Confused about sign of mutually induced EMF when applying KVL

Question

Suppose I have several circuits with inductors like so:

and I want to find the current $ i_1, i_2, i_3 $ in each circuit. I will use Faraday's law to find the induced EMF and then use Kirchhoff's voltage law to write the equations. I've drawn positive current directions in the blue arrows (am I correct in understanding these can be arbitrary?). My confusion right now is about the signs when I apply Kirchhoff's law, in other words whether the voltages due to self and mutually induced EMFs should be positive or negative.

My current thought process is as follows.

Across each component in the center circuit:

• Battery: $ +\xi $
• $ R_2 $: negative, since it's dissipating energy. $ -i_2 R_2 $
• Self inductance: negative, since it opposes change in current $ -L_2 \dfrac{di_2}{dt} $
• Mutual inductance from circuit 1: If $ i_1 $ increases, then the left inductor generates an increasing magnetic field to the right. By Lenz's law the center inductor will want to generate a magnetic field to the left, so the induced EMF will want to increase the current in the positive direction. $ +M_{12} \dfrac{di_1}{dt} $
• Mutual inductance from circuit 2: If $ i_3 $ increases, then the right inductor generates an increasing magnetic field to the right. By Lenz's law the center inductor will want to generate a magnetic field to the left, so the induced EMF will want to increase the current in the positive direction. $ +M_{23} \dfrac{di_3}{dt} $

So by KVL their sum should be 0:

$$ \xi - i_2 R_2 - L_2 \frac{di_2}{dt} + M_{12} \frac{di_1}{dt} + M_{23} \frac{di_3}{dt} = 0 $$

For left circuit

• $ R_1 $: negative, since it's dissipating energy. $ -i_1 R_1 $
• Self inductance: negative, since it opposes change in current. $ -L_1 \dfrac{di_1}{dt} $
• Mutual inductance: if $ i_2 $ increases, then the center inductor generates a magnetic field pointing left. Then the left inductor will want to generate a magnetic field pointing right, so the induced EMF wants to increase the current in the positive direction. I think of the induced EMF acting as the "battery" in the left circuit, so it should be positive. $ + M_{12} \dfrac{di_2}{dt} $

Applying KVL:

$$ -i_1 R_1 - L_1 \frac{di_1}{dt} + M_{12} \dfrac{di_2}{dt} = 0 $$

However, according to some other forums (such as this one or this one) both self inductance and mutual inductance EMFs should be negative? Another answer here also uses the same sign for self and mutual inductance. If I were to rewrite the equations using negative signs for both self and mutual inductance then it would look like

$$ \xi - i_2 R_2 - L_2 \frac{di_2}{dt} - M_{12} \frac{di_1}{dt} - M_{23} \frac{di_3}{dt} = 0 $$

$$ -i_1 R_1 - L_1 \frac{di_1}{dt} - M_{12} \frac{di_2}{dt} = 0 $$

I can't wrap my head around which one of these is correct, since for this second set of equations I have a hard time understanding why the signs are the way they are. What am I misunderstanding here?

Answer 1

I finally have some time to type this up, and show you how and why KVL is just the wrong way to deal with this.

Faraday's Law is, with minus sign as the Lenz's Law, $$\tag1\oint_{\partial S}\vec E\cdot\vec{\mathrm d\ell} =-\frac{\mathrm d\ }{\mathrm dt}\iint_S\vec B\cdot\vec{\mathrm d^2A}$$ Note that the LHS is an integration over the electric field, and the RHS is an integration over the magnetic field. This is a statement of a relationship between two different phenomena, and should be kept as such.

Now, let us apply Faraday's Law to the middle circuit. Integrating clockwise starting from the switch, the first contribution is the electric field inside the battery, and we are traversing from the negative terminal to the positive terminal. A battery is a chemical soup that maintains the EMF voltage difference between its terminals, but for our current integration purposes, it is the same as if a capacitor. We are going against the electric field, and so it gives us $-\mathscr E$ as the contribution. After the battery, we are going from positive terminal to negative via the usual route, and the current $I_2$ that is drawn, is in the correct direction, i.e. $I_2>0$. Going down the resistor, the electric field integration will be $+I_2R_2$. Finally, we integrate the negligible electric field along the loops of the solenoid, and return to the switch. The inductance is a magnetic effect, and it appears on the RHS. I will not pretend to know the directions involved; we will fix them later. $$\tag2\therefore\qquad-\mathscr E+I_2R_2=-L_2\dot{I_2}+M_{21}\dot{I_1}+M_{23}\dot{I_3}$$ Rearranging the terms, we have $L_2\dot{I_2}+I_2R_2=\mathscr E+M_{21}\dot{I_1}+M_{23}\dot{I_3}$; the LHS of this equation reproduces the usual expression for LRC circuits and so must be correct, thereby telling us that the sign choice is correct. Now, note that KVL, strictly speaking, should not have any magnetic terms appearing. It is only via the brutal hacking away at Faraday's Law, rearranging terms haphazardly, that we would have KVL appearing to work. Such hacking is very easily going to make up confusing experimental results. For example, if you put in a voltmeter inside this circuit, it is possible to engineer things so that the voltmeter reading can differ depending upon the location of the voltmeters and the wires connecting them. Here is a video of simultaneous measurement of the same two points, with the oscilloscopes put on opposite sides. And it is very annoying that YouTube is not automatically recommending the part two of that same video.

