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See more @ teodorickbarry.multiply.com PHYSICS DIVISION

2. CONDUCTORS AND INSULATORS • Conductors

are materials, where electrons are free to move about the entire material (ex. Cu and other metals)

• Insulators are materials, where electrons are bound to a nearby atom, rendering no motion (ex.

Wood and glass) • Ion

An atom where electron/(s) is/are added or removed. Normally, a conductor is electrically neutral due to a balance between

positive and negative charges. So in order to create a net charge, free electrons are added or removed from the lattice.

A macroscopic object can be…

Net Charge Property Process

Electrically Neutral p = e- None

Positively Charged p > e- Remove electron

Negatively Charged p < e- Add electron

• Only electrons can be transferred due to the atomic structure, and the minimal amount of energy required.

• Protons are bound by very “strong forces” so their removal is very hard to accomplish. THE ELECTROSCOPE • The Electroscope

Is a device for detecting electric charges • The Diverging Leaves:

Two gold “leaves” diverge when a charge is placed near or in contact with the bob.

The leaves return to normal, when charges are no longer present in the bob

CHARGING BY INDUCTION AND CONDUCTION • By Conduction- charging by contact

Implements an effective transfer of electrons • By Induction – charging without contact, only by placing objects

close to each other Implements only motion of charges within a material

• How to produce a NET charge? • RUB!!!

ARBITRARY, BARI-ABLE • Question1:

When a glass rod is rubbed by silk, which of the two materials acquire a net positive charge? • Answer 1:

Any of the two, as long as the other gets the opposite. We can not know for certain which charge is which. We can only arbitrarily assign a charge • Question 2:

If you walk across a rug and scuff electrons from your feet, are you negatively or positively charged? • Answer 2:

You are positively charge, since electrons were scuffed off/from your feet!

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THE ELECTRIC FIELD AND COULOMB’S LAW • Since Force and the Field are both vectors they follow superposition principle!

• Coulomb’s Law:

• Obtaining the Electric Field

Note: We do not need to know the magnitude of the test charge to calculate the electric field, all we need to know is the magnitude of the charge that produces it, and where we will measure the electric field! ELECTRIC DIPOLES • Electric dipoles are systems composed of two equal and opposite charges q, separated by a

small distance L. • Electric dipole moment describes the strength and orientation of electric dipoles.

5. ELECTRIC FIELD LINES • We can picture the electric field by drawing lines to indicated its direction. • At any given point, the field vector E is tangent to the lines, because they show the direction of

the force exerted on a positive test charge! • At any point near the positive charge, the electric field points radially away from the charge. • Similarly the electric field lines converge toward a point occupied by a negative charge!

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PART THREE: THE ELECTRIC POTENTIAL OUTLINE OBJECTIVES

1. Potential Difference

Continuity of V

Units

Potential and Electric Field Lines

2. Potential due to a System of Point Charges

3. Finding the Electric Field from the Potential

General Relation between E and V

4. V of Continuous Charge Distributions

5. Equipotential Surfaces

The Van de Graff Generator

Dielectric Breakdown

After this chapter you should be able to:

1. Define and differentiate electric potential difference, electric potential, and electrostatic potential energy.

2. Calculate the potential difference between two points, given the electric field in the region.

3. Define of the electron-volt (eV) energy and the conversion factor between eV and the joule.

4. Calculate the electric potential of discrete and continuous charge distributions

__________________________________________________________________ THE CONSERVATIVE ELECTRIC FORCE

Electric Force between two charges is directed along the line of charges and depends on the inverse square of their separation (this is the same as the gravitational force between two mass, Recall Physics 3)

Like Gravitational Force, electric force is conservative! Now, when we say that a force is conservative, there is always a potential energy function U

associated with it! ELECTRIC POTENTIAL

If we place a test charge q0 in an electric field, its potential energy is proportional to q0. The potential energy per unit charge is a function of the position in space of the charge and its

called ELECTRIC POTENTIAL!

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S DIVISION

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CONTINU

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to the positive

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point charget the origin is

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paration?

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e terminal of

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OTENTIAL”

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to U= 0 at inf

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two lighter npton nucleus

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E S | 21

S DIVISION

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VAN DE GRAFF GENERATOR How does a Van de Graff Generator works?

Topics that can explain: 1. Potential 2. Electric Field Lines 3. Conductor Property

DIELECTRIC BREAKDOWN

Happens when non-conducting material become ionized when exposed to very high electric fields and become conductors

Dielectric Strength The magnitude of the electric field for which dielectric breakdown occurs in a material Emax,air = 3 x 106 V/m = 3MN/C

Arc Discharge The discharge through the conducting air resulting from dielectric breakdown. An example is the electric shock you receive when you touch the metal door knob

after walking across a rug on a dry day.

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INTRODU

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UNIT TWOUTLINE

1. Electrosta

2. Capacita

3. The Stora

and Electro

4. Combina

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Energy StoDielectric

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Magnitude

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UCTION When we briWE MUST DOvicinity. The work doThe electrosta system. When a chaThe ratio of tWhen a cha(Why???) The ratio of tA useful devA capacitor When a capconductors cThe ratio of between the

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age of Electri

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lectric Effect

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ne is stored atatic potentia

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the charge tovice for storing consists of 2

pacitor is attacarry equal a the magnite conductors

many uses! attachment

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y.multiply.c

ECTRIC E

l Energy

apacitors

cal Energy

Energy

pacitors:

acitors

presence

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t is your camde the suddecommunicati phones allo

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ENERGYOBJ

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At thbe a

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tic potential a system of

ted conductial is called thated conduc

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source of poe charges. charge on e

e capacitanc

mera uses a en flash of lighion devicesowing them

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Y AND CJECTIVES

he end of thable to:

efine Electros

efine Capaccept to exk;

xplain how ed;

Define a ectrics store

olve probleacitance,

gnitude of thrge;

y to a region at the final po

energy. charges is th

tor, the potenhe capacitactor, the pote

he capacitahe capacitoced but insulatential differe

either conduce of the cap

capacitor toht. such as tto operate

ng near-deaase the right

L E C T U R

CAPACI

his chapter,

static Energy

citance andxplain how

electrostatic

Dielectric electrostat

ms that caelectrostatic

he bound an

where otherosition due to

he total work

ntial of the cnce of the cential of the

nce of the cr. ated from eaence (such a

uctor to the pacitor.

o store the

the radios, at certain

ath patients amount of

R E N O T E

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ITANCE

you should

;

d apply the capacitors

c energy is

and how tic energy;

alculate for c energy, nd the free

r charges areo other charg

needed to a

onductor inconductor. conductor in

onductor.

ach other. as a battery)

potential d

E S | 23

S DIVISION

e present, ges in the

assemble

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ncreases.

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ROSTATIC POThe Electrostcharges fromThe electrost

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CITANCE The potentiacharge Q isconductor.

