ch 30 notes - fcps 30 note… · the induced emf through the inductor or choke. an inductor in a...

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley

PowerPoint® Lectures forUniversity Physics, Twelfth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

Chapter 30

Inductance


Goals for Chapter 30

• To introduce and illustrate mutual inductance

• To consider self-inductance

• To calculate magnetic-field energy

• To describe and study R-L circuits

• To describe and study L-C circuits

• To describe and study L-R-C circuits\

• HW: 1, 9, 17, 21, 29, 33, 41, 65, 67


Introduction

• A charged coil can create a field that will induce a current in a neighboring coil.

• Inductance can allow a sensor to trigger the traffic light to change when the car arrives at an intersection.


Mutual inductanceA coil in one device generates a field that creates a current in a neighboring coil. Check out Figure 30.1The emf in coil 2:

The flux through coil 2 and i1 are proportional. Introducing the proportionality constant M21 called the mutual inductance of the two coils:

With a changing current and hence, flux:

And the emf is now:

dtdN b2

22Φ

−=ε

dtdiM 1

212 −=εdtdiM

dtdN b 1

212

2 =Φ

12122 iMN B =Φ


Mutual InductanceA change in the current i1 in coil 1 induces an emf in coil 2 that is directly proportional to the rate of change of i1.We can write the definition of mutual inductance as:

The mutual inductance M21 is a constant that depends only on the geometry of the two coils. We will assume that if any magnetic material is present, it has a constant relative permeability Km so that flux is directly proportional to current and M21 depends on geometry only.If we consider what happens when we reverse our discussion, then a changing current in coil 2 i2 would induce an emf in coil 1 that is proportional to i2 by M12. It turns out that M21 is exactly equal to M12 and we no longer have to distinguish between them. We can write our mutually induced emfs and mutual inductance as:

1

2221 i

NM BΦ=

2

11

1

22

iN

iNM BB Φ

=Φ

=dtdiM 2

1 −=εdtdiM 1

2 −=ε


Inductance and the henryThe unit for inductance is called the henry (1H) after Joseph Henry.

Just as the farad was a large unit of capacitance, the henry is a large unit of inductance. A typical value of inductance is the millihenry or microhenry.

Mutual inductance can cause some unwanted and possible dangerous emfs in sensitive circuits. Fortunately, mutual inductance also has many useful applications. It is the main principle that allows for long range transfer of electrical power. We’ll learn about transformers in the next chapter.

211111 AJsA

sVA

WbH =⋅Ω=⋅==


Mutual inductance—examples• Refer to Example 30.1.

• See Figure 30.3 below.

• Follow Example 30.2, you need the evaluate of 30.1.


Self-inductance Any circuit that carries a varying current has an emf induced in it by the variation in its own magnetic field. Such an emf is called a self-induced emf.By Lenz’s law the induced emf always opposes the chance, and it is difficult for a change to take place.These currents occure in any circuit, but the effect is enhanced if the coil has many turns. With N number of turns, an average flux of ΦB and varying current i, the self inductance L is:

The unit for self-inductance is the henry.

iNL BΦ

=


Self-inductance The derivation to the left turns equation 30.6 into 30.7. Equation 30.7 shows that it is relatively simple to calculate an unknown inductance by taking the ratio of a known change in current to the induced emf through the inductor or choke.An inductor in a direct current circuit helps to maintain a steady current despite any fluctuation in the applied emf; in an alternating current circuit, an inductor tends to suppress variations of the current that are more rapid than desired.To understand and analyze how an inductor affects a circuit we need to develop a principle similar to Kirchhoff’s loop rule. The loop rule states that the sum of all potential differences around a closed loop is zero. This is only true for conservative electric fields, and the induced emf is not conservative.

dtdiL

dtdN

dtdiL

NLi

B

B

−=

Φ=

Φ=

ε


Applications and calculationsConsider figure 30.5 to the right. Lets assume that we can control the variable current i and that our inductor has a very small resistance such that Ec + En = 0. From Faraday’s Law:But the field is different from zero only within the inductor, so we only need to integrate from a to b:Because the total electric field within the inductor is zero we can state:This is also just the potential difference between aand b:We conclude that there is an actual potential difference across the inductor, so we are justified in using Kirchhoff’s rules.Note that the self-induced emf does not oppose the current i, only any change in the current di/dt.

dtdiLldEn −=⋅∫

dtdiLldE

b

an −=⋅∫

dtdiLldE

b

ac =⋅∫

dtdiLVVV baab =−=


Applications and calculationsWe will assume that the inductors in the following examples are enclosing a vacuum or some other non-magnetic material. If the material is magnetic with permeability μ, we replace μo with μ = Kmμo. If the material is diamagnetic or paramagnetic Km is very close to 1 and makes no significant change.Try examples 30.3 and 30.4.


