Transcript Electro-Magnetic Induction PowerPoint

Electro-Magnetic Induction
© David Hoult 2009
Magnetic flux
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© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
If the magnitude of the flux density is B then the
magnitude of the magnetic flux (f) linked with the
area A is defined to be
© David Hoult 2009
If the magnitude of the flux density is B then the
magnitude of the magnetic flux (f) linked with the
area A is defined to be
f = AB
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f = AB
units of flux
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f = AB
units of flux T m2
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f = AB
units of flux T m2
1 T m2 = 1 Weber (Wb)
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© David Hoult 2009
Now, the magnitude of the component of the flux
density perpendicular to the area is B cos q
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Now, the magnitude of the component of the flux
density perpendicular to the area is B cos q
so the magnetic flux (f) linked with the area is now
f = A B cos q
alternatively
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In practice, the area in question is usually
surrounded by a conductor, often a coil of wire.
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In practice, the area in question is usually
surrounded by a conductor, often a coil of wire.
If the coil of wire has N turns, we define the flux
linkage as follows
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In practice, the area in question is usually
surrounded by a conductor, often a coil of wire.
If the coil of wire has N turns, we define the flux
linkage as follows
Flux linkage = N f
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EMF induced in a conductor moving through a
uniform magnetic field
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The wire moves distance Ds in time Dt.
In this time, a charge Dq moves past any point in
the wire.
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The wire moves distance Ds in time Dt.
In this time, a charge Dq moves past any point in
the wire.
work done = F Ds
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The wire moves distance Ds in time Dt.
In this time, a charge Dq moves past any point in
the wire.
work done = F Ds
F Ds
work done per unit charge =
Dq
© David Hoult 2009
The wire moves distance Ds in time Dt.
In this time, a charge Dq moves past any point in
the wire.
work done = F Ds
F Ds
work done per unit charge =
Dq
work done per unit charge is the induced emf
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If the wire moves at constant speed, the force F
must be
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If the wire moves at constant speed, the force F
must be equal but opposite to the force acting on
it due to the current I, induced in it
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If the wire moves at constant speed, the force F
must be equal but opposite to the force acting on
it due to the current I, induced in it
F = -ILB
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If the wire moves at constant speed, the force F
must be equal but opposite to the force acting on
it due to the current I, induced in it
F = -ILB
F Ds
E=
Dq
© David Hoult 2009
If the wire moves at constant speed, the force F
must be equal but opposite to the force acting on
it due to the current I, induced in it
F = -ILB
F Ds
E=
Dq
-ILB Ds
E=
Dq
© David Hoult 2009
I=
© David Hoult 2009
I=
Dq
Dt
© David Hoult 2009
I=
Dq
Dt
- (Dq/Dt) L B Ds
E=
Dq
and
© David Hoult 2009
I=
Dq
Dt
- (Dq/Dt) L B Ds
E=
Dq
and
Ds
Dt
=
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I=
Dq
Dt
- (Dq/Dt) L B Ds
E=
Dq
and
Ds
Dt
= v
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I=
Dq
Dt
- (Dq/Dt) L B Ds
E=
Dq
and
Ds
Dt
= v
E = -BLv
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The Laws of Electro-magnetic Induction
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Lenz’s Law
When e.m.i. occurs, any induced current will flow
in such a direction as to
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Lenz’s Law
When e.m.i. occurs, any induced current will flow
in such a direction as to oppose the change
producing it
© David Hoult 2009
Lenz’s Law
When e.m.i. occurs, any induced current will flow
in such a direction as to oppose the change
producing it
It should be clear that Lenz’s law is an “electromagnetic version” of
© David Hoult 2009
Lenz’s Law
When e.m.i. occurs, any induced current will flow
in such a direction as to oppose the change
producing it
It should be clear that Lenz’s law is an “electromagnetic version” of the law of conservation of
energy
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Lenz’s Law
When e.m.i. occurs, any induced current will flow
in such a direction as to oppose the change
producing it
Faraday’s Law
The induced emf is directly proportional to the
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Lenz’s Law
When e.m.i. occurs, any induced current will flow
in such a direction as to oppose the change
producing it
Faraday’s Law
The induced emf is directly proportional to the rate
of change of flux linking the conductor
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The sense of the induced current can be predicted
using Fleming’s RIGHT hand rule
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The sense of the induced current can be predicted
using Fleming’s RIGHT hand rule
which is pretty much like Fleming’s left hand rule
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The sense of the induced current can be predicted
using Fleming’s RIGHT hand rule
which is pretty much like Fleming’s left hand rule
except, guess what...
© David Hoult 2009
The sense of the induced current can be predicted
using Fleming’s RIGHT hand rule
which is pretty much like Fleming’s left hand rule
except, guess what... using the right hand instead
of the left hand !
