ELECTROMAGNETIC INDUCTION
12.1
Induced electromotive force (emf )
12.2
Alternating current
12.3
Transmission of electric power
I
n the chapter on electrostatics, and in this chapter on
electromagnetism, mention is made of Michael Faraday.
he laws of electricity and magnetism owe more perhaps to
the experimental work of Faraday than any other person.
here were great theoreticians like Ampère and Maxwell
but Faraday was a real experimenter. He invented the irst
dynamo, electric motor and transformer. It was Faraday
who originated the use of electric ields lines that he called
“lines of force” even before the concept of the electric
ield was clearly understood. He along with Joseph Henry
discovered electromagnetic induction, and this concept will
be expanded on in this chapter. Electromagnetic induction
has revolutionised the way we live. his phenomenon has
had a huge impact on society and it has become the basis
for the generation of electric power that we so oten take for
granted in our everyday life.
Michael Faraday (1791-1867), the son of a blacksmith,
was born in Newington, Surrey. He had little formal
education as a child and at the age of 14, he took up an
apprenticeship as a bookbinder. While rebinding a copy
of the Encyclopaedia Britannica, he happened to read an
article on electricity, and to his own admission, this article
gave him a lifelong fascination with science.
He started to attend lectures given by Sir Humphry Davy,
a famous electrochemist and publicist. Faraday became
interested in electrolysis and he prepared a set of lecture
notes that greatly impressed Davy. By good fortune or
misfortune, when Davy was temporarily blinded in a
laboratory accident at the Royal Institution in 1812,
he needed a laboratory assistant and he requested that
Faraday be given the position. During this time, Faraday
discovered and described the organic compound benzene
12
as well as some other chloro-carbon compounds. He did
research on steel, optical glass and the liquefaction of
gases. Faraday’s work was impressive, and he eventually
became director of the Royal Institution.
Faraday had a great talent for explaining his ideas to both
children and adults. He gave many “wizz-bang” lectures
to the young, and his book addressed at their level called
“he Chemical History of the Candle” is still in print.
He introduced the Friday Evening Discourses and the
Christmas lectures for children at the Royal Institution,
and these lectures still continue to this day. In 1865, he
retired from the Royal Institution ater 50 years service.
In the 1830s, Faraday became interested in electrochemistry
and he was the irst to use the term “electrolysis” in
1832. Furthermore, he introduced the use of the terms
‘electrolyte’, ‘cell’, ‘electrodes’ and ‘electrochemical reaction’
so commonly used in the subject of electrochemistry.
He subjected electrolysis to the irst quantitative
experimentation and in 1834 was able to establish that
the amount of chemical compound decomposed at the
electrodes was proportional to the amount of electricity
used – Faraday’s First and Second Law of Electrolysis.
He devised the terminology, “ions”, for the part of the
compound discharged at the electrodes.
Once Oersted had discovered that a current lowing in a
conductor produced a magnetic ield in 1819, scientists
were convinced that a moving magnetic ield should be able
to produce a current in the conductor. It took eleven years
before, in 1831, the American, Joseph Henry (1797-1878),
and the Englishman, Michael Faraday (1791-1867), while
working independently, explained the cause and efect of
313
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ELECTROMAGNETIC
INDUCTION
CHAPTER 12
an induced current/emf being produced by a changing
magnetic ield. Henry is credited with the discovery but
Faraday was the irst to publish, introducing the concept
of line of magnetic lux in his explanations.
In his notebooks in the early 1830s, Faraday described
how he placed wires near magnets looking for current
in the wire but without success. However, as he moved
the apparatus he noticed a brief pulse of current but the
current immediately fell back to zero. Perhaps the missing
ingredient was motion.
AHL
he solution came in 1831 when he set up an apparatus
similar to that in Figure 1201 that he called an “induction
ring”. he apparatus may look familiar to us with its battery,
coils and galvanometer (a meter to detect current). We
also know that a sot iron core increases the strength of a
magnetic ield.
change in magnetic lux. Faraday presented his indings
to the Royal Society in November 1831 and January 1832
in his ‘Experimental researches into electricity’ in which he
gave his “Law which governs the evolution of electricity by
magneto-electric induction” – a change in magnetic lux
through any surface bounded by closed lines causes an
ε.m.f around the lines.
Within no time, the dynamo, the generator and the
transformer were invented by this brilliant experimental
scientist. We will expand on the principles of electromagnetic induction in the remainder of this chapter.
12.1 INDUCED
ELECTROMOTIVE
FORCE (EMF)
switch
G
soft iron torus
Figure 1201
Faraday’s induction ring
To his initial disappointment, when he closed the switch to
allow steady current to low, only a slight twitch was observed
in the galvanometer before the needle fell back to zero.
his twitch could have been due to mechanical vibration.
However, using his intuition, he noticed that when he slowly
opened and closed the switch, a current was produced in
one direction, then fell to zero, then a current was produced
in the opposite direction. He called the current produced by
a changing magnetic lux an induced current, and he called
the general phenomenon electromagnetic induction. He
assumed that the magnetic lux must be changing, but how?
Was the iron ring really necessary to produce induction or
did it merely strengthen an efect? Was it necessary to have
two coils or could an induced current be produced simply
by moving a magnet in and out of a single coil of wire?
Remembering back to his earlier experiment where he
considered that motion could be a factor, he quickly set
up experiments using a coil and a magnet, and proved
that the iron ring was not essential, and that motion inside
one coil could produce an induced current. Furthermore,
he found that a rotating copper disc inserted between
the poles of a magnet could be used (instead of a coil) to
produce an induced current. It didn’t matter whether the
magnet was in motion or the coil (or disc) was in motion,
an induced current was produced provided there was a
314
12.1.1 Describe the inducing of an emf by relative
motion between a conductor and a
magnetic field.
12.1.2 Derive the formula for the emf induced in
a straight conductor moving in a magnetic
field.
12.1.3 Define magnetic flux and magnetic flux
linkage.
12.1.4 Describe the production of an induced emf
by a time-changing magnetic flux.
12.1.5 State Faraday’s law and Lenz’s law.
12.1.6 Solve electromagnetic induction problems.
© IBO 2007
12.1.1 INDUCED EMF bETWEEN A
CONDUCTOR AND A MAGNETIC
FIELD
Faraday used an apparatus similar to that in Figure 1202.
