alternating currentv3

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    MagnetismAlternating-Current Circuits

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    Alternating Current Generators

    Alternating current motor

    A coil of area A and N turns rotating with constant angular velocity in a uniform

    magnetic field produces a sinusoidal emf

    Instead of mechanically rotating, we can apply an ac potential difference

    generated by other ac generator to the coil. This produces an ac current in the

    coil, and the magnetic field exerts forces on the wires producing a torque that

    rotates the coil.

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    The magnetic field.Magnetic forces on moving charges.

    Magnetic forces on a current element

    Torques on current loops and magnets

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    The Magnetic Field

    Magnets

    The Earth is a natural magnet with

    magnetic poles near the north and

    south geographic poles.

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    The Magnetic Field

    Magnets Does an isolated magnetic

    pole exist?

    The SI units of magnetic

    field is the tesla [T]

    Earth magnetic field isabout 10-4 T

    Powerful laboratories

    produce fields of 1 -2 T as a

    maximum

    1 Gauss [G] = 10-4 Tesla [T]

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    The Magnetic Field

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    The Magnetic Field. Magnetic force on a moving charge

    A proton is moving in a region of crossed fields E =

    2x105 N/C and B = 3000 G, as shown in the figure.

    (a) What is the speed of the proton if it is not

    deflected. (b) If the electric field is disconnected,

    draw the path of the proton

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    The Magnetic Field. Magnetic force on a current element

    In the case of a straight segment

    of length L

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    The Magnetic Field. Torques on Current loops and

    Magnets

    A circular loop of radius 2 cm has 10 turns of wire and carries a current of 3 A. The

    axis of the loop makes an angle of 30º with a magnetic field of 8000 G. Find the

    magnitude of the torque on the loop.

    B

    A current-carrying loopexperiences no net force in a

    uniform magnetic field, but it does

    experience a torque that tends to

    twist the loop

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    Sources of Magnetic Field

    The Magnetic Field of Moving Charges

    The Magnetic Field of Currents. The Biot-

    Savart Law

    The Magnetic Field Due to a Current loop

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    Sources of Magnetic Field

    Moving Point Charges are the source of Magnetic Field

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    A solenoid is a wire tightlywounded into a helix of

    closely space turns . A

    solenoid is used to produce a

    strong, uniform magnetic

    field in the region surrounded

    by the loops

    For a long

    solenoid Find the magnetic field at the center of asolenoid of 600 turns, length 20 cm;

    radius 1.4 cm that carries a current of 4 A

    Sources of Magnetic Field

    A solenoid

    In this figure, the length is ten

    times longer than the radius

    n = N/L; N number of turns;

    L : length of solenoid

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    Sources of Magnetic Field: Solenoid and magnets

    Left: Magnetic field lines of a solenoid; right Magnetic field lines of a bar magnet

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    Magnetic Induction

    Magnetic Flux

    Induced EMF and Faraday´s Law

    Lenz´s Law

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    MAGNETIC INDUCTION

    In 1830, Michel Faraday in England and Joseph Henry in the USA

    independently discovered that in a changing magnetic field, a changing

    magnetic flux through a surface bounded by a stationary loop of wire induces

    a current in the wire: emf induced and induced current. This process is

    known as induction. 

    In a static magnetic field, a changing magnetic flux through a surface bounded

    by a moving loop of wire induces an emf in the wire: motional emf 

    Magnetic flux through a surface bounded by a loop of wire.

    The SI unit for magnetic flux is

    weber [Wb] 1Wb= 1 T·m2 Find the magnetic flux through a 40 cm longsolenoid with a 2.5 cm radius and 600 turns

    carrying a current of 7.5 A.

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    The induced emf is in such direction as to

    oppose, or tend to oppose, the change that

    produces it. Lenz´s Law

    MAGNETIC INDUCTION

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    The coil with many turns of wire gives

    a large flux for a given current in thecircuit. Thus, when the current

    changes, there is a large emf induced

    in the coil opposing the change. This

    self-induced emf  is called a back emf 

    MAGNETIC INDUCTION

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    Eddy Currents

    Heat produced by eddy currents constitute a power loss in a

    transformer. But eddy currents have some practical applications:

    damping mechanical oscillations, magnetic braking system

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    InductanceSelf-inductance The SI unit of

    inductance is

    the henry [H]

    1 H = 1 Wb/A= 1 T.m2

    .A-1

    Find the self-inductance of a solenoid of length 10 cm,

    area 5 cm2, and 100 turns

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    Magnetic Energy

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    Alternating Current Generators

    Alternating current motor

    A coil of area A and N turns rotating with constant angular velocity in a uniform

    magnetic field produces a sinusoidal emf

    Instead of mechanically rotating, we can apply an ac potential difference

    generated by other ac generator to the coil. This produces an ac current in the

    coil, and the magnetic field exerts forces on the wires producing a torque that

    rotaes the coil.

