Generating electricity - Homepage | WileyCHAPTER 7 Generating electricity 167...
Transcript of Generating electricity - Homepage | WileyCHAPTER 7 Generating electricity 167...
c07GeneratingElectricity 166 9 June 2016 5:11 PM
REMEMBER
Before beginning this chapter, you should be able to:
describe how a magnetic eld exerts a force on a current
describe the operation of a simple DC motor, including the role of the commutator.
KEY IDEAS
After completing this chapter, you should be able to:
determine the amount of magnetic ux passing through an area
explain how a moving conductor in a magnetic eld generates a voltage drop
describe how the magnetic ux through a rotating coil changes with time
explain how a rotating loop in a magnetic eld generates a voltage that varies as a sine wave — that is, an AC voltage
determine the average induced voltage in a loop from the ux change and the time in which the change took place
determine the direction of the induced current in a loop, using Lenz’s Law
calculate the average induced voltage for more than one loop
describe and determine the following properties of an AC voltage: frequency, period, amplitude, peak-to-peak voltage, peak-to-peak current, RMS voltage and RMS current
interpret RMS in terms of the DC supply that delivers the same power as the AC supply
describe the operation of an alternator with the use of slip rings to produce AC, and the operation of a generator with a split-ring commutator to produce uctuating DC.
CHAPTER
7 Generating electricity
A generator inside a wind turbine
UNCORRECTED voltage that varies as a sine wave — that is, an AC voltage
UNCORRECTED voltage that varies as a sine wave — that is, an AC voltage
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UNCORRECTED PAGE AC voltage: frequency, period, amplitude, peak-to-
PAGE AC voltage: frequency, period, amplitude, peak-to-peak voltage, peak-to-peak current, RMS voltage and
PAGE peak voltage, peak-to-peak current, RMS voltage and RMS current
PAGE RMS current
interpret RMS in terms of the DC supply that delivers the
PAGE interpret RMS in terms of the DC supply that delivers the same power as the AC supply
PAGE same power as the AC supply
describe the operation of an alternator with the use
PAGE describe the operation of an alternator with the use
PAGE of slip rings to produce AC, and the operation of a PAGE of slip rings to produce AC, and the operation of a generator with a split-ring commutator to produce PAGE
generator with a split-ring commutator to produce uctuating DC.PAGE
uctuating DC.
PROOFSdetermine the average induced voltage in a loop from
PROOFSdetermine the average induced voltage in a loop from the ux change and the time in which the change took
PROOFSthe ux change and the time in which the change took
determine the direction of the induced current in a loop,
PROOFSdetermine the direction of the induced current in a loop,
calculate the average induced voltage for more than one
PROOFScalculate the average induced voltage for more than one
describe and determine the following properties of an PROOFS
describe and determine the following properties of an AC voltage: frequency, period, amplitude, peak-to-PROOFS
AC voltage: frequency, period, amplitude, peak-to-peak voltage, peak-to-peak current, RMS voltage and PROOFS
peak voltage, peak-to-peak current, RMS voltage and
167CHAPTER 7 Generating electricity
c07GeneratingElectricity 167 9 June 2016 5:11 PM
Making electricityChapter 6 describes how a magnetic eld exerts a force on a moving charge, either in a wire as part of an electric current or as a free charge. is chapter applies this idea to new situations to produce or generate electricity. In doing this, a new concept, magnetic ux, will be developed to explain how a gener-ator works.
Generating voltage with a magnetic eldWhat should happen when a metal rod moves through a magnetic eld? Imagine a horizontal rod falling down through a magnetic eld as shown in the gure at left.
As the rod falls, the electrons and the positively charged nuclei in the rod are both moving down through the magnetic eld. As was explained in the last chapter, the magnetic eld will therefore exert a magnetic force on the elec-trons, and on the nuclei.
In which direction will the magnetic force act on the electrons and the nuclei?
e hand rules from chapter 6 can be used for both the electrons and the nuclei, keeping in mind that the hand rules use conventional current, so elec-trons moving down are equivalent to positive charges moving up.
e force on the electrons will be towards the far end of the rod, while the force on the nuclei will be to the near end of the rod, as is shown in the gure below.
++
++
N S
N S
+
–– –
–
–
F
(a)
(b)
V
V
The magnetic eld forces electrons to the far end of the falling rod.
e atomic structure of the metal restricts the movement of the positively charged nuclei. e negatively charged electrons, on the other hand, are free to move. e electrons move towards the far end of the rod, leaving the near end short of electrons and thus positively charged.
Not all electrons move to the far end. As the far end becomes more negative, there will be an increasingly repulsive force on any extra electrons. Similarly, there will be an increasingly attractive force from the positively charged near end, attempting to keep the remaining electrons at that end. is process is similar to the charging of a capacitor.
e movement of the metal rod through the magnetic eld has resulted in the separation of charge, causing a voltage between the ends. is is called induced voltage. As long as the rod keeps moving, the charges will remain
N SN S
V
A metal rod falling down through a magnetic eld
Induced voltage is a voltage that is caused by the separation of charge due to the presence of a magnetic eld.
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PAGE nuclei, keeping in mind that the hand rules use conventional current, so elec-
PAGE nuclei, keeping in mind that the hand rules use conventional current, so elec-trons moving down are equivalent to positive charges moving up.
PAGE trons moving down are equivalent to positive charges moving up. e force on the electrons will be towards the far end of the rod, while the
PAGE e force on the electrons will be towards the far end of the rod, while the
force on the nuclei will be to the near end of the rod, as is shown in the gure
PAGE force on the nuclei will be to the near end of the rod, as is shown in the gure
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PAGE PROOFS
What should happen when a metal rod moves through a magnetic eld?
PROOFSWhat should happen when a metal rod moves through a magnetic eld? Imagine a horizontal rod falling down through a magnetic eld as shown in
PROOFSImagine a horizontal rod falling down through a magnetic eld as shown in
As the rod falls, the electrons and the positively charged nuclei in the rod
PROOFSAs the rod falls, the electrons and the positively charged nuclei in the rod
are both moving down through the magnetic eld. As was explained in the last
PROOFSare both moving down through the magnetic eld. As was explained in the last chapter, the magnetic eld will therefore exert a magnetic force on the elec-
PROOFSchapter, the magnetic eld will therefore exert a magnetic force on the elec-
In which direction will the magnetic force act on the electrons and the
PROOFSIn which direction will the magnetic force act on the electrons and the
e hand rules from chapter 6 can be used for both the electrons and the PROOFS
e hand rules from chapter 6 can be used for both the electrons and the nuclei, keeping in mind that the hand rules use conventional current, so elec-PROOFS
nuclei, keeping in mind that the hand rules use conventional current, so elec-trons moving down are equivalent to positive charges moving up.PROOFS
trons moving down are equivalent to positive charges moving up.
UNIT 3 168
c07GeneratingElectricity 168 9 June 2016 5:11 PM
separated. As soon as the rod stops falling, the magnetic force is reduced to zero; electrons are then attracted back to the positive end and soon the elec-trons in the rod are distributed evenly.
e charge in the moving rod is separated by the magnetic eld, but the charge has nowhere to go. A source of voltage, an emf (electromotive force), has been produced. It is like a DC battery with one end positive and the other negative.
What determines the size of this induced emf? e size depends on the number of electrons shifted to one end. e electrons are shifted by the magnetic force until their own repulsion balances this force. So, the larger the magnetic force pushing the electrons, the more there will be at the end.
e size of this pushing magnetic force, as seen in chapter 6, depends on the size of the magnetic eld and the current. In this case, the size depends on how fast the electrons are moving down with the rod (which is, of course, how fast the rod is falling). So the faster the rod falls, the larger the emf.
An expression for the induced emf can be obtained by combining the expression from the end of the last chapter for the force on a moving charge with the denition of voltage from book 1. When the rod is moving down with speed (v) each electron experiences a sideways force along the rod equal to Bqv. is force pushes the electron along the length (l) of the rod and so is doing work in separating charge. e amount of work done is equal to the force times the distance and so equals Bqvl and is measured in joules. However the denition of emf or the voltage drop across the rod is energy supplied per unit of charge, measured in joules per coulomb or volts. So the induced emf (ε) is given by Bqvl
q, which gives:
ε = Blv
whereε is the induced emf measured in voltsB is the magnetic eld strength in teslal is the length of the rod or wire in metres that is in the magnetic eld v is the speed in metres per second at which the rod or wire is moving across the magnetic eld.
