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Copyright © by Holt, Rinehart and Winston. All rights reserved.

PHYSICS IN ACTION

In an electric guitar, the vibrations of the

strings are converted to an electric signal,

which is then amplified outside the guitar

and heard as sound from a loudspeaker.

Yet the electric guitar is not plugged

directly into an electric source. Instead, it

generates electric current by a process

called induction. By changing the magnetic

field near a coil of wire, an electric current

can be induced in the coil by the vibrations

of the guitar strings.

This chapter explores how induction

produces and changes alternating currents.

• Why does the generated current last only aslong as the string vibrates?

• Why must the strings be ferromagnetic inorder for an electric guitar to work?

CONCEPT REVIEW

Resistance (Section 19–2)

emf (Section 20–1)

Magnetic force (Section 21–3)

CHAPTER 22

Induction andAlternatingCurrent

Induction and Alternating Current 793Copyright © by Holt, Rinehart and Winston. All rights reserved.

Copyright © by Holt, Rinehart and Winston. All rights reserved.Chapter 22794

MAGNETIC FIELDS AND INDUCED EMFS

Recall that in Chapter 20 you were asked if it was possible to produce an elec-

tric current using only wires and no battery. So far, all electric circuits that you

have studied have used a battery or an electrical power supply to create a

potential difference within a circuit. In both of these cases, an emf increases

the electrical potential energy of charges in the circuit, causing them to move

through the circuit and create a current.

It is also possible to induce a current in a circuit without the use of a battery

or an electrical power supply. Just as a magnetic field can be formed by a cur-

rent in a circuit, a current can be formed by moving a portion of a closed elec-

tric circuit through an external magnetic field, as indicated in Figure 22-1.The process of inducing a current in a circuit with a changing magnetic field

is called electromagnetic induction.Consider a circuit consisting of only a resistor in the vicinity of a magnet.

There is no battery to supply a current. If neither the magnet nor the circuit is

moving with respect to the other, no current will be present in the circuit. But

if the circuit moves toward or away from the magnet or the magnet moves

toward or away from the circuit, a current is induced. As long as there is rela-

tive motion between the two, a current can form in the circuit.

The separation of charges by the magnetic force induces an emf

It may seem strange that there can be an induced emf and a corresponding

induced current without a battery or similar source of electrical energy. Recall

that in the previous chapter a moving charge can be deflected by a magnetic

field. This deflection can be used to explain how an emf occurs in a wire that

moves through a magnetic field.

22-1Induced current

22-1 SECTION OBJECTIVES

• Describe how the change inthe number of magnetic fieldlines through a circuit loopaffects the magnitude anddirection of the inducedcurrent.

• Apply Lenz’s law todetermine the direction of aninduced current.

• Calculate the induced emfand current using Faraday’slaw of induction.

electromagnetic induction

the production of an emf in aconducting circuit by a change inthe strength, position, or orienta-tion of an external magnetic field

Figure 22-1When the circuit loopcrosses the lines of themagnetic field, a current is induced in the circuit,as indicated by the movement of the galvanometer needle.

100

200

300

400

500

600

700

800

900

100

200

300

400

500

600

700

800

900

1000

Magnetic fieldGalvanometer

N

S

0

Current

Copyright © by Holt, Rinehart and Winston. All rights reserved.795Induction and Alternating Current

Consider a conducting wire pulled through a magnetic field, as shown on the

left in Figure 22-2. You learned in Chapter 21 that positive charges moving with

a velocity at an angle to the magnetic field will experience a magnetic force.

According to the right-hand rule, this force will be directed perpendicular to

both the magnetic field and the motion of the charges. For positive charges in the

wire, the force is directed downward along the wire. For negative charges, the

force is upward. This effect is equivalent to replacing the segment of wire and

the magnetic field with a battery that has a potential difference, or emf,

between its terminals, as shown on the right in Figure 22-2. As long as the

conducting wire moves through the magnetic field, the emf will be maintained.

The polarity of the induced emf—and thus the direction of the induced

current in the circuit—depends on the direction in which the wire is moved

through the magnetic field. For instance, if the wire in Figure 22-2 is moved

to the right, the right-hand rule predicts that the positive charges will be

pushed downward. If the wire is moved to the left, the positive charges will be

pushed upward. The magnitude of the induced emf depends on the velocity

with which the wire is moved through the magnetic field, the length of the

wire, and the strength of the magnetic field.

The angle between a magnetic field and a circuit affects induction

To induce an emf in a wire loop, part of the loop must move through a mag-

netic field or the entire loop must pass into or out of the magnetic field. No

emf is induced if the loop is static or the magnetic field is constant.

The magnitude of the induced emf and current depend partly on how the

loop is oriented to the magnetic field, as shown in Figure 22-3. The induced

current is largest when the plane of the loop is perpendicular to the magnetic

field, as in (a); it is less when the plane is tilted into the field, as in (b); and it is

zero when the plane is parallel to the field, as in (c).The role of orientation on induced current can be understood in terms of

the force exerted by a magnetic field on charges in the moving loop. The small-

er the component of the magnetic field perpendicular to both the plane and

the motion of the loop, the smaller the magnetic force on the charges in the

loop. When the area of the loop is moved parallel to the magnetic field, there is

no magnetic field component perpendicular to the plane of the loop and there-

fore no force that moves the charges through the circuit.

B (out of page)

v=

−

+

–+

Figure 22-2The separation of positive and nega-tive moving charges by the magneticforce creates a potential difference(emf ) between the ends of the conductor.

vv v

(a) (b) (c)

Figure 22-3The induced emf and current are largest when the plane of theloop is perpendicular to the magnetic field (a), smaller when theplane of the loop is tilted into the field (b), and zero when theplane of the loop and the magnetic field are parallel (c).

TOPIC: Magnetic fieldsGO TO: www.scilinks.orgsciLINKS CODE: HF2221

NSTA


In 1996, the space shuttle Columbiaattempted to use a 20.7 km con-ducting tether to study Earth’s mag-netic field in space. The plan was todrag the tether through the mag-netic field, inducing an emf in thetether. The magnitude of the emfwould directly vary with thestrength of the magnetic field.Unfortunately, the tether brokebefore it was fully extended, so theexperiment was abandoned.

Chapter 22796

Change in the number of magnetic field lines induces a current

So far, you have learned that moving a circuit loop into or out of a magnetic

field can induce an emf and a current in the circuit. Changing the size of the

loop or the strength of the magnetic field also will induce an emf in the circuit.

One way to predict whether a current will be induced in a given situation

involves the concept of changes in magnetic field lines. For example, moving

the circuit into the magnetic field causes some lines to move into the loop.

Changing the size of the circuit loop or rotating the loop changes the number

of field lines passing through the loop, as does changing the magnetic field’s

strength. Table 22-1 summarizes these three ways of inducing a current.

CHARACTERISTICS OF INDUCED CURRENT

Suppose a bar magnet is pushed into a coil of wire. As the magnet moves into

the coil, the strength of the magnetic field within the coil increases, and a cur-

rent is induced in the circuit. This induced current in turn produces its own

magnetic field, whose direction can be found by using the right-hand rule. If

you were to apply this rule for several cases, you would notice that the induced

magnetic field direction depends on the change in the applied field.

As the magnet approaches, the magnetic field passing through the coil

increases in strength. The induced current in the coil must be in a direction

that produces a magnetic field that opposes the increasing strength of the

approaching field. The induced magnetic field is therefore in the direction

opposite that of the approaching magnetic field.

Table 22-1 Ways of inducing a current in a circuit

Description Before After

Circuit is moved into or out of magneticfield (either circuit or magnet moving).

Circuit is rotated in the magnetic field(angle between area of circuit andmagnetic field changes).

Intensity of magnetic field is varied.

v

B

B

B

v

I

I B

BI


Approachingmagnetic field

Magnetic field frominduced currentInduced current

Wire

NS

v

N S

Recedingmagnetic field

Magnetic field frominduced currentInduced current

Wire

v

N SSN

Figure 22-4The approaching magnetic field isopposed by the induced magneticfield, which behaves like a bar mag-net with the orientation shown.

Figure 22-5The receding magnetic field isattracted by the induced magneticfield, which behaves like a bar mag-net with the orientation shown.

1. Falling magnet A bar magnet is droppedtoward the floor, on which lies a large ring of con-ducting metal. The magnet’s length—and thus thepoles of the magnet—is parallel to the direction ofmotion. Disregarding air resistance, does the magnetduring its fall toward the ring move with the sameconstant acceleration as a freely falling body? Explainyour answer.

2. Induction in a bracelet Suppose you arewearing a bracelet that is an unbroken ring of copper. Ifyou walk briskly into a strong magnetic field while wearing the bracelet, how would you hold your wrist with respect to the magnetic field in order to avoid inducing a current in the bracelet?

The induced magnetic field is like the field of a bar magnet oriented as

shown in Figure 22-4. Note that the coil and magnet repel each other.

If the magnet is moved away from the coil, the magnetic field passing

through the coil decreases in strength. Again, the direction of current induced

in the coil must be such that it produces a magnetic field that opposes the

decreasing strength of the receding field. This means that the magnetic field

set up by the coil must be in the same direction as the receding magnetic field.

