# How can we derive the polarity of the induced voltage in an inductor?

As we know, the voltage-current relation of a two-terminal constant-inductance inductor is:

$$v_L(t) = L \, \dfrac{\mathrm di_L(t)}{\mathrm dt} \tag 1$$

I've managed to prove that equation without considering the signs nor the polarity and direction of the voltage and current. However, in many textbooks on circuit analysis and physics that equation is valid for any sign of the voltage, current, and rate of change of current, as long as the reference polarity of the voltage and the reference direction of the conventional current are chosen as shown in the following figure. Figure 1. Reference direction for the current and reference polarity for the voltage in an inductor. Image source: Engineering Circuit Analysis (8th ed.) by Hayt, Kemmerly and Durbin.

So I'm trying to prove if equation (1) indeed considers the signs and the reference polarity and reference direction of the voltage, current, and rate of change of current. Specifically, as you'll see below, I have troubles while determining the polarity of the induced EMF or voltage (such that its numerical value is positive), given the current and its rate of change.

# Basics

Before showing my attempt, I'll briefly mention two basic facts I'll use: the right-hand rule and the direction of the induced magnetic field.

## Right-hand rule

As we know there are many right-hand rules used in electromagnetism (for Lorentz's force law, for Ampère's law, Flemming's right-hand rule, etc.). The one I'm going to use is the one used to determine the direction of the magnetic field induced by a positive conventional current through a cylindrical inductor/solenoid/coil, as shown in the following figure. Figure 2. The right-hand rule #1: for the magnetic field induced by a positive conventional current in a cylindrical inductor. Image source: link.

The previous rule can be used to determine the direction of the magnetic field if given the direction of the current, or the vice versa.

## Change in the magnetic field and the direction of the induced magnetic field

As we know, Faraday's law is based on the change in the magnetic flux/field, and Lenz's law says the induced EMF will be such that it opposes the change in the magnetic flux/field, or in other words the induced magnetic field is the negative of the change in the original magnetic field. The change in some quantity is defined as the final value minus the initial value. And to find the difference $$\mathbf u − \mathbf v$$ of two vectors geometrically, first we find the negative $$− \mathbf v$$ of the second vector $$\mathbf v$$, then move it such that its tail is at the head of the first vector $$\mathbf u$$, and then the difference of the two original vectors is the sum $$\mathbf u + (− \mathbf v)$$, which is the vector that points form the tail of the first vector $$\mathbf u$$ to the head of the vector $$− \mathbf v$$. With these facts we can determine the direction of the induced magnetic field (the one that opposes the original magnetic field), as shown below. Figure 3. Lenz’s law: direction of induced magnetic field due to an increasing magnetic field. Image source: own. Figure 4. Lenz’s law: direction of induced magnetic field due to a decreasing magnetic field. Image source: own.

# My attempt

We’re going to use the following notation: $$i(t)$$ will be the conventional current through the inductor imposed by the external active circuit connected to the terminals of the inductor, whose direction and numerical sign will be assumed as known/given; $$\mathbf B(t)$$ will be the magnetic field produced by the current $$i$$ as described by Ampère’s law, whose direction is described by the RHR (figure 2); $$v_\text{ind}(t)$$, $$i_\text{ind}(t)$$ and $$\mathbf B_\text{ind}(t)$$ will be the induced voltage/emf, induced conventional current and induced magnetic fields, respectively, by the change in the magnetic field $$\mathbf B$$ as described by Faraday’s law, and whose polarities/directions are described by Lenz’s law (figures 3 and 4) and the RHR (figure 2); $$v_L(t)$$ and $$i_L(t)$$ will be the voltage across and current through the inductor respectively such that their reference polarity and reference direction are as shown in figure 1.

I think many of us will agree that an intuitive way to think about the order in which the "events" occur is as follows: the current $$i$$ creates a changing magnetic field $$\mathbf B$$ that induces a voltage $$v_\text{ind}$$, which produces a current $$i_\text{ind}$$, which produces a magnetic field $$\mathbf B_\text{ind}$$. So, in my attempt I'll proceed as follows: given $$i$$ and its rate of change, I find the $$\mathbf B$$, then I find the change in $$\mathbf B$$, then I find $$\mathbf B_\text{ind}$$, then I find $$i_\text{ind}$$, and then I find $$v_\text{ind}$$. As you'll see below, I managed to find all of those quantities and understand why they had the direction they had, except $$v_\text{ind}$$.