Sample problem 7.1
A 5.0 cm metal rod moves at right angles across a magnetic eld of strength 0.25 T at a speed of 40 cm s−1. What is the size of the induced emf across the ends of the rod?
l = 5.0 cm = 5.0 × 10−2 m, v = 40 cm s−1 = 0.4 m s−1, B = 0.25 Tε = Blv
= 0.25 T × 5.0 × 10−2 m × 0.4 m s−1
= 5.0 × 10−3 V = 5.0 mV
Revision question 7.1
At what speed would the rod need to move to induce an emf of 1.0 V?
Generating a currentEmfs can be used to produce a current. e experimental design illustrated in the rst gure for ‘Generating voltage with a magnetic eld’ (page 167) can be modied to produce a current by attaching a wire to each end of the metal rod
An emf is a source of voltage that can cause an electric current to ow.
Solution:
UNCORRECTED is the induced emf measured in volts
UNCORRECTED is the induced emf measured in volts is the magnetic eld strength in tesla
UNCORRECTED is the magnetic eld strength in tesla
is the length of the rod or wire in metres that is in the magnetic eld
UNCORRECTED is the length of the rod or wire in metres that is in the magnetic eld is the speed in metres per second at which the rod or wire is moving across
UNCORRECTED is the speed in metres per second at which the rod or wire is moving across
UNCORRECTED
UNCORRECTED
UNCORRECTED the magnetic eld.
UNCORRECTED the magnetic eld.
Sample problem 7.1
UNCORRECTED
Sample problem 7.1
A 5.0
UNCORRECTED
A 5.0
Solution:
UNCORRECTED
Solution:
PAGE force pushes the electron along the length (
PAGE force pushes the electron along the length (
PAGE in separating charge. e amount of work done is equal to the force times the
PAGE in separating charge. e amount of work done is equal to the force times the and is measured in joules. However the denition
PAGE and is measured in joules. However the denition of emf or the voltage drop across the rod is energy supplied per unit of charge,
PAGE of emf or the voltage drop across the rod is energy supplied per unit of charge, measured in joules per coulomb or volts. So the induced emf (
PAGE measured in joules per coulomb or volts. So the induced emf (
PROOFSuntil their own repulsion balances this force. So, the larger the magnetic force
PROOFSuntil their own repulsion balances this force. So, the larger the magnetic force
e size of this pushing magnetic force, as seen in chapter 6, depends on the
PROOFSe size of this pushing magnetic force, as seen in chapter 6, depends on the size of the magnetic eld and the current. In this case, the size depends on how
PROOFSsize of the magnetic eld and the current. In this case, the size depends on how fast the electrons are moving down with the rod (which is, of course, how fast
PROOFSfast the electrons are moving down with the rod (which is, of course, how fast the rod is falling). So the faster the rod falls, the larger the emf.
PROOFSthe rod is falling). So the faster the rod falls, the larger the emf.
An expression for the induced emf can be obtained by combining the
PROOFSAn expression for the induced emf can be obtained by combining the
expression from the end of the last chapter for the force on a moving charge with
PROOFSexpression from the end of the last chapter for the force on a moving charge with the denition of voltage from book 1. When the rod is moving down with speed
PROOFSthe denition of voltage from book 1. When the rod is moving down with speed
) each electron experiences a sideways force along the rod equal to PROOFS
) each electron experiences a sideways force along the rod equal to force pushes the electron along the length ( PROOFS
force pushes the electron along the length (l PROOFS
l) of the rod and so is doing work PROOFS
) of the rod and so is doing work in separating charge. e amount of work done is equal to the force times the PROOFS
in separating charge. e amount of work done is equal to the force times the
169CHAPTER 7 Generating electricity
c07GeneratingElectricity 169 9 June 2016 5:11 PM
and connecting these wires outside the magnetic eld. (See the gure at left.) Now the electrons have the path of a low-resistance conductor to go around to the positively charged end.
Once the electrons reach the positive end, they will be back in the magnetic eld, falling down with the metal rod, and will again experience a magnetic force pushing them to the far end of the rod. e electrons will then move around the circuit for a second time.
e electrons will continue to go around as long as the wire is falling through the magnetic eld. An electric current has been generated!
The source of a current’s electrical energyElectric current has electrical energy. Where did this energy come from? Before the rod (discussed earlier) was released, it had gravitational potential energy. If it is dropped outside the magnetic eld (see gure (b) below), this gravitational potential energy is converted into kinetic energy. If it is dropped inside the magnetic eld (see gure (a) below), some electrical energy is pro-duced. Since energy is conserved (that is, it cannot be created or destroyed), there must be less kinetic energy in the rod falling in the magnetic eld. at is, the rod in the magnetic eld is falling slower. Why?
N S
(a) (b)
kinetic energy
Distance fallen
gravitationalpotential energy
E
Distance fallen
kinetic energyelectrical energy
gravitational potential energy
E
– –
–
(a) Inside the magnetic eld, the gravitational potential energy of the falling rod is converted into both kinetic energy and electrical energy, whereas (b) outside the magnetic eld it is converted only into kinetic energy.
e induced current in the falling rod means that when the electrons are in the rod they are moving in two directions — downwards with the rod and along the rod.
e downward movement produced the sideways force along the rod that keeps the current going. But if the electrons are also moving along the rod, how does the magnetic eld respond to this?
e movement of electrons along the rod is also at right angles to the mag-netic eld so the eld exerts a second force on the electrons. e direction of this force is once again given by the hand rule and is directed upwards. is magnetic force opposes the accelerating force of the weight of the rod. (See the gure at left.)
N S
++
++
direction of electron ow
–
–––
–
–
The accumulated electrons at the far end of a rod move to the positive near end of the rod through the connecting wire.
N SI
weight of rod
F magneticforce onfalling electron
–
– –
magnetic force on rod
The magnetic force opposes the weight of the rod.
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED gravitational
UNCORRECTED gravitational potential
UNCORRECTED potential
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
magnetic
UNCORRECTED
magnetic force on rodUNCORRECTED
force on rod
PAGE the rod in the magnetic eld is falling slower. Why?
PAGE the rod in the magnetic eld is falling slower. Why?
PAGE
PAGE SPAGE SPAGE
PAGE
PAGE
PAGE
PAGE
PAGE –
PAGE – –
PAGE –
PROOFSThe source of a current’s electrical energy
PROOFSThe source of a current’s electrical energyElectric current has electrical energy. Where did this energy come from?
PROOFSElectric current has electrical energy. Where did this energy come from? Before the rod (discussed earlier) was released, it had gravitational potential
PROOFSBefore the rod (discussed earlier) was released, it had gravitational potential energy. If it is dropped outside the magnetic eld (see gure (b) below), this
PROOFSenergy. If it is dropped outside the magnetic eld (see gure (b) below), this gravitational potential energy is converted into kinetic energy. If it is dropped
PROOFSgravitational potential energy is converted into kinetic energy. If it is dropped inside the magnetic eld (see gure (a) below), some electrical energy is pro-
PROOFSinside the magnetic eld (see gure (a) below), some electrical energy is pro-duced. Since energy is conserved (that is, it cannot be created or destroyed),
PROOFSduced. Since energy is conserved (that is, it cannot be created or destroyed), there must be less kinetic energy in the rod falling in the magnetic eld. at is, PROOFS
there must be less kinetic energy in the rod falling in the magnetic eld. at is, the rod in the magnetic eld is falling slower. Why?PROOFS
the rod in the magnetic eld is falling slower. Why?
UNIT 3 170
c07GeneratingElectricity 170 9 June 2016 5:11 PM
e size of the upward magnetic force depends on the size of the current. is current will depend, in turn, on the size of the voltage between the ends of the rod. Voltage will increase as the rod moves faster.
When the rod rst starts falling, the magnetic force opposing the weight is small, but as the rod falls faster the opposing magnetic force increases until it equals the weight of the rod. At this point the rod has reached a maximum steady speed. is situation is identical to the terminal velocity experienced by objects falling through the air.
As the metal rod falls through the magnetic eld at constant speed, the loss in gravitational potential energy is converted to electrical energy as the gener-ated emf drives the current through the resistance of the circuit.
is e ect is di cult to demonstrate in practice. (A magnetic eld large enough for the rod to achieve terminal velocity is too di cult to construct.) However, it is possible to drop a magnet through a cylindrical conductor. With a su ciently strong magnet, measurable slowing-down against the accel-eration due to gravity can be observed.
Faraday’s discovery of electromagnetic inductionMichael Faraday was aware of the magnetic e ect of a current and he spent six years searching for the reverse e ect — that is, the electrical e ect of magnetism.
