wires with parallel currents pointing in opposite directions are shown. The direction of the magnetic field due to each wire isindicated.

Parallel wires with currents in opposite directions repel, as you can see if we isolate the magnetic field lines of wire 2influencing the current in wire 1. Right-hand rule 1 tells us the direction of the resulting magnetic force.


Figure 22.48 The same two wires with opposite currents are shown, but now only a part of the magnetic field due to wire 2 is shown in order todemonstrate how the magnetic force from wire 2 affects wire 1.

This force is responsible for the pinch effect in electric arcs and plasmas. The force exists whether the currents are in wires ornot. In an electric arc, where currents are moving parallel to one another, there is an attraction that squeezes currents into asmaller tube. In large circuit breakers, like those used in neighborhood power distribution systems, the pinch effect canconcentrate an arc between plates of a switch trying to break a large current, burn holes, and even ignite the equipment. Anotherexample of the pinch effect is found in the solar plasma, where jets of ionized material, such as solar flares, are shaped bymagnetic forces.

The operational definition of the ampere is based on the force between current-carrying wires. Note that for parallel wiresseparated by 1 meter with each carrying 1 ampere, the force per meter is

(22.33)Fl =

⎛⎝4π×10−7 T ⋅ m/A⎞

⎠(1 A)2

(2π)(1 m) = 2×10−7 N/m.

Since µ0 is exactly 4π×10−7 T ⋅ m/A by definition, and because 1 T = 1 N/(A ⋅ m) , the force per meter is exactly

2×10−7 N/m . This is the basis of the operational definition of the ampere.

The Ampere

The official definition of the ampere is:

One ampere of current through each of two parallel conductors of infinite length, separated by one meter in empty space

free of other magnetic fields, causes a force of exactly 2×10−7 N/m on each conductor.

Infinite-length straight wires are impractical and so, in practice, a current balance is constructed with coils of wire separated by afew centimeters. Force is measured to determine current. This also provides us with a method for measuring the coulomb. Wemeasure the charge that flows for a current of one ampere in one second. That is, 1 C = 1 A ⋅ s . For both the ampere and thecoulomb, the method of measuring force between conductors is the most accurate in practice.

22.11 More Applications of Magnetism


• Describe some applications of magnetism.

Mass SpectrometryThe curved paths followed by charged particles in magnetic fields can be put to use. A charged particle moving perpendicular toa magnetic field travels in a circular path having a radius r .

(22.34)r = mvqB

It was noted that this relationship could be used to measure the mass of charged particles such as ions. A mass spectrometer isa device that measures such masses. Most mass spectrometers use magnetic fields for this purpose, although some of themhave extremely sophisticated designs. Since there are five variables in the relationship, there are many possibilities. However, ifv , q , and B can be fixed, then the radius of the path r is simply proportional to the mass m of the charged particle. Let us

examine one such mass spectrometer that has a relatively simple design. (See Figure 22.49.) The process begins with an ion



source, a device like an electron gun. The ion source gives ions their charge, accelerates them to some velocity v , and directs abeam of them into the next stage of the spectrometer. This next region is a velocity selector that only allows particles with aparticular value of v to get through.

Figure 22.49 This mass spectrometer uses a velocity selector to fix v so that the radius of the path is proportional to mass.

The velocity selector has both an electric field and a magnetic field, perpendicular to one another, producing forces in oppositedirections on the ions. Only those ions for which the forces balance travel in a straight line into the next region. If the forcesbalance, then the electric force F = qE equals the magnetic force F = qvB , so that qE = qvB . Noting that q cancels, we

see that

(22.35)v = EB

is the velocity particles must have to make it through the velocity selector, and further, that v can be selected by varying E and

B . In the final region, there is only a uniform magnetic field, and so the charged particles move in circular arcs with radiiproportional to particle mass. The paths also depend on charge q , but since q is in multiples of electron charges, it is easy to

determine and to discriminate between ions in different charge states.

Mass spectrometry today is used extensively in chemistry and biology laboratories to identify chemical and biological substancesaccording to their mass-to-charge ratios. In medicine, mass spectrometers are used to measure the concentration of isotopesused as tracers. Usually, biological molecules such as proteins are very large, so they are broken down into smaller fragmentsbefore analyzing. Recently, large virus particles have been analyzed as a whole on mass spectrometers. Sometimes a gaschromatograph or high-performance liquid chromatograph provides an initial separation of the large molecules, which are theninput into the mass spectrometer.

Cathode Ray Tubes—CRTs—and the LikeWhat do non-flat-screen TVs, old computer monitors, x-ray machines, and the 2-mile-long Stanford Linear Accelerator have incommon? All of them accelerate electrons, making them different versions of the electron gun. Many of these devices usemagnetic fields to steer the accelerated electrons. Figure 22.50 shows the construction of the type of cathode ray tube (CRT)found in some TVs, oscilloscopes, and old computer monitors. Two pairs of coils are used to steer the electrons, one verticallyand the other horizontally, to their desired destination.


Figure 22.50 The cathode ray tube (CRT) is so named because rays of electrons originate at the cathode in the electron gun. Magnetic coils are usedto steer the beam in many CRTs. In this case, the beam is moved down. Another pair of horizontal coils would steer the beam horizontally.

Magnetic Resonance ImagingMagnetic resonance imaging (MRI) is one of the most useful and rapidly growing medical imaging tools. It non-invasivelyproduces two-dimensional and three-dimensional images of the body that provide important medical information with none of thehazards of x-rays. MRI is based on an effect called nuclear magnetic resonance (NMR) in which an externally appliedmagnetic field interacts with the nuclei of certain atoms, particularly those of hydrogen (protons). These nuclei possess their ownsmall magnetic fields, similar to those of electrons and the current loops discussed earlier in this chapter.

When placed in an external magnetic field, such nuclei experience a torque that pushes or aligns the nuclei into one of two newenergy states—depending on the orientation of its spin (analogous to the N pole and S pole in a bar magnet). Transitions fromthe lower to higher energy state can be achieved by using an external radio frequency signal to “flip” the orientation of the smallmagnets. (This is actually a quantum mechanical process. The direction of the nuclear magnetic field is quantized as is energy inthe radio waves. We will return to these topics in later chapters.) The specific frequency of the radio waves that are absorbed andreemitted depends sensitively on the type of nucleus, the chemical environment, and the external magnetic field strength.Therefore, this is a resonance phenomenon in which nuclei in a magnetic field act like resonators (analogous to those discussedin the treatment of sound in Oscillatory Motion and Waves) that absorb and reemit only certain frequencies. Hence, thephenomenon is named nuclear magnetic resonance (NMR).

NMR has been used for more than 50 years as an analytical tool. It was formulated in 1946 by F. Bloch and E. Purcell, with the1952 Nobel Prize in Physics going to them for their work. Over the past two decades, NMR has been developed to producedetailed images in a process now called magnetic resonance imaging (MRI), a name coined to avoid the use of the word“nuclear” and the concomitant implication that nuclear radiation is involved. (It is not.) The 2003 Nobel Prize in Medicine went toP. Lauterbur and P. Mansfield for their work with MRI applications.

The largest part of the MRI unit is a superconducting magnet that creates a magnetic field, typically between 1 and 2 T instrength, over a relatively large volume. MRI images can be both highly detailed and informative about structures and organfunctions. It is helpful that normal and non-normal tissues respond differently for slight changes in the magnetic field. In mostmedical images, the protons that are hydrogen nuclei are imaged. (About 2/3 of the atoms in the body are hydrogen.) Theirlocation and density give a variety of medically useful information, such as organ function, the condition of tissue (as in thebrain), and the shape of structures, such as vertebral disks and knee-joint surfaces. MRI can also be used to follow themovement of certain ions across membranes, yielding information on active transport, osmosis, dialysis, and other phenomena.With excellent spatial resolution, MRI can provide information about tumors, strokes, shoulder injuries, infections, etc.

An image requires position information as well as the density of a nuclear type (usually protons). By varying the magnetic fieldslightly over the volume to be imaged, the resonant frequency of the protons is made to vary with position. Broadcast radiofrequencies are swept over an appropriate range and nuclei absorb and reemit them only if the nuclei are in a magnetic field withthe correct strength. The imaging receiver gathers information through the body almost point by point, building up a tissue map.The reception of reemitted radio waves as a function of frequency thus gives position information. These “slices” or crosssections through the body are only several mm thick. The intensity of the reemitted radio waves is proportional to theconcentration of the nuclear type being flipped, as well as information on the chemical environment in that area of the body.Various techniques are available for enhancing contrast in images and for obtaining more information. Scans called T1, T2, orproton density scans rely on different relaxation mechanisms of nuclei. Relaxation refers to the time it takes for the protons toreturn to equilibrium after the external field is turned off. This time depends upon tissue type and status (such as inflammation).

While MRI images are superior to x rays for certain types of tissue and have none of the hazards of x rays, they do notcompletely supplant x-ray images. MRI is less effective than x rays for detecting breaks in bone, for example, and in imagingbreast tissue, so the two diagnostic tools complement each other. MRI images are also expensive compared to simple x-rayimages and tend to be used most often where they supply information not readily obtained from x rays. Another disadvantage ofMRI is that the patient is totally enclosed with detectors close to the body for about 30 minutes or more, leading toclaustrophobia. It is also difficult for the obese patient to be in the magnet tunnel. New “open-MRI” machines are now available inwhich the magnet does not completely surround the patient.

Over the last decade, the development of much faster scans, called “functional MRI” (fMRI), has allowed us to map thefunctioning of various regions in the brain responsible for thought and motor control. This technique measures the change inblood flow for activities (thought, experiences, action) in the brain. The nerve cells increase their consumption of oxygen whenactive. Blood hemoglobin releases oxygen to active nerve cells and has somewhat different magnetic properties whenoxygenated than when deoxygenated. With MRI, we can measure this and detect a blood oxygen-dependent signal. Most of thebrain scans today use fMRI.



Ampere’s law:

B-field:

Biot-Savart law:

Curie temperature:

direction of magnetic field lines:

domains:

electromagnet:

electromagnetism:

ferromagnetic:

gauss:

Hall effect:

Hall emf:

Lorentz force:

magnetic field:

Other Medical Uses of Magnetic FieldsCurrents in nerve cells and the heart create magnetic fields like any other currents. These can be measured but with some

difficulty since their strengths are about 10−6 to 10−8 less than the Earth’s magnetic field. Recording of the heart’s magneticfield as it beats is called a magnetocardiogram (MCG), while measurements of the brain’s magnetic field is called amagnetoencephalogram (MEG). Both give information that differs from that obtained by measuring the electric fields of theseorgans (ECGs and EEGs), but they are not yet of sufficient importance to make these difficult measurements common.

In both of these techniques, the sensors do not touch the body. MCG can be used in fetal studies, and is probably more sensitivethan echocardiography. MCG also looks at the heart’s electrical activity whose voltage output is too small to be recorded bysurface electrodes as in EKG. It has the potential of being a rapid scan for early diagnosis of cardiac ischemia (obstruction ofblood flow to the heart) or problems with the fetus.

MEG can be used to identify abnormal electrical discharges in the brain that produce weak magnetic signals. Therefore, it looksat brain activity, not just brain structure. It has been used for studies of Alzheimer’s disease and epilepsy. Advances ininstrumentation to measure very small magnetic fields have allowed these two techniques to be used more in recent years. Whatis used is a sensor called a SQUID, for superconducting quantum interference device. This operates at liquid heliumtemperatures and can measure magnetic fields thousands of times smaller than the Earth’s.

Finally, there is a burgeoning market for magnetic cures in which magnets are applied in a variety of ways to the body, frommagnetic bracelets to magnetic mattresses. The best that can be said for such practices is that they are apparently harmless,unless the magnets get close to the patient’s computer or magnetic storage disks. Claims are made for a broad spectrum ofbenefits from cleansing the blood to giving the patient more energy, but clinical studies have not verified these claims, nor isthere an identifiable mechanism by which such benefits might occur.

PhET Explorations: Magnet and Compass

Ever wonder how a compass worked to point you to the Arctic? Explore the interactions between a compass and barmagnet, and then add the Earth and find the surprising answer! Vary the magnet's strength, and see how things change bothinside and outside. Use the field meter to measure how the magnetic field changes.

Figure 22.51 Magnet and Compass (http://cnx.org/content/m55403/1.2/magnet-and-compass_en.jar)

Glossarythe physical law that states that the magnetic field around an electric current is proportional to the current;

each segment of current produces a magnetic field like that of a long straight wire, and the total field of any shapecurrent is the vector sum of the fields due to each segment

another term for magnetic field

a physical law that describes the magnetic field generated by an electric current in terms of a specificequation

the temperature above which a ferromagnetic material cannot be magnetized

the direction that the north end of a compass needle points

regions within a material that behave like small bar magnets

an object that is temporarily magnetic when an electrical current is passed through it

the use of electrical currents to induce magnetism

materials, such as iron, cobalt, nickel, and gadolinium, that exhibit strong magnetic effects

G, the unit of the magnetic field strength; 1 G = 10–4 T

the creation of voltage across a current-carrying conductor by a magnetic field

the electromotive force created by a current-carrying conductor by a magnetic field, ε = Blv

the force on a charge moving in a magnetic field

the representation of magnetic forces


http://cnx.org/content/m55403/1.2/magnet-and-compass_en.jar

magnetic field lines:

magnetic field strength (magnitude) produced by a long straight current-carrying wire:

magnetic field strength at the center of a circular loop:

magnetic field strength inside a solenoid:

magnetic force:

magnetic monopoles:

magnetic resonance imaging (MRI):

magnetized:

magnetocardiogram (MCG):

magnetoencephalogram (MEG):

Maxwell’s equations:

meter:

motor:

north magnetic pole:

nuclear magnetic resonance (NMR):

permeability of free space:

right hand rule 1 (RHR-1):

right hand rule 2 (RHR-2):

solenoid:

south magnetic pole:

tesla:

the pictorial representation of the strength and the direction of a magnetic field

defined as B = µ0 I2πr , where I

is the current, r is the shortest distance to the wire, and µ0 is the permeability of free space

defined as B = µ0 I2R where R is the radius of the loop

defined as B = µ0nI where n is the number of loops per unit length of the

solenoid (n = N / l , with N being the number of loops and l the length)

the force on a charge produced by its motion through a magnetic field; the Lorentz force

an isolated magnetic pole; a south pole without a north pole, or vice versa (no magnetic monopole hasever been observed)

a medical imaging technique that uses magnetic fields create detailed images ofinternal tissues and organs

to be turned into a magnet; to be induced to be magnetic

a recording of the heart’s magnetic field as it beats

a measurement of the brain’s magnetic field

a set of four equations that describe electromagnetic phenomena

common application of magnetic torque on a current-carrying loop that is very similar in construction to a motor; bydesign, the torque is proportional to I and not θ , so the needle deflection is proportional to the current

loop of wire in a magnetic field; when current is passed through the loops, the magnetic field exerts torque on theloops, which rotates a shaft; electrical energy is converted to mechanical work in the process

the end or the side of a magnet that is attracted toward Earth’s geographic north pole

a phenomenon in which an externally applied magnetic field interacts with the nuclei ofcertain atoms

the measure of the ability of a material, in this case free space, to support a magnetic field; the

constant µ0 = 4π×10−7 T ⋅ m/A

the rule to determine the direction of the magnetic force on a positive moving charge: when thethumb of the right hand points in the direction of the charge’s velocity v and the fingers point in the direction of the

magnetic field B , then the force on the charge is perpendicular and away from the palm; the force on a negative chargeis perpendicular and into the palm

a rule to determine the direction of the magnetic field induced by a current-carrying wire: Point thethumb of the right hand in the direction of current, and the fingers curl in the direction of the magnetic field loops

a thin wire wound into a coil that produces a magnetic field when an electric current is passed through it

the end or the side of a magnet that is attracted toward Earth’s geographic south pole

T, the SI unit of the magnetic field strength; 1 T = 1 NA ⋅ m

Section Summary

22.1 Magnets• Magnetism is a subject that includes the properties of magnets, the effect of the magnetic force on moving charges and

currents, and the creation of magnetic fields by currents.• There are two types of magnetic poles, called the north magnetic pole and south magnetic pole.• North magnetic poles are those that are attracted toward the Earth’s geographic north pole.• Like poles repel and unlike poles attract.• Magnetic poles always occur in pairs of north and south—it is not possible to isolate north and south poles.



22.2 Ferromagnets and Electromagnets• Magnetic poles always occur in pairs of north and south—it is not possible to isolate north and south poles.• All magnetism is created by electric current.• Ferromagnetic materials, such as iron, are those that exhibit strong magnetic effects.• The atoms in ferromagnetic materials act like small magnets (due to currents within the atoms) and can be aligned, usually

in millimeter-sized regions called domains.• Domains can grow and align on a larger scale, producing permanent magnets. Such a material is magnetized, or induced

to be magnetic.• Above a material’s Curie temperature, thermal agitation destroys the alignment of atoms, and ferromagnetism disappears.• Electromagnets employ electric currents to make magnetic fields, often aided by induced fields in ferromagnetic materials.

22.3 Magnetic Fields and Magnetic Field Lines• Magnetic fields can be pictorially represented by magnetic field lines, the properties of which are as follows:

1. The field is tangent to the magnetic field line.

2. Field strength is proportional to the line density.

3. Field lines cannot cross.

4. Field lines are continuous loops.

22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field• Magnetic fields exert a force on a moving charge q, the magnitude of which is

F = qvB sin θ,where θ is the angle between the directions of v and B .

• The SI unit for magnetic field strength B is the tesla (T), which is related to other units by

1 T = 1 NC ⋅ m/s = 1 N

A ⋅ m.

• The direction of the force on a moving charge is given by right hand rule 1 (RHR-1): Point the thumb of the right hand in thedirection of v , the fingers in the direction of B , and a perpendicular to the palm points in the direction of F .

• The force is perpendicular to the plane formed by v and B . Since the force is zero if v is parallel to B , chargedparticles often follow magnetic field lines rather than cross them.

22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications• Magnetic force can supply centripetal force and cause a charged particle to move in a circular path of radius

r = mvqB,

where v is the component of the velocity perpendicular to B for a charged particle with mass m and charge q .

22.6 The Hall Effect• The Hall effect is the creation of voltage ε , known as the Hall emf, across a current-carrying conductor by a magnetic field.• The Hall emf is given by

ε = Blv (B, v, and l, mutually perpendicular)for a conductor of width l through which charges move at a speed v .

22.7 Magnetic Force on a Current-Carrying Conductor• The magnetic force on current-carrying conductors is given by

F = IlB sin θ,where I is the current, l is the length of a straight conductor in a uniform magnetic field B , and θ is the angle between

I and B . The force follows RHR-1 with the thumb in the direction of I .

22.8 Torque on a Current Loop: Motors and Meters• The torque τ on a current-carrying loop of any shape in a uniform magnetic field. is

τ = NIAB sin θ,where N is the number of turns, I is the current, A is the area of the loop, B is the magnetic field strength, and θ isthe angle between the perpendicular to the loop and the magnetic field.


22.9 Magnetic Fields Produced by Currents: Ampere’s Law• The strength of the magnetic field created by current in a long straight wire is given by

B = µ0 I2πr (long straight wire),

I is the current, r is the shortest distance to the wire, and the constant µ0 = 4π × 10−7 T ⋅ m/A is the permeability of

free space.• The direction of the magnetic field created by a long straight wire is given by right hand rule 2 (RHR-2): Point the thumb of

the right hand in the direction of current, and the fingers curl in the direction of the magnetic field loops created by it.• The magnetic field created by current following any path is the sum (or integral) of the fields due to segments along the

path (magnitude and direction as for a straight wire), resulting in a general relationship between current and field known asAmpere’s law.

• The magnetic field strength at the center of a circular loop is given by

B = µ0 I2R (at center of loop),

R is the radius of the loop. This equation becomes B = µ0nI / (2R) for a flat coil of N loops. RHR-2 gives the direction

of the field about the loop. A long coil is called a solenoid.• The magnetic field strength inside a solenoid is

B = µ0nI (inside a solenoid),where n is the number of loops per unit length of the solenoid. The field inside is very uniform in magnitude and direction.

22.10 Magnetic Force between Two Parallel Conductors• The force between two parallel currents I1 and I2 , separated by a distance r , has a magnitude per unit length given by

Fl = µ0 I1 I2

2πr .• The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.

22.11 More Applications of Magnetism• Crossed (perpendicular) electric and magnetic fields act as a velocity filter, giving equal and opposite forces on any charge

with velocity perpendicular to the fields and of magnitude

v = EB.

Conceptual Questions

22.1 Magnets1. Volcanic and other such activity at the mid-Atlantic ridge extrudes material to fill the gap between separating tectonic platesassociated with continental drift. The magnetization of rocks is found to reverse in a coordinated manner with distance from theridge. What does this imply about the Earth’s magnetic field and how could the knowledge of the spreading rate be used to giveits historical record?

22.3 Magnetic Fields and Magnetic Field Lines2. Explain why the magnetic field would not be unique (that is, not have a single value) at a point in space where magnetic fieldlines might cross. (Consider the direction of the field at such a point.)

3. List the ways in which magnetic field lines and electric field lines are similar. For example, the field direction is tangent to theline at any point in space. Also list the ways in which they differ. For example, electric force is parallel to electric field lines,whereas magnetic force on moving charges is perpendicular to magnetic field lines.

4. Noting that the magnetic field lines of a bar magnet resemble the electric field lines of a pair of equal and opposite charges, doyou expect the magnetic field to rapidly decrease in strength with distance from the magnet? Is this consistent with yourexperience with magnets?

5. Is the Earth’s magnetic field parallel to the ground at all locations? If not, where is it parallel to the surface? Is its strength thesame at all locations? If not, where is it greatest?

22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field6. If a charged particle moves in a straight line through some region of space, can you say that the magnetic field in that region isnecessarily zero?

22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications



7. How can the motion of a charged particle be used to distinguish between a magnetic and an electric field?

8. High-velocity charged particles can damage biological cells and are a component of radiation exposure in a variety of locationsranging from research facilities to natural background. Describe how you could use a magnetic field to shield yourself.

9. If a cosmic ray proton approaches the Earth from outer space along a line toward the center of the Earth that lies in the planeof the equator, in what direction will it be deflected by the Earth’s magnetic field? What about an electron? A neutron?

10. What are the signs of the charges on the particles in Figure 22.52?

Figure 22.52

11. Which of the particles in Figure 22.53 has the greatest velocity, assuming they have identical charges and masses?

Figure 22.53

12. Which of the particles in Figure 22.53 has the greatest mass, assuming all have identical charges and velocities?

13. While operating, a high-precision TV monitor is placed on its side during maintenance. The image on the monitor changescolor and blurs slightly. Discuss the possible relation of these effects to the Earth’s magnetic field.

22.6 The Hall Effect14. Discuss how the Hall effect could be used to obtain information on free charge density in a conductor. (Hint: Consider howdrift velocity and current are related.)

22.7 Magnetic Force on a Current-Carrying Conductor15. Draw a sketch of the situation in Figure 22.31 showing the direction of electrons carrying the current, and use RHR-1 toverify the direction of the force on the wire.

16. Verify that the direction of the force in an MHD drive, such as that in Figure 22.33, does not depend on the sign of thecharges carrying the current across the fluid.

