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MagnetismAlternatingCurrent 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 104 T
Powerful laboratories
produce fields of 1 2 T as a
maximum
1 Gauss [G] = 104 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 currentcarrying 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
selfinduced 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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InductanceSelfinductance The SI unit of
inductance is
the henry [H]
1 H = 1 Wb/A= 1 T.m2
.A1
Find the selfinductance 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
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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RootMeanSquare 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
V
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 40mHinductor 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
+
=→−==
=
⇒====
t
C
V I t CV
dt
dQ I
t CV Q
t V t C
QV
peak C
peak C
peak C
peak C C
=
=
C
V I
C
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 closedloop 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
C
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 120V rms 60Hz 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
d
dt
d
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 9095%
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.