If a conductor is held stationary in a magnetic ield of
a U-shaped magnet that is connected in series with a
very sensitive, zero-centred galvanometer, no reading
is observed. However, if the conductor is moved across
the magnetic ield, then a delection occurs in the needle
of the galvanometer in one direction. Ater a very short
period of time, the needle returns to zero on the scale.
ELECTROMAGNETIC INDUCTION
N
S
1.
he speed of the movement
2.
he strength of the magnetic lux density
3.
he number of turns on the coil
he current produced is called an induced current.
4.
he area of the coil
As work is done in moving the current from one end of the
conductor to the other, an electrical potential diference
exists, and an induced emf is produced.
Faraday realised that the magnitude of the induced ε.m.f
was not proportional to the rate of change of the magnetic
ield B but rather proportional to the rate of change of
magnetic lux Φ for a straight conductor or lux linkage NΦ.
his will be discussed further in sectin 12.1.4.
sensitive
galvanometer
Figure 1202 Producing an induced current
If the conductor is then moved in the opposite direction,
the needle of the galvanometer delects in the opposite
direction before then falling to zero again. If the conductor
is moved in the same direction as the magnetic ield then
no delection occurs.
he direction of the induced current can be obtained by
using the let-hand palm rule (refer to the palm rules
discussed for the motor efect in Chpater 6). Using the
magnetic ield and direction of movement of the wire, if
the palm of your let hand points in the direction of motion
of the conductor, and your ingers point in the direction of
the magnetic ield, then the thumb gives the direction of
the induced current. In this case, the current is in an anticlockwise direction.
Alternatively, you can continue to use the right-hand
palm rule BUT your palm points in the opposite direction
to the applied force. Fleming’s right-hand rule can also
be used. he right-hand palm rule for the direction of an
induced current and Fleming’s right-hand rule are shown
in Figures 1203 (a) and (b) respectively.
Thumb points in direction
of induced current
Fingers points in direction
First finger points in
of magnetic field
direction of field
Palm points in direction
of opposing force
(a)
Example
Determine the direction of the induced current for each
situation given below.
a.
b.
c.
d.
Solution
Using a hand rule, the direction of the induced current for
each situation is indicated by the arrows as shown in the
diagram below:
a.
b.
c.
d.
Thumb points in
direction of movement
Second finger points in
direction of current
(b)
Figure 1203 (a) and (b) Palm rules
for electromagnetic induction.
315
AHL
he simple apparatus in Figure 1202 can detect the
induced current, but the readings on the galvanometer are
small (a zero-centred micro-ammeter is better). Faraday
improved the apparatus by moving diferent magnetic
lux densities into and out of diferent sized solenoids at
diferent speeds. He found that the strength of the induced
emf was dependent on
motion
CHAPTER 12
12.1.2 DERIVATION OF AN INDUCED
12.1.3 DEFINE MAGNETIC FLUx AND
EMF IN A CONDUCTOR
Consider a conductor of length l that moves with velocity
v perpendicular to a magnetic lux density or induction B
as shown in Figure 1206.
B into page
MAGNETIC FLUx LINkAGE
Consider a small planar coil of a conductor for simplicity
as shown in Figure 1208 (it could be any small shape).
Now imagine it is cut by magnetic lines of lux. It would
be reasonable to deduce that the number of lines per
unit cross-sectional area is equal to the magnitude of the
magnetic lux density B ×ε the cross-sectional area A. his
product is the magnetic lux Φ.
l
A
B
v
AHL
A
θ
B
Figure 1206 Cause of an induced emf.
Figure 1208 (a) and (b) Flux through a small, plane surface
When the wire conductor moves in the magnetic ield, the
free electrons experience a force because they are caused to
move with velocity v as the conductor moves in the ield.
F = e ×v ×B
his force causes the electrons to drit from one end of the
conductor to the other, and one end builds-up an excess of
electrons and the other a deiciency of electrons. his means
that there is a potential diference or emf between the ends.
Eventually, the emf becomes large enough to balance the
magnetic force and thus stop electrons from moving.
ev×B = e ×E ⇔ E = B ×v
If the potential diference (emf) between the ends of the
conductor is ε then
ε = E ×l
he magnetic lux Φ through a small plane surface is the
product of the lux density normal to the surface and the
area of the surface.
Φ = BA
he unit of magnetic lux is the weber Wb.
Rearranging this equation it can be seen that:
B = Φ / A which helps us understand why B can be called
the lux density. So the unit for lux density can be the tesla T,
or the weber per square metre Wbm-2. So, 1T = 1 Wbm-2.
If the normal shown by the dotted line in Figure 1208 (b)
to the area makes an angle θ with B, the the magnetic lux
is given by:
Φ = B A cos
By substitution, we have,
ε = B ×l ×v
If the conducting wire was a tightly wound coil of N turns
of wire the equation becomes:
ε = NB l v
Φ the region and θ is the angle of
where A is the area of
movement between the magnetic ield and a line drawn
perpendicular to the area swept out. (Be careful that you
choose the correct vector component and angle because
questions on past IB examinations give the correct answer
of BA sin θ).
If Φ is the lux density through a cross-sectional area of a
conductor with N coils,
Φ the total lux density will be given
by:
Φ = N B A cos
316
ELECTROMAGNETIC INDUCTION
his is called the lux linkage.
So it should now be Φ
obvious that we can increase the
magnetic lux by:
•
•
•
Increasing the conductor area
Increasing the magnetic lux density B
Keeping the lux density normal to the surface of
the conductor
12.1.4 INDUCED EMF IN A TIMEΦ
CHANGING
MAGNETIC FLUx
B into page
Faraday’s Law can therefore be stated as:
he magnitude of the induced emf in a circuit is directly
proportional to rate of change of magnetic lux or luxlinkage.
Figure 1210 shows some relative movements of a bar
magnet at various positions relative to a coil with many
turns.
S
N
magnet moving inwards
current flow
AHL
Now examine Figure 1209 that shows the shaded area of
the magnetic lux density swept out in one second by a
conductor of length l moving from the top to the bottom
of the igure through a distance d.
12.1.5 FARADAy’s LAW AND LENz’s
LAW
l
initial position of wire
S
area swept out by wire in one
second
v
no current flow
position of wire 1 second later
Figure 1209
Rate of area swept out.
