Induction

INDUCTIONSpring ‘09

DIS HERE WEEKWe begin the study of magnetic

inductionThere will be a quiz on FridayThere is a new WebAssign.Exam #3 is Wednesday4/8

It will include the material covered through April 6th.

The end is in sight …Check the website for the Final Exam

Schedule

AMPERE’S LAW

)( 210

0

iid

id enclosed

sB

sB

USE THE RIGHT HAND RULE IN THESE CALCULATIONS

LAST TIME: A SIMPLE EXAMPLE

FIELD AROUND A LONG STRAIGHT WIRE

enclosedid 0 sB

riB

irB

2

2

0

0

THE FIGURE BELOW SHOWS A CROSS SECTION OF AN INFINITE CONDUCTING SHEET CARRYING A CURRENT PER UNIT X-LENGTH OF L; THE CURRENT EMERGES PERPENDICULARLY OUT OF THE PAGE. (A) USE THE BIOT–SAVART LAW AND SYMMETRY TO SHOW THAT FOR ALL POINTS P ABOVE THE SHEET, AND ALL POINTS P´ BELOW IT, THE MAGNETIC FIELD B IS PARALLEL TO THE SHEET AND DIRECTED AS SHOWN. (B) USE AMPERE'S LAW TO FIND B AT ALL POINTS P AND P´.

FIRST PART

Vertical ComponentsCancel

APPLY AMPERE TO CIRCUIT

Infinite Extent

B

B

Li

: thereforeis loop theinsideCurrent

lengthunit per current

L

THE “MATH”

Infinite Extent

B

B

20

0

0

B

LBLBL

id enclosedsB

Bds=0

Infinite Sheet of Charge

Infinite sheet of current

2

2

0

0

B

E

COIL OR SOLENOID

Switches

Valve Application

THE MAGNETIC FIELD

A PHYSICAL SOLENOID/COIL MODEL

INSIDE THE SOLENOID

For an “INFINITE” (long) solenoid the previous problem and SUPERPOSITION suggests that

the field OUTSIDE this solenoid is small!

MORE ON LONG SOLENOIDField is ZERO far from coil!

Field is ZERO far from coil

Field looks “UNIFORM”

THE REAL THING…..

Weak Field

Stronger - Leakage

Fairly Uniform field

Finite Length

ANOTHER WAY

0

:

0

0

0

niBnihBhh

id

Ampere

enclosed

sB

Far away

n=number of turns per unit length. nh=total number of turns.

THE LENGTH OF A SOLENOID IS DOUBLED BUT THE NUMBER OF TURNS REMAINS THE SAME.

A. The magnetic field is doubled.B. The magnetic field is cut to a quarter.C. The magnetic field does not change.D. The magnetic field is quadrupled.E. The magnetic field is cut in half.

THE NUMBER OF TURNS IS DOUBLED BUT THE LENGTH REMAINS THE SAME.

A. The magnetic field is quadrupled.B. The magnetic field is cut to a quarter.C. The magnetic field is doubled.D. The magnetic field does not change.E. The magnetic field is cut in half.

APPLICATION Creation of Uniform Magnetic Field Region Minimal field outside

except at the ends!

TWO COILS

“REAL” HELMHOLTZ COILSUsed for experiments.

Can be aligned to cancelout the Earth’s magneticfield for critical measurements.

THE TOROID

Slightly lessdense than

inner portion

THE TOROID

rNiB

so

totalNirBd

Ampere

2

turns)# (N 2

:nintegratio ofpath thein contained coil INNER about the

only worry need Weagain.

0

0

sB

NEW CONCEPT – WELL SORTA NEW

MAGNETIC FLUX

MAGNETIC FLUX

MdAB

SurfaceOpen - Gauss Like AB dM

MAGNETIC FLUX IS A

A. VectorB. ScalarC. Tensor

MAGNETIC FLUX

SurfaceOpen - Gauss Like AB dM

For a CLOSED Surface we might expect this to be equal to some constant times the

enclosed poles … but there ain’t no such thing!

0 AB dCLOSED SURFACECLOSED SURFACE

A PUZZLEMENT ..

enclosedpathclosed id 0 sB

Let’s apply this to the gap of a capacitor.

CONSIDER THE POOR LITTLE CAPACITOR…

i i

CHARGING OR DISCHARGING …. HOW CAN CURRENTFLOW THROUGH THE GAP

In a FIELD description??

THROUGH WHICH SURFACE DO WE MEASURE THE CURRENT FOR AMPERE’S LAW?

I=0

IN THE GAP… DISPLACEMENT CURRENT

dtd

dtdqI

ntDisplaceme

Idtdq

Letdtdq

dtd

qEA

Ed

d

E

E

0

0

0

2

Current

gap)(in

1

S through FLUX ELECTRIC The

SO FAR .. We found that currents create magnetic fields. Stationary charges do not. Static magnetic fields do not create currents. What about CHANGING magnetic fields??

FROM THE DEMO ..

FARADAY’S EXPERIMENTS

??

INSERT MAGNET INTO COIL

REMOVE COIL FROM FIELD REGION

Summary

THAT’S STRANGE …..