Now, I'll explain my attempt.

Consider a two-terminal constant-inductance cylindrical inductor. Let’s consider four different scenarios: when the positive conventional current is flowing in one direction and then in the other direction, and when the positive conventional current is increasing and then decreasing. The inductor is connected to an external active circuit that establishes a time-varying current $$i(t)$$ in each of the four cases. We assume such current is conventional, is positive, we know its direction and we know whether it is increasing or decreasing. This is shown in the following figure. Figure 5. Image source: own.

Since there is current flowing through the inductor, from Ampère’s law a magnetic field $$\mathbf B(t)$$ is produced due to that current $$i$$, whose direction we can determine applying the RHR (figure 2) to each of the four cases; to do this, we assume we grab the inductor with our right hand, such that our fingers point in the direction the current $$i$$ is flowing throughout the inductor, then our thumb points in the direction of the magnetic field $$\mathbf B$$ inside the inductor. Applying this rule to each of the four cases of the previous figure:

• In cases a) and b), when we look at the inductor as shown in the previous figure (and not, say, from the top or from the bottom or from the rear side), we see the current $$i$$ is flowing from left to right, so we grab the inductor with our right hand such that our fingers point to the right. Then our thumb will point upwards, so that’s the direction the magnetic field $$\mathbf B$$ points inside the inductor.

• In cases c) and d), when we look at the inductor as shown in the previous figure, we see the current $$i$$ is flowing from right to left, so we grab the inductor with our right hand such that our fingers point to the left. Then our thumb will point downwards, so that’s the direction the magnetic field $$\mathbf B$$ points inside the inductor.

Thus we get the direction of the magnetic field shown in the following figure. Figure 6. Image source: own.

Because we assume the current $$i$$ is time-varying, the magnetic field $$\mathbf B$$ produced by it is also time-varying, so the magnetic flux $$\Phi(t)$$ is also time-varying, so by Faraday’s law an EMF $$v_\text{ind}(t)$$ is induced across the inductor. To find the polarity of such EMF or voltage, we first find the direction of the current $$i_\text{ind}(t)$$ such EMF tries to establish, which we determine using Lenz’s law. So while our goal is to find the polarity of the induced EMF, we first find the direction of the induced magnetic field, secondly we find the direction of the induced current that produced such induced magnetic field, and thirdly we find the polarity of the induced EMF produced by the original changing magnetic field. First, we find the change $$\mathrm d \mathbf B$$ in the magnetic field.

• In case a), the magnetic field points upwards inside the inductor, and the current is increasing so the magnetic field is also increasing, to the change points upwards (figure 3).

• In case b), the magnetic field points upwards inside the inductor, and the current is decreasing so the magnetic field is also decreasing, to the change points downwards (figure 4).

• In case c), the magnetic field points downwards inside the inductor, and the current is increasing so the magnetic field is also increasing, to the change points downwards (figure 3).

• In case d), the magnetic field points downwards inside the inductor, and the current is decreasing so the magnetic field is also decreasing, to the change points upwards (figure 4).

Thus we get the direction of the change in the magnetic field shown in the following figure. Figure 7. Image source: own.

The voltage $$v_\text{ind}$$ induced across the inductor according to Faraday’s law due to the time-varying magnetic field $$\mathbf B$$ will try to establish a current $$i_\text{ind}$$, that from Ampère’s law would produce a magnetic field $$\mathbf B_\text{ind}(t)$$ whose direction according to Lenz’s law will try to oppose the change $$\mathrm d \mathbf B$$ in the original magnetic field $$\mathbf B$$. In order for $$\mathbf B_\text{ind}$$ to oppose $$\mathbf B$$, the former must point in the opposite direction of the latter.

• In case a), the change in the original magnetic field points upwards, so the induced magnetic field points downwards.

• In case b), the change in the original magnetic field points downwards, so the induced magnetic field points upwards.

• In case c), the change in the original magnetic field points downwards, so the induced magnetic field points upwards.