The induced magnetic field is like the field of a bar magnet oriented as

shown in Figure 22-5. Note that the coil and magnet attract each other.


The rule for finding the direction of the induced current is called Lenz’s law

and is expressed as follows:

The magnetic field of the induced current opposes the change in the applied

magnetic field.

Note that the field of the induced current does not oppose the applied field

but rather the change in the applied field. If the applied field changes, the

induced field attempts to keep the total field strength constant, according to

the principle of energy conservation.

Faraday’s law of induction

Because of the principle of energy conservation, Lenz’s law allows you to deter-

mine the direction of an induced current in a circuit. Lenz’s law does not provide

information on the magnitude of the induced current or the induced emf. To

calculate the magnitude of the induced emf, you must use Faraday’s law of mag-

netic induction. For a single loop of a circuit, this may be expressed as follows:

emf = − ∆[AB

∆(c

t

os q)]

Certain features of this equation should be noted. First, a change with time

of any of the three variables—applied magnetic field strength, B, circuit area,

A, or angle of orientation, q—can give rise to an induced emf. The term

B (cos q) represents the component of the magnetic field perpendicular to the

plane of the loop. The angle q is measured between the applied magnetic field

and the normal to the plane of the loop, as indicated in Figure 22-6.The minus sign in front of the equation is included to indicate the polarity

of the induced emf. It indicates that the induced magnetic field opposes the

changing applied magnetic field according to Lenz’s law.

If a circuit contains a number, N, of tightly wound loops, the average

induced emf is simply N times the induced emf for a single loop. The equation

thus takes the general form of Faraday’s law of magnetic induction.

In this chapter, N is always assumed to be a whole number.

Recall that the SI unit for magnetic field strength is the tesla (T), which equals

one newton per ampere-meter, or N/(A•m). When calculating induced emf,

express the tesla in units of one volt-second per meter squared, or (V•s)/m2.

FARADAY’S LAW OF MAGNETIC INDUCTION

emf = −N∆[AB

∆(c

t

os q)]

average induced emf = −the number of loops in the circuit × the rate of change of (circuit loop area × magnetic field component

normal to the plane of loop)

B

Normal toplane of loop

Loop

θ

B (cos )θ

Figure 22-6The angle q is defined as the anglebetween the magnetic field and thenormal to the plane of the loop.B (cos q ) equals the strength of themagnetic field perpendicular to theplane of the loop.


SAMPLE PROBLEM 22A

Induced emf and current

P R O B L E MA coil with 25 turns of wire is wrapped around a hollow tube with an areaof 1.8 m2. Each turn has the same area as the tube. A uniform magneticfield is applied at a right angle to the plane of the coil. If the field increasesuniformly from 0.00 T to 0.55 T in 0.85 s, find the magnitude of the inducedemf in the coil. If the resistance in the coil is 2.5 Ω, find the magnitude ofthe induced current in the coil.

S O L U T I O NGiven: ∆t = 0.85 s A = 1.8 m2 q = 0.0° N = 25 turns

Bi = 0.00 T = 0.00 V•s/m2 Bf = 0.55 T = 0.55 V•s/m2

R = 2.5 Ω

Unknown: emf = ? I = ?

Diagram: Show the coil before and after the change in the magnetic

field.

Choose an equation(s) or situation: Use Faraday’s law of magnetic induc-

tion to find the induced emf in the coil.

emf = −N ∆[AB

∆(c

t

os q)]

Substitute the induced emf into the definition of resistance to determine the

induced current in the coil.

I = em

R

f

Rearrange the equation(s) to isolate the unknown(s): In this example, only

the magnetic field strength changes with time. The other components (the coil

area and the angle between the magnetic field and the coil) remain constant.

emf = −NA (cos q) ∆∆

B

t

N = 25 turns

B = 0.00 T at t = 0.00 s B = 0.55 T at t = 0.85 s

A = 1.8 m2

R = 2.5 Ω

N = 25 turns

A = 1.8 m2

R = 2.5 Ω

1. DEFINE

2. PLAN

continued onnext page


Substitute the values into the equation(s) and solve:

emf = −(25)(1.8 m2)(cos 0.0°) = −29 V

I = −2

2

.5

9

ΩV

= −12 A

The induced emf, and therefore the induced current, are directed through the

coil so that the magnetic field produced by the induced current opposes the

change in the applied magnetic field. For the diagram shown on page 799, the

induced magnetic field is directed to the right and the current that produces

it is directed from left to right through the resistor.

emf = −29 V

I = −12 A

(0.55 − 0.00) V

m

•2s

(0.85 s)

3. CALCULATE

4. EVALUATE

CALCULATOR SOLUTION

Because the minimum number of significant figures for the data is two,the calculator answer, 29.11764706,should be rounded to two digits.

1. A single circular loop with a radius of 22 cm is placed in a uniform exter-

nal magnetic field with a strength of 0.50 T so that the plane of the coil is

perpendicular to the field. The coil is pulled steadily out of the field in

0.25 s. Find the average induced emf during this interval.

2. A coil with 205 turns of wire, a total resistance of 23 Ω, and a cross-

sectional area of 0.25 m2 is positioned with its plane perpendicular to the

field of a powerful electromagnet. What average current is induced in the

coil during the 0.25 s that the magnetic field drops from 1.6 T to 0.0 T?

3. A circular wire loop with a radius of 0.33 m is located in an external

magnetic field of strength +0.35 T that is perpendicular to the plane of

the loop. The field strength changes to −0.25 T in 1.5 s. (The plus and

minus signs for a magnetic field refer to opposite directions through the

coil.) Find the magnitude of the average induced emf during this interval.

4. A 505-turn circular-loop coil with a diameter of 15.5 cm is initially

aligned so that its plane is perpendicular to the Earth’s magnetic field. In

2.77 ms the coil is rotated 90.0° so that its plane is parallel to the Earth’s

magnetic field. If an average emf of 0.166 V is induced in the coil, what is

the value of the Earth’s magnetic field?

Induced emf and current

PRACTICE 22A


APPLICATIONS OF INDUCTION

In electric circuits, the need often arises for a temporary or continuously

changing current to be produced. This kind of current can be generated

through electromagnetic induction.

Door bells

Certain types of door bells chime when the button is briefly

pushed. A hint to how these door bells work is the small electric

light bulb often found behind the doorbell button. When you

press the button, the light briefly goes out. This indicates that

the door bell’s circuit has been opened and that the current has

been temporarily discontinued.

You can see the effect of opening the circuit in Figure 22-7.While the current is constant in the first circuit, the magnetic

field in the coil of this circuit is steady. As long as this field does

not change, no current is induced in the coil of the second cir-

cuit. When the current in the first circuit is interrupted by press-

ing the doorbell button, the magnetic field in the first coil drops

rapidly, inducing a current in the second circuit and causing a

magnetic field to quickly rise along the axis of the second coil.

This induced magnetic field is strong enough to push the iron

plunger against the chime.

Tape recorders

One common use of induced currents and emfs is found in the

tape recorder. Many different types of tape recorders are made,

but the same basic principles apply for all. A magnetic tape moves

past a recording head and a playback head. The tape is a plastic

ribbon coated with either iron oxide or chromium dioxide.

A microphone transforms a sound wave into a fluctuating

electric current. This current is amplified and allowed to pass

through a wire coiled around an iron ring, as shown in Figure22-8, which functions as the recording head. The iron ring

and wire constitute an electromagnet; the lines of the magnet-

ic field are contained completely inside the iron except at the

point where a gap is cut in the ring. The magnetic tape is

passed over this gap.

Because the magnetic field does not pass as easily through

the air of the gap as it does through the nearby small pieces of

metal oxide in the tape, the magnetic field passes around the

gap and magnetizes the metal oxide particles. As the tape moves

past the slot, it becomes magnetized in a pattern that corre-

sponds to both the frequency and the intensity of the sound sig-

nal entering the microphone.

–

Decreasingcurrentemf sources

Increasingmagnetic field

Decreasingmagnetic field

Inducedcurrent

Door bell switchCoil Coil

Chime

Iron plunger

+

Figure 22-7The sudden drop in current in the first circuit induces a current in the second circuit. The resulting magneticfield pushes the plunger against the door-bell chime.

Magnetic tape

Changingcurrent

v

Changing magnetic field

Figure 22-8The signal fluctuations in the current induce a fluctuat-ing magnetic field, which is recorded along the length of the tape. During playback, the changing magnetic fieldof the tape will induce in the coil a fluctuating currentthat can be converted to reproduce the originalrecorded sound.


In the recording process, the magnetic field is produced by an applied cur-

rent. During the playback process, induction is used to create a current from a

changing magnetic field. The sound signal is reconstructed by passing the tape

over the playback head, which is an iron ring with wire wound around it. In

some recorders, one head is used for both playback and recording.

When the tape moves past the playback head, the varying magnetic fields

on the tape produce changing magnetic-field lines through the wire coil,

inducing a current in the coil. The current corresponds to the current in the

recording head that originally produced the recording on the tape. This

changing electric current can be amplified and used to drive a speaker. Play-

back is thus an example of induction of a current by a moving magnet.