His equipment consisted of two coils of insulated wire, wrapped around a wooden ring. One coil was connected to a battery, the other to a galvanometer, a sensitive current detector. Faraday observed that the galvanometer needle gave a little kick when the battery switch was closed and a little kick the oppo-site way when the switch was opened. e rest of the time, either with the switch open or closed, the needle was stationary, reading zero. e current was momentary, not the constant current he was looking for. What Faraday had observed came to be called electromagnetic induction.
switch
battery
galvanometer
0
+ –
When the switch in the battery circuit is opened or closed, there is a momentary current through the galvanometer.
Investigating further, Faraday found that using an iron ring instead of a wooden one increased the size of the current. He concluded that when the magnetic eld of the battery coil was changing, there was a current induced in the other coil.
He therefore replaced the battery coil with a magnet. Moving the magnet through the other coil changed the magnetic eld and produced a current. e faster the magnet moved, the larger the current. When the magnet was moved back away from the coil, current owed in the opposite direction.
magnetic force
weight force
velocity ofmagnet
A magnet falling through a metal tube falls with an acceleration less than 9.8 m s–2 because it experiences a retarding magnetic force.
A galvanometer is an instrument used to detect small electric currents.
Electromagnetic induction is the generation of an electric current in a coil as a result of a changing magnetic eld or as a result of the movement of the coil within a constant magnetic eld.
Digital docInvestigation 7.1Inducing a currentThe induced current in a solenoid is demonstrated by moving a magnet in and out of a solenoid.doc-18544
UNCORRECTED gave a little kick when the battery switch was closed and a little kick the oppo-
UNCORRECTED gave a little kick when the battery switch was closed and a little kick the oppo-site way when the switch was opened. e rest of the time, either with the
UNCORRECTED site way when the switch was opened. e rest of the time, either with the switch open or closed, the needle was stationary, reading zero. e current was
UNCORRECTED switch open or closed, the needle was stationary, reading zero. e current was momentary, not the constant current he was looking for. What Faraday had
UNCORRECTED momentary, not the constant current he was looking for. What Faraday had observed came to be called
UNCORRECTED observed came to be called
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
+ –
UNCORRECTED
+ –
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED PAGE electromagnetic induction
PAGE electromagnetic inductionMichael Faraday was aware of the magnetic e ect of a current and he spent
PAGE Michael Faraday was aware of the magnetic e ect of a current and he spent six years searching for the reverse e ect — that is, the electrical e ect of
PAGE six years searching for the reverse e ect — that is, the electrical e ect of
His equipment consisted of two coils of insulated wire, wrapped around a
PAGE His equipment consisted of two coils of insulated wire, wrapped around a
wooden ring. One coil was connected to a battery, the other to a
PAGE wooden ring. One coil was connected to a battery, the other to a a sensitive current detector. Faraday observed that the galvanometer needle PAGE a sensitive current detector. Faraday observed that the galvanometer needle gave a little kick when the battery switch was closed and a little kick the oppo-PAGE
gave a little kick when the battery switch was closed and a little kick the oppo-
PROOFSin gravitational potential energy is converted to electrical energy as the gener-
PROOFSin gravitational potential energy is converted to electrical energy as the gener-ated emf drives the current through the resistance of the circuit.
PROOFSated emf drives the current through the resistance of the circuit. is e ect is di cult to demonstrate in practice. (A magnetic eld large
PROOFS is e ect is di cult to demonstrate in practice. (A magnetic eld large enough for the rod to achieve terminal velocity is too di cult to construct.)
PROOFSenough for the rod to achieve terminal velocity is too di cult to construct.) However, it is possible to drop a magnet through a cylindrical conductor. With
PROOFSHowever, it is possible to drop a magnet through a cylindrical conductor. With a su ciently strong magnet, measurable slowing-down against the accel-
PROOFSa su ciently strong magnet, measurable slowing-down against the accel-
Faraday’s discovery of PROOFS
Faraday’s discovery of electromagnetic inductionPROOFS
electromagnetic induction
171CHAPTER 7 Generating electricity
c07GeneratingElectricity 171 9 June 2016 5:11 PM
II
0 0
NNS NNS
(a) (b)
Magnet (a) moving into a coil and (b) away again
If there was an induced current, then there must have been an induced emf. An emf gives energy to a charge to move it through the wire, and the resistance of the wire limits the size of the current. So it is more correct to say that the changing magnetic eld induced an emf.
Magnetic uxMagnetic ux is the amount of magnetic eld passing through an area, such as a coil. It is the change in the magnetic ux that will help explain electromag-netic induction.
e stronger the magnetic eld going through an area, the larger the amount of magnetic ux. Similarly, the larger the area the magnetic eld is going through, the larger the magnetic ux.
is is summarised in the de nition of magnetic ux:
amount of magnetic ux (ΦB) = strength of magnetic eld (B) × the area (A)
ΦB = BA.
(a) (b)
Area A
B
Magnetic ux is the amount of magnetic eld passing through an area. In (a) it is the maximum BA; in (b) the value is less, as fewer eld lines pass through the coil.
Magnetic ux is measured in webers. One weber (Wb) is the amount of magnetic ux from a uniform magnetic eld with a strength of 1.0 tesla passing through an area of 1.0 square metre. e magnetic ux can also take on posi-tive and negative values, depending on which side of the area the magnetic eld is coming from.
Magnetic ux is a measure of the amount of magnetic eld passing through an area. It is measured in webers (Wb).
Unit 3 Magnetic uxSummary screen and practice questions
AOS 2
Topic 1
Concept 1
Unit 3
See moreMagnetic ux
AOS 2
Topic 1
Concept 1
UNCORRECTED as a coil. It is the change in the magnetic ux that will help explain electromag-
UNCORRECTED as a coil. It is the change in the magnetic ux that will help explain electromag-
e stronger the magnetic eld going through an area, the larger the amount
UNCORRECTED e stronger the magnetic eld going through an area, the larger the amount of magnetic ux. Similarly, the larger the area the magnetic eld is going
UNCORRECTED of magnetic ux. Similarly, the larger the area the magnetic eld is going through, the larger the magnetic ux.
UNCORRECTED through, the larger the magnetic ux.
is is summarised in the de nition of magnetic ux:
UNCORRECTED is is summarised in the de nition of magnetic ux:
amount of magnetic ux (
UNCORRECTED amount of magnetic ux (
(a)
UNCORRECTED
(a)
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
Magnetic ux
UNCORRECTED
Magnetic uxSummary screen
UNCORRECTED
Summary screen and practice
UNCORRECTED
and practice questions
UNCORRECTED
questions
UNCORRECTED
Concept
UNCORRECTED
Concept 1
UNCORRECTED
1
UNCORRECTED PAGE An emf gives energy to a charge to move it through the wire, and the resistance
PAGE An emf gives energy to a charge to move it through the wire, and the resistance of the wire limits the size of the current. So it is more correct to say that the
PAGE of the wire limits the size of the current. So it is more correct to say that the changing magnetic eld induced an emf.
PAGE changing magnetic eld induced an emf.
Magnetic ux
PAGE Magnetic ux
is the amount of magnetic eld passing through an area, such PAGE is the amount of magnetic eld passing through an area, such
as a coil. It is the change in the magnetic ux that will help explain electromag-PAGE
as a coil. It is the change in the magnetic ux that will help explain electromag-
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
If there was an induced current, then there must have been an induced emf. PROOFS
If there was an induced current, then there must have been an induced emf. An emf gives energy to a charge to move it through the wire, and the resistance PROOFS
An emf gives energy to a charge to move it through the wire, and the resistance of the wire limits the size of the current. So it is more correct to say that the PROOFS
of the wire limits the size of the current. So it is more correct to say that the
UNIT 3 172
c07GeneratingElectricity 172 9 June 2016 5:11 PM
is description has assumed that the magnetic eld is at right angles to the area, as shown in gure (a). If the magnetic eld went through the area at an angle less than 90º (as shown in gure (b)), the amount of magnetic ux passing through the area would be less. In fact, if the magnetic eld is parallel to the area, the amount of magnetic ux will be zero, as none of the magnetic eld lines pass through the area from one side to the other.
A more correct de nition of magnetic ux would therefore be:
amount of magnetic ux (ΦB) = component of magnetic eld strength perpendicular to the area (B⊥) × the area (A)
ΦB = B⊥ × A.
Sample problem 7.2
Calculate the magnetic ux in each of the following situations.