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ThuMb
Motion
First finger
Field
SeCond finger
Current
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Both the laws of e.m.i. can be combined in a single
mathematical statement
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Both the laws of e.m.i. can be combined in a single
mathematical statement
Faraday
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Both the laws of e.m.i. can be combined in a single
mathematical statement
Faraday
E a
Df
Dt
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Both the laws of e.m.i. can be combined in a single
mathematical statement
Faraday
E a
Df
Dt
Faraday + Lenz
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Both the laws of e.m.i. can be combined in a single
mathematical statement
Faraday
E a
Df
Dt
Faraday + Lenz
E a -
Df
Dt
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Using the S.I. system of units the constant of
proportionality is 1 so, if (as is often the case) the
conductor is a coil of wire having N turns, we have
© David Hoult 2009
Using the S.I. system of units the constant of
proportionality is 1 so, if (as is often the case) the
conductor is a coil of wire having N turns, we have
E = -N
Df
Dt
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Using the S.I. system of units the constant of
proportionality is 1 so, if (as is often the case) the
conductor is a coil of wire having N turns, we have
E = -N
Df
Dt
This equation is often referred to as Neumann’s
Law
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For a wire moving at 90° to a field with speed v, the
induced emf is given by
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For a wire moving at 90° to a field with speed v, the
induced emf is given by E = - B L v
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E = -BLv
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E = -BLv = -BL
Ds
Dt
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E = -BLv = -BL
Ds
Dt
= -B
DA
Dt
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E = -BLv = -BL
Ds
Dt
= -B
DA
Dt
=-
Df
Dt
© David Hoult 2009
E = -BLv = -BL
Ds
Dt
= -B
DA
Dt
=-
Df
Dt
This suggests that change of flux linking a
conductor and flux cutting by a conductor are
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equivalent actions
Simple a.c. generator
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Simple a.c. generator
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Simple a.c. generator
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Simple a.c. generator
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Simple a.c. generator
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Simple a.c. generator
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Simple a.c. generator
Fleming’s right hand rule gives the sense of the
current to be...
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Simple a.c. generator
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Simple a.c. generator
Fleming’s right hand rule gives the sense of the
current to be...
which means that, at the instant shown in the
diagram, terminal Q is the positive terminal of the
generator
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Simple a.c. generator
Fleming’s right hand rule gives the sense of the
current to be
which means that, at the instant shown in the
diagram, terminal Q is the positive terminal of the
generator
When the coil has turned through 180°, P will be
the positive terminal so the generator gives us
alternating current
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The Transformer
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weak induced
alternating emf
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iron core
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The Ideal Transformer
If a transformer is described as ideal we mean it is
100% efficient
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The Ideal Transformer
If a transformer is described as ideal we mean it is
100% efficient
In other words, power output (at secondary coil)
equals power input (to primary coil)
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The Ideal Transformer
Therefore, in an ideal transformer:
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The Ideal Transformer
Therefore, in an ideal transformer:
i) the coils have zero resistance
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The Ideal Transformer
Therefore, in an ideal transformer:
i) the coils have zero resistance
ii) all the magnetic flux f produced by the primary
current Ip is linked with the secondary coil
© David Hoult 2009
The Ideal Transformer
Therefore, in an ideal transformer:
i) the coils have zero resistance
ii) all the magnetic flux f produced by the primary
current Ip is linked with the secondary coil
iii) no current flows in the iron core
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The Ideal Transformer
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The Ideal Transformer
When Ip changes, f changes.
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The Ideal Transformer
When Ip changes, f changes.
When f changes, an emf is induced in both coils.
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At the primary coil, the magnitude of the induced
emf is given by
© David Hoult 2009
At the primary coil, the magnitude of the induced
emf is given by
Df
Ep = - Np
Dt
© David Hoult 2009
At the primary coil, the magnitude of the induced
emf is given by
Df
Ep = - Np
Dt
as we are assuming zero resistance coils, the
supply voltage must also have this magnitude in
order to maintain the flow of current
© David Hoult 2009
At the primary coil, the induced emf is given by
Ep = - Np
Df
Dt
as we are assuming zero resistance coils, the
supply voltage must also have this magnitude in
order to maintain the flow of current
At the secondary coil, the magnitude of the
induced emf is given by
Df
Es = - Ns
Dt
© David Hoult 2009
Ep
Es
=
Np
Ns
© David Hoult 2009
Ep
Es
=
Np
Ns
here we are assuming that f is the same for both
coils
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The power input to the primary coil is given by
power in =
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The power input to the primary coil is given by
power in = Ep Ip
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The power input to the primary coil is given by
power in = Ep Ip
The power output from the secondary coil is given
by
power out =
© David Hoult 2009
The power input to the primary coil is given by
power in = Ep Ip
The power output from the secondary coil is given
by
power out = Es Is
© David Hoult 2009
The power input to the primary coil is given by
power in = Ep Ip
The power output from the secondary coil is given
by
power out = Es Is
as we are considering an ideal transformer
© David Hoult 2009
The power input to the primary coil is given by
power in = Ep Ip
The power output from the secondary coil is given
by
power out = Es Is
as we are considering an ideal transformer
Ep Ip = Es Is
© David Hoult 2009
Ep Ip = Es Is
Therefore, with an ideal transformer, when the
secondary coil is open circuit (not connected to
anything), there is no net energy taken from the
supply.
© David Hoult 2009
Ep Ip = Es Is
Therefore, with an ideal transformer, when the
secondary coil is open circuit (not connected to
anything), there is no net energy taken from the
supply.
Energy is stored in the magnetic field during the
time the current is increasing but is recovered
from the field when it “collapses”.
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Transmission of Electrical Energy
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If the voltage across the bulb is 6V the current will
be
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If the voltage across the bulb is 6V the current will
be 1A
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If the voltage across the bulb is 6V the current will
be 1A
Power wasted in R = R I2 = R
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© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
If the voltage across the bulb is still 6V the current
flowing through the bulb is still 1A
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If the step down transformer is 100% efficient the
current flowing through R is
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If the step down transformer is 100% efficient the
current flowing through R is 1/8 A
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If the step down transformer is 100% efficient the
current flowing through R is 1/8 A
Power wasted in R = R I2 =
© David Hoult 2009
If the step down transformer is 100% efficient the
current flowing through R is 1/8 A
Power wasted in R =
R I2
R
=
64
© David Hoult 2009
64 times less energy wasted !
© David Hoult 2009