Suppose we have a changing magnetic field in the $z$ direction and a conductive ring of radius r inside the magnetic field lying in the $xy$ plane. Then we have an induced emf: $$ ε= -{{dΦ} \over {dt} }$$

This emf causes a current along the conductor-ring: $ I = \frac{ε}{R} $ , R its resistance. (1)

However we know that all points inside a conductor have the same electric potential. (2)

How can both (1) and (2) be true?

Some thoughts:This is a pure Faraday field and the induced electric field is of course not conservative. So how can there even be a potential? If $E$ is not conservative then we can't write $E=- \nabla V$ , right? However we do have some sort of potential from which we get E: the emf. What's the difference between the induced emf and the electric potential?

Also: what would happen if I cut the ring at one point? Would I have potential difference between the 2 ends?

  • 4
    $\begingroup$ Hopefully, someone comes along with a more complete answer, but I think that there only exists a potential V such that E= - grad(V) for the case of electrostatics. The situation you described doesn't fall into that category. See en.wikipedia.org/wiki/Electric_potential and look at the sections on "Electrostatics" and "Generalization to electrodynamics". $\endgroup$
    – user93237
    Sep 10, 2017 at 23:17
  • 1
    $\begingroup$ This is why EMF is distinct from electric potential. $\endgroup$
    – garyp
    Sep 10, 2017 at 23:43

2 Answers 2


You're right. When you have a conservative $E$ field, you can define an electric potential $V$. And when you don't, you can't define such a potential (just as you can't define a potential energy $U$ for a nonconservative force $F$ - nonconservative forces still do work - but there is no associated potential energy function for such forces).

General Remarks

Electric potential is measured in volts and defined by $$V(x) = -\int_{\mathcal{O}}^x \vec{E}\cdot d\vec{l} \tag{1}$$ A voltage is defined by (in the textbooks) as a difference in potential $$\Delta V = V_b - V_a = -\int_a^b \vec{E}\cdot d\vec{l} \tag{2}$$

However this is too restrictive. In general, anything that takes the form $$\int \frac{\vec{F}}{q}\cdot d\vec{l} \tag{3}$$ can be called voltage and is measured in volts. So yes $\text{(2)}$ above is a voltage and even $\text{(1)}$ is a voltage if you like (as $V(x)$ is secretly $V(x) - V(\mathcal{O}) = V(x) - 0$ and therefore a difference in potential). But note that $\vec{F}$ can be anything. It doesn't have to be a conservative electric force. Again, anything that takes the form of $\text{(3)}$ is measured in volts and can be called a voltage. EMF takes the form $\text{(3)}$ [I explain this further down]. So too does potential difference. This is one reason why EMF and potential difference get mixed up: they take similar forms and hence both are measured in volts and can be called voltage (or induced voltage or whatever). And actually, this is great. If you are doing anything in the lab or talking to engineers, it doesn't really matter whether voltage means potential difference or EMF. It does, but all we really care about is energy. Voltage (equation $\text{(3)}$)is energy. EMF and potential difference/potential are energy. Energy is energy, whether it be EMF or potential difference. Voltage is just a general term for energy. If you are ever in a situation where you don't know whether to say "the EMF is 5 volts" or the "potential difference is 5 volts", just say "5 volts" or "the voltage/induced voltage/whatever is 5 volts" and you are safe.

What is EMF

In order to get current to flow around a circuit, you need some force pushing charges around the wire. Let's call this the driving force $\vec{F}$. EMF $\mathcal{E}$ is defined as

$$ \mathcal{E} = \oint \frac{\vec{F}}{q}\cdot d\vec{l} \tag{4}$$ where the integration is taken around the loop. There are two main forces that drive current around a circuit: a "source" force from say a battery and a conservative electric field which pushes charges around the wire. $\vec{F} = \vec{F}_s + q\vec{E}$. Therefore, $\text{(4)}$ can be written

$$\mathcal{E} = \int_a^b \frac{\vec{F}_s}{q}\cdot d\vec{l} $$ as $\vec{F}_s$ is usually confined to a section of the loop and $E$ is conservative so it integrates to 0 (started where we left off - once around the loop). $\vec{F}_s$ can be anything. It can be a chemical force, some temperature gradient thing, pressure on a crystal, a magnetic force, a nonconservative $E$ field, etc. So consider a battery. A conservative electric field goes from the positive terminal, around the loop, to the negative terminal, as well as from the positive terminal to the negative terminal inside the battery. Using the last equation, assuming the battery is ideal so that the chemical force is equal and opposite to the electric force,

$$ \mathcal{E} = \int_a^b \frac{\vec{F}_{\text{chemical}}}{q}\cdot d\vec{l} = -\int_a^b \vec{E}\cdot d\vec{l} = V$$ The EMF of the battery is equal to the potential difference across its terminals. But this does not mean that EMF is potential difference. It just happens to be so in this case. Most simple circuits turn out to be this way but realize again that EMF and potential difference are totally different. In the first place, you can't have a potential difference without an EMF generating that separation of charge in the battery. EMF generates a potential difference that happens to match the numerical value of the EMF (which you can think of as energy conservation). Then if you have a resistor connected to this battery, current $I = V/R$. I could also say $I = \mathcal{E}/R$ as they are numerically equivalent. But it's more appropriate in my opinion to use $I = V/R$ as the energy drop is coming as electric potential energy in a conservative $E$ field (which exists throughout the wire doing the pushing). Here we begin to see, as in the next section, that All circuits require an EMF to function.