These two coils are perpendicular to each otherThese two coils are perpendicular to each other

REMEMBER THE DEFINITION OF TOTAL REMEMBER THE DEFINITION OF TOTAL ELECTRIC FLUX THROUGH A ELECTRIC FLUX THROUGH A CLOSEDCLOSED SURFACE:SURFACE:

dA

is surface aLEAVING FieldElectric theofFlux Total

outE

surface EE d

nE

MAGNETIC FLUX - REMINDER Applies to an OPEN SURFACE only. “Quantity” of magnetism that goes through a

surface. A Scalar

AB dBsurface

CONSIDER A LOOP Magnetic field passing through the loop is CHANGING.

FLUX is changing. There must be an emf

developed around the loop. A current develops (as

we saw in demo) Work has to be done to

move a charge completely around the loop.

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FARADAY’S LAW (MICHAEL FARADAY) Again, for a current

to flow around the circuit, there must be an emf.

(An emf is a voltage) The voltage is found

to increase as the rate of change of flux increases.


FARADAY’S LAW (MICHAEL FARADAY)

dtdemf

Law sFaraday'xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

We will get to the minus sign in a short time.

FARADAY’S LAW (THE MINUS SIGN)


Using the right hand rule, wewould expect the directionof the current to be in thedirection of the arrow shown.

FARADAY’S LAW (MORE ON THE MINUS SIGN)


The minus sign means that the current goes the other way.

This current will produce a magnetic field that would be coming OUT of the page.

The Induced Current therefore creates a magnetic field that OPPOSES the attempt to INCREASE the magnetic field! This is referred to as Lenz’s Law.

HOW MUCH WORK?


dtddVqW

sE/

ChargeWork/Unit

A magnetic field and an electric field areintimately connected.)

emf

MAGNETIC FLUX

This is an integral over an OPENOPEN Surface. Magnetic Flux is a Scalar The UNIT of FLUX is the weber

1 weber = 1 T-m2

AB dB

WE FINALLY STATED

dtddVemf

sE

FARADAY’s LAW

FROM THE EQUATION

AB dB

dtddVemf

sE

LentzLentz

FLUX CAN CHANGE

If B changes If the AREA of the loop changes Changes cause emf s and currents and consequently

there are connections between E and B fields

These are expressed in Maxwells Equations

AB dB

MAXWELL’S FOUR EQUATIONS

0dABsurfaceclosed

Gauss

Faraday

Ampere’s Law

0 AB dNo Monopoles

ANOTHER VIEW OF THAT DAMNED MINUS SIGN AGAIN …..SUPPOSE THAT B BEGINS TO INCREASE ITS MAGNITUDE INTO THE PAGE

The Flux into the page begins to increase.

An emf is induced around a loop

A current will flow That current will create a

new magnetic field. THAT new field will

change the magnetic flux.


THE STRANGE WORLD OF DR. LENTZ

LENZ’S LAWInduced Magnetic Fields always FIGHT to stop what you are trying to do!i.e... Murphy’s Law for Magnets

EXAMPLE OF NASTY LENZ

The induced magnetic field opposes thefield that does the inducing!

DON’T HURT YOURSELF!

The current i induced in the loop has the directionsuch that the current’s magnetic field Bi opposes thechange in the magnetic field B inducing the current.

Let’s do theLentz Warp

again !

AGAIN: LENZ’S LAW

An induced current has a directionsuch that the magnetic field due tothe current opposes the change in the magnetic flux that induces thecurrent. (The result of the negative sign!) …

ORThe toast will always fall buttered side down!

AN EXAMPLE

The field in the diagramcreates a flux given byB=6t2+7t in milliWebersand t is in seconds.

(a)What is the emf whent=2 seconds?

(b) What is the directionof the current in the resistor R?

THIS IS AN EASY ONE …

mVemf

tdtdemf

ttB

31724seconds 2at t

712

76 2

Direction? B is out of the screen and increasing.Current will produce a field INTO the paper (LENZ). Therefore current goes clockwise and R to left in the resistor.

The diagram shows two parallel loops of wire having a common axis. The smaller loop (radius r) is above the larger loop (radius R) by a distance x >> R. Consequently, the magnetic field due to the current i in the larger loop is nearly constant throughout the smaller loop. Suppose that x is increasing at the constant rate of dx/dt = v. (a) Determine the magnetic flux through the area bounded by the smaller loop as a function of x. In the smaller loop, find (b) the induced emf and (c) the direction of the induced current.

v

DOING IT

B is assumed to be constant through the center of the small loop and caused by the large one.

THE CALCULATION OF BZ

2/322

20

2/122220

2/122

220

2

4

cos

4coscos

xRiRB

RddsxRR

xRidsdB

xRR

xRidsdBdB

z

z

z

MORE WORK

In the small loop:

VxiRr

dtdemf

xiRr

xRiRrBrAB zz

4

20

2

3

20

2

2/322

20

22

23

2

)prescribed asAway (Far RFor x2

dx/dt=v

WHICH WAY IS CURRENT IN SMALL LOOP EXPECTED TO FLOW??

