digested handout in electricity and magnetism
DESCRIPTION
A compilation of topics under electricity and magnetism.TRANSCRIPT
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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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See more @ teodorickbarry.multiply.com PHYSICS DIVISION
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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See more @ teodorickbarry.multiply.com PHYSICS DIVISION
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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See more @ teodorickbarry.multiply.com PHYSICS DIVISION
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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If κ=1 (Meaning there is no dielectric), the bound charge density, σb, is zero If κ=∞ (Meaning a conducting slab is inserted between the plate), the bound charge density, σb, is equal to the free charge density, σf!
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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See more @ teodorickbarry.multiply.com PHYSICS DIVISION
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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ntains the cu. re segment,
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E S | 42
S DIVISION
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NETIC FIELD LI
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E S | 43
S DIVISION
which is magnetic
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4. MOTIOCASE 1:
CASE 2:
CASE 3:
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ON OF A POINCYCLOTRONThe magnetmagnetic fieparticle. The magnetbut not its mTherefore, mchange theirIn the specperpendiculparticle und
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E S | 44
S DIVISION
ot entirely
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agnetic forcetric force if th
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ATIONS The velocity during the laIn this course
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ged particle es and direct
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th between t
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R E N O T E
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the earth’s m
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ve particles)
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nce only for
erse the cros
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ected in the
hat were dis
E S | 45
S DIVISION
magnetic
anced by
and the
must be
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1. Thoms
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EXAMPLEElectronV/m andIf the plaFind the 2. The M
MASS SP
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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
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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
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R E N O T E
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the particle
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E S | 46
S DIVISION
have the
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ld is 3000
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EXAMPLEA 58Ni ioand def(a) Find (b) Find 58/60.) 3. The Cy
PART 2: BRIEF HIS
1. THE M
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E S | 47
S DIVISION
ce of 3kV
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5T and a
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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
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4. AMPER
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C-MAGNETICThe two equHowever, theE is in the direB is in the dir
B DUE TO
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RE’S LAW
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R E N O T E
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PENDIX A
E S | 48
S DIVISION
the x-axis
3m)
Law.
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P H Y S I C S 1 3 L E C T U R E N O T E S | 49
See more @ teodorickbarry.multiply.com PHYSICS DIVISION
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.
P H Y S I C S 1 3 L E C T U R E N O T E S | 50
See more @ teodorickbarry.multiply.com PHYSICS DIVISION
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.
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INTRODU
1. MAGN
Exercise
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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
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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
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T FIVE: M
raday’s Law
uctance and
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caused by d induced cured to as magg of an elecmall spark. Th
ield through d.
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MAGNEOBJ
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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
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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
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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
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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
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LAW
Henry and oflux through a
agnitude to thmf by: rrent in the ccurrent.
emfs that wminals of the
an be consid
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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
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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
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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
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e circuit
d
eased.
he rate of ch
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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
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3. LENZ’S
“Theprod
There is a
APPLY LE
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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
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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
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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
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e will be “no
mf!
E S | 53
S DIVISION
which can
ange the
rrent. The
o overall”
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4. EDDY
How are
5. INDUC 5.1 SELF-
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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
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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
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to I at every
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EXAMPLE A rectangin a magnportion of of the coil Find the mmoved wit
uced by chaets up circulaof changing much current c
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gh a circuit i
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magnitude anth a speed o
anging flux weating currentmagnetic fluconstitutes a
because powdissipated. he resistanceable.
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0 turns, 20 cm = 0.8T direc
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nd direction of 2m/s (a) to
ere set up in ts, called Eddx. power loss in
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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
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ible paths of
al applicationetic transit t
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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
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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
P H Y S
com
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
dorickbarry
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
P H Y S
com
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
N
See mo
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
dorickbarry
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
P H Y S
com
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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See mo
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
dorickbarry
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
P H Y S
com
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
See mo
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
dorickbarry
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
P H Y S
com
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
N
See mo
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
dorickbarry
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
P H Y S
com
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
dorickbarry
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
P H Y S
com
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
See mo
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
dorickbarry
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.
P H Y S
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
N
See mo
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!
dorickbarry
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
P H Y S
com
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
N
See mo
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
dorickbarry
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
N
See mo
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
dorickbarry
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
P H Y S
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
N
See mo
EXAMPLEA doorbconnectcurrent i
ore @ teod
If there are n
E: bell requires 0ted to a 120n the primary
dorickbarry
no losses, due
0.4A at 6V. It-V ac line. (ay?
_____
y.multiply.c
e to Joule He
t is connectea) How man
___________
P H Y S
com
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