Transcript phys1444-lec13

PHYS 1444 – Section 004
Lecture #13
Thursday July 19, 2012
Ian Howley
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Chapter 29
Generator
Transformer
Thursday, July 19, 2012
Test next Thursday 7/26
HW due Tuesday 7/24
PHYS 1444 - Ian Howley
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Induction of EMF
How can we induce an emf?
Let’s look at the formula for magnetic flux
 B  B  dA  B cos q dA


What do you see? What are the things that can change
with time to result in change of magnetic flux?
– Magnetic field
– The area of the loop
– The angle q between the
field and the area vector
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Example 29 – 5
Pulling a coil from a magnetic field. A square coil of wire with side
5.00cm contains 100 loops and is positioned perpendicular to a
uniform 0.600-T magnetic field. It is quickly and uniformly pulled
from the field (moving perpendicular to B) to a region where B drops
abruptly to zero. At t=0, the right edge of the coil is at the edge of
the field. It takes 0.100s for the whole coil to reach the field-free
region. Find (a) the rate of change in flux through the coil, (b) the emf and current induced,
and (c) how much energy is dissipated in the coil if its resistance is 100W. (d) what was the
average force required?
What should be computed first? The initial flux at t=0.
2
2
The flux at t=0 is  B  B  A  BA  0.600T   5  10 m   1.50  103 Wb
The change of flux is  B  0  1.50  103 Wb  1.50  103 Wb
Thus the rate of change of the flux is
 B 1.50  103 Wb
 1.50  102 Wb s

0.100s
t
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Example 29 – 2, cnt’d
Thus the total emf induced in this period is
d B
  N
 100  1.50  102 Wb s  1.5V
dt


The induced current in this period is
I

1.5V

 1.50  102 A  15.0mA
R 100W
Clockwise to
compensate for
the loss of flux
through coil
Which direction would the induced current flow?
The total energy dissipated is

2
E  Pt  I Rt  1.50  10 A
2
Force for each coil is F  Il  B


2
 100W  0.100s  2.25  103 J
Force for N coil is F  NIl  B


F  NIlB  100  1.50  102 A  4  5  102  0.600T  0.045 N
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EMF Induced on a Moving Conductor
• Another way of inducing emf is using a U shaped
conductor with a movable rod resting on it.
• As the rod moves at a speed v, it travels vdt in
time dt, changing the area of the loop by dA=lvdt.
• Using Faraday’s law, the induced emf for this loop is
d  B BdA Blvdt


 Blv
 
dt
dt
dt
–This equation is valid as long as B, l and v are
perpendicular to each other.
•An emf induced on a conductor moving in a
magnetic field is called a motional emf
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Electric Generator (Dynamo)
• An electric generator transforms mechanical
energy into electrical energy
• It consists of many coils of wires wound on an
armature that can be rotated in a magnetic
field
• An emf is induced in the rotating coil
• Electric current is the output of a generator
• Which direction does the output current flow when the
armature rotates counterclockwise?
– Initially the current flows as shown in figure to reduce flux through
the loop
– After half a revolution, the current flow is reversed
• Thus a generator produces alternating current
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How does an Electric Generator work?
• Let’s assume the loop is rotating in a uniform B field with
constant angular velocity w. The induced emf is
•    d  B   d  B  dA   d  BA cos q 
dt
dt
dt
• What is the variable that changes above?
–The angle q. What is dq/dt?
•The angular speed w.
–So qq0+wt
–If we choose q0=0, we obtain
–    BA d cos t   BA sin  t
dt

–If the coil contains N loops:
–What is the shape of the output?
N
d B
 NBA sin  t   0 sin  t
dt
•Sinusoidal w/ amplitude 0=NBAw
• US AC frequency is 60Hz. Europe uses 50Hz
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Example 29 – 9
An AC generator. The armature of a 60-Hz AC generator
rotates in a 0.15-T magnetic field. If the area of the coil is
2.0x10-2m2, how many loops must the coil contain if the peak
output is to be 0=170V?
 0  NBA
The maximum emf of a generator is
Solving for N
Since   2 f
0
N
BA
We obtain
0
170V
N
 150turns