B = 0.05 T
Area = 0.3 m2
(a)
B = 0.2 T
(b)
8 cm
B = 1.7 T
(c)
15 cm
(a) ΦB = B⊥ × A = 0.05 T × 0.3 m2
= 0.015 Wb
(b) First calculate area A. (Don’t forget to convert the radius to metres.) A = π r 2
= π × (0.08 m)2
= 0.020 106 m2 (Don’t round o the nal answer.)
Now calculate the ux:ΦB = B⊥ × A = 0.2 T × 0.020 106 m2
= 0.004 Wb
(c) Note that the plane of the loop is parallel to the magnetic eld,
B⊥ = 0.ΦB = B⊥ × A
= 0 × A = 0 Wb
B
Zero magnetic ux, as no eld lines ‘thread’ the loop
Solution:
UNCORRECTED = 0.05 T
UNCORRECTED = 0.05 T
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
Solution:
UNCORRECTED
Solution:
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE Area = 0.3 m
PAGE Area = 0.3 m2
PAGE 2 PROOFS
PROOFS
PROOFSCalculate the magnetic ux in each of the following situations.
PROOFSCalculate the magnetic ux in each of the following situations.
(b) PROOFS
(b) PROOFS
173CHAPTER 7 Generating electricity
c07GeneratingElectricity 173 9 June 2016 5:11 PM
Revision question 7.2
Estimate the maximum amount of magnetic ux passing through an earring when placed near a typical school magnet.
Induced EMFNow the concept of magnetic ux can be used to explain the induced emf. e two principles are described here.1. An emf is induced in a coil when the amount of magnetic ux passing
through the coil changes.2. e size of the emf depends on how quickly the amount of magnetic ux
changes. ese two statements can be written formally as:
temf , .average
Bε Φ= ∆∆
is statement is known as Faraday’s Law. e word ‘average’ is included because the change in magnetic ux took place over a nite interval of time.
Lenz’s Law states: e direction of the induced current is such that its mag-netic eld is in the opposite direction to the change in magnetic ux. It can be incorporated in the above equation as a minus sign:
temf, .Bε Φ= −∆
∆If the coil consists of several turns of wire, the equation can be generalised
further:N
temf, Bε Φ= − ∆
∆where N is the number of turns in the coil.
In part (a) of the following gure there is no magnetic ux passing through the loop. When the magnet approaches the coil ( gure (b)), there is an increase in the amount of magnetic eld passing through it from left to right. e loop has experienced a change in the magnetic ux passing through it (c), and the direc-tion of this change is from left to right. e direction of the induced magnetic eld (d) from the induced current in the loop (e) will be such that its magnetic e ect will oppose the change in the magnetic ux (c). is means its direction will be from right to left.
(a) (b) (c) (d) (e)
NN N
before after changein ux
inducedmagnetic
eld
induced current(check using
right-hand-grip rule)
The loop (a) before and (b) after; (c) change in ux, (d) direction of induced eld and (e) direction of current
To achieve an induced magnetic eld from right to left, the induced current, using the right-hand-grip rule, must be travelling up the front of the loop.
Unit 3 Faraday’s LawSummary screen and practice questions
AOS 2
Topic 1
Concept 2
Unit 3
Do moreGenerating an emf
AOS 2
Topic 1
Concept 2
eLessonMagnetic ux and Lenz’s Laweles-0026InteractivityMagnetic ux and Lenz’s Lawint-0050
Unit 3 Induced emf from a ux–time graphSummary screen and practice questions
AOS 2
Topic 1
Concept 3
Unit 3 Lenz’s LawSummary screen and practice questions
AOS 2
Topic 1
Concept 4
Unit 3
See moreMagnetic ux and Lenz’s Law
AOS 2
Topic 1
Concept 4
UNCORRECTED is the number of turns in the coil.
UNCORRECTED is the number of turns in the coil.
In part (a) of the following gure there is no magnetic ux passing through the
UNCORRECTED In part (a) of the following gure there is no magnetic ux passing through the loop. When the magnet approaches the coil ( gure (b)), there is an increase in
UNCORRECTED loop. When the magnet approaches the coil ( gure (b)), there is an increase in the amount of magnetic eld passing through it from left to right. e loop has
UNCORRECTED the amount of magnetic eld passing through it from left to right. e loop has experienced a change in the magnetic ux passing through it (c), and the direc-
UNCORRECTED
experienced a change in the magnetic ux passing through it (c), and the direc-tion of this change is from left to right. e direction of the induced magnetic
UNCORRECTED
tion of this change is from left to right. e direction of the induced magnetic eld (d) from the induced current in the loop (e) will be such that its magnetic
UNCORRECTED
eld (d) from the induced current in the loop (e) will be such that its magnetic e ect will oppose the change in the magnetic ux (c). is means its direction
UNCORRECTED
e ect will oppose the change in the magnetic ux (c). is means its direction
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Lenz’s Law
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Lenz’s LawSummary screen
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Summary screen and practice
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and practice questions
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questions
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Topic
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Topic 1
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1
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Concept UNCORRECTED
Concept 4UNCORRECTED
4UNCORRECTED PAGE e direction of the induced current is such that its mag-
PAGE e direction of the induced current is such that its mag-netic eld is in the opposite direction to the change in magnetic ux
PAGE netic eld is in the opposite direction to the change in magnetic uxincorporated in the above equation as a minus sign:
PAGE incorporated in the above equation as a minus sign:
If the coil consists of several turns of wire, the equation can be generalised
PAGE If the coil consists of several turns of wire, the equation can be generalised
PROOFS1. An emf is induced in a coil when the amount of magnetic ux passing
PROOFS1. An emf is induced in a coil when the amount of magnetic ux passing
2. e size of the emf depends on how quickly the amount of magnetic ux
PROOFS2. e size of the emf depends on how quickly the amount of magnetic ux
is statement is known as Faraday’s Law. e word ‘average’ is included
PROOFS is statement is known as Faraday’s Law. e word ‘average’ is included
because the change in magnetic ux took place over a nite interval of time.PROOFS
because the change in magnetic ux took place over a nite interval of time. e direction of the induced current is such that its mag-PROOFS
e direction of the induced current is such that its mag-netic eld is in the opposite direction to the change in magnetic uxPROOFS
netic eld is in the opposite direction to the change in magnetic ux
UNIT 3 174
c07GeneratingElectricity 174 9 June 2016 5:11 PM
e coil responds in such a way as to keep its magnetic environment con-stant. In this example, there is increasing ux from left to right, so the induced magnetic eld goes from right to left. When the magnet is pulled back, the ux that is still going from left to right is decreasing this time, so the induced mag-netic eld adds to the existing ux to compensate for the loss, and this eld points from left to right.
Sample problem 7.3
e rectangular loop shown takes 2.0 s to fully enter a perpendicular magnetic eld of 0.66 T strength.(a) What is the magnitude of the emf induced in the loop?(b) In which direction does the current ow around the loop?
(a) First calculate the area of the loop.A = 0.25 m × 0.3 m
= 0.075 m2
Now nd the change in ux.ΔΦB = ΦB nal − ΦB initial = (BA)nal − (BA)initial = (0.66 T × 0.075 m2) − (0 T × 0.075 m2)
(e initial eld strength through the coil is zero.) = (0.05 T m2) − (0 T m2) = 0.05 Wb into the page
Finally, using Faraday’s Law:
emf, ε = N
tBΦ− ∆
∆
= −1 × 0.05 Wb
2.0 s
= −0.025 V
So the magnitude of the induced voltage is 0.025 V. e minus sign is there to indicate that the induced emf opposes the change in magnetic ux.
(b) Change in ux = nal − initial = ux into the page
Direction of induced magnetic eld = out of the page (Lenz’s Law)
Direction of induced current = anticlockwise (right-hand-grip rule)
Revision question 7.3
A spring is bent into a circle and stretched out to a radius of 5.0 cm. It is then placed in a magnetic eld of strength 0.55 T. e spring is released and con-tracts down to a circle of radius 3.0 cm. is happens in 0.15 seconds.(a) What is magnitude of the induced emf?(b) In what direction does the current move?
Unit 3
Do moreMagnetic ux and Lenz’s Law
AOS 2
Topic 1
Concept 4
0.25 m
0.3 m B = 0.66 T
Solution:
UNCORRECTED
UNCORRECTED t
UNCORRECTED t
B
UNCORRECTED BΦ
UNCORRECTED ΦBΦB
UNCORRECTED BΦB− ∆
UNCORRECTED − ∆N− ∆N
UNCORRECTED N− ∆N Φ− ∆Φ
UNCORRECTED Φ− ∆Φ∆
UNCORRECTED ∆
=
UNCORRECTED = −
UNCORRECTED −1
UNCORRECTED 1
So the magnitude of the induced voltage is 0.025
UNCORRECTED
So the magnitude of the induced voltage is 0.025e min
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e min
PAGE (0 T
PAGE (0 T ×
PAGE × 0.075
PAGE 0.075(e initial eld strength through the coil is zero.)
