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31. 7. 20031 IV–2 Inductance

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31. 7. 20032 Main Topics Transporting Energy. Counter Torque, EMF and Eddy Currents. Self Inductance Mutual Inductance

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31. 7. 20033 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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31. 7. 20034 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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31. 7. 20035 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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31. 7. 20036 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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31. 7. 20037 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.integrate

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31. 7. 20038 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 = BIl 2 /2 = I, which is again exactly the power dissipated on the resistor.

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31. 7. 20039 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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31. 7. 200310 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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31. 7. 200311 Counter EMF III Before the rod (or any other electro motor) moves the current is the greatest I 0 = 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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31. 7. 200312 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 I w < I 0.

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31. 7. 200313 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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31. 7. 200314 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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31. 7. 200315 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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31. 7. 200316 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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31. 7. 200317 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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31. 7. 200318 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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31. 7. 200319 The Self Inductance IV 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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31. 7. 200320 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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31. 7. 200321 The Self Inductance VI If N and L are constant we can integrate and get the inductance: The unit for magnetic flux is 1 weber 1 Wb = 1 Tm 2 The unit for the inductance is 1 henry 1H = Vs/A = Tm 2 /A = Wb/A

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31. 7. 200322 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 : In electronics compomemts having inductance inductors are needed and are produced.

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31. 7. 200323 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 N i, I i 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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31. 7. 200324 The Mutual Inductance II Then we define the mutual inductance M 21 as total flux in all loops in the coil 2 per the unit of current (1 ampere) in the coil 1: M 21 = N 2 21 /I 1 I 1 M 21 = N 2 21 EMF in the coil 2 from the Faraday’s law: 2 = - N 2 d 21 /dt = - M 21 dI 1 /dt M 21 depends on geometry of both coils.

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31. 7. 200325 The Mutual Inductance III It can be shown that the mutual inductance of both coils is the same M 21 = M 12. 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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31. 7. 200326 Homework No homework today!

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31. 7. 200327 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: ^ The total torque is:

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