Instead of working out what the signs of the mutual inductances should be right away, let us consider the easier circuits on the side. We know that $I_1$ is definitely going to be increasing from zero. This means that a new magnetic $\vec B$ field will be created in the solenoid $L_2$ with North pole pointing towards $L_1$. This means that Lenz's Law says that the solenoid $L_1$ will be creating a $\vec B$ field that with North pole pointing towards $L_2$, attempting to negate what is being thrown at it. Thus, we can write down that the left loop must be having $$\tag3I_1R_1=-L_1\dot{I_1}+M_{12}\dot{I_2}$$ where we are finally, for the first time, fully aware that this mutual inductance term is chosen with the correct sign. The right loop is not even a loop, so we can only guess, using similar reasoning, that it should be $$\tag40?=-L_3\dot{I_3}+M_{32}\dot{I_2}$$ unless we know what is to be connected between those terminals, replacing the $0?$.

Now, we try to fix the signs on Equation (2). If we imagine taking away the battery, and only consider increasing $I_1$ and $I_3$ independently from zero, then it is clear that the signs there are chosen correctly. Of course, I am assuming that all of $M_{12},M_{23},L_1,L_2,L_3,\mathscr E>0$ and we know by definition that $M_{21}=M_{12}\ \wedge\ M_{32}=M_{23}$; all resistances are positive, or else they won't even make sense. Mind you, the blue arrows defined the positive directions of all the currents, and we now know that all three of them are going to increase from zero when the switch is initially closed.

Note that I have given you not just the correct solution, but that the scheme is completely general: the only way to make sure all the signs are correct, is to actually consider the electric fields and magnetic fields in a theoretically coherent manner. Otherwise, it will have too many possible avenues for making a mistake.

Answer 2

I have used the excellent answers from @LPZ, @JánLalinský and @naturallyInconsistent to produce a little more detail about the solutions.

The first is related to Faraday's law, $\displaystyle\oint_{\partial \Sigma}{\vec E}\cdot{\mathrm d\vec l} =-\frac{\mathrm d\ }{\mathrm dt}\iint_\Sigma {\vec B}\cdot \hat n\, {\mathrm dA}$.

To use this equation a convention is needed which relates ${\mathrm d{\vec l}}$ and $\hat n$. The direction of the normal to the surface, $\hat n$, used to work out the magnetic flux and direction of the incremental path (circulation), ${\mathrm d}\vec l$, used to work out the line integral conform to the right-hand rule as shown in the diagram below.

Using the current arrow labels to define the line integral direction, normals $\hat n_1,\,\hat n_2,\,\hat n_3$ can be defined.

$\vec B_1,\,\vec B_2,\,\vec B_3$ are the directions of the magnetic fields produced by the three currents $i_1,\,i_2,\,i_3$.
Using @LPZ's suggestion, the magnetic flux can be evaluated with the signs consistent with the directions of the normals.

Coil 1, $\phi_1=\phi_{11}+\phi_{12} = L_1i_1 - M_{12}i_2$

Coil 2, $\phi_2=\phi_{21}+\phi_{22}+\phi_{23} = - M_{21}i_1+ L_2i_2 - M_{23}i_3$

Coil 3, $\phi_3=\phi_{11}+\phi_{12} = - M_{32}i_2+L_3i_3 $

Using Faraday's law gives the induced emfs.

Coil 1, $\mathcal E_1= L_1\dfrac{{\rm d}i_1}{\rm{d}t} - M_{12}\dfrac{{\rm d}i_2}{\rm{d}t}$

Coil 2, $\mathcal E_2= - M_{21}\dfrac{{\rm d}i_1}{\rm{d}t}+ L_2\dfrac{{\rm d}i_2}{\rm{d}t} - M_{23}\dfrac{{\rm d}i_3}{\rm{d}t}$

Coil 3, $\mathcal E_3= - M_{32}\dfrac{{\rm d}i_2}{\rm{d}t}+L_3\dfrac{{\rm d}i_3}{\rm{d}t} $

Back to the original diagram and using $\sum_m \mathcal{E}_m = \sum_{k=1}^n R_kI_k$ in the direction of the current labels, as explained by @JánLalinský, yields,

Coil 1, $\mathcal E_1 = i_1R_1$

Coil 2, $\mathcal {E_2 + E}= i_2R_2$

Coil 3, $\mathcal E_3 = v_3$ with, as a possibility, $v_3$ being the potential difference across the capacitor formed by the two ends.