17. Why would a magnetohydrodynamic drive work better in ocean water than in fresh water? Also, why would superconductingmagnets be desirable?

18. Which is more likely to interfere with compass readings, AC current in your refrigerator or DC current when you start yourcar? Explain.

22.8 Torque on a Current Loop: Motors and Meters19. Draw a diagram and use RHR-1 to show that the forces on the top and bottom segments of the motor’s current loop inFigure 22.35 are vertical and produce no torque about the axis of rotation.

22.9 Magnetic Fields Produced by Currents: Ampere’s Law20. Make a drawing and use RHR-2 to find the direction of the magnetic field of a current loop in a motor (such as in Figure22.35). Then show that the direction of the torque on the loop is the same as produced by like poles repelling and unlike polesattracting.

22.10 Magnetic Force between Two Parallel Conductors21. Is the force attractive or repulsive between the hot and neutral lines hung from power poles? Why?


22. If you have three parallel wires in the same plane, as in Figure 22.54, with currents in the outer two running in oppositedirections, is it possible for the middle wire to be repelled by both? Attracted by both? Explain.

Figure 22.54 Three parallel coplanar wires with currents in the outer two in opposite directions.

23. Suppose two long straight wires run perpendicular to one another without touching. Does one exert a net force on the other?If so, what is its direction? Does one exert a net torque on the other? If so, what is its direction? Justify your responses by usingthe right hand rules.

24. Use the right hand rules to show that the force between the two loops in Figure 22.55 is attractive if the currents are in thesame direction and repulsive if they are in opposite directions. Is this consistent with like poles of the loops repelling and unlikepoles of the loops attracting? Draw sketches to justify your answers.

Figure 22.55 Two loops of wire carrying currents can exert forces and torques on one another.

25. If one of the loops in Figure 22.55 is tilted slightly relative to the other and their currents are in the same direction, what arethe directions of the torques they exert on each other? Does this imply that the poles of the bar magnet-like fields they create willline up with each other if the loops are allowed to rotate?

26. Electric field lines can be shielded by the Faraday cage effect. Can we have magnetic shielding? Can we have gravitationalshielding?

22.11 More Applications of Magnetism27. Measurements of the weak and fluctuating magnetic fields associated with brain activity are called magnetoencephalograms(MEGs). Do the brain’s magnetic fields imply coordinated or uncoordinated nerve impulses? Explain.

28. Discuss the possibility that a Hall voltage would be generated on the moving heart of a patient during MRI imaging. Alsodiscuss the same effect on the wires of a pacemaker. (The fact that patients with pacemakers are not given MRIs is significant.)

29. A patient in an MRI unit turns his head quickly to one side and experiences momentary dizziness and a strange taste in hismouth. Discuss the possible causes.

30. You are told that in a certain region there is either a uniform electric or magnetic field. What measurement or observationcould you make to determine the type? (Ignore the Earth’s magnetic field.)

31. An example of magnetohydrodynamics (MHD) comes from the flow of a river (salty water). This fluid interacts with the Earth’smagnetic field to produce a potential difference between the two river banks. How would you go about calculating the potentialdifference?



32. Draw gravitational field lines between 2 masses, electric field lines between a positive and a negative charge, electric fieldlines between 2 positive charges and magnetic field lines around a magnet. Qualitatively describe the differences between thefields and the entities responsible for the field lines.


Problems & Exercises

22.4 Magnetic Field Strength: Force on aMoving Charge in a Magnetic Field1. What is the direction of the magnetic force on a positivecharge that moves as shown in each of the six cases shownin Figure 22.56?

Figure 22.56

2. Repeat Exercise 22.1 for a negative charge.

3. What is the direction of the velocity of a negative chargethat experiences the magnetic force shown in each of thethree cases in Figure 22.57, assuming it movesperpendicular to B?

Figure 22.57

4. Repeat Exercise 22.3 for a positive charge.

5. What is the direction of the magnetic field that produces themagnetic force on a positive charge as shown in each of thethree cases in the figure below, assuming B is perpendicularto v ?

Figure 22.58

6. Repeat Exercise 22.5 for a negative charge.

7. What is the maximum force on an aluminum rod with a0.100-μC charge that you pass between the poles of a

1.50-T permanent magnet at a speed of 5.00 m/s? In whatdirection is the force?

8. (a) Aircraft sometimes acquire small static charges.Suppose a supersonic jet has a 0.500-μC charge and flies

due west at a speed of 660 m/s over the Earth’s south

magnetic pole, where the 8.00×10−5-T magnetic fieldpoints straight up. What are the direction and the magnitudeof the magnetic force on the plane? (b) Discuss whether thevalue obtained in part (a) implies this is a significant ornegligible effect.

9. (a) A cosmic ray proton moving toward the Earth at

5.00×107 m/s experiences a magnetic force of

1.70×10−16 N . What is the strength of the magnetic field if

there is a 45º angle between it and the proton’s velocity? (b)Is the value obtained in part (a) consistent with the knownstrength of the Earth’s magnetic field on its surface? Discuss.

10. An electron moving at 4.00×103 m/s in a 1.25-Tmagnetic field experiences a magnetic force of

1.40×10−16 N . What angle does the velocity of theelectron make with the magnetic field? There are twoanswers.

11. (a) A physicist performing a sensitive measurement wantsto limit the magnetic force on a moving charge in her

equipment to less than 1.00×10−12 N . What is thegreatest the charge can be if it moves at a maximum speed of30.0 m/s in the Earth’s field? (b) Discuss whether it would bedifficult to limit the charge to less than the value found in (a)by comparing it with typical static electricity and noting thatstatic is often absent.

22.5 Force on a Moving Charge in a MagneticField: Examples and ApplicationsIf you need additional support for these problems, see MoreApplications of Magnetism.

12. A cosmic ray electron moves at 7.50×106 m/sperpendicular to the Earth’s magnetic field at an altitude

where field strength is 1.00×10−5 T . What is the radius ofthe circular path the electron follows?

13. A proton moves at 7.50×107 m/s perpendicular to amagnetic field. The field causes the proton to travel in acircular path of radius 0.800 m. What is the field strength?

14. (a) Viewers of Star Trek hear of an antimatter drive on theStarship Enterprise. One possibility for such a futuristicenergy source is to store antimatter charged particles in avacuum chamber, circulating in a magnetic field, and thenextract them as needed. Antimatter annihilates with normalmatter, producing pure energy. What strength magnetic field

is needed to hold antiprotons, moving at 5.00×107 m/s in acircular path 2.00 m in radius? Antiprotons have the samemass as protons but the opposite (negative) charge. (b) Isthis field strength obtainable with today’s technology or is it afuturistic possibility?



15. (a) An oxygen-16 ion with a mass of 2.66×10−26 kg

travels at 5.00×106 m/s perpendicular to a 1.20-Tmagnetic field, which makes it move in a circular arc with a0.231-m radius. What positive charge is on the ion? (b) Whatis the ratio of this charge to the charge of an electron? (c)Discuss why the ratio found in (b) should be an integer.

16. What radius circular path does an electron travel if itmoves at the same speed and in the same magnetic field asthe proton in Exercise 22.13?

17. A velocity selector in a mass spectrometer uses a 0.100-Tmagnetic field. (a) What electric field strength is needed to

select a speed of 4.00 x 106 m/s? (b) What is the voltagebetween the plates if they are separated by 1.00 cm?

18. An electron in a TV CRT moves with a speed of

6.00×107 m/s , in a direction perpendicular to the Earth’s

field, which has a strength of 5.00×10−5 T . (a) Whatstrength electric field must be applied perpendicular to theEarth’s field to make the electron moves in a straight line? (b)If this is done between plates separated by 1.00 cm, what isthe voltage applied? (Note that TVs are usually surroundedby a ferromagnetic material to shield against externalmagnetic fields and avoid the need for such a correction.)

19. (a) At what speed will a proton move in a circular path ofthe same radius as the electron in Exercise 22.12? (b) Whatwould the radius of the path be if the proton had the samespeed as the electron? (c) What would the radius be if theproton had the same kinetic energy as the electron? (d) Thesame momentum?

20. A mass spectrometer is being used to separate commonoxygen-16 from the much rarer oxygen-18, taken from asample of old glacial ice. (The relative abundance of theseoxygen isotopes is related to climatic temperature at the timethe ice was deposited.) The ratio of the masses of these two

ions is 16 to 18, the mass of oxygen-16 is 2.66×10−26 kg,

and they are singly charged and travel at 5.00×106 m/s ina 1.20-T magnetic field. What is the separation between theirpaths when they hit a target after traversing a semicircle?

21. (a) Triply charged uranium-235 and uranium-238 ions arebeing separated in a mass spectrometer. (The much rareruranium-235 is used as reactor fuel.) The masses of the ions

are 3.90×10−25 kg and 3.95×10−25 kg , respectively,

and they travel at 3.00×105 m/s in a 0.250-T field. What isthe separation between their paths when they hit a targetafter traversing a semicircle? (b) Discuss whether thisdistance between their paths seems to be big enough to bepractical in the separation of uranium-235 from uranium-238.

22.6 The Hall Effect22. A large water main is 2.50 m in diameter and the averagewater velocity is 6.00 m/s. Find the Hall voltage produced if

the pipe runs perpendicular to the Earth’s 5.00×10−5-Tfield.

23. What Hall voltage is produced by a 0.200-T field appliedacross a 2.60-cm-diameter aorta when blood velocity is 60.0cm/s?

24. (a) What is the speed of a supersonic aircraft with a17.0-m wingspan, if it experiences a 1.60-V Hall voltagebetween its wing tips when in level flight over the north

magnetic pole, where the Earth’s field strength is

8.00×10−5 T? (b) Explain why very little current flows as aresult of this Hall voltage.

25. A nonmechanical water meter could utilize the Hall effectby applying a magnetic field across a metal pipe andmeasuring the Hall voltage produced. What is the averagefluid velocity in a 3.00-cm-diameter pipe, if a 0.500-T fieldacross it creates a 60.0-mV Hall voltage?

26. Calculate the Hall voltage induced on a patient’s heartwhile being scanned by an MRI unit. Approximate theconducting path on the heart wall by a wire 7.50 cm long thatmoves at 10.0 cm/s perpendicular to a 1.50-T magnetic field.

27. A Hall probe calibrated to read 1.00 μV when placed in

a 2.00-T field is placed in a 0.150-T field. What is its outputvoltage?

28. Using information in Example 20.6, what would the Hallvoltage be if a 2.00-T field is applied across a 10-gaugecopper wire (2.588 mm in diameter) carrying a 20.0-Acurrent?

29. Show that the Hall voltage across wires made of the samematerial, carrying identical currents, and subjected to thesame magnetic field is inversely proportional to theirdiameters. (Hint: Consider how drift velocity depends on wirediameter.)

30. A patient with a pacemaker is mistakenly being scannedfor an MRI image. A 10.0-cm-long section of pacemaker wiremoves at a speed of 10.0 cm/s perpendicular to the MRIunit’s magnetic field and a 20.0-mV Hall voltage is induced.What is the magnetic field strength?

22.7 Magnetic Force on a Current-CarryingConductor31. What is the direction of the magnetic force on the currentin each of the six cases in Figure 22.59?

Figure 22.59

32. What is the direction of a current that experiences themagnetic force shown in each of the three cases in Figure22.60, assuming the current runs perpendicular to B ?


Figure 22.60

33. What is the direction of the magnetic field that producesthe magnetic force shown on the currents in each of the threecases in Figure 22.61, assuming B is perpendicular to I ?

Figure 22.61

34. (a) What is the force per meter on a lightning bolt at theequator that carries 20,000 A perpendicular to the Earth’s

3.00×10−5-T field? (b) What is the direction of the force ifthe current is straight up and the Earth’s field direction is duenorth, parallel to the ground?

35. (a) A DC power line for a light-rail system carries 1000 A

at an angle of 30.0º to the Earth’s 5.00×10−5 -T field.What is the force on a 100-m section of this line? (b) Discusspractical concerns this presents, if any.

36. What force is exerted on the water in an MHD driveutilizing a 25.0-cm-diameter tube, if 100-A current is passedacross the tube that is perpendicular to a 2.00-T magneticfield? (The relatively small size of this force indicates theneed for very large currents and magnetic fields to makepractical MHD drives.)

37. A wire carrying a 30.0-A current passes between thepoles of a strong magnet that is perpendicular to its field andexperiences a 2.16-N force on the 4.00 cm of wire in the field.What is the average field strength?

38. (a) A 0.750-m-long section of cable carrying current to acar starter motor makes an angle of 60º with the Earth’s

5.50×10−5 T field. What is the current when the wire

experiences a force of 7.00×10−3 N ? (b) If you run thewire between the poles of a strong horseshoe magnet,subjecting 5.00 cm of it to a 1.75-T field, what force is exertedon this segment of wire?

39. (a) What is the angle between a wire carrying an 8.00-Acurrent and the 1.20-T field it is in if 50.0 cm of the wireexperiences a magnetic force of 2.40 N? (b) What is the forceon the wire if it is rotated to make an angle of 90º with thefield?

40. The force on the rectangular loop of wire in the magneticfield in Figure 22.62 can be used to measure field strength.The field is uniform, and the plane of the loop is perpendicularto the field. (a) What is the direction of the magnetic force onthe loop? Justify the claim that the forces on the sides of theloop are equal and opposite, independent of how much of theloop is in the field and do not affect the net force on the loop.(b) If a current of 5.00 A is used, what is the force per tesla onthe 20.0-cm-wide loop?

Figure 22.62 A rectangular loop of wire carrying a current isperpendicular to a magnetic field. The field is uniform in the regionshown and is zero outside that region.

22.8 Torque on a Current Loop: Motors andMeters41. (a) By how many percent is the torque of a motordecreased if its permanent magnets lose 5.0% of theirstrength? (b) How many percent would the current need to beincreased to return the torque to original values?

42. (a) What is the maximum torque on a 150-turn squareloop of wire 18.0 cm on a side that carries a 50.0-A current ina 1.60-T field? (b) What is the torque when θ is 10.9º?43. Find the current through a loop needed to create amaximum torque of 9.00 N ⋅ m. The loop has 50 squareturns that are 15.0 cm on a side and is in a uniform 0.800-Tmagnetic field.

44. Calculate the magnetic field strength needed on a200-turn square loop 20.0 cm on a side to create a maximumtorque of 300 N ⋅ m if the loop is carrying 25.0 A.

45. Since the equation for torque on a current-carrying loop isτ = NIAB sin θ , the units of N ⋅ m must equal units of

A ⋅ m2 T . Verify this.

46. (a) At what angle θ is the torque on a current loop 90.0%of maximum? (b) 50.0% of maximum? (c) 10.0% ofmaximum?

47. A proton has a magnetic field due to its spin on its axis.The field is similar to that created by a circular current loop

0.650×10−15 m in radius with a current of 1.05×104 A(no kidding). Find the maximum torque on a proton in a2.50-T field. (This is a significant torque on a small particle.)



48. (a) A 200-turn circular loop of radius 50.0 cm is vertical,with its axis on an east-west line. A current of 100 A circulatesclockwise in the loop when viewed from the east. The Earth’sfield here is due north, parallel to the ground, with a strength

of 3.00×10−5 T . What are the direction and magnitude ofthe torque on the loop? (b) Does this device have anypractical applications as a motor?

49. Repeat Exercise 22.41, but with the loop lying flat on theground with its current circulating counterclockwise (whenviewed from above) in a location where the Earth’s field isnorth, but at an angle 45.0º below the horizontal and with a

strength of 6.00×10−5 T .

22.10 Magnetic Force between Two ParallelConductors50. (a) The hot and neutral wires supplying DC power to alight-rail commuter train carry 800 A and are separated by75.0 cm. What is the magnitude and direction of the forcebetween 50.0 m of these wires? (b) Discuss the practicalconsequences of this force, if any.

51. The force per meter between the two wires of a jumpercable being used to start a stalled car is 0.225 N/m. (a) Whatis the current in the wires, given they are separated by 2.00cm? (b) Is the force attractive or repulsive?

52. A 2.50-m segment of wire supplying current to the motorof a submerged submarine carries 1000 A and feels a 4.00-Nrepulsive force from a parallel wire 5.00 cm away. What is thedirection and magnitude of the current in the other wire?

53. The wire carrying 400 A to the motor of a commuter train

feels an attractive force of 4.00×10−3 N/m due to aparallel wire carrying 5.00 A to a headlight. (a) How far apartare the wires? (b) Are the currents in the same direction?

54. An AC appliance cord has its hot and neutral wiresseparated by 3.00 mm and carries a 5.00-A current. (a) Whatis the average force per meter between the wires in the cord?(b) What is the maximum force per meter between the wires?(c) Are the forces attractive or repulsive? (d) Do appliancecords need any special design features to compensate forthese forces?

55. Figure 22.63 shows a long straight wire near arectangular current loop. What is the direction and magnitudeof the total force on the loop?

Figure 22.63

56. Find the direction and magnitude of the force that eachwire experiences in Figure 22.64(a) by, using vector addition.

Figure 22.64

57. Find the direction and magnitude of the force that eachwire experiences in Figure 22.64(b), using vector addition.

22.11 More Applications of Magnetism58. Indicate whether the magnetic field created in each of thethree situations shown in Figure 22.65 is into or out of thepage on the left and right of the current.

Figure 22.65

59. What are the directions of the fields in the center of theloop and coils shown in Figure 22.66?

Figure 22.66

60. What are the directions of the currents in the loop andcoils shown in Figure 22.67?

Figure 22.67

61. To see why an MRI utilizes iron to increase the magneticfield created by a coil, calculate the current needed in a400-loop-per-meter circular coil 0.660 m in radius to create a1.20-T field (typical of an MRI instrument) at its center with noiron present. The magnetic field of a proton is approximately

like that of a circular current loop 0.650×10−15 m in radius

carrying 1.05×104 A . What is the field at the center of sucha loop?

62. Inside a motor, 30.0 A passes through a 250-turn circularloop that is 10.0 cm in radius. What is the magnetic fieldstrength created at its center?


63. Nonnuclear submarines use batteries for power whensubmerged. (a) Find the magnetic field 50.0 cm from astraight wire carrying 1200 A from the batteries to the drivemechanism of a submarine. (b) What is the field if the wires toand from the drive mechanism are side by side? (c) Discussthe effects this could have for a compass on the submarinethat is not shielded.

64. How strong is the magnetic field inside a solenoid with10,000 turns per meter that carries 20.0 A?

65. What current is needed in the solenoid described in

Exercise 22.58 to produce a magnetic field 104 times the

Earth’s magnetic field of 5.00×10−5 T ?

66. How far from the starter cable of a car, carrying 150 A,must you be to experience a field less than the Earth’s

(5.00×10−5 T)? Assume a long straight wire carries the

current. (In practice, the body of your car shields thedashboard compass.)

67. Measurements affect the system being measured, suchas the current loop in Figure 22.62. (a) Estimate the field theloop creates by calculating the field at the center of a circularloop 20.0 cm in diameter carrying 5.00 A. (b) What is thesmallest field strength this loop can be used to measure, if itsfield must alter the measured field by less than 0.0100%?

68. Figure 22.68 shows a long straight wire just touching aloop carrying a current I1 . Both lie in the same plane. (a)

What direction must the current I2 in the straight wire have

to create a field at the center of the loop in the directionopposite to that created by the loop? (b) What is the ratio ofI1 / I2 that gives zero field strength at the center of the loop?

(c) What is the direction of the field directly above the loopunder this circumstance?

Figure 22.68

69. Find the magnitude and direction of the magnetic field atthe point equidistant from the wires in Figure 22.64(a), usingthe rules of vector addition to sum the contributions from eachwire.

70. Find the magnitude and direction of the magnetic field atthe point equidistant from the wires in Figure 22.64(b), usingthe rules of vector addition to sum the contributions from eachwire.

71. What current is needed in the top wire in Figure 22.64(a)to produce a field of zero at the point equidistant from thewires, if the currents in the bottom two wires are both 10.0 Ainto the page?

72. Calculate the size of the magnetic field 20 m below a highvoltage power line. The line carries 450 MW at a voltage of300,000 V.

73. Integrated Concepts

(a) A pendulum is set up so that its bob (a thin copper disk)swings between the poles of a permanent magnet as shown

in Figure 22.69. What is the magnitude and direction of themagnetic force on the bob at the lowest point in its path, if ithas a positive 0.250 μC charge and is released from a

height of 30.0 cm above its lowest point? The magnetic fieldstrength is 1.50 T. (b) What is the acceleration of the bob atthe bottom of its swing if its mass is 30.0 grams and it is hungfrom a flexible string? Be certain to include a free-bodydiagram as part of your analysis.

Figure 22.69


(a) What voltage will accelerate electrons to a speed of

6.00×10−7 m/s ? (b) Find the radius of curvature of thepath of a proton accelerated through this potential in a0.500-T field and compare this with the radius of curvature ofan electron accelerated through the same potential.


Find the radius of curvature of the path of a 25.0-MeV protonmoving perpendicularly to the 1.20-T field of a cyclotron.


To construct a nonmechanical water meter, a 0.500-Tmagnetic field is placed across the supply water pipe to ahome and the Hall voltage is recorded. (a) Find the flow ratein liters per second through a 3.00-cm-diameter pipe if theHall voltage is 60.0 mV. (b) What would the Hall voltage be forthe same flow rate through a 10.0-cm-diameter pipe with thesame field applied?


(a) Using the values given for an MHD drive in Exercise22.59, and assuming the force is uniformly applied to the

fluid, calculate the pressure created in N/m2 . (b) Is this asignificant fraction of an atmosphere?


(a) Calculate the maximum torque on a 50-turn, 1.50 cmradius circular current loop carrying 50 μA in a 0.500-T field.

(b) If this coil is to be used in a galvanometer that reads50 μA full scale, what force constant spring must be used, if

it is attached 1.00 cm from the axis of rotation and isstretched by the 60º arc moved?




A current balance used to define the ampere is designed sothat the current through it is constant, as is the distancebetween wires. Even so, if the wires change length withtemperature, the force between them will change. Whatpercent change in force per degree will occur if the wires arecopper?


(a) Show that the period of the circular orbit of a chargedparticle moving perpendicularly to a uniform magnetic field isT = 2πm / (qB) . (b) What is the frequency f ? (c) What is

the angular velocity ω ? Note that these results areindependent of the velocity and radius of the orbit and, hence,of the energy of the particle. (Figure 22.70.)

Figure 22.70 Cyclotrons accelerate charged particles orbiting in amagnetic field by placing an AC voltage on the metal Dees, betweenwhich the particles move, so that energy is added twice each orbit. Thefrequency is constant, since it is independent of the particle energy—theradius of the orbit simply increases with energy until the particlesapproach the edge and are extracted for various experiments andapplications.


A cyclotron accelerates charged particles as shown in Figure22.70. Using the results of the previous problem, calculate thefrequency of the accelerating voltage needed for a proton in a1.20-T field.


(a) A 0.140-kg baseball, pitched at 40.0 m/s horizontally and

perpendicular to the Earth’s horizontal 5.00×10−5 T field,has a 100-nC charge on it. What distance is it deflected fromits path by the magnetic force, after traveling 30.0 mhorizontally? (b) Would you suggest this as a secrettechnique for a pitcher to throw curve balls?