PAGE (e initial eld strength through the coil is zero.)
(0
PAGE (0
PAGE T
PAGE T
PAGE m
PAGE m2
PAGE 2)
PAGE )
b into the page
PAGE b into the page
Finally, using Faraday’s Law:PAGE Finally, using Faraday’s Law:
PROOFSo fully enter a perpendicular magnetic
PROOFSo fully enter a perpendicular magnetic
n which direction does the current ow around the loop?
PROOFSn which direction does the current ow around the loop?
175CHAPTER 7 Generating electricity
c07GeneratingElectricity 175 9 June 2016 5:11 PM
Rotating a loopA magnet moving in and out of a coil to generate a current is not a very e cient means of converting the mechanical energy of the moving magnet into elec-trical energy of a current in the coil. It does not have much technological potential; an alternative is needed.
Another way of changing the amount of magnetic ux passing through a loop is to rotate a loop in a magnetic eld.
When the loop is ‘face on’ to the magnetic eld, the maximum amount of magnetic ux is passing through the loop. As the loop turns, the amount decreases. When it has turned 90º, there is no ux passing through it at all. As the loop continues to turn, the magnetic eld passes through the loop from the other side: a negative amount of ux, from the point of view of the loop.
As the loop turns further still, the amount of magnetic ux passing through the loop reaches a negative maximum, then comes back to zero, and nally passes through the original face of the loop.
S SN
slip rings
loopaxis of rotation
P
Q side-on view of loop
N
A loop ‘face on’ to a magnetic eld has maximum magnetic ux.
e amount of magnetic ux passing through the loop varies like a sine wave. e induced emf across the ends of the loop is equal to the change of magnetic ux with time. In mechanics, the velocity is de ned as the change of displace-ment over time and it is shown as the gradient of the displacement-time graph. Similarly the induced emf is shown as the gradient of a magnetic ux–time graph, which is also a sine wave.
e emf graph is the same shape as the ux graph (see the gure at left) but shifted sideways, so that when the ux is a maximum, the emf is zero. (At this point the ux–time graph is at, so the gradient is zero.)
Similarly, when the ux is zero, the ux–time graph is steepest, so the gradient is a maximum and the emf is a maximum.
Which way does the current travel in the loop? From which connection, P or Q, does the current leave the loop to go around the external circuit? is is not easy to determine. It can be worked out using Lenz’s Law or using the mag-netic force of electrons in the loop. is is shown below.
Using Lenz’s LawAs the loop passes through the horizontal plane the magnetic ux changes from passing through one side to passing through the other.
In part (a) of the following gure, the magnetic ux is entering the loop from above. In part (b), it enters from below. e change in magnetic ux is therefore upwards. e induced magnetic eld will then be down at this point. To produce this eld, the conventional current needs to run in the order ABCD.
Time
loop
ux ФB emf
B
Flux–time graphUNCORRECTED
UNCORRECTED
UNCORRECTED A loop ‘face on’ to a magnetic eld has maximum magnetic ux.
UNCORRECTED A loop ‘face on’ to a magnetic eld has maximum magnetic ux.
UNCORRECTED e amount of magnetic ux passing through the loop varies like a sine wave.
UNCORRECTED e amount of magnetic ux passing through the loop varies like a sine wave.
e induced emf across the ends of the loop is equal to the change of magnetic
UNCORRECTED e induced emf across the ends of the loop is equal to the change of magnetic ux with time. In mechanics, the velocity is de ned as the change of displace-
UNCORRECTED
ux with time. In mechanics, the velocity is de ned as the change of displace-ment over time and it is shown as the gradient of the displacement-time graph.
UNCORRECTED
ment over time and it is shown as the gradient of the displacement-time graph. Similarly the induced emf is shown as the gradient of a magnetic ux–time
UNCORRECTED
Similarly the induced emf is shown as the gradient of a magnetic ux–time graph, which is also a sine wave.
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graph, which is also a sine wave.
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PAGE slip rings
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PAGE PROOFS
of magnetic ux is passing through the loop. As the loop turns, the amount
PROOFSof magnetic ux is passing through the loop. As the loop turns, the amount
, there is no ux passing through it at all. As
PROOFS, there is no ux passing through it at all. As the loop continues to turn, the magnetic eld passes through the loop from
PROOFSthe loop continues to turn, the magnetic eld passes through the loop from the other side: a negative amount of ux, from the point of view of the loop.
PROOFSthe other side: a negative amount of ux, from the point of view of the loop.As the loop turns further still, the amount of magnetic ux passing through
PROOFSAs the loop turns further still, the amount of magnetic ux passing through
the loop reaches a negative maximum, then comes back to zero, and nally
PROOFSthe loop reaches a negative maximum, then comes back to zero, and nally
UNIT 3 176
c07GeneratingElectricity 176 9 June 2016 5:11 PM
S
axis of rotation
A
C
D
I
SB
C
D
A
N SN
P(–)
Q(+)
side-on view of loop
B(b)
N SN
axis of rotation
P
Qside-on view of loop
(a)
Direction of current ow as loop passes through the horizontal position
At this point in the rotation, the current will enter the external circuit from the slip ring at Q and return to the loop by the slip ring at P. So, for the time being, Q is the positive terminal and P the negative.
In the diagrams above, the wire from A is attached to the front metal ring, the one connected to P, and the wire from B is attached to the back ring, the one connected to Q. ese connections are xed. When the loop rotates about its axis, the two slip rings also rotate about the same axis. e black blocks are made of graphite. ey are being held in place against the spinning slip rings by the springs. Graphite is used because it not only conducts electricity but is also a lubricant. e spinning slip rings easily slide past the xed block. e blocks are also called ‘brushes’ because early designs used thin metal strips that rested against the slip rings.
Using magnetic force on the charges in the wireAs the loop passes through the horizontal plane, the left side of the loop, AB (see the gure at left), is moving up and the right side, CD, is moving down. e force of the magnetic eld on the positive charges in AB will be towards B, while the force on the electrons in AB will be towards A.
Similarly, the positive charges in CD will be pushed to D, while the electrons will be pushed towards C.
is means that conventional current will ow ABCD, while the electrons will travel around the loop in the order DCBA. e conventional current will leave the external circuit from D and return to the loop by A. is is the same result obtained as with the previous method.
WeblinkGenerator applet
S
B C
DA
N
P
QI
Fmagnetic eld
F — direction of force on positive charges
I — direction positive charges move due to rotation
Legend
Using your left-hand rule to determine current direction in a rotating loop
UNCORRECTED At this point in the rotation, the current will enter the external circuit from
UNCORRECTED At this point in the rotation, the current will enter the external circuit from the slip ring at Q and return to the loop by the slip ring at P. So, for the time
UNCORRECTED the slip ring at Q and return to the loop by the slip ring at P. So, for the time being, Q is the positive terminal and P the negative.
UNCORRECTED being, Q is the positive terminal and P the negative.
In the diagrams above, the wire from A is attached to the front metal ring,
UNCORRECTED In the diagrams above, the wire from A is attached to the front metal ring,
the one connected to P, and the wire from B is attached to the back ring, the
UNCORRECTED the one connected to P, and the wire from B is attached to the back ring, the one connected to Q. ese connections are xed. When the loop rotates about
UNCORRECTED one connected to Q. ese connections are xed. When the loop rotates about its axis, the two slip rings also rotate about the same axis. e black blocks are
UNCORRECTED its axis, the two slip rings also rotate about the same axis. e black blocks are made of graphite. ey are being held in place against the spinning slip rings
UNCORRECTED
made of graphite. ey are being held in place against the spinning slip rings by the springs. Graphite is used because it not only conducts electricity but is
UNCORRECTED
by the springs. Graphite is used because it not only conducts electricity but is also a lubricant. e spinning slip rings easily slide past the xed block. e
UNCORRECTED
also a lubricant. e spinning slip rings easily slide past the xed block. e blocks are also called ‘brushes’ because early designs used thin metal strips
UNCORRECTED
blocks are also called ‘brushes’ because early designs used thin metal strips
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PAGE Direction of current ow as loop passes through the horizontal position
PAGE Direction of current ow as loop passes through the horizontal position
PAGE
At this point in the rotation, the current will enter the external circuit from PAGE
At this point in the rotation, the current will enter the external circuit from
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFSN
PROOFSN
Q
PROOFSQ( PROOFS
(+PROOFS
+)PROOFS
)
177CHAPTER 7 Generating electricity
c07GeneratingElectricity 177 9 June 2016 5:11 PM
e sinusoidal emf drives current through the external circuit rst one way, then the opposite way, and is thus called alternating current (AC).
is design of a rotating coil in a magnetic eld is called a generator. If the ends of the coil are connected to slip rings, then the voltage across the external connections is alternating in direction, producing an alternating current. e device is now called an alternator.
