Transcript VI-2
IV–2 Inductance
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Main Topics
•
•
•
•
Transporting Energy.
Counter Torque, EMF and Eddy Currents.
Self Inductance
Mutual Inductance
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Transporting Energy I
• The electromagnetic induction is a basis of
generating and transporting electric energy.
• The trick is that power is delivered at power
stations, transported by means of electric
energy (which is relatively easy) and used
elsewhere, perhaps in a very distant place.
• To show the principle lets revisit our rod.
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Moving Conductive Rod VIII
• If the rails are not connected (or there are no
rails), no work in done on the rod after the
equilibrium voltage is reached since there
is no current.
• If we don’t move the rod but there is a
current I flowing through it, there will be a
force pointing to the left acting on it. We
have already shown that F = BIl.
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Moving Conductive Rod IX
• If we move the rod and connect the rails by
a resistor R, there will be current I = /R
from Ohm’s law. Since the principle of
superposition is valid, there will also be the
force due to the current and we have to
deliver power to move the rod against this
force: P = Fv = BIlv = I, which is exactly
the power dissipated on the resistor R.
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Counter Torque I
• We can expect that the same what is valid
for a rod which makes a translation
movement in a magnetic field is also true
for rotation movement.
• We can show this on rotating conductive
rod. We have to exchange the translation
qualities for the rotation ones:
P = Fv = T
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Counter Torque II
• First let us show that if we run current I
through a rod of the length l which can
rotate around one of its ends in uniform
magnetic field B, torque appears.
• There is clearly a force on every dr of the
rod. But to calculate the torque also r the
distance from the rotation center must be
taken into account, so we must integrate.
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Counter Torque III
• If we rotate the rod and connect a circular
rail with the center by a resistor R, there
will be current I = /R. Due to the principle
of superposition, there will be the torque
due to the current and we have to deliver
power to rotate the rod against this torque: P
= T = BIl2/2 = I, which is again
exactly the power dissipated on the resistor.
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Counter EMF I
• From the previous we know that the same
conclusions are valid for linear as well as
for rotating movement. So we return to our
rod, linearly moving on rails for simplicity.
• Let us connect some input voltage to the
rails. There will be current given by this
voltage and resistance in the circuit and
there will be some force due to it.
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Counter EMF II
• After the rod moves also EMF appears in
the circuit. It depends on the speed and it
has opposite polarity that the input voltage
since the current due to this EMF must,
according to the Lenz’s law, oppose the
initial current. We call this counter EMF.
• The result current is superposition of the
original current and that due to this EMF.
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Counter EMF III
• Before the rod (or any other electro motor)
moves the current is the greatest I0 = V/R.
• When the rod moves the current is given
from the Kirchhof’s law by the difference of
the voltages in the circuit and resistance:
I = (V - )/R = (V – vBl)/R
• The current apparently depends on the
speed of the rod.
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Counter EMF IV
• If the rod was without any load, if would
accelerate until the induced EMF equals to the
input voltage. At this point the current disappears
and so does the force on the rod so there is no
further acceleration.
• So the final speed v depends on the applied voltage V.
• Now, we also understand that an over-loaded motor,
when it slows too much or stops, can burn-out due to
large current. Motors are constructed to work at some
speed and withstand a certain current Iw < I0.
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Eddy Currents I
• So far we dealt with one-dimensional rods
totally immersed in the uniform magnetic
field.
• But if the conductor must be considered as
two or three dimensional and/or it is not
completely immersed in the field or the
field is non- uniform a new effect, called
eddy currents appears.
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Eddy Currents II
• The change is that now the induced currents
can flow within the conductor. They cause a
forces opposing the movement so the
movement is attenuated or power has to be
delivered to maintain it.
• Eddy currents can be used for some
purposes e.g. smooth braking of hi-tech
trains or other movements.
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Eddy Currents III
• But eddy currents produce heat so they are
source of power loses and in most cases
they have to be eliminated as much as
possible by special construction of
electromotor frames or transformer cores
e.g. laminating.
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The Self Inductance I
• We have shown that if we connect some
input voltage to a free conductive rod
immersed in external magnetic field an
EMF appears which has the opposite
polarity then the input voltage.
• But even a circuit of conducting wire
without any external field will behave
qualitatively the very same way.
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The Self Inductance II
• If some current already flows through such a wire,
the wire is actually immersed in the magnetic field
produced by its own current.
• If we now try to change the current we are
changing this magnetic field and thereby the
magnetic flux and so an EMF is induced in a
direction opposing the change.
• If we make N loops in our circuit, the effect is
increased N times!
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The Self Inductance III
• We can expect that the induced EMF in this
general case depends on the:
• geometry of the wire and material properties of
the surrounding space
• rate of the change of the current
• It is convenient to separate these effects and
concentrate the former into one parameter
called the (self) inductance L.
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The Self Inductance IV
dI
L
dt
• Then we can simply write:
• We are in a similar situation as we were in
electrostatics. We used capacitors to set up known
electric field in a given region of space. Now we
use coils or inductors to set up known magnetic
field in a specified region.
• As a prototype coil we usually use a solenoid (part
near its center) or a toroid.
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The Self Inductance V
• Let’s have a long solenoid with N loops.
• If some current I is flowing through it there will be
the same flux m1 passing through each loop.
• If there is a change in the flux, there will be EMF
induced in each loop and since the loops are in
series the total EMF induced in the solenoid will
be N times the EMF induced in each loop.
• We use Faraday’s law slightly modified for this
situation and previous definition of inductance.
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The Self Inductance VI
d m1
dI
N
L
dt
dt
• If N and L are constant we can integrate and get
the inductance:
N m1
N m1 LI L
I
• The unit for magnetic flux is 1 weber
1 Wb = 1 Tm2
• The unit for the inductance is 1 henry
1H = Vs/A = Tm2/A = Wb/A
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The Self Inductance VII
• The flux through the loops of a solenoid depends
on the current and the field produced by it and the
geometry. In the case of a solenoid of the length l
and cross section A and core material with r:
N m1 NA
0 r NI
L
AN
2
l
l
• In electronics compomemts having inductance
inductors are needed and are produced.
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The Mutual Inductance I
• In a similar way we can describe mutual
influence of two inductances more
accurately total flux in one as a function of
current in the other.
• Let us have two coils Ni, Ii on a common
core or close to each other.
• Let 21 be the flux in each loop of coil 2
due to the current in the coil 1.
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The Mutual Inductance II
• Then we define the mutual inductance M21
as total flux in all loops in the coil 2 per the
unit of current (1 ampere) in the coil 1:
M21 = N221/I1 I1M21 = N221
• EMF in the coil 2 from the Faraday’s law:
2 = - N2d21/dt = - M21 dI1/dt
• M21 depends on geometry of both coils.
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The Mutual Inductance III
• It can be shown that the mutual inductance
of both coils is the same M21 = M12 .
• The fact that current in one loop induces
EMF in other loop or loops has practical
applications. It is e.g. used to power supply
pacemakers so it is not necessary to lead
wires through human tissue. But the most
important use is in transformers.
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Homework
• No homework today!
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Things to read and learn
• This lecture covers:
Chapter 29 – 5, 6; 30 – 1, 2
• Advance reading:
Chapter 26-4; 29 – 6; 30 – 3, 4, 5, 6
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Rotating Conductive Rod
• Torque on a piece dr which is in a distance r
from the center of rotation of a conductive
rod l with a current I in magnetic field B is:
dT rdF BIrdr
• The total torque is:
l
BIl 2
T BIrdr
2
0
^