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acitance of a

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acitor was th

OTENTIAL ENERtatic Potentiam an infinite static potentia

he potential ue for continu

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the charge Q

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rical conduct

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tor of radius R

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s such as the

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L E C T U R

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to all the othof conducto

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onductor is ca

for a given pratio does N

s called a fa

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out with gold

k needed to

her charges. ors.

conductor cae and shap

harge it can

charge Q is

alled its capa

potential diffeNOT depend

rad (F) after

d (1 μF = 10-6

st slide, that t

E S | 24

S DIVISION

foil.

bring the

arrying a e of the

carry for

acitance

erence. on either

r Michael

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the SI unit

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HOW BIG1. Find thAnswer, 2. A sphenew capC1 = C2. tripled, tshape o CAPACITIs a syste

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ore @ teod

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ere of capacpacitance C The capacitthe potentialf the conduc

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1. W 2. Fi

betw PARALL

separa

EL PLATOS

allel-plate caculate the cas capacitor is88.5 pF b. 1.0

arge would t

3 x 108 m2, wh

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ctor and of th

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ctor to the oth +Q and the ance of this do calculate t

We place equind the poteween them.

LEL PLATE CAParallel plaIn practice

ated and insuThe capac

apacitor has apacitance os charged to 6 nC)

the plates ha

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onductor thaout 1400 time

arries a char

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When a cap charge is t potential di

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much charge

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acitance of of the earth!

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s on the condirst finding t

n capacitor e thin metaby a thin plas

ate capacitor

pacitor is coransferred frfference bet across the b of charge tr

cm separated

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tance to be

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e is increased

otential differnce depends

m th

ductors, thenhe electric

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d by 1mm.

d?

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m)

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d to 60μC, w

rence. If the cs only on the

n field E

at are

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nals. Q = CV

e to be sepa

E S | 25

S DIVISION

hat is the

charge is e size and

as shown the other quals the

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Example1. A 15-μ2. How mconduct PP EXAMA parallea batterThe battQuestion1. What 2. How m3. By how THE ELEC

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RICAL CAPACA cylindricalconcentric cA coaxial ccapacitor.

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es: μF capacitor much energtor using a po

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much energyw much is the

CTRIC FIELD ENIn the proceThe work reqthe electric fThat is, we ccalled ELECTThe quantitythe energy s

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acitor with sqed to 12V. isconnected

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radius r1 and

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DENSITY (ue)y

E S | 26

S DIVISION

a larger,

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U = V(∆q),

, we can

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nected to

3.5mm.

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ctric field,

, which is

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PP EXAM4. Calcu4. COMB

1. PARAL

2. SERIES

ore @ teod

MPLE PART TWulate the eneBINATIONS OTwo or moreTypes of Com LLEL

S

PA

Notes for C

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F CAPACITO capacitors a

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ARALLEL CAP

CAPACITORS

e voltagference)

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ference of th

e charge apacitor totored in all of e circuit

you increasapacitors in crease Ceq.

eq is always dividual capapacitors in th

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he upper platd are therefoted togethern parallel, tors!

two capacitors are equa

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ACITORS

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ge (poteacross

he same athe pote

he source

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tors are coal. potential diffof the capac

ential each nd is ential

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n the of the

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2

3

4

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tion is 2.0mm

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o capacitorsmmon potena common p

al difference

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m

s are connecntial Va, and potential Vb. e Va-Vb is t

that the c

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CAPACITOR

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rges stored aors in the circ

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R E N O T E

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cted by a co the lower p the same a

charge on

e individual

S

.

IES

cross all the cuit are the

ence across totals the of source.

citors in the es.

s than the nces of the

E S | 27

S DIVISION

nducting plates are

across all

the two

potential

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5. DIELEC

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CTRICS

Dielectric

Reasons for t

κ (Kappa)

CASE 1:

CASE 2:

In either case

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When a The cap

dielectrithe Increase

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Thus, for the Cap

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uations found

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s are air, glas

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etween the

rence is redu

tric is inserteden removed rtion of the dous slide”, an

otal charge while the batt

harge to the

ious slide, ex

E S | 28

S DIVISION

ss, paper,

c of the

plates is

uced and

d from the

dielectric. nd noting

after the tery is still

plates to

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Dielectri1. The 8(a) Find (b) Find battery. 2. Thedisconne(a) The c(b) the v(c) the c(THIS IS C ENERGY

6. MOLEC

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PIEZOELECTRIC EFFECT

In certain crystals that contain polar molecules such as quartz, tourmaline, and topaz, a mechanical stress applied to the crystal produces polarization of the molecules!

As we all know, again, these polarization mean an electric field is produced, thus a potential difference across the crystals!

USES OF PIEZOELECTRIC EFFECT Transducers in microphone, phonograph pickups, and vibration-sensing devices

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S DIVISIONN



UNIT FOUR: THE MAGNETIC FIELD AND ITS SOURCES OUTLINE OBJECTIVES 1. THE FORCE EXERTED BY A MAGNETIC FIELD 2. MOTION OF A POINT CHARGE IN A MAGNETIC FIELD 3.THE MAGNETIC FIELD OF MOVING POINT CHARGES 4. THE MAGNETIC FIELD OF CURRENTS 5. GAUSS’S LAW FOR MAGNETISM 6. AMPERE’S LAW: LIMIT AND CORRECTION 7. MAGNETISM IN MATTER

At the end of this chapter you must be able to: 1. Calculate the force exerted by a magnetic field; 2. Calculate the magnetic field from various field-source configurations; 3. Calculate parameters from velocity-selector applications; 4. Define the Ampere; 5. Apply Gauss’s Law for Magnetism; and 6. Apply Ampere’s Law;

PART 1: THE MAGNETIC FIELD BRIEF HISTORY

Ancient Greeks (around 2000 years ago) were aware that magnetite attracts pieces of iron. There are written references to the use of magnets for navigation during the 12th century. In 1269, Pierre de Maricourt discovered using simple observations, the existence of magnetic

poles. Note that like poles repel and unlike poles attract. In 1600, William Gilbert discovered that the earth itself is a natural magnet. Although electric charges and magnetic poles are similar in many respects, there is an

important difference: Magnetic Poles always exist as pairs. No isolated magnetic poles were ever observed. 1. THE FORCE EXERTED BY A MAGNETIC FIELD In this course, we will examine the force exerted by a magnetic field on a

› Moving point charge › Current-Carrying Wire

Before we proceed, there are something we need to convene with F is force, q is charge, v is velocity, B is the magnetic field F is force, I is current, ∆L is the length vector, B is the magnetic field We will also, exhaustively discuss, the right-hand rule, to determine the direction of the vector

that results from a cross product. Magnetic Force on a Moving Point Charge

Experimental observations reveal that magnetic force on a moving point charge › Is proportional to q and v, and to the sine of the angle between v and B. › Surprisingly perpendicular to both the velocity and the field.

The abovementioned observations are summarized as the equation below

This is the force exerted on a point charge moving with a velocity v in a magnetic field Since F is perpendicular to both v and B, it is perpendicular to the plane defined by this two

vectors. The direction of F is given by the right-hand rule as v is rotated into B, as illustrated below.