If the slip rings are replaced by a split ring used in a DC motor, the current reverses every half-cycle, and so the alternating current is converted into pul-sating direct current (DC). e device is now called a DC generator.
SN
+–
(a)
commutator
(b)emf output
emf in loop
Time
AC voltage coming from loop, and DC coming from commutator
Peak, RMS and peak-to-peak voltages e voltage output of an AC generator varies with time, producing a sinusoidal signal. is signal, shown in the gure below, can be described in terms of the physical quantities described below.
Voltage
Time
Vpeak
–Vpeak
T–2
T
Sinusoidal signal from voltage output of an AC generator
• e period, T, is the time taken for one complete cycle.• e frequency, f, is the number of full cycles completed in one second. e
frequency is related to the period by the equation:
=Tf
1.
e frequency of the power supplied to households is 50 Hz (1 hertz is one cycle per second). e period is therefore 1
50 per second = 0.02 s.
• e amplitude is the maximum variation of the voltage output from zero. It is called the peak voltage, Vpeak. Similarly, the amplitude of the current is called the peak current, Ipeak.
• e RMS (root mean square) voltage, VRMS, is the value of the constant DC voltage that would produce the same power as the AC voltage across
An alternating current is an electric current that reverses direction at short, regular intervals.
A generator is a device in which a rotating coil in a magnetic eld is used to produce a voltage.
A direct current is an electric current that ows in one direction only.
Unit 3 Principles of an electricity generatorSummary screen and practice questions
AOS 2
Topic 1
Concept 5
e period, T, of a periodic wave is the time it takes a source to produce a complete wave. is is the same as the time taken for a complete wave to pass a given point.
e frequency, f, of a periodic wave is the number of times that it repeats itself every second.
e amplitude of a periodic disturbance is the maximum variation from zero.
e peak voltage, Vpeak, is the amplitude of an alternating voltage.
e peak current, Ipeak, is the amplitude of an alternating current.
e RMS (root mean square) voltage, VRMS, is the value of the constant DC voltage that would produce the same power as AC voltage across the same resistance.
UNCORRECTED Peak, RMS and peak-to-peak voltages
UNCORRECTED Peak, RMS and peak-to-peak voltages e voltage output of an AC generator varies with time, producing a sinusoidal
UNCORRECTED e voltage output of an AC generator varies with time, producing a sinusoidal signal. is signal, shown in the gure below, can be described in terms of the
UNCORRECTED signal. is signal, shown in the gure below, can be described in terms of the physical quantities described below.
UNCORRECTED physical quantities described below.
a complete wave to pass a given
UNCORRECTED
a complete wave to pass a given
, of a periodic
UNCORRECTED
, of a periodic wave is the number of times that it
UNCORRECTED
wave is the number of times that it repeats itself every second.
UNCORRECTED
repeats itself every second.
UNCORRECTED
amplitude
UNCORRECTED
amplitude of a periodic
UNCORRECTED
of a periodic disturbance is the maximum
UNCORRECTED
disturbance is the maximum variation from zero.
UNCORRECTED
variation from zero.
UNCORRECTED
peak voltageUNCORRECTED
peak voltageamplitude of an alternating UNCORRECTED
amplitude of an alternating
PAGE
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PAGE emf output
PAGE emf output
emf in loop
PAGE emf in loop
PAGE
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PAGE
PAGE AC voltage coming from loop, and DC coming from commutator
PAGE AC voltage coming from loop, and DC coming from commutator
PAGE
Peak, RMS and peak-to-peak voltagesPAGE
Peak, RMS and peak-to-peak voltages
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS+
PROOFS+
commutator
PROOFS
commutator
emf outputPROOFS
emf output
UNIT 3 178
c07GeneratingElectricity 178 9 June 2016 5:11 PM
the same resistance. e RMS voltage is related to the peak voltage by the equation:
=VV
2.RMS
peak
e peak voltage of a 230 V RMS household power supply is 325 V. A 230 V RMS output from a generator delivers the same amount of power as a 230 V DC power supply across the same resistance. Similarly, IRMS is the value of a DC current that generates the same power as an AC current through the same resistance:
=II
2.RMS
peak
• e peak-to-peak voltage, Vp−p, is the di erence recorded between the maximum and minimum voltages. In the case of a symmetrical AC voltage:
Vp−p = 2Vpeak
Similarly:
Ip−p = 2Ipeak
Sample problem 7.4
A digital multimeter gives a measurement of 6.3 V for the RMS value of an AC voltage. A CRO is used to measure the peak-to-peak voltage. What value do you expect?
VRMS = 6.3 V
V V V2 2 2p p peak RMS= = × ×−
= × ×2 2 6.3 V = 17.8 V = 18 V
Revision question 7.4
A toaster is rated at 230 V RMS and 1800 W. What are the values of the RMS and peak currents?
Producing a greater EMF e AC voltage produced by a generator has a sub-stantial technological application because it is easy to make things spin. In hydroelectricity, electricity isproduced when water falls under gravity through pipes and hits the vanes of a propeller connected to a generator. In coal and gas- red turbines, the burning fuel heats up water to a high temperature and pressure to direct against the vanes of the turbine.
e emf that is produced by a generator has a fre-quency the same as the frequency of the rotation of a coil in a magnetic eld.
Using the Faraday equation for average emf:
=− ∆
∆N BA
temf
( )
Unit 3 The alternator — an AC generatorSummary screen and practice questions
AOS 2
Topic 1
Concept 6
e peak-to-peak voltage, Vp−p, is the di erence between the maximum and minimum voltages of a DC voltage.
Unit 3 The DC generatorSummary screen and practice questions
AOS 2
Topic 1
Concept 7
Solution:
Vanes of a turbine at a coal plant
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED V
UNCORRECTED V2 2
UNCORRECTED 2 2 2
UNCORRECTED 22 2= =2 2
UNCORRECTED 2 2= =2 2 × ×
UNCORRECTED × ×× ×
UNCORRECTED × ×2× ×2
UNCORRECTED 2× ×2
UNCORRECTED × ×
UNCORRECTED × ×2 2
UNCORRECTED 2 22 2
UNCORRECTED 2 22 2
UNCORRECTED 2 22 2
UNCORRECTED 2 2× ×2 2× ×
UNCORRECTED × ×2 2× ×× ×2 2× ×
UNCORRECTED × ×2 2× × 6.3 V
UNCORRECTED 6.3 V
17.8 V
UNCORRECTED 17.8 V
=
UNCORRECTED = 18 V
UNCORRECTED 18 V
Revision question 7.4
UNCORRECTED
Revision question 7.4
A toaster is rated at 230 V RMS and 1800 W. What are the values of the RMS and
UNCORRECTED
A toaster is rated at 230 V RMS and 1800 W. What are the values of the RMS and peak currents?
UNCORRECTED
peak currents?
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
Vanes of a turbine at a coal plant
UNCORRECTED
Vanes of a turbine at a coal plant
UNCORRECTED PAGE
PAGE
PAGE A digital multimeter gives a measurement of 6.3 V for the RMS value of an AC
PAGE A digital multimeter gives a measurement of 6.3 V for the RMS value of an AC voltage. A CRO is used to measure the peak-to-peak voltage. What value do
PAGE voltage. A CRO is used to measure the peak-to-peak voltage. What value do
PROOFSthe value of a DC current that generates the same power as an AC current
PROOFSthe value of a DC current that generates the same power as an AC current
, is the di erence recorded between the
PROOFS, is the di erence recorded between the
maximum and minimum voltages. In the case of a symmetrical AC voltage:
PROOFSmaximum and minimum voltages. In the case of a symmetrical AC voltage:
179CHAPTER 7 Generating electricity
c07GeneratingElectricity 179 9 June 2016 5:11 PM
and ignoring the − sign (which relates to direction), we can deduce the following ways to produce a larger emf:• increase the number of turns• increase the strength of the magnetic eld• increase the area of each coil• decrease the time for one turn (that is, increase the frequency of rotation).