We have already derived that ε = B l v
N
no movement
S
N
magnet moving outwards
current flow
he area swept out in a given time is given by (l × d) / t.
But v = d / t. So that the area swept out = lv / t.
Figure 1210 Lenz’s Law applied to a solenoid.
hat is,
ε = Δ- (- BA )Δt
where A is the area in m2.
For a single conductor in the magnetic lux density, it can
be seen that:
∆Φ
Φ
ε∝∆
-------- ⇔ ε = –-------∆t
∆t
where the constant equals –1. he negative sign will be
explained in the next section.
When the north pole is moved toward the core of the
solenoid, an induced current lows in the external circuit
as indicated by a zero-centred galvanometer or a microammeter. he pointer moves to the right meaning that the
conventional induced current is lowing anti-clockwise at
the end of the solenoid nearest the magnet. his end is
acting like a north pole.When the magnet is stationary the
meter reads zero. his suggests that the induced current is
dependent on the speed of the movement. When the bar
magnet is removed from the solenoid, the induced current
lows in the opposite direction, and a south pole is created
in the end that was previously a north pole.
If there are N number of coils, then:
Φ
ε = –N × ∆
-------∆t
In 1834, a Russian physicist Heinrich Lenz (1804-1865)
applied the Law of Conservation of Energy to determine the
direction of the induced emf for all types of conductors. It
is known as the Second Law of Electromagnetic Induction
and it can be stated as:
317
CHAPTER 12
he direction of the induced emf is such that the current
it causes to low opposes the change producing it.
In the above case, the current induced in the coil creates
a north pole to oppose the incoming north pole of the
magnet. Similarly, when the magnet is withdrawn its
north pole creates a south pole in the solenoid to oppose
the change.
AHL
It can be reasoned that the Law of Conservation of Energy
must apply. If the solenoid in Figure 1211 had an induced
south pole when the north pole of the magnet was moved
towards it, the magnet would accelerate as it would
experience a force of attraction. More induced current
would be produced creating more acceleration. he
kinetic energy would increase indeinitely – energy would
be created. As this is impossible, it makes sense that the
induced current must oppose the change producing it.
Lenz’s Law can be applied to straight conductors as well as
solenoids. Figure 1211 shows the magnetic lines of force
for a bar magnet and a current-carrying wire directed
into the page before and during interaction. Suppose the
conductor is carrying an induced current initially.
After interaction
Before interaction
magnetic field
due to conductor
N
S
12.1.6 ELECTROMAGNETIC INDUCTION
pRObLEMs
Example 1
A metal conductor 2.5 m long moves at right angles to a
magnetic ield of 4.0 × 10–3 T with a velocity of 35 m s–1.
Calculate the emf of the conductor.
Solution
Using the formula, ε = B l v, we have,
ε = ( 4.0 × 10
= 0.35 V
–3
–1
T ) × ( 2.5 m ) × ( 35 ms )
he potential diference between the ends of the conductor
is 0.35 V.
Example 2
magnetic field interaction
produces an opposing force
N
interaction of the two
magnetic fields causes
‘tension’ on lower side
S
force due to hand
pushing down
A square solenoid with 120 turns and sides of 5.0 cm is
placed in air with each turn perpendicular to a uniform
magnetic lux density of 0.60 T. Calculate the induced emf
if the ield decreases to zero in 3.0 s.
Figure 1211 Lenz’s Law in a straight conductor.
Solution
he straight conductor is then pushed downwards say with
your hand. Your energy source induces the current but
the combined magnetic ields tend to push the conductor
upwards (a force is applied in the direction from the region
of most lux density to the region of least lux density).
herefore, the induced current will be in such a direction
that tries to stop the conductor through the ield.
his time we need to use the formula, Φ = B A, so that
2
Φ = ( 0.05 m ) × (0.60 T )
= 1.5 × 10
–3
Wb
Next, we make use of the formula,
If we now combine Faraday’s and Lenz’s Laws of
electromagnetic induction into the equation, we can now
understand the signiicance of the negative sign.
Φ
ε = –N × ∆
-------∆t
318
–3
( 0 – 1.5 × 10 Wb )
∆Φ
ε = – N × -------- = – 120 × ------------------------------------------∆t
3.0 s
= 0.060 V
he induced emf is 6.0 × 10-2 V.
ELECTROMAGNETIC INDUCTION
2.
Example 3
(a)
(b)
he magnetic lux Φ through a coil having 200
turns varies with time t as shown below.
5.0
A coil with 20 turns has an area of 2.0 × 10-1 m2.
It is placed in a uniform magnetic ield of lux
density 1.0 × 10-1 T so that the lux links the turns
normally. Calculate the average induced emf in the
coil if it is removed from the ield in 0.75 s.
4.0
he same coil is turned from its normal position
through an angle of 30° in 0.3 s in the ield.
Calculate the average induced emf.
3.0
–2
Φ/ 10 Wb
2.0
Solution
1.0
his time we need to use the formula, Φ = NB A, so
that
0.0
Φ = 20 turns × 2.0 × 10-1 m2 × 1.0 × 10-1 T = 4.0 × 10-1 Wb
0
0.5
Next, we make use of the formula,
1 1.5 2
t / 10–2 s
AHL
(a)
2.5
he magnitude of the emf induced in the coil is:
emf = - (Δ Φ / Δt) = 0 - 4.0 × 10-1 Wb / 0.75 s
A.
B.
C.
D.
= 0.533 V
he induced emf is 0.53 V.
(b)
he lux change through the coil = NBA – NBAcosθ
3.
0.5 V
2V
100 V
500 V
A metal ring falls over a bar magnet as shown
= 4.0 × 10-1 Wb - 4.0 × 10-1 Wb cos 30 = 0.054 Wb
Average induced emf = 0.54 Wb / 0.3 s = 0.179 V
S
he induced emf is 0.18 V.
Exercise
N
12.1
he induced current is directed
1.
Consider a coil of length l, cross-sectional area A,
number of turns n, in which a current I is lowing.
he magnetic lux density of the coil depends on
A.
B.
C.
D.
I, l, n but not A
I, n, A but not l
I, A, l but not n
A, l, n but not I
A.
B.
C.
D.
always opposite to the direction of the
arrow.
always in the same direction as the arrow.
irst in the opposite direction to the arrow,
then as shown by the arrow.
irst as shown by the arrow, then in the
opposite direction to the arrow.
