Transcript Document

Things to know
(a)deduce from Faraday’s 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
1
(where needed)
• (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
2
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
3
A stationary
magnet is near
the coil
Hollow paper
or plastic tube
N
S
-1 0 1
-2
2
Sensitive
Galvanometer
In this experiment, no battery is connected to the
coil. Hence no e.m.f. is found in the coil.
4
motion
Hollow paper
or plastic tube
N
-1 0 1
-2
2
G
S
Induced
I
When the magnet is moving towards the coil, an
electric current is induced simultaneously.
5
Faster motion
Hollow paper
or plastic tube
N
-1 0 1
-2
2
G
S
Induced
I more
When the magnet is moving faster, the induced
current is more.
6
Not moving
Hollow paper
or plastic tube
N
-1 0 1
-2
2
S
No
current
G
When the magnet is not moving, no current is
induced even though the magnetic flux is linked
with the coil closely.
7
Faraday’s 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.
8
Self Test
Question
Not moving
N
S
-1 0 1
-2
2
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.
9
What law did you apply when you answer
the question in the previous slide?
10
Self Test Question
Moving constantly
E
D
C
B
N
AS
-1 0 1
-2
2
G
Deflection of G or
emf induced
E
D
C
B
A
11
Self Test Question: Sketch the graph as the magnet moves from A to E
stant speed moving
N
E
D
S
C
B
A
B
A
-1 0 1
-2
2
G
Deflection of G or
emf induced
Playing back of the
graph
E
`
D
C
12
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.
13
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?
14
Lenz’s 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!
15
Beware of a different way the coil
can be wound:
The paper tube can be
taken away to test you
16
Can you spot the difference of winding?
Note: The dotted parts are at
the back. The solid lines are at
the front.
17
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
18
Please mark + or – signs at the points X and Y to show the
induced e.m.f. :
Motion
N
S
X
+
Y
19
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.
20
Coil is stationary
N
S
G
21
Coil is in motion,
approaching the
magnet
Induced
current
Induced
current
N
N
S
G
22
Coil is in motion,
approaching the magnet
Induced
current
N
N
S
23
G
Coil is approaching the
magnet
N
Existing
flux
Induced
current
N
S
Can you see the Right
Hand Rule ?
G
This is also called the
Dynamo Rule
24
Coil is stationary again
N
Existing
flux
S
No more induced
current here
25
G
Fleming’s 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 Lenz’s Law
So, sometimes you use the right hand rule instead of
Lenz’s Law.
26
Straight
wire
The straight
moving
vertically
wire stops
tomoving.
magnetic field.
Current
is is
No current
induced
inducedaway
at all.
from you
This straight wire is
This
straight
wire
moving
along
theis
not moving
magnetic
field
No current is
induced
Wire
moving
Wire
stops
moving
vertically to the flux
Current
is induced
No current
is
towardsatyou
induced
all
27
28
29
Weak induced
current
Strong induced
current
wire cutting
flux vertically
wire cutting flux
obliquely
No current is induced in
the wire
wire moving alongside flux
No current is induced in
wire
wire moving alongside
flux
30
N
Magnetic flux
S
31
N
S
32
B
C
Motion
N
S
Motion
A
D
33
B
Motion
N
Magnetic
Flux
C
S
A
Motion
D
34
B
motion
N
S
A
C
motion
No induced current
D
35
B
N
S
A
C
D
36
Motion
C
B
N
S
D
A
37
C
N
Motion
Magnetic
Flux
B
S
D
A
38
C
motion
Magnetic flux
N
S
D
B
No induced current
A
39
C
N
S
D
B
A
40
B
C
Motion
N
S
Motion
A
D
41
B
C
N
S
A
D
Red arrows
represents magnetic
flux
Induced
current
Time
A D
42
B
N
C
A
S
Red arrows
represents magnetic
flux
D
Induced
current
Time
43
B
motion
N
A
motion C
S
D
Induced
current
Time
44
motion C
N
D
B
Flux
S
A
Induced
current
Time
45
D A
C
motion
N
D
B
S
A
Induced
current
Time
46
B
C
N
S
A
D
Induced
current
Time
A D
47
Induced current / emf
Time
48
Induced emf
2.5V
Time
-2.5V
49
How would you demonstrate electromagnetic
induction here?