Section Review

1. A circular current loop made of flexible wire is located in a magnetic

field. Describe three ways an emf can be induced in the loop.

2. A spacecraft orbiting Earth has a coil of wire in it. An astronaut measures

a small current in the coil, even though there is no battery connected to it

and there are no magnets on the spacecraft. What is causing the current?

3. A bar magnet is positioned near a coil of

wire, as shown in Figure 22-9. What is the

direction of the current through the resis-

tor when the magnet is moved to the

left, as in (a)? to the right, as in (b)?

4. A 256-turn coil with a cross-sectional area of 0.0025 m2 is placed in a

uniform external magnetic field of strength 0.25 T so that the plane of

the coil is perpendicular to the field. The coil is pulled steadily out of the

field in 0.75 s. Find the average induced emf during this interval.

5. Physics in Action Electric guitar strings are made of ferromag-

netic materials that can be magnetized. The strings lie closely over and

perpendicular to a coil of wire. Inside the coil are permanent magnets

that magnetize the segments of the strings overhead. Using this arrange-

ment, explain how the vibrations of a plucked string produce an electric

signal at the same frequency as the vibration of the string.

6. Physics in Action The magnetic field strength of a magnetized

electric guitar string is 9.0 × 10−4 T. The pickup coil consists of 5200 turns

of wire and has an effective area for each string of 5.4 × 10−5 m2. If the

string vibrates with a frequency of 440 Hz, what is the average induced

emf? (Hint: Assume the magnetic field strength varies from a minimum

value of 0.0 T to its maximum value during one fourth of the string’s

vibration cycle.)

R

v

v(a)

(b)

S N

Figure 22-9


GENERATORS AND ALTERNATING CURRENT

In the previous section you learned that a current can be induced in a circuit

either by changing the magnetic field strength or by moving the circuit loop

in or out of the magnetic field. Another way to induce a current is to change

the orientation of the loop with respect to the magnetic field.

This second approach to inducing a current represents a practical means of gen-

erating electrical energy. In effect, the mechanical energy used to turn the loop is

converted to electrical energy.A device that does this is called an electric generator.In most commercial power plants, mechanical energy is provided in the form

of rotational motion. For example, in a hydroelectric plant, falling water directed

against the blades of a turbine causes the turbine to turn; in a coal or natural-gas-

burning plant, energy produced by burning fuel is used to convert water to

steam, and this steam is directed against the turbine blades to turn the turbine.

Basically, a generator uses the turbine’s rotary motion to turn a wire loop in

a magnetic field. A simple generator is shown in Figure 22-10. As the loop

rotates, the effective area of the loop changes with time, inducing an emf and a

current in an external circuit connected to the ends of the loop.

A generator produces a continuously changing emf

Consider a single loop of wire that is rotated with a constant angular fre-

quency in a uniform magnetic field. The loop can be thought of as four con-

ducting wires similar to the conducting wire discussed on page 795 of Section

22-1. In this example, the loop is rotating counterclockwise within a magnetic

field directed to the left.

22-2Alternating current, generators,

and motors


• Calculate the maximum emffor an electric generator.

• Calculate rms current andpotential difference for accircuits.

• Describe how an electricmotor relates to an electricgenerator.

generator

a device that uses induction toconvert mechanical energy toelectrical energy

Figure 22-10In a simple generator, the rotation of conducting loops(located on the left end of the axis) through a constantmagnetic field induces an alternating current in the loops.


When the area of the loop is perpendicular to the magnetic field lines, as

shown in Figure 22-11(a), every segment of wire in the loop is moving paral-

lel to the magnetic field lines. At this instant, the magnetic field does not exert

force on the charges in any part of the wire, so the induced emf in each seg-

ment is therefore zero.

As the loop rotates away from this position, segments a and c cross magnetic

field lines, so the magnetic force on the charges in these segments, and thus

the induced emf, increases. The magnetic force on the charges in segments b

and d is directed outside of the wire, so the motion of these segments does not

contribute to the emf or the current. The greatest magnetic force on the

charges and the greatest induced emf occur at the instant when segments a

and c move perpendicularly to the magnetic field lines, as in Figure 22-11(b).This occurs when the plane of the loop is parallel to the field lines.

Because segment a moves downward through the field while segment c moves

upward, their emfs are in opposite directions, but both produce a counter-

clockwise current. As the loop continues to rotate, segments a and c cross fewer

lines, and the emf decreases. When the plane of the loop is perpendicular to the

magnetic field, the motion of segments a and c is again parallel to the magnetic

lines and the induced emf is again zero, as shown in Figure 22-11(c). Segments a

and c now move in directions opposite those in which they moved from their posi-

tions in (a) to those in (b). As a result, the polarity of the induced emf and the

direction of the current are reversed, as shown in Figure 22-11(d).

Figure 22-11For a rotating loop in a magneticfield, the induced emf is zero whenthe loop is perpendicular to themagnetic field, as in (a) and (c), andis at a maximum when the loop isparallel to the field, as in (b) and (d).

0 +–

Induced emf

a

d

cb

B

c

a0 +–

Induced emf

d

b

B

a

dc

b

B

0 +–

Induced emf

a d

cb

B

0 +–

Induced emf

(a) (b)

(c) (d)

TOPIC: Alternating currentGO TO: www.scilinks.orgsciLINKS CODE: HF2222

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emf

Time

Maximumemf

Figure 22-12The change with time of the inducedemf is depicted by a sine wave.

SAMPLE PROBLEM 22B

Induction in generators

P R O B L E MA generator consists of exactly eight turns of wire, each with an area A = 0.095 m2 and a total resistance of 12 Ω. The loop rotates in a magneticfield of 0.55 T at a constant frequency of 60.0 Hz. Find the maximuminduced emf and maximum induced current in the loop.

S O L U T I O NGiven: f = 60.0 Hz A = 0.095 m2 R = 12 Ω

B = 0.55 T = 0.55 V•s/m2 N = 8

Unknown: maximum emf = ? Imax = ?

Diagram:

N = 8 turns

B = 0.55 T

R = 12 Ω

A = 0.095 m2

f = 60.0 Hz

1. DEFINE


A graph of the change in emf versus time as the loop rotates is shown in

Figure 22-12. Note the similarities between this graph and a sine curve.

The induced emf is the result of the steady change in the angle between the

magnetic field lines and the normal to the loop. The following equation for

the emf produced by a generator can be derived from Faraday’s law of induc-

tion. The derivation is not shown here because it requires the use of calculus.

In this equation, the angle of orientation, q, has been replaced with the equiva-

lent expression wt, where w is the angular frequency of rotation (2p f ).

emf = NABw (sin wt)

The equation describes the sinusoidal variation of emf with time, as graphed in

Figure 22-12.The emf has a maximum value when the plane of the loop is parallel to the

magnetic field, that is, when sin (wt) = 1, which occurs when wt = q = 90°. In

this case, the expression above reduces to the following:

MAXIMUM EMF FOR A GENERATOR

maximum emf = NABw

maximum induced emf = number of loops × cross-sectional area ofloops × magnetic field strength × angular frequency of rotation of loops


2. PLAN

3. CALCULATE

4. EVALUATE

1. In a model generator, a 510-turn rectangular coil 0.082 m by 0.25 m

rotates with an angular frequency of 12.8 rad/s in a uniform magnetic

field of 0.65 T. What is the maximum emf induced in the coil?

2. A circular coil with a radius of R = 0.22 m and 17 turns is rotated in a

uniform magnetic field of 1.7 T. The coil rotates at a constant frequency

of 2.0 Hz. Determine the maximum value of the emf induced in the coil.

3. A square coil with an area of 0.045 m2 consists of 120 turns of wire. The

coil rotates about a vertical axis at 157 rad/s. The horizontal component

of Earth’s magnetic field at the location of the loop is 2.0 × 10–5 T. Calcu-

late the maximum emf induced in the coil.

4. A maximum emf of 90.4 V is induced in a generator coil rotating with a

frequency of 65 Hz. If the coil has an area of 230 cm2 and rotates in a

magnetic field of 1.2 T, how many turns are in the coil?

Induction in generators

PRACTICE 22B

Choose an equation(s) or situation: Use the maximum emf equation for a

generator. Use the definition of angular frequency to convert f to w. The

maximum current can be obtained from the definition for resistance.

maximum emf = NABw w = 2pf

Imax = maxim

R

um emf

Rearrange the equation(s) to isolate the unknown(s): Substitute the angu-

lar frequency expression into the maximum emf equation.

maximum emf = NAB(2pf )


maximum emf = (8)(0.095 m2)(0.55 T)(2p)(60.0 Hz)

Imax = 1.6

1

×2

1

Ω02 V = 13 A

By expressing the units used in the calculation in terms of equivalent units

(1 T = 1 (V•s/m2) and 1 Hz = 1/s), you can see which terms cancel. In this

case, m2 and s cancel to leave the answer for emf in units of volts.

Imax = 13 A

maximum emf = 1.6 × 102 V CALCULATOR SOLUTION

Because the minimum number of sig-nificant figures for the data is two,the calculator answer, 157.5822875,should be rounded to two digits.