319
CHAPTER 12
4.
he magnitude of an induced emf produced by the
relative motion between a solenoid and a magnetic
ield is dependent upon:
A.
B.
C.
D.
10.
What efect would the following have on the
magnitude of the induced emf in a conductor
moving perpendicular to a magnetic ield?
(a)
the strength of the magnetic lux density
the number of turns on the coil
the area of the coil
all of the above
(b)
(c)
5.
Which of the following is a suitable unit to
measure magnetic lux density?
A.
B.
C.
D.
AHL
6.
11.
Explain in detail the diference between magnetic
lux density and magnetic lux.
12.
he magnetic lux through a coil of wire
containing 5 loops changes from –25 Wb to
+ 15 Wb in 0.12 s. What is the induced emf in the
coil?
Faraday’s law of electromagnetic induction states
that the induced emf is
13.
equal to the change in magnetic lux
equal to the change in magnetic lux linkage
proportional to the change in magnetic lux
linkage
proportional to the rate of change of
magnetic lux linkage
he wing of a Jumbo jet is 9.8 m long. It is lying
at 840 km h–1. If it is lying in a region where the
earth’s magnetic ield has a vertical component
of 7.2 × 10–4 T, what potential diference could be
produced across the wing?
14.
A uniform magnetic ield of strength B completely
links a coil of area A. he ield makes an angle θ to
the plane of the coil.
Find the total lux through an area of 0.04 m2
perpendicular to a uniform magnetic lux density
of 1.25 T.
15.
If the total lux threading an area of 25 cm2 is
1.74 × 10–2 Wb, what would be the magnetic lux
density?
16.
A coil of area 5 cm2 is in a uniform magnetis ield
of lux density 0.2 T. Determine the magnetic lux
in the coil when:
A.
B.
C.
D.
7.
A m N-1
Kg A-1 s-2
A N-1 m-1
T m-1
B
θ
area A
(a)
(b)
(c)
he magnetic lux linking the coil is
A.
B.
C.
D.
BAcosθ
BA
BAsinθ
BAtanθ
8.
What factors determine the magnitude of an
induced emf?
9.
Refer to Figure 1204. Use Lenz’s Law to explain
what would happen if the solenoid was moved
rather than the magnet.
320
Doubling the velocity of movement of the
conductor.
Halving the magnetic lux density and
velocity.
Changing the conductor from copper to
iron.
he coil is normal to the magnetic ield
he coil is parallel to the magnetic ield
he normal to the coil and the ield have an
angle of 60°
17.
A metal conductor 2.5 m long moves at right
angles to a magnetic ield of 4.0 .10-3 T with
a velocity of 35 m.s-1.calculate the emf of the
conductor.
18.
A square solenoid with 120 turns and sides of
5.0 cm is placed in air with each turn
perpendicular to a uniform magnetic lux density
of 0.60 T. Calculate the induced emf if the ield
decreases to zero in 3.0 s.
ELECTROMAGNETIC INDUCTION
20.
A coil with 1500 turns and a mean area of 45 cm2
is placed in air with each turn perpendicular to
a uniform magnetic ield of 0.65 T. Calculate the
induced emf if the ield decreases to zero in 5.0 s.
he radius of the copper ring is 0.15 m and
its resistance is 2.0 × 10–2 Ω. A magnetic ield
strength is increasing at rate of 1.8 × 10–3 T s–1.
Calculate the value of the induced current in the
copper ring.
12.2 ALTERNATING
CURRENT
12.2.1 Describe the emf induced in a coil rotating
within a uniform magnetic field.
12.2.2 Explain the operation of a basic alternating
current (ac) generator.
12.2.3 Describe the effect on the induced emf of
changing the generator frequency.
12.2.4 Discuss what is meant by the root mean
squared (rms) value of an alternating
current or voltage.
12.2.5 State the relation between peak and rms
values for sinusoidal currents and voltages.
12.2.1 EMF IN A COIL ROTATED WITHIN
A UNIFORM MAGNETIC FIELD
he induced emf in a coil rotated within a uniform
magnetic ield is sinusoidal if the rotation is at constant
speed.
he most important practical application of the Laws of
Electromagnetic Induction was the development of the
electric generator or dynamo.
he frequency of the generator cycle used in power
stations can be investigated with the use of a cathode ray
oscilloscope. A C.R.O. can be used to measure the voltage
output of an ac source. If a low safe ac voltage (9V) from
a power pack is connected to the C.R.O. then its source
is the power stations of your community supplier. If the
time-base is adjusted to obtain a sine curve trace on the
screen, the mains supply frequency can be determined.
Example
In Figure 1215, if the potentiometer is set on 2 V/division
and the time base is set a 5 ms/cm, what is the voltage and
frequency of the ac generator?
Solution
12.2.6 Solve problems using peak and rms values.
he amplitude of the wave is 3 divisions and each division is
2 V. herefore, the emf would be 6 V.
12.2.7 Solve ac circuit problems for ohmic
resistors.
12.2.8 Describe the operation of an ideal
transformer.
12.2.9 Solve problems on the operation of ideal
transformers.
© IBO 2007
Figure 1215
CRO trace
Between the two dots there are 6 divisions. herefore, the
wavelength is equivalent to 12 divisions. Now there are
5 milliseconds/ cm. So the period of the wave is 60 ms, that
is, T = 60 ms.
321
AHL
19.
CHAPTER 12
herefore, the frequency of the source is given by,
he magnitude of the emf and current varies with time
as shown in Figure 1217. Consider a coil ABCD rotating
clockwise initially in the horizontal position. From
the graph of current versus time, you can see that the
current reaches a maximum when the coil is horizontal
and a minimum when the coil is vertical. If more lines of
magnetic lux are being cut, then the induced current will
be greater. his occurs to the greatest extent when the coil
is moving at right angles to the magnetic ield. When the
coil moves parallel to the ield, no current lows.
1
1
f = --- = --------------------3 = 16.67 Hz
–
T
60 × 10
he frequency of the source is 17 Hz.