• In case d), the change in the original magnetic field points upwards, so the induced magnetic field points downwards.

Thus we get the direction of the induced magnetic field shown in the following figure. Figure 8. Image source: own.

Secondly we find the direction of the induced current, by applying the RHR (figure 2), except that now, unlike figure 5, we know the direction of the magnetic field and we want to find the direction of the current. To apply this rule, we assume we grab the inductor with our right hand, such that our thumb points in the direction of the induced magnetic field $$\mathbf B_\text{ind}$$ inside the inductor, then our fingers point in the direction the induced current $$i_\text{ind}$$ is flowing throughout the inductor. Applying this rule to each of the four cases of the previous figure:

• In cases a) and d), when we look at the inductor as shown in the previous figure, we see the induced magnetic field $$\mathbf B_\text{ind}$$ is flowing downwards inside the inductor, so we grab the inductor with our right hand such that our thumb points downwards. Then our fingers will point to the left, so that’s the direction the induced current $$i_\text{ind}$$ is flowing throughout the inductor.

• In cases b) and c), when we look at the inductor as shown in the previous figure, we see the induced magnetic field $$\mathbf B_\text{ind}$$ is flowing upwards inside the inductor, so we grab the inductor with our right hand such that our thumb points upwards. Then our fingers will point to the right, so that’s the direction the induced current $$i_\text{ind}$$ is flowing throughout the inductor.

Thus we get the direction of the induced current shown in the following figure. Figure 9. Image source: own.

So far so good; I don't have any doubts. The only step missing is to find the polarity of the induced voltage $$v_\text{ind}$$. I don't know how to do this, this is where I'm stuck.

However, I realized/observed something. It looks like we must assign the reference polarity of $$v_\text{ind}$$ such that the induced current $$i_\text{ind}$$ flows from higher potential (the terminal marked with the “$$+$$” sign of $$v_\text{ind}$$) to lower potential (the terminal marked with the “$$-$$” sign of $$v_\text{ind}$$) through the external active circuit. (Equivalently, the induced current $$i_\text{ind}$$ flows from lower potential [the terminal marked with the “$$-$$” sign of $$v_\text{ind}$$] to higher potential [the terminal marked with the “$$+$$” sign of $$v_\text{ind}$$] through the inductor.) But I don't know why it is correct, and I'd like to know it.

You may ask how I'm sure my observation above is correct. The reason is because if we assume it is correct, then, as I'll show below, equation (1) indeed considers the sign of $$v$$ and the sign of the rate of change of $$i$$.

So let's suppose the above observation is correct. Then, in the previous figure:

• In cases a) and d), the induced current $$i_\text{ind}$$ flows into the active circuit through the upper terminal, so the induced voltage $$i_\text{ind}$$ has the positive reference polarity at the upper terminal.

• In cases b) and c), the induced current $$i_\text{ind}$$ flows out of the active circuit through the upper terminal, so it flows into the active circuit through the lower terminal, so the induced voltage $$i_\text{ind}$$ has the positive reference polarity at the lower terminal.

In this way we get the polarity of the induced voltage shown in the following figure. Figure 10. Image source: own.

Now that we’ve obtained the polarity of the induced voltage $$v_\text{ind}$$ (such that its numerical value is positive) and we know the direction of the current $$i$$ (such that its numerical value is positive) as well as the sign of the rate of change of $$i$$, let’s get rid of the irrelevant details in the previous figure (magnetic fields, induced current). Also, let’s change the reference direction of the current (which also changes its sign) and the reference polarity of the voltage (which also changes its sign) such that the reference direction of the current always points towards the inductor in the upper terminal and such that the reference polarity of the voltage is positive at the upper terminal. Thus we get the following figure. Figure 11. Image source: own.

Notice that each case of the previous figure has the reference direction for the current and the reference polarity for the voltage such that the arrowhead points into the inductor towards the positive reference terminal. In other words, the four cases satisfy the reference direction and reference polarity shown in figure 1.

• In case a), $$i_L(t) = i(t)$$ and $$v_L = v_\text{ind}(t)$$.

• In case b), $$i_L(t) = i(t)$$ and $$v_L = -v_\text{ind}(t)$$.