(a) What is the direction of the force on a wire carrying acurrent due east in a location where the Earth’s field is duenorth? Both are parallel to the ground. (b) Calculate the forceper meter if the wire carries 20.0 A and the field strength is

3.00×10−5 T . (c) What diameter copper wire would haveits weight supported by this force? (d) Calculate theresistance per meter and the voltage per meter needed.


One long straight wire is to be held directly above another byrepulsion between their currents. The lower wire carries 100A and the wire 7.50 cm above it is 10-gauge (2.588 mm

diameter) copper wire. (a) What current must flow in theupper wire, neglecting the Earth’s field? (b) What is the

smallest current if the Earth’s 3.00×10−5 T field is parallelto the ground and is not neglected? (c) Is the supported wirein a stable or unstable equilibrium if displaced vertically? Ifdisplaced horizontally?

85. Unreasonable Results

(a) Find the charge on a baseball, thrown at 35.0 m/s

perpendicular to the Earth’s 5.00×10−5 T field, thatexperiences a 1.00-N magnetic force. (b) What isunreasonable about this result? (c) Which assumption orpremise is responsible?


A charged particle having mass 6.64×10−27 kg (that of a

helium atom) moving at 8.70×105 m/s perpendicular to a1.50-T magnetic field travels in a circular path of radius 16.0mm. (a) What is the charge of the particle? (b) What isunreasonable about this result? (c) Which assumptions areresponsible?


An inventor wants to generate 120-V power by moving a

1.00-m-long wire perpendicular to the Earth’s 5.00×10−5 Tfield. (a) Find the speed with which the wire must move. (b)What is unreasonable about this result? (c) Whichassumption is responsible?


Frustrated by the small Hall voltage obtained in blood flowmeasurements, a medical physicist decides to increase theapplied magnetic field strength to get a 0.500-V output forblood moving at 30.0 cm/s in a 1.50-cm-diameter vessel. (a)What magnetic field strength is needed? (b) What isunreasonable about this result? (c) Which premise isresponsible?


A surveyor 100 m from a long straight 200-kV DC power linesuspects that its magnetic field may equal that of the Earthand affect compass readings. (a) Calculate the current in the

wire needed to create a 5.00×10−5 T field at this distance.(b) What is unreasonable about this result? (c) Whichassumption or premise is responsible?

90. Construct Your Own Problem

Consider a mass separator that applies a magnetic fieldperpendicular to the velocity of ions and separates the ionsbased on the radius of curvature of their paths in the field.Construct a problem in which you calculate the magnetic fieldstrength needed to separate two ions that differ in mass, butnot charge, and have the same initial velocity. Among thethings to consider are the types of ions, the velocities theycan be given before entering the magnetic field, and areasonable value for the radius of curvature of the paths theyfollow. In addition, calculate the separation distance betweenthe ions at the point where they are detected.


Consider using the torque on a current-carrying coil in amagnetic field to detect relatively small magnetic fields (lessthan the field of the Earth, for example). Construct a problemin which you calculate the maximum torque on a current-carrying loop in a magnetic field. Among the things to be


considered are the size of the coil, the number of loops it has,the current you pass through the coil, and the size of the fieldyou wish to detect. Discuss whether the torque produced islarge enough to be effectively measured. Your instructor mayalso wish for you to consider the effects, if any, of the fieldproduced by the coil on the surroundings that could affectdetection of the small field.

Test Prep for AP® Courses

22.2 Ferromagnets and Electromagnets1. A bar magnet is oriented so that the north pole of the barmagnet points north. A compass needle is placed to the northof the bar magnet. In which direction does the north pole ofthe compass needle point?

a. Northb. Eastc. Southd. West

2. Assume for simplicity that the Earth’s magnetic north poleis at the same location as its geographic north pole. If you arein an airplane flying due west along the equator, as you crossthe prime meridian (0° longitude) facing west and look downat a compass you are carrying, you see that the compassneedle is perpendicular to your direction of motion, and thenorth pole of the needle dipole points to your right. As youcontinue flying due west, describe how and why theorientation of the needle will (or will not) change.

3. Describe how the magnetic domains in an unmagnetizediron rod will respond to the presence of a strong externalmagnetic field.

a. The domains will split into monopoles.b. The domains will tend to align with the external field.c. The domains will tend to orient themselves

perpendicular to the external field.d. The domains will tend to align so as to cancel out the

external field.

4. Describe what steps must be undertaken in order toconvert an unmagnetized iron rod into a permanentlymagnetized state. As part of your answer, explain what amagnetic domain is and how it responds to the stepsdescribed.

5. Iron is ferromagnetic and lead is diamagnetic, which meansits magnetic domains respond in the opposite direction offerromagnets but many orders of magnitude more weakly.The two blocks are placed in a magnetic field that points tothe right. Which of the following best represents theorientations of the dipoles when the field is present?

Iron Lead

(a)

Figure 22.71 Figure 22.72

(b)


(c)


(d)


6. A weather vane is some sort of directional arrow parallel tothe ground that may rotate freely in a horizontal plane. Atypical weather vane has a large cross-sectional areaperpendicular to the direction the arrow is pointing, like a“One Way” street sign. The purpose of the weather vane is toindicate the direction of the wind. As wind blows past theweather vane, it tends to orient the arrow in the samedirection as the wind. Consider a weather vane’s response toa strong wind. Explain how this is both similar to and differentfrom a magnetic domain’s response to an external magneticfield. How does each affect its surroundings?

22.4 Magnetic Field Strength: Force on aMoving Charge in a Magnetic Field7. A proton moves in the –x-direction and encounters auniform magnetic field pointing in the +x-direction. In whatdirection is the resulting magnetic force on the proton?

a. The proton experiences no magnetic force.b. +x-directionc. −y-directiond. +y-direction

8. A proton moves with a speed of 240 m/s in the +x-directioninto a region of a 4.5-T uniform magnetic field directed 62°above the +x-direction in the x,y-plane. Calculate themagnitude of the magnetic force on the proton.

22.5 Force on a Moving Charge in a MagneticField: Examples and Applications9. A wire oriented north-south carries current south. The wireis immersed in the Earth’s magnetic field, which is alsooriented north-south (with a horizontal component pointingnorth). The Earth’s magnetic field also has a vertical



component pointing down. What is the direction of themagnetic force felt by the wire?

a. Westb. Eastc. Upd. North

22.6 The Hall Effect10. An airplane wingspan can be approximated as aconducting rod of length 35 m. As the airplane flies due north,it is flying at a rate of 82 m/s through the Earth’s magneticfield, which has a magnitude of 45 μT toward the north in adirection 57° below the horizontal plane. (a) Which end of thewingspan is positively charged, the east or west end?Explain. (b) What is the Hall emf along the wingspan?

22.9 Magnetic Fields Produced by Currents:Ampere’s Law11. An experimentalist fires a beam of electrons, creating avisible path in the air that can be measured. The beam is firedalong a direction parallel to a current-carrying wire, and theelectrons travel in a circular path in response to the wire’smagnetic field. Assuming the mass and charge of theelectrons is known, what quantities would you need tomeasure in order to deduce the current in the wire?

a. the radius of the circular pathb. the average distance between the electrons and the

wirec. the velocity of the electronsd. two of the abovee. all of the above

12. Electrons starting from rest are accelerated through apotential difference of 240 V and fired into a region of uniform3.5-mT magnetic field generated by a large solenoid. Theelectrons are initially moving in the +x-direction upon enteringthe field, and the field is directed into the page. Determine (a)the radius of the circle in which the electrons will move in thisuniform magnetic field and (b) the initial direction of themagnetic force the electrons feel upon entering the uniformfield of the solenoid.

13. In terms of the direction of force, we use the left-handrule. Pointing your thumb in the +x-direction with the velocityand fingers of the left hand into the page reveals that themagnetic force points down toward the bottom of the page inthe –y-direction.

A wire along the y-axis carries current in the +y-direction. Inwhat direction is the magnetic field at a point on the +x-axisnear the wire?

a. away from the wireb. vertically upwardc. into the paged. out of the page

14. Imagine the xy coordinate plane is the plane of the page.A wire along the z-axis carries current in the +z-direction (outof the page, or ⊙ ). Draw a diagram of the magnetic field in

the vicinity of this wire indicating the direction of the field.Also, describe how the strength of the magnetic field variesaccording to the distance from the z-axis.

22.10 Magnetic Force between Two ParallelConductors15. Two parallel wires carry equal currents in the samedirection and are separated by a small distance. What is the

direction of the magnetic force exerted by the two wires oneach other?

a. No force since the wires are parallel.b. No force since the currents are in the same direction.c. The force is attractive.d. The force is repulsive.

16. A wire along the y-axis carries current in the +y-direction.An experimenter would like to arrange a second wire parallelto the first wire and crossing the x-axis at the coordinatex = 2.0 m so that the total magnetic field at the coordinate

x = 1.0 m is zero. In what direction must the current flow inthe second wire, assuming it is equal in magnitude to thecurrent in the first wire? Explain.




23 ELECTROMAGNETIC INDUCTION, ACCIRCUITS, AND ELECTRICALTECHNOLOGIES

Figure 23.1 This wind turbine in the Thames Estuary in the UK is an example of induction at work. Wind pushes the blades of the turbine, spinning ashaft attached to magnets. The magnets spin around a conductive coil, inducing an electric current in the coil, and eventually feeding the electrical grid.(credit: phault, Flickr)

Chapter Outline23.1. Induced Emf and Magnetic Flux

23.2. Faraday’s Law of Induction: Lenz’s Law

23.3. Motional Emf

23.4. Eddy Currents and Magnetic Damping

23.5. Electric Generators

23.6. Back Emf

23.7. Transformers

23.8. Electrical Safety: Systems and Devices

23.9. Inductance

23.10. RL Circuits

23.11. Reactance, Inductive and Capacitive

23.12. RLC Series AC Circuits

Chapter 23 | Electromagnetic Induction, AC Circuits, and Electrical Technologies 1015

Connection for AP® CoursesNature’s displays of symmetry are beautiful and alluring. As shown in Figure 23.2, a butterfly’s wings exhibit an appealingsymmetry in a complex system. The laws of physics display symmetries at the most basic level – these symmetries are a sourceof wonder and imply deeper meaning. Since we place a high value on symmetry, we look for it when we explore nature. Theremarkable thing is that we find it.

Figure 23.2 A butterfly with symmetrically patterned wings is resting on a flower. Physics, like this butterfly, has inherent symmetries. (credit: ThomasBresson).

This chapter supports Big Idea 4, illustrating how electric and magnetic changes can take place in a system due to interactionswith other systems. The hint of symmetry between electricity and magnetism found in the preceding chapter will be elaboratedupon in this chapter. Specifically, we know that a current creates a magnetic field. If nature is symmetric in this case, thenperhaps a magnetic field can create a current. Historically, it was very shortly after Oersted discovered that currents causemagnetic fields that other scientists asked the following question: Can magnetic fields cause currents? The answer was soonfound by experiment to be yes. In 1831, some 12 years after Oersted’s discovery, the English scientist Michael Faraday(1791–1862) and the American scientist Joseph Henry (1797–1878) independently demonstrated that magnetic fields canproduce currents. The basic process of generating emfs (electromotive forces), and hence currents, with magnetic fields isknown as induction; this process is also called “magnetic induction” to distinguish it from charging by induction, which utilizes theCoulomb force.

Today, currents induced by magnetic fields are essential to our technological society. The ubiquitous generator – found inautomobiles, on bicycles, in nuclear power plants, and so on – uses magnetism to generate current. Other devices that usemagnetism to induce currents include pickup coils in electric guitars, transformers of every size, certain microphones, airportsecurity gates, and damping mechanisms on sensitive chemical balances. Explanations and examples in this chapter will helpyou understand current induction via magnetic interactions in mechanical systems (Enduring Understanding 4.E, EssentialKnowledge 4.E.2). You will also learn how the behavior of AC circuits depends strongly on the effect of magnetic fields oncurrents.

The content of this chapter supports:

Big Idea 4 Interactions between systems can result in changes in those systems.

Enduring Understanding 4.E The electric and magnetic properties of a system can change in response to the presence of, orchanges in, other objects or systems.

Essential Knowledge 4.E.2 Changing magnetic flux induces an electric field that can establish an induced emf in a system.

23.1 Induced Emf and Magnetic Flux


• Calculate the flux of a uniform magnetic field through a loop of arbitrary orientation.• Describe methods to produce an electromotive force (emf) with a magnetic field or a magnet and a loop of wire.


• 4.E.2.1 The student is able to construct an explanation of the function of a simple electromagnetic device in which aninduced emf is produced by a changing magnetic flux through an area defined by a current loop (i.e., a simplemicrophone or generator) or of the effect on behavior of a device in which an induced emf is produced by a constantmagnetic field through a changing area. (S.P. 6.4)

The apparatus used by Faraday to demonstrate that magnetic fields can create currents is illustrated in Figure 23.3. When theswitch is closed, a magnetic field is produced in the coil on the top part of the iron ring and transmitted to the coil on the bottom

1016 Chapter 23 | Electromagnetic Induction, AC Circuits, and Electrical Technologies


part of the ring. The galvanometer is used to detect any current induced in the coil on the bottom. It was found that each time theswitch is closed, the galvanometer detects a current in one direction in the coil on the bottom. (You can also observe this in aphysics lab.) Each time the switch is opened, the galvanometer detects a current in the opposite direction. Interestingly, if theswitch remains closed or open for any length of time, there is no current through the galvanometer. Closing and opening theswitch induces the current. It is the change in magnetic field that creates the current. More basic than the current that flows is theemfthat causes it. The current is a result of an emf induced by a changing magnetic field, whether or not there is a path forcurrent to flow.

Figure 23.3 Faraday’s apparatus for demonstrating that a magnetic field can produce a current. A change in the field produced by the top coil inducesan emf and, hence, a current in the bottom coil. When the switch is opened and closed, the galvanometer registers currents in opposite directions. Nocurrent flows through the galvanometer when the switch remains closed or open.

An experiment easily performed and often done in physics labs is illustrated in Figure 23.4. An emf is induced in the coil when abar magnet is pushed in and out of it. Emfs of opposite signs are produced by motion in opposite directions, and the emfs arealso reversed by reversing poles. The same results are produced if the coil is moved rather than the magnet—it is the relativemotion that is important. The faster the motion, the greater the emf, and there is no emf when the magnet is stationary relative tothe coil.

Figure 23.4 Movement of a magnet relative to a coil produces emfs as shown. The same emfs are produced if the coil is moved relative to the magnet.The greater the speed, the greater the magnitude of the emf, and the emf is zero when there is no motion.

The method of inducing an emf used in most electric generators is shown in Figure 23.5. A coil is rotated in a magnetic field,producing an alternating current emf, which depends on rotation rate and other factors that will be explored in later sections.Note that the generator is remarkably similar in construction to a motor (another symmetry).


Figure 23.5 Rotation of a coil in a magnetic field produces an emf. This is the basic construction of a generator, where work done to turn the coil isconverted to electric energy. Note the generator is very similar in construction to a motor.

So we see that changing the magnitude or direction of a magnetic field produces an emf. Experiments revealed that there is acrucial quantity called the magnetic flux, Φ , given by

(23.1)Φ = BA cos θ,

where B is the magnetic field strength over an area A , at an angle θ with the perpendicular to the area as shown in Figure23.6. Any change in magnetic flux Φ induces an emf. This process is defined to be electromagnetic induction. Units of

magnetic flux Φ are T ⋅ m2 . As seen in Figure 23.6, B cos θ = B ⊥, which is the component of B perpendicular to the area

A . Thus magnetic flux is Φ = B⊥A , the product of the area and the component of the magnetic field perpendicular to it.

Figure 23.6 Magnetic flux Φ is related to the magnetic field and the area over which it exists. The flux Φ = BA cos θ is related to induction; any

change in Φ induces an emf.

All induction, including the examples given so far, arises from some change in magnetic flux Φ . For example, Faraday changed

B and hence Φ when opening and closing the switch in his apparatus (shown in Figure 23.3). This is also true for the bar

magnet and coil shown in Figure 23.4. When rotating the coil of a generator, the angle θ and, hence, Φ is changed. Just how

great an emf and what direction it takes depend on the change in Φ and how rapidly the change is made, as examined in thenext section.



23.2 Faraday’s Law of Induction: Lenz’s Law


• Calculate emf, current, and magnetic fields using Faraday’s law.• Explain the physical results of Lenz’s law.

Faraday’s and Lenz’s LawFaraday’s experiments showed that the emf induced by a change in magnetic flux depends on only a few factors. First, emf isdirectly proportional to the change in flux ΔΦ . Second, emf is greatest when the change in time Δt is smallest—that is, emf is

inversely proportional to Δt . Finally, if a coil has N turns, an emf will be produced that is N times greater than for a single coil,

so that emf is directly proportional to N . The equation for the emf induced by a change in magnetic flux is

(23.2)emf = −N ΔΦΔt .

This relationship is known as Faraday’s law of induction. The units for emf are volts, as is usual.

The minus sign in Faraday’s law of induction is very important. The minus means that the emf creates a current I and magneticfield B that oppose the change in flux ΔΦ —this is known as Lenz’s law. The direction (given by the minus sign) of the emfis soimportant that it is called Lenz’s law after the Russian Heinrich Lenz (1804–1865), who, like Faraday and Henry,independentlyinvestigated aspects of induction. Faraday was aware of the direction, but Lenz stated it so clearly that he is credited for itsdiscovery. (See Figure 23.7.)

Figure 23.7 (a) When this bar magnet is thrust into the coil, the strength of the magnetic field increases in the coil. The current induced in the coilcreates another field, in the opposite direction of the bar magnet’s to oppose the increase. This is one aspect of Lenz’s law—induction opposes anychange in flux. (b) and (c) are two other situations. Verify for yourself that the direction of the induced Bcoil shown indeed opposes the change in flux

and that the current direction shown is consistent with RHR-2.

Problem-Solving Strategy for Lenz’s Law

To use Lenz’s law to determine the directions of the induced magnetic fields, currents, and emfs:

1. Make a sketch of the situation for use in visualizing and recording directions.

2. Determine the direction of the magnetic field B.

3. Determine whether the flux is increasing or decreasing.

4. Now determine the direction of the induced magnetic field B. It opposes the change in flux by adding or subtractingfrom the original field.

5. Use RHR-2 to determine the direction of the induced current I that is responsible for the induced magnetic field B.


6. The direction (or polarity) of the induced emf will now drive a current in this direction and can be represented as currentemerging from the positive terminal of the emf and returning to its negative terminal.

For practice, apply these steps to the situations shown in Figure 23.7 and to others that are part of the following text material.

Applications of Electromagnetic InductionThere are many applications of Faraday’s Law of induction, as we will explore in this chapter and others. At this juncture, let usmention several that have to do with data storage and magnetic fields. A very important application has to do with audio andvideo recording tapes. A plastic tape, coated with iron oxide, moves past a recording head. This recording head is basically around iron ring about which is wrapped a coil of wire—an electromagnet (Figure 23.8). A signal in the form of a varying inputcurrent from a microphone or camera goes to the recording head. These signals (which are a function of the signal amplitudeand frequency) produce varying magnetic fields at the recording head. As the tape moves past the recording head, the magneticfield orientations of the iron oxide molecules on the tape are changed thus recording the signal. In the playback mode, themagnetized tape is run past another head, similar in structure to the recording head. The different magnetic field orientations ofthe iron oxide molecules on the tape induces an emf in the coil of wire in the playback head. This signal then is sent to aloudspeaker or video player.

Figure 23.8 Recording and playback heads used with audio and video magnetic tapes. (credit: Steve Jurvetson)

Similar principles apply to computer hard drives, except at a much faster rate. Here recordings are on a coated, spinning disk.Read heads historically were made to work on the principle of induction. However, the input information is carried in digital ratherthan analog form – a series of 0’s or 1’s are written upon the spinning hard drive. Today, most hard drive readout devices do notwork on the principle of induction, but use a technique known as giant magnetoresistance. (The discovery that weak changes ina magnetic field in a thin film of iron and chromium could bring about much larger changes in electrical resistance was one of thefirst large successes of nanotechnology.) Another application of induction is found on the magnetic stripe on the back of yourpersonal credit card as used at the grocery store or the ATM machine. This works on the same principle as the audio or videotape mentioned in the last paragraph in which a head reads personal information from your card.

Another application of electromagnetic induction is when electrical signals need to be transmitted across a barrier. Consider thecochlear implant shown below. Sound is picked up by a microphone on the outside of the skull and is used to set up a varyingmagnetic field. A current is induced in a receiver secured in the bone beneath the skin and transmitted to electrodes in the innerear. Electromagnetic induction can be used in other instances where electric signals need to be conveyed across various media.

Figure 23.9 Electromagnetic induction used in transmitting electric currents across mediums. The device on the baby’s head induces an electricalcurrent in a receiver secured in the bone beneath the skin. (credit: Bjorn Knetsch)

Another contemporary area of research in which electromagnetic induction is being successfully implemented (and withsubstantial potential) is transcranial magnetic simulation. A host of disorders, including depression and hallucinations can betraced to irregular localized electrical activity in the brain. In transcranial magnetic stimulation, a rapidly varying and verylocalized magnetic field is placed close to certain sites identified in the brain. Weak electric currents are induced in the identifiedsites and can result in recovery of electrical functioning in the brain tissue.

Sleep apnea (“the cessation of breath”) affects both adults and infants (especially premature babies and it may be a cause ofsudden infant deaths [SID]). In such individuals, breath can stop repeatedly during their sleep. A cessation of more than 20seconds can be very dangerous. Stroke, heart failure, and tiredness are just some of the possible consequences for a personhaving sleep apnea. The concern in infants is the stopping of breath for these longer times. One type of monitor to alert parentswhen a child is not breathing uses electromagnetic induction. A wire wrapped around the infant’s chest has an alternating currentrunning through it. The expansion and contraction of the infant’s chest as the infant breathes changes the area through the coil.A pickup coil located nearby has an alternating current induced in it due to the changing magnetic field of the initial wire. If thechild stops breathing, there will be a change in the induced current, and so a parent can be alerted.



Making Connections: Conservation of Energy

Lenz’s law is a manifestation of the conservation of energy. The induced emf produces a current that opposes the change influx, because a change in flux means a change in energy. Energy can enter or leave, but not instantaneously. Lenz’s law is aconsequence. As the change begins, the law says induction opposes and, thus, slows the change. In fact, if the induced emfwere in the same direction as the change in flux, there would be a positive feedback that would give us free energy from noapparent source—conservation of energy would be violated.

Example 23.1 Calculating Emf: How Great Is the Induced Emf?

Calculate the magnitude of the induced emf when the magnet in Figure 23.7(a) is thrust into the coil, given the followinginformation: the single loop coil has a radius of 6.00 cm and the average value of B cos θ (this is given, since the barmagnet’s field is complex) increases from 0.0500 T to 0.250 T in 0.100 s.

Strategy

To find the magnitude of emf, we use Faraday’s law of induction as stated by emf = −N ΔΦΔt , but without the minus sign

that indicates direction:

(23.3)emf = N ΔΦΔt .

Solution

We are given that N = 1 and Δt = 0.100 s , but we must determine the change in flux ΔΦ before we can find emf.Since the area of the loop is fixed, we see that

(23.4)ΔΦ = Δ(BA cos θ) = AΔ(B cos θ).