30.1 – 30.2 Summary and HomeworkWhen a changing current i in one circuit causes a changing magnetic flux in a second circuit, an emf is induced in the second circuit. Likewise, a changing current in the second circuit induces an emf in the first circuit. The mutual inductance M depends on the geometry of the two coils and the material between them. If the circuits are coils of wire with N1 and N2turns, M can be expressed in terms of the average flux through each turnof coil 2 that is caused by the current in coil 1, or in terms of the average flux through each turn of coil 1 that is caused by the current in coil 2. The SI unit of mutual inductance is the henry, abbreviated H.A changing current i in any circuit causes a self-induced emf. The inductance (or self-inductance) depends on the geometry of the circuit and the material surrounding it. The inductance of a coil of N turns is related to the average flux through each turn caused by the current i in the coil. An inductor is a circuit device, usually including a coil of wire, intended to have a substantial inductance.Page 1174: 1, 3, 7, 9


Magnetic field energyEstablishing a current in an inductor requires an input of energy, and an inductor carrying a current has energy stored in it. An increasing current i in the inductor causes an emf between its terminals and a corresponding potential difference between the terminals of the source. Thus the source must be adding energy to the inductor, and the instantaneous power, the rate of energy transfer into the inductor, is P = Vabi. We can calculate the amount of energy U needed to establish a current Iin an inductor with inductance L:

The energy dU supplied to the inductor durring an infinitesimal time interval dt is dU = Pdt:

The total energy supplied while the current increases from zero to I:

When the current decreases from I to zero, the inductor acts as a source.

dtdiLii

dtdiLiVP ab ===

LididtdtdiLiidt

dtdiLidtVPdtdU ab =====

221

0

LIidiLUI

== ∫


CAUTIONIt’s important not to confuse the behavior of resistors and inductors where energy is concerned. Energy flows into a resistor whenever a current passes through it, whether the current is steady or varying; this energy is dissipated in the form of heat. By contrast, energy flows into an ideal, zero-resistance inductor only when the current in the inductor increases. This energy is not dissipated; it is stored in the inductor and released when the current decreases. When a steady current flows through and inductor, there is no energy flow in or out.


Magnetic energy density: 1 of 2The energy stored in an inductor is stored in the magnetic fieldof the inductor, like the energy in a capacitor is stored in theelectric field between its plates. To achieve relations for themagnetic field energy analogous to those used for electric fieldenergy, lets assume an ideal toroidal inductor.The volume enclosed is approximately equal to the circumference multiplied by the area: V = 2πrA. From example 30.3 the inductance is:

From equation 30.9 the energy stored at a current I is:

The energy per unit volume is:

rANL o

πμ

2

2

=

( )222

221

2 rIN

rAUu o π

μπ

==

22

221

221 I

rANLIU o

πμ

==


Magnetic energy density: 2 of 2We can express this in terms of the magnitude of the magnetic field inside the toroidal solenoid.

We can substitute B2 into the equation for u to get the final expression of the magnetic field density in a vacuum:

When the material inside the toroid is not a vacuum but a material with a constant magnetic permeability:

This expression, derived for a specific situation, turns out to be true for any magnetic field configuration with constant permeability.

rNIB o

πμ2

=

μ2

2Bu =

( )2222

2

2 rINB o

πμ

=

o

Buμ2

2

=


Magnetic field energy• Consider Figure 30.10, below center.

• Refer to Example 30.5.

• Refer to Example 30.6.


30.3 Summary and HomeworkAn inductor with inductance L carrying current I has energy U associated with the inductor’s magnetic field. The magnetic energy density u(energy per unit volume) is proportional to the square of the magnetic field magnitude.

Page 1174: #17


The R-L circuitHaving an inductor in a circuit makes changes in current harder to achieve. The inductor tends to “smooth out” the current. The problem solving strategy on page 1159 gives us some ideas on how to deal with R-L circuits.See figure 30.11. At time t = 0 we close S1, the current grows at a rate that depends on L. Let i be the current at some time after S1 has been closed. vab = iR and vbc = Ldi/dtIf the current is in the direction as shown, there is a drop from b to c.Using Kirchhoff’s loop rule:Solving for di/dt we find that the increase in current is:When S1 is initially closed, i = 0 and the initial rate of charge:After a long time, the rate of current increase goes to zero and:

0=−− dtdiLiRε

iLR

LLiR

dtdi −== − εε

( ) Ldtdi

initialε=

( ) ILR

Ldtdi

final−== ε0 RI ε=


The R-L circuitThe final current I does not depend on the inductance L; it is the same as it would be if the inductor was not there.The behavior of the current as a function of time is shown to the right.To derive the equation for this we rearrange equation 30.13:

