After a switch has been closed or opened for a ver- 123docz.net

The fi rst rule above can be understood from the process shown in Figure 21.16.

Whenever there is an attempt to change the current through an inductor (i.e., by closing or opening a switch), the induced emf across the inductor always opposes this change. Momentarily, this opposition prevents the current through the inductor from changing, and the current just after a switch is thrown is equal to the current just before. This result applies only for the instant after a switch is thrown, and is simply a statement of Lenz’s law as applied to inductors.

After some time, the current through the inductor will change, eventually approaching a constant value. If I is constant, I/t in Equation 21.23 is zero and all currents and magnetic fi elds are then constant. Hence, the voltage across the inductor is zero, leaving us with DC circuit rule 2.

These two rules apply only to DC circuits; in these circuits, the only time- dependent changes are initiated by the closing or opening of switches. Another important class of circuits, called AC circuits, contain power sources whose electric potentials vary sinusoidally with time. We’ll discuss those circuits in Chapter 22, when we generalize our circuit rules to apply to both AC and DC circuits.

Let’s use our two circuit analysis rules to study the behavior of the DC circuit sketched in Figure 21.18, which contains a battery, a switch, two resistors, and an inductor. The presence of the resistors and inductor make it an “RL circuit.” The general circuit symbol for an inductor is a solenoid-like coil. In Figure 21.18A, the switch has been open for a very long time, so the current is zero through both branches of the circuit. At t 0, the switch is closed, inducing a potential across the inductor that is equivalent to a battery with emf E VL (Fig. 21.18B). To fi nd the values of I1 (the current through resistor 1) and I2 (the current through resistor 2) just after the switch is closed, we apply rule 1 from the boldfaced list above. At the instant after t 0, the induced emf keeps the current through the inductor at the value it had before the switch was closed. Hence, I2 0 just after t 0; at that moment, the entire circuit is equivalent to the lower circuit in Figure 21.18B, and

General behavior of RL circuits General behavior of RL circuits

Figure 21.17 Two coils wound around each other exhibit a mutual inductance. Changes in the current in one coil (coil 1) induce an emf in the other (coil 2).

ⴙ ⴚ

I2 I2

Current induced in coil 2

Field produced by coil 1

the circuit behaves as if L and R2 have been removed. The result is a circuit with a battery connected to a single resistor carrying current I1 V/R1.

Now let’s fi nd the currents after the switch in Figure 21.18 has been closed for a very long time. We apply rule 2 highlighted above and fi nd that the voltage across the inductor is now zero, which is equivalent to replacing the inductor by a simple wire that has zero resistance (lower circuit in Fig. 21.18C). The voltage across each resistor is now the battery emf V, so the currents in the two circuit branches are

I15 V

R1 and I25 V

R2 (21.25)

21.5 | RL CIRCUITS 705

V L

I1 0 I2 0 I2 0

Symbol for an inductor

No current in this branch Just after switch is closed

ⴙ ⴚ

R2 R2

R1 V

I1 L ⴙ

ⴚ

E VL

V I1

ⴙ

ⴚ R1

After the switch has been closed for a long time

R2 ⴙ

ⴚ R1 VL 0

V I1 I2

ⴙ

ⴚ R1

A B C

Figure 21.18 A An RL circuit. B Just after the switch is closed, the emf induced in the inductor opposes any increase in current through it. At this instant, the current I2 is zero. The equivalent circuit just after the switch is closed is shown below. C After the switch has been closed for a long time, the magnetic fl ux through the inductor is constant (not changing with time), so the induced emf is zero. The inductor then acts as a wire (a short circuit).

E X A M P L E 2 1 . 6 Current in an RL Circuit

Consider again the circuit in Figure 21.18. Now assume the switch has been closed for a very long time, so the currents through R1 and R2 are as given in Equation 21.25 with the directions shown in Figure 21.18C. Find the directions of the currents through R1 and R2 the instant after the switch is reopened.

RECOGNIZE T HE PRINCIPLE

Whenever a switch is opened or closed, the induced emf across an inductor always acts to keep the current the same as it was just before the switch was thrown. This principle is circuit rule 1 of “Qualitative behavior of DC circuits with inductors,” page 704.

SK E TCH T HE PROBLEM

The current I2 through the inductor remains at the value it had just prior to the change of the switch (Eq. 21.25), and it will also keep the same direction. Hence, at the instant the switch is opened the current through the branch containing R2 and the inductor is directed clockwise as sketched in Figure 21.19B.

IDENT IF Y T HE REL AT IONSHIPS A ND SOLV E

Applying the equivalent circuit in Figure 21.19B, this current must fl ow “upward”

through R1, opposite to the direction found when the switch was closed in Figure 21.18B.

Figure 21.19 Example 21.6.

A ⴙ ⴚ

This current flows clockwise around the circuit loop.