Now Δ(B cos θ) = 0.200 T , since it was given that B cos θ changes from 0.0500 to 0.250 T. The area of the loop is

A = πr2 = (3.14...)(0.060 m)2 = 1.13×10−2 m2 . Thus,

(23.5)ΔΦ = (1.13×10−2 m2)(0.200 T).

Entering the determined values into the expression for emf gives

(23.6)Emf = N ΔΦ

Δt = (1.13×10−2 m2)(0.200 T)0.100 s = 22.6 mV.

Discussion

While this is an easily measured voltage, it is certainly not large enough for most practical applications. More loops in thecoil, a stronger magnet, and faster movement make induction the practical source of voltages that it is.

PhET Explorations: Faraday's Electromagnetic Lab

Play with a bar magnet and coils to learn about Faraday's law. Move a bar magnet near one or two coils to make a light bulbglow. View the magnetic field lines. A meter shows the direction and magnitude of the current. View the magnetic field linesor use a meter to show the direction and magnitude of the current. You can also play with electromagnets, generators andtransformers!

Figure 23.10 Faraday's Electromagnetic Lab (http://cnx.org/content/m55407/1.2/faraday_en.jar)

23.3 Motional Emf


• Calculate emf, force, magnetic field, and work due to the motion of an object in a magnetic field.


http://cnx.org/content/m55407/1.2/faraday_en.jar

As we have seen, any change in magnetic flux induces an emf opposing that change—a process known as induction. Motion isone of the major causes of induction. For example, a magnet moved toward a coil induces an emf, and a coil moved toward amagnet produces a similar emf. In this section, we concentrate on motion in a magnetic field that is stationary relative to theEarth, producing what is loosely called motional emf.

One situation where motional emf occurs is known as the Hall effect and has already been examined. Charges moving in amagnetic field experience the magnetic force F = qvB sin θ , which moves opposite charges in opposite directions and

produces an emf = Bℓv . We saw that the Hall effect has applications, including measurements of B and v . We will now seethat the Hall effect is one aspect of the broader phenomenon of induction, and we will find that motional emf can be used as apower source.

Consider the situation shown in Figure 23.11. A rod is moved at a speed v along a pair of conducting rails separated by a

distance ℓ in a uniform magnetic field B . The rails are stationary relative to B and are connected to a stationary resistor R .The resistor could be anything from a light bulb to a voltmeter. Consider the area enclosed by the moving rod, rails, and resistor.B is perpendicular to this area, and the area is increasing as the rod moves. Thus the magnetic flux enclosed by the rails, rod,and resistor is increasing. When flux changes, an emf is induced according to Faraday’s law of induction.

Figure 23.11 (a) A motional emf = Bℓv is induced between the rails when this rod moves to the right in the uniform magnetic field. The magnetic

field B is into the page, perpendicular to the moving rod and rails and, hence, to the area enclosed by them. (b) Lenz’s law gives the directions of theinduced field and current, and the polarity of the induced emf. Since the flux is increasing, the induced field is in the opposite direction, or out of thepage. RHR-2 gives the current direction shown, and the polarity of the rod will drive such a current. RHR-1 also indicates the same polarity for the rod.(Note that the script E symbol used in the equivalent circuit at the bottom of part (b) represents emf.)

To find the magnitude of emf induced along the moving rod, we use Faraday’s law of induction without the sign:

(23.7)emf = N ΔΦΔt .



Here and below, “emf” implies the magnitude of the emf. In this equation, N = 1 and the flux Φ = BA cos θ . We have

θ = 0º and cos θ = 1 , since B is perpendicular to A . Now ΔΦ = Δ(BA) = BΔA , since B is uniform. Note that the area

swept out by the rod is ΔA = ℓ Δ x . Entering these quantities into the expression for emf yields

(23.8)emf = BΔAΔt = BℓΔx

Δt .

Finally, note that Δx / Δt = v , the velocity of the rod. Entering this into the last expression shows that

(23.9)emf = Bℓv (B, l, and v perpendicular)

is the motional emf. This is the same expression given for the Hall effect previously.

Making Connections: Unification of Forces

There are many connections between the electric force and the magnetic force. The fact that a moving electric fieldproduces a magnetic field and, conversely, a moving magnetic field produces an electric field is part of why electric andmagnetic forces are now considered to be different manifestations of the same force. This classic unification of electric andmagnetic forces into what is called the electromagnetic force is the inspiration for contemporary efforts to unify other basicforces.

To find the direction of the induced field, the direction of the current, and the polarity of the induced emf, we apply Lenz’s law asexplained in Faraday's Law of Induction: Lenz's Law. (See Figure 23.11(b).) Flux is increasing, since the area enclosed isincreasing. Thus the induced field must oppose the existing one and be out of the page. And so the RHR-2 requires that I becounterclockwise, which in turn means the top of the rod is positive as shown.

Motional emf also occurs if the magnetic field moves and the rod (or other object) is stationary relative to the Earth (or someobserver). We have seen an example of this in the situation where a moving magnet induces an emf in a stationary coil. It is therelative motion that is important. What is emerging in these observations is a connection between magnetic and electric fields. Amoving magnetic field produces an electric field through its induced emf. We already have seen that a moving electric fieldproduces a magnetic field—moving charge implies moving electric field and moving charge produces a magnetic field.

Motional emfs in the Earth’s weak magnetic field are not ordinarily very large, or we would notice voltage along metal rods, suchas a screwdriver, during ordinary motions. For example, a simple calculation of the motional emf of a 1 m rod moving at 3.0 m/s

perpendicular to the Earth’s field gives emf = Bℓv = (5.0×10−5 T)(1.0 m)(3.0 m/s) = 150 μV . This small value is

consistent with experience. There is a spectacular exception, however. In 1992 and 1996, attempts were made with the spaceshuttle to create large motional emfs. The Tethered Satellite was to be let out on a 20 km length of wire as shown in Figure23.12, to create a 5 kV emf by moving at orbital speed through the Earth’s field. This emf could be used to convert some of theshuttle’s kinetic and potential energy into electrical energy if a complete circuit could be made. To complete the circuit, thestationary ionosphere was to supply a return path for the current to flow. (The ionosphere is the rarefied and partially ionizedatmosphere at orbital altitudes. It conducts because of the ionization. The ionosphere serves the same function as the stationaryrails and connecting resistor in Figure 23.11, without which there would not be a complete circuit.) Drag on the current in thecable due to the magnetic force F = IℓB sin θ does the work that reduces the shuttle’s kinetic and potential energy and allowsit to be converted to electrical energy. The tests were both unsuccessful. In the first, the cable hung up and could only beextended a couple of hundred meters; in the second, the cable broke when almost fully extended. Example 23.2 indicatesfeasibility in principle.


Example 23.2 Calculating the Large Motional Emf of an Object in Orbit

Figure 23.12 Motional emf as electrical power conversion for the space shuttle is the motivation for the Tethered Satellite experiment. A 5 kV emfwas predicted to be induced in the 20 km long tether while moving at orbital speed in the Earth’s magnetic field. The circuit is completed by areturn path through the stationary ionosphere.

Calculate the motional emf induced along a 20.0 km long conductor moving at an orbital speed of 7.80 km/s perpendicular to

the Earth’s 5.00×10−5 T magnetic field.

Strategy

This is a straightforward application of the expression for motional emf— emf = Bℓv .

Solution

Entering the given values into emf = Bℓv gives

(23.10)emf = Bℓv= (5.00×10−5 T)(2.0×104 m)(7.80×103 m/s)= 7.80×103 V.

Discussion

The value obtained is greater than the 5 kV measured voltage for the shuttle experiment, since the actual orbital motion ofthe tether is not perpendicular to the Earth’s field. The 7.80 kV value is the maximum emf obtained when θ = 90º and

sin θ = 1 .

23.4 Eddy Currents and Magnetic Damping


• Explain the magnitude and direction of an induced eddy current, and the effect this will have on the object it is inducedin.

• Describe several applications of magnetic damping.

Eddy Currents and Magnetic DampingAs discussed in Motional Emf, motional emf is induced when a conductor moves in a magnetic field or when a magnetic fieldmoves relative to a conductor. If motional emf can cause a current loop in the conductor, we refer to that current as an eddycurrent. Eddy currents can produce significant drag, called magnetic damping, on the motion involved. Consider the apparatusshown in Figure 23.13, which swings a pendulum bob between the poles of a strong magnet. (This is another favorite physicslab activity.) If the bob is metal, there is significant drag on the bob as it enters and leaves the field, quickly damping the motion.If, however, the bob is a slotted metal plate, as shown in Figure 23.13(b), there is a much smaller effect due to the magnet.



There is no discernible effect on a bob made of an insulator. Why is there drag in both directions, and are there any uses formagnetic drag?

Figure 23.13 A common physics demonstration device for exploring eddy currents and magnetic damping. (a) The motion of a metal pendulum bobswinging between the poles of a magnet is quickly damped by the action of eddy currents. (b) There is little effect on the motion of a slotted metal bob,implying that eddy currents are made less effective. (c) There is also no magnetic damping on a nonconducting bob, since the eddy currents areextremely small.

Figure 23.14 shows what happens to the metal plate as it enters and leaves the magnetic field. In both cases, it experiences aforce opposing its motion. As it enters from the left, flux increases, and so an eddy current is set up (Faraday’s law) in thecounterclockwise direction (Lenz’s law), as shown. Only the right-hand side of the current loop is in the field, so that there is anunopposed force on it to the left (RHR-1). When the metal plate is completely inside the field, there is no eddy current if the fieldis uniform, since the flux remains constant in this region. But when the plate leaves the field on the right, flux decreases, causingan eddy current in the clockwise direction that, again, experiences a force to the left, further slowing the motion. A similaranalysis of what happens when the plate swings from the right toward the left shows that its motion is also damped whenentering and leaving the field.

Figure 23.14 A more detailed look at the conducting plate passing between the poles of a magnet. As it enters and leaves the field, the change in fluxproduces an eddy current. Magnetic force on the current loop opposes the motion. There is no current and no magnetic drag when the plate iscompletely inside the uniform field.

When a slotted metal plate enters the field, as shown in Figure 23.15, an emf is induced by the change in flux, but it is lesseffective because the slots limit the size of the current loops. Moreover, adjacent loops have currents in opposite directions, andtheir effects cancel. When an insulating material is used, the eddy current is extremely small, and so magnetic damping oninsulators is negligible. If eddy currents are to be avoided in conductors, then they can be slotted or constructed of thin layers ofconducting material separated by insulating sheets.


Figure 23.15 Eddy currents induced in a slotted metal plate entering a magnetic field form small loops, and the forces on them tend to cancel, therebymaking magnetic drag almost zero.

Applications of Magnetic DampingOne use of magnetic damping is found in sensitive laboratory balances. To have maximum sensitivity and accuracy, the balancemust be as friction-free as possible. But if it is friction-free, then it will oscillate for a very long time. Magnetic damping is a simpleand ideal solution. With magnetic damping, drag is proportional to speed and becomes zero at zero velocity. Thus the oscillationsare quickly damped, after which the damping force disappears, allowing the balance to be very sensitive. (See Figure 23.16.) Inmost balances, magnetic damping is accomplished with a conducting disc that rotates in a fixed field.

Figure 23.16 Magnetic damping of this sensitive balance slows its oscillations. Since Faraday’s law of induction gives the greatest effect for the mostrapid change, damping is greatest for large oscillations and goes to zero as the motion stops.

Since eddy currents and magnetic damping occur only in conductors, recycling centers can use magnets to separate metals fromother materials. Trash is dumped in batches down a ramp, beneath which lies a powerful magnet. Conductors in the trash areslowed by magnetic damping while nonmetals in the trash move on, separating from the metals. (See Figure 23.17.) This worksfor all metals, not just ferromagnetic ones. A magnet can separate out the ferromagnetic materials alone by acting on stationarytrash.



Figure 23.17 Metals can be separated from other trash by magnetic drag. Eddy currents and magnetic drag are created in the metals sent down thisramp by the powerful magnet beneath it. Nonmetals move on.

Other major applications of eddy currents are in metal detectors and braking systems in trains and roller coasters. Portable metaldetectors (Figure 23.18) consist of a primary coil carrying an alternating current and a secondary coil in which a current isinduced. An eddy current will be induced in a piece of metal close to the detector which will cause a change in the inducedcurrent within the secondary coil, leading to some sort of signal like a shrill noise. Braking using eddy currents is safer becausefactors such as rain do not affect the braking and the braking is smoother. However, eddy currents cannot bring the motion to acomplete stop, since the force produced decreases with speed. Thus, speed can be reduced from say 20 m/s to 5 m/s, butanother form of braking is needed to completely stop the vehicle. Generally, powerful rare earth magnets such as neodymiummagnets are used in roller coasters. Figure 23.19 shows rows of magnets in such an application. The vehicle has metal fins(normally containing copper) which pass through the magnetic field slowing the vehicle down in much the same way as with thependulum bob shown in Figure 23.13.

Figure 23.18 A soldier in Iraq uses a metal detector to search for explosives and weapons. (credit: U.S. Army)

Figure 23.19 The rows of rare earth magnets (protruding horizontally) are used for magnetic braking in roller coasters. (credit: Stefan Scheer,Wikimedia Commons)


Induction cooktops have electromagnets under their surface. The magnetic field is varied rapidly producing eddy currents in thebase of the pot, causing the pot and its contents to increase in temperature. Induction cooktops have high efficiencies and goodresponse times but the base of the pot needs to be ferromagnetic, iron or steel for induction to work.

23.5 Electric Generators


• Calculate the emf induced in a generator.• Calculate the peak emf that can be induced in a particular generator system.


• 4.E.2.1 The student is able to construct an explanation of the function of a simple electromagnetic device in which aninduced emf is produced by a changing magnetic flux through an area defined by a current loop (i.e., a simplemicrophone or generator) or of the effect on behavior of a device in which an induced emf is produced by a constantmagnetic field through a changing area. (SP.6.4)

Electric generators induce an emf by rotating a coil in a magnetic field, as briefly discussed in Induced Emf and MagneticFlux. We will now explore generators in more detail. Consider the following example.

Example 23.3 Calculating the Emf Induced in a Generator Coil

The generator coil shown in Figure 23.20 is rotated through one-fourth of a revolution (from θ = 0º to θ = 90º ) in 15.0ms. The 200-turn circular coil has a 5.00 cm radius and is in a uniform 1.25 T magnetic field. What is the average emfinduced?

Figure 23.20 When this generator coil is rotated through one-fourth of a revolution, the magnetic flux Φ changes from its maximum to zero,inducing an emf.

Strategy

We use Faraday’s law of induction to find the average emf induced over a time Δt :

(23.11)emf = −N ΔΦΔt .

We know that N = 200 and Δt = 15.0 ms , and so we must determine the change in flux ΔΦ to find emf.

Solution



Since the area of the loop and the magnetic field strength are constant, we see that

(23.12)ΔΦ = Δ(BA cos θ) = ABΔ(cos θ).

Now, Δ(cos θ) = −1.0 , since it was given that θ goes from 0º to 90º . Thus ΔΦ = −AB , and

(23.13)emf = N ABΔt .

The area of the loop is A = πr2 = (3.14...)(0.0500 m)2 = 7.85×10−3 m2 . Entering this value gives

(23.14)emf = 200(7.85×10−3 m2)(1.25 T)

15.0×10−3 s= 131 V.

Discussion

This is a practical average value, similar to the 120 V used in household power.

The emf calculated in Example 23.3 is the average over one-fourth of a revolution. What is the emf at any given instant? It varieswith the angle between the magnetic field and a perpendicular to the coil. We can get an expression for emf as a function of timeby considering the motional emf on a rotating rectangular coil of width w and height ℓ in a uniform magnetic field, as illustratedin Figure 23.21.

Figure 23.21 A generator with a single rectangular coil rotated at constant angular velocity in a uniform magnetic field produces an emf that variessinusoidally in time. Note the generator is similar to a motor, except the shaft is rotated to produce a current rather than the other way around.

Charges in the wires of the loop experience the magnetic force, because they are moving in a magnetic field. Charges in thevertical wires experience forces parallel to the wire, causing currents. But those in the top and bottom segments feel a forceperpendicular to the wire, which does not cause a current. We can thus find the induced emf by considering only the side wires.Motional emf is given to be emf = Bℓv , where the velocity v is perpendicular to the magnetic field B . Here the velocity is at an

angle θ with B , so that its component perpendicular to B is v sin θ (see Figure 23.21). Thus in this case the emf induced on

each side is emf = Bℓv sin θ , and they are in the same direction. The total emf around the loop is then

(23.15)emf = 2Bℓv sin θ.This expression is valid, but it does not give emf as a function of time. To find the time dependence of emf, we assume the coilrotates at a constant angular velocity ω . The angle θ is related to angular velocity by θ = ωt , so that

(23.16)emf = 2Bℓv sin ωt.

Now, linear velocity v is related to angular velocity ω by v = rω . Here r = w / 2 , so that v = (w / 2)ω , and

(23.17)emf = 2Bℓw2 ω sin ωt = (ℓw)Bω sin ωt.

Noting that the area of the loop is A = ℓw , and allowing for N loops, we find that

(23.18)emf = NABω sin ωt

is the emf induced in a generator coil of N turns and area A rotating at a constant angular velocity ω in a uniform magnetic

field B . This can also be expressed as

(23.19)emf = emf0 sin ωt,

where


(23.20)emf0 = NABω

is the maximum (peak) emf. Note that the frequency of the oscillation is f = ω / 2π , and the period is T = 1 / f = 2π / ω .

Figure 23.22 shows a graph of emf as a function of time, and it now seems reasonable that AC voltage is sinusoidal.

Figure 23.22 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 afunction of time. emf0 is the peak emf. The period is T = 1 / f = 2π / ω , where f is the frequency. Note that the script E stands for emf.

The fact that the peak emf, emf0 = NABω , makes good sense. The greater the number of coils, the larger their area, and the

stronger the field, the greater the output voltage. It is interesting that the faster the generator is spun (greater ω ), the greater theemf. This is noticeable on bicycle generators—at least the cheaper varieties. One of the authors as a juvenile found it amusing toride his bicycle fast enough to burn out his lights, until he had to ride home lightless one dark night.

Figure 23.23 shows a scheme by which a generator can be made to produce pulsed DC. More elaborate arrangements ofmultiple coils and split rings can produce smoother DC, although electronic rather than mechanical means are usually used tomake ripple-free DC.

Figure 23.23 Split rings, called commutators, produce a pulsed DC emf output in this configuration.

Example 23.4 Calculating the Maximum Emf of a Generator

Calculate the maximum emf, emf0 , of the generator that was the subject of Example 23.3.

Strategy

Once ω , the angular velocity, is determined, emf0 = NABω can be used to find emf0 . All other quantities are known.



Solution

Angular velocity is defined to be the change in angle per unit time:

(23.21)ω = ΔθΔt .

One-fourth of a revolution is π/2 radians, and the time is 0.0150 s; thus,

(23.22)ω = π / 2 rad0.0150 s

= 104.7 rad/s.104.7 rad/s is exactly 1000 rpm. We substitute this value for ω and the information from the previous example into

emf0 = NABω , yielding

(23.23)emf0 = NABω

= 200(7.85×10−3 m2)(1.25 T)(104.7 rad/s)= 206 V

.

Discussion

The maximum emf is greater than the average emf of 131 V found in the previous example, as it should be.

In real life, electric generators look a lot different than the figures in this section, but the principles are the same. The source ofmechanical energy that turns the coil can be falling water (hydropower), steam produced by the burning of fossil fuels, or thekinetic energy of wind. Figure 23.24 shows a cutaway view of a steam turbine; steam moves over the blades connected to theshaft, which rotates the coil within the generator.

Figure 23.24 Steam turbine/generator. The steam produced by burning coal impacts the turbine blades, turning the shaft which is connected to thegenerator. (credit: Nabonaco, Wikimedia Commons)

Generators illustrated in this section look very much like the motors illustrated previously. This is not coincidental. In fact, a motorbecomes a generator when its shaft rotates. Certain early automobiles used their starter motor as a generator. In Back Emf, weshall further explore the action of a motor as a generator.

23.6 Back Emf


• Explain what back emf is and how it is induced.

It has been noted that motors and generators are very similar. Generators convert mechanical energy into electrical energy,whereas motors convert electrical energy into mechanical energy. Furthermore, motors and generators have the sameconstruction. When the coil of a motor is turned, magnetic flux changes, and an emf (consistent with Faraday’s law of induction)is induced. The motor thus acts as a generator whenever its coil rotates. This will happen whether the shaft is turned by anexternal input, like a belt drive, or by the action of the motor itself. That is, when a motor is doing work and its shaft is turning, anemf is generated. Lenz’s law tells us the emf opposes any change, so that the input emf that powers the motor will be opposedby the motor’s self-generated emf, called the back emf of the motor. (See Figure 23.25.)


Figure 23.25 The coil of a DC motor is represented as a resistor in this schematic. The back emf is represented as a variable emf that opposes theone driving the motor. Back emf is zero when the motor is not turning, and it increases proportionally to the motor’s angular velocity.

Back emf is the generator output of a motor, and so it is proportional to the motor’s angular velocity ω . It is zero when the motoris first turned on, meaning that the coil receives the full driving voltage and the motor draws maximum current when it is on butnot turning. As the motor turns faster and faster, the back emf grows, always opposing the driving emf, and reduces the voltageacross the coil and the amount of current it draws. This effect is noticeable in a number of situations. When a vacuum cleaner,refrigerator, or washing machine is first turned on, lights in the same circuit dim briefly due to the IR drop produced in feederlines by the large current drawn by the motor. When a motor first comes on, it draws more current than when it runs at its normaloperating speed. When a mechanical load is placed on the motor, like an electric wheelchair going up a hill, the motor slows, theback emf drops, more current flows, and more work can be done. If the motor runs at too low a speed, the larger current can

overheat it (via resistive power in the coil, P = I 2R ), perhaps even burning it out. On the other hand, if there is no mechanicalload on the motor, it will increase its angular velocity ω until the back emf is nearly equal to the driving emf. Then the motoruses only enough energy to overcome friction.

Consider, for example, the motor coils represented in Figure 23.25. The coils have a 0.400 Ω equivalent resistance and are

driven by a 48.0 V emf. Shortly after being turned on, they draw a current I = V/R = (48.0 V)/(0.400 Ω ) = 120 A and,

thus, dissipate P = I 2R = 5.76 kW of energy as heat transfer. Under normal operating conditions for this motor, suppose theback emf is 40.0 V. Then at operating speed, the total voltage across the coils is 8.0 V (48.0 V minus the 40.0 V back emf), andthe current drawn is I = V/R = (8.0 V)/(0.400 Ω ) = 20 A . Under normal load, then, the power dissipated is

P = IV = (20 A) / (8.0 V) = 160 W . The latter will not cause a problem for this motor, whereas the former 5.76 kW would

burn out the coils if sustained.

23.7 Transformers


• Explain how a transformer works.• Calculate voltage, current, and/or number of turns given the other quantities.



Transformers do what their name implies—they transform voltages from one value to another (The term voltage is used ratherthan emf, because transformers have internal resistance). For example, many cell phones, laptops, video games, and powertools and small appliances have a transformer built into their plug-in unit (like that in Figure 23.26) that changes 120 V or 240 VAC into whatever voltage the device uses. Transformers are also used at several points in the power distribution systems, suchas illustrated in Figure 23.27. Power is sent long distances at high voltages, because less current is required for a given amountof power, and this means less line loss, as was discussed previously. But high voltages pose greater hazards, so thattransformers are employed to produce lower voltage at the user’s location.



