0
$\begingroup$

Assume the following conductor, with current flowing, powered from a source $S$: enter image description here

Now, I've introduced a magnetic field($B) to the conductor covering a small area of it like so: enter image description here

The change in magnetic field is in a way that would induced an $\epsilon$ that would be additive to the conductor's own $V$ like so:

enter image description here

Is the following formula correct: $V_t = V_s + \epsilon $ from orientation of things?

Where $V_t$ is the total EMF of the conductor, does that mean the induced EMF($\epsilon$) is in series with the conductor?

EDIT: Varying magnetic field over the same area(B(

EDIT(2): enter image description here If KVL fails due to the varying magnetic field(and I assume in a varying magnetic flux as well), if instead I had to conductors parallel to each other connected to a closed circuit, $+V(S)$ will split in parallel at divided currents from the supplied power source(assume all equal resistance) then the two wires experiencing the same change in $B$ at the same change in $t$, they induce the same voltage, however, now parallel to one another the induced current increases. Is this correct?

$\endgroup$
  • 1
    $\begingroup$ emf induced will be on the inductor. So i think the induced v will be out of phase with 90 degree. So vt = (vs^2 + e^2)^(1/2) $\endgroup$ – Kishan Kumar Jul 18 '15 at 4:09
  • $\begingroup$ @KishanKumar What do you mean by out of phase? $\endgroup$ – Pupil Jul 18 '15 at 5:02
  • $\begingroup$ See, here you say that there is a conductor upon which the voltage is applied.And then to produce induced emf, we will need an inductor and voltages of inductor and conductor will be out of phases by 90. $\endgroup$ – Kishan Kumar Jul 18 '15 at 8:19
1
+50
$\begingroup$

You are partially correct. You are correct about the part that the total voltage in the conductor is the addition of the voltage (S) and induced electromotive force, assuming that: 1. The applied magnetic field doesn't affect the flowing current due to the voltage (S). 2. The induced current due to the time varying magnetic field doesn't affect the magnetic field that generated it.

Taking those conditions into account will over complicate the problem, from your question we can assume they are valid assumptions.

The bit where you are wrong is when you say the voltage (S) is in series with the induced electromotive force. Please note that Kirchoff's circuit law (KVL in particular) is derived for cases where the magnetic field is static. For time varying magnetic fields (as the one you describe in this problem), KVL doesn't apply any more. Please have a look at the limitations section for KVL in this page. A nice explanation of why KVL breaks when there is a time varying magnetic field can be found in this presentation

EDIT: Lenz law (the law describing the generated EMF) is based on the magnetic flux variation, which could be either due to time varying field or due to changing area as well, or even both. So yes KVL would fail in the case of moving area as well.

In your edit, it is tricky to tell how the total current will behave. The top and bottom wires are connected to the power supply (S), so now you have three meshes in the circuit:

  1. A mesh consisting of V(S) which is connected to the wire to the LEFT hand side.
  2. A mesh consisting of V(S) which is connected to the wire to the RIGHT hand side.
  3. A mesh consisting of the RIGHT hand side wire and the LEFT hand side wire.

Now, depending on which meshes enclose the magnetic field, an induced current will flow in that mesh. For example, if the source V(S) is connected to the right of the shown figure, the magnetic field on the left wire will induce a current in mesh number 1 and mesh number 3, while the magnetic field on the right wire will induce a current in mesh number 1 and number 2. So in general, you need to do some calculation to find out whether the total current increase or not.

To make this concept clearer to you, please keep in mind that a current induced by EMF should always flow in circles in closed loops, which is a consequent of Lenz law, before you made your edit, there was no closed loops, so there was no closed loops and as a result there was EMF without a current. Now there are three closed loops, and you need to track the current in everyone of them to see their total effect.

$\endgroup$
  • $\begingroup$ It certainly did! Would KVL/KCL fail with any form of changing magnetic flux(i.e change in area due to motion as well)? Or only due to a varying magnetic field? I've added an edit in regards to the series question. $\endgroup$ – Pupil Jul 22 '15 at 1:06
  • $\begingroup$ @Key please have a look at my edit. It is tricky to explain clearly, I am happy to emphasize it more if you had a particular point that is not clear $\endgroup$ – Gotaquestion Jul 22 '15 at 18:13
  • $\begingroup$ It is indeed tricky, what if we focused only on the induced-EMF of both wires in their respective fields and assume the behavior of current? Looking back at diagram(3) you and I both agree that the total EMF there is V(S) + Induced EMF, however, introducing another wire parallel to it like diagram(4) having a change in magnetic flux at the same rate they induce the same EMF, however, they must be parallel to one another. Meaning the current increases at the same voltage... I know this won't work because of KVL failing, but it's worth an assumption. $\endgroup$ – Pupil Jul 22 '15 at 19:21
  • $\begingroup$ I have a better diagram for you here: i.imgur.com/0QMp1K5.png , At point $V_a$, the voltages are additive, then we can simply apply ohms law, the current is increased to the increase in applied voltages? $\endgroup$ – Pupil Jul 28 '15 at 21:24
  • 1
    $\begingroup$ V(S) is the battery, and eps is equal to the time derivative of the magnetic flux through the surface, as described by Lenz law, keep in mind that is true in the EQUIVALENT circuit. In the real circuit, Va is ill-defined $\endgroup$ – Gotaquestion Jul 29 '15 at 23:01
0
$\begingroup$

I'm assuming the bottom of the conductor is grounded as you've indicated a flowing current?

EMF occurs in a conductor when submitted to a time varying magnetic field as governed by Faraday's law, is the field in your problem static or time-varying, or is your conductor in motion? In this case the solid conductor would experience eddy currents, circles of currents created by the time varying magnetic field as it hits the conductor.

In the DC case, you have a constant magnetic field acting on current (moving charge) which will create a force given by

$ F = Q\vec{v} \times \vec{B}$

By definition, this force is perpendicular to velocity and therefore the direction of current. Force is then going towards the left of your conductor initially for those charges (assuming a rather large field). This force is then not additive to the initial voltage as you've implied, but perpendicular.

$\endgroup$
  • $\begingroup$ I've added an edit, that it's a changing magnetic field that induces it and it's stationary connected to a circuit. The change in magnetic flux, is in such a way that would induce EMF in the polarity shown in the final diagram. $\endgroup$ – Pupil Jul 21 '15 at 11:33
  • $\begingroup$ Great, so now we have an AC field hitting part of a conductor carrying a current. The EMF generated will just create eddy currents in the conductor since there isn't a closed loop shown (such as in the case of an inductor) that would make a circuit. The left side of Faraday's law shows the EMF must occur over a closed loop: $ \oint_C {E \cdot d\ell = - \frac{d}{{dt}}} \int_S {B_n dA} $ The Eddy Currents will be circular in the conductor subjected to a field. link. The EMF you've indicated would not appear. $\endgroup$ – Kthaxt Jul 21 '15 at 12:52
  • $\begingroup$ That's true that EMF would not be true because it's an open circuit, however, the conductors are a part of a loop(hence connection to the exterior source ($+V(S)$). Even though the diagrams are terrible(bear with me). The diagram in Edit(2) could clarify things. $\endgroup$ – Pupil Jul 22 '15 at 3:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.