Things to know

1
Things to know

• deduce from Faradays experiments on
• electromagnetic induction or other
  appropriate experiments
• (i) that a changing magnetic field can induce
  an
• e.m.f. in a circuit
• (ii) that the direction of the induced e.m.f.
  
• opposes the change producing it
• (iii) the factors affecting the magnitude of
  the
• induced e.m.f.
• (b) describe a simple form of a.c. generator
  (rotating
• coil or rotating magnet) and the use of slip
  rings (where needed)

2

• (c) sketch a graph of voltage output against time
  
• for a simple a.c. generator
• (d) describe the structure and principle of
• operation of a simple iron-cored
  transformer
• as used for voltage transformations
• (e) recall and apply the equations
• VP / Vs NP / Ns and VPIP VsIs to new
  
• situations or to solve related problems
  (for an
• ideal transformer)
• (f) describe the energy loss in cables and
• deduce the advantages of high voltage
• transmission

3
Electromagnetic Induction
Definition Electromagnetic induction is the
production of electricity using magnetism.

• Need to know Describe an experiment which shows
  that a changing magnetic field can induce an
  e.m.f. in a circuit

4
A stationary magnet is near the coil
N S
SensitiveGalvanometer
In this experiment, no battery is connected to
the coil. Hence no e.m.f. is found in the coil.
5
motion
N S
Induced I
G
When the magnet is moving towards the coil, an
electric current is induced simultaneously.
6
Faster motion
N S
When the magnet is moving faster, the induced
current is more.
7
Not moving
N S
When the magnet is not moving, no current is
induced even though the magnetic flux is linked
with the coil closely.
8
Faradays Law of Electromagnetic Induction
The magnitude (how strong) of the induced emf (or
induced current) is directly proportional to the
rate of change of the magnetic flux linked with
the coil or the rate at which the magnetic flux
and wire are cutting each other.
This means that when the magnetic field is not
moving in relation to the coil, there will be NO
induced emf at all.
9
Self Test Question
Not moving
G
There is plenty of magnetic flux linkage with the
coil, but there is no motion. Is there any
induced current in the coil now? Answer _________
Please draw the needle of the galvanometer.
10
What law did you apply when you answer the
question in the previous slide?
11
Self Test Question
Deflection of G or emf induced
12
Self Test Question Sketch the graph as the
magnet moves from A to E
Deflection of G or emf induced
Playing back of the graph

13
By now, you have learned that the size or
strength of the induced current (or induced
e.m.f.) is determined by the speed of change of
the magnetic flux linkage with the coil.
There is still one more thing about
electromagnetic induction you need to
investigate. Look at the next slide.
14
When a current is induced in a coil, it has to
flow in the certain direction.
What factor determines the direction of the
induced current?
15
Lenzs Law of electromagnetic induction The
direction of the induced current is such that its
own magnetic effect always opposes the change
producing it.
This law is actually related to the Law of
Conservation of Energy. The coil needs to oppose
something in order to obtain energy from it. The
coil itself cannot CREATE energy!
16
Beware of a different way the coil can be wound
The paper tube can be taken away to test you
17
Can you spot the difference of winding?
Note The dotted parts are at the back. The solid
lines are at the front.
18
Please mark one arrow on the left end of the coil
and one arrow through the bulb to show how the
induced current should flow
Motion
N S
Induced CURRENT
19
Please mark or signs at the points X and Y to
show the induced e.m.f.
Motion
N S
X
Y