Figure 23.26 The plug-in transformer has become increasingly familiar with the proliferation of electronic devices that operate on voltages other thancommon 120 V AC. Most are in the 3 to 12 V range. (credit: Shop Xtreme)

Figure 23.27 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 over 200 kV—sometimes as great as 700 kV—to limit energy losses. Local power distribution toneighborhoods or industries goes through a substation and is sent short distances at voltages ranging from 5 to 13 kV. This is reduced to 120, 240, or480 V for safety at the individual user site.

The type of transformer considered in this text—see Figure 23.28—is based on Faraday’s law of induction and is very similar inconstruction to the apparatus Faraday used to demonstrate magnetic fields could cause currents. The two coils are called theprimary and secondary coils. In normal use, the input voltage is placed on the primary, and the secondary produces thetransformed output voltage. Not only does the iron core trap the magnetic field created by the primary coil, its magnetizationincreases the field strength. Since the input voltage is AC, a time-varying magnetic flux is sent to the secondary, inducing its ACoutput voltage.

Figure 23.28 A typical construction of a simple transformer has two coils wound on a ferromagnetic core that is laminated to minimize eddy currents.The magnetic field created by the primary is mostly confined to and increased by the core, which transmits it to the secondary coil. Any change incurrent in the primary induces a current in the secondary.


For the simple transformer shown in Figure 23.28, the output voltage Vs depends almost entirely on the input voltage Vp and

the ratio of the number of loops in the primary and secondary coils. Faraday’s law of induction for the secondary coil gives itsinduced output voltage Vs to be

(23.24)Vs = −NsΔΦΔt ,

where Ns is the number of loops in the secondary coil and ΔΦ / Δt is the rate of change of magnetic flux. Note that the output

voltage equals the induced emf ( Vs = emfs ), provided coil resistance is small (a reasonable assumption for transformers). The

cross-sectional area of the coils is the same on either side, as is the magnetic field strength, and so ΔΦ / Δt is the same on

either side. The input primary voltage Vp is also related to changing flux by

(23.25)V p = −NpΔΦΔt .

The reason for this is a little more subtle. Lenz’s law tells us that the primary coil opposes the change in flux caused by the inputvoltage Vp , hence the minus sign (This is an example of self-inductance, a topic to be explored in some detail in later sections).

Assuming negligible coil resistance, Kirchhoff’s loop rule tells us that the induced emf exactly equals the input voltage. Taking theratio of these last two equations yields a useful relationship:

(23.26)VsVp

= NsNp

.

This is known as the transformer equation, and it simply states that the ratio of the secondary to primary voltages in atransformer equals the ratio of the number of loops in their coils.

The output voltage of a transformer can be less than, greater than, or equal to the input voltage, depending on the ratio of thenumber of loops in their coils. Some transformers even provide a variable output by allowing connection to be made at differentpoints on the secondary coil. A step-up transformer is one that increases voltage, whereas a step-down transformerdecreases voltage. Assuming, as we have, that resistance is negligible, the electrical power output of a transformer equals itsinput. This is nearly true in practice—transformer efficiency often exceeds 99%. Equating the power input and output,

(23.27)Pp = IpVp = IsVs = Ps.

Rearranging terms gives

(23.28)VsVp

=IpIs

.

Combining this withVsVp

= NsNp

, we find that

(23.29)IsIp

=NpNs

is the relationship between the output and input currents of a transformer. So if voltage increases, current decreases. Conversely,if voltage decreases, current increases.

Example 23.5 Calculating Characteristics of a Step-Up Transformer

A portable x-ray unit has a step-up transformer, the 120 V input of which is transformed to the 100 kV output needed by thex-ray tube. The primary has 50 loops and draws a current of 10.00 A when in use. (a) What is the number of loops in thesecondary? (b) Find the current output of the secondary.

Strategy and Solution for (a)

We solveVsVp

= NsNp

for Ns , the number of loops in the secondary, and enter the known values. This gives

(23.30)Ns = NpVsVp

= (50)100,000 V120 V = 4.17×104 .

Discussion for (a)



A large number of loops in the secondary (compared with the primary) is required to produce such a large voltage. Thiswould be true for neon sign transformers and those supplying high voltage inside TVs and CRTs.

Strategy and Solution for (b)

We can similarly find the output current of the secondary by solvingIsIp

=NpNs

for Is and entering known values. This gives

(23.31)Is = Ip

NpNs

= (10.00 A) 504.17×104 = 12.0 mA.

Discussion for (b)

As expected, the current output is significantly less than the input. In certain spectacular demonstrations, very large voltagesare used to produce long arcs, but they are relatively safe because the transformer output does not supply a large current.Note that the power input here is Pp = IpVp = (10.00 A)(120 V) = 1.20 kW . This equals the power output

Pp = IsVs = (12.0 mA)(100 kV) = 1.20 kW , as we assumed in the derivation of the equations used.

The fact that transformers are based on Faraday’s law of induction makes it clear why we cannot use transformers to change DCvoltages. If there is no change in primary voltage, there is no voltage induced in the secondary. One possibility is to connect DCto the primary coil through a switch. As the switch is opened and closed, the secondary produces a voltage like that in Figure23.29. This is not really a practical alternative, and AC is in common use wherever it is necessary to increase or decreasevoltages.

Figure 23.29 Transformers do not work for pure DC voltage input, but if it is switched on and off as on the top graph, the output will look something likethat on the bottom graph. This is not the sinusoidal AC most AC appliances need.

Example 23.6 Calculating Characteristics of a Step-Down Transformer

A battery charger meant for a series connection of ten nickel-cadmium batteries (total emf of 12.5 V DC) needs to have a15.0 V output to charge the batteries. It uses a step-down transformer with a 200-loop primary and a 120 V input. (a) Howmany loops should there be in the secondary coil? (b) If the charging current is 16.0 A, what is the input current?


You would expect the secondary to have a small number of loops. SolvingVsVp

= NsNp

for Ns and entering known values

gives


(23.32)Ns = NpVsVp

= (200)15.0 V120 V = 25.


The current input can be obtained by solvingIsIp

=NpNs

for Ip and entering known values. This gives

(23.33)Ip = IsNsNp

= (16.0 A) 25200 = 2.00 A.

Discussion

The number of loops in the secondary is small, as expected for a step-down transformer. We also see that a small inputcurrent produces a larger output current in a step-down transformer. When transformers are used to operate large magnets,they sometimes have a small number of very heavy loops in the secondary. This allows the secondary to have low internalresistance and produce large currents. Note again that this solution is based on the assumption of 100% efficiency—orpower out equals power in ( Pp = Ps )—reasonable for good transformers. In this case the primary and secondary power is

240 W. (Verify this for yourself as a consistency check.) Note that the Ni-Cd batteries need to be charged from a DC powersource (as would a 12 V battery). So the AC output of the secondary coil needs to be converted into DC. This is done usingsomething called a rectifier, which uses devices called diodes that allow only a one-way flow of current.

Transformers have many applications in electrical safety systems, which are discussed in Electrical Safety: Systems andDevices.

PhET Explorations: Generator

Generate electricity with a bar magnet! Discover the physics behind the phenomena by exploring magnets and how you canuse them to make a bulb light.

Figure 23.30 Generator (http://cnx.org/content/m55414/1.2/generator_en.jar)

23.8 Electrical Safety: Systems and Devices


• Explain how various modern safety features in electric circuits work, with an emphasis on how induction is employed.



Electricity has two hazards. A thermal hazard occurs when there is electrical overheating. A shock hazard occurs when electriccurrent passes through a person. Both hazards have already been discussed. Here we will concentrate on systems and devicesthat prevent electrical hazards.

Figure 23.31 shows the schematic for a simple AC circuit with no safety features. This is not how power is distributed in practice.Modern household and industrial wiring requires the three-wire system, shown schematically in Figure 23.32, which hasseveral safety features. First is the familiar circuit breaker (or fuse) to prevent thermal overload. Second, there is a protectivecase around the appliance, such as a toaster or refrigerator. The case’s safety feature is that it prevents a person from touchingexposed wires and coming into electrical contact with the circuit, helping prevent shocks.



http://cnx.org/content/m55414/1.2/generator_en.jar

Figure 23.31 Schematic of a simple AC circuit with a voltage source and a single appliance represented by the resistance R . There are no safetyfeatures in this circuit.

Figure 23.32 The three-wire system connects the neutral wire to the earth at the voltage source and user location, forcing it to be at zero volts andsupplying an alternative return path for the current through the earth. Also grounded to zero volts is the case of the appliance. A circuit breaker or fuseprotects against thermal overload and is in series on the active (live/hot) wire. Note that wire insulation colors vary with region and it is essential tocheck locally to determine which color codes are in use (and even if they were followed in the particular installation).

There are three connections to earth or ground (hereafter referred to as “earth/ground”) shown in Figure 23.32. Recall that anearth/ground connection is a low-resistance path directly to the earth. The two earth/ground connections on the neutral wire forceit to be at zero volts relative to the earth, giving the wire its name. This wire is therefore safe to touch even if its insulation, usuallywhite, is missing. The neutral wire is the return path for the current to follow to complete the circuit. Furthermore, the two earth/ground connections supply an alternative path through the earth, a good conductor, to complete the circuit. The earth/groundconnection closest to the power source could be at the generating plant, while the other is at the user’s location. The third earth/ground is to the case of the appliance, through the green earth/ground wire, forcing the case, too, to be at zero volts. The live orhot wire (hereafter referred to as “live/hot”) supplies voltage and current to operate the appliance. Figure 23.33 shows a morepictorial version of how the three-wire system is connected through a three-prong plug to an appliance.

Figure 23.33 The standard three-prong plug can only be inserted in one way, to assure proper function of the three-wire system.


A note on insulation color-coding: Insulating plastic is color-coded to identify live/hot, neutral and ground wires but these codesvary around the world. Live/hot wires may be brown, red, black, blue or grey. Neutral wire may be blue, black or white. Since thesame color may be used for live/hot or neutral in different parts of the world, it is essential to determine the color code in yourregion. The only exception is the earth/ground wire which is often green but may be yellow or just bare wire. Striped coatings aresometimes used for the benefit of those who are colorblind.

The three-wire system replaced the older two-wire system, which lacks an earth/ground wire. Under ordinary circumstances,insulation on the live/hot and neutral wires prevents the case from being directly in the circuit, so that the earth/ground wire mayseem like double protection. Grounding the case solves more than one problem, however. The simplest problem is worninsulation on the live/hot wire that allows it to contact the case, as shown in Figure 23.34. Lacking an earth/ground connection(some people cut the third prong off the plug because they only have outdated two hole receptacles), a severe shock is possible.This is particularly dangerous in the kitchen, where a good connection to earth/ground is available through water on the floor or awater faucet. With the earth/ground connection intact, the circuit breaker will trip, forcing repair of the appliance. Why are someappliances still sold with two-prong plugs? These have nonconducting cases, such as power tools with impact resistant plasticcases, and are called doubly insulated. Modern two-prong plugs can be inserted into the asymmetric standard outlet in only oneway, to ensure proper connection of live/hot and neutral wires.

Figure 23.34 Worn insulation allows the live/hot wire to come into direct contact with the metal case of this appliance. (a) The earth/ground connectionbeing broken, the person is severely shocked. The appliance may operate normally in this situation. (b) With a proper earth/ground, the circuit breakertrips, forcing repair of the appliance.

Electromagnetic induction causes a more subtle problem that is solved by grounding the case. The AC current in appliances caninduce an emf on the case. If grounded, the case voltage is kept near zero, but if the case is not grounded, a shock can occur as



pictured in Figure 23.35. Current driven by the induced case emf is called a leakage current, although current does notnecessarily pass from the resistor to the case.

Figure 23.35 AC currents can induce an emf on the case of an appliance. The voltage can be large enough to cause a shock. If the case is grounded,the induced emf is kept near zero.

A ground fault interrupter (GFI) is a safety device found in updated kitchen and bathroom wiring that works based onelectromagnetic induction. GFIs compare the currents in the live/hot and neutral wires. When live/hot and neutral currents are notequal, it is almost always because current in the neutral is less than in the live/hot wire. Then some of the current, again called aleakage current, is returning to the voltage source by a path other than through the neutral wire. It is assumed that this pathpresents a hazard, such as shown in Figure 23.36. GFIs are usually set to interrupt the circuit if the leakage current is greaterthan 5 mA, the accepted maximum harmless shock. Even if the leakage current goes safely to earth/ground through an intactearth/ground wire, the GFI will trip, forcing repair of the leakage.

Figure 23.36 A ground fault interrupter (GFI) compares the currents in the live/hot and neutral wires and will trip if their difference exceeds a safevalue. The leakage current here follows a hazardous path that could have been prevented by an intact earth/ground wire.

Figure 23.37 shows how a GFI works. If the currents in the live/hot and neutral wires are equal, then they induce equal andopposite emfs in the coil. If not, then the circuit breaker will trip.


Figure 23.37 A GFI compares currents by using both to induce an emf in the same coil. If the currents are equal, they will induce equal but oppositeemfs.

Another induction-based safety device is the isolation transformer, shown in Figure 23.38. Most isolation transformers haveequal input and output voltages. Their function is to put a large resistance between the original voltage source and the devicebeing operated. This prevents a complete circuit between them, even in the circumstance shown. There is a complete circuitthrough the appliance. But there is not a complete circuit for current to flow through the person in the figure, who is touching onlyone of the transformer’s output wires, and neither output wire is grounded. The appliance is isolated from the original voltagesource by the high resistance of the material between the transformer coils, hence the name isolation transformer. For current toflow through the person, it must pass through the high-resistance material between the coils, through the wire, the person, andback through the earth—a path with such a large resistance that the current is negligible.

Figure 23.38 An isolation transformer puts a large resistance between the original voltage source and the device, preventing a complete circuitbetween them.

The basics of electrical safety presented here help prevent many electrical hazards. Electrical safety can be pursued to greaterdepths. There are, for example, problems related to different earth/ground connections for appliances in close proximity. Manyother examples are found in hospitals. Microshock-sensitive patients, for instance, require special protection. For these people,currents as low as 0.1 mA may cause ventricular fibrillation. The interested reader can use the material presented here as abasis for further study.

23.9 Inductance


• Calculate the inductance of an inductor.• Calculate the energy stored in an inductor.• Calculate the emf generated in an inductor.

InductorsInduction is the process in which an emf is induced by changing magnetic flux. Many examples have been discussed so far,some more effective than others. Transformers, for example, are designed to be particularly effective at inducing a desiredvoltage and current with very little loss of energy to other forms. Is there a useful physical quantity related to how “effective” agiven device is? The answer is yes, and that physical quantity is called inductance.

Mutual inductance is the effect of Faraday’s law of induction for one device upon another, such as the primary coil intransmitting energy to the secondary in a transformer. See Figure 23.39, where simple coils induce emfs in one another.



Figure 23.39 These coils can induce emfs in one another like an inefficient transformer. Their mutual inductance M indicates the effectiveness of thecoupling between them. Here a change in current in coil 1 is seen to induce an emf in coil 2. (Note that " E2 induced" represents the induced emf in

coil 2.)

In the many cases where the geometry of the devices is fixed, flux is changed by varying current. We therefore concentrate onthe rate of change of current, ΔI/Δt , as the cause of induction. A change in the current I1 in one device, coil 1 in the figure,

induces an emf2 in the other. We express this in equation form as

(23.34)emf2 = −MΔI1

Δt ,

where M is defined to be the mutual inductance between the two devices. The minus sign is an expression of Lenz’s law. The

larger the mutual inductance M , the more effective the coupling. For example, the coils in Figure 23.39 have a small Mcompared with the transformer coils in Figure 23.28. Units for M are (V ⋅ s)/A = Ω ⋅ s , which is named a henry (H), after

Joseph Henry. That is, 1 H = 1 Ω ⋅ s .

Nature is symmetric here. If we change the current I2 in coil 2, we induce an emf1 in coil 1, which is given by

(23.35)emf1 = −MΔI2

Δt ,

where M is the same as for the reverse process. Transformers run backward with the same effectiveness, or mutual inductance

M .

A large mutual inductance M may or may not be desirable. We want a transformer to have a large mutual inductance. But anappliance, such as an electric clothes dryer, can induce a dangerous emf on its case if the mutual inductance between its coilsand the case is large. One way to reduce mutual inductance M is to counterwind coils to cancel the magnetic field produced.(See Figure 23.40.)


Figure 23.40 The heating coils of an electric clothes dryer can be counter-wound so that their magnetic fields cancel one another, greatly reducing themutual inductance with the case of the dryer.

Self-inductance, the effect of Faraday’s law of induction of a device on itself, also exists. When, for example, current through acoil is increased, the magnetic field and flux also increase, inducing a counter emf, as required by Lenz’s law. Conversely, if thecurrent is decreased, an emf is induced that opposes the decrease. Most devices have a fixed geometry, and so the change influx is due entirely to the change in current ΔI through the device. The induced emf is related to the physical geometry of thedevice and the rate of change of current. It is given by

(23.36)emf = −LΔIΔt ,

where L is the self-inductance of the device. A device that exhibits significant self-inductance is called an inductor, and giventhe symbol in Figure 23.41.

Figure 23.41

The minus sign is an expression of Lenz’s law, indicating that emf opposes the change in current. Units of self-inductance arehenries (H) just as for mutual inductance. The larger the self-inductance L of a device, the greater its opposition to any change

in current through it. For example, a large coil with many turns and an iron core has a large L and will not allow current to

change quickly. To avoid this effect, a small L must be achieved, such as by counterwinding coils as in Figure 23.40.

A 1 H inductor is a large inductor. To illustrate this, consider a device with L = 1.0 H that has a 10 A current flowing through it.

What happens if we try to shut off the current rapidly, perhaps in only 1.0 ms? An emf, given by emf = −L(ΔI / Δt) , will

oppose the change. Thus an emf will be induced given by emf = −L(ΔI / Δt) = (1.0 H)[(10 A) / (1.0 ms)] = 10,000 V .

The positive sign means this large voltage is in the same direction as the current, opposing its decrease. Such large emfs cancause arcs, damaging switching equipment, and so it may be necessary to change current more slowly.

There are uses for such a large induced voltage. Camera flashes use a battery, two inductors that function as a transformer, anda switching system or oscillator to induce large voltages. (Remember that we need a changing magnetic field, brought about by achanging current, to induce a voltage in another coil.) The oscillator system will do this many times as the battery voltage isboosted to over one thousand volts. (You may hear the high pitched whine from the transformer as the capacitor is beingcharged.) A capacitor stores the high voltage for later use in powering the flash. (See Figure 23.42.)



Figure 23.42 Through rapid switching of an inductor, 1.5 V batteries can be used to induce emfs of several thousand volts. This voltage can be used tostore charge in a capacitor for later use, such as in a camera flash attachment.

It is possible to calculate L for an inductor given its geometry (size and shape) and knowing the magnetic field that it produces.

This is difficult in most cases, because of the complexity of the field created. So in this text the inductance L is usually a givenquantity. One exception is the solenoid, because it has a very uniform field inside, a nearly zero field outside, and a simpleshape. It is instructive to derive an equation for its inductance. We start by noting that the induced emf is given by Faraday’s lawof induction as emf = −N(ΔΦ / Δt) and, by the definition of self-inductance, as emf = −L(ΔI / Δt) . Equating these yields

(23.37)emf = −N ΔΦΔt = −LΔI

Δt .

Solving for L gives

(23.38)L = N ΔΦΔI .

This equation for the self-inductance L of a device is always valid. It means that self-inductance L depends on how effective

the current is in creating flux; the more effective, the greater ΔΦ / ΔI is.

Let us use this last equation to find an expression for the inductance of a solenoid. Since the area A of a solenoid is fixed, the

change in flux is ΔΦ = Δ(BA) = AΔB . To find ΔB , we note that the magnetic field of a solenoid is given by

B = µ0nI = µ0NIℓ . (Here n = N / ℓ , where N is the number of coils and ℓ is the solenoid’s length.) Only the current

changes, so that ΔΦ = AΔB = µ0NAΔIℓ . Substituting ΔΦ into L = N ΔΦ

ΔI gives

(23.39)L = N ΔΦ

ΔI = Nµ0 NAΔI

ℓΔI .

This simplifies to

(23.40)L = µ0 N 2 A

ℓ (solenoid).

This is the self-inductance of a solenoid of cross-sectional area A and length ℓ . Note that the inductance depends only on thephysical characteristics of the solenoid, consistent with its definition.

Example 23.7 Calculating the Self-inductance of a Moderate Size Solenoid

Calculate the self-inductance of a 10.0 cm long, 4.00 cm diameter solenoid that has 200 coils.

Strategy

This is a straightforward application of L = µ0 N 2 Aℓ , since all quantities in the equation except L are known.

Solution

Use the following expression for the self-inductance of a solenoid:

(23.41)L = µ0 N 2 A

ℓ .


The cross-sectional area in this example is A = πr2 = (3.14...)(0.0200 m)2 = 1.26×10−3 m2 , N is given to be 200,

and the length ℓ is 0.100 m. We know the permeability of free space is µ0 = 4π×10−7 T ⋅ m/A . Substituting these into

the expression for L gives

(23.42)L = (4π×10−7 T ⋅ m/A)(200)2(1.26×10−3 m2)

0.100 m= 0.632 mH.

Discussion

This solenoid is moderate in size. Its inductance of nearly a millihenry is also considered moderate.

One common application of inductance is used in traffic lights that can tell when vehicles are waiting at the intersection. Anelectrical circuit with an inductor is placed in the road under the place a waiting car will stop over. The body of the car increasesthe inductance and the circuit changes sending a signal to the traffic lights to change colors. Similarly, metal detectors used forairport security employ the same technique. A coil or inductor in the metal detector frame acts as both a transmitter and areceiver. The pulsed signal in the transmitter coil induces a signal in the receiver. The self-inductance of the circuit is affected byany metal object in the path. Such detectors can be adjusted for sensitivity and also can indicate the approximate location ofmetal found on a person. (But they will not be able to detect any plastic explosive such as that found on the “underwearbomber.”) See Figure 23.43.

Figure 23.43 The familiar security gate at an airport can not only detect metals but also indicate their approximate height above the floor. (credit:Alexbuirds, Wikimedia Commons)

Energy Stored in an InductorWe know from Lenz’s law that inductances oppose changes in current. There is an alternative way to look at this opposition thatis based on energy. Energy is stored in a magnetic field. It takes time to build up energy, and it also takes time to deplete energy;hence, there is an opposition to rapid change. In an inductor, the magnetic field is directly proportional to current and to theinductance of the device. It can be shown that the energy stored in an inductor Eind is given by

(23.43)Eind = 12LI 2.

This expression is similar to that for the energy stored in a capacitor.

Example 23.8 Calculating the Energy Stored in the Field of a Solenoid

How much energy is stored in the 0.632 mH inductor of the preceding example when a 30.0 A current flows through it?

Strategy

The energy is given by the equation Eind = 12LI 2 , and all quantities except Eind are known.

Solution

Substituting the value for L found in the previous example and the given current into Eind = 12LI 2 gives

(23.44)Eind = 12LI 2

= 0.5(0.632×10−3 H)(30.0 A)2 = 0.284 J.



