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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley PowerPoint ® Lectures for University Physics, Twelfth Edition – Hugh D. Young and Roger A. Freedman Lectures by James Pazun Chapter 30 Inductance Modified by P. Lam 8_6_2008

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Page 1: 30 Lecture Lam - hawaii.edu

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

PowerPoint® Lectures for

University Physics, Twelfth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

Chapter 30

Inductance

Modified by P. Lam 8_6_2008

Page 2: 30 Lecture Lam - hawaii.edu

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

Topics for Chapter 30

• Mutual inductance

• Self-inductance

• Magnetic-field energy

• R-L circuits

• Intermission

• L-C circuits

• L-R-C circuits

Page 3: 30 Lecture Lam - hawaii.edu

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

Mutual inductance

• Faraday’s Law of Induction states

that changing the current in one

coil can induce a current in a

nearby coil.

• This opens up a whole new way

of transferring signal or energy

from one coil (one device) to

another device without contact!

• To quantify this coupling, we

define a “mutual inductance” M:

2 = M21

di1dt

Likewise : 1 = M12

di2dt

It can shown that M21 = M12; the multual

inductance depends on the geometry of coil 1

and coil 2.

Page 4: 30 Lecture Lam - hawaii.edu

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Example - Mutual inductance of two concentric solenoids

• Refer to Example 30.1. To calculate the mutual inductance

M, we imagine putting a current i1(t) through coil 1.

Use the long solenoid formula : B1(t) = μoN1li1(t)

The magnetic flux through coil 2 : 2 = N2B1(t)A

The induced emf in coil 2 : 2 = -d 2

dt= μo

N1N2

lAdi1dt

M = μoN1N2

lA

Page 5: 30 Lecture Lam - hawaii.edu

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Numerical Example - Mutual inductance

Numerical example : l = 0.5m,A =10 3m2,

N1 =1,000,N2 =10

M = μoN1N2

lA = 4 x10 7( )

1,000( ) 10( )0.5

10 3( )

25x10 6(unit ?)

M = 25x10 6 Henry = 25μH

Suppose i1(t) = (2x106A

s)t, find 2 .

2 = Mdi1dt

= 25μH( )(2x106A

s) = 50volts

Suppose instead of putting the current through

coil 1, we put the current through coil 2, i.e.

i2(t) = (2x106A

s)t. What is the induced emf on coil 1?

Page 6: 30 Lecture Lam - hawaii.edu

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Self-inductance (L)

= Ldi

dt

Even when there is only one coil, there is an induced emf the

coil if the current in the coil is changing with time.

For a long solenoid, the self - inductance is L =μoN

2A

l.

For a straight conductor, L = 0.

Page 7: 30 Lecture Lam - hawaii.edu

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How to increase Self-inductance and Mutual inductance

Recall vaccum (or air) gap capacitor has very small capacitance :

C = o

A

d because o is a very small number.

We can increase the capacitance by filling the gap with a dielectric

material (electrically polarizable material),

C = K o

A

d; K = dielectric constant.

Simliarly, coils in vacuum (or air) have very small self and mutual

inductances because μo is a very small number.

We can increase the inductance by filling the space inside the coil

with a magnetically polarizable material such as iron and the

inductance can be increased by several thousand times.

Page 8: 30 Lecture Lam - hawaii.edu

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Voltage across an inductor (a coil)

+ -

Emf points from - to +;

= Ldi

dt

Vab = = Ldi

dt

Page 9: 30 Lecture Lam - hawaii.edu

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Magnetic field energy

A charge capacitor stores energy : U =12CV 2

so does an inductor with a current : U =12LI 2

Derivation :

Power =dU

dt= iV = iL

di

dt

U =dU

dt0

I

dt = iLdi

dt0

I

dt =12Li2

0

I

=12LI2

Page 10: 30 Lecture Lam - hawaii.edu

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The R-L circuit

The swith is closed at t = 0.

Before t = 0, the current is zero.

Since the inductor (L) opposes change

current, the current CANNOT change

abruptly thus the current increases

gradually (In RC circuit, the current can

change abruptly but the voltage across

the capacitor has to change gradually

because charge has to build up)

+ -+ -

After a long time, the current stops changing

(di

dt= 0), the voltage across the inductor = L

di

dt= 0

the voltage across R is

Ifinal = R

Page 11: 30 Lecture Lam - hawaii.edu

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The R-L circuit - Differential equation

Kirchoff 's loop rule :

- iR - Ldi

dt= 0

Ldi

dt+ Ri =

(Inhomogeneous, first order linear DE)

We have solved this type of DE when we

consider charging a capacitor :RdQ

dt+1CQ =

The solution consists of two parts :

The steady state (or inhomogeneous solution)

+ the transient (or homogeneous solution)

i( t) =R

+ Ae t / ; =L

RA is determined by the initial current.

If i( t = 0) = 0, then A = -R

i(t) =R1 e t /[ ]

Page 12: 30 Lecture Lam - hawaii.edu

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The R-L circuit -Mechanical analog

Kirchoff 's loop rule :

- iR - Ldi

dt= 0

Ldi

dt+ Ri =

i velocity

di

dtacceleration

L mass

constant force (say, due to gravity)

Ri air - resistance force

Drop a mass in the air,

the velocity increases gradually until

it reaches terminal velocity.

Page 13: 30 Lecture Lam - hawaii.edu

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The R-L circuit - “discharging the current”

+ - - +

Kirchoff 's Rule :

-(iR) + (-Ldi

dt) = 0

Ldi

dt+ iR = 0

i(t) = Ae t / ; =L

R

When S2 is open, the "initial" current is R

i(t) =Re t /

Page 14: 30 Lecture Lam - hawaii.edu

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The R-L circuit - “discharging the current” -II

• Follow Example 30.7.

Page 15: 30 Lecture Lam - hawaii.edu

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Intermission

Page 16: 30 Lecture Lam - hawaii.edu

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The L-C circuit

• Consider Figure 30.14.

Page 17: 30 Lecture Lam - hawaii.edu

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Compare LC circuit with mass-spring system

• Consider Figure 30.15.

• Use Table 30.1.

• Follow Example 30.9.

• Follow Example 30.10.

Page 18: 30 Lecture Lam - hawaii.edu

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The L-R-C circuit - damped harmonic oscillator

• An application of the dampedoscillator.

• Consider Figure 30.16.

• Consider Figure 30.17.

• Follow Example 30.11.