Although the light intensity from a60 W incandescent light bulbappears to be constant, the currentin the bulb fluctuates 60 times eachsecond between −0.7 1 A and 0.7 1 A.The light appears to be steadybecause the fluctuations are toorapid for our eyes to perceive.

807Induction and Alternating Current

Alternating current changes direction at a constant frequency

The output emf of a typical generator has a sinusoidal pattern, as you can see

in the graph in Figure 22-12 on page 805. Note that the emf alternates from

positive to negative. As a result, the output current from the generator

changes its direction at regular intervals. This variety of current is called alter-nating current, or, more commonly, ac.

The rate at which the coil in an ac generator rotates determines the maxi-

mum generated emf. The frequency of the alternating current can differ from

country to country. In the United States, Canada, and Central America, the

frequency of rotation for commercial generators is 60 Hz. This means that the

emf undergoes one full cycle of changing direction 60 times each second. In

the United Kingdom, Europe, and most of Asia and Africa, 50 Hz is used.

(Recall that w = 2p f , where f is the frequency in Hz.)

Resistors can be used in either alternating- or direct-current applications.

A resistor resists the motion of charges regardless of whether they move in

one continuous direction or shift direction periodically. Thus, if the definition

for resistance holds for circuit elements in a dc circuit, it will also hold for the

same circuit elements with alternating currents and emfs.

Effective current and potential difference are measured in ac circuits

An ac circuit consists of combinations of circuit elements and an ac generator

or an ac power supply, which provides the alternating current. As shown

earlier, the emf produced by a typical ac generator is sinusoidal and varies

with time. The induced emf equals the instantaneous ac potential difference,

which is written as ∆v. The quantity for the maximum emf can be written as

the maximum potential difference ∆Vmax, and the emf produced by a genera-

tor can be expressed as follows:

∆v = ∆Vmax(sin wt)

Because all circuits have some resistance, a simple ac circuit can be treated

as an equivalent resistance and an ac source. In a circuit diagram, the ac

source is represented by the symbol , as shown in Figure 22-13.The instantaneous current that changes with the potential difference can

be determined using the definition for resistance. The instantaneous current,

i, is related to maximum current by the following expression:

i = Imax(sin wt)

The rate at which electrical energy is converted to internal energy in the

resistor (the power, P) has the same form as in the case of direct current. The

electrical energy converted to internal energy in a resistor is proportional to

the square of the current and is independent of the direction—or the change

of direction—of the current. However, the energy produced by an alternating

current with a maximum value of Imax is not the same as that produced by a

direct current of the same value. This is because the alternating current is at its

maximum value for only a very brief instant of time during a cycle.

v

Req

ac source∆

Figure 22-13An ac circuit represented sche-matically consists of an ac sourceand an equivalent resistance.

alternating current

an electric current that changesdirection at regular intervals


The important quantity for current in an ac circuit is the rms current. The

rms (or root-mean-square) current is the same as the amount of direct current

that would dissipate the same energy in a resistor as is dissipated by the

instantaneous alternating current over a complete cycle.

Figure 22-14 shows a graph in which instantaneous and rms currents are

compared. Table 22-2 summarizes the notations used in this chapter for these

and other ac quantities.

The equation for the power dissipated in an ac circuit has the same form as

the equation for power dissipated in a dc circuit except that the dc current I is

replaced by the rms current (Irms).

P = (I rms)2R

This equation is identical in form to the one for direct current. However,

the power dissipated in the ac circuit equals half the power dissipated in a dc

circuit when the dc current equals Imax.

P = (I rms)2 R = 1

2(I max)2R

From this equation you may note that the rms current is related to the

maximum value of the alternating current by the following equation:

(I rms)2 =

(Im

2ax)2

Irms = Ima

2x = 0.707 Imax

This equation says that an alternating current with a maximum value of 5 A

produces the same heating effect in a resistor as a direct current of (5/2 ) A,

or about 3.5 A.

Alternating potential differences are also best discussed in terms of rms

potential differences, with the relationship between rms and maximum values

being identical to the one for currents. The rms potential difference, ∆Vrms, is

related to the maximum value of the potential difference, ∆Vmax, as follows:

∆Vrms = ∆

Vm

2ax = 0.707 Vmax

ImaxIrms

Cur

rent

Time

Figure 22-14The rms current is a little morethan two-thirds as large as the maximum current.

rms current

the amount of direct current thatdissipates as much energy in aresistor as an instantaneousalternating current does during acomplete cycle

Table 22-2 Notation used for ac circuits

Potential difference Current

instantaneous values ∆v i

maximum values ∆Vmax Imax

rms values ∆Vrms = ∆

Vm

2ax Irms =

Imax2


The ac potential difference of 120 V measured from an electric outlet is

actually an rms potential difference of 120 V. A quick calculation shows that

such an ac potential difference has a maximum value of about 170 V.

A resistor limits the current in an ac circuit just as it does in a dc circuit. If

the definition of resistance is valid for an ac circuit, the rms potential differ-

ence across a resistor equals the rms current multiplied by the resistance.

Thus, all maximum and rms values can be calculated if only one current or

emf value and the circuit resistance are known.

Because ac ammeters and voltmeters measure rms values, all values of

alternating current and alternating potential difference in this chapter will be

given as rms values unless otherwise noted. The equations for ac circuits have

the same form as those for dc circuits when rms values are used.

SAMPLE PROBLEM 22C

rms currents and potential differences

P R O B L E MA generator with a maximum output emf of 205 V is connected to a 115 Ωresistor. Calculate the rms potential difference. Find the rms currentthrough the resistor. Find the maximum ac current in the circuit.

S O L U T I O NGiven: ∆Vmax = 205 V R = 115 Ω

Unknown: ∆Vrms = ? Irms = ? Imax = ?

Diagram:

Choose an equation(s) or situation: Use the equation

for the rms potential difference to find ∆Vrms.

∆Vrms = 0.707 ∆Vmax

Rearrange the definition for resistance to calculate Irms.

Irms = ∆V

Rrms

Use the equation for rms current to find Imax.

Irms = 0.707 Imax

Rearrange the equation(s) to isolate the unknown(s):Rearrange the equation relating rms current to maximum current so that

maximum current is calculated.

Imax = 0

I

.r

7m

0s

7

1. DEFINE

2. PLAN


R = 115 Ω

Vmax = 205 V

Vrms = ?

Irms = ? Imax = ?

∆∆



∆Vrms = (0.707)(205 V) = 145 V

Irms = 1

1

1

4

5

5

ΩV

= 1.26 A

Imax = 1

0

.

.

2

7

6

07

A = 1.78 A

The rms values for potential difference and current are a little more than

two-thirds the maximum values, as expected.

∆Vrms = 145 V

Irms = 1.26 A

Imax = 1.78 A

4. EVALUATE

3. CALCULATE

CALCULATOR SOLUTION

Because the minimum number of significant figures for the data is three,the calculator solution for the rmspotential difference, 144.935, shouldbe rounded to three digits.

1. What is the rms current in a light bulb that has a resistance of 25 Ω and

an rms potential difference of 120 V? What are the maximum values for

current and potential difference?

2. The current in an ac circuit is measured with an ammeter. The meter

gives a reading of 5.5 A. Calculate the maximum ac current.

3. A toaster is plugged into a source of alternating potential difference with

an rms value of 110 V. The heating element is designed to convey a cur-

rent with a maximum value of 10.5 A. Find the following:

a. the rms current in the heating element

b. the resistance of the heating element

4. An audio amplifier provides an alternating rms potential difference of

15.0 V. A loudspeaker connected to the amplifier has a resistance of

10.4 Ω. What is the rms current in the speaker? What are the maximum

values of the current and the potential difference?

5. An ac generator has a maximum potential difference output of 155 V.

a. Find the rms potential difference output.

b. Find the rms current in the circuit when the generator is connected to

a 53 Ω resistor.

6. The largest potential difference that can be placed across a certain capaci-

tor at any instant is 451 V. What is the largest rms potential difference

that can be placed across the capacitor without damaging it?

rms currents and potential differences

PRACTICE 22C


Alternating current can be converted to direct current

The conducting loop in an ac generator must be free to rotate through the

magnetic field. Yet it must also be part of an electric circuit at all times. To

accomplish this, the ends of the loop are connected to conducting rings, called

slip rings, that rotate with the loop. Connections to the external circuit are

made by stationary graphite strips, called brushes, that make continuous con-

tact with the slip rings. Because the current changes direction in the loop, the

output current through the brushes alternates direction as well.

By varying this arrangement slightly, an ac generator can be converted to a

dc generator. Note in Figure 22-15 that the components of a dc generator are

essentially the same as those of the ac generator except that the contacts to the

rotating loop are made by a single split slip ring, called a commutator.

At the point in the loop’s rotation when the current has dropped to zero

and is about to change direction, each half of the commutator comes into

contact with the brush that was previously in contact with the other half of

the commutator. The reversed current in the loop changes directions again so

that the output current has the same direction as it originally had, although it

still changes from a maximum value to zero. A plot of this pulsating direct

current is shown in Figure 22-16.A steady direct current can be produced by using many loops and commu-

tators distributed around the rotation axis of the dc generator. The sinusoidal

currents from each loop overlap. The superposition of the currents produces a

direct current output that is almost entirely free of fluctuations.