Discussion

This amount of energy is certainly enough to cause a spark if the current is suddenly switched off. It cannot be built upinstantaneously unless the power input is infinite.

23.10 RL Circuits


• Calculate the current in an RL circuit after a specified number of characteristic time steps.• Calculate the characteristic time of an RL circuit.• Sketch the current in an RL circuit over time.

We know that the current through an inductor L cannot be turned on or off instantaneously. The change in current changes flux,inducing an emf opposing the change (Lenz’s law). How long does the opposition last? Current will flow and can be turned off,but how long does it take? Figure 23.44 shows a switching circuit that can be used to examine current through an inductor as afunction of time.

Figure 23.44 (a) An RL circuit with a switch to turn current on and off. When in position 1, the battery, resistor, and inductor are in series and a currentis established. In position 2, the battery is removed and the current eventually stops because of energy loss in the resistor. (b) A graph of currentgrowth versus time when the switch is moved to position 1. (c) A graph of current decay when the switch is moved to position 2.

When the switch is first moved to position 1 (at t = 0 ), the current is zero and it eventually rises to I0 = V/R , where R is the

total resistance of the circuit. The opposition of the inductor L is greatest at the beginning, because the amount of change isgreatest. The opposition it poses is in the form of an induced emf, which decreases to zero as the current approaches its finalvalue. The opposing emf is proportional to the amount of change left. This is the hallmark of an exponential behavior, and it canbe shown with calculus that

(23.45)I = I0(1 − e−t / τ) (turning on),

is the current in an RL circuit when switched on (Note the similarity to the exponential behavior of the voltage on a chargingcapacitor). The initial current is zero and approaches I0 = V/R with a characteristic time constant τ for an RL circuit, given

by

(23.46)τ = LR,

where τ has units of seconds, since 1 H=1 Ω·s . In the first period of time τ , the current rises from zero to 0.632I0 , since

I = I0(1 − e−1) = I0(1 − 0.368) = 0.632I0 . The current will go 0.632 of the remainder in the next time τ . A well-known

property of the exponential is that the final value is never exactly reached, but 0.632 of the remainder to that value is achieved inevery characteristic time τ . In just a few multiples of the time τ , the final value is very nearly achieved, as the graph in Figure23.44(b) illustrates.

The characteristic time τ depends on only two factors, the inductance L and the resistance R . The greater the inductance L ,the greater τ is, which makes sense since a large inductance is very effective in opposing change. The smaller the resistance

R , the greater τ is. Again this makes sense, since a small resistance means a large final current and a greater change to get

there. In both cases—large L and small R —more energy is stored in the inductor and more time is required to get it in andout.


When the switch in Figure 23.44(a) is moved to position 2 and cuts the battery out of the circuit, the current drops because ofenergy dissipation by the resistor. But this is also not instantaneous, since the inductor opposes the decrease in current byinducing an emf in the same direction as the battery that drove the current. Furthermore, there is a certain amount of energy,

(1/2)LI02 , stored in the inductor, and it is dissipated at a finite rate. As the current approaches zero, the rate of decrease slows,

since the energy dissipation rate is I 2 R . Once again the behavior is exponential, and I is found to be

(23.47)I = I0e−t / τ (turning off).

(See Figure 23.44(c).) In the first period of time τ = L / R after the switch is closed, the current falls to 0.368 of its initial value,

since I = I0e−1 = 0.368I0 . In each successive time τ , the current falls to 0.368 of the preceding value, and in a few

multiples of τ , the current becomes very close to zero, as seen in the graph in Figure 23.44(c).

Example 23.9 Calculating Characteristic Time and Current in an RL Circuit

(a) What is the characteristic time constant for a 7.50 mH inductor in series with a 3.00 Ω resistor? (b) Find the current5.00 ms after the switch is moved to position 2 to disconnect the battery, if it is initially 10.0 A.

Strategy for (a)

The time constant for an RL circuit is defined by τ = L / R .

Solution for (a)

Entering known values into the expression for τ given in τ = L / R yields

(23.48)τ = LR = 7.50 mH

3.00 Ω = 2.50 ms.

Discussion for (a)

This is a small but definitely finite time. The coil will be very close to its full current in about ten time constants, or about 25ms.

Strategy for (b)

We can find the current by using I = I0e−t / τ , or by considering the decline in steps. Since the time is twice the

characteristic time, we consider the process in steps.

Solution for (b)

In the first 2.50 ms, the current declines to 0.368 of its initial value, which is

(23.49)I = 0.368I0 = (0.368)(10.0 A)= 3.68 A at t = 2.50 ms.

After another 2.50 ms, or a total of 5.00 ms, the current declines to 0.368 of the value just found. That is,

(23.50)I′ = 0.368I = (0.368)(3.68 A)= 1.35 A at t = 5.00 ms.

Discussion for (b)

After another 5.00 ms has passed, the current will be 0.183 A (see Exercise 23.69); so, although it does die out, the currentcertainly does not go to zero instantaneously.

In summary, when the voltage applied to an inductor is changed, the current also changes, but the change in current lags thechange in voltage in an RL circuit. In Reactance, Inductive and Capacitive, we explore how an RL circuit behaves when asinusoidal AC voltage is applied.

23.11 Reactance, Inductive and Capacitive


• Sketch voltage and current versus time in simple inductive, capacitive, and resistive circuits.• Calculate inductive and capacitive reactance.• Calculate current and/or voltage in simple inductive, capacitive, and resistive circuits.



Many circuits also contain capacitors and inductors, in addition to resistors and an AC voltage source. We have seen howcapacitors and inductors respond to DC voltage when it is switched on and off. We will now explore how inductors and capacitorsreact to sinusoidal AC voltage.

Inductors and Inductive ReactanceSuppose an inductor is connected directly to an AC voltage source, as shown in Figure 23.45. It is reasonable to assumenegligible resistance, since in practice we can make the resistance of an inductor so small that it has a negligible effect on thecircuit. Also shown is a graph of voltage and current as functions of time.

Figure 23.45 (a) An AC voltage source in series with an inductor having negligible resistance. (b) Graph of current and voltage across the inductor asfunctions of time.

The graph in Figure 23.45(b) starts with voltage at a maximum. Note that the current starts at zero and rises to its peak after thevoltage that drives it, just as was the case when DC voltage was switched on in the preceding section. When the voltagebecomes negative at point a, the current begins to decrease; it becomes zero at point b, where voltage is its most negative. Thecurrent then becomes negative, again following the voltage. The voltage becomes positive at point c and begins to make thecurrent less negative. At point d, the current goes through zero just as the voltage reaches its positive peak to start anothercycle. This behavior is summarized as follows:

AC Voltage in an Inductor

When a sinusoidal voltage is applied to an inductor, the voltage leads the current by one-fourth of a cycle, or by a 90ºphase angle.

Current lags behind voltage, since inductors oppose change in current. Changing current induces a back emf V = −L(ΔI / Δt). This is considered to be an effective resistance of the inductor to AC. The rms current I through an inductor L is given by aversion of Ohm’s law:

(23.51)I = VXL

,

where V is the rms voltage across the inductor and XL is defined to be

(23.52)XL = 2π fL,

with f the frequency of the AC voltage source in hertz (An analysis of the circuit using Kirchhoff’s loop rule and calculus actually

produces this expression). XL is called the inductive reactance, because the inductor reacts to impede the current. XL has

units of ohms ( 1 H = 1 Ω ⋅ s , so that frequency times inductance has units of (cycles/s)( Ω ⋅ s) = Ω ), consistent with its

role as an effective resistance. It makes sense that XL is proportional to L , since the greater the induction the greater its

resistance to change. It is also reasonable that XL is proportional to frequency f , since greater frequency means greater

change in current. That is, ΔI/Δt is large for large frequencies (large f , small Δt ). The greater the change, the greater the

opposition of an inductor.

Example 23.10 Calculating Inductive Reactance and then Current

(a) Calculate the inductive reactance of a 3.00 mH inductor when 60.0 Hz and 10.0 kHz AC voltages are applied. (b) What isthe rms current at each frequency if the applied rms voltage is 120 V?

Strategy


The inductive reactance is found directly from the expression XL = 2π fL . Once XL has been found at each frequency,

Ohm’s law as stated in the Equation I = V / XL can be used to find the current at each frequency.

Solution for (a)

Entering the frequency and inductance into Equation XL = 2π fL gives

(23.53)XL = 2π fL = 6.28(60.0 / s)(3.00 mH) = 1.13 Ω at 60 Hz.

Similarly, at 10 kHz,

(23.54)XL = 2π fL = 6.28(1.00×104 /s)(3.00 mH) = 188 Ω at 10 kHz.

Solution for (b)

The rms current is now found using the version of Ohm’s law in Equation I = V / XL , given the applied rms voltage is 120

V. For the first frequency, this yields

(23.55)I = VXL

= 120 V1.13 Ω = 106 A at 60 Hz.


(23.56)I = VXL

= 120 V188 Ω = 0.637 A at 10 kHz.

Discussion

The inductor reacts very differently at the two different frequencies. At the higher frequency, its reactance is large and thecurrent is small, consistent with how an inductor impedes rapid change. Thus high frequencies are impeded the most.Inductors can be used to filter out high frequencies; for example, a large inductor can be put in series with a soundreproduction system or in series with your home computer to reduce high-frequency sound output from your speakers orhigh-frequency power spikes into your computer.

Note that although the resistance in the circuit considered is negligible, the AC current is not extremely large because inductivereactance impedes its flow. With AC, there is no time for the current to become extremely large.

Capacitors and Capacitive ReactanceConsider the capacitor connected directly to an AC voltage source as shown in Figure 23.46. The resistance of a circuit like thiscan be made so small that it has a negligible effect compared with the capacitor, and so we can assume negligible resistance.Voltage across the capacitor and current are graphed as functions of time in the figure.

Figure 23.46 (a) An AC voltage source in series with a capacitor C having negligible resistance. (b) Graph of current and voltage across the capacitoras functions of time.

The graph in Figure 23.46 starts with voltage across the capacitor at a maximum. The current is zero at this point, because thecapacitor is fully charged and halts the flow. Then voltage drops and the current becomes negative as the capacitor discharges.At point a, the capacitor has fully discharged ( Q = 0 on it) and the voltage across it is zero. The current remains negative

between points a and b, causing the voltage on the capacitor to reverse. This is complete at point b, where the current is zeroand the voltage has its most negative value. The current becomes positive after point b, neutralizing the charge on the capacitorand bringing the voltage to zero at point c, which allows the current to reach its maximum. Between points c and d, the currentdrops to zero as the voltage rises to its peak, and the process starts to repeat. Throughout the cycle, the voltage follows what thecurrent is doing by one-fourth of a cycle:



AC Voltage in a Capacitor

When a sinusoidal voltage is applied to a capacitor, the voltage follows the current by one-fourth of a cycle, or by a 90ºphase angle.

The capacitor is affecting the current, having the ability to stop it altogether when fully charged. Since an AC voltage is applied,there is an rms current, but it is limited by the capacitor. This is considered to be an effective resistance of the capacitor to AC,and so the rms current I in the circuit containing only a capacitor C is given by another version of Ohm’s law to be

(23.57)I = VXC

,

where V is the rms voltage and XC is defined (As with XL , this expression for XC results from an analysis of the circuit

using Kirchhoff’s rules and calculus) to be

(23.58)XC = 12π fC ,

where XC is called the capacitive reactance, because the capacitor reacts to impede the current. XC has units of ohms

(verification left as an exercise for the reader). XC is inversely proportional to the capacitance C ; the larger the capacitor, the

greater the charge it can store and the greater the current that can flow. It is also inversely proportional to the frequency f ; the

greater the frequency, the less time there is to fully charge the capacitor, and so it impedes current less.

Example 23.11 Calculating Capacitive Reactance and then Current

(a) Calculate the capacitive reactance of a 5.00 mF capacitor when 60.0 Hz and 10.0 kHz AC voltages are applied. (b) Whatis the rms current if the applied rms voltage is 120 V?

Strategy

The capacitive reactance is found directly from the expression in XC = 12π fC . Once XC has been found at each

frequency, Ohm’s law stated as I = V / XC can be used to find the current at each frequency.

Solution for (a)

Entering the frequency and capacitance into XC = 12π fC gives

(23.59)XC = 12π fC

= 16.28(60.0 / s)(5.00 µF) = 531 Ω at 60 Hz.


(23.60)XC = 12π fC = 1

6.28(1.00×104 / s)(5.00 µF)= 3.18 Ω at 10 kHz

.

Solution for (b)

The rms current is now found using the version of Ohm’s law in I = V / XC , given the applied rms voltage is 120 V. For the

first frequency, this yields

(23.61)I = VXC

= 120 V531 Ω = 0.226 A at 60 Hz.


(23.62)I = VXC

= 120 V3.18 Ω = 37.7 A at 10 kHz.

Discussion

The capacitor reacts very differently at the two different frequencies, and in exactly the opposite way an inductor reacts. Atthe higher frequency, its reactance is small and the current is large. Capacitors favor change, whereas inductors opposechange. Capacitors impede low frequencies the most, since low frequency allows them time to become charged and stop


the current. Capacitors can be used to filter out low frequencies. For example, a capacitor in series with a soundreproduction system rids it of the 60 Hz hum.

Although a capacitor is basically an open circuit, there is an rms current in a circuit with an AC voltage applied to a capacitor.This is because the voltage is continually reversing, charging and discharging the capacitor. If the frequency goes to zero (DC),XC tends to infinity, and the current is zero once the capacitor is charged. At very high frequencies, the capacitor’s reactance

tends to zero—it has a negligible reactance and does not impede the current (it acts like a simple wire). Capacitors have theopposite effect on AC circuits that inductors have.

Resistors in an AC CircuitJust as a reminder, consider Figure 23.47, which shows an AC voltage applied to a resistor and a graph of voltage and currentversus time. The voltage and current are exactly in phase in a resistor. There is no frequency dependence to the behavior ofplain resistance in a circuit:

Figure 23.47 (a) An AC voltage source in series with a resistor. (b) Graph of current and voltage across the resistor as functions of time, showing themto be exactly in phase.

AC Voltage in a Resistor

When a sinusoidal voltage is applied to a resistor, the voltage is exactly in phase with the current—they have a 0º phaseangle.

23.12 RLC Series AC Circuits


• Calculate the impedance, phase angle, resonant frequency, power, power factor, voltage, and/or current in an RLCseries circuit.

• Draw the circuit diagram for an RLC series circuit.• Explain the significance of the resonant frequency.

ImpedanceWhen alone in an AC circuit, inductors, capacitors, and resistors all impede current. How do they behave when all three occurtogether? Interestingly, their individual resistances in ohms do not simply add. Because inductors and capacitors behave inopposite ways, they partially to totally cancel each other’s effect. Figure 23.48 shows an RLC series circuit with an AC voltagesource, the behavior of which is the subject of this section. The crux of the analysis of an RLC circuit is the frequencydependence of XL and XC , and the effect they have on the phase of voltage versus current (established in the preceding

section). These give rise to the frequency dependence of the circuit, with important “resonance” features that are the basis ofmany applications, such as radio tuners.



Figure 23.48 An RLC series circuit with an AC voltage source.

The combined effect of resistance R , inductive reactance XL , and capacitive reactance XC is defined to be impedance, an

AC analogue to resistance in a DC circuit. Current, voltage, and impedance in an RLC circuit are related by an AC version ofOhm’s law:

(23.63)I0 = V0

Z or Irms = VrmsZ .

Here I0 is the peak current, V0 the peak source voltage, and Z is the impedance of the circuit. The units of impedance are

ohms, and its effect on the circuit is as you might expect: the greater the impedance, the smaller the current. To get anexpression for Z in terms of R , XL , and XC , we will now examine how the voltages across the various components are

related to the source voltage. Those voltages are labeled VR , VL , and VC in Figure 23.48.

Conservation of charge requires current to be the same in each part of the circuit at all times, so that we can say the currents inR , L , and C are equal and in phase. But we know from the preceding section that the voltage across the inductor VL leads

the current by one-fourth of a cycle, the voltage across the capacitor VC follows the current by one-fourth of a cycle, and the

voltage across the resistor VR is exactly in phase with the current. Figure 23.49 shows these relationships in one graph, as well

as showing the total voltage around the circuit V = VR + VL + VC , where all four voltages are the instantaneous values.

According to Kirchhoff’s loop rule, the total voltage around the circuit V is also the voltage of the source.

You can see from Figure 23.49 that while VR is in phase with the current, VL leads by 90º , and VC follows by 90º . Thus

VL and VC are 180º out of phase (crest to trough) and tend to cancel, although not completely unless they have the same

magnitude. Since the peak voltages are not aligned (not in phase), the peak voltage V0 of the source does not equal the sum of

the peak voltages across R , L , and C . The actual relationship is

(23.64)V0 = V0R2 +(V0L − V0C)2,

where V0R , V0L , and V0C are the peak voltages across R , L , and C , respectively. Now, using Ohm’s law and definitions

from Reactance, Inductive and Capacitive, we substitute V0 = I0Z into the above, as well as V0R = I0R , V0L = I0XL ,

and V0C = I0XC , yielding

(23.65)I0 Z = I02 R2 + (I0 XL − I0XC )2 = I0 R2 + (XL − XC)2.

I0 cancels to yield an expression for Z :

(23.66)Z = R2 + (XL − XC)2,

which is the impedance of an RLC series AC circuit. For circuits without a resistor, take R = 0 ; for those without an inductor,

take XL = 0 ; and for those without a capacitor, take XC = 0 .


Figure 23.49 This graph shows the relationships of the voltages in an RLC circuit to the current. The voltages across the circuit elements add to equalthe voltage of the source, which is seen to be out of phase with the current.

Example 23.12 Calculating Impedance and Current

An RLC series circuit has a 40.0 Ω resistor, a 3.00 mH inductor, and a 5.00 μF capacitor. (a) Find the circuit’s impedance

at 60.0 Hz and 10.0 kHz, noting that these frequencies and the values for L and C are the same as in Example 23.10 and

Example 23.11. (b) If the voltage source has Vrms = 120 V , what is Irms at each frequency?

Strategy

For each frequency, we use Z = R2 + (XL − XC)2 to find the impedance and then Ohm’s law to find current. We can

take advantage of the results of the previous two examples rather than calculate the reactances again.

Solution for (a)

At 60.0 Hz, the values of the reactances were found in Example 23.10 to be XL = 1.13 Ω and in Example 23.11 to be

XC = 531 Ω . Entering these and the given 40.0 Ω for resistance into Z = R2 + (XL − XC)2 yields

(23.67)Z = R2 + (XL − XC)2

= (40.0 Ω )2 + (1.13 Ω − 531 Ω )2

= 531 Ω at 60.0 Hz.

Similarly, at 10.0 kHz, XL = 188 Ω and XC = 3.18 Ω , so that

(23.68)Z = (40.0 Ω )2 + (188 Ω − 3.18 Ω )2

= 190 Ω at 10.0 kHz.Discussion for (a)

In both cases, the result is nearly the same as the largest value, and the impedance is definitely not the sum of the individualvalues. It is clear that XL dominates at high frequency and XC dominates at low frequency.

Solution for (b)

The current Irms can be found using the AC version of Ohm’s law in Equation Irms = Vrms / Z :

Irms = VrmsZ = 120 V

531 Ω = 0.226 A at 60.0 Hz

Finally, at 10.0 kHz, we find

Irms = VrmsZ = 120 V

190 Ω = 0.633 A at 10.0 kHz

Discussion for (a)



The current at 60.0 Hz is the same (to three digits) as found for the capacitor alone in Example 23.11. The capacitordominates at low frequency. The current at 10.0 kHz is only slightly different from that found for the inductor alone inExample 23.10. The inductor dominates at high frequency.

Resonance in RLC Series AC CircuitsHow does an RLC circuit behave as a function of the frequency of the driving voltage source? Combining Ohm’s law,

Irms = Vrms / Z , and the expression for impedance Z from Z = R2 + (XL − XC)2 gives

(23.69)Irms = Vrms

R2 + (XL − XC)2.

The reactances vary with frequency, with XL large at high frequencies and XC large at low frequencies, as we have seen in

three previous examples. At some intermediate frequency f0 , the reactances will be equal and cancel, giving Z = R —this is a

minimum value for impedance, and a maximum value for Irms results. We can get an expression for f0 by taking

(23.70)XL = XC.

Substituting the definitions of XL and XC ,

(23.71)2πf0 L = 12πf0 C .

Solving this expression for f0 yields

(23.72)f0 = 12π LC

,

where f0 is the resonant frequency of an RLC series circuit. This is also the natural frequency at which the circuit would

oscillate if not driven by the voltage source. At f0 , the effects of the inductor and capacitor cancel, so that Z = R , and Irms is

a maximum.

Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined to be a forced oscillation—in thiscase, forced by the voltage source—at the natural frequency of the system. The receiver in a radio is an RLC circuit thatoscillates best at its f0 . A variable capacitor is often used to adjust f0 to receive a desired frequency and to reject others.

Figure 23.50 is a graph of current as a function of frequency, illustrating a resonant peak in Irms at f0 . The two curves are for

two different circuits, which differ only in the amount of resistance in them. The peak is lower and broader for the higher-resistance circuit. Thus the higher-resistance circuit does not resonate as strongly and would not be as selective in a radioreceiver, for example.

Figure 23.50 A graph of current versus frequency for two RLC series circuits differing only in the amount of resistance. Both have a resonance at f0 ,

but that for the higher resistance is lower and broader. The driving AC voltage source has a fixed amplitude V0 .

Example 23.13 Calculating Resonant Frequency and Current

For the same RLC series circuit having a 40.0 Ω resistor, a 3.00 mH inductor, and a 5.00 μF capacitor: (a) Find the

resonant frequency. (b) Calculate Irms at resonance if Vrms is 120 V.

Strategy


The resonant frequency is found by using the expression in f0 = 12π LC

. The current at that frequency is the same as if

the resistor alone were in the circuit.

Solution for (a)

Entering the given values for L and C into the expression given for f0 in f0 = 12π LC

yields

(23.73)f0 = 12π LC

= 12π (3.00×10−3 H)(5.00×10−6 F)

= 1.30 kHz.

Discussion for (a)

We see that the resonant frequency is between 60.0 Hz and 10.0 kHz, the two frequencies chosen in earlier examples. Thiswas to be expected, since the capacitor dominated at the low frequency and the inductor dominated at the high frequency.Their effects are the same at this intermediate frequency.

Solution for (b)

The current is given by Ohm’s law. At resonance, the two reactances are equal and cancel, so that the impedance equalsthe resistance alone. Thus,

(23.74)Irms = VrmsZ = 120 V

40.0 Ω = 3.00 A.

Discussion for (b)

At resonance, the current is greater than at the higher and lower frequencies considered for the same circuit in thepreceding example.

Power in RLC Series AC CircuitsIf current varies with frequency in an RLC circuit, then the power delivered to it also varies with frequency. But the average poweris not simply current times voltage, as it is in purely resistive circuits. As was seen in Figure 23.49, voltage and current are out ofphase in an RLC circuit. There is a phase angle ϕ between the source voltage V and the current I , which can be found from

(23.75)cos ϕ = RZ .

