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This is not homework question,just a thought experiment about a general question i have about induction. Let's suppose that we have a closed circuit with only two resistors in series.We also have a changing magnetic flux going through the circuit(we do not care about exact numbers here).
As we know,this produces an emf due to induction(changing magnetic flux causes induced current).
So,we make our calculations and we find that we have an emf=5V(we do not care about its sign because we have already figured out which way the induced current flows with Lenz's law). The question is,that emf is the potential difference between which two points in the circuit? I mean,when we have an emf in a battery for example,we know that the battery has say 6V and we know that it is the potential difference between its two ends.So,in this case,between which two points is the induced potential difference?

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Electric fields produces a difference of potential on two points with different distances of the field source. Magnetic fields induces current on a closed loop if the loop is not on parallel in relation of the lines of field and the magnitude of the field does have to change (you have to have a flux).

If you have a magnetic field interfering on your circuit, you will have a constant inducted current flowing on it (which depends of its magnitude, the radius of your circuit and the tax of change of your field). So, the voltage drop will only depend of where is the point you are measuring, since it will depends of only the resistance between that two points. (Ohm's law - constant current)

If you have an electrical field interfering on your circuit, you will have a change of voltage on your circuit which will depends of the distance of the electrical field and also the distance between two points. So you have only to know the magnitude of the field at the two points you chose and the distance between them.

To consider both effects you have only to sum them.

God already stated all EM effects on four simple equations.

Also, we have an Electrical Engineering SE on which probably you will get more (and better) answers of related questions.

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Take the simplest case of a uniform magnetic field and a circular wire perpendicular to the magnetic field. You know that a changing magnetic field induces an electric field. Because the magnetic field is uniform and the wire perpendicular, the electric field has the same magnitude on all points of the wire. You can find the total emf from the change in flux. But remember the definition of emf for a closed loop. $$ emf= \oint_C \vec{E}\cdot dl$$ This is also equal to $-\frac{d\Phi_B}{dt}$ for the same closed loop. Now say you calculate from the magnetic flux that the induced emf is 6V. This also means taking the line integral of the $\vec{E}$ field around the entire closed loop gives you 6V. Now instead of going around the whole loop, integrate around half the loop. Our $\vec{E}$ field is the same magnitude at every point, so we get half the emf, 3V. If we take two points that are only 10% of the total circumference of the loop, then between those two points there will be a .6V emf.

In other words, unlike a battery where the emf is confined to one point on the circuit, in the case of an induced emf, there is a voltage gain across any two points of the circuit where $\int_A^B\vec{E}\cdot dl$ is nonzero.

In order to then find the voltage difference between two points, you need to add the contributions from $\int_A^B\vec{E}\cdot dl$ and from the resistors. You know the induced emf is equal to the voltage drop across the resistors. So if the emf is 6V, the voltage across both resistors is -6V. So lets say again use the simplest example where the field is the same magnitude at every point. Choose two points so that they have two resistors and 25% of the wire between them. In the part of the loop with no resistors and 75% of the wire loop, you would see a 4.5V gain. In the section with 25% of the wire and both resistors, you would see a 1.5V - 6V = -4.5V gain.

The absolute voltage at any point would be difficult to calculate without a reference point. You would at the very least need to know if the wire was originally at 0V or some other $V_0$, and then I think you would be able to calculate it. But normally in these kinds of cases you would ground a portion of the circuit.

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  • $\begingroup$ Good answer.But i have one question.When you go around the wire,the voltage ramps up and at the resistors it falls.In this we agree.But where is the point that the voltage is zero and from there and on it starts growing(until it drops at the resistors)? $\endgroup$ – TheQuantumMan Apr 2 '15 at 16:17
  • $\begingroup$ I'm not completely sure. In certain situations you can use symmetry to figure it out. For example, if both resistors are the same size, you can split the loop down the middle between the resistors and along that line there should be zero voltage. For a general configuration where the $\vec{E}$ isn't circular and the initial voltage is $V_0$, I think you would need to make sure $\oint Vdl=V_0$. If your loop does not have a reference voltage anywhere, this integral will be difficult. $\endgroup$ – MonkeysUncle Apr 2 '15 at 16:50
  • $\begingroup$ And in reality, almost all circuits will have a reference voltage at some point(usually ground). The reason is wires without a reference voltage can develop a floating voltage. If something at 2V touches the circuit, the circuit will now be stuck at 2V. This can cause problems if certain circuit elements break down outside of specified voltage ranges. $\endgroup$ – MonkeysUncle Apr 2 '15 at 16:53
  • $\begingroup$ Yes,i understand.Great answer although i did ask something just a bit different(about where is OV and 5V in the system).I think it is a difficult question. $\endgroup$ – TheQuantumMan Apr 2 '15 at 16:58
  • $\begingroup$ @LandosAdam There's no 0V and 5V. What matters is the 5V of difference of potential. Have you heard about isolation transformer? If not, search for it. For example, suppose you have not 5V on your loop, but now 5kV. Do you expect to be electrified if you touch one point of your circuit? (Considering your feet is grounded on earth) $\endgroup$ – Pedro Quadros Apr 2 '15 at 17:26

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