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magnetic fie

a in terms o

E S | 42

S DIVISION

rrent, will

and the

e and its

he unit of

eld of 1T

f a more

N

See mo

EXAMPLE1. A prodirectedforce on 2. A wirethat is inmagneti

3. MAGN

ore @ teod

ES oton is movind downward n the proton.

e segment 3mn the xy-planic force exer

NETIC FIELD LI

Just as the Ebe representIn both case

› (1) D› (2) M

There are holines:

› (1) Emag

› (2) magnort

dorickbarry

ng with a ve and northw

mm long carne and makerted on the w

INES

Electric Field ted by Magn

es Direction of thMagnitude ofowever, two

Electric field gnetic field linElectric field

gnetic field lh pole and e

y.multiply.c

elocity of 10ward, making

rries a currenes an angle

wire segment

E can be repnetic Field Lin

he field is indf the field is in important di

lines are in thnes are perpd lines begilines form cl

enters the sou

P H Y S

com

0Mm/s. It expg an angle o

t of 3A in theof 30o with t.

presented bynes.

dicated by thndicated by ifferences be

he direction pendicular to

n on positivlosed loops! uth pole!

I C S 1 3

periences a of 70o with t

e x-direction.the x-axis, as

y Electric Fiel

e direction o the density oetween elect

of the electr the magnetive charges Note: Magn

L E C T U R

magnetic fihe horizonta

. It lies in a ms shown in th

ld Lines, the

of the lines of the lines tric field line

ric force on aic force on a and end o

netic field lin

R E N O T E

PHYSICS

ield of 0.6G al. Find the m

magnetic fieldhe figure. Wh

Magnetic Fie

es and magn

a positive ch moving chan negative

nes emerge

E S | 43

S DIVISION

which is magnetic

d of 0.02T hat is the

eld B can

netic field

arge, the arge

charges; from the

N

See mo

4. MOTIOCASE 1:

CASE 2:

CASE 3:

ore @ teod

ON OF A POINCYCLOTRONThe magnetmagnetic fieparticle. The magnetbut not its mTherefore, mchange theirIn the specperpendiculparticle und

In this specprovides thfor the ccircular mo We use Nethe quantit T and f aperiod and The cyclodepend oq/m but athe particle

THE HELIX PASuppose thaperpendiculThe velocity

› (1) t› (2) t

The trajector

THE MAGNETThe motion oThe figure bfield is weak

dorickbarry

NT CHARGE IN tic force on eld is alway

tic force, thuagnitude.

magnetic fieldr kinetic enecial case, war to a unifergoes unifo

cial case, thehe centripetacentripetal

otion.

ewton’s Secoties.

are known ad frequency r

tron period on the charre independe!

ATH at a chargedar to the field vector is resohe v| experiehe v|| experiry is then call

TIC BOTTLE of particles inbelow shows at the cente

y.multiply.c

N A MAGNET

a charged s perpendic

us changes th

ds do no worgy. where the form field arm circular m

e magnetic al force nece

acceleratio

ond Law to r

as the cyclrespectively.

and frequrge-to-mass

dent of r and

d particle end B. olved into twences magneiences no maled a helix, w

n non-uniform a magneticer and strong

P H Y S

com

TIC FIELD

particle movular to the

he direction

ork on partic

velocity of s shown in

motion (UCM)

force essary on in

relate

lotron

uency ratio

d v of

nters a unifor

wo componenetic force, anagnetic force

which is illustra

m magnetic fic bottle. This g at both end

I C S 1 3

ving throughvelocity of t

of the veloc

cles and do n

a particlethe figure, t).

rm magnetic

nts nd is thus acce, and remaated below.

ields can be interesting c

ds.

L E C T U R

h a the

city

not

is the

c field with a

celerated. ins constant.

quite complconfiguration

R E N O T E

PHYSICS

a velocity no

lex. n happens w

E S | 44

S DIVISION

ot entirely

when the

N

See mo

5. THE VE “The maan elect

Thbwpp Cth(tg Ifmm Th

APPLICA

ore @ teod

This configurA similar phepoles in the V

ELOCITY SELE

agnetic forcetric force if th

Since the elmagnetic foThe electric perpendiculSuch regions

he Figure sbetween thewhere there perpendiculaplane of the p

Consider a pahis space frotermed Loren

given by:

f q is positimagnitude magnetic forc

he two force

For given mawith the speeThe arrangeAny particlespace undef

› A pdirec

› A paof th

ATIONS The velocity during the laIn this course

dorickbarry

ration is used enomenon isVan Allen Be

CTOR AND A

e on a charghe magnitude

ectric force orce is perpen and magnear to each os is said to ha

hows a rege plates ofis an electr

ar magnetic paper.

article of chaom the left, ntz force) on

ve, the eleqE is dow

ce of magnit

es balances if

agnitudes of ed v = E/B. ment of the f

e with this spflected.

particle with ction of the marticle with lehe electric fie

selector for ate 19th and ee, three appli

y.multiply.c

to trap denss the oscillati

elts

APPLICATIONS

ged particle es and direct

is in the direndicular to thetic fields in other if the forave crossed f

gion of spaf a capaciric field and field into t

arge q enteri the net forn the particle

ectric force wn and ttude qvB is up

f qE = qvB or

the electric

fields gives upeed, regard

greater spemagnetic fielesser speed teld.

r crossed-fieldearly 20th cenications will b

P H Y S

com

se beams of on of ions b

S

moving in a tions of the tw

ection of thehe magnetic the region thrces are to bfields.

ce tor

d a the

ing rce e is

of the p.

and magnet

s the velocitydless of its m

eed than theld. than the velo

ds has very ntury. be discussed

I C S 1 3

plasmas. ack and fort

uniform mawo fields are

e electric fie field. hrough whic

balance!

tic fields, the

y selector v =mass or charg

e velocity s

ocity selector

important ap

:

L E C T U R

th between t

gnetic field c chosen prop

ld (for positiv

ch the partic

e forces bala

= E/B. ge will trave

elector will

r will be defle

pplications t

R E N O T E

PHYSICS

the earth’s m

can be balaperly.”

ve particles)

le is moving

nce only for

erse the cros

be deflecte

ected in the

hat were dis

E S | 45

S DIVISION

magnetic

anced by

and the

must be

particles

ssed-field

ed in the

direction

scovered

N

See mo

1. Thoms

The

EXAMPLEElectronV/m andIf the plaFind the 2. The M

MASS SP

ore @ teod

son’s MeasurJJ Thomson,q/m of electIn his experimcan be defcharged paBy measuringsame q/m. Thomson alssource, whicconstituent o

vo is the veloq/m can be magnetic fie

E s pass undef

d there is a cates are 4cm deflection o

ass SpectromThe mass sp

PECTROMETER

dorickbarry

rement of q/m, in 1897, illutrons. ment, he shoflected by Erticles. g the deflect

so showed thch only meaof all matters

ocity selector determined

eld is only intr

flected throurossed magn long and th

on the screen

meter pectrometer

Aston inof meas

importapresencin natur

has bee10.1% 25

approxi

photog

R EQUATION S

y.multiply.c

m for Electronstrated the

owed that thE and B field

tions of these

hat particles ans that the.

r (E/B) from the eq

roduced at th

ugh the platenetic field of e ends of the

n when the m

, first designn 1919, was dsuring the maSuch mea

ant way of ce of isotopere.