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    Alternating Current GeneratorsA coil of area A and N turns rotating with constant angular velocity in a uniform

    magnetic field produces a sinusoidal emf

      f    f    frequency

    t  NBAdt d 

    t  NBA

    t and  NBA

     peak 

    m

    m

    m

    π ω 

    δ ω ε ε 

    δ ω ω φ ε 

    δ ω φ 

    δ ω θ θ φ 

    2;

    )sin(

    )sin(

    )cos(

    cos

    =

    +=

    +=−=

    +=+==

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    Alternating Current Circuits:Alternating current in a Resistor

    Inductors in Alternating CurrentsCapacitors in Alternating Currents

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    Potential drop across the resistor, VRCurrent in the resistor I

    Power dissipated in the resistor, P

    Average power dissipated in the

    resistor Paverage

    Alternating currents in a Resistor

     R

    V  I t 

     R

    V  I 

    t V t V 

     peak  R

     peak 

     peak  R

     peak  R R

    ,,

    ,max

    cos

    coscos

    =⇒=

    ===

    ω 

    ω ω ε ε 

     Rt  I  R I  P 

     Rt  I  R I  P 

    av peak avav

     peak 

    ))(cos()(

    )(cos

    222

    222

    ω 

    ω 

    ==

    ==

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    Root-Mean-Square Values

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    Inductors in Alternating Current Circuits

    The potential drop across the inductor

    led the current 90º (out of phase)

    Instantaneous power delivered by the emf to the inductor is not zero

    The average power delivered by the emf to the inductor is zero.

    )2

    cos(sincos

    coscos

    ,,,

    ,max

    π ω 

    ω ω 

    ω ω 

    ω ω ε ε 

    −==→=

    ====

    t  L

    V t 

     L

    V  I t 

     L

    dt 

    dI 

    t V t dt 

    dI  LV 

     peak  L peak  L peak  L

     peak  L L

    INDUCTIVE

    REACTANCE   L

    V  I 

     L

    V  I 

      rms L

    rms

     peak  L

     peak ω ω 

    ,,;   ==

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    Inductors in Alternating Current Circuits

    Instantaneous power delivered by the emf to the inductor is not zero

    The average power delivered by the emf to the inductor is zero.

    INDUCTIVE

    REACTANCE

     L

    V  I 

     L

    V  I 

      rms L

    rms

     peak  L

     peak ω ω 

    ,,;   ==

    The potential drop across a 40-mHinductor is sinusoidal with a peak

    potential drop of 120 V. Find the

    inductive reactance and the peak current

    when the frequency is (a) 60 Hz, and

    (b) 2000 Hz

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    Capacitors in Alternating Current CircuitsThe

    potential

    drop lagsthe

    current

    by 90º

    Power delivered by the emf in the capacitor: Instantaneous and average

    CAPACITIVE

    REACTANCE

    )2

    cos(1

    sin

    cos

    coscos

    ,

    ,

    ,

    ,max

    π ω 

    ω 

    ω ω 

    ω 

    ω ω ε ε 

    +

      

     

     

     =→−==

    =

    ⇒====

    V  I t CV 

    dt 

    dQ I 

    t CV Q

    t V t C 

    QV 

     peak C 

     peak C 

     peak C 

     peak C C 

       

      

    =   

      

    =

    V  I 

    V  I 

      rmsC 

    rms

     peak C 

     peak 

    ω ω 1

    ;1

    ,,

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    Driven RLC  Circuits

    Series RLC circuit  The Kirchhoff´s rules governthe behavior of potentialdrops and current across the

    circuit.