AC Generator DC Generator

Slip rings

Brush

Brush

N

S

N

S

Brush

BrushCommutator

Figure 22-15A simple dc generator (shown on the right) employs the samedesign as an ac generator (shown on the left). A split slip ringconverts alternating current to direct current.

Time

Cur

rent

Figure 22-16The output current for a dc genera-tor is a sine wave with the negativeparts of the curve made positive.

TOPIC: Electrical safetyGO TO: www.scilinks.orgsciLINKS CODE: HF2223

NSTA


MOTORS

Motors are devices that convert electrical energy to mechanical energy.

Instead of a current being generated by a rotating loop in a magnetic field, a

current is supplied to the loop by an emf source and the magnetic force on the

current loop causes it to rotate (see Figure 22-17).

A motor is almost identical in construction to a dc generator. The coil of

wire is mounted on a rotating shaft and is positioned between the poles of a

magnet. Brushes make contact with a commutator, which alternates the cur-

rent in the coil. This alternation of the current causes the magnetic field pro-

duced by the current to regularly reverse and thus always be repelled by the

fixed magnetic field. Thus, the coil and the shaft are kept in continuous rota-

tional motion.

A motor can perform useful mechanical work when a shaft connected to

its rotating coil is attached to some external device. As the coil in the motor

rotates, however, the changing normal component of the magnetic field

through it induces an emf that acts to reduce the current in the coil. If this

were not the case, Lenz’s law would be violated. This induced emf is called

the back emf.The back emf increases in magnitude as the magnetic field changes at a

higher rate. In other words, the faster the coil rotates, the greater the back

emf becomes. The potential difference available to supply current to the

motor equals the difference between the applied potential difference and the

back emf. Consequently, the current in the coil is also reduced because of

the presence of back emf. The faster the motor turns, the smaller the net

potential difference across the motor, and the smaller the net current in the

coil, becomes.

DC Motor

Brush

Brush

CommutatorN

S

emf

+

Figure 22-17In a motor, the current in the coil interactswith the magnetic field, causing the coil and theshaft on which the coil is mounted to turn.

back emf

the emf induced in a motor’s coilthat tends to reduce the currentpowering the motor


A person can receive an electric shock by touchinga conducting or “live” wire while in contact with a

lower electric potential, or ground. The groundcontact might be made by touching a waterpipe (which is normally at zero potential)or by standing on the floor with wet feet

(because impure water is a good conductor).

Electric shock can result in fatal burns or cancause the muscles of vital organs, such as the

heart, to malfunction. The degree of damage tothe body depends on the magnitude of the current,

the length of time it acts, and the part of the bodythrough which it passes. A current of

100 milliamps (mA) can be fatal. If thecurrent is larger than about 10 mA,the hand muscles contract andthe person may be unable to let go of the wire.

To prevent electrocution, any wiresdesigned to have such currents in them

are wrapped in insulation, usually plastic orrubber. However, with frequent use, electric

cords can fray, exposing some of the conductors. Tofurther prevent electrocution in these and other situ-ations in which electrical contact can be made,devicescalled a Ground Fault Circuit Interrupter (GFCI) anda Ground Fault Interrupter (GFI) are mounted in elec-trical outlets and individual appliances.

GFCIs and GFIs provide protection by comparingthe current in one side of the electrical outlet sock-et to the current in the other socket. If there is evena 5 mA difference, the interrupter opens the circuitin a few milliseconds (thousandths of a second). Ifyou accidentally touch a bare wire and create analternate conducting path through you to ground,the device detects this redirection of current andyou get only a small shock or tingle.

Despite these safety devices, you can still be elec-trocuted. Never use electrical appliances nearwater or with wet hands. Use a battery-poweredradio near water because batteries cannot supplyenough current to harm you. It is also a good ideato replace old outlets with GFCI-equipped units orto install GFI-equipped circuit breakers.

Avoiding Electrocution

Section Review

1. A loop with an area of 0.33 m2 is rotating at 281 rad/s with its axis of

rotation perpendicular to a uniform magnetic field. The magnetic field

has a strength of 0.035 T. If the loop contains 37 turns of wire, what is

the maximum potential difference induced in the loop?

2. A generator develops a maximum emf of 2.8 V. If the generator coil has

25 turns of wire, a cross-sectional area of 36 cm2, and rotates with a con-

stant frequency of 60.0 Hz, what is the strength of the magnetic field in

which the coil rotates?

3. What is the purpose of a commutator in a motor? Explain what would

happen if a commutator were not used in a motor.

4. Physics in Action The rms current produced by a single coil of an

electric guitar is 0.025 mA. How large is the maximum instantaneous

current? If the coil’s resistance is 4.3 kΩ, what are the rms and maximum

potential differences produced by the coil?


22-3Inductance

MUTUAL INDUCTANCE

The basic principle of electromagnetic induction was first demonstrated by

Michael Faraday. His experimental apparatus, which resembled the arrange-

ment shown in Figure 22-18, used a coil connected to a switch and a battery

instead of a magnet to produce a magnetic field. This coil is called the primary

coil, and its circuit is called the primary circuit. The magnetic field is strength-

ened by the magnetic properties of the iron ring around which the primary

coil is wrapped.

A second coil is wrapped around another part of the iron ring and is con-

nected to a galvanometer. An emf is induced in this coil, called the secondary

coil, when the magnetic field of the primary coil is changed. When the switch in

the primary circuit is closed, the galvanometer in the secondary circuit deflects

in one direction and then returns to zero. When the switch is opened, the gal-

vanometer deflects in the opposite direction and again returns to zero. When

there is a steady current in the primary circuit, the galvanometer reads zero.

The magnitude of this emf is predicted by Faraday’s law of induction.

However, Faraday’s law can be rewritten so that the induced emf is propor-

tional to the changing current in the primary coil. This can be done because of

the direct proportionality between the magnetic field produced by a current

in a coil, or solenoid, and the current itself. The form of Faraday’s law in terms

of changing primary current is as follows:

emf = −N∆[AB

∆(c

t

os q)] = −M

∆∆

I

t

The constant, M, is called the mutual inductance of the two-coil system. The

mutual inductance depends on the geometrical properties of the coils and


• Describe how mutualinduction occurs in circuits.

• Calculate the potentialdifference from a step-up orstep-down transformer.

• Describe how self-inductionoccurs in an electric circuit.

Ironring

Primarycoil

Secondarycoil

Galvanometer

+

100

200

300

400

500

600

700

800

900

100

200

300

400

500

600

700

800

900

1000

0 +

Figure 22-18Faraday’s electromagnetic-inductionexperiment used a changing currentin one circuit to induce a current inanother circuit.

mutual inductance

a measure of the ability of one cir-cuit carrying a changing current toinduce an emf in a nearby circuit


their orientation to each other. Because these properties are kept constant, it

follows that a changing current in the secondary coil can also induce an emf in

the primary circuit. The equation holds as long as the coils remain unchanged

with respect to each other, so that the mutual inductance is constant.

By changing the number of turns of wire in the secondary coil, the induced

emf in the secondary circuit can be changed. This arrangement is the basis of

an extremely useful electrical device: the transformer.

TRANSFORMERS

It is often desirable or necessary to change a small ac potential difference to a

larger one or to change a large potential difference to a smaller one. The

device that makes these conversions possible is the transformer.In its simplest form, an ac transformer consists of two coils of wire wound

around a core of soft iron, like the apparatus for the Faraday experiment. The

coil on the left in Figure 22-19 has N1 turns and is connected to the input ac

potential difference source. This coil is called the primary winding, or the pri-

mary. The coil on the right, which is connected to a resistor R and consists of

N2 turns, is the secondary. As in Faraday’s experiment, the iron core provides a

medium for nearly all magnetic field lines passing through the two coils.

Because the strength of the magnetic field in the iron core and the cross-

sectional area of the core are the same for both the primary and secondary

windings, the potential differences across the two windings differ only

because of the different number of turns of wire for each. The ac potential dif-

ference that gives rise to the changing magnetic field in the primary is related

to that changing field by Faraday’s law of induction.

∆V1 = −N1 ∆(AB

∆c

t

os q)

Similarly, the induced potential difference (emf) across the secondary coil is

∆V2 = −N2 ∆(AB

∆c

t

os q)

Taking the ratio of ∆V1 to ∆V2 causes all terms on the right side of both equa-

tions except for N1 and N2 to cancel. This result is the transformer equation.

TRANSFORMER EQUATION

∆V2 = N

N2

1 ∆V1

potential difference in secondary =

number of turns in secondarynumber of turns in primary

potential difference in primary

Soft iron core

Primary(input)

∆V1 ∆V2N1 N2 R

Secondary(output)

Figure 22-19A transformer uses the alternatingcurrent in the primary circuit toinduce an alternating current in thesecondary circuit.

transformer

a device that changes one acpotential difference to a differentac potential difference

TOPIC: TransformersGO TO: www.scilinks.orgsciLINKS CODE: HF2224

NSTA


Another way to express this equation is to equate the ratio of the potential dif-

ferences to the ratio of the number of turns.

∆∆

V

V2

1 =

N

N2

1

When N2 is greater than N1, the secondary potential difference is greater

than that of the primary, and the transformer is called a step-up transformer.