I Switch is

opened

What does it mean?

The induced emf in an inductor always opposes changes in the current. In this example, the initial current through L has a nonzero value; the induced emf maintains this nonzero current momentarily, even after the switch is opened. At that instant, the battery is disconnected leaving the circuit shown in Figure 21.19B with a nonzero current even though there is no battery! Where does the energy supplying this current come from? We’ll explain that in Section 21.6 when we consider how an inductor can store energy.

CO N C E P T C H E C K 2 1 . 5 | Analyzing an RL Circuit (1)

Consider the circuit in Figure 21.20. The current is initially zero in both branches. The switch is then closed. Is the current through R1 just after closing the switch (a) I5V/R1, (b) I5V/R2, (c) I5V/1R11R22, or (d) I 0?

CO N C E P T C H E C K 2 1 .6 | Analyzing an RL Circuit (2)

Consider the circuit in Figure 21.21. The switch is in position 1 for a very long time and then changed to position 2. What is the direction of the current through the inductor just after the switch is changed?

(a) It is to the right.

(b) It is to the left.

(d) The direction depends on the values of R1, R2, and R3.

Quantitative Behavior of an RL Circuit

Our qualitative rules for dealing with inductors in DC circuits tell how to fi nd the current the moment just after a switch is closed or opened and how to fi nd the current after a switch has been closed or open for a very long time. The current between these two times can also be calculated using Kirchhoff’s rules for circuit analysis. Figure 21.22A shows the simplest RL circuit, with a battery connected through a switch to a resistor R and an inductor L. The switch is open for a long time and then is closed at t 0. The quantitative behavior of the current through and voltage across the inductor is shown in parts B and C of Figure 21.22. The current starts at I 0 because the induced emf keeps the current through the inductor momentarily at the value it had before the switch was thrown (in accord with our circuit rule 1). After a very long time, the voltage across the inductor falls to zero (in accord with rule 2) and the current approaches I V/R. Mathematically, the current is given by

I5 V

R112e2t/t2 (21.26) where ex is the exponential function and t is called the time constant of the circuit.

For an RL circuit in which a single resistor is in series with a single inductor as in Figure 21.22, the time constant is

t 5 L

R (21.27) The corresponding behavior of the voltage across the inductor is

VL5Ve2t/t (21.28)

The time constant t is the approximate time scale over which the current and voltage change in an RL circuit. From the properties of the exponential function (see Appendix B), the current in Equation 21.26 reaches approximately 63% of its fi nal value when t t and the voltage across the inductor (Eq. 21.28) falls to approximately 37% of its initial value after this time. After three time constants Figure 21.20 Concept Check

21.5.

Vⴙ L ⴚ

Figure 21.21 Concept Check 21.6.

Vⴙ ⴚ

R1 L

R3 R2

Switch

2 1

Figure 21.22 A An RL circuit with the switch closed. B Current through the circuit as a function of time. The switch was closed at t 0. C Voltage across the inductor as a function of time. When this voltage is positive, it opposes the fl ow of current through the inductor.

B V/R

0 2 3 4

0.2 0 0.4 0.6 0.8 1.0

0 2 3 4

0.2 0 0.4 0.6 0.8 1.0 I

VL/V t t A V L

I VL

Switch closed at t 0

t t

t t 0.63

0.37

(t 3t), the current reaches 95% of its fi nal value. The time constant t in Equa- tion 21.27 applies only to an RL circuit. It is different from the time constant we encountered for RC circuits in Chapter 19.

E X A M P L E 2 1 . 7 Value of t for a Typical RL Circuit

Consider an RL circuit with R 4000 and L 3.0 mH; these values are typical of those used in many electronic circuits such as radios and computers. Find the time constant for a circuit that uses these components.

RECOGNIZE T HE PRINCIPLE

The time constant for an RL circuit is just the ratio L/R. The purpose of this example is to determine a “typical” value of the time constant t. Is it hours, seconds, or micro- seconds? The value of t is important in applications (see below).

SK E TCH T HE PROBLEM No fi gure is necessary.

IDENT IF Y T HE REL AT IONSHIPS A ND SOLV E

We evaluate the time constant t from Equation 21.27 using the given values of R and L:

t 5 L

R5 3.031023 H

4000 V 57.531027 s5 0.75 ms What does it mean?

For this circuit, the time constant is approximately 1 microsecond (1 ms). In many applications of RL circuits (Chapter 22), this time constant is associated with the period of an oscillation, so the period of this circuit would correspond to a frequency of about

f51

t < 1

131026 s513106 Hz

(1 megahertz). This frequency lies in the middle of the “AM band” on your radio, so such an RL circuit could be useful in an AM radio.