2
2

1
2 BAf 2   0.15T    2.0  10 m    60 s 
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A DC Generator
• A DC generator is almost the same as an ac
generator except the slip rings are replaced by splitring commutators
Smooth output using
many windings
• Dips in output voltage can be reduced by using
a capacitor, or more commonly, by using many
armature windings
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Transformer
• What is a transformer?
– A device for increasing or decreasing an AC voltage
– Examples, the complete power chain from generator to
your house, high voltage electronics
– A transformer consists of two coils of wires known as the
primary and secondary
– The two coils can be interwoven or linked by a laminated
soft iron core to reduce eddy current losses
• Transformers are designed so
that all magnetic flux produced
by the primary coil pass
through the secondary
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How does a transformer work?
• When an AC voltage is applied to the primary, the
changing B it produces will induce voltage of the
same frequency in the secondary wire
• So how would we make the voltage different?
– By varying the number of loops in each coil
– From Faraday’s law, the induced emf in the secondary is
d B
– VS  N S
dt
–The input primary voltage is
d B
– VP  N P
dt
–Since dB/dt is the same, we obtain
– VS N S
Transformer

Thursday,V
July 19, 2012
Equation 11
NP
P
PHYS 1444 - Ian Howley
Transformer Equation
• The transformer equation does not work for DC current
since there is no change of magnetic flux
• If NS>NP, the output voltage is greater than the input so
it is called a step-up transformer while NS<NP is called
step-down transformer
• Now, it looks like energy conservation is violated since
we can get a larger emf from a smaller ones, right?
–
–
–
–
–
Wrong! Energy is always conserved!
A well designed transformer can be more than 99% efficient
The power output is the same as the input:
VP I P  VS I S
I S VP N P
Thursday, July
19, 2012 
I P VS N S
The output current for step-up transformer will be lower than the
input, while it is larger for a step-down transformer than the input.
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Example 29 – 12
Cell Phone Charger. The charger for a cell phone contains a transformer
that reduces 120-V ac to 5.0V ac to charge the 3.7V battery. The
secondary contains 30 turns, and the charger supplies 700mA. Calculate
(a) the number of turns in the primary; (b) the current in the primary; and
(c) the power transformed.
(a) What kind of a transformer is this? A step-down transformer
VP N P
V

Since
We obtain N P  N S P 
VS N S
V
S
IS
(b) Also from the
transformer equation I
P
VP

VS
(c) Thus the power transformed is
We obtain
I P  I S VS 
VP
P  I S VS 
How about the input power?
Thursday, July 19, 2012
The same assuming 100% efficiency.
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Example 29 – 13: Power Transmission
Transmission lines. An average of 120kW of electric power is sent to
a small town from a power plant 10km away. The transmission lines
have a total resistance of 0.4W. Calculate the power loss if the power
is transmitted at (a) 240V and (b) 24,000V.
We cannot use P=V2/R since we do not know the voltage along the
transmission line. We, however, can use P=I2R.
P 120  103
I 
 500 A.
240
V
(a) If 120kW is sent at 240V, the total current is
Thus the power loss due to the transmission line is
P I 2 R   500 A   0.4W  100kW
P 120  103
. 
 5.0 A.
(b) If 120kW is sent at 24,000V, the total current is I  V
3
24  10
2
Thus the power loss due to transmission line is
P  I 2 R   5 A   0.4W  10W
2
The higher the transmission voltage, the smaller the current, causing less loss of energy.
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This is why power is transmitted w/ HV, as high as 170kV.
E Field due to Magnetic Flux Change
• When electric current flows through a wire, there is an
electric field in the wire that moves electrons
• We saw, however, that changing magnetic flux
induces a current in the wire. What does this mean?
– There must be an electric field induced by the changing
magnetic flux.
• In other words, a changing magnetic flux produces an
electric field
• This result applies not just to wires but to any
conductor or any region in space
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Generalized Form of Faraday’s Law
• What is the relation between belectric field and the
potential difference Vab  E  dl
a
• The induced emf in a circuit is equal to the work done
per unit charge by the induced electric field
•   E  dl