• In case c), $$i_L(t) = -i(t)$$ and $$v_L = -v_\text{ind}(t)$$.

• In case d), $$i_L(t) = -i(t)$$ and $$v_L = v_\text{ind}(t)$$.

Thus we get the following figure. Figure 12. Image source: own.

At last, from equation (1), since $$L$$ is always a positive parameter, the sign of the right-hand side depends only on (and is the same as) the sign of $$\mathrm di_L/ \mathrm dt$$. Thus this equation requires that $$v_L$$ must have the same sign as $$\mathrm di_L/ \mathrm dt$$. Let’s see if this is satisfied in each case of the previous figure:

• In case a), $$\mathrm di_L(t)/\mathrm dt > 0$$ and $$v_L > 0$$.

• In case b), $$\mathrm di_L(t)/\mathrm dt < 0$$ and $$v_L < 0$$.

• In case c), $$\mathrm di_L(t)/\mathrm dt > 0$$ and $$v_L < 0$$.

• In case d), $$\mathrm di_L(t)/\mathrm dt < 0$$ and $$v_L > 0$$.

Notice indeed $$v_L$$ and $$\mathrm di_L(t)/\mathrm dt$$ have the same sign for each case. Therefore, the equation $$v_L(t) = L \, \mathrm di_L(t)/ \mathrm dt$$ indeed takes into account the signs of $$v_L$$, $$i_L$$ and $$\mathrm di_L/ \mathrm dt$$, provided the reference direction for $$i_L$$ and the reference polarity for $$v_L$$ are chosen as shown in figure 1.

By the way, notice in figure 10 that another way to formulate my question is: why does the current $$i$$ flow from higher potential to lower potential through the inductor when it is increasing (cases a) and c)), and why does it flow from lower potential to higher potential through the inductor when it is decreasing (cases b) and d))?

I found this question, which is essentially the same as mine. The user Farcher explained in his answer how to obtain the direction of $$i_\text{ind}$$ (such that its numerical value would be positive). They assumed an increasing (and positive) increasing current $$i$$, and got the same result as me (figure 10, cases a) and c)), namely that the current $$i$$ flows from positive to negative inside the inductor. When obtaining the polarity of $$v_\text{ind}$$ (such that its numerical value is positive), they said in the comments the induced current $$i_\text{ind}$$ flows from negative to positive inside the inductor, whether the current $$i$$ is increasing or decreasing, which is the same as my result.

Their explanation for the polarity of the induced voltage was based on an analogy with a battery: that inside a battery, current flows from negative to positive. However, that's true only when the battery is supplying energy to the circuit (it's been discharged/de-energized), it's not true when the battery is consuming energy from the circuit (it's been recharged/re-energized) in which case current flows from positive to negative inside the battery. While in the case of the inductor, in cases a) and c) the inductor consumes/stores energy (and $$i_\text{ind}$$ flows from negative to positive inside the inductor, as expected from the comparison with the battery), but in cases b) and d) the inductor supplies/releases energy (and yet $$i_\text{ind}$$ stills flows from negative to positive inside the inductor, unlike in the battery). So I'm not satisfied/convinced with Farcher's reasoning.

• This is way too long for a question to be put here. Please mind the other users of this site and ask concise questions that are likely to be of interest to other users. Dec 11, 2021 at 3:42
• @JánLalinský Thanks for your opinion, but... Re: "This is way too long for a question to be put here." The question is short: "How can we derive the polarity of the induced voltage in an inductor?"; maybe you meant the answer would be long. // Re: "Please [...] ask concise questions that are likely to be of interest to other users." The question is of interest to people who like/need to know about electric circuits; it is also concise, reread the title and the first four paragraphs of the post. Dec 11, 2021 at 4:45
• The post is too long and if you wrote a much shorter one, more people would be interested. Dec 12, 2021 at 11:00

Your thought train is fine up to Fig.6. But then you get into a needlessly complicated descriptions of signs. You state

"by Faraday’s law an EMF $$v_\text{ind}(t)$$ is induced across the inductor. To find the polarity of such EMF or voltage, we first find the direction of the current $$i_\text{ind}(t)$$ such EMF tries to establish, which we determine using Lenz’s law. So while our goal is to find the polarity of the induced EMF"

It seems you are using the word "voltage" and symbol $$v_{ind}$$ for two different concepts: 1) induced EMF in the ideal inductor - that is due to induced electric field present in the coils; 2) potential drop across the ideal inductor terminals (when moving from one terminal to the other in the designated positive direction) - that is due to the fact that electric field has electrostatic component, and potential drop is integral of this electrostatic field.