When N2 is less than N1, the secondary potential difference is less than that

of the primary, and the transformer is called a step-down transformer.

It may seem that a transformer provides something for nothing. For

example, a step-up transformer can change an input potential difference

from 10 V to 100 V. However, the power input at the primary must equal the

power output at the secondary. An increase in potential difference at the sec-

ondary means that there must be a proportional decrease in current. If the

potential difference at the secondary is 10 times that at the primary, then the

current at the secondary is reduced by a factor of 10.

Real transformers are not perfectly efficient

The transformer equation assumes that there are no

power losses between the transformer’s primary and its

secondary. Real transformers typically have efficiencies

ranging from 90 percent to 99 percent. Power losses

occur because of the small currents induced by changing

magnetic fields in the iron core of the transformer and

because of resistance in the wires of the windings.

When electric power is transmitted over large dis-

tances, it is economical to use a high potential difference

and a low current. This is because the power lost to resis-

tive heating in the transmission lines varies as I2R. By

reducing the current by a factor of 10, the power loss is

reduced by a factor of 100. In practice, potential differ-

ence is stepped up to around 230 000 V at the generating

station, then stepped down to 20 000 V at a regional dis-

tribution station, and finally stepped down to 120 V at the

customer’s utility pole. The high potential difference in

long-distance transmission lines makes them especially

dangerous when they are knocked down by high winds.

Coils in gasoline engines are transformers

An automobile ignition system uses a transformer, or

ignition coil, to convert the car battery’s 12 dc volts to a

potential difference that is large enough to cause sparking

between the gaps of the spark plugs. The diagram in Fig-ure 22-20 shows the type of ignition system that has been

–+

12 V battery

Step-up transfomer (ignition coil)

Spark plug

Ignitionswitch

Computer

Crank anglesensor

Figure 22-20The transformer in an automobile engine raises the potential difference across the gap in a spark plug so thatsparking occurs.


used in automobiles since 1990. In this arrangement, each cylinder has its own

transformer, and a photoelectric detector called a crank angle sensor determines

from the crankshaft’s position which cylinder’s contents are at maximum com-

pression. Upon receiving this signal, the computer closes the primary circuit to

the cylinder’s coil, causing the current in the primary to rapidly increase and

the magnetic field in the transformer to change. This, in turn, induces a poten-

tial difference of about 20 000 V across the secondary.

SAMPLE PROBLEM 22D

Transformers

P R O B L E MA step-up transformer is used on a 120 V line to provide a potential differ-ence of 2400 V. If the primary has 75 turns, how many turns must the sec-ondary have?

S O L U T I O NGiven: ∆V1 = 120 V ∆V2 = 2400 V N1 = 75 turns

Unknown: N2 = ?

Diagram:

Choose an equation(s) or situation: Use the transformer equation.

∆V2 = N

N2

1 ∆V1

Rearrange the equation(s) to isolate the unknown(s):

N2 = ∆∆

V

V2

1 N1


N2 = 2142000V

V 75 turns = 1500 turns

The greater number of turns in the secondary accounts for the increase in

the potential difference in the secondary. The step-up factor for the trans-

former is 20:1.

N2 = 1500 turns

∆V2 = 2400 V∆V1 = 120 V

N1 =75 turns N2 = ?

1. DEFINE

2. PLAN

4. EVALUATE

3. CALCULATE

INTERACTIV

E•

T U T O RPHYSICSPHYSICS

Module 20“Induction and Transformers”provides an interactive lessonwith guided problem-solvingpractice to teach you aboutinduction and transformers.


SELF-INDUCTION

Consider a circuit consisting of a switch, a resistor, and a source of emf, as

shown in Figure 22-21. When the switch is closed, the current does not imme-

diately change from zero to its maximum value, Imax. Instead, the current

increases with time, and the magnetic field through the loop due to this cur-

rent also increases. The increasing field induces an emf in the circuit to oppose

the change in magnetic field, according to Lenz’s law, and so the polarity of the

induced emf must oppose the direction of the original current. The induced

emf is in the direction indicated by the dashed battery. The net potential differ-

ence across the resistor is the emf of the battery minus the induced emf.

Chapter 22818

1. A step-down transformer providing electricity for a residential neighbor-

hood has exactly 2680 turns in its primary. When the potential difference

across the primary is 5850 V, the potential difference at the secondary is

120 V. How many turns are in the secondary?

2. A step-up transformer used in an automobile has a potential difference

across the primary of 12 V and a potential difference across the secondary

of 2.0 × 104 V. If the number of turns in the primary is 21, what is the

number of turns in the secondary?

3. A step-up transformer for long-range transmission of electric power is used

to create a potential difference of 119 340 V across the secondary. If the

potential difference across the primary is 117 V and the number of turns in

the secondary is 25 500, what is the number of turns in the primary?

4. A potential difference of 0.750 V is needed to provide a large current for

arc welding. If the potential difference across the primary of a step-down

transformer is 117 V, what is the ratio of the number of turns of wire on

the primary to the number of turns on the secondary?

5. A television picture tube requires a high potential difference, which in

older models is provided by a step-up transformer. The transformer has

12 turns in its primary and 2550 turns in its secondary. If 120 V is placed

across the primary, what is the output potential difference?

6. A step-down transformer has 525 turns in its secondary and 12 500 turns

in its primary. If the potential difference across the primary is 3510 V,

what is the potential difference across the secondary?

Transformers

PRACTICE 22D

BSwitch

Inducedemf

emf

I

R

Figure 22-21The changing magnetic field that isproduced by changing currentinduces an emf that opposes theapplied emf.


As the current increases, the rate of increase lessens and the induced emf

decreases, as shown in Figure 22-22. The decrease in the induced emf results

in a gradual increase in the current. Similarly, when the switch is opened,

the current gradually decreases to zero. This effect is called self-inductionbecause the changing magnetic field through the circuit arises from the cur-

rent in the circuit itself. The induced emf in this case is called self-induced emf.

An example of self-induction is seen in a coil wound on a cylindrical iron

core. (A practical device would have several hundred turns.) When the cur-

rent is in the direction shown in Figure 22-23(a), a magnetic field forms

inside the coil. As the current changes with time, the field through the coil

changes and induces an emf in the coil.

Lenz’s law indicates that this induced emf opposes the change in the current.

For increasing current, the induced emf is as pictured in Figure 22-23(b), and

for decreasing current, the induced emf is as shown in Figure 22-23(c).Faraday’s law of induction takes the same form for self-induction as it does

for mutual induction except that the symbol for mutual inductance, M, is

replaced with the symbol for self-inductance, L.

emf = −N∆[AB

∆(c

t

os q)] = −L

∆∆

I

t

The self-inductance of a coil depends on the number of turns of wire in the

coil, the coil’s cross-sectional area, and other geometric factors. The SI unit of

inductance is the henry (H), which is equal to 1 volt-second per ampere.

Cur

rent

Time

Figure 22-22The self-induced emf reduces therate of increase of the current inthe circuit immediately after the circuit is closed.

I

B

Lenz’s law emffor increasing I

Lenz’s law emffor decreasing I

(a) (b) (c)

+ −− +

Figure 22-23The polarity of the self-induced emfis such that it can produce a currentwhose direction is opposite thechange in the current in the coil.

Section Review

1. The centers of two circular loops are separated by a fixed distance. For

what relative orientation of the loops will their mutual inductance be a

maximum? For what orientation will it be a minimum?

2. A step-up transformer has exactly 50 turns in its primary and exactly

7000 turns in its secondary. If the potential difference across the primary

is 120 V, what is the potential difference across the secondary?

3. Does the self-inductance (L) of a coil depend on the current in the coil?

self-induction

the process by which a changingcurrent in a circuit induces anemf in that same circuit


CHAPTER 22Summary

KEY IDEAS

Section 22-1 Induced current• Changing the magnetic field strength near a conductor induces an emf.

• The direction of an induced current in a circuit is such that its magnetic

field opposes the change in the applied magnetic field.

• Induced emf can be calculated using

Faraday’s law of magnetic induction.

Section 22-2 Alternating current, generators, and motors• Generators use induction to convert mechanical energy into electrical energy.

• Alternating current is measured in terms of rms current.

• Motors use an arrangement similar to that of generators to convert electri-

cal energy into mechanical energy.

Section 22-3 Inductance• Mutual inductance involves the induction of a current in one circuit by

means of a changing current in a nearby circuit.

• Transformers change the potential difference of an alternating current.

• Self-induction occurs when the changing current in a circuit induces an

emf in the same circuit.

KEY TERMS

alternating current (p. 807)

back emf (p. 812)

electromagnetic induction(p. 794)

generator (p. 803)

mutual inductance (p. 814)

rms current (p. 808)

self-induction (p. 819)

transformer (p. 815)

emf = −N ∆[AB

∆(c

t

os q)]

Variable symbols

Quantities Units

N number of turns (unitless)

∆Vmax maximum potential difference V volt

∆Vrms rms potential difference V volt

Imax maximum current A ampere

Irms rms current A ampere

M mutual inductance H henry = vo

a

lt

m

•s

p

e

e

c

r

o

e

nd

L self-inductance H henry = vo

a

lt

m

•s

p

e

e

c

r

o

e

nd

Diagram symbols

Induced emf

ac generator/alternating currentemf source

+−


INDUCED CURRENT

Review questions

1. Suppose you have two circuits. One consists of anelectromagnet, a dc emf source, and a variable resis-tor that permits you to control the strength of themagnetic field. In the second circuit, you have a coilof wire and a galvanometer. List three ways that youcan induce a current in the second circuit.