20
Please mark or signs at the points X and Y to
show the presence of induced e.m.f.
Motion
N S
X
Y

Induced emf
This induced emf is still there as long as the
magnet is moving, even though the circuit is
broken and the induced current cannot flow.
21
Coil is stationary
22
Coil is in motion, approaching the magnet
Induced current
coil motion
N
Induced current
23
Coil is in motion, approaching the magnet
coil motion
N
Induced current
24
Coil is approaching the magnet
coil motion
N
Existing flux
Induced current
Motion of wire
Can you see the Right Hand Rule ?
This is also called the Dynamo Rule
25
Coil is stationary again
Existing flux
No more induced current here
26
Flemings Right Hand Rule is also called the
Dynamo Rule.
thuMb -- the Motion of the wire
First finger -- magnetic Flux (Field)
seCond finger -- induced Current
This is actually the result of Lenzs Law
So, sometimes you use the right hand rule instead
of Lenzs Law.
27
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28
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29
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30
Weak induced current
Strong induced current
wire cutting flux obliquely
wire cutting flux vertically
No current is induced in the wire
No current is induced in wire
wire moving alongside flux
wire moving alongside flux
31
N
Magnetic flux
S
32
N
S
33
B
C
N
Motion
S
Motion
A
D
34
B
Induced current
Motion
N
Magnetic Flux
S
C
A
Motion
D
35
B
motion
No induced current
N
S
No pole is needed
A
C
motion
No induced current
D
36
B
N
motion
S
A
motion
C
D
37
C
B
Motion
N
S
D
A
38
C
Induced current
Motion
N
Magnetic Flux
S
B
D
A
39
C
No induced current
motion
N
Magnetic flux
S
D
B
No induced current
A
40
C
N
motion
S
D
motion
B
A
41
B
C
N
Motion
S
Motion
A
D
42
Induced current
Time
43
Induced current
Time
44
Induced current
Time
A
D
45
motion
Flux
Induced current
Time
A
D
46
Induced current
Time
D
A
47
Induced current
Time
48
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49
Induced emf
2.5V
Time
-2.5V
50
How would you demonstrate electromagnetic
induction here?
G
The iron core
51
Ns turns
output emf
Np turns
Input emf
AC Source
The iron core
Insulated copper wire Secondary windings
Insulated copper wire Primary windings
52
Ns turns
Secondary emf
Np turns
Primary emf
AC Source
The iron core
Ns
emf secondary

Np
emf primary
53
Ns turns
Secondary emf
Np turns
Primary emf
AC Source
emf
Time
54
Secondary Alternating emf Peak 18V 50 Hz
Primary Alternating emf Peak 12V, 50 Hz
emf
Time
55
Ns turns
Secondary emf
Np turns
Primary emf
AC Source
The iron core
Ns
emf secondary
1

Np
emf primary
Step-up Transformer
56
Ns turns
Secondary emf
Np turns
Primary emf
AC Source
The iron core
Ns
emf secondary
1

Np
emf primary
Step-down Transformer
57
60?
output emf
Np 200 turns
Ns 600 turns
Input emf 150V
AC Source
Calculate (i) the output emf (ii) the induced
current
(iii) the primary current
58
60?
output emf
Np 200 turns
Ns 600 turns
Input emf 150V
AC Source
Calculate (i) the output emf
Vs
Ns
Vs
600


Np
Vp
200
150
Vs 450 V peak
59
60?
output emf
Np 200 turns
Ns 600 turns
Input emf 150V
AC Source
Calculate (i) the output emf
Vs 450 V peak
(ii) the induced current
V
I
450
I
R
7.5 A Peak
60
60
Secondary Alternating emf Peak 450 V,
freq 50 Hz
emf , current
450V
7.5A
Time
61
60?
output emf
Np 200 turns
Ns 600 turns
Input emf 150V
AC Source
Calculate (i) the output emf
(ii) Is 7.5 A
Vs 450 V peak
(iii) the primary current
Output power Input power
Power VI,
(assuming that the transformer is 100 effecient)
450x 7.5 150x Ip
Vsx Is Vpx Ip
,
62
60?
output emf
Np 200 turns
Ns 600 turns
Input emf 150V
AC Source
Calculate (i) the output emf
Vs 450 V peak
(iii) the primary current
Np
Is
provided that the transformer has an efficiency
of 100

Ip
Ns
63
Secondary Alternating emf Peak 450V,
freq 50 Hz
emf , current
22.5A
450V
7.5A
150V
Time
64
Ns turns
Secondary emf
Np turns
Primary emf
AC Source
The iron core
eddy current
Power lost in a transformer, Textbook Page 350
Power losses are
(i) due to the electrical resistance in both the
windings
Heat is generated in the wires unnecessarily
(ii) due to the production of the eddy currents
in the iron core.
65
Ns turns
Secondary emf
Np turns
Primary emf
AC Source
The iron core
Such power losses are minimized
(i) by using thicker copper wires in the
windings.
Less heat is generated in the wires.
(ii) by using a laminated iron core.
66
Power Loss in Cables
67
Heating power in the transmitting cables ?
Heating power in the transmitting cables I2 R
P V I
2,000000 10000 x I
I 200 A
Heating power in cables 2002 x 40 1,600000 W
1.6 MW A huge loss
Power transmission
2MW
0.4MW
L
10KV
R 20?
P.S.
N
0V
R 20 ?
68
Heating power in the transmitting cables ?
Heating power in the transmitting cables I2 R
P V I
2,000000 20000 x I
I 100 A
Heating power in cables 1002 x 40 400,000 W
0.4 MW A smaller loss
Power transmission
2MW
1.6MW
L
20KV
R 20?
P.S.
N
0V
R 20 ?