(Note that turning the coil twice as fast doubles both the induced emf and the frequency — that is, it halves the period.)
Other technological strategies can also increase the emf. ese are described below.• e pole ends of the magnet can be curved so that the coils are close to the
magnets for more of the rotation.• An iron core can be placed inside the coils to strengthen the magnetic eld.• e coils can be wound onto the iron core in grooves cut into the outer
surface so that the iron core is as close as possible to the magnetic poles to increase the magnetic eld.
N S
–+ slip rings
N
S
brushes
Improvements to the design of a DC motor and an alternator.
coil turned twice as fast
Time
Emf
Doubling the frequency doubles the induced emf.
Unit 3
Do moreProducing AC in alternators
AOS 2
Topic 1
Concept 6
Unit 3
See moreDescribing induced AC voltage
AOS 2
Topic 2
Concept 1
Unit 3 Characteristics of the AC waveformSummary screen and practice questions
AOS 2
Topic 2
Concept 1
Unit 3 RMS values and AC powerSummary screen and practice questions
AOS 2
Topic 2
Concept 2
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED Improvements to the design of a DC motor and an alternator.
UNCORRECTED Improvements to the design of a DC motor and an alternator.
UNCORRECTED
UNCORRECTED
UNCORRECTED
Producing AC in
UNCORRECTED
Producing AC in alternators
UNCORRECTED
alternators
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
Unit
UNCORRECTED
Unit 3
UNCORRECTED
3
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
AOSUNCORRECTED
AOS 2UNCORRECTED
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2UNCORRECTED
2
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PAGE N
PAGE N PROOFS e pole ends of the magnet can be curved so that the coils are close to the
PROOFS e pole ends of the magnet can be curved so that the coils are close to the
An iron core can be placed inside the coils to strengthen the magnetic eld.
PROOFSAn iron core can be placed inside the coils to strengthen the magnetic eld. e coils can be wound onto the iron core in grooves cut into the outer
PROOFS e coils can be wound onto the iron core in grooves cut into the outer surface so that the iron core is as close as possible to the magnetic poles to
PROOFSsurface so that the iron core is as close as possible to the magnetic poles to
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
UNIT 3 180
c07GeneratingElectricity 180 9 June 2016 5:11 PM
Chapter review
Unit 3 Generation of electricity Describing AC electricity
Sit Topic test
AOS 2
Topics 1 & 2
Summary A metal rod moving across a magnetic eld experi-
ences an induced voltage across its ends. e induced voltage across the ends of a moving
conductor in a magnetic eld will produce an elec-tric current if the ends are connected by a wire out-side the magnetic eld.
Magnetic ux is a measure of the amount of mag-netic eld passing through an area. It is measured in webers (Wb). Its magnitude is the product of the component of the magnetic eld strength, B, that is perpendicular to the area and the area, A.
An emf is induced in a loop if the magnetic ux passing through the loop changes. e emf induced
in a single loop is given by emf = t
BΦ∆∆ , where ΦB
is the magnetic ux. e negative sign in the equation acknowledges Lenz’s Law, which states that the induced current (and hence emf) is such that it creates a magnetic eld that opposes the change in ux.
e emf generated in N loops threaded by a magnetic
ux, ΦB, is given by Nt
emf BΦ= − ∆∆ .
In an alternator, a coil rotates in a magnetic eld to induce a sinusoidal voltage and therefore an alter-nating current. Slip rings are used at the end of the coil to allow the alternating current to ow in an external circuit.
In a DC generator, the slip rings are replaced with a commutator to allow a direct current to ow in an external circuit.
e voltage output of an AC generator can be described in terms of its amplitude, frequency and period. e amplitude of the voltage output is known as the peak voltage, Vpeak. e peak-to-peak voltage, Vp−p, is the dierence between the maximum and minimum voltages of the output.
e RMS (root mean square) voltage, VRMS, is the value of the constant DC voltage that would pro-duce the same power as AC voltage across the same resistance. Similarly, IRMS is the value of the con-stant direct current that would produce the same power as alternating current through the same resistance.
e emf produced by a generator can be increased by increasing the number of turns in the coil, increasing the strength of the magnetic eld, increasing the area of each coil or increasing the frequency of rotation of the coil.
QuestionsMagnetic ux 1. What is the dierence between magnetic ux and
magnetic eld strength? 2. Why did Faraday use coils with many turns of
copper wire? 3. Calculate the maximum magnetic ux passing
through:(a) a single coil of area 0.050 m2 in a magnetic eld
of strength 3.0 T(b) a single coil of area 4.5 cm2 in a magnetic eld
of strength 0.4 T(c) a coil of 50 turns, 12 cm2 in area in a magnetic
eld of strength 0.025 T. 4. Draw a graph of the magnetic ux passing through
a loop which is turning anticlockwise, from the position shown in the diagram below.
SN
5. As the metal rod shown falls through the magnetic eld, charge is separated and a voltage is established between the two ends of the rod. is requires energy. Where did the energy come from?
6. A magnet falling through a metal tube can achieve terminal velocity. Why?
7. (a) Explain what happens to the voltage between the ends of the rod in question 5 as the rod falls faster.
(b) How does this process dier from charging a capacitor?
Induced emf 8. e loop of wire shown on the next page is quickly
withdrawn from the magnetic eld. Which way does the current ow in the loop?
UNCORRECTED loops threaded by a magnetic
UNCORRECTED loops threaded by a magnetic
In an alternator, a coil rotates in a magnetic eld to
UNCORRECTED In an alternator, a coil rotates in a magnetic eld to induce a sinusoidal voltage and therefore an alter
UNCORRECTED induce a sinusoidal voltage and therefore an alter-
UNCORRECTED -
nating current. Slip rings are used at the end of the
UNCORRECTED nating current. Slip rings are used at the end of the coil to allow the alternating current to ow in an
UNCORRECTED coil to allow the alternating current to ow in an
In a DC generator, the slip rings are replaced with a
UNCORRECTED
In a DC generator, the slip rings are replaced with a commutator to allow a direct current to ow in an
UNCORRECTED
commutator to allow a direct current to ow in an
e voltage output of an AC generator can be
UNCORRECTED
e voltage output of an AC generator can be described in terms of its amplitude, frequency and
UNCORRECTED
described in terms of its amplitude, frequency and period. e amplitude of the voltage output is known
UNCORRECTED
period. e amplitude of the voltage output is known as the peak voltage,
UNCORRECTED
as the peak voltage, V
UNCORRECTED
Vpeak
UNCORRECTED
peakVpeakV
UNCORRECTED
VpeakV . e peak-to-peak voltage,
UNCORRECTED
. e peak-to-peak voltage, , is the dierence between the maximum and
UNCORRECTED
, is the dierence between the maximum and minimum voltages of the output.
UNCORRECTED
minimum voltages of the output.e RMS (root mean square) voltage,
UNCORRECTED
e RMS (root mean square) voltage, value of the constant DC voltage that would pro
UNCORRECTED
value of the constant DC voltage that would produce the same power as AC voltage across the same UNCORRECTED
duce the same power as AC voltage across the same esistance. Similarly, UNCORRECTED
esistance. Similarly,
PAGE eld of strength 0.025
PAGE eld of strength 0.025aw a graph of the magnetic ux passing through
PAGE aw a graph of the magnetic ux passing through a loop which is turning anticlockwise, from the
PAGE a loop which is turning anticlockwise, from the position shown in the diagram below.
PAGE position shown in the diagram below.
PAGE
PAGE
PAGE PROOFShy did Faraday use coils with many turns of
PROOFShy did Faraday use coils with many turns of
alculate the maximum magnetic ux passing
PROOFSalculate the maximum magnetic ux passing
ingle coil of area 0.050
PROOFSingle coil of area 0.050
PROOFS m
PROOFSm2
PROOFS2 in a magnetic eld
PROOFS in a magnetic eld
ingle coil of area 4.5
PROOFSingle coil of area 4.5
PROOFS
PROOFScm
PROOFScm2
PROOFS2 in a magnetic eld
PROOFS in a magnetic eld
of strength 0.4
PROOFSof strength 0.4 T
PROOFST
a coil of 50 t PROOFS
a coil of 50 turns, 12PROOFS
urns, 12 PROOFS
cmPROOFS
cmeld of strength 0.025PROOFS
eld of strength 0.025aw a graph of the magnetic ux passing through PROOFS
aw a graph of the magnetic ux passing through
181CHAPTER 7 Generating electricity
c07GeneratingElectricity 181 9 June 2016 5:11 PM
9. Two coils are placed one on top of the other with their centres in line as shown in the diagram below.(a) If a battery is switched on in the bottom coil,
producing a clockwise current seen from above, what happens in the top coil?
