I have trouble understanding Faraday's law when there is an induced current which in turn induces another current in the same circuit. I shall illustrate my confusion with an homework problem and I will try to formulate it by steps to really try to pin point the confusion.

For the following circuit with a single loop with an resistor and a time-dependent current $I_{2}(t)$ Faraday's law gives:

\begin{equation}\oint \vec E \cdot \vec {dl}=-L_{2}\frac{dI_2}{dt}=R_2I_2 \end{equation}

where $L_2$ is the self-inductance of the loop.

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Next let's look at the following circuit:

enter image description here

Now for this circuit $I_2$ is the induced current from $I_1$ (this is given as a fact). The loop now experiences induced current from the long wire and that induced current in turn gives rise to an self-induced current. I imagine the self-induced current happen immediately when the induced current from the wire occur which means one has two account for both those currents (or voltages) when evaluating the closed loop integral. I am a bit unsure about my own reasoning here and could easily try to reason the opposite that one should not already account for the self-induced current. However with the previous stated then Faraday's law should give:

\begin{equation}\oint \vec E \cdot \vec {dl}=-L_{12}\frac{dI_1}{dt}=R_2I_2 - L_2\frac{dI_2}{dt} \end{equation}

But this is wrong according to my professor who wrote $$ L_{12}\frac{dI_1}{dt}=R_2I_2 + L_2\frac{dI_2}{dt}$$

to what he called "Kirchoff's voltage law" implying that these three terms are voltages which confuses me since how can there be three voltages when there is only two currents in the loop (with a single resistor), the induced $I_2$ and the self-induced from $I_2$. The loop does not actually have three voltages? I'm guessing that one should not see the above equation as three voltages in the loop but instead as two voltages in the loop which has to equal some voltage that is not the loop due to induction? Not only do the voltages confuse me but also the signs, I believe $R_2I_2$ should have opposite sign to the two other terms. Where am I wrong and how can one resolve my confusion?

Thanks in advance for taking your time to read and perhaps even giving some clarification.


2 Answers 2


Faraday's law, in the first example, gives us value of self-induced EMF as

$$ -L_2 \frac{dI_2}{dt}; $$ this is in terms of

  • rate of change of current $I_2$ in the loop and
  • self-inductance constant $L_2$, which is always positive.

It does not tell us that

$$ -L_2\frac{dI_2}{dt} = R_2I_2; $$ this latter equation is a result of Kirchhoff's second circuital law, which is sometimes confusingly called Kirchhoff's voltage law (KVL).

This law actually states the following:

For any simple closed path (loop) made of conducting elements $k=1..N$, sum of all terms $R_k I_k$, where $I_k$ is current in element $k$, and $R_k$ is its ohmic resistance, equals sum of all electromotive forces $\mathscr{E}_i$, $i=1..M$, acting on the loop:

$$ \sum_{k=1}^N R_kI_k = \sum_{i=1}^M \mathscr{E}_i. $$

This law can be viewed as a generalization of Ohm's law that states potential difference equals current times resistance; the generalization is in going from simple element to a loop, and in replacing the potential difference by an electromotive force, a more general concept.

This law is valid not only when all the EMFs are due to chemical cells/batteries in the loop, but also when some or all EMFs are due to EM induction.

That's why the name "Kirchhoff's voltage law" and its formulation in terms of voltages (sum of voltages in loop equals zero) is often confusing people: although it is always true, often there are no relevant voltages (=potential differences) available to use, but instead there are EMFs. In the present case, it is necessary to use the original version of the law phrasing in terms of EMFs.

In your simple case of one circuit, sum of all electromotive forces there is, due to Faraday's law, just the self-induced EMF:

$$ \mathscr{E} = -L_2 \frac{dI_2}{dt}. $$

So Kirchhoff's second law then implies

$$ -L_2 \frac{dI_2}{dt} = R_2 I_2. $$

In the second case where we have two circuits, and the first has total current $I_1$ and the second has total current $I_2$, the sum of all electromotive forces in the circuit 2 now has also new term due to action of the circuit 1:

$$ \mathscr{E} = -L_2 \frac{dI_2}{dt} - L_{12}\frac{dI_1}{dt}. $$

The sign in front of the term is again put conventionally as minus; this is the most natural choice because when the two circuits are very similar shape, are put on top of each other and have the same current with same rate of change, induced EMF due to circuit 1 in the circuit 2 has the same direction as self-induced EMF in circuit 2, so it is best to put $L_{12}$ as positive and keep the minus sign in front of the term.

However, in your second example the first circuit is not "similar shape and on top of the other circuit". Instead, although the first circuit is not completely specified, we can see/assume that the closest part of the circuit 1 is on the left-hand side of the circuit 2. We will still use the same convention and put minus in front of the EMF term, but now it is possible that $L_{12}$ may be negative. It can also be positive.