For examplen found to 5Mg, and 11.

These isotopimate ratio oIons ejected

raphic film a

SET

P H Y S

com

ns technique fo

e rays of a cds, thus they

e particles, Th

with this q/mese particles

uation belowhe entrance

es of Thomso1.40G. e plates are 3

magnetic field

ned by Frandeveloped aasses of isotoasurements determining

es and their a

le, natural m consist of 72% 26Mg. pes have ma

of 24:25:26. d from the so

at P2.

I C S 1 3

or measuring

cathode-ray y must consi

homson, show

m can be o (now called

w. .

on’s apparat

30cm from thd is turned of

ncis William as a means opes.

are an g both the abundance

magnesium 78.7% 24Mg,

asses in the

ource move i

L E C T U R

g the

tube ist of

wed that all

obtained usind electrons)

tus when the

he screen. f.

n a semi-circ

R E N O T E

PHYSICS

the particle

ng any mate are a fund

e electric fiel

cular orbit an

E S | 46

S DIVISION

have the

erial for a damental

ld is 3000

d strike a

N

See mo

EXAMPLEA 58Ni ioand def(a) Find (b) Find 58/60.) 3. The Cy

PART 2: BRIEF HIS

1. THE M

ore @ teod

E on of charge lected in a mthe radius of the differenc

yclotron

: SOURCES

STORY Permanent MOersted annJean Baptista magnet neAndre-Marieexperience each other.

AGNETIC FIEL

When a poinproduces a

(For Left) Whμ0 is a const

(For Right) θ

dorickbarry

+e and mamagnetic fieldf curvature ofce in the rad

Lawrenacceleto high CYCLO

EXAMPA cycmaxim(a) Wh(b) Wh

OF THE MA

Magnets wernounced his de Biot and Feear a long cu

e Ampere exa force in th

LD OF MOVIN

nt charge q magnetic fie

here r is calleant of propo

is the angle

y.multiply.c

ss 9.62 x 10-2

d of 0.12T. f the orbit of

dii of curvatu

The cyclonce and erate particleh kinetic ener

OTRON EQUAT

PLE lotron for ac

mum radius ofhat is the cychat is the kine

AGNETIC F

re the earliesdiscovery thaelix Savart anurrent-carryinxtended thee presence

NG POINT CH

moves with aeld B in space

ed the positioortionality ca

between r a

P H Y S

com

26 kg is acce

the ion. ure of 58Ni ion

otron was M.S. Livings

es such as prgies*.

TION SET

ccelerating pf 0.5m lotron freque

etic energy o

IELD

t known sourat a compassnnounced theng wire and aese experime

of a magne

HARGES

a velocity v, e given by:

on vector thaalled the perm

nd v.

I C S 1 3

lerated throu

ns and 60Ni io

invented bston in 19rotons or de

protons has

ency? f the protons

rces of magns needle is dee results of thanalyzed resuents and shotic field and

it

at points frommeability of f

L E C T U R

ugh a poten

ons. (Assume

y E.O. 934 to uterons

a magnetic

s when they e

netism. eflected by aheir measuremults in terms oowed that c that two cu

m the chargeree space, w

R E N O T E

PHYSICS

tial differenc

that the ma

c field of 1.5

emerge?

an electric cments of theof the magnecurrent elemrrents exert f

e to the fieldwhich has the

E S | 47

S DIVISION

ce of 3kV

ass ratio is

5T and a

urrent. force on

etic field. ents also forces on

d point P. e value

N

See mo

EXAMPLEA point along thFind the (1) at the(2) at the(3) at theAns: (1) 2. THE M The calccan be electric The equ

ELECTRIC

3. GAUSS

4. AMPER

ore @ teod

E charge of m

he line y = 3m magnetic fiee origin e point (0,3me point (0, 6m3.89 x 10-10 T

AGNETIC FIEL

culation of thextended tocurrent in a w

ation below

C-MAGNETICThe two equHowever, theE is in the direB is in the dir

B DUE TO

S’S LAW FOR

We know thaMagnetic fieThe magnetpole. Gauss’s Law

That is, no m

RE’S LAW

Ampere’s LaIt relates theloop (called Ampere’s Lasymmetry. It is stated as

dorickbarry

magnitude q m. elds produce

m) m) T in the paper

LD OF CURRE

he magnetico calculate twire.

is known as t

C ANALOGY uations used tere is a distinection of theection perpe

O DIFFERE

MAGNETISM

at magnetic eld lines form ic equivalen

w for Magnetis

agnetic mon

aw is very anae magnetic f Amperian Loaw works for

s:

y.multiply.c

= 4.5 nC is m

ed by this cha

r. (2) 0 (Why?

NTS: BIOT-SA

c field causethe magnetic

the Biot-Sava

to calculate ct difference

e force. endicular to t

ENT CON

field lines dif closed loops

nt of the elec

sm is stated a

nopoles!

alogous to Gield to the coop). r configuratio

P H Y S

com

moving with

arge when th

?). (3) 3.89 x 1

AVART LAW

d by a poinc field cause

art Law.

the magnetie in the direc

the force.

NFIGURA

ffer from elecs.

ctric charge

as:

Gauss’s Law focurrent enclo

ons that hav

I C S 1 3

speed v = 3

he charge is a

10-10 T out the

t charge ed by an

ic field are actional aspec

ATIONS—

ctric field line

is called a m

or Electricity. sed by an im

ve a high de

L E C T U R

3.6 x 107 m/s

at the point

e the paper

nalogous to cts.

—See APP

s.

magnetic

maginary

egree of

R E N O T E

PHYSICS

s parallel to t

(x = -4m, y =

Coulomb’s L

PENDIX A

E S | 48

S DIVISION

the x-axis

3m)

Law.

A

N



EXAMPLES: 1. Find the magnetic field caused by a wire that carries a current I at a point P which is at a perpendicular distance R from the wire. 2. A long straight wire of radius R carries a current I that is distributed uniformly over he cross-sectional area of the wire. Find the magnetic field both inside and outside the wire. LIMITATIONS OF AMPERE’S LAW Ampere’s Law will only work if and only if the following statements hold: 1. The configuration has a very high level of symmetry 2. The current is continuous everywhere in space. Therefore, there are only three cases where Ampere’s Law can be used: 1. Long straight lines 2. Long, tightly wound solenoids 3. Toroids 5. MAGNETISM IN MATTER

Unlike the E field and the dipole moment p, magnetic moments inside all materials tend to increase the magnetic field during alignment.

Materials fall into three categories: (1) Paramagnetic (2) Diamagnetic (3) Ferromagnetic PARAMAGNETISM

Paramagnetism arises from partial alignment of the electron spins (in metals) or of atomic or molecular magnetic moments by an applied magnetic field in the direction of the field.