(b) Would the eect be dierent if the battery was connected to the top coil?
(c) Would the eect be dierent if the battery was switched o?
10. Two coils are placed side by side on a page with their centres in line.(a) If a battery is switched on in the left coil,
producing a clockwise current (seen from the left), what happens in the right coil?
(b) Would the eect be dierent if the current was anticlockwise?
11. e diagram below shows a conned uniform magnetic eld coming out of the page with a wire coil in the plane of the page. Is there an induced current in the coil as it is moved in direction:(a) A(b) B(c) C(d) D?Give a reason for each answer. If there is a current, indicate the direction.
Answer key:
A –
B –
C –
D –
into page
out of page
12. Calculate the average induced emf in each of the following situations.(a) A circular loop of wire of 5.0 cm radius is
removed from a magnetic eld of strength 0.40 T in a time of 0.2 s.
(b) e magnetic ux through a coil changes from 60 Wb to 35 Wb in 1.5 s.
(c) e magnetic ux through a coil changes from 60 Wb to −35 Wb in 2.5 s.
13. Calculate the average induced current in each of the following situations.(a) A circular loop of wire, 10 cm long with a
resistance 0.4 Ω, is removed from a magnetic eld of strength 0.60 T in a time of 0.3 s.
(b) e magnetic eld strength perpendicular to a square loop, of side length 0.26 m and resistance 2.5 Ω, is increased from 0.2 T to 1.2 T in 0.5 s.
(c) A stretched circular spring coil of 8 cm radius and resistance 0.2 Ω is threaded by a perpendicular magnetic eld of strength 2.0 T. e coil shrinks back to a radius of 4 cm in 0.8 s.
14. A coil with an area of 0.04 m2 of 100 turns spins at a rate of 50 Hz in a magnetic eld of strength 2.5 T.(a) What is the average emf induced as the
coil turns from parallel to the eld to perpendicular to the eld?
(b) What is the average emf as the coil does one complete turn?
15. How can a motor operate as a DC generator? 16. (a) Use the relationship for the size of induced
emf to show that the unit for the magnetic eld, the tesla, can be written as volt × second × metre−2.
(b) Now use Ohm’s Law and the denition of electric current to show that the tesla can also be written as ohm × coulomb × metre−2.
(c) Now use the denition of magnetic ux to show that the unit for magnetic ux, the weber, can be written as ohm × coulomb.
17. An orbiting satellite has a small module tethered to it by a 5.0 km conducting cable. As the satellite and its module orbit Earth, they cut across Earth’s magnetic eld at right angles.(a) If the pair are travelling at a speed of
6000 m s−1, how far do they travel in 1.0 s?(b) What area does the conducting cable cross
during the 1.0 s period?(c) If the strength of Earth’s magnetic eld at
this distance is 0.1 mT, what is the size of the induced emf?
18. A bar magnet, with its north end down, is dropped through a horizontal wire loop.(a) What is the direction of the induced current
when the magnet is: (i) just above the loop (ii) halfway through the loop (iii) just below the loop?(b) Draw the graph of the induced current against
time.
UNCORRECTED producing a clockwise current (seen from the
UNCORRECTED producing a clockwise current (seen from the
ould the eect be dierent if the current was
UNCORRECTED ould the eect be dierent if the current was
agram below shows a conned uniform
UNCORRECTED agram below shows a conned uniform
magnetic eld coming out of the page with a wire
UNCORRECTED magnetic eld coming out of the page with a wire coil in the plane of the page. Is there an induced
UNCORRECTED coil in the plane of the page. Is there an induced current in the coil as it is moved in direction:
UNCORRECTED
current in the coil as it is moved in direction:
Give a reason for each answer. If there is a current,
UNCORRECTED
Give a reason for each answer. If there is a current, indicate the direction.
UNCORRECTED
indicate the direction.
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED PAGE
PAGE H
PAGE Hhat is the average emf induced as the
PAGE hat is the average emf induced as the coil turns from parallel to the eld to
PAGE coil turns from parallel to the eld to perpendicular to the eld?
PAGE perpendicular to the eld?
(b)
PAGE (b) W
PAGE What is the average emf as the coil does one
PAGE hat is the average emf as the coil does one What is the average emf as the coil does one W
PAGE What is the average emf as the coil does one Wcomplete turn?
PAGE complete turn?
15.PAGE 15. HPAGE
How can a motor operate as a DC generator?PAGE ow can a motor operate as a DC generator?
16.PAGE 16. (a)PAGE
(a)
PROOFSagnetic eld strength perpendicular
PROOFSagnetic eld strength perpendicular
to a square loop, of side length 0.26
PROOFSto a square loop, of side length 0.26
PROOFS m and
PROOFSm and , is increased from 0.2
PROOFS, is increased from 0.2
PROOFS T t
PROOFST t
tretched circular spring coil of 8
PROOFStretched circular spring coil of 8
adius and resistance 0.2
PROOFSadius and resistance 0.2
PROOFS Ω
PROOFSΩ is threaded by a
PROOFS is threaded by a
perpendicular magnetic eld of strength 2.0
PROOFSperpendicular magnetic eld of strength 2.0e coil shrinks back to a radius of 4
PROOFSe coil shrinks back to a radius of 4
ith an area of 0.04PROOFS
ith an area of 0.04PROOFS
z in a magnetic eld of strength 2.5PROOFS
z in a magnetic eld of strength 2.5hat is the average emf induced as the PROOFS
hat is the average emf induced as the
UNIT 3 182
c07GeneratingElectricity 182 9 June 2016 5:11 PM
(c) Where did the electrical energy of the induced current come from?
(d) If the magnet falls from a very long distance above the loop to a very long distance below the loop, what is the overall change in magnetic ux through the loop? What does this imply about the area under the current–time graph?
(e) If the magnet accelerates under gravity, how will the induced current in the coil compare in size and duration when the magnet is above and then below the loop?
19. How much charge, in coulombs, ows in a loop of wire of area 1.6 × 10−3 m2 and resistance 0.2 Ω when it is totally withdrawn from a magnetic eld of strength 3.0 T?
20. A magnet passes through two loops, one wire and the other plastic. Compare the induced emfs and the induced currents of the two loops.
21. Lenz’s Law is an illustration of the conservation of energy. Explain why the reverse of Lenz’s Law (the direction of the induced current reinforces the change in magnetic ux) contravenes the law about the conservation of energy. Use the example of a north end of a magnet approaching a loop of conducting wire (as shown below).
NS
22. A DC motor has a coil rotating in a magnetic eld. is rotation produces a ‘back emf’ that opposes the current from the battery.(a) How does the back emf vary with the speed of
the motor?(b) How then would the current vary with the
speed of the motor?
(c) If the DC motor is used to lift masses, the speed of the motor is less for heavier masses. Why is there a risk that a heavy mass would burn out the motor?
23. Calculate the average emf in the axle of a car travelling at 120 kph if the vertical component of the Earth’s magnetic eld is 40 μT and the length of the axle is 1.5 m. (Hint: Calculate the area covered by the axle in one second.)
AC voltage and current 24. In the past electronic valves were powered by 6.3 V
RMS AC. What was the maximum voltage received by the valve?
25. A CRO shows the following trace. e settings are:Y: 10 mV per divisionX: 5 ms per division.
What are the:(a) period(b) frequency(c) peak voltage(d) peak-to-peak voltage(e) RMS voltageof this AC signal?
26. Some appliances are designed to run o either AC or batteries. e size of the batteries is equivalent to the peak of the AC voltage. If the appliance can run o 9 V DC, what RMS voltage would it also run o?
UNCORRECTED
C motor has a coil rotating in a magnetic eld.
UNCORRECTED
C motor has a coil rotating in a magnetic eld. is rotation produces a ‘back emf’ that opposes
UNCORRECTED
is rotation produces a ‘back emf’ that opposes the current from the battery.
UNCORRECTED
the current from the battery.ow does the back emf vary with the speed of
UNCORRECTED
ow does the back emf vary with the speed of
ow then would the current vary with the
UNCORRECTED
ow then would the current vary with the speed of the motor?
UNCORRECTED
speed of the motor?
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PAGE PROOFSn the past electronic valves were powered by 6.3
PROOFSn the past electronic valves were powered by 6.3C. What was the maximum voltage received
PROOFSC. What was the maximum voltage received
O shows the following trace. e settings are:
PROOFSO shows the following trace. e settings are:
er division.
PROOFSer division.
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