This (using "voltage" for two different concepts and getting confused as a result) is common in many people's understanding of electricity, even those with high credentials. I think mostly because many textbooks and teachers do not understand this either, because they don't understand the general concept of electromotive force and its variants. Old papers and textbooks on electricity (pre-WWII) did not suffer from this confusion.

Understanding the difference between EMF and potential difference in physics is important. It also resolves the question of sign of potential drop across the ideal inductor which you are interested in.

Both concepts - EMF and potential drop - are valid and useful, and both depend on the sign convention. The sign convention is that current intensity $$i$$ is positive when it flows in the designated positive direction in the loop, and emf and potential drops are positive when their effect is to act on current in the circuit element to increase it in that same direction.

Induced EMF

Induced emf for the path defined by coils of the inductor is defined as integral of induced electric field over that path.

When we have single inductor of self-inductance $$L$$ whose interaction with other currents/inductors can be neglected, induced EMF always obeys the equation (whether the inductor is ideal or not):

$$emf = -L\frac{di}{dt},$$ the minus sign making sure that emf acts against changes of current. This follows from Faraday's law and the mentioned sign convention for current and emf. It is not a good idea to call this quantity "voltage", but it is often being done, with various variations (electromotive voltage, induced voltage, etc). Although induced EMF has the unit Volt just as voltage in electrostatics has, EMF is a path-dependent quantity and in general one cannot associate two points of space (such as terminals of an inductor) with unique EMF. We need to state the path as well, but because of this, it is always better to say EMF instead. It is also almost never the quantity of interest when measuring in practice; instead, we want to measure potential differences, which do not depend on paths.

Potential drop

Potential drop across the inductor has (unfortunately) become a more complicated thing to explain. It is not always the same as EMF, not even in magnitude. Potential drop is integral of the electrostatic part of electric field when going from one terminal to the other in the positive direction, the exact path not being important. This quantity is not, in general, determined merely by induced part of electric field; the value of current $$i$$, resistance of the inductor $$R_i$$ and its internal capacitance $$C_i$$ is also important. So in general, potential drop across inductor cannot be determined from induced EMF alone.

However, in the special case where the inductor is ideal (zero internal resistance $$R$$, capacitance $$C$$), there is a simple relation: potential drop is exactly minus induced EMF. This is because electric field in ideal inductor coils has to be zero, and this implies that electrostatic field completely cancels the induced field inside the conductive coils, and that implies integral of electrostatic field from one terminal to the other has to be minus integral of induced field over path from one terminal to the other that goes inside the coils. So, for a perfect inductor, we have

$$potential~drop = - emf = L\frac{di}{dt}.$$

This quantity is also often called voltage drop across the inductor, or simply voltage across the inductor. This quantity is useful when writing down the so-called Kirchhoff's voltage law for any closed loop in a lumped model of AC circuit. Then it is simply denoted $$V$$ or $$v$$. This is the preferred meaning of the word "voltage"; it is used in electrostatics, and is used also in AC circuits. It is also what we often want to measure in a complicated real circuit via oscilloscope. To make sure we really measure this and not some path dependent quantity like EMF, the probes and wires have to be made with good field insulation (coax cables) and during measurements, we prevent the wires from arranging themselves in loops.

From the above expression all signs are evident; when current increases in positive direction, emf is negative so the potential drop has to be positive, to counteract the induced EMF. When current decreases in that same direction, the induced EMF is positive, so the potential drop is negative to counteract the induced EMF.

What would be different for a real inductor with internal resistance? Here, net electric field must exist in the coil to push against resistance, so we cannot assume effect of potential drop cancels effect of the induced EMF. Let the real inductor be connected directly to a source of varying potential drop (more often called "source of voltage") $$V(t)$$:

     ---------------
|      ->       |
|               )
(V)              )  real inductor
|               )
|      <-       |
---------------


Then by assumption, potential drop on the inductor is $$V(t)$$. We cannot find induced EMF or current $$i$$ from these assumptions alone.