d B
E  dl  
dt
• The integral is taken around a path enclosing the
area through which the magnetic flux B is changing.
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Ch 30 Inductance, Electromagnetic
Oscillations, AC Circuits
Thursday July 19, 2012
Ian Howley
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Chapter 30
Mutual Inductance
Self Inductance
Thursday, July 19, 2012
This material is NOT on
the test next week…but
you should still pay
attention!
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Inductance
• Changing the magnetic flux through a circuit
induces an emf in that circuit
• An electric current produces a magnetic field
• From these, we can deduce
– A changing current in one circuit must induce an emf in
a nearby circuit  Mutual inductance
– Or induce an emf in itself  Self inductance
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Mutual
Inductance
If two coils of wire are placed near each other, a changing
current in one will induce an emf in the other.
What is the induced emf, 2, in coil 2 proportional to?
– Rate of change of the magnetic flux passing through it
This flux is due to current I1 in coil 1
If 21 is the magnetic flux in each loop of coil 2 created by coil1 and N2
is the number of closely packed loops in coil 2, then N221 is the total
flux passing through coil2.
If the two coils are fixed in space, N221 is proportional to the current I1
in coil 1,
N 2  21  M 21 I
1
• The proportionality constant for this is called the Mutual
Inductance and defined by M 21  N 2 21 I1
• The emf induced in coil2 due to the changing current in
d  N 2  21 
d  21
dI1
coil1 is
  N

 M
2
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dt
dt
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dt
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Mutual Inductance
• The mutual induction of coil2 with respect to coil1, M21,
– is a constant and does not depend on I1.
– depends only on “geometric” factors such as the size, shape,
number of turns and relative position of the two coils, and whether a
ferromagnetic material is present
– The further apart the two coils are the less flux passes through coil
2, so M21 will be less.
– Typically the mutual inductance is determined experimentally
• Just as a changing current in coil 1 will induce an emf in coil
2, a changing current in coil 2 will induce an emf in coil 1
1   M12
dI 2
dt
dI 2
dI1
1  M
and  2  M
–We can put M=M12=M21 and obtain
dt
dt
–SI unit for mutual inductance is henry (H)
1H  1V  s A  1W  s
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Example 30 – 1
Solenoid and coil. A long thin solenoid of length l and
cross-sectional area A contains N1 closely packed
turns of wire. Wrapped around it is an insulated coil of
N2 turns. Assume all the flux from coil1 (the solenoid)
passes through coil2, and calculate the mutual
inductance.
First we need to determine the flux produced by the solenoid.
What is the magnetic field inside the solenoid? B  0 N1 I1
l
Since the solenoid is closely packed, we can assume that the field lines
are perpendicular to the surface area of the coils 2. Thus the flux
0 N1 I1
through coil2 is
 21  BA 
Thus the mutual
M 21
inductance of coil2 is
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A
l
0 N1 N 2
N 2  21 N 2 0 N1 I1


A
A
I1
I1
l
l
PHYS 1444 - Ian Howley
Note that M21 only depends on geometric factors!
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Self Inductance
• The concept of inductance applies to a single isolated coil of
N turns. How does this happen?
–
–
–
–
When a changing current passes through a coil
A changing magnetic flux is produced inside the coil
The changing magnetic flux in turn induces an emf in the same coil
This emf opposes the change in flux. Whose law is this?
• Lenz’s law
• What would this do?
– When the current through the coil is increasing?
• The increasing magnetic flux induces an emf that opposes the original current
• This tends to impede the increased current
– When the current through the coil is decreasing?
• The decreasing flux induces an emf in the same direction as the current
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Self Inductance
• Since the magnetic flux B passing through an N turn
coil is proportional to current I in the coil, N  B  L I
• We define self-inductance, L:
N B
Self Inductance
L
I
•The induced emf in a coil of self-inductance L is
d
dI
B



N
 L
–
dt
dt
–What is the unit for self-inductance? 1H  1V  s
A 1W  s
•What does magnitude of L depend on?
–Geometry and the presence of a ferromagnetic material
•Self inductance can be defined for any circuit or part
of a circuit
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