Transcript Chapter 29 Electromagnetic Induction and Faraday`s Law

Chapter 29
Electromagnetic Induction
and Faraday’s Law
Copyright © 2009 Pearson Education, Inc.
29-6 Transformers and Transmission of
Power
I P  BP   P  N P BP A
BP  I P   P  N P   I P  A
B
BS  BP !
  S  N S BS A  N S B P A  N S   I P  A
dI
dP
 N P A P
dt
dt
dS
dI
 N S A P
VS  
dt
dt
N
V
 S  S
VP N P
VP  
Conservation of Energy:
I PVP  I SVS

VS I P

VP I S

B
B

B
29-6 Transformers and Transmission
of Power
Example 29-12: Cell phone charger.
The charger for a cell phone contains a
transformer that reduces 120-V ac to 5.0-V ac
to charge the 3.7-V battery. (It also contains
diodes to change the 5.0-V ac to 5.0-V dc.)
Suppose the secondary coil contains 30 turns
and the charger supplies 700 mA. Calculate
(a) the number of turns in the primary coil, (b)
the current in the primary, and (c) the power
transformed.
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29-6 Transformers and Transmission
of Power
Transformers work only if the current is
changing; this is one reason why electricity
is transmitted as ac.
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29-6 Transformers and Transmission
of Power
Example 29-13: Transmission lines.
An average of 120 kW of electric power is sent
to a small town from a power plant 10 km away.
The transmission lines have a total resistance
of 0.40 Ω. Calculate the power loss if the power
is transmitted at (a) 240 V and (b) 24,000 V.
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ConcepTest 29.12b Transformers II
1) 1/4 A
Given that the intermediate
2) 1/2 A
current is 1 A, what is the
3) 1 A
current through the
4) 2 A
lightbulb?
5) 5 A
1 A
120 V
240 V
120 V
ConcepTest 29.12b Transformers II
1) 1/4 A
Given that the intermediate
current is 1 A, what is the
current through the
lightbulb?
2) 1/2 A
3) 1 A
4) 2 A
5) 5 A
Power in = Power out
240 V  1 A = 120 V  ???
1 A
The unknown current is 2 A.
120 V
240 V
120 V
ConcepTest 29.12c Transformers III
A 6 V battery is connected to
one side of a transformer.
Compared to the voltage drop
1) greater than 6 V
2) 6 V
across coil A, the voltage
3) less than 6 V
across coil B is:
4) zero
A
6V
B
ConcepTest 29.12c Transformers III
A 6 V battery is connected to
1) greater than 6 V
one side of a transformer.
2) 6 V
Compared to the voltage drop
across coil A, the voltage
3) less than 6 V
across coil B is:
4) zero
The voltage across B is zero.
Only a changing magnetic flux
induces an emf. Batteries can
provide only dc current.
A
6V
B
29-7 A Changing Magnetic Flux
Produces an Electric Field
A changing magnetic flux induces an electric
field; this is a generalization of Faraday’s
law. The electric field will exist regardless of
whether there are any conductors around:
 