For example, at the resonant frequency or in a purely resistive circuit Z = R , so that cos ϕ = 1 . This implies that ϕ = 0 ºand that voltage and current are in phase, as expected for resistors. At other frequencies, average power is less than atresonance. This is both because voltage and current are out of phase and because Irms is lower. The fact that source voltage

and current are out of phase affects the power delivered to the circuit. It can be shown that the average power is

(23.76)Pave = IrmsVrms cos ϕ,

Thus cos ϕ is called the power factor, which can range from 0 to 1. Power factors near 1 are desirable when designing an

efficient motor, for example. At the resonant frequency, cos ϕ = 1 .

Example 23.14 Calculating the Power Factor and Power

For the same RLC series circuit having a 40.0 Ω resistor, a 3.00 mH inductor, a 5.00 μF capacitor, and a voltage source

with a Vrms of 120 V: (a) Calculate the power factor and phase angle for f = 60.0Hz . (b) What is the average power at

50.0 Hz? (c) Find the average power at the circuit’s resonant frequency.


The power factor at 60.0 Hz is found from

(23.77)cos ϕ = RZ .

We know Z= 531 Ω from Example 23.12, so that

(23.78)cos ϕ = 40.0 Ω531 Ω = 0.0753 at 60.0 Hz.



This small value indicates the voltage and current are significantly out of phase. In fact, the phase angle is

(23.79)ϕ = cos−1 0.0753 = 85.7º at 60.0 Hz.

Discussion for (a)

The phase angle is close to 90º , consistent with the fact that the capacitor dominates the circuit at this low frequency (a

pure RC circuit has its voltage and current 90º out of phase).


The average power at 60.0 Hz is

(23.80)Pave = IrmsVrms cos ϕ.

Irms was found to be 0.226 A in Example 23.12. Entering the known values gives

(23.81)Pave = (0.226 A)(120 V)(0.0753) = 2.04 W at 60.0 Hz.

Strategy and Solution for (c)

At the resonant frequency, we know cos ϕ = 1 , and Irms was found to be 6.00 A in Example 23.13. Thus,

Pave = (3.00 A)(120 V)(1) = 360 W at resonance (1.30 kHz)

Discussion

Both the current and the power factor are greater at resonance, producing significantly greater power than at higher andlower frequencies.

Power delivered to an RLC series AC circuit is dissipated by the resistance alone. The inductor and capacitor have energy inputand output but do not dissipate it out of the circuit. Rather they transfer energy back and forth to one another, with the resistordissipating exactly what the voltage source puts into the circuit. This assumes no significant electromagnetic radiation from theinductor and capacitor, such as radio waves. Such radiation can happen and may even be desired, as we will see in the nextchapter on electromagnetic radiation, but it can also be suppressed as is the case in this chapter. The circuit is analogous to thewheel of a car driven over a corrugated road as shown in Figure 23.51. The regularly spaced bumps in the road are analogousto the voltage source, driving the wheel up and down. The shock absorber is analogous to the resistance damping and limitingthe amplitude of the oscillation. Energy within the system goes back and forth between kinetic (analogous to maximum current,and energy stored in an inductor) and potential energy stored in the car spring (analogous to no current, and energy stored in theelectric field of a capacitor). The amplitude of the wheels’ motion is a maximum if the bumps in the road are hit at the resonantfrequency.

Figure 23.51 The forced but damped motion of the wheel on the car spring is analogous to an RLC series AC circuit. The shock absorber damps themotion and dissipates energy, analogous to the resistance in an RLC circuit. The mass and spring determine the resonant frequency.

A pure LC circuit with negligible resistance oscillates at f0 , the same resonant frequency as an RLC circuit. It can serve as a

frequency standard or clock circuit—for example, in a digital wristwatch. With a very small resistance, only a very small energyinput is necessary to maintain the oscillations. The circuit is analogous to a car with no shock absorbers. Once it startsoscillating, it continues at its natural frequency for some time. Figure 23.52 shows the analogy between an LC circuit and a masson a spring.


back emf:

capacitive reactance:

Figure 23.52 An LC circuit is analogous to a mass oscillating on a spring with no friction and no driving force. Energy moves back and forth betweenthe inductor and capacitor, just as it moves from kinetic to potential in the mass-spring system.

PhET Explorations: Circuit Construction Kit (AC+DC), Virtual Lab

Build circuits with capacitors, inductors, resistors and AC or DC voltage sources, and inspect them using lab instrumentssuch as voltmeters and ammeters.

Figure 23.53 Circuit Construction Kit (AC+DC), Virtual Lab (http://cnx.org/content/m55420/1.2/circuit-construction-kit-ac-virtual-lab_en.jar)

Glossarythe emf generated by a running motor, because it consists of a coil turning in a magnetic field; it opposes the

voltage powering the motor

the opposition of a capacitor to a change in current; calculated by XC = 12π fC



http://cnx.org/content/m55420/1.2/circuit-construction-kit-ac-virtual-lab_en.jar

http://cnx.org/content/m55420/1.2/circuit-construction-kit-ac-virtual-lab_en.jar

characteristic time constant:

eddy current:

electric generator:

electromagnetic induction:

emf induced in a generator coil:

energy stored in an inductor:

Faraday’s law of induction:

henry:

impedance:

inductance:

induction:

inductive reactance:

inductor:

Lenz’s law:

magnetic damping:

magnetic flux:

mutual inductance:

peak emf:

phase angle:

power factor:

resonant frequency:

self-inductance:

shock hazard:

step-down transformer:

step-up transformer:

thermal hazard:

three-wire system:

transformer:

denoted by τ , of a particular series RL circuit is calculated by τ = LR , where L is the

inductance and R is the resistance

a current loop in a conductor caused by motional emf

a device for converting mechanical work into electric energy; it induces an emf by rotating a coil in amagnetic field

the process of inducing an emf (voltage) with a change in magnetic flux

emf = NABω sin ωt , where A is the area of an N -turn coil rotated at a constant

angular velocity ω in a uniform magnetic field B , over a period of time t

self-explanatory; calculated by Eind = 12LI 2

the means of calculating the emf in a coil due to changing magnetic flux, given by

emf = −N ΔΦΔt

the unit of inductance; 1 H = 1 Ω ⋅ s

the AC analogue to resistance in a DC circuit; it is the combined effect of resistance, inductive reactance, and

capacitive reactance in the form Z = R2 + (XL − XC)2

a property of a device describing how efficient it is at inducing emf in another device

(magnetic induction) the creation of emfs and hence currents by magnetic fields

the opposition of an inductor to a change in current; calculated by XL = 2π fL

a device that exhibits significant self-inductance

the minus sign in Faraday’s law, signifying that the emf induced in a coil opposes the change in magnetic flux

the drag produced by eddy currents

the amount of magnetic field going through a particular area, calculated with Φ = BA cos θ where B is the

magnetic field strength over an area A at an angle θ with the perpendicular to the area

how effective a pair of devices are at inducing emfs in each other

emf0 = NABω

denoted by ϕ , the amount by which the voltage and current are out of phase with each other in a circuit

the amount by which the power delivered in the circuit is less than the theoretical maximum of the circuit due tovoltage and current being out of phase; calculated by cos ϕ

the frequency at which the impedance in a circuit is at a minimum, and also the frequency at which the

circuit would oscillate if not driven by a voltage source; calculated by f0 = 12π LC

how effective a device is at inducing emf in itself

the term for electrical hazards due to current passing through a human

a transformer that decreases voltage

a transformer that increases voltage

the term for electrical hazards due to overheating

the wiring system used at present for safety reasons, with live, neutral, and ground wires

a device that transforms voltages from one value to another using induction


transformer equation: the equation showing that the ratio of the secondary to primary voltages in a transformer equals the

ratio of the number of loops in their coils;VsVp

= NsNp

Section Summary

23.1 Induced Emf and Magnetic Flux• The crucial quantity in induction is magnetic flux Φ , defined to be Φ = BA cos θ , where B is the magnetic field strength

over an area A at an angle θ with the perpendicular to the area.

• Units of magnetic flux Φ are T ⋅ m2 .

• Any change in magnetic flux Φ induces an emf—the process is defined to be electromagnetic induction.

23.2 Faraday’s Law of Induction: Lenz’s Law• Faraday’s law of induction states that the emfinduced by a change in magnetic flux is

emf = −N ΔΦΔt

when flux changes by ΔΦ in a time Δt .

• If emf is induced in a coil, N is its number of turns.

• The minus sign means that the emf creates a current I and magnetic field B that oppose the change in flux ΔΦ —thisopposition is known as Lenz’s law.

23.3 Motional Emf• An emf induced by motion relative to a magnetic field B is called a motional emf and is given by

emf = Bℓv (B, ℓ, and v perpendicular),where ℓ is the length of the object moving at speed v relative to the field.

23.4 Eddy Currents and Magnetic Damping• Current loops induced in moving conductors are called eddy currents.• They can create significant drag, called magnetic damping.

23.5 Electric Generators• An electric generator rotates a coil in a magnetic field, inducing an emfgiven as a function of time by

emf = NABω sin ωt,where A is the area of an N -turn coil rotated at a constant angular velocity ω in a uniform magnetic field B .

• The peak emf emf0 of a generator is

emf0 = NABω.

23.6 Back Emf• Any rotating coil will have an induced emf—in motors, this is called back emf, since it opposes the emf input to the motor.

23.7 Transformers• Transformers use induction to transform voltages from one value to another.• For a transformer, the voltages across the primary and secondary coils are related by

VsVp

= NsNp

,

where Vp and Vs are the voltages across primary and secondary coils having Np and Ns turns.

• The currents Ip and Is in the primary and secondary coils are related byIsIp

=NpNs

.

• A step-up transformer increases voltage and decreases current, whereas a step-down transformer decreases voltage andincreases current.

23.8 Electrical Safety: Systems and Devices• Electrical safety systems and devices are employed to prevent thermal and shock hazards.



• Circuit breakers and fuses interrupt excessive currents to prevent thermal hazards.• The three-wire system guards against thermal and shock hazards, utilizing live/hot, neutral, and earth/ground wires, and

grounding the neutral wire and case of the appliance.• A ground fault interrupter (GFI) prevents shock by detecting the loss of current to unintentional paths.• An isolation transformer insulates the device being powered from the original source, also to prevent shock.• Many of these devices use induction to perform their basic function.

23.9 Inductance• Inductance is the property of a device that tells how effectively it induces an emf in another device.• Mutual inductance is the effect of two devices in inducing emfs in each other.• A change in current ΔI1 / Δt in one induces an emf emf2 in the second:

emf2 = −MΔI1Δt ,

where M is defined to be the mutual inductance between the two devices, and the minus sign is due to Lenz’s law.

• Symmetrically, a change in current ΔI2 / Δt through the second device induces an emf emf1 in the first:

emf1 = −MΔI2Δt ,

where M is the same mutual inductance as in the reverse process.• Current changes in a device induce an emf in the device itself.• Self-inductance is the effect of the device inducing emf in itself.• The device is called an inductor, and the emf induced in it by a change in current through it is

emf = −LΔIΔt ,

where L is the self-inductance of the inductor, and ΔI / Δt is the rate of change of current through it. The minus signindicates that emf opposes the change in current, as required by Lenz’s law.

• The unit of self- and mutual inductance is the henry (H), where 1 H = 1 Ω ⋅ s .

• The self-inductance L of an inductor is proportional to how much flux changes with current. For an N -turn inductor,

L = N ΔΦΔI .

• The self-inductance of a solenoid is

L = µ0 N 2 Aℓ (solenoid),

where N is its number of turns in the solenoid, A is its cross-sectional area, ℓ is its length, and

μ0 = 4π×10−7 T ⋅ m/A is the permeability of free space.

• The energy stored in an inductor Eind is

Eind = 12LI 2.

23.10 RL Circuits• When a series connection of a resistor and an inductor—an RL circuit—is connected to a voltage source, the time variation

of the current is

I = I0(1 − e−t / τ) (turning on).where I0 = V / R is the final current.

• The characteristic time constant τ is τ = LR , where L is the inductance and R is the resistance.

• In the first time constant τ , the current rises from zero to 0.632I0 , and 0.632 of the remainder in every subsequent time

interval τ .• When the inductor is shorted through a resistor, current decreases as

I = I0e−t / τ (turning off).Here I0 is the initial current.

• Current falls to 0.368I0 in the first time interval τ , and 0.368 of the remainder toward zero in each subsequent time τ .


23.11 Reactance, Inductive and Capacitive• For inductors in AC circuits, we find that when a sinusoidal voltage is applied to an inductor, the voltage leads the current

by one-fourth of a cycle, or by a 90º phase angle.• The opposition of an inductor to a change in current is expressed as a type of AC resistance.• Ohm’s law for an inductor is

I = VXL

,

where V is the rms voltage across the inductor.

• XL is defined to be the inductive reactance, given by

XL = 2π fL,with f the frequency of the AC voltage source in hertz.

• Inductive reactance XL has units of ohms and is greatest at high frequencies.

• For capacitors, we find that when a sinusoidal voltage is applied to a capacitor, the voltage follows the current by one-fourthof a cycle, or by a 90º phase angle.

• Since a capacitor can stop current when fully charged, it limits current and offers another form of AC resistance; Ohm’s lawfor a capacitor is

I = VXC

,

where V is the rms voltage across the capacitor.

• XC is defined to be the capacitive reactance, given by

XC = 12π fC .

• XC has units of ohms and is greatest at low frequencies.

23.12 RLC Series AC Circuits• The AC analogy to resistance is impedance Z , the combined effect of resistors, inductors, and capacitors, defined by the

AC version of Ohm’s law:

I0 = V0Z or Irms = Vrms

Z ,

where I0 is the peak current and V0 is the peak source voltage.

• Impedance has units of ohms and is given by Z = R2 + (XL − XC)2 .

• The resonant frequency f0 , at which XL = XC , is

f0 = 12π LC

.

• In an AC circuit, there is a phase angle ϕ between source voltage V and the current I , which can be found from

cos ϕ = RZ ,

• ϕ = 0º for a purely resistive circuit or an RLC circuit at resonance.

• The average power delivered to an RLC circuit is affected by the phase angle and is given by

Pave = IrmsVrms cos ϕ,cos ϕ is called the power factor, which ranges from 0 to 1.

Conceptual Questions

23.1 Induced Emf and Magnetic Flux1. How do the multiple-loop coils and iron ring in the version of Faraday’s apparatus shown in Figure 23.3 enhance theobservation of induced emf?

2. When a magnet is thrust into a coil as in Figure 23.4(a), what is the direction of the force exerted by the coil on the magnet?Draw a diagram showing the direction of the current induced in the coil and the magnetic field it produces, to justify yourresponse. How does the magnitude of the force depend on the resistance of the galvanometer?



3. Explain how magnetic flux can be zero when the magnetic field is not zero.

4. Is an emf induced in the coil in Figure 23.54 when it is stretched? If so, state why and give the direction of the inducedcurrent.

Figure 23.54 A circular coil of wire is stretched in a magnetic field.

23.2 Faraday’s Law of Induction: Lenz’s Law5. A person who works with large magnets sometimes places her head inside a strong field. She reports feeling dizzy as shequickly turns her head. How might this be associated with induction?

6. A particle accelerator sends high-velocity charged particles down an evacuated pipe. Explain how a coil of wire wrappedaround the pipe could detect the passage of individual particles. Sketch a graph of the voltage output of the coil as a singleparticle passes through it.

23.3 Motional Emf7. Why must part of the circuit be moving relative to other parts, to have usable motional emf? Consider, for example, that therails in Figure 23.11 are stationary relative to the magnetic field, while the rod moves.

8. A powerful induction cannon can be made by placing a metal cylinder inside a solenoid coil. The cylinder is forcefully expelledwhen solenoid current is turned on rapidly. Use Faraday’s and Lenz’s laws to explain how this works. Why might the cylinder getlive/hot when the cannon is fired?

9. An induction stove heats a pot with a coil carrying an alternating current located beneath the pot (and without a hot surface).Can the stove surface be a conductor? Why won’t a coil carrying a direct current work?

10. Explain how you could thaw out a frozen water pipe by wrapping a coil carrying an alternating current around it. Does itmatter whether or not the pipe is a conductor? Explain.

23.4 Eddy Currents and Magnetic Damping11. Explain why magnetic damping might not be effective on an object made of several thin conducting layers separated byinsulation.

12. Explain how electromagnetic induction can be used to detect metals? This technique is particularly important in detectingburied landmines for disposal, geophysical prospecting and at airports.

23.5 Electric Generators13. Using RHR-1, show that the emfs in the sides of the generator loop in Figure 23.23 are in the same sense and thus add.

14. The source of a generator’s electrical energy output is the work done to turn its coils. How is the work needed to turn thegenerator related to Lenz’s law?

23.6 Back Emf15. Suppose you find that the belt drive connecting a powerful motor to an air conditioning unit is broken and the motor is runningfreely. Should you be worried that the motor is consuming a great deal of energy for no useful purpose? Explain why or why not.

23.7 Transformers16. Explain what causes physical vibrations in transformers at twice the frequency of the AC power involved.

23.8 Electrical Safety: Systems and Devices17. Does plastic insulation on live/hot wires prevent shock hazards, thermal hazards, or both?

18. Why are ordinary circuit breakers and fuses ineffective in preventing shocks?

19. A GFI may trip just because the live/hot and neutral wires connected to it are significantly different in length. Explain why.

23.9 Inductance20. How would you place two identical flat coils in contact so that they had the greatest mutual inductance? The least?

21. How would you shape a given length of wire to give it the greatest self-inductance? The least?


22. Verify, as was concluded without proof in Example 23.7, that units of T ⋅ m2 / A = Ω ⋅ s = H .

23.11 Reactance, Inductive and Capacitive23. Presbycusis is a hearing loss due to age that progressively affects higher frequencies. A hearing aid amplifier is designed toamplify all frequencies equally. To adjust its output for presbycusis, would you put a capacitor in series or parallel with the hearingaid’s speaker? Explain.

24. Would you use a large inductance or a large capacitance in series with a system to filter out low frequencies, such as the 100Hz hum in a sound system? Explain.

25. High-frequency noise in AC power can damage computers. Does the plug-in unit designed to prevent this damage use alarge inductance or a large capacitance (in series with the computer) to filter out such high frequencies? Explain.

26. Does inductance depend on current, frequency, or both? What about inductive reactance?

27. Explain why the capacitor in Figure 23.55(a) acts as a low-frequency filter between the two circuits, whereas that in Figure23.55(b) acts as a high-frequency filter.

Figure 23.55 Capacitors and inductors. Capacitor with high frequency and low frequency.

28. If the capacitors in Figure 23.55 are replaced by inductors, which acts as a low-frequency filter and which as a high-frequency filter?

23.12 RLC Series AC Circuits29. Does the resonant frequency of an AC circuit depend on the peak voltage of the AC source? Explain why or why not.

30. Suppose you have a motor with a power factor significantly less than 1. Explain why it would be better to improve the powerfactor as a method of improving the motor’s output, rather than to increase the voltage input.



Problems & Exercises

23.1 Induced Emf and Magnetic Flux1. What is the value of the magnetic flux at coil 2 in Figure23.56 due to coil 1?

Figure 23.56 (a) The planes of the two coils are perpendicular. (b) Thewire is perpendicular to the plane of the coil.

2. What is the value of the magnetic flux through the coil inFigure 23.56(b) due to the wire?

23.2 Faraday’s Law of Induction: Lenz’s Law3. Referring to Figure 23.57(a), what is the direction of thecurrent induced in coil 2: (a) If the current in coil 1 increases?(b) If the current in coil 1 decreases? (c) If the current in coil 1is constant? Explicitly show how you follow the steps in theProblem-Solving Strategy for Lenz's Law.

Figure 23.57 (a) The coils lie in the same plane. (b) The wire is in theplane of the coil

4. Referring to Figure 23.57(b), what is the direction of thecurrent induced in the coil: (a) If the current in the wireincreases? (b) If the current in the wire decreases? (c) If thecurrent in the wire suddenly changes direction? Explicitlyshow how you follow the steps in the Problem-SolvingStrategy for Lenz’s Law.

5. Referring to Figure 23.58, what are the directions of thecurrents in coils 1, 2, and 3 (assume that the coils are lying inthe plane of the circuit): (a) When the switch is first closed?(b) When the switch has been closed for a long time? (c) Justafter the switch is opened?

Figure 23.58

6. Repeat the previous problem with the battery reversed.

7. Verify that the units of ΔΦ / Δt are volts. That is, show

that 1 T ⋅ m2 / s = 1 V .

8. Suppose a 50-turn coil lies in the plane of the page in auniform magnetic field that is directed into the page. The coil

originally has an area of 0.250 m2 . It is stretched to have noarea in 0.100 s. What is the direction and magnitude of theinduced emf if the uniform magnetic field has a strength of1.50 T?

9. (a) An MRI technician moves his hand from a region ofvery low magnetic field strength into an MRI scanner’s 2.00 Tfield with his fingers pointing in the direction of the field. Findthe average emf induced in his wedding ring, given itsdiameter is 2.20 cm and assuming it takes 0.250 s to move itinto the field. (b) Discuss whether this current wouldsignificantly change the temperature of the ring.


Referring to the situation in the previous problem: (a) Whatcurrent is induced in the ring if its resistance is 0.0100 Ω ?(b) What average power is dissipated? (c) What magneticfield is induced at the center of the ring? (d) What is thedirection of the induced magnetic field relative to the MRI’sfield?

11. An emf is induced by rotating a 1000-turn, 20.0 cm

diameter coil in the Earth’s 5.00×10−5 T magnetic field.What average emf is induced, given the plane of the coil isoriginally perpendicular to the Earth’s field and is rotated to beparallel to the field in 10.0 ms?

12. A 0.250 m radius, 500-turn coil is rotated one-fourth of arevolution in 4.17 ms, originally having its plane perpendicularto a uniform magnetic field. (This is 60 rev/s.) Find themagnetic field strength needed to induce an average emf of10,000 V.


Approximately how does the emf induced in the loop inFigure 23.57(b) depend on the distance of the center of theloop from the wire?


(a) A lightning bolt produces a rapidly varying magnetic field.If the bolt strikes the earth vertically and acts like a current ina long straight wire, it will induce a voltage in a loop alignedlike that in Figure 23.57(b). What voltage is induced in a 1.00

m diameter loop 50.0 m from a 2.00×106 A lightning strike,


if the current falls to zero in 25.0 μs ? (b) Discuss

circumstances under which such a voltage would producenoticeable consequences.

23.3 Motional Emf15. Use Faraday’s law, Lenz’s law, and RHR-1 to show thatthe magnetic force on the current in the moving rod in Figure23.11 is in the opposite direction of its velocity.

16. If a current flows in the Satellite Tether shown in Figure23.12, use Faraday’s law, Lenz’s law, and RHR-1 to show thatthere is a magnetic force on the tether in the directionopposite to its velocity.

17. (a) A jet airplane with a 75.0 m wingspan is flying at 280m/s. What emf is induced between wing tips if the vertical

component of the Earth’s field is 3.00×10−5 T ? (b) Is anemf of this magnitude likely to have any consequences?Explain.

18. (a) A nonferrous screwdriver is being used in a 2.00 Tmagnetic field. What maximum emf can be induced along its12.0 cm length when it moves at 6.00 m/s? (b) Is it likely thatthis emf will have any consequences or even be noticed?