2 1 . 6 | E N E R G Y S T O R E D I N A M A G N E T I C F I E L D

The RL circuit in Figure 21.23A is the same circuit we analyzed in Figures 21.18 and 21.19. We know that if the switch has been closed for a long time, currents I1 and I2 are established through the resistors as given in Equation 21.25. If the switch is then opened at t 0, there is a current in the circuit loop formed by R1, R2, and L (Fig. 21.23B); the time dependence of this current is shown qualitatively in Figure 21.23C. After the switch is opened, the battery is completely disconnected from this circuit. Current thus exists even with the battery removed, and electrical

Insight 21.2

BUILDING A REAL INDUCTOR We calculated L for a simple solenoid inductor in Equation 21.21, but most practical inductors are constructed by wrapping a wire coil around a magnetic material. From Chapter 20, we know that a small magnetic fi eld can align the domains of a magnetic material, producing a much larger magnetic fi eld. Filling a coil with magnetic material greatly increases the magnetic fl ux through the coil and therefore increases the induced emf.

According to the defi nition of inductance Equation 21.22, the presence of magnetic material thus increases the inductance. For this reason, most inductors contain a magnetic material inside, which produces a larger value of L in a smaller package.

A I1 L

Vⴙ ⴚ

R1 R1

B C

L R2

V ⴙ

ⴚ I2

t Switch opened here

Figure 21.23 A If the switch in this RL circuit is closed for a long time, there will be current in the inductor. B When the switch is opened, the current through the inductor decays with time as plotted in C.

21.6 | ENERGY STORED IN A MAGNETIC FIELD 707

energy is dissipated as heat in the resistors during this time. Where does this energy come from if the battery has been disconnected? The answer is that this energy was originally (at t ⫽ 0) stored in the magnetic fi eld of the inductor.

One way to analyze the energy stored in this magnetic fi eld is to consider the energy required to establish the current in an inductor in the fi rst place. Suppose the voltage across the inductor in Figure 21.24A has a constant value V. To keep V constant, the current must increase with time. We can see that from Equation 21.23; the voltage drop across the inductor is V5L DI/Dt. Solving for the change in current, we have

DI5V

L Dt (21.29)

Hence, if V is constant and the current starts from zero at t ⫽ 0, Equation 21.29 means that the current through the inductor I must increase linearly with t (Fig.

21.24B), and

I5 V

L t (21.30)

The power P dissipated in a circuit element is equal to the product of the voltage across that element and the current (Eq. 19.15). Hence, the instantaneous power in the inductor—the rate at which energy is delivered to the inductor at time t—is

P5VI5VaV

L tb 5 aV2

L bt (21.31)

This linear relation between P and t is plotted in Figure 21.24C. We can use the result of Equation 21.31 to fi nd the total energy that has been delivered to the inductor when the current reaches a particular value I0 at time t0. We denote this energy as PEind since we will soon see that it is a kind of potential energy. This total potential energy PEind is equal to the average of the power in Figure 21.24C multi- plied by the total time t0. The instantaneous power P varies linearly with time, so the average power is half the maximum power. Using Equation 21.31, we get

Pave5 1 2 V2t0

PEind5Pavet05V2t20

2L (21.32) From Equation 21.30, we fi nd I05Vt0/L, which we rearrange to give t05I0L/V.

Inserting into our expression for PEind leads to PEind5 V2

2LaI0L V b251

2 LI20

where I0 is the fi nal current through the inductor; we can just call it I and then write our fi nal result as

PEind512LI2 (21.33)

Equation 21.33 is the general result for the magnetic energy PEind stored in an inductor. It has a form that is very similar to the electric energy stored in a capacitor, PEcap512CV2 (Eq. 18.33). In the case of a capacitor, we saw that this energy is stored in the electric fi eld that exists between the capacitor plates. In a similar way, the energy PEind is stored in the magnetic fi eld of the inductor, and we can rewrite Equation 21.33 to make this more apparent. The inductance of a solenoid is L5 m0N2A/, (Eq. 21.21). Inserting into our expression for PEind (Eq. 21.33) gives

PEind5 1

2LI25 1

2am0N2A , bI2

Energy stored in an inductor Energy stored in an inductor

Figure 21.24 A Voltage across and current through an inductor.

B If the voltage across the inductor is constant, the current increases linearly with time (Eq. 21.30).

C The power delivered to the inductor also increases linearly with time.

I I

V L

C I I0

t0 t

P P0

Pave

t0 t

which can be rearranged as PEind5 1

2am0N2A

, bI25 1

2m0am0NI , b21A,2

We have rewritten the potential energy in this way because the factor involving I in parentheses is the solenoid’s magnetic fi eld (B5 m0NI/,, Eq. 21.19), whereas the last factor A, is the volume of the solenoid (the length times the cross-sectional area). So we get our fi nal result,

PEmag5 1

2m0 B21volume2 (21.34) In Equation 21.34, we have now denoted the energy as PEmag. It is the energy contained in the magnetic fi eld of the inductor and applies not only to an inductor but to any region of space that contains a magnetic fi eld. This energy is stored in the magnetic fi eld. Because a magnetic fi eld can exist in a vacuum, this potential energy can exist even in “empty” regions containing no matter! This fact, which also applies to electric fi elds and electric potential energy, will be important in understanding electromagnetic waves (Chapter 23).