    (a)When any closed-loop is

    traversed, the algebraic sum

    of the changes of potentialmust equal zero (loops rule)

    (b)At any junction (branch

     point) in a circuit where the

    current can be divided, the

    sum of the currents into the

     junction must equal the sum

    of the currents out of the

     junction (junction rule)

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    )cos(

    cos

    ;cos

    ,

    2

    ,

    δ ω 

    ω 

    ω 

    −=

    =++

    =++=

    t  I  I 

    t V C 

    Q R

    dt 

    dQ

    dt 

    Qd  L

    dt 

    dQ

     I C 

    Q

     IRdt 

    dI 

     Lt V 

     peak 

     peak app

     peak app

    Series RLC circuits

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    Power delivered to the series RLC circuit

    δ δ 

    δ 

    δ ε δ ε 

    π ω ω ε ε 

    coscos2

    1

    /cos/

    coscos2

    1

    )2

    cos(cos

    ,,

    ,

    2

    2

    ,

    2

    2

    rmsrmsapp peak  peak app

     peak app peak 

    rmsapprmsav

    rmsrms peak  peak av

     peak  peak 

     I V  I V  P 

     Z V  I and  Z  Ras Z 

     RV  RI  P 

    resistor theindissipated  RI  P 

     I  I  P 

    t  I t  I  P 

    ==

    ==

    ==

    =

    ==

    −==

    Power factor:cosδ

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    Phasors

    Potential drop across a resistor can be

    represented by a vector VR, which is called a

    phasor. Then, the potential drop across the

    resistor IR, is the x component of vector VR,

    Potential drop across aseries RLC circuit

    Q IR

    dt 

    dI  Lt V app   ++=ω cos

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    In the circuit shown in the figure, the

    ac generator produces an rms voltageof 115 V when operated at 60 Hz. (a)

    What is the rms current in the circuit

    (b) What is the power delivered by

    the ac generator (c) What is the rms

    voltage across: points AB; points

    BC; points CD; points AC; pointsBD?.

    A certain electrical device draws 10 A rms and has an average power of 720 W

    when connected to a 120-V rms 60-Hz power line. (a ) What is the impedance

    of the device? (b) What series combination of resistance and reactance is thisdevice equivalent to? (c) If the current leads the emf, is the reactance inductive

    or capacitive?

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    The Transformer

    Because of the iron core, there is a large

    magnetic flux through each coil, even when

    the magnetizing current Im in the primary

    circuit is very small .

    The primary circuit consists of an ac

    generator and a pure inductance (we considera negligible resistance for the coil). Then the

    average power dissipated in the primary coil

    is zero. Why?: The magnetizing current in

    the primary coil and the voltage drop across

    the primary coil are out of phase by 90º1

    1

    2

    222

    11

    V  N 

     N V  N V 

     N V 

    dt 

    dt 

    turn

    turn

    =→=

    =

    φ 

    φ 

    Secondary coil open circuit 

    The potential drop across the

    primary coil is

    If there is no flux leakage out of the ironcore, the flux through each turn is the

    same for both coils, and then

    A transformer is a device to raise or lower

    the voltage in a circuit without an

    appreciable loss of power. Power losses

    arise from Joule heating in the smallresistances in both coils, or in currents

    loops (eddy currents) within the iron core.

    An ideal transformer is that in which these

    losses do not occur, 100% efficiency.

    Actual transformers reach 90-95%

    efficiency

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    The Transformer

    rmsrmsrmsrms  I V  I V 

     I  N  I  N 

    ,2,2,1,1

    2211

    =

    =

    However, the potential drop in the primary is determined by the generator emfAccording to this, the total flux in the iron core must be the same as

    when there is no load in the secondary. The primary coil thus draws an additional current

    I1 to maintain the original fluxΦturn. The flux through each turn produced by this

    additional current is proportional to N1I1. Since this flux equals – Φ́ turn, the additional

    current I1 in the primary is related to the current I2 in the secondary by

    These curents are 180 º out of phase and produce counteracting fluxes. Since I2 is in

    phase with V2, the additional current I1 is in phase with the potential drop across the

    primary. Then, if there are no losses

    A resistance R, load resistance, in the

    secondary circuit

    A current I2 will be in the secondary coil, which is

    in phase with the potential drop V2 across theresistance. This current sets up and additional flux

    Φ´ turn through each turn, which is proportional to

    N2I2. This flux opposes the original flux sets up

    by the original magnetizing current Im in the

    primary.