12.2.2 OpERATION OF A bAsIC AC
AHL
GENERATOR
A generator is essentially a device for producing electrical
energy from mechanical energy. (Remember that an electric
motor did the opposite energy conversion). Generators
use mechanical rotational energy to provide the force to
turn a coil of wire, called an armature, in a magnetic ield
(the magnet can also be turned while the coil remains
stationary). As the armature cuts the magnetic lux
density, ε.m.fs are induced in the coil. As the sides of the
coil reverse direction every half turn, the ε.m.fs alternate in
polarity. If there is a complete circuit, alternating current
ac is produced.
current
B
maximum current is produced when wire cut magnetic flux lines at 90°
C
A
C
B
D
C
B
A
A
D
D
C
time
B
D
C
A
B
D
A
zero current is produced when no wires are cutting magnetic flux line
Figure 1217 The changes of current with time
he induced currents are conducted in and out by way of
the slip-rings and the carbon brushes.
Each complete cycle of the sinusoidal graph corresponds
to one complete revolution of the generator.
Turbines driven by steam or water are the commonest
devices used for the generation of electricity in modern
day society.
12.2.3 CHANGING FREqUENCy OF THE
GENERATOR
Figure 1216 shows a simple generator in which a coil is
rotating clockwise. he circuit is completed with a lamp
acting as the load.
F
I
N
S
slip rings
N
I
F
S
he emf of a rotating coil can be calculated at a given time.
If a coil of N turns has an area A, and its normal makes an
angle θ with the magnetic ield B, then the lux-linkage Φ
is given by:
Φ = N ×A× B cos θ
I
he emf varies sinusoidally (sin and cos graphs have the
same shape) with time and can be calculated using
brush contacts
Figure 1216
AC generator
To determine the direction of the induced current
produced as the coil rotates, we must apply Lenz’s Law.
As the let hand side of the coil (nearest the north-pole of
the magnet) moves upward, a downward magnetic force
must be exerted to oppose the rotation. By applying the
right-hand palm rule for electromagnetic induction, you
can determine the direction of the induced current on that
side of the coil. he direction of the current in the rightside of the coil can also be determined.
322
∆ cos θ
∆Φ
ε = – -------- = –N × A × B × --------------∆t
∆t
Using calculus and diferentiating cos θ, this relationship
becomes
∆θ
ε = N × A × B × sin θ × -----∆t
ELECTROMAGNETIC INDUCTION
Remember from your knowledge of rotational motion that
Example 1
Δθ ÷ Δt = the angular velocity in rad s = 2 π f
-1
Also
Calculate the peak voltage of a simple generator if the
square armature has sides of 5.40 cm and it contains 120
loops. It rotates in a magnetic ield of 0.80 T at the rate of
110 revolutions per second.
θ = ωt = 2 πft
so that
Solution
ε = ω × N × A × B × sin ( ωt)
So that,
Making use of the formula above, that is,
ε = ω N A B sin ( ωt )
ε = 2 πf ×N × A × B × sin (2 πft )
We can see that the maximum emf will occur when sin ω
t = 1, so that, εmax = ωNAB
But, ω = 2 π f, so that
ε0 = ω NBA
ε0 = (2π) × (110.0 Hz) × (120 turns) × (5.4 × 10-4 m2) × (0.80 T)
herefore:
= 35.8 V
ε = ε0 sin ω t
hat is, the output voltage is 36 V.
he frequency of rotation in North America is 60 Hz but
the main frequency used by many other countries is 50 Hz.
Example 2
Note that if the speed of the coil is doubled then the
frequency and the magnitude of the emf will both increase
as shown in Figures 1218 and 1219 respectively.
ε/V
0
Suppose a coil with 1200 turns has an area of 2.0 × 10-2 m2
and is rotating at 50 revolutions per second in a magnetic
ield of magnitude 0.50 T. Draw graphs to show how
the magnetic lux, the emf and the current change as a
function of time. (Assume the current lows in a circuit
with a resistance of 25 Ω).
Solution
Figure 1218
Normal frequency
he magnetic lux in the coil changes over time as shown
in Figure 1220.
ε/V
t /ms
0
Φ = NBA = 1200 turns × 0.5 T × 2 × 10-2 m2 = 12 Wb
50 revolutions per second would have 1 revolution in 0.02
seconds = 20 ms.
Figure 1219
Doubled frequency.
323
AHL
= 2πf N A B sin ( 2 πft )
When the plane of the coil is parallel to the magnetic
ield, sin ωt will have its maximum value as ωt = 90°, so
sin ωt = 1. his maximum value for the emf ε0 is called the
peak voltage, and is given by:
CHAPTER 12
Φ/Wb
12.2.4-12.2.5
pEAk AND RMs VOLTAGE
12
0
10
20
30
40
An alternating current varies sinusoidally and can be
represented by the equation
-12
Figure 1220 Changing flux linkage over time
I = I p sin ( ωt)
We can see that the maximum emf will occur when
sin ω t = 1, so that,
where I0 is the maximum current called the peak current
as shown in Figure 1223 for a 50 Hz mains supply.
εmax = ωNAB
I/A
I0
But, ω = 2 π f, so that
AHL
ε0 = (2π) × (50 Hz) × (1200 turns) × (2 × 10-2 m2) × (0.50 T)
0
10
20
30
40
t
= 75.4V = 75V
Figure 1223 Peak current and current over time
he appropriate graph is shown in Figure 1221.
In commercial practice, alternating currents are expressed
in terms of their root-mean-square (r.m.s.) value.
75
ε/V
0
10
20
30
40
t
-75
Consider 2 identical resistors each of resistance R, one
carrying d.c. and the other a.c. in an external circuit.
Suppose they are both dissipating the same power as
thermal energy.
he r.m.s. value of the alternating current that produces
the power is equal to the d.c. value of the direct current.
Figure 1221
Induced emf over time
he current lows in a circuit with a resistance of 25 Ω.
I = ε/R = ε0 sin ωt / R = I0 sin ωt where I0 75.4 V / 25 Ω =
3.0 A
he appropriate graph is shown in Figure 1222.
For the maximum value in a.c., the power dissipated is
given by
2
P = V0 sin ( ωt ) × I0 sin ( ωt ) = V0 I0 sin ( ωt )
his means that the power supplied to the resistor in time
by an alternating current is equal to the average value of
I 2R multiplied by time.
3
P =I 2ave × R = I02 R sin2 ωt
I/A
0
10
20
30
40
-3
Figure 1222
324
Induced current over time
t
Because the current is squared the, the value for the power
dissipated is always positive as shown in Figure 1224.
