addendum – chapter 21

ADDENDUM – CHAPTER 21

Mutual inductance –

Circulation of currents in one coil can generate a field in the coil that will extend to a second, close by device.

Suppose i1 CHANGES

Flux Changes

Current (emf) isinduced in 2nd

coil.

Mutual Inductance

i1 creates a field that (partially) passes through the second coil.

As i1 changes, the flux through coil 2 changes and an emf (and current i2) are created.

The two coils are mutually linked by what we call an “inductance”

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3

i2

Watch Out!

Exam #2 one week from today. Chapters 20 & 21 Same format but possibly one set of multiple choice

questions that you hate. You should already be studying.

QUIZ on Friday – Chapter #21 Today we continue with the chapter. We should

finish it on Friday. Maybe. No study session on Monday next week We will have a study session on Tuesday morning

like last time. Details to follow.

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4

04/21/23Induction5

This schedule is now in effect:This schedule is now in effect:

PHY2054 Problem Solving/Office Hours ScheduleRoom MAP-318

Monday Tuesday Wednesday Thursday FridayBindell 8:30-9:15AM 8:30-9:15AM 8:30-9:15AM

Bindell 11:00-12:00PM10:30 - 11:15

AM* 10:30-11:30AMDubey 12:00-1:00PM 1:30-2:45PM

These sessions will be used both for office hours and problem solving. Students from anysection of 2054 are invited to stop by for assistance in course materials (problems, etc.)

Note: There will be times when the room may not be available. In that case we will use our individual offices.

* In Office Dr. Dubey's hours are for problem

solving only.

Mutual Inductance

04/21/23Induction

6

i2

12

2

~

#1. coil

incurrent the todue #2 coil

influx magnetic The

iB

B

t

iM

tNemf

iMN

ki

B

B

121

2122

12122

12

mutual Inductance

Note the form:

04/21/23Induction

7

i

NM something

Think of this when we define INDUCTANCEINDUCTANCE (L) ofa small coil in the next section.

UNIT: henry

The two coils

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Remember – the magneticfield outside of the solenoidis pretty much zero.

Two fluxes (fluxi?) are the same!Two fluxes (fluxi?) are the same!

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One solenoid is centered inside another. The outer one has a length of 50.0 cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and 0.120 cm in diameter and contains 15 coils. The current in the outer solenoid is changing at 37.5 A/s. (a) What is the mutual inductance of these solenoids? (b) Find the emf induced in the inner solenoid

Length = 0.5 metersN=6750 coils

n=6750/.5=1.35E04 turn/meter

Magnetic field INSIDE the smaller coilis the same as in the larger coil andis given by:

iiniB 7.11035.1104 4701

04/21/23Induction10


iniB 7.101

24

22

1013.1

)100/106.0(

mArea

cmmcmrAreacoilsmaller

mVVt

Nemf

VVt

i

t

iBA

B

B

B

11102.715

102.75.371092.11092.1

1092.1

3222

3442

42

04/21/23Induction11


iniB 7.101

mH

ti

emfM 297.0

37

1011 32

Self-inductance –

Any circuit which carries a varying current self-induced from it’s own magnetic field is said to

have INDUCTANCE (L).

An inductor resists CHANGESCHANGES in the current going through it.

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Inductance Defined

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i

NL B

If the FLUX changes a bit during a short time t, then the current will change by a small amount i.

t

iL

tN

NLi

B

B

This is actually acalculus equation

Faraday says this is the emf!

So …

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t

iLemf

E=

The UNIT of “Inductance – L” of a coil is the henry.

SYMBOL:

There should bea (-) sign but weuse Lenz’s Lawinstead!

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Consider “AC” voltage

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19

V1

Maximum Change@t

Minimum Change@t

The transformer

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20

FLUX is the same throughboth coils (windings). 2

2

1

1

222

111

N

V

N

V

tNVemf

tNVemf

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Input/Output Impedance (Resistance)

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22

!resistance

inputan like looks

)/(

(Lossless)

2121

1

2211

1

2

1

2

NN

R

I

V

So

VIVI

PowerPower

N

N

V

V

outin

Remember that a Capacitor stored ENERGY? U=(1/2)CV2

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23

interval

iLiU

iLiUt

iLiP

t

iLV

ViPower

i

i

LiLI

i

Li

DU

IInduction

U=Area=(1/2)LI2

SO …

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24

Energy Stored in a capacitorThe energy stored in a capacitor with capacitance C and a voltage V is

U=(1/2)LI2

The Energy stored is in the Magnetic Field

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Consider a solenoid with N turns that is very long. We assume that the field is uniform throughout its length, ignoring any “end effects”. For a long enough solenoid, we can get away with it for the following argument. Maybe.

il

NniB 00

Energy Storage in Inductor

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0

0

22

2

1

2

1

2

1

BlNi

il

NB

NiBAU

BA

il

NLiU

l

NL

B

B

B

20

0

2

0

2

0

2

0

2

1ED

2

1

V

UDensity

2

1

2

1

2

1

ECapacitor

BEnergy

VBlABBA

BlU

Back to Circuits

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27

Induction

Series LR Circuit

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RL or LR Series Circuit

Switch is open .. no current flows for obvious reasons.

Switch closed for a long time: Steady current,

voltage across the inductor is zero. All voltage (E) is across the resistor.

i=E/R04/21/23Induction

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RL or LR Series Circuit

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i

t

E/R

When the switch opens, current change is high and back emf from L is maximum.

As the current increases, more voltage is across R, the rate of change of I decreasesand as the current increases, it increases more slowly.

RL Circuit

When L=0, the current rises very rapidly (almost instantly)

As L increases, it takes longer for the current to get to its maximum.

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RL Circuit - Kirchoff Stuff

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0

i

iLiRemf

constant) (time

)1(

)1(

/

)/(

R

L

eR

i

eR

i

Solution

t

tLR

E

E

The Graphic Result – Current Growth

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33

e= 2.71828…

63.01

1

e63% of

maximum}

Decay – Short out the battery

Magnetic field begins to collapse, sending its energy into driving the current.

The energy is dissipated in the resistor.

i begins at maximum (E/R) and decays.04/21/23Induction

34

Solution

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)( /teR

i E

Up and Down and Up and Down and …..

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Induction37NEXT: AC Circuits

addendum – chapter 21

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