WHAT HAPPENS HERE? Begin to move

handle as shown. Flux through the

loop decreases. Current is induced

which opposed this decrease – current tries to re-establish the B field.

MOVING THE BAR

RBLv

Remfi

BLvdtdxBL

dtdemf

BLxBAFlux

sign... minus theDropping

MOVING THE BAR TAKES WORK

v

RvLBP

vRvLBP

FvFxdtd

dtdWPOWER

RvLBF

orRBLvBLBiLF

222

22

22

WHAT ABOUT A SOLID LOOP??

METAL Pull

Energy is LOSTBRAKING SYSTEM

Back to Circuits for a bit ….

DEFINITION

Current in loop produces a magnetic fieldin the coil and consequently a magnetic flux.

If we attempt to change the current, an emfwill be induced in the loops which will tend tooppose the change in current.

This this acts like a “resistor” for changes in current!

REMEMBER FARADAY’S LAW

dtddVemf

sE

Lentz

LOOK AT THE FOLLOWING CIRCUIT:

Switch is open NO current flows in the circuit. All is at peace!

CLOSE THE CIRCUIT…

After the circuit has been close for a long time, the current settles down.

Since the current is constant, the flux through the coil is constant and there is no Emf.

Current is simply E/R (Ohm’s Law)


When switch is first closed, current begins to flow rapidly.

The flux through the inductor changes rapidly. An emf is created in the coil that opposes the

increase in current. The net potential difference across the resistor is

the battery emf opposed by the emf of the coil.


dtdemf

0

)(

dtdiRV

notationVEbattery

MOVING RIGHT ALONG …

0

solonoid, aFor N. turns,ofnumber the toas wellas

current the toalproportion isflux The

0

)(

dtdiLiRV

dtdiL

dtd

NLii

dtdiRV

notationVE

B

battery

DEFINITION OF INDUCTANCE L

iNL B

UNIT of Inductance = 1 henry = 1 T- m2/A

is the flux near the center of one of the coilsmaking the inductor

CONSIDER A SOLENOID

n turns per unit length niBor

nliBl

id enclosed

0

0

0

sBl

SO….

AnlL

orAlnL

oriniAnl

inlBA

iNL B

2

20

0

lengthunit inductance/

Depends only on geometry just like C andis independent of current.

INDUCTIVE CIRCUIT Switch to “a”. Inductor seems like a

short so current rises quickly.

Field increases in L and reverse emf is generated.

Eventually, i maxes out and back emf ceases.

Steady State Current after this.

i

THE BIG INDUCTION As we begin to increase the current in the coil The current in the first coil produces a

magnetic field in the second coil Which tries to create a current which will

reduce the field it is experiences And so resists the increase in current.

BACK TO THE REAL WORLD…

i

0

equationcapacitor theas form same

0

:0 drops voltageof sum

dtdqR

CqE

dtdiLiRE

Switch to “a”

SOLUTION

RL

eREi LRt

constant time

)1( /

SWITCH POSITION “B”

/

0

0

teREi

iRdtdiL

E

Max Current Rate ofincrease = max emfVR=iR

~current

constant) (time

)1( /

RL

eREi LRt

Solve the loop equation.

IMPORTANT QUESTION Switch closes. No emf Current flows for a while It flows through R Energy is conserved (i2R)

WHERE DOES THE ENERGY COME FROM??

FOR AN ANSWERRETURN TO THE BIG C

We move a charge dq from the (-) plate to the (+) one.

The (-) plate becomes more (-)

The (+) plate becomes more (+).

dW=Fd=dq x E x d+q -q

E=0A/d

+dq

THE CALC

20

202

0

2

00

22

0

2

00

00

21

eunit volumenergy

21

21

21)(

2

2

)()()(

E

E

u

AdAdAdAA

dW

or

qAdqdq

AdW

dAqdqddqEddqdW

The energy is inthe FIELD !!!

WHAT ABOUT POWER??

RidtdiLiiE

i

iRdtdiLE

2

:

powerto

circuit

powerdissipatedby resistor

Must be dWL/dt

SO

2

2

21

21

CVW

LiidiLW

dtdiLi

dtdW

C

L

L

Energystoredin the

Capacitor

WHERE IS THE ENERGY??

l

AlNiBA

lNi

niBnilBll

id enclosed

0

0

0

0

0

B

or

0

sB

REMEMBER THE INDUCTOR??

turn.onegh flux throu MagneticΦcurrent.

inductorin turnsofNumber

iN

iNL

SO …

lAiN

lNiANiW

lNiA

iNiNiLiW

LNi

iNL

2220

0

0

0

22

21

21

21

21

21

2

0

2

0

22

0

0

2220

0

21

or

(volume) 2

12

1

B

:before From2

1

BVWu

VBlAlBW

lNi

lAiNW

ENERGY IN THEFIELD TOO!

IMPORTANT CONCLUSION A region of space that contains either a

magnetic or an electric field contains electromagnetic energy.

The energy density of either is proportional to the square of the field strength.

Induction

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Transcript of Induction