19. At what speed must the sliding rod in Figure 23.11 moveto produce an emf of 1.00 V in a 1.50 T field, given the rod’slength is 30.0 cm?

20. The 12.0 cm long rod in Figure 23.11 moves at 4.00 m/s.What is the strength of the magnetic field if a 95.0 V emf isinduced?

21. Prove that when B , ℓ , and v are not mutually

perpendicular, motional emf is given by emf = Bℓvsin θ . If

v is perpendicular to B , then θ is the angle between ℓand B . If ℓ is perpendicular to B , then θ is the angle

between v and B .

22. In the August 1992 space shuttle flight, only 250 m of theconducting tether considered in Example 23.2 could be letout. A 40.0 V motional emf was generated in the Earth’s

5.00×10−5 T field, while moving at 7.80×103 m/s . Whatwas the angle between the shuttle’s velocity and the Earth’sfield, assuming the conductor was perpendicular to the field?


Derive an expression for the current in a system like that inFigure 23.11, under the following conditions. The resistancebetween the rails is R , the rails and the moving rod are

identical in cross section A and have the same resistivity ρ .

The distance between the rails is l, and the rod moves atconstant speed v perpendicular to the uniform field B . At

time zero, the moving rod is next to the resistance R .


The Tethered Satellite in Figure 23.12 has a mass of 525 kgand is at the end of a 20.0 km long, 2.50 mm diameter cablewith the tensile strength of steel. (a) How much does thecable stretch if a 100 N force is exerted to pull the satellite in?(Assume the satellite and shuttle are at the same altitudeabove the Earth.) (b) What is the effective force constant ofthe cable? (c) How much energy is stored in it when stretchedby the 100 N force?


The Tethered Satellite discussed in this module is producing5.00 kV, and a current of 10.0 A flows. (a) What magneticdrag force does this produce if the system is moving at 7.80km/s? (b) How much kinetic energy is removed from thesystem in 1.00 h, neglecting any change in altitude or velocityduring that time? (c) What is the change in velocity if themass of the system is 100,000 kg? (d) Discuss the long termconsequences (say, a week-long mission) on the spaceshuttle’s orbit, noting what effect a decrease in velocity hasand assessing the magnitude of the effect.

23.4 Eddy Currents and Magnetic Damping26. Make a drawing similar to Figure 23.14, but with thependulum moving in the opposite direction. Then useFaraday’s law, Lenz’s law, and RHR-1 to show that magneticforce opposes motion.

27.

Figure 23.59 A coil is moved into and out of a region of uniformmagnetic field. A coil is moved through a magnetic field asshown in Figure 23.59. The field is uniform inside therectangle and zero outside. What is the direction of theinduced current and what is the direction of the magneticforce on the coil at each position shown?

23.5 Electric Generators28. Calculate the peak voltage of a generator that rotates its200-turn, 0.100 m diameter coil at 3600 rpm in a 0.800 Tfield.

29. At what angular velocity in rpm will the peak voltage of agenerator be 480 V, if its 500-turn, 8.00 cm diameter coilrotates in a 0.250 T field?

30. What is the peak emf generated by rotating a 1000-turn,

20.0 cm diameter coil in the Earth’s 5.00×10−5 T magneticfield, given the plane of the coil is originally perpendicular tothe Earth’s field and is rotated to be parallel to the field in 10.0ms?

31. What is the peak emf generated by a 0.250 m radius,500-turn coil is rotated one-fourth of a revolution in 4.17 ms,originally having its plane perpendicular to a uniform magneticfield. (This is 60 rev/s.)

32. (a) A bicycle generator rotates at 1875 rad/s, producingan 18.0 V peak emf. It has a 1.00 by 3.00 cm rectangular coilin a 0.640 T field. How many turns are in the coil? (b) Is thisnumber of turns of wire practical for a 1.00 by 3.00 cm coil?


This problem refers to the bicycle generator considered in theprevious problem. It is driven by a 1.60 cm diameter wheelthat rolls on the outside rim of the bicycle tire. (a) What is thevelocity of the bicycle if the generator’s angular velocity is1875 rad/s? (b) What is the maximum emf of the generatorwhen the bicycle moves at 10.0 m/s, noting that it was 18.0 Vunder the original conditions? (c) If the sophisticatedgenerator can vary its own magnetic field, what field strengthwill it need at 5.00 m/s to produce a 9.00 V maximum emf?

34. (a) A car generator turns at 400 rpm when the engine isidling. Its 300-turn, 5.00 by 8.00 cm rectangular coil rotates inan adjustable magnetic field so that it can produce sufficient



voltage even at low rpms. What is the field strength needed toproduce a 24.0 V peak emf? (b) Discuss how this requiredfield strength compares to those available in permanent andelectromagnets.

35. Show that if a coil rotates at an angular velocity ω , the

period of its AC output is 2π/ω .

36. A 75-turn, 10.0 cm diameter coil rotates at an angularvelocity of 8.00 rad/s in a 1.25 T field, starting with the planeof the coil parallel to the field. (a) What is the peak emf? (b) Atwhat time is the peak emf first reached? (c) At what time isthe emf first at its most negative? (d) What is the period of theAC voltage output?

37. (a) If the emf of a coil rotating in a magnetic field is zero att = 0 , and increases to its first peak at t = 0.100 ms , whatis the angular velocity of the coil? (b) At what time will its nextmaximum occur? (c) What is the period of the output? (d)When is the output first one-fourth of its maximum? (e) Whenis it next one-fourth of its maximum?


A 500-turn coil with a 0.250 m2 area is spun in the Earth’s

5.00×10−5 T field, producing a 12.0 kV maximum emf. (a)At what angular velocity must the coil be spun? (b) What isunreasonable about this result? (c) Which assumption orpremise is responsible?

23.6 Back Emf39. Suppose a motor connected to a 120 V source draws10.0 A when it first starts. (a) What is its resistance? (b) Whatcurrent does it draw at its normal operating speed when itdevelops a 100 V back emf?

40. A motor operating on 240 V electricity has a 180 V backemf at operating speed and draws a 12.0 A current. (a) Whatis its resistance? (b) What current does it draw when it is firststarted?

41. What is the back emf of a 120 V motor that draws 8.00 Aat its normal speed and 20.0 A when first starting?

42. The motor in a toy car operates on 6.00 V, developing a4.50 V back emf at normal speed. If it draws 3.00 A at normalspeed, what current does it draw when starting?


The motor in a toy car is powered by four batteries in series,which produce a total emf of 6.00 V. The motor draws 3.00 Aand develops a 4.50 V back emf at normal speed. Eachbattery has a 0.100 Ω internal resistance. What is theresistance of the motor?

23.7 Transformers44. A plug-in transformer, like that in Figure 23.29, supplies9.00 V to a video game system. (a) How many turns are in itssecondary coil, if its input voltage is 120 V and the primarycoil has 400 turns? (b) What is its input current when itsoutput is 1.30 A?

45. An American traveler in New Zealand carries atransformer to convert New Zealand’s standard 240 V to 120V so that she can use some small appliances on her trip. (a)What is the ratio of turns in the primary and secondary coils ofher transformer? (b) What is the ratio of input to outputcurrent? (c) How could a New Zealander traveling in theUnited States use this same transformer to power her 240 Vappliances from 120 V?

46. A cassette recorder uses a plug-in transformer to convert120 V to 12.0 V, with a maximum current output of 200 mA.(a) What is the current input? (b) What is the power input? (c)Is this amount of power reasonable for a small appliance?

47. (a) What is the voltage output of a transformer used forrechargeable flashlight batteries, if its primary has 500 turns,its secondary 4 turns, and the input voltage is 120 V? (b)What input current is required to produce a 4.00 A output? (c)What is the power input?

48. (a) The plug-in transformer for a laptop computer puts out7.50 V and can supply a maximum current of 2.00 A. What isthe maximum input current if the input voltage is 240 V?Assume 100% efficiency. (b) If the actual efficiency is lessthan 100%, would the input current need to be greater orsmaller? Explain.

49. A multipurpose transformer has a secondary coil withseveral points at which a voltage can be extracted, givingoutputs of 5.60, 12.0, and 480 V. (a) The input voltage is 240V to a primary coil of 280 turns. What are the numbers ofturns in the parts of the secondary used to produce the outputvoltages? (b) If the maximum input current is 5.00 A, what arethe maximum output currents (each used alone)?

50. A large power plant generates electricity at 12.0 kV. Its oldtransformer once converted the voltage to 335 kV. Thesecondary of this transformer is being replaced so that itsoutput can be 750 kV for more efficient cross-countrytransmission on upgraded transmission lines. (a) What is theratio of turns in the new secondary compared with the oldsecondary? (b) What is the ratio of new current output to oldoutput (at 335 kV) for the same power? (c) If the upgradedtransmission lines have the same resistance, what is the ratioof new line power loss to old?

51. If the power output in the previous problem is 1000 MWand line resistance is 2.00 Ω , what were the old and newline losses?


The 335 kV AC electricity from a power transmission line isfed into the primary coil of a transformer. The ratio of thenumber of turns in the secondary to the number in the primaryis Ns / Np = 1000 . (a) What voltage is induced in the

secondary? (b) What is unreasonable about this result? (c)Which assumption or premise is responsible?


Consider a double transformer to be used to create very largevoltages. The device consists of two stages. The first is atransformer that produces a much larger output voltage thanits input. The output of the first transformer is used as input toa second transformer that further increases the voltage.Construct a problem in which you calculate the output voltageof the final stage based on the input voltage of the first stageand the number of turns or loops in both parts of bothtransformers (four coils in all). Also calculate the maximumoutput current of the final stage based on the input current.Discuss the possibility of power losses in the devices and theeffect on the output current and power.

23.8 Electrical Safety: Systems and Devices54. Integrated Concepts

A short circuit to the grounded metal case of an applianceoccurs as shown in Figure 23.60. The person touching thecase is wet and only has a 3.00 kΩ resistance to earth/ground. (a) What is the voltage on the case if 5.00 mA flows


through the person? (b) What is the current in the short circuitif the resistance of the earth/ground wire is 0.200 Ω ? (c)Will this trigger the 20.0 A circuit breaker supplying theappliance?

Figure 23.60 A person can be shocked even when the case of anappliance is grounded. The large short circuit current produces a voltageon the case of the appliance, since the resistance of the earth/groundwire is not zero.

23.9 Inductance55. Two coils are placed close together in a physics lab todemonstrate Faraday’s law of induction. A current of 5.00 A inone is switched off in 1.00 ms, inducing a 9.00 V emf in theother. What is their mutual inductance?

56. If two coils placed next to one another have a mutualinductance of 5.00 mH, what voltage is induced in one whenthe 2.00 A current in the other is switched off in 30.0 ms?

57. The 4.00 A current through a 7.50 mH inductor isswitched off in 8.33 ms. What is the emf induced opposingthis?

58. A device is turned on and 3.00 A flows through it 0.100ms later. What is the self-inductance of the device if aninduced 150 V emf opposes this?

59. Starting with emf2 = −MΔI1Δt , show that the units of

inductance are (V ⋅ s)/A = Ω ⋅ s .

60. Camera flashes charge a capacitor to high voltage byswitching the current through an inductor on and off rapidly. Inwhat time must the 0.100 A current through a 2.00 mHinductor be switched on or off to induce a 500 V emf?

61. A large research solenoid has a self-inductance of 25.0 H.(a) What induced emf opposes shutting it off when 100 A ofcurrent through it is switched off in 80.0 ms? (b) How muchenergy is stored in the inductor at full current? (c) At what ratein watts must energy be dissipated to switch the current off in80.0 ms? (d) In view of the answer to the last part, is itsurprising that shutting it down this quickly is difficult?

62. (a) Calculate the self-inductance of a 50.0 cm long, 10.0cm diameter solenoid having 1000 loops. (b) How muchenergy is stored in this inductor when 20.0 A of current flowsthrough it? (c) How fast can it be turned off if the induced emfcannot exceed 3.00 V?

63. A precision laboratory resistor is made of a coil of wire1.50 cm in diameter and 4.00 cm long, and it has 500 turns.(a) What is its self-inductance? (b) What average emf isinduced if the 12.0 A current through it is turned on in 5.00 ms(one-fourth of a cycle for 50 Hz AC)? (c) What is itsinductance if it is shortened to half its length and counter-wound (two layers of 250 turns in opposite directions)?

64. The heating coils in a hair dryer are 0.800 cm in diameter,have a combined length of 1.00 m, and a total of 400 turns.(a) What is their total self-inductance assuming they act like asingle solenoid? (b) How much energy is stored in them when6.00 A flows? (c) What average emf opposes shutting themoff if this is done in 5.00 ms (one-fourth of a cycle for 50 HzAC)?

65. When the 20.0 A current through an inductor is turned offin 1.50 ms, an 800 V emf is induced, opposing the change.What is the value of the self-inductance?

66. How fast can the 150 A current through a 0.250 Hinductor be shut off if the induced emf cannot exceed 75.0V?


A very large, superconducting solenoid such as one used inMRI scans, stores 1.00 MJ of energy in its magnetic fieldwhen 100 A flows. (a) Find its self-inductance. (b) If the coils“go normal,” they gain resistance and start to dissipatethermal energy. What temperature increase is produced if allthe stored energy goes into heating the 1000 kg magnet,given its average specific heat is 200 J/kg·ºC ?


A 25.0 H inductor has 100 A of current turned off in 1.00 ms.(a) What voltage is induced to oppose this? (b) What isunreasonable about this result? (c) Which assumption orpremise is responsible?

23.10 RL Circuits69. If you want a characteristic RL time constant of 1.00 s,and you have a 500 Ω resistor, what value of self-inductance is needed?

70. Your RL circuit has a characteristic time constant of 20.0ns, and a resistance of 5.00 MΩ . (a) What is the inductanceof the circuit? (b) What resistance would give you a 1.00 nstime constant, perhaps needed for quick response in anoscilloscope?

71. A large superconducting magnet, used for magneticresonance imaging, has a 50.0 H inductance. If you wantcurrent through it to be adjustable with a 1.00 s characteristictime constant, what is the minimum resistance of system?

72. Verify that after a time of 10.0 ms, the current for thesituation considered in Example 23.9 will be 0.183 A asstated.

73. Suppose you have a supply of inductors ranging from1.00 nH to 10.0 H, and resistors ranging from 0.100 Ω to

1.00 MΩ . What is the range of characteristic RL timeconstants you can produce by connecting a single resistor toa single inductor?

74. (a) What is the characteristic time constant of a 25.0 mHinductor that has a resistance of 4.00 Ω ? (b) If it isconnected to a 12.0 V battery, what is the current after 12.5ms?



75. What percentage of the final current I0 flows through an

inductor L in series with a resistor R , three time constantsafter the circuit is completed?

76. The 5.00 A current through a 1.50 H inductor is dissipatedby a 2.00 Ω resistor in a circuit like that in Figure 23.44 withthe switch in position 2. (a) What is the initial energy in theinductor? (b) How long will it take the current to decline to5.00% of its initial value? (c) Calculate the average powerdissipated, and compare it with the initial power dissipated bythe resistor.

77. (a) Use the exact exponential treatment to find how muchtime is required to bring the current through an 80.0 mHinductor in series with a 15.0 Ω resistor to 99.0% of its finalvalue, starting from zero. (b) Compare your answer to theapproximate treatment using integral numbers of τ . (c)Discuss how significant the difference is.

78. (a) Using the exact exponential treatment, find the timerequired for the current through a 2.00 H inductor in serieswith a 0.500 Ω resistor to be reduced to 0.100% of itsoriginal value. (b) Compare your answer to the approximatetreatment using integral numbers of τ . (c) Discuss howsignificant the difference is.

23.11 Reactance, Inductive and Capacitive79. At what frequency will a 30.0 mH inductor have areactance of 100 Ω ?

80. What value of inductance should be used if a 20.0 kΩreactance is needed at a frequency of 500 Hz?

81. What capacitance should be used to produce a2.00 MΩ reactance at 60.0 Hz?

82. At what frequency will an 80.0 mF capacitor have areactance of 0.250 Ω ?

83. (a) Find the current through a 0.500 H inductor connectedto a 60.0 Hz, 480 V AC source. (b) What would the current beat 100 kHz?

84. (a) What current flows when a 60.0 Hz, 480 V AC sourceis connected to a 0.250 μF capacitor? (b) What would the

current be at 25.0 kHz?

85. A 20.0 kHz, 16.0 V source connected to an inductorproduces a 2.00 A current. What is the inductance?

86. A 20.0 Hz, 16.0 V source produces a 2.00 mA currentwhen connected to a capacitor. What is the capacitance?

87. (a) An inductor designed to filter high-frequency noisefrom power supplied to a personal computer is placed inseries with the computer. What minimum inductance should ithave to produce a 2.00 kΩ reactance for 15.0 kHz noise?(b) What is its reactance at 60.0 Hz?

88. The capacitor in Figure 23.55(a) is designed to filter low-frequency signals, impeding their transmission betweencircuits. (a) What capacitance is needed to produce a100 kΩ reactance at a frequency of 120 Hz? (b) Whatwould its reactance be at 1.00 MHz? (c) Discuss theimplications of your answers to (a) and (b).

89. The capacitor in Figure 23.55(b) will filter high-frequencysignals by shorting them to earth/ground. (a) Whatcapacitance is needed to produce a reactance of 10.0 mΩfor a 5.00 kHz signal? (b) What would its reactance be at 3.00

Hz? (c) Discuss the implications of your answers to (a) and(b).


In a recording of voltages due to brain activity (an EEG), a10.0 mV signal with a 0.500 Hz frequency is applied to acapacitor, producing a current of 100 mA. Resistance isnegligible. (a) What is the capacitance? (b) What isunreasonable about this result? (c) Which assumption orpremise is responsible?


Consider the use of an inductor in series with a computeroperating on 60 Hz electricity. Construct a problem in whichyou calculate the relative reduction in voltage of incominghigh frequency noise compared to 60 Hz voltage. Among thethings to consider are the acceptable series reactance of theinductor for 60 Hz power and the likely frequencies of noisecoming through the power lines.

23.12 RLC Series AC Circuits92. An RL circuit consists of a 40.0 Ω resistor and a 3.00

mH inductor. (a) Find its impedance Z at 60.0 Hz and 10.0

kHz. (b) Compare these values of Z with those found inExample 23.12 in which there was also a capacitor.

93. An RC circuit consists of a 40.0 Ω resistor and a

5.00 μF capacitor. (a) Find its impedance at 60.0 Hz and

10.0 kHz. (b) Compare these values of Z with those found inExample 23.12, in which there was also an inductor.

94. An LC circuit consists of a 3.00 mH inductor and a

5.00 µF capacitor. (a) Find its impedance at 60.0 Hz and

10.0 kHz. (b) Compare these values of Z with those found inExample 23.12 in which there was also a resistor.

95. What is the resonant frequency of a 0.500 mH inductorconnected to a 40.0 μF capacitor?

96. To receive AM radio, you want an RLC circuit that can bemade to resonate at any frequency between 500 and 1650kHz. This is accomplished with a fixed 1.00 μH inductor

connected to a variable capacitor. What range of capacitanceis needed?

97. Suppose you have a supply of inductors ranging from1.00 nH to 10.0 H, and capacitors ranging from 1.00 pF to0.100 F. What is the range of resonant frequencies that canbe achieved from combinations of a single inductor and asingle capacitor?

98. What capacitance do you need to produce a resonantfrequency of 1.00 GHz, when using an 8.00 nH inductor?

99. What inductance do you need to produce a resonantfrequency of 60.0 Hz, when using a 2.00 μF capacitor?

100. The lowest frequency in the FM radio band is 88.0 MHz.(a) What inductance is needed to produce this resonantfrequency if it is connected to a 2.50 pF capacitor? (b) Thecapacitor is variable, to allow the resonant frequency to beadjusted to as high as 108 MHz. What must the capacitancebe at this frequency?

101. An RLC series circuit has a 2.50 Ω resistor, a 100 μHinductor, and an 80.0 μF capacitor.(a) Find the circuit’s

impedance at 120 Hz. (b) Find the circuit’s impedance at 5.00


kHz. (c) If the voltage source has Vrms = 5.60 V , what is

Irms at each frequency? (d) What is the resonant frequency

of the circuit? (e) What is Irms at resonance?

102. An RLC series circuit has a 1.00 kΩ resistor, a

150 μH inductor, and a 25.0 nF capacitor. (a) Find the

circuit’s impedance at 500 Hz. (b) Find the circuit’simpedance at 7.50 kHz. (c) If the voltage source hasVrms = 408 V , what is Irms at each frequency? (d) What

is the resonant frequency of the circuit? (e) What is Irms at

resonance?

103. An RLC series circuit has a 2.50 Ω resistor, a 100 μHinductor, and an 80.0 μF capacitor. (a) Find the power factor

at f = 120 Hz . (b) What is the phase angle at 120 Hz? (c)

What is the average power at 120 Hz? (d) Find the averagepower at the circuit’s resonant frequency.

104. An RLC series circuit has a 1.00 kΩ resistor, a

150 μH inductor, and a 25.0 nF capacitor. (a) Find the

power factor at f = 7.50 Hz . (b) What is the phase angle at

this frequency? (c) What is the average power at thisfrequency? (d) Find the average power at the circuit’sresonant frequency.

105. An RLC series circuit has a 200 Ω resistor and a 25.0

mH inductor. At 8000 Hz, the phase angle is 45.0º . (a) Whatis the impedance? (b) Find the circuit’s capacitance. (c) IfVrms = 408 V is applied, what is the average power

supplied?

106. Referring to Example 23.14, find the average power at10.0 kHz.

Test Prep for AP® Courses

23.1 Induced Emf and Magnetic Flux1. To produce current with a coil and bar magnet you can:

a. move the coil but not the magnet.b. move the magnet but not the coil.c. move either the coil or the magnet.d. It is not possible to produce current.

2. Calculate the magnetic flux for a coil of area 0.2 m2 placedat an angle of θ=60º (as shown in the figure above) to amagnetic field of strength 1.5×10-3 T. At what angle will theflux be at its maximum?

23.5 Electric Generators3. The emf induced in a coil that is rotating in a magnetic fieldwill be at a maximum when

a. the magnetic flux is at a maximum.b. the magnetic flux is at a minimum.c. the change in magnetic flux is at a maximum.d. the change in magnetic flux is at a minimum.

4. A coil with circular cross section and 20 turns is rotating ata rate of 400 rpm between the poles of a magnet. If themagnetic field strength is 0.6 T and peak voltage is 0.2 V,what is the radius of the coil? If the emf of the coil is zero at t= 0 s, when will it reach its peak emf?

23.7 Transformers

5. Which of the following statements is true for a step-downtransformer? Select two answers.

a. Primary voltage is higher than secondary voltage.b. Primary voltage is lower than secondary voltage.c. Primary current is higher than secondary current.d. Primary current is lower than secondary current.

6. An ideal step-up transformer with turn ratio 1:30 is suppliedwith an input power of 120 W. If the output voltage is 210 V,calculate the output power and input current.

23.8 Electrical Safety: Systems and Devices7. Which of the following statements is true for an isolationtransformer?

a. It has more primary turns than secondary turns.b. It has fewer primary turns than secondary turns.c. It has an equal number of primary and secondary turns.d. It can have more, fewer, or an equal number of primary

and secondary turns.

8. Explain the working of a ground fault interrupter (GFI).



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