In paramagnetic materials, the magnetic dipoles do not interact strongly with each other and are randomly oriented.

In the presence of an external magnetic field, the dipoles are partially aligned in the direction of the field, thereby increasing the field.

However, in external magnetic fields of ordinary strength at ordinary temperatures, only a small fraction of the molecules are aligned. The total increase in the field is therefore small.

DIAMAGNETISM

Diamagnetism arises from the orbital magnetic dipole moments induced by an applied magnetic field.

These magnetic moments are opposite the direction of the applied magnetic field so they decrease the total magnetic field B

This effect actually happens to all material, but because of the induced magnetic moments are very small compared to the permanent magnetic moments, diamagnetism is masked by paramagnetic or ferromagnetic moments.

Diamagnetism is thus only observed in materials that have no permanent magnetic moments.



FERROMAGNETISM

Ferromagnetism is much more complicated than paramagnetism because of a strong interaction between neighboring magnetic dipoles.

A high degree of alignment occurs even in weak external magnetic fields, thus causing a great increase in the total field.

Even when there is no external field, ferromagnetic materials may have its dipoles aligned and have its own magnetic field just like a permanent magnet.

See mo

INTRODU

1. MAGN

Exercise

ore @ teod

OUTLINE 1. Magnetic2. Induced 3. Lenz’s La4. Motional5. Eddy Cu6. InductanMutual Indu7. RL Circui

UCTION 1830’s – Mindependencurrent in theThe emfs ancalled inducThe process When you psometimes o

NETIC FLUX

The flux of athe flux of anThe magneti

The unit of fweber (Wb) 1 Wb = 1 T•m

e: Show that a

We are oftcontaining sIf the coil cotimes the flux

dorickbarry

UNI

c Flux EMF and Far

aw l EMF rrents

nce: Self-Induuctance ts

ichael Faradntly discoveree wire. nd currents

ced emfs anditself, is referrpull the plugobserved a sm

a magnetic fin electric fielic flux Φm is d

flux is that of m2

a weber per

ten interesteeveral turns oontains N turx through ea

y.multiply.c

T FIVE: M

raday’s Law

uctance and

day (Englaned that chan

caused by d induced cured to as magg of an elecmall spark. Th

ield through d.

defined as

f a magnetic

second is a v

ed in the fof wire. rns, the flux tch turn.

P H Y S

com

MAGNEOBJ

At thable1. Dflux; 2. Dcalcseve3. Dcalcinduc4. Chredu5. Co6. Lemag7. CRL C

nd) and Jonging magne

changing murrents. gnetic inducctric cord frohis phenomen

a surface is

c field times

volt.

lux through

hrough the c

I C S 1 3

ETIC INDJECTIVES he end of thi to:

Define and c

efine and uulate for t

eral configuraDefine and ulate for tced current iharacterize ace Eddy Cur

ompute Induearn means

gnetic energyompute for ircuits.

seph Henry etic field indu

magnetic fiel

ction. om its sockenon is explain

defined simi

area, tesla-m

a coil

coil is N

L E C T U R

DUCTIO

is chapter, y

compute fo

utilize Faradathe inducedations;

utilize Lenzthe directioin Faraday’s and enumerarrents; ctances; s and waysy; and circuital par

(USA) uces a

ds are

et, you ned by magn

larly to

meter square

R E N O T E

PHYSICS

N

you must be

r magnetic

ay’s Law to d emf for

z’s Law to on of the Law; ate ways to

s in storing

rameters of

netic inductio

ed, which is

E S | 51

S DIVISION

on!

called a

N

See mo

EXERCISFind thecarries a 2. INDUC

1. 2. 3. 4. 5.

In everymagneti

EXA1.

2.

3.

ore @ teod

E magnetic fl

a current of 7

CED EMF AND

Experiments If the

emf We usually d

Obsand

BEFORE: We

betwHOWEVER:

Indu

How to chanThe current pPermanent MCircuit itself mThe orientatiThe area of t

y case, an eic flux.

Figure at theIf the flux throSince emf is the emf. The F/q is theE fields that work done aE fields tha(Meaning, wThese finding

The negativinduced em MPLES: A uniform ma radius of 4in the coil. An 80-turn perpendiculA solenoid o600 G that mthrough the magnetic fie

dorickbarry

lux through a.5A

D FARADAY’S

by Faraday, e magnetic f equal in ma

detect the emerving a cur there is no c

considered ween the ter

uced emfs ca

nge magneticproducing thMagnets maymay be movon of the circthe circuit in

mf is induce

e right shows ough the loo W/q, there m

e E, which in resulted from

across a closeat resulted fwork done acgs are summa

ve sign in Faf, which we w

agnetic field4 cm. The fiel

coil has a ar magnetic

of length 25 cmakes an ansolenoid. (b)

eld is reduced

y.multiply.c

a solenoid th

LAW

Henry and oflux through a

agnitude to thmf by: rrent in the ccurrent.

emfs that wminals of the

an be consid

c flux??? he magnetic y be moved

ved toward ocuit may be a fixed mag

ed in the circ

a single loopop is changinmust be force

this case is inm static elected curve is zefrom changi

cross a closedarized as Fara

araday’s Lawwill discuss sh

d makes an ad changes a

radius of 5 field to prodcm and radingle of 50o ) Find the mad to zero in 1

P H Y S

com

hat is 40cm lo

others showean area bouhe rate of ch

circuit, but it’s

were localizee battery

dered to be d

field may be toward the c

or away from changed netic field m

cuit that is e

p of wire in a g, an emf is ie exerted on

nduced by thtric charges ero) ing magnetd curve is NOaday’s Law, w

w has to dohortly!

angle of 30o wat a rate of 8

.0cm and aduce a curreius 0.8cm witwith the axis

agnitude of t.4s.

I C S 1 3

ong, has a r

d that: nded by a c

hange of the

s present eve

ed in a spec

distributed thr

e increased ocircuit or awa the source o

ay be increa

qual in mag

magnetic fieinduced in th

n the charge

he changing are conserva

ic flux is nOT ZERO!)

which is give

o with the d

with the axis 85T/s. Find the

a resistance nt of 4.0A in tth 400 turns is of the solethe emf indu

L E C T U R

radius of 2.5c

circuit chang flux is induce

en when the

cific part of

rough out the

or decreaseday from it of the flux

ased or decre

gnitude to th

eld. he loop. associated w

flux. ative! (Mean

nonconservat

en below:

direction of

of a circulare magnitude

of 30Ω. Atthe coil? s in an externoid. (a) Findced in the so

R E N O T E

PHYSICS

cm, as 600 tu

ed by any med in the circ

e circuit is inc

f the circuit,

e circuit

d

eased.

he rate of ch

with

ning,

tive!

the

r coil of 300 te of the indu

t what rate

rnal magnetid the magniolenoid if the

E S | 52

S DIVISION

urns, and

means, an uit!

complete

such as

hange of

turns and uced emf

e must a

c field of itude flux

e external

N

See mo

3. LENZ’S

“Theprod

There is a

APPLY LE

ore @ teod

S LAW

The negativebe found fro

e induced emduces them.”