However, if we assume this real inductor behaves as ideal inductor with self-inductance $$L$$ with resistor $$R$$ in series (ignoring capacitance effects), we can use Kirchhoff's second circuital law (for its formulation, see my answer here: Using Faraday's law twice ). The replacement lumped model circuit looks like this:

     ---------------
|      ->       |
|               |
(V)              L
|               |
|               |
|               R
|               |
|      <-       |
---------------


Now we can write down Kirchhoff's second circuital law for this circuit:

$$Ri = V - L\frac{di}{dt}.$$

Here our voltage source contributes to the circuit effective electromotive force of magnitude V, and the other r.h.s. term is induced emf.

From this equation, we can find that potential drop on the real inductor is $$V = Ri + L\frac{dI}{dt} = Ri - emf.$$ So potential drop on the inductor is not the same as induced EMF, not merely due to opposite sign but also due to magnitude. It depends on emf and current $$i$$ and internal resistance of the inductor $$R$$. In general it has different phase from EMF, and in special cases it can have the same sign as induced EMF has, something that can't happen with ideal inductor.

This was still a simplification, and a more realistic model would include contribution due to capacitive interactions between the inductor's coils (ideal capacitor $$C$$ in parallel with the ideal inductor $$L$$).

• “This is because electric field in ideal inductor coils has to be zero” // Why, though? Mar 3, 2022 at 3:35
• @alejnavab operational definition of ideal conductor. In reality the better the conductor, the lower the electric field. We use the limit of ideal conductor to simplify analysis. Mar 5, 2022 at 16:27

Okay, I managed to answer my question (at least from one point of view) by thinking in terms of energy/power.

As we know, it can be proven if the voltage reference polarity and the conventional current reference direction are defined as shown in figure 1 (i.e. such that the arrowhead points into the inductor towards the positive reference terminal), then the product $$v_L(t) \, i_L(t)$$ is the instantaneous power consumed by the inductor; this is discussed in the passive sign convention. As we also know, it can be proven the "instantaneous" energy stored in the inductor is $$w(t) = 0.5 \, L \, i^2(t)$$. From this:

• In cases a) and c) of figure 9, the numerically positive conventional current $$i$$ through the inductor (with the direction shown) is increasing, so the "instantaneous" energy $$w$$ stored is also increasing, so energy is flowing from the circuit to the inductor, so the instantaneous power $$p(t)$$ consumed by the inductor is positive. Thus:

• In case a) of figure 9, in order for the instantaneous power consumed $$p(t) = v_\text{ind}(t) \, i(t)$$ to be positive, since $$i$$ is positive, $$v_\text{ind}$$ must be positive when measured at the upper terminal with respect to the lower terminal. Thus we get the reference polarity shown in case a) of figure 10.

• In case c) of figure 9, in order for the instantaneous power consumed $$p(t) = v_\text{ind}(t) \, i(t)$$ to be positive, since $$i$$ is positive, $$v_\text{ind}$$ must be positive when measured at the lower terminal with respect to the upper terminal. Thus we get the reference polarity shown in case c) of figure 10.

• In cases b) and d) of figure 9, the numerically positive conventional current $$i$$ through the inductor (with the direction shown) is decreasing, so the "instantaneous" energy $$w$$ stored is also decreasing, so energy is flowing from the inductor to the circuit, so the instantaneous power $$p(t)$$ consumed by the inductor is negative. Thus:

• In case b) of figure 9, in order for the instantaneous power consumed $$p(t) = v_\text{ind}(t) \, i(t)$$ to be positive, since $$i$$ is positive, $$v_\text{ind}$$ must be positive when measured at the lower terminal with respect to the upper terminal. Thus we get the reference polarity shown in case b) of figure 10.

• In case d) of figure 9, in order for the instantaneous power consumed $$p(t) = v_\text{ind}(t) \, i(t)$$ to be positive, since $$i$$ is positive, $$v_\text{ind}$$ must be positive when measured at the upper terminal with respect to the lower terminal. Thus we get the reference polarity shown in case d) of figure 10.