G
The iron core
50
Input
emf
Np
turns
Ns
turns
output
emf
AC Source
The iron core
Insulated copper wire
Insulated copper wire
Primary windings
Secondary windings
51
Primary
emf
Ns
turns
Np
turns
Secondary
emf
AC Source
The iron core
emf secondary
emf primary
=
Ns
Np
52
Primary
emf
Np
turns
Ns
turns
Secondary
emf
AC Source
emf
Time
53
emf
Primary Alternating
emf Peak 12V, 50 Hz
Secondary
Alternating emf Peak
18V 50 Hz
Time
54
Primary
emf
Ns
turns
Np
turns
Secondary
emf
AC Source
The iron core
emf secondary
emf primary
=
Ns
Np
Step-up Transformer
1
55
Primary
emf
Ns
turns
Np
turns
Secondary
emf
AC Source
The iron core
emf secondary
emf primary
=
Ns
Np
Step-down Transformer
1
56
Input emf
150V
AC Source
Np 200
turns
Calculate (i) the output emf
(iii) the primary current
Ns 600
turns
output
emf
60
(ii) the induced current
57
Np 200
turns
Input emf
150V
AC Source
Ns 600
turns
output
emf
Vs
600
60
Calculate (i) the output emf
Vs
Vp
=
Ns
Np
Vs = 450 V peak
150
=
200
58
Np 200
turns
Input emf
150V
AC Source
Calculate (i) the output emf
output
emf
Ns 600
turns
60
Vs = 450 V peak
(ii) the induced current
I=
V
R
I=
450
60
= 7.5 A Peak
59
emf , current
450V
Secondary
Alternating emf Peak
450 V,
freq 50
Hz
7.5A
Time
60
Np 200
turns
Input emf
150V
AC Source
Calculate (i) the output emf
(iii) the primary current
Power= VI,
Ns 600
turns
Vs = 450 V peak
output
emf
60
(ii) Is = 7.5 A
Output power = Input power
(assuming that the transformer is 100% effecient)
Vsx Is = Vpx Ip
,
450x 7.5 = 150x Ip
61
Np 200
turns
Input emf
150V
AC Source
Calculate (i) the output emf
Ns 600
turns
output
emf
60
Vs = 450 V peak
(iii) the primary current
Is
Ip
Np
=
Ns
provided that the transformer
has an efficiency of 100%
62
emf , current
22.5A
450V
Secondary
Alternating emf Peak
450V,
freq 50
Hz
7.5A
150V
Time
63
AC Source
Primary
emf
Np
turns
The iron core
Ns
turns
Secondary
emf
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.
64
AC Source
Primary
emf
Np
turns
Ns
turns
Secondary
emf
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.
65
Power Loss in Cables
66
Heating
Heatingpower
powerininthe
thetransmitting
transmittingcables
cables== I2? R
P=VI
2,000000 = 10000 x I
I = 200 A
Heating power in cables = 2002 x 40 =
1,600000 W = 1.6 MW A huge loss
2MW
10KV
P.S.
0V
Power transmission
R = 20
R = 20 
0.4MW
L
N
67
Heating
Heatingpower
powerininthe
thetransmitting
transmittingcables
cables== I2? R
P=VI
2,000000 = 20000 x I
I = 100 A
Heating power in cables = 1002 x 40 =
400,000 W = 0.4 MW A smaller loss
2MW
20KV
P.S.
0V
Power transmission
R = 20
R = 20 
1.6MW
L
N
68