dB
E d  
dt
But! But! Isn’t the integral of E around a
closed path ZERO? Kirchoff’s Loop Rule?
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29-7 A Changing Magnetic Flux
Produces an Electric Field
Basic requirement of Conservative Potential:
V  V  r   V r ,t 
 F  V  F  t 
  not a Conservative Force!
 Kirchoff's Rules do not apply
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29-7 A Changing Magnetic Flux
Produces an Electric Field
Example 29-14: E produced by
changing B
B.
A magnetic field B between the pole
faces of an electromagnet is nearly
uniform at any instant over a circular
area of radius r0. The current in the
windings of the electromagnet is
increasing in time so that B changes in
time at a constant rate dB/dt
B at each
point. Beyond the circular region (r > r0),
we assume B
B = 0 at all times. Determine
the electric field E at any point P a
distance r from the center of the
circular area due to the changing B
B.
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29-8 Applications of Induction:
Sound Systems, Computer Memory,
Seismograph, GFCI
This microphone works by induction; the
vibrating membrane induces an emf in the coil.
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29-8 Applications of Induction:
Sound Systems, Computer Memory,
Seismograph, GFCI
Differently magnetized
areas on an audio tape
or disk induce signals in
the read/write heads.
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29-8 Applications of Induction:
Sound Systems, Computer Memory,
Seismograph, GFCI
A seismograph has a fixed coil and a magnet
hung on a spring (or vice versa), and records
the current induced when the Earth shakes.
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Summary of Chapter 29
• Magnetic flux:
• Changing magnetic flux induces emf:
• Induced emf produces current that
opposes original flux change.
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Summary of Chapter 29
• Changing magnetic field produces an electric
field.
• General form of Faraday’s law:
.
• Electric generator changes mechanical
energy to electrical energy; electric motor
does the opposite.
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Summary of Chapter 29
• Transformer changes magnitude of
voltage in ac circuit; ratio of currents is
inverse of ratio of voltages:
and
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Chapter 30
Inductance, Electromagnetic
Oscillations, and AC Circuits
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Units of Chapter 30
• Mutual Inductance
• Self-Inductance
• Energy Stored in a Magnetic Field
• LR Circuits
• LC Circuits and Electromagnetic Oscillations
• LC Circuits with Resistance (LRC Circuits)
• AC Circuits with AC Source
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Units of Chapter 30
• LRC Series AC Circuit
• Resonance in AC Circuits
• Impedance Matching
• Three-Phase AC
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30-1 Mutual Inductance
B1  N1I1
B1   21
dB1
dI1

 N1
dt
dt
d  21
dI1
 N2
 N1 N 2
  2
dt
dt
dI1
  2   N1 N 2
  Geometry 
dt
dI1
dI 2
  2   M 21
and 1   M 12
dt
dt
M 12  M 21  M
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30-1 Mutual Inductance
Unit of inductance: the henry, H:
1 H = 1 V·s/A = 1 Ω·s.
A transformer is an
example of mutual
inductance.
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30-1 Mutual Inductance
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 coil 1 (the solenoid)
passes through coil 2, and calculate the mutual
inductance.
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30-2 Self-Inductance
A changing current in a coil will also induce
an emf in itself:
Here, L is called the self-inductance:
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30-2 Self-Inductance
Example 30-3: Solenoid inductance.
(a) Determine a formula for the self-inductance
L of a tightly wrapped and long solenoid
containing N turns of wire in its length l and
whose cross-sectional area is A.
(b) Calculate the value of L if N = 100, l = 5.0 cm,
A = 0.30 cm2, and the solenoid is air filled.
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30-2 Self-Inductance
Conceptual Example 30-4: Direction of emf in
inductor.
Current passes through a coil from left to right
as shown. (a) If the current is increasing with
time, in which direction is the induced emf?
(b) If the current is decreasing in time, what
then is the direction of the induced emf?
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30-2 Self-Inductance
Example 30-5: Coaxial cable
inductance.
Determine the inductance per unit
length of a coaxial cable whose
inner conductor has a radius r1 and
the outer conductor has a radius r2.
Assume the conductors are thin
hollow tubes so there is no
magnetic field within the inner
conductor, and the magnetic field
inside both thin conductors can be
ignored. The conductors carry equal
currents I in opposite directions.
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30-3 Energy Stored in a Magnetic Field
Just as we saw that energy can be stored in
an electric field:
2
Energy 1
E 
 0 E
Volume 2
energy can be stored in a magnetic field as
well, in an inductor, for example:
2
Energy
1
B 

B
Volume 2 0
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30-3 Energy Stored in a Magnetic Field
Example: a solenoid
N
Bs  0   I  0 nI  L  0 N 2 A
 
Let I  t    t
dEB
 dI 
2
Pext  t   I  t    t    t   L   L t 
dt
 dt 
I t 
t
 EB  t    Pext  t '   L t ' dt '  L
L
0
0
2
2
2
EB LI 2 0 N 2 A
1
B
2
 B 


I2 
 0nI   s
V
2V
2 0
2 0
2A
t
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t
2
2
2
2
Questions?
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Have a great
Spring(?) Break!
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