ELECTROMAGNETIC INDUCTION
Power
12.2.6 sOLVING pRObLEMs UsING
RMs AND pEAk VALUEs
I 0 2R
½ I 0 2R
10
20
30
40
t
Figure 1224 Power delivered to a resistor in an
alternating current circuit.
Example 1
In the USA, the r.m.s value of the “standard line voltage”
is 110 V and in some parts of Europe, it is 230 V. Calculate
the peak voltage for each region.
he value of sin2ωt will therefore vary between 0 and 1.
herefore its average value
Solution
= (0 + 1) / 2 = ½ .
V0 = √2 Vrms
AHL
herefore the average power that dissipates in the resistor
equals:
In the USA = 1.414 × 110 V = 170 V
Pave = ½ I02 R = I0 / √2 × I0 / √2 × R or
In Europe = 1.414 × 230 V = 325 V.
or
Pave = ½ V02 / R = (V0 / √2 × V0 / √2) / R.
So the current dissipated in a resistor in an a.c. circuit that
varies between I0 and - I0 would be equal to a current I0
/√2 dissipated in a d.c circuit. his d.c current is known as
r.m.s. equivalent current to the alternating current.
Example 2
he domestic standard line voltage in Australia is 240 V.
Calculate the current and resistance in a 1200 W electric
jug and compare these values with the same electric jug
used in the USA
It can be shown that
Ir.m.s = I 0 × 2
and
Vr.m.s = V0 × 2
Provided a circuit with alternating current only contains
resistance components, it can be treated like a direct
current circuit.
Solution
Irms = Pave / Vrms and R = Vrms / Irms
In Australia:
I = 1200 W / 240 V = 5.0 A and R = 240 V / 5.0 A = 48 Ω
In the USA:
I = 1200 W / 110 V = 11 A and R = 110 V / 10.9 A = 10 Ω
he current would have a greater heating efect in the USA
than Australia but the element in the electric jug would need
to be made of a conductor with a a greater cross-sectional
area.
325
CHAPTER 12
12.2.7 sOLVING AC CIRCUIT pRObLEMs
FOR OHMIC REsIsTORs
a.c. input
a.c. output
circuit symbol for
simple transformer
primary coil
secondary coil
Example
Figure 1225 A simple transformer
he V0 value for a circuit containing a 35 Ω resistor is 45 V.
Calculate the current and the power dissipated in the
resistor.
Solution
AHL
Vrms = V0 / √2 = 45 / 1.414 = 31.82 V.
Irms = Vrms / R = 31.82 / 35 = 0.91 A.
When an ac voltage is applied to the primary coil, an ac
voltage of the same frequency is induced in the secondary
coil. his frequency in most countries is 50 Hz.
When a current lows in the primary coil, a magnetic ield
is produced around the coil. It grows quickly and cuts the
secondary coil to induce a current and thus to induce a
magnetic ield also. When the current falls in the primary
coil due to the alternating current, the magnetic ield
collapses in the primary coil and cuts the secondary coil
producing an induced current in the opposite direction.
Pave = I2 rms × R = 0.912 A2 × 35 Ω = 29 W.
he size of the voltage input/output depends on the
number of turns on each coil. It is found that
12.2.8 OpERATION OF AN IDEAL
Np
Vp
Is
----- = ------ = ---Ns
Vs
Ip
Where N = the number of turns on a designated coil and I
is the current in each coil.
TRANsFORMER
A useful device that makes use of electromagnetic
induction is the ac transformer as it can be used for
increasing or decreasing ac voltages and currents.
It consists of two coils of wire known as the primary and
secondary coils. Each coil has a laminated (thin sheets
fastened together) sot iron core to reduce eddy currents
(currents that reduce the eiciency of transformers). he
coils are then enclosed with top and bottom sot iron bars
that increase the strength of the magnetic ield. Figure
1225 shows a typical circuit for a simple transformer
together with the recommended circuit symbol.
It can be seen that if Ns is greater than Np then the
transformer is a step-up transformer. If the reverse
occurs and Ns is less than Np it will be a step-down
transformer.
If a transformer was 100% eicient, the power produced
in the secondary coil should equal to the power input of
the primary coil. In practice the eiciency is closer to 98%
because of eddy currents.
I
V
VpIp = Vs Is ⇔ -----p- = ---sVs
Ip
his means that if the voltage is stepped-up by a certain
ratio, the current in the secondary coil is stepped-down
by the same ratio.
326
ELECTROMAGNETIC INDUCTION
P = 240 × 150 × 10
12.2.9 sOLVING pRObLEMs ON
(e)
THE OpERATION OF IDEAL
he igure below shows a step-down transformer that is
used to light a ilament globe of resistance 4.0 Ω under
operating conditions.
the reading on the voltmeter with S open
the current in the secondary coil with an
efective resistance of 0.2 Ω with S closed
the power dissipated in the lamp
the power taken from the supply if the
primary current is 150 mA
the eiciency of the transformer.
(c)
(d)
(e)
12.3.2 Explain the use of high-voltage step-up and
step-down transformers in the transmission
of electric power.
12.3.3 Solve problems on the operation of real
transformers and power transmission.
12.3.4 Suggest how extra-low-frequency
electromagnetic fields, such as those
created by electrical appliances and power
lines, induce currents within a human body.
Solution
4.0 Ω
V
12.3.5 Discuss some of the possible risks involved
in living and working near high-voltage
power lines.
240 V ac
S
© IBO 2007
50 turns
1000 turns
Using the formula, VP / VS = Np / Ns, with Vp =
240 V, Np = 1000 turns and Ns = 50 turns we have,
240
50
--------- = 1000
----------- ⇒ Vs = 240 × ----------- = 12
Vs
50
1000
hat is,
Vs = 12 V
Total resistance = 0.2 Ω + 4 Ω = 4.2 Ω.
From the formula, I = V / R , we have
I=
(c)
12 V
= 2.86 A
4.2 Ω
From the formula, P = VI = (IR) × I = I2R, we have
that
2
P = 2.86 × 4.0 = 32.7 W
(d)
Using, P = VI, we have,
12.3.1 pOWER LOssEs IN TRANsMIssION
LINEs AND REAL TRANsFORMERs
here are a number of reasons for power losses in
transmission lines such as:
•
•
•
•
Heating efect of a current
Resistance of the metal used
Dielectric losses
Self-inductance
he main heat loss is due to the heating efect of a current.