2. Explain how Lenz’s law allows you to determine thedirection of an induced current.

3. What four factors affect the magnitude of the inducedemf in a coil of wire?

4. If you have a fixed magnetic field and a length ofwire, how can you increase the induced emf acrossthe ends of the wire?

Conceptual questions

5. Rapidly inserting the north pole of a bar magnetinto a coil of wire connected to a galvanometercauses the needle of the galvanometer to deflect tothe right. What will happen to the needle if you dothe following?

a. pull the magnet out of the coilb. let the magnet sit at rest in the coilc. thrust the south end of the magnet into the coil

6. Explain how Lenz’s law illustrates the principle ofenergy conservation.

7. Does dropping a bar magnet down a long coppertube induce a current in the tube? If so, how mustthe magnet be oriented with respect to the tube?

8. Two bar magnets are placed side by side so thatthe north pole of one magnet is next to the southpole of the other magnet. If these magnets arethen pushed toward a coil of wire, would youexpect an emf to be induced in the coil? Explainyour answer.

9. An electromagnet is placed next to a coil of wire in thearrangement shown in Figure 22-24. According toLenz’s law, what will be the direction of the inducedcurrent in the resistor R in the following cases?

a. the magnetic field suddenly decreases afterthe switch is opened

b. the coil is moved closer to the electromagnet

Practice problems

10. A flexible loop of conducting wire has a radius of0.12 m and is perpendicular to a uniform magnetic fieldwith a strength of 0.15 T, as in Figure 22-25(a). Theloop is grasped at opposite ends and stretched until itcloses to an area of 3 × 10−3 m2, as in Figure 22-25(b).If it takes 0.20 s to close the loop, find the magnitude ofthe average emf induced in the loop during this time.(See Sample Problem 22A.)

11. A rectangular coil 0.055 m by 0.085 m is positionedso that its cross-sectional area is perpendicular tothe direction of a magnetic field, B. If the coil has 75turns and a total resistance of 8.7 Ω and the fielddecreases at a rate of 3.0 T/s, what is the magnitudeof the induced current in the coil?(See Sample Problem 22A.)

Figure 22-25

(a) (b)

Figure 22-24

CHAPTER 22Review and Assess

Electromagnet

emf

Coil

Switch− +

R

B


22. A bar magnet is attached perpendicular to a rotat-ing shaft. It is then placed in the center of a coil ofwire. In which of the arrangements shown in Figure 22-26 could this device be used as an electricgenerator? Explain your choice.

Practice problems

23. A generator can be made using the component ofEarth’s magnetic field that is parallel to Earth’s sur-face. A 112-turn square wire coil with an area of4.41 × 10−2 m2 is mounted on a shaft so that thecross-sectional area of the coil is perpendicular to the ground. The shaft then rotates with a fre-quency of 25.0 Hz. The horizontal component ofEarth’s magnetic field at the location of the loop is5.00 × 10−5 T. Calculate the maximum emf inducedin the coil by Earth’s magnetic field.(See Sample Problem 22B.)

24. An ac generator consists of 45 turns of wire withan area of 0.12 m2. The loop rotates in a magne-tic field of 0.118 T at a constant frequency of60.0 Hz. The generator is connected across a cir-cuit load with a total resistance of 35 Ω. Find thefollowing:

a. the maximum induced emfb. the maximum induced current

(See Sample Problem 22B.)

Figure 22-26

(a)

(b) (c)

R

R SN S

N

R

N

S

12. A 52-turn coil with an area of 5.5 × 10−3 m2 isdropped from a position where B = 0.00 T to a newposition where B = 0.55 T. If the displacementoccurs in 0.25 s and the area of the coil is perpen-dicular to the magnetic field lines, what is theresulting average emf induced in the coil?(See Sample Problem 22A.)

ALTERNATING CURRENT,GENERATORS, AND MOTORS

Review questions

13. List the essential components of an electric genera-tor, and explain the role of each component in gen-erating an alternating emf.

14. A student turns the handle of a small generatorattached to a lamp socket containing a 15 W bulb.The bulb barely glows. What should the student doto make the bulb glow more brightly?

15. What is meant by the term frequency in reference toan alternating current?

16. How can an ac generator be converted to a dc gen-erator? Explain your answer.

17. In what way is the output of a dc generator differentfrom the output of a battery?

18. What is meant by back emf? How is it induced in anelectric motor?


19. When the plane of a rotating loop of wire is parallelto the magnetic field lines, the number of lines pass-ing through the loop is zero. Why is the current at amaximum at this point in the loop’s rotation?

20. The faster the coil of loops, or armature, of an acgenerator rotates, the harder it is to turn the arma-ture. Use Lenz’s law to explain why this happens.

21. Voltmeters and ammeters that measure ac quanti-ties measure the rms values of emf and current,respectively. Why would this be preferred to mea-suring the maximum emf or current? (Hint: Thinkabout what a meter reading would look like if acquantities other than rms values were measured.)


25. The rms potential difference across high-voltagetransmission lines in Great Britain is 220 000 V.What is the maximum potential difference?(See Sample Problem 22C.)

26. The maximum potential difference across certainheavy-duty appliances is 340 V. If the total resistanceof an appliance is 120 Ω, calculate the following:

a. the rms potential differenceb. the rms current

(See Sample Problem 22C.)

27. The maximum current that can pass through a lightbulb filament is 0.909 A when its resistance is 182 Ω.

a. What is the rms current conducted by the fila-ment of the bulb?

b. What is the rms potential difference across thebulb’s filament?

c. How much power does the light bulb use?(See Sample Problem 22C.)

28. A 996 W hair dryer is designed to carry a maximumcurrent of 11.8 A.

a. How large is the rms current in the hair dryer? b. What is the rms potential difference across the

hair dryer?(See Sample Problem 22C.)

INDUCTANCE

Review questions

29. Describe how mutual induction occurs.

30. What is the difference between a step-up transformerand a step-down transformer?

31. Does a step-up transformer increase power? Explainyour answer.

32. Describe how self-induction occurs.


33. In many transformers, the wire around one windingis thicker, and therefore has lower resistance, than thewire around the other winding. If the thicker wire iswrapped around the secondary winding, is the devicea step-up or a step-down transformer? Explain.

34. Would a transformer work with pulsating directcurrent? Explain your answer.

Practice problems

35. A transformer is used to convert 120 V to 9.0 V foruse in a portable CD player. If the primary, which isconnected to the outlet, has 640 turns, how manyturns does the secondary have?(See Sample Problem 22D.)

36. A transformer is used to convert 120 V to 6.3 V inorder to power a toy electric train. If there are 210turns in the primary, how many turns should therebe in the secondary?(See Sample Problem 22D.)

MIXED REVIEW PROBLEMS

37. A student attempts to make a simple generator bypassing a single loop of wire between the poles of ahorseshoe magnet with a 2.5 × 10−2 T field. Thearea of the loop is 7.54 × 10−3 m2 and is moved per-pendicular to the magnetic field lines. In whattime interval will the student have to move theloop out of the magnetic field in order to inducean emf of 1.5 V? Is this a practical generator?

38. The same student in item 37 modifies the simplegenerator by wrapping a much longer piece ofwire around a cylinder with about one-fourth thearea of the original loop (1.886 × 10−3 m2). Againusing a uniform magnetic field with a strength of2.5 × 10−2 T, the student finds that by removing thecoil perpendicular to the magnetic field lines during0.25 s, an emf of 149 mV can be induced. Howmany turns of wire are wrapped around the coil?

39. A coil of 325 turns and an area of 19.5 × 10−4 m2 isremoved from a uniform magnetic field at an angleof 45° in 1.25 s. If the induced emf is 15 mV, what isthe magnetic field’s strength?

40. A transformer has 22 turns of wire in its primaryand 88 turns in its secondary.

a. Is this a step-up or step-down transformer?b. If 110 V ac is applied to the primary, what is

the output potential difference?

41. The potential difference in the lines that carry elec-tric power to homes is typically 20.0 kV. What is theratio of the turns in the primary to the turns in thesecondary of the transformer if the output potentialdifference is 117 V?


42. A bolt of lightning, such as the one shown on the left inFigure 22-27, behaves like a vertical wire conductingelectric current.As a result, it produces a magnetic fieldwhose strength varies with the distance from the light-ning. A 105-turn circular coil is oriented perpendicularto the magnetic field, as shown on the right in Figure22-27. The coil has a radius of 0.833 m. If the magneticfield at the coil drops from 4.72 × 10–3 T to 0.00 T in10.5 ms, what is the average emf induced in the coil?

First make sure your calculator is in radian mode

by pressing m ∂ ∂ ¬.