We can find which of these two possibilities is the case, by analyzing the effect of current $I_1$ increase on magnetic flux through the circuit 2 (positive direction is from screen towards the eyes). If the effect is the same as the effect of increase of $I_1$, then $L_{12}$ is positive just as $L_2$ is. If the effect is opposite to that of increase of $I_2$, then $L_{12}$ has to be negative.

In your picture where the closest circuit 1 current element is on the left of the circuit 2, we see that increase in current $I_1$ increases magnetic flux in direction from eyes towards the screen, which is opposite to what increase in current $I_2$ does. So the mutual inductance $L_{12}$ is negative.

However, if we put the circuit 1 element to the right of the circuit 2, the direction of magnetic field due to circuit 1 in region of circuit 2 would change to "from the screen towards the eyes" and so mutual inductance $L_{12}$ would be positive.

  • 1
    $\begingroup$ Your answer was very helpful. Does $L_2\frac{dI_2}{dt}$ and $L_{12}\frac{dI_1}{dt}$ have opposite signs due to the flux for each is in opposite direction as in accordance to Lenz's law? If that is the case and let's assume the signs are as following $\oint \vec E \cdot \vec{dl}=-L_{12}\frac{dI_1}{dt}+L_2\frac{dI_2}{dt}$ how should one now determine that this should be equal to $-R_2I_2$ and not $R_2I_2$ ? $\endgroup$
    – ludz
    Commented Aug 15, 2021 at 14:23
  • $\begingroup$ No, both terms have to be written with minus sign in the front, however it is possible $L_{12}$ may be negative number. So total emf is always $-L_{12}\frac{dI_1}{dt} - L_2 \frac{dI_2}{dt}$, and sometimes $L_{12}$ is negative, depending on position of the circuits in space. Now use Kirchhoff's second law as given above. We obtain $-L_{12}\frac{dI_1}{dt} - L_2 \frac{dI_2}{dt} = R_2I_2$. $\endgroup$ Commented Aug 15, 2021 at 21:25

Sorry for my poor english. My native language is french.

I think that there is a difficulty related to the orientation of the circuits.

In general, you should write: $\oint \vec E \cdot \vec {dl}=- L_2\frac{dI_2}{dt}-L_{12}\frac{dI_1}{dt}=R_2I_2 \ $

But, with the orientations chosen, the coefficient $L_{12}$ is negative. If we impose it positive, then we must change the sign : $\oint \vec E \cdot \vec {dl}=- L_2\frac{dI_2}{dt}+L_{12}\frac{dI_1}{dt}=R_2I_2 \ $

  • $\begingroup$ Good point about the general case. I just realized one should do a superposition of the magnetic field from $I_1$ and $I_2$ and that will be on the right hand side of Faraday's law, $\frac{d}{dt}(\phi_{B1}+\phi_{B2})=-L_2\frac{dI_2}{dt}-L_{12}\frac{dI_1}{dt}$. Correct me if I'm wrong. However I still cannot see why $L_{12}$ should be negative with this orientation (making it positive on the left hand side as you did). And by orientation do you mean the clockwise vs counterclockwise of closed integral? $\endgroup$
    – ludz
    Commented Aug 15, 2021 at 12:32
  • $\begingroup$ Actually shouldn't $\phi_{B_1}$ and $\phi_{B_2}$ have opposite signs in accordance with Lenz's law? This will then I believe fix the sign problem $\endgroup$
    – ludz
    Commented Aug 15, 2021 at 14:10
  • 1
    $\begingroup$ Imagine a positive current $I_1$. Then the magnetic field would be towards the back of the screen. Whereas, with the positive direction for $I_2$ indicated, the normal to the circuit (2) is towards us. So for a positive current $I_1$, the flux would be negative and so $L_{12}$ is négative. $\endgroup$ Commented Aug 15, 2021 at 15:54
  • $\begingroup$ Exactly, the flux from $I_1$ and $I_2$ is in opposite directions and hence opposite sign. But how does one now determine the sign on $R_2I_2$? $\endgroup$
    – ludz
    Commented Aug 15, 2021 at 20:19
  • $\begingroup$ The rule is simple: the circulation of the electric field along the positive direction for the current (emf) is equal to $Ri$. This is an immediate consequence of Ohm's Law. In induction problems, it is convenient to have rules that apply without having to think about it. Obviously, thinking is useful and checking Lenz's law is a good idea. But, sometimes, we need to do it quickly without risk of error. It is also useful to keep in mind that the mutual inductance has a sign which depends on the choice of the relative orientations of the two circuits. Afterwards, write the equation is easy. $\endgroup$ Commented Aug 16, 2021 at 6:04

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