Change the variables and integrate:

Take the exponentials of both sides and solve for i:

( ) dtLR

Ridi

−=− ε

( ) ∫∫ ′−=−′

′ ti

o

tdLR

Riid

0ε

⎟⎠⎞⎜

⎝⎛ −=

− tLR

eR

i )(1ε


Time constant and other R-L considerationsIf we take the derivative of equation 30.14, we get:

What happens at the limits t = 0 and t = ∞?The time constant of the circuit τ = L/RThe instantaneous rate at which the source

delivers power to the circuit is P = εi. The instantaneous rate of dissipation by the resistor is i2R, and the rate at which energy is stored in the inductor is ivbc = Lidi/dt.Part of the energy delivered to the circuit is dissipated by the resistor and part is stored in the inductor:

Try Example 30.7dtdiLiRii += 2ε

( )RL tdi e

dt Lε −=


Current Decay in an R-L CircuitIf we open S1 while closing S2, we get the combination shown to the right. At a time t after S2 has been closed the instantaneous current i in the circuit is:

The energy needed to maintain this current decay is from the energy stored in the inductor. In this case the source supplies no energy, so the energy that is stored in the inductor will be dissipated by the resistor:

tLRoeIi )/(−=

dtdiLiRi += 20


R-L circuit II

• Consider Figure 30.13 at right.

• Follow Example 30.8.


30.4 Summary and HomeworkIn a circuit containing a resistor R an inductor L and a source of emf, the growth and decay of current are exponential. The time constant τ is the time required for the current to approach within a fraction 1/e of its final value.

Page 1175: 19, 21


The L-C circuitAs capacitor discharges, energy is stored in the inductor. Energy leaves the inductor, in an induced current that charges the capacitor with opposite polarity. The process continues, if no energy loss, indefinitely.


Applications and comparisons Apply Kirchhoff’s loop rule to the circuit below:

Since i = dq/dt, di/dt = d2q/dt2. Substitute and divide by –L:

This has the same form as equation 13.4, which was derived for SHM:

The capacitor charge q plays the role of displacement x, and the current is analagous to the particles velocity. The inductance is similar to the object’s mass and 1/C is like the force constant k.Recall that the angular frequency ω=2πf is equal to (k/m)1/2, and the position is given by equation 13.13:

Following out analogy:

0di qLdt C

− − =

2

2

1 0d q qdt LC

+ =

2

2 0d x k xdt m

+ =

cos( )x A tω φ= +cos( )

1

q Q t

LC

ω φ

ω

= +

=


Applications and comparisons When you verify that equation 30.21 satisfies the loop equation, you will find that the instantaneous current in the circuit is found by:

Q and φ are determined by the initial conditions. If at time t = 0 the left-hand plate in figure 30.15 has its maximum charge Q and the current i is zero, then φ = 0. If q = 0 at time t = 0, then φ = +/- π/2 rad.We can also analyze the circuit using an energy approach.The L-C circuit is a conservative system. Let Q be the maximum capacitor charge. The magnetic-field enregy ½Li2 in the inductor and the electric-field energy q2/2C in the capacitor add to get the total energy of the system: Q2/2C

solving for i:

sin( )cos( )

i Q tq Q t

ω ω φω φ

= − += +

2 21i Q qLC

= ± −2 2

212 2 2

q QLiC C

+ =


Applications and comparisons

• Consider Figure 30.15.

• Use Table 30.1.




The L-R-C circuit




2

2

14R

LC Lω′ = −


30.5 & 30.6 Summary and HomeworkA circuit that contains inductance L and capacitance C undergoes electrical oscillations with an angular frequency ω that depends on L and C. Such a circuit is analogous to a mechanical harmonic oscillator, with inductance L analogous to mass m, the reciprocal of capacitance 1/C to force constant k, charge q to displacement x, and current i to velocity vx.A circuit that contains inductance, resistance, and capacitance undergoes damped oscillations for sufficiently small resistance. The frequency ω’ of damped oscillations depends on the values of L, R and C. As R increases, the damping increases; if R is greater than a certain value, the behavior becomes overdamped and no longer oscillates. The cross-over between underdamping and overdamping occurs when R = 4L/C; when this condition is satisfied, the oscillations are critically damped.