Note: We distatement w

an alternativ“For a chanchange in thThis “counte

ENZ’S LAW IN

dorickbarry

e sign in Faraom a general

mf and induc”

idn’t specify was left vague

e statement ge in magne

he flux!” r flux” will the

N EACH

y.multiply.c

aday’s law hal physical prin

ced current

just what kie to cover a

to Lenz’s Lawetic flux, a c

en give the d

P H Y S

com

as to do withnciple known

are in such

nd of chang variety of co

w to make it counter flux i

direction of th

I C S 1 3

h the direction as Lenz’s La

a direction

ge causes thonditions we w

more operats produced

he induced c

L E C T U R

on of the induaw

so as to opp

he induced ewill now illust

tional! so that there

current or em

R E N O T E

PHYSICS

uced emf, w

pose the cha

emf and currate.

e will be “no

mf!

E S | 53

S DIVISION

which can

ange the

rrent. The

o overall”

N

See mo

4. EDDY

How are

5. INDUC 5.1 SELF-

ore @ teod

CURRENTS Previously, cOften a chamove througThe heat pro

e Eddies prod

Eddy currentcurrent, andPower loss is Eddy currentEddies are oEddies are a

CTANCE

-INDUCTANCEThe magnetother, nearbThe current proportional

dorickbarry

urrents produanging flux segh a region ooduced by suduced?

ts are usually that heat its reduced by ts are not alw

often used to also used in th

E tic flux throug

by circuits* produces a

to I at every

y.multiply.c

EXAMPLE A rectangin a magnportion of of the coil Find the mmoved wit

uced by chaets up circulaof changing much current c

y unwanted bself must be d increasing thways undesira lessen unwahe magnetic

gh a circuit i

a magnetic y point.

P H Y S

com

ular coil of 80netic field B the coil in th is 30Ω.

magnitude anth a speed o

anging flux weating currentmagnetic fluconstitutes a

because powdissipated. he resistanceable.

anted oscillat breaking sys

is related to

field B that

I C S 1 3

0 turns, 20 cm = 0.8T direc

he region of t

nd direction of 2m/s (a) to

ere set up in ts, called Eddx. power loss in

wer is lost in th

e on the poss

ions in severastem of mag

the current

varies from

L E C T U R

m wide and 3cted into ththe magnetic

of the induce the right, (b)

definite circudy currents in

n the conduc

he form of he

ible paths of

al applicationetic transit t

in that circu

point to po

R E N O T E

PHYSICS

30 cm long, ise page, witc field. The re

ed current if t) up, and (c)

uits. n conductor

ctor and the s

eat generate

f the eddies.

ns trains.

it and the c

oint, but B i

E S | 54

S DIVISION

s located th only a esistance

the coil is down.

rs as they

system.

ed by the

urrents in

is always

N

See mo

SI UNIT O

CALCULA

EXAMPLE1. Find th2. At wh20V? The Equa

ore @ teod

The magnetiinductance!

Where L is a The self-indu

OF INDUCTANFrom the eqof flux divide1 H = 1 Wb/AAfter Josepthoroughly d

ATING SELF-IN

ES: he self-induchat rate must

ation that rela

dorickbarry

ic flux throug

constant cactance depe

NCE quation that ded by the uniA = 1 Tm2/A, h Henry, wh

during the 19t

NDUCTANCE

tance of a sot the current

ates Faraday

y.multiply.c

gh the coil is t

lled the self-iends on the

defines self-int of current. (H) is called tho also discth century.

olenoid of lent in the sole

y’s Law with I

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therefore also

nductance ogeometric sh

nductance,

the henry. covered and

ngth 10 cm, anoid in the e

nductance

I C S 1 3

o proportiona

of the coil. hape of the c

we see that

d studied th

area 5 cm2, example abo

L E C T U R

al to I, hence

coil.

the unit of in

he phenome

and 100 turnove change

R E N O T E

PHYSICS

e the definitio

nductance is

enon of ind

ns. to induce a

E S | 55

S DIVISION

on of self-

s the unit

ductance

an emf of

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5.2 MUTU

SEE DERINOTES H

6. MAGN

ore @ teod

UAL INDUCTA

When two othrough oneThe flux, for mutual induc

The mutual inMutual Induc

VATION ON TERE:

NETIC ENERGYAn inductor in it, just as aConsider theThe energy s

The magneti

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ANCE

or more circue circuit does example thctance M(2,1)

nductance Mctances, dep

THE BOARD O

Y AND THE INstores magn

a capacitor ste circuit at thstored in an i

ic energy de

y.multiply.c

uits are close not dependrough circuit and current

M21 = M12, wepend on the

ON HOW TO C

NDUCTOR etic energy tores electrice right. nductor carr

ensity is given

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e to each othd only to its owt 2, is due tot I.

e drop the sugeometric a

CALCULATE FO

through the cal energy.

rying a curren

by:

I C S 1 3

her, as in thewn, but the oo it’s self-ind

bscript and carrangement

OR MUTUAL I

current build

nt I is given b

L E C T U R

e figure abovother’s contriuctance an

call it M. of circuits!

NDUCTANCE

ding up

by:

R E N O T E

PHYSICS

ve, the magbution as wed current I2,

ES:

E S | 56

S DIVISION

netic flux ell. and the

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7. RL CIR

THE GRO

EXAMPLEA coil obattery o(A) Wha(B) What(C) How THE DECA

ore @ teod

RCUITS RL Circuit- cthe one in thFor all RL Cirsolve for the Current I, flowith time. RL Circuits ischarge/discWe just want

OWTH OF I IN When the swbuild up instaIt grows expountil it reachτ is called timWhen curren

E: of self-inductaof negligible t is the final ct is the curren

w much energ

AY OF I IN RLThe circuit aof I”. However, wethe battery aHere, we let Initially the c

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circuit contaihe right. cuits in Physic circuital par

ows in a sing

s very similaharge RL circt to know the

RL CIRCUITS witch is closantaneously.onentially viaes the final c

me constant, nt reaches its

ance 5.0mH internal resiscurrent? nt after 100μsgy is stored in

L CIRCUITS bove is very

e place addand, R1 to pr the circuit at

current I0 = ξ0/

y.multiply.c

ning a resisto

cs 13, we carameters. gle direction

r with RC Ccuits. e behavior of

sed, current . a the equatiocurrent If = ξ0/ τ = L/R, whics maximum, t

and a resisttance.

s? n this inducto

similar to the

ditional switcrotect the battain If, then /R then it stea

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or and an in

an apply Kirc

but it’s valu

Circuits, howe

f currents in t

does not

on above, /R. ch is the time the inductor a

tance of 150

r when the fi

e circuit for t

ches in order attery from suwe proceedadily decays

I C S 1 3

ductor such

hhoff’s Rules

ue is changi

ever, we do

them.

it takes the cacts as a “sh

0 Ω is place

nal current h

he “growth

to remove urge and sho with the dec

s until it is neg

L E C T U R

as

to

ng

on’t

circuit to reachort” or just a

ed across the

has been atta

rt. caying procegligible.