By keeping the current as low as possible, the heating efect
can be reduced. he resistance in a wire due to the low of
electrons over long distances also has a heating efect. If
the thickness of the copper wire used in the core of the
transmission line is increased, then the resistance can be
decreased. However, there are practical considerations
such as weight and the mechanical and tensional strength
327
AHL
12.3.1 Outline the reasons for power losses in
transmission lines and real transformers.
(a)
(b)
(b)
Eiciency =
12.3 TRANsMIssION
OF ELECTRIC
pOWER
Calculate
(a)
= 36 W
32.7 + 1.6 × 100 % = 95%
= ----------------------36
TRANsFORMERs
Example
–3
CHAPTER 12
that have to be taken into account. he copper wire is
usually braided (lots of copper wires wound together)
and these individual wires are insulated. he insulation
material has a dielectric value which can cause some power
loss. Finally, the changing electric and magnetic ields of
the electrons can encircle other electrons and retard their
movement on the outer surface of the wire through selfinductance. his is known as the ‘skin efect’. he size of the
power loss depends on the magnitude of the transmission
voltage, and power losses of the order of magnitude of
105 watts per kilometre are common.
Power losses in real transformers are due to factors such as:
AHL
•
•
•
•
•
•
•
Eddy currents
Resistance of the wire used for the windings
Hysteresis
Flux leakage
Physical vibration and noise of the core and
windings
Electromagnetic radiation
Dielectric loss in materials used to insulate the core
and windings.
Hysteresis is derived from the Greek word that means
“lagging behind” and it becomes an important factor in
the changes in lux density as a magnetic ield changes
in ferromagnetic materials. Transformer coils are subject
to many changes in lux density. As the magnetic ield
strength increases in the positive direction, the lux
density increases. If the ield strength is reduced to zero,
the iron remains strongly magnetised due to the retained
lux density. When the magnetic ield is reversed the lux
density is reduced to zero. So in one cycle the magnetisation
lags behind the magnetising ield and we have another
iron loss that produces heat. Hysteresis is reduced again by
using silicon iron cores.
he capacity for the primary coil to carry current is limited
by the insulation and air gaps between the turnings of the
copper wire and this leads to lux leakage. his can be
up to 50% of the total space in some cases. Because the
power is being delivered to the transformer at 50Hz, you
can oten hear them making a humming noise. Minimal
energy is lost in the physical vibration and noise of the
core and windings.
Modern transformers are up to 99% eicient.
As already mentioned, any conductor that moves in a
magnetic ield has emf induced in it, and as such current,
called eddy currents, will also be induced in the conductor.
his current has a heating efect in the sot iron core of the
transformer which causes a power loss termed an iron loss.
here is also a magnetic efect in that the created magnetic
ields will oppose the lux change that produces them
according to Lenz’s Law. his means that eddy currents
will move in the opposite direction to the induced current
causing a braking efect. Eddy currents are considerably
reduced by alloying the iron with 3% silicon that increases
the resistivity of the core. To reduce the heating efect due
to eddy currents, the sot-iron core is made of sheets of iron
called laminations that are insulated from each other by
an oxide layer on each lamination. his insulationprevents
currents from moving from one lamination to the next.
Copper wire is used as the windings on the sot-iron
core because of its low resistivity and good electrical
conductivity. Real transformers used for power
transmission reach temperatures well above room
temperature and are cooled down by transformer oil.
his oil circulates through the transformer and serves
not only as a cooling luid but also as a cleaning and anticorrosive agent. However, power is lost due resistance and
temperature commonly referred to as ‘copper loss’.
328
Long-distance AC alternating current transmission
is afected by a transmission line’s reactive power that
is actually 90 degrees out of phase with the low of real
current to a load at the other end of the line. For short
transmission lines, the efect is not as signiicant. Direct
current transmission does not have reactive power once
the voltage has been raised to the normal level. he power
losses are considerably less than alternating current.
TOk The dangers of EM radiation
Students should be aware that current experimental
evidence suggests that low-frequency fields do not
harm genetic material. Students should appreciate
that the risks attached to the inducing of current in
the body are not fully understood. These risks are
likely to be dependent on current (density), frequency
and length of exposure.
The use of risk assessment in making scientific
decisions can be discussed here. The issues of
correlation and cause, and the limitations of data, are
also relevant here.
© IBO 2007
ELECTROMAGNETIC INDUCTION
For economic reasons, there is no ideal value of voltage
for electrical transmission. Electric power is generated
at approximately 11 000 V and then it is stepped-up to
the highest possible voltage for transmission. Alternating
current transmission of up to 765 kV are quite common.
For voltages higher than this, direct current transmission
at up to 880 kV is used. A.c. can be converted to d.c.
using rectiiers and this is what is done in electric train
and tram operations. D.c. can be converted to a.c. using
inverters. here a number of d.c. transmision lines such
as the underground cross-channel link between the UK
and France. he New Zealand high-voltage direct current
scheme has around 610 km of overhead and submarine
transmission lines.
here are 3 conductors on a transmission line to maximize
the amount of power that can be generated. Each high
voltage circuit has three phases. he generators at the
power station supplying the power system have their coils
connected through terminals at 120° to each other.
12.3.3 sOLVING pRObLEMs ON pOWER
TRANsMIssION
Example
An average of 120 kW of power is delivered to a suburb
from a power plant 10 km away. he transmission lines
have a total resistance of 0.40 Ω. Calculate the power loss
if the transmission voltage is
(a)
(b)
240 V
24 000 V
Solution
(a)
AHL
12.3.2 pOWER TRANsMIssION
From the formula, P = VI, we have that
5
1.2 × 10 W
P
I = --- = ----------------------------- = 500 A
V
240 V
When each generator at the power station rotates through
a full rotation, the voltages and the currents rise and fall in
each terminal in a synchronized manner.
So that
Once the voltage has been stepped-up, it is transmitted
into a national supergrid system from a range of power
stations. As it nears a city or town it is stepped-down into a
smaller grid. As it approaches heavy industry, it is stepped
down to around 33 – 132 kV in the UK, and when it arrives
at light industry it is stepped-down to 11-33 kV. Finally,
cities and farms use a range of values down to 240V from
a range of power stations.