Answer 3

There are two distinct rules that can be used to analyze effect of induced EMFs on a current in a circuit.

The standard, modern-textbook rule is the so-called Kirchhoff Voltage Law (KVL), which states that sum of potential differences in a closed path equals zero; for $n$ elements, and $V_k$ being potential drop in positive direction on $k$-th element, we have

$$ V_1 + V_2 + ... + V_n = 0. $$

This rule is always true (even in your example), but is useful only when we can easily express the potential drops $V_k$ in terms of something else such as charge, current or time derivative of current, such as in case of isolated RLC circuits which do not experience external influences and where all magnetic elements are lumped (their induced electric field is "concentrated" in them and does not affect other components in the circuit). In such circuits, potential decrease on a lumped inductor is always minus the self-induced EMF on the inductor, thus $V = LdI/dt$.

In your example, KVL is thus not very useful, because due to the external influences of the other circuits, we cannot express the potential differences in the main circuit in terms of current and its derivatives immediately, without further analysis based on the rule below.

The other rule is the original Kirchhoff second circuital law, which is very different: it states that sum of all EMFs acting in a closed conductive path (e.g. due to a battery, due to induced electric field of elements in the circuit, due to induced electric field of bodies outside the circuit) equals sum of terms $R_k I_k$ for all circuit elements making up the closed conductive path:

$$ \sum_m \mathscr{E}_m = \sum_{k=1}^n R_kI_k. $$

This original rule (really, an approximate law) is more appropriate to your example, because it does not refer to potential differences at all, it is not limited to lumped element circuits and thus can easily take into account external EMFs on the left-hand side, such as those due to the other solenoids nearby. From a modern point of view, this Kirchhoff rule is a result/combination of the Faraday law, superposed effect of non-magnetic EMFs (batteries), and Ohm's law, so one can also understand it as generalized Ohm's law for circuits.

So you do this: write down the Kirchhoff second circuital law equation for all three circuits (thus, three equations), taking into account external influences. Whether external EMF on the left-hand side has a plus or a minus sign in any single equation has to be determined (if you haven't memorized some specialized rule giving the answer off-hand, which is most physicists) using the Faraday law (including the Lenz law) based on the spatial relations of the two circuits, taking into account also details such as left-handedness/right-handedness (helicity) of both windings. When induced field of the external circuit 2 acts in the main circuit 1 in such a way that it supports current in the designated positive sense of circulation in circuit 1, its external EMF is a positive quantity in the equation for circuit 1; when it acts against the designated positive sense of circulation, that EMF is a negative quantity.

Answer 4

You don't need Lenz' law, just magnetostatics to get the relative sign between $M$ and $L$. You usually use Lenz' law to get the correct sign of the electromotive force from Faraday's law.

The sign of the influence coefficients depends on the orientations of the two currents, it is conventional. The current orientation convention needs to be included in the definition of $M$ to make the result unambiguous. What matters is that $M_{ij}$ and $M_{ji}$ have the same sign so that $M_{ij}=M_{ji}$. Your two conventions amount to changing the relative sign convention between $i_2$ and $i_1,i_3$.

Assuming that the circuits are spatially displayed as in the drawing with the currents drawn as you did, taking $M_{ij}>0$, you'd get: $$ \phi_2 = \phi_{21}+\phi_{22}+\phi_{23}\\ \begin{align} \phi_{21} &= -M_{21}i_1 & \phi_{22} &= L_2i_2 & \phi_{23} &= -M_{23}i_3 \end{align} $$ since the normals of circuits $1,2,3$ point to the right, left, right respectively and so do the magnetic fields using the right hand rule (or the fact that $L_i>0$). Basically, since you chose the current $i_2$ to flow in the reverse direction compared to $i_1,i_3$, you have the extra relative sign between $M$ and $L$.

Similarly, you should also get: $$ \phi_1 = \phi_{11}+\phi_{12}\\ \begin{align} \phi_{11} &= L_1i_1 & \phi_{12} &= -M_{12}i_2 \end{align} $$

Had you chosen $i_2$ in the opposite direction so that it's normal also points to the right, you would have gotten: $$ \begin{align} \phi_{21} &= M_{21}i_1 & \phi_{22} &= L_2i_2 & \phi_{23} &= M_{23}i_3\\ \phi_{11} &= L_1i_1 & \phi_{12} &= M_{12}i_2 \end{align} $$

Hope this helps.