Another way to express the potential energy in Equation 21.34 is to say that there is a certain energy density (the energy per unit volume) in the magnetic fi eld.

From Equation 21.34, we fi nd

energy density5umag5 PEmag

volume 5 1

2m0 B2 (21.35)

This expression for the magnetic energy density is very similar in form to the energy density contained in an electric fi eld. In Chapter 18 we found that uelec512e0E2 (Equation 18.48), so there is an interesting parallel between our results for magnetic and electric energies. In both cases, the energy density is proportional to the square of the fi eld.

Energy stored in a magnetic ﬁ eld Energy stored in a magnetic ﬁ eld

21.6 | ENERGY STORED IN A MAGNETIC FIELD 709

E X A M P L E 2 1 . 8 Magnetic Field Energy: How Large Is It?

To get a feeling for the magnitude of the energy stored in a magnetic fi eld, consider the energy stored in a magnet used for magnetic resonance imaging (MRI) (Fig. 21.25).

A typical MRI magnet is a solenoid providing a fi eld of about B 1.5 T. (a) What is the approximate energy stored in this magnetic fi eld? (b) If the current through the magnet during normal operation is I 100 A, what is its inductance?

RECOGNIZE T HE PRINCIPLE

The motivation for this example is to understand how the potential energy stored in a magnetic fi eld compares with cases familiar from mechanics such as the gravitational potential energy or the kinetic energy of a moving object. To fi nd the energy stored in an MRI magnet, we need to know its magnetic fi eld and its inside volume. The volume is not given, but we can estimate its value because a person must be able to fi t inside (Fig. 21.25).

SK E TCH T HE PROBLEM

Figure 21.25 shows a typical MRI magnet. We can estimate the volume from this photo.

IDENT IF Y T HE REL AT IONSHIPS

We can fi nd the stored magnetic energy using our result for the potential energy PEmag (Eq. 21.34).

SOLV E

(a) From personal experience (and the photo in Fig. 21.25), we estimate that a typical MRI magnet is about 1.5 m long with an inside diameter of about 0.5 m so that

Figure 21.25 Example 21.8.

The magnets used for magnetic resonance imaging (MRI) are very large, with a large amount of stored magnetic energy.

Alvis Upitis/Brand X Pictures/Jupiter

a person can fi t inside. We have volume5A,5 1pr22,, and inserting our estimates for r and 艎 gives

volume5 1pr22,5 p10.25 m2211.5 m250.3 m3 Using this volume in Equation 21.34 along with the given value of B, we fi nd

PEmag5 1

2m0 B21volume2 5 1

214p 31027 T#m/A211.5 T2

210.3 m32 PEmag5 3 105 J

(b) The magnetic energy found in part (a) is stored in an inductor because the MRI magnet is just a solenoid. Therefore, PEmag PEind is related to the inductance through Equation 21.33. Using our calculated value of PEmag gives

2LI25PEmag L52PEmag

I2 5 2133105 J2

1100 A22 5 60 H

This inductance is large, much larger than we found for the small coil in Exam- ple 21.5.

What does it mean?

The amount of energy found in part (a) is quite large; it is approximately equal to the kinetic energy of a car of mass 1000 kg with a velocity of 50 mi/h (23 m/s)!

An MRI magnet is carefully designed to be sure this energy is dealt with safely. In addition, patients undergoing an MRI exam as well as MRI technicians must take precautions such as removing all magnetic objects from their pockets.

2 1 . 7 | A P P L I C AT I O N S

In this section, we consider a few important applications of Faraday’s law.

Bicycle Odometers

Most serious bicyclists use an odometer to monitor their speed and distance trav- eled. Figure 21.26A shows a photo of the odometer control unit, and a sketch of the working elements is shown in Figure 21.26B. A small, permanent magnet is

Figure 21.26 A This bicycle odometer uses Faraday’s law to count the rotations of a bicycle wheel. B A small, permanent magnet is attached to a spoke, and a pickup coil is mounted on the bicycle frame. C When the magnet passes near the pickup coil, the time-varying magnetic fl ux induces an emf in the coil. D Qualitative sketch of the fl ux through the pickup coil as a function of time. This time-dependent fl ux produces an induced emf that varies with time as a series of pulses. The odometer counts these pulses.

Pickup coil

Large flux FB 0 Permanent

magnet attached to spoke

B C D

FB 0

t FB

t Dt E

After a switch has been closed or opened for a very long time, the induced emfs are zero