R E N O T E

PHYSICS

ch maximum wire.

e terminals o

ained?

ess of I.

E S | 57

S DIVISION

m current.

of a 12-V

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INTRODU

THE CHA

1. AC GE

ore @ teod

UOUTLINE 1. AC GENE 2. ALTERNARESISTORS & INDUCTOR 3. PHASORS 4. LC, RLC CLC and RLCSeries RLC WParallel RLC 5. TRANSFO

UCTION More than 9by electricalAC’s advantransported currents to reAC can thelower and sfor everyday

ANGE IN WAVHere are som

Constants:ωThese formul

ENERATOR anFigure belowIt consists offield. The ends of tThey make e

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UNIT SIX

ERATORS

ATING CURREN& RMS VALUE

RS AND CAPA

S

CIRCUITS C Without a GWith a Gener

C With a Gen

ORMERS

9% of the elel generators i

ntage over Dover long d

educe energen be transfoafer voltage

y use!

VE FUNCTIONSme of the bas

is angular frelas are very im

nd the GENERw shows a simf a coil of ar

the coil are celectrical con

y.multiply.c

X: ALTER

NT: ES, ACITORS

Generator rator erator

ectrical energin the form o

DC because istances at v

gy losses due ormed, with es and corre

S sic formulas f

equency, δ ismportant in o

RATION OF Ample AC generea A and N

connected tontact through

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RNATINGOBJAt thable1. UworkEMF;2. CAlterCapcalc3. relatdiffe4. Coinvol(or n5.Cochar

gy used todaof alternating electrical every high vo to Joule heaalmost no espondingly h

for obtaining

s phase diffeobtaining the

LTERNATING erator.

N turns rotatin

o rings (calleh stationary c

I C S 1 3

G CURRJECTIVES he end of thi to: nderstand h

k and comp

Comprehendrnating curacitor, andulate for circDefine Ph

ionships betwrences in ACompute for Aved in an LCot) by an AC

ompute for thracteristics.

ay is produce current (ac)

energy can bltage and lo

at! energy loss, higher curren

g changes in

rence

e value of so

CURRENT

ng (with freq

d slip rings) thconducting b

L E C T U R

ENT CIR

is chapter, y

how an AC pute for the

d the beharrent in ad an Inducuital paramehasors andween circuit

C; AC-circuital

C and RLC CirC generator; he Transforme

ed .

be ow

to nts

wave functio

me AC circu

quency ω) in

hat rotate wibrushes in co

R E N O T E

PHYSICS

RCUITS

you must be

Generator e maximum

vior of an a Resistor, uctor and eters; d identify al potential

parameters rcuits driven

er’s

ons:

uital paramet

n a uniform m

ith the coil. ontact with th

E S | 58

S DIVISION

ters.

magnetic

he rings.

N

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EXAMPLEA 250-tu

ore @ teod

The emf in th

Or

Where

We can thusmagnetic fieAs we all knoWith an Alte

E: rn coil has an

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he coil will the

s produce a seld. ow, with an inrnating EMF,

n area of 3cm

y.multiply.c

en be:

sinusoidal em

nduced emf, there is also

m2. If it rotate

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mf in a coil by

, there is also an alternatin

es in a magne

I C S 1 3

y rotating it w

o an inducedng current!

etic field of 0

L E C T U R

with constant

current.

0.4T at 60Hz, w

R E N O T E

PHYSICS

t angular vel

what is ξmax?

E S | 59

S DIVISION

ocity in a

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2. ALTERN 2.1 RESIS

THE POW

RMS VAL

ore @ teod

NATING CUR

STORS IN AC

WER DISSIPATE

LUES Most ac amcurrent and RULE: The RMquantity divi*The rms curthe actual aExample: The

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RENT IN CIRC

ED IN A RESIST

meters and v voltage rath

MS value of aded by √2. rrent equals t

ac current. e rms value o

y.multiply.c

CUITAL ELEME

TOR

voltmeters arer than the m

any quantity

the steady d

of a current, I

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NTS

re designed maximum or that varies s

dc current th

Irms is given b

I C S 1 3

to measure r peak valuesinusoidally e

at would pro

by:

L E C T U R

root-mean-ss! equals the ma

oduce the sa

R E N O T E

PHYSICS

quare (rms) v

aximum valu

ame Joule he

E S | 60

S DIVISION

values of

ue of that

eating as

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RMS EXE 1. Find P2. Find P3. Find P4. Find Irm5. A 12-Ωcurrent, Note: Inresistor isof VR,rms! 2.2 ALTER

INDUCTO

VL Leads

EXAMPLEA 40mH reactana) 60 Hz

ore @ teod

ERCISE:

av in terms ofav in terms ofav in terms ofms in terms of Ω resistor is c(b) the avera

a circuit, ths not usually

RNATING CU

Alternating cWhen a capan open circBut if the cuand at highelike a short cConversely, for dc. But when thfrequencies,

ORS IN AC CI

s I by 90o In the previoThis functionWe say that This is illustratAs with earlieE inductor is pce and the m

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f Irms and R f ξmax and Ima

f ξrms and Irms ξrms and R

connected aage power, (

hat consists oequal to the

RRENT IN IND

current behavpacitor becocuit. urrent alternaer frequencie

circuit! an inductor

he current is the back em

IRCUITS

ous set of slideal difference I is “out of phted by the pler technique

laced acrossmaximum cu

y.multiply.c

ax

across a sinus(c) the maxim

of more thae generator v

DUCTORS AND

ves differentlmes fully cha

ates, charge es, the capac

coil usually h

alternating, mf is so large,

es, we see the is due to thehase” with VL

ot at the righs we can tra

s an ac generrent when th

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soidal emf thmum power.

n a generatvoltage, so w

D CAPACITOR

ly than directarged in a dc

continually citor, will hard

has a very sm

a back em, the inducto

he functional e current I’s pL, more preciht. nsform the e

erator that hahe frequency

I C S 1 3

hat has a pe

tor and a rewe write volta

RS

t current in inc circuit, it st

flows onto odly impede c

mall resistanc

f is generateor acts like an

difference ophase differeisely VL leads

quation to p

as a maximuy is

L E C T U R

eak value of

esistor, the voage drop ac

nductors andtops the curre

or off the placurrent at all,

ce and is esse

ed in an indn open circui

of VL and I. ence with vol current by 9

prove this “lea

m emf of 120

R E N O T E

PHYSICS

48V. Find (a

oltage drop ross a resisto

capacitors. ent, that is, it

ates of the c, which mea

entially a sho

ductor, and at!

tage VL. 90o.