Execute “Chap22” on the p menu and press

e to begin the program. Enter the values for the

frequency and maximum current (shown below),

pressing e after each value.

The calculator will provide graphs of the instan-

taneous current and root-mean-square current ver-

sus time in various ac circuits. (If the graphs are not

visible, press w and change the settings for the

graph window, then press g.)

Press ◊ and use the arrow keys to trace along

the curve. The x value corresponds to the time in

seconds, and the y value of the sine curve corre-

sponds to the instantaneous current in amperes.

The y value of the straight line gives the root-mean-

square value for the current in amperes. Use the ¨

and ∂ keys to toggle between the two graphs.

Determine the instantaneous and root-mean-

square values for the current in the following

ac circuits:

b. an ac circuit with a maximum current of 0.250 A

and a frequency of 60.0 Hz at t = 3.27 s

c. the same circuit at t = 5.14 s

d. an ac circuit with a maximum current of 0.110 A

and a frequency of 110.0 Hz at t = 2.50 s

e. the same circuit at t = 4.24 s

f. Would you expect the graph of Y2 to be above the

maximum instantaneous current (Y1)?

Press @ q to stop graphing. Press e to

input new values or ı to end the program.

Graphing calculatorsRefer to Appendix B for instructions on download-

ing programs for your calculator. The program

“Chap22” allows you to analyze graphs of instanta-

neous and root-mean-square currents versus time

in various ac circuits.

Instantaneous current and rms current, as you

learned earlier in this chapter, are related to the

maximum current in an ac circuit by the following

equations:

i = Imax (sin wt) and Irms =

The program “Chap22” stored on your graphing

calculator makes use of the equations for instanta-

neous and rms currents. Once the “Chap22” pro-

gram is executed, your calculator will ask for the

frequency and maximum current. The graphing cal-

culator will use the following equations to create

graphs of the instantaneous current (Y1) versus the

time (X) and the rms current (Y2) versus the time

(X). The relationships in these equations are the

same as those in the equations shown above.

Y1 = I sin (2πFX) and Y2 = I/√

2

a. What value in the calculator equations equals the

amount of direct current that would dissipate the

same energy in a resistor as is dissipated by the

instantaneous ac current during a full cycle?

Imax√2

0.833

m

Figure 22-27


43. A generator supplies 5.0 × 103 kW of power. The out-put potential difference is 4500 V before it is steppedup to 510 kV. The electricity travels 410 mi (6.44 ×105 m) through a transmission line that has a resis-tance per unit length of 4.5 × 10−4 Ω/m.

a. How much power is lost through transmissionof the electrical energy along the line?

b. How much power would be lost throughtransmission if the generator’s output poten-tial difference were not stepped up? What doesthis answer tell you about the role of largepotential differences in power transmission?

44. The alternating potential difference of a generator isrepresented by the equation emf = (245 V) sin 560t,where emf is in volts and t is in seconds. Use thesevalues to find the frequency of the potential differ-ence and the maximum potential difference outputof the source.

45. A pair of adjacent coils has a mutual inductance of1.06 H. Determine the average emf induced in thesecondary circuit when the current in the primarycircuit changes from 0 A to 9.50 A in a time intervalof 0.0336 s.

Induction and Alternating Current 825

Performance assessment1. Identify the chain of electromagnetic energy trans-

formations involved in making the blades of a ceiling

fan spin. Include the fan’s motor, the transformers

bringing electricity to the house, and the turbines

generating the electricity.

2. Two identical magnets are dropped simultaneously

from the same point. One of them passes through a

coil of wire in a closed circuit. Predict whether the

two magnets will hit the ground at the same time.

Explain your reasoning, then plan an experiment to

test which of the following variables measurably

affect how long each magnet takes to fall: magnetic

strength, coil cross-sectional area, and the number

of loops the coil has. What measurements will you

make? What are the limits of precision in your mea-

surements? If your teacher approves your plan,

obtain the necessary materials and perform the

experiments. Report your results to the class,

describing how you made your measurements, what

you concluded, and what additional questions need

to be investigated.

3. What do adapters do to potential difference, current,

frequency, and power? Examine the input/output

information on several adapters to find out. Do they

contain step-up or step-down transformers? How

does the output current compare to the input? What

happens to the frequency? What percentage of the

energy do they transfer? What are they used for?

Portfolio projects4. Research the debate between the proponents of

alternating current and those who favored direct

current in the 1880–1890s. How were Thomas

Edison and George Westinghouse involved in the

controversy? What advantages and disadvantages

did each side claim? What uses of electricity were

anticipated? What kind of current was finally gener-

ated in the Niagara Falls hydroelectric plant? Had

you been in a position to fund these projects at that

time, which projects would you have funded? Pre-

pare your arguments to re-enact a meeting of busi-

nesspeople in Buffalo in 1887.

5. Research the history of telecommunication. Who

invented the telegraph? Who patented it in Eng-

land? Who patented it in the United States?

Research the contributions of Charles Wheatstone,

Joseph Henry, and Samuel Morse. How did each of

these men deal with issues of fame, wealth, and

credit to other people’s ideas? Write a summary of

your findings, and prepare a class discussion about

the effect patents and copyrights have had on mod-

ern technology.

Alternative Assessment

Chapter 22826

ELECTROMAGNETIC INDUCTIONIn this laboratory, you will use a magnet, a conductor, and a galvanometer to

explore electromagnets and the principle of self-induction.

PREPARATION

1. Read the entire lab, and plan what measurements you will take.

2. In your lab notebook, prepare an observation table with three wide

columns. Label the columns Sketch of setup, Experiment, and Observation.

For each part of the lab, you will sketch the apparatus and label the poles

of the magnet, write a brief description, and record your observations.

PROCEDURE

Induction with a permanent magnet

3. Connect the ends of the smaller coil to the galvanometer. Hold the magnet

still and move the coil quickly over the north pole of the magnet, as shown

in Figure 22-28. Remove the coil quickly. Observe the galvanometer.

4. Repeat, moving the coil more slowly. Observe the galvanometer.

5. Turn the magnet over, and repeat steps 3 and 4, moving the coil over

the south pole of the magnet. Observe the galvanometer.

6. Hold the coil stationary and quickly move the north pole of the magnet

in and out of the coil. Repeat slowly. Turn the magnet, and repeat both

quickly and slowly with the south pole. Observe the galvanometer.

CHAPTER 22Laboratory Exercise

OBJECTIVES

•Use a galvanometer to detect an inducedcurrent.

•Determine the relation-ship between the mag-netic field of a magnetand the current inducedin a conductor.

•Determine what factorsaffect the direction andmagnitude of an inducedcurrent.

MATERIALS LIST galvanometer insulated connecting wires momentary contact switch pair of bar magnets or a large

horseshoe magnet power supply rheostat (10 Ω) or

potentiometer student primary and sec-

ondary coil set with iron core

SAFETY

• Never close a circuit until it has been approved by your teacher. Neverrewire or adjust any element of a closed circuit. Never work with elec-tricity near water; be sure the floor and all work surfaces are dry.

• If the pointer on any kind of meter moves off scale, open the circuitimmediately by opening the switch.

• Do not attempt this exercise with any batteries, electrical devices, ormagnets other than those provided by your teacher for this purpose.


827Induction and Alternating Current

Induction with an electromagnet

7. Connect the larger coil to the galvanometer. Connect the small coil in

series with a switch, battery, and rheostat. Slip the smaller coil inside the

larger coil, so that the arrangement resembles that shown in Figure 22-29.Close the switch. Adjust the rheostat so that the galvanometer reading reg-

isters on the scale. Observe the galvanometer.

8. Open the switch to interrupt the current in the small coil. Observe the

galvanometer.

9. Close the switch again, and open it after a few seconds. Observe the

galvanometer.

10. Adjust the rheostat to increase the current in the small coil. Close the

switch, and observe the galvanometer.

11. Decrease the current in the circuit, and observe the galvanometer. Open

the switch.

12. Reverse the direction of the current by reversing the battery connections.

Close the switch, and observe the galvanometer.

13. Place an iron rod inside the small coil. Open and close the switch while

observing the galvanometer. Record all observations in your notebook.

14. Clean up your work area. Put equipment away safely so that it is ready to

be used again.

ANALYSIS AND INTERPRETATION

Calculations and data analysis

1. Organizing information Based on your observations from the first

part of the lab, did the speed of the motion have any effect on the

galvanometer?

2. Organizing results In the first part of the lab, did it make any difference

whether the coil or the magnet moved? Explain why or why not.

Conclusions

3. Applying ideas Explain what the galvanometer readings revealed to

you about the magnet and the wire coil.

4. Evaluating results Based on your observations, what conditions are

required to induce a current in a wire?

5. Evaluating results Based on your observations, what factors influence

the direction and magnitude of the induced current?

Figure 22-28Step 3: Connect the coil to thegalvanometer. Holding the magnetstill, move the coil over the magnetquickly.Step 4: Holding the magnet still,move the coil over the magnetslowly.Step 6: Repeat the procedure, buthold the coil still while moving themagnet.

Figure 22-29Step 7: Connect the larger coil tothe galvanometer, and connect thesmaller coil in series with the bat-tery, switch, and rheostat. Place thesmaller coil inside the larger coil.


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