When the current lows in the cables, some energy is lost
to the surroundings as heat. Even good conductors such
as copper still have a substantial resistance because of the
signiicant length of wire needed for the distribution of
power via the transmission cables. To minimise energy
losses the current must be kept low.
2
2
P = I × R = ( 500 A ) × ( 0.40 Ω ) = 100 kW
(b) Again, using P = VI, we have,
5
P
1.2 × 10 W
I = --- = ------------------------------ = 5.0 A
4
V
2.4 × 10 V
So that,
2
2
P = I × R = ( 5.0 A ) × ( 0.40 Ω ) = 10 W
329
CHAPTER 12
12.3.4 ExTRA-LOW-FREqUENCy
2.
An alternating current with a root-mean-square
value of 2 A is compared with the direct current I
lowing through a given resistor. If both currents
generate heat at the same rate, the value of I would be
ELECTROMAGNETIC FIELDs
AHL
We have all seen though the media patients being given
shock treatment through 2 electrodes to try and get the
heart beat at its natural frequency. he human body is a
conducting medium so any alternating magnetic ield
produced at the extra-low frequency will induce an electric
ield which in turn produces a very small induced current
in the body. Using a model calculation in a human of body
radius 0.2 m and a conductivity of 0.2 Sm-1 (sieverts per
metre), it has been shown that a magnetic ield of 160 μT
can induce a body surface current density of 1 mA m-2.
It is currently recommended that current densities to the
head, neck and body trunk should not be greater than
10 mA m-2.
12.3.5 pOssIbLE RIsks OF HIGH-
A.
B.
C.
D.
3.
An ideal transformer has a primary coil of 5000
turns and a secondary coil of 250 turns. he
primary voltage produced is 240 V. If a 24 W lamp
connected to the secondary coil operates at this
power rating, the current in the primary coil is:
A.
B.
C.
D.
4.
Exercise
0.05 A
0.1 A
12 A
20 A
If an alternating e.m.f has a peak value of
12 V then the root-mean-square value of this
alternating e.m.f would be:
VOLTAGE pOWER LINEs
Current experimental evidence suggests that low-frequency
ields do not harm genetic material in adults but there is
some evidence that there could be a link to infant cancer
rates due to low-frequency ields. he risks attached to the
inducing of current in the body are not fully understood.
But these risks are likely to be dependent on current
(density), frequency and length of exposure. IBO 2007
4A
2√2A
2A
√2A
A.
B.
C.
D.
5.
0V
√6 V
3.5 V
6√2 V
he igure below below shows the variation with
time t of the emf ε generated in a coil rotating in a
uniform magnetic ield.
12.2
ε0
ε/V
1.
Which of the following could correctly describe a
step-up transformer?
A.
B.
C.
D.
330
power
supply
dc
dc
ac
ac
Core
Steel
Iron
Steel
Iron
primary
coil
10 turns
100 turns
10 turns
100 turns
secondary
coil
100 turns
10 turns
100 turns
10 turns
0
T/ 2
T
- ε0
What is the root-mean-square value ε rms of the emf
and also the frequency f of rotation of the coil?
ε rms
f
A.
ε0
B.
e0
C.
e0 / √2
D.
e0 / √2
2
T
1
T
2
T
1
T
ELECTROMAGNETIC INDUCTION
A load resistor is connected in series with an
alternating current supply of negligible internal
resistance. If the peak value of the supply voltage is
Vo and the peak value of the current in the resistor
is I0, the average power dissipation in the resistor
would be:
A.
his question deals with the production and
transmission of electric power, electricity costs
and eiciency, and fuse systems.
(a)
V0 I 0
B.
2
V0 I 0
C.
2
2 V0 I 0 .
D.
V0 I 0 .
7.
Explain why a sot-iron core is used in the
construction of a transformer.
8.
Explain why a transformer will not work with
direct current.
9.
If there are 1200 turns in the primary coil of an
ac transformer with a primary voltage of 240 V,
calculate the secondary voltage if the secondary
coil has
(a)
(b)
(c)
13.
(b)
(c)
(d)
300 turns
900 turns
1800 turns
10.
he armature of a 30 Hz a.c. generator contains
120 loops. he area of each loop is 2.0 × 10-2 m2 .
It produces a peak output voltage of 120 V when it
rotates in a magnetic ield. Calculate the strength
of the magnetic ield.
11.
Calculate the r.m.s value of the following currents:
(e)
(f)
2A, 4A, 6A, 3A, -5A, 1A, 6A, 8A, -9A and 10A.
12.
Calculate the peak current in a 2.4 × 103 Ω resistor
connected to a 230 V domestic a.c source.
14.
Is it feasible to transmit power from a
power station over long distances using
direct current rather than alternating
current? Justify your answer.
An aluminum transmission cable has a
resistance of 5.0 Ω when 10 kW of power
is transmitted in the cable. Justify why it is
better to transmit the power at 100 000V
rather than 1000 V by comparing the power
that would be wasted in the transmission at
both of these voltages.
Many step-up and step-down transformers
are used in the electricity transmission from
the power station to the home. In order to
increase the eiciency of the transformers,
eddy currents have to minimised. Describe
how this achieved in the transformer
design.
If the fuse controlling the maximum power
for lighting in your house is rated at 8 A,
calculate the maximum number of 60 W
light bulbs that can be operated in parallel
with a 110 V power supply so as not to blow
the fuse?
A stainless steel calorimeter with a mass of
720 g was used to heat 2.5 kg of water. If
the current / voltage in a heating element
supplying the power was 30.2 A / 110V,
and it took 2.5 minutes to heat the water
from 25 °C to 98 °C, determine the speciic
heat capacity of the steel. (Assume no heat
loss to the surroundings. he speciic heat
capacity of water is 4.18 × 103 J kg-1K-1).
How much does it cost to run the following
appliances at the same time for one hour if
electricity costs 10.5 cent per kilowatt-hour:
a 6 kW oven, two 300 W colour televisions
and ive 100 W light globes?
Name four factors that afect the magnitude of an
induced emf in a generator.
331
AHL
6.
CHAPTER 12
AHL
15.
332
If there are 4 laminations in a transformer core,
what fraction of the lux is in each lamination and
what fraction the power dissipated in each?