ading” pheno

0V. Find the i

E S | 61

S DIVISION

) the rms

across a r in terms

t acts like

capacitor ns, it acts

ort circuit

at higher

omenon

inductive

N

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b) 2000 HWhat ca CAPACIT

VC Lags

EXAMPLEA 20-μF reactanA) 60 Hz B) 5000 HWhat ca 3. PHASO

ore @ teod

Hz an you concl

TORS IN AC C

I by 90o In the previoThis functionWe say that This is illustratAs with earlieE: capacitor is ce and the m Hz an you concl

ORS The phase rinductor canPhasors rotcounterclocWhen severathey are conMeaning, ccircuital paraaddition, usinConsider a capacitor Cseries. Since they acurrent, whicof the currenThe voltagdefinitions wand inductiv

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ude about th

CIRCUITS

ous set of slideal difference I is “out of phted by the pler technique

placed acromaximum cu

ude about th

relations betn be representates counkwise in the Cal componennected in p

complicationsameters canng phasors! circuit con

C, and a re

are in seriesch is represent phasor I. es are ob

which includeve reactance

y.multiply.c

he relation of

es, we see the is due to thehase” with VC

ot at the righs we can tra

oss a generarrent when th

he relation of

tween the cnted by two terclockwiseCCS). nts are connarallel, their cs in the co

n be simplified

taining an sistor R, all

, they all caented as the

btained usines the resistives.

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com

f inductive re

he functional e current I’s pC, more precht. nsform the e

ator that has he frequency

f capacitive

current and dimensional

e (since in

nected togetcurrent add.

omputation d using vect

inductor L, connected

arry the sam x-compone

ng the prie, capacitiv

I C S 1 3

eactance an

difference ophase differecisely VC lags

quation to p

a maximum y is

reactance a

the voltage vectors calle

ncreasing a

ther in series of or

a in

me nt

or e,

L E C T U R

nd current?

of VC and I. ence with vol current by 90

prove this “lag

emf of 100V

and current?

drop in a ed phasors. angular de

circuit, their

R E N O T E

PHYSICS

tage VC. 0o.

gging” pheno

V. Find the ca

resistor, cap

grees are

r voltages ad

E S | 62

S DIVISION

omenon

apacitive

pacitor or

moving

dd, when

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4. RLC C 4.1 LC C

There ar

EXAMPLA 2-μF cfrequenc 4.2 RLC C

ore @ teod

CIRCUITS

IRCUITS WITHFigure to theIn an LC CircWhen the swThe effect is current in thinductor. This circuit is

e important Angular “Na

Current:

E: capacitor is cy of oscillati

CIRCUITS WITIf we includinductor, weIt is basicallcharging/disIt is a sprforces!

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HOUT A GENERe right shows cuit, we assumwitch is closed that, the cae inductor, w

very similar to

parameters iatural” Freque

charged to ion? (b) Wha

THOUT A GENde a resistoe have an RLCy the same scharging dring system

y.multiply.c

RATOR an LC Circuitme that the cd, charge be

apacitor is bewhich in turn

o a mass atta

involving LC ency:

20V and is at is the maxim

ERATOR r in series wC Circuit. as an LC c

does not ha that enco

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t. capacitor caegins to flow teing discharg, re-charges

ached to a s

Circuits witho

then connecmum value o

with a

circuit, appen ounters

I C S 1 3

arries an initiathrough the i

ged by the lo the capacit

spring.

out a genera

cted across of the current

L E C T U R

al charge Q0inductor. oss of chargetor and deca

ator:

a 6-μH indut?

R E N O T E

PHYSICS

.

e, then it incrays the curre

uctor. (a) Wh

capac

howev“forev

E S | 63

S DIVISION

rease the ent in the

hat is the

citor and

ver, the ver”

frictional

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4.3 SERIE

RESONA

EXAMPLE1. A serie100 V anresonanc2. A serie100 V ancapacito3. A resipeak voC=14.7μ

ore @ teod

ES RLC WITH A

NCE IN SERIE

Resonance iThere conditAt resonanc

ES: es RLC Circuind a variablce, (c) the pes RLC Circuind a variablor. istor R and c

oltage of 220F, Find Vout,rm

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A GENERATOR

ES RLC

in circuit is whtions for resone, the power

it with L = 2H,e frequencyhase angle δit with L = 2H,e frequency

capacitor C 0V, 60Hz, as

ms.

y.multiply.c

R

hen the impenance are gir factor is 1.

, C = 2μF, any. Find (a) thδ, (d) the pow, C = 2μF, an

y. Find the m

are in seriesshown in the

P H Y S

com

edance is at iven in the rig

d R = 20Ω is de resonancewer factor, ad R = 20Ω is d

maximum volt

s with a gene Figure. If R

I C S 1 3

its smallest, aght.

driven by a ge frequency and (e) the avdriven by a gtage across

nerator with R=20 Ω and

L E C T U R

and the curre

generator wit(f0), (b) the verage powegenerator witthe resistor, t

R E N O T E

PHYSICS

ent is at its gr

th a maximu maximum cer delivered.th a maximuthe inductor

E S | 64

S DIVISION

eatest.

m emf of current at m emf of

r and the

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4.4 PARA

RESONA

1) 2)

5. TRANS

THE TRAN

ore @ teod

ALLEL RLC WIT

NCE IN PARA

Conditions aparallel RLC:Impedance The currents they are oppthe current in

SFORMERS A transformea circuit withoA simple tracommon ironThe coil carryThe coil carryThe transformcircuit induceinductances The iron corethat nearly acoil.

NSFORMER EQFor a transfsecondary, generator em

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TH A GENERA

ALLEL RLC

are basically : is a maximum in the inducposite, so then the resistor.

r is a device uout an apprensformer con

n core. ying the input ying the outpumer operates o

es an alterna of the two cir increases the

all the magne

QUATIONS former with the voltagemf across the

y.multiply.c

ATOR

the same, ho

m, current is ctor and capey cancel, th.

used to raise ciable loss of

nsisting of two

power is calleut power is caon the principating emf in rcuits. e magnetic fieetic flux throu

N1 turns in e across the e primary coi

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com

owever, we n

a minimum pacitor are ehe total curr

or lower the v power o wire coils a

ed the primaralled the secople that an aa nearby cir

eld fir a givenugh one coil

the primary secondary il by:

I C S 1 3

note some im

qual, but ent is just

voltage in

around a

ry. ondary. lternating currcuit due to t

n current and goes through

y and N2 tucoil is relat

L E C T U R

mportant fea

rrent in one the mutual

guides it so h the other

urns in the ed to the

R E N O T E

PHYSICS

atures of reso

E S | 65

S DIVISION

onance in

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See mo

EXAMPLEA doorbconnectcurrent i

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If there are n

E: bell requires 0ted to a 120n the primary

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no losses, due

0.4A at 6V. It-V ac line. (ay?

_____

y.multiply.c

e to Joule He

t is connectea) How man

___________

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eating (which

ed to a transy turns shoul

____end___

I C S 1 3

h is due to ne

former whosd there be i

___________

L E C T U R

egligible resist

e primary con the second

______

R E N O T E

PHYSICS

tance in the

ontaining 200dary? (b) Wh

E S | 66

S DIVISION

coils),

00turns, is hat is the

N

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