Page 1175:29, 31, 33, 39, 41, 65, 67


Homework and chapter review1, 9, 17, 21, 29, 33, 41, 65, 67

Chapter Review Follows:


A small, circular ring of wire (shown in blue) is inside a larger loop of wire that carries a current Ias shown. The small ring and the larger loop both lie in the same plane. If I increases, the current that flows in the small ring

Q30.1

A. is clockwise and caused by self-inductance.

B. is counterclockwise and caused by self-inductance.

C. is clockwise and caused by mutual inductance.

D. is counterclockwise and caused by mutual inductance.

Small ring

I

I

Large loop


A small, circular ring of wire (shown in blue) is inside a larger loop of wire that carries a current Ias shown. The small ring and the larger loop both lie in the same plane. If I increases, the current that flows in the small ring

A30.1

A. is clockwise and caused by self-inductance.

B. is counterclockwise and caused by self-inductance.

C. is clockwise and caused by mutual inductance.

D. is counterclockwise and caused by mutual inductance.

Small ring

I

I

Large loop


A current i flows through an inductor L in the direction from point b toward point a. There is zero resistance in the wires of the inductor. If the current is decreasing,

Q30.2

A. the potential is greater at point a than at point b.

B. the potential is less at point a than at point b.

C. the answer depends on the magnitude of di/dt compared to the magnitude of i.

D. The answer depends on the value of the inductance L.

E. both C. and D. are correct.


A current i flows through an inductor L in the direction from point b toward point a. There is zero resistance in the wires of the inductor. If the current is decreasing,

A30.2

A. the potential is greater at point a than at point b.

B. the potential is less at point a than at point b.

C. the answer depends on the magnitude of di/dt compared to the magnitude of i.

D. The answer depends on the value of the inductance L.

E. both C. and D. are correct.


A steady current flows through an inductor. If the current is doubled while the inductance remains constant, the amount of energy stored in the inductor

Q30.3

A. increases by a factor of .

B. increases by a factor of 2.

C. increases by a factor of 4.

D. increases by a factor that depends on the geometry of the inductor.

E. none of the above

2


A steady current flows through an inductor. If the current is doubled while the inductance remains constant, the amount of energy stored in the inductor

A30.3

A. increases by a factor of .

B. increases by a factor of 2.

C. increases by a factor of 4.

D. increases by a factor that depends on the geometry of the inductor.

E. none of the above

2


An inductance L and a resistance R are connected to a source of emf as shown. When switch S1 is closed, a current begins to flow. The final value of the current is

Q30.4

A. directly proportional to RL.

B. directly proportional to R/L.

C. directly proportional to L/R

D. directly proportional to 1/(RL).

E. independent of L.


An inductance L and a resistance R are connected to a source of emf as shown. When switch S1 is closed, a current begins to flow. The final value of the current is

A30.4



C. directly proportional to L/R.




Q30.5An inductance L and a resistance R are connected to a source of emf as shown. When switch S1 is closed, a current begins to flow. The time required for the current to reach one-half its final value is







A30.5An inductance L and a resistance R are connected to a source of emf as shown. When switch S1 is closed, a current begins to flow. The time required for the current to reach one-half its final value is







Q30.6An inductance L and a resistance R are connected to a source of emf as shown. Initially switch S1is closed, switch S2 is open, and current flows through L and R. When S2 is closed, the rate at which this current decreases

A. remains constant.

B. increases with time.

C. decreases with time.

D. not enough information given to decide


A30.6An inductance L and a resistance R are connected to a source of emf as shown. Initially switch S1is closed, switch S2 is open, and current flows through L and R. When S2 is closed, the rate at which this current decreases

A. remains constant.

B. increases with time.

C. decreases with time.

D. not enough information given to decide


An inductor (inductance L) and a capacitor (capacitance C) are connected as shown.

If the values of both L and Care doubled, what happens to the time required for the capacitor charge to oscillate through a complete cycle?

Q30.7

A. It becomes 4 times longer. B. It becomes twice as long.

C. It is unchanged. D. It becomes 1/2 as long.

E. It becomes 1/4 as long.


An inductor (inductance L) and a capacitor (capacitance C) are connected as shown.

If the values of both L and Care doubled, what happens to the time required for the capacitor charge to oscillate through a complete cycle?

A30.7

A. It becomes 4 times longer. B. It becomes twice as long.

C. It is unchanged. D. It becomes 1/2 as long.

E. It becomes 1/4 as long.


An inductor (inductance L) and a capacitor (capacitance C) are connected as shown. The value of the capacitor charge qoscillates between positive and negative values. At any instant, the potential difference between the capacitor plates is

Q30.8

A. proportional to q. B. proportional to dq/dt.

C. proportional to d2q/dt2. D. both A and C.

E. all of A, B, and C.


An inductor (inductance L) and a capacitor (capacitance C) are connected as shown. The value of the capacitor charge qoscillates between positive and negative values. At any instant, the potential difference between the capacitor plates is

A30.8

A. proportional to q. B. proportional to dq/dt.

C. proportional to d2q/dt2. D. both A and C.

E. all of A, B, and C.

ch 30 notes - fcps 30 note… · the induced emf through the inductor or choke. an inductor in a...

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