Let's Look at your example

There is no such thing as an electric potential in your example. There's an $E$ field, but it's not conservative. Therefore, don't say potential. There is an EMF however. You can say there's a voltage or an induced voltage if you like (from the above discussion). But there's definitely not a potential difference/potential present. For this specific example, the EMF is given by

$$ \mathcal{E} = -\frac{d\phi}{dt} = \oint \vec{E} \cdot d\vec{l}$$ where the driving force is that nonconservative $E$ field. The current as you say is $I = \mathcal{E}/R$. Here again we see a true instance of the following: all circuits require an EMF. The idea that all points in a conductor are at the same potential is equivalent to saying that there is no $E$ field in a conductor. Note that this idea of there being no $E$ field in a conductor only holds for electrostatics + no time varying external magnetic fields (having a time varying $B$ prevents electrostatics anyways so saying "+ no time varying $B$" was redundant). Having an $E$ field in a conductor is completely fine. Turn on an $E$ field in a conductor. There is definitely an $E$ field in the conductor until electrostatics is reached. This is why conducting wires in simple circuits can have $E$ fields in them. This $E$ field is essential for driving current around even though it's through a conductor. It's just that the conductor can never reach electrostatics when it's part of a circuit. It desperately tries, but the battery prevents the wire from coming to a static situation. And with time-varying $B$ fields, nothing wrong with having an $E$ field in a conductor. When you stop varying $B$, you'll stop changing the $E$ field and things will reach statics. While you are changing the $B$ field, $E$ is changing with time. The conductor is trying to reach statics, but can never do so. So there will always be an $E$ present and hence the conductor won't be an equipotential (albeit, in simple circuits, you can take the wires to be equipotentials because $E$ is so so tiny).

Too Much Theory, What to know about EMF

From equation $\text{(4)}$, because of the closed line integral, EMF does not care about conservative forces while electric potential crucially depends on a conservative E field. EMF and potential are both instances of equation $\text{(3)}$, and therefore both tell you about energy. Potential is energy in a conservative E field. EMF is energy added to your circuit through "nonconservative" driving forces. In order for circuits to work, you need to pump energy into them so that charges will flow back down to low energy, making a circuit. EMF tells you how much energy driving forces give to a unit charge in one trip around the loop. Conservative forces don't give any net energy to a charge after one complete loop (started where you stopped). "Nonconservative" forces will give you some nonzero value to equation $\text{(4)}$. EMF tells you how much energy was added by driving forces and hence how much energy must be dropped by dissipative/"friction" forces in one trip around the loop. An EMF of 5 volts means 5 volts must be dropped by every unit of charge. If you have a battery providing 2 volts of EMF and a changing magnetic field providing 6 volts of EMF, 8 volts must be dropped

[by the way, If you ever see an example circuit out there with both a battery and a changing flux enclosed by the loop, more than likely their derived equations have wrong explanations. Right equation. Wrong explanation (which is basically just as bad as not knowing what you are doing). What they do is say $-d\phi/dt = \oint \vec{E}\cdot d\vec{l}$. This is Faraday's Law, true in anycase. But what actually does that $\vec{E}$ mean? It's the net $E$ field on your loop. In the case of a battery and a changing flux, that $E$ has both a conservative and a nonconservative component. The conservative component integrates to zero, leaving only the nonconservative $E$ providing the $-d\phi/dt$. Therefore, if you have this simple battery + changing flux + resistor circuit, $-d\phi/dt = I_{\phi}R$ where $I_{\phi}R$ is the integral of nonconservative $E$ around the loop. Now we can add a constant to each side of the equation. Since EMF battery $V_0 = I_0R$, we can say $V_0 - d\phi/dt = (I_{\phi} + I_0)R = IR$. Or if you want, you can write out $-d\phi/dt = I_{\phi}R + \oint \vec{E}_{\text{conserv}} \cdot d\vec{l} = I_{\phi}R - V_0 + I_0R$].

  • $\begingroup$ The topic of self-inductance has confused me for a while. This post helps alot since it answers one of my biggest doubts on induced emf. However, one thing I was not able to understand was the non-electrostatic component of the Field. Upon taking the line integral of the non-electrostatic component and dividing it by resistance we get the current flowing in the loop. But isn't this result (ohms law) only applicable to POTENTIAL drop and not emf. $\endgroup$ Feb 18, 2020 at 18:29

Wasn't sure if this should just be a comment but relevant and lengthy.

You can apply a potential to the problem, but not the electric potential.

There's a theorem that for any vector field whose divergence is zero in all space, there exists a vector field who's curl is the first field. Given Gauss' Law for magnetic fields then:

$$\nabla \cdot \vec{B}=0 \implies \vec{B}=\nabla \times \vec{A}$$

Where $\vec{B}$ is the magnetic field and $\vec{A} $ is the Vector or Magnetic Potential.

With Maxwell's Laws and vector identities, it can be shown that $\vec{E}=-\frac{\partial \vec{A}}{\partial t}$

By definition $\Phi_B=\int\vec{B}\cdot\hat{n}dA$ where $\hat{n}$ is the unit normal to the surface contained by the current loop, and $dA$ is an infinitesimal area element.

But $\vec{B}=\nabla\times \vec{A}$. By Stoke's Theorem: $\Phi_B=\int (\nabla \times \vec{A})\cdot \hat{n} \ dA=\int \vec{A}\cdot \vec{dl}$.

$$- \frac{d\Phi_B}{dt}=\int \frac{-\partial\vec{A}}{\partial t}\cdot \vec{dl}=\epsilon$$

So we have an EMF anywhere we have a time varying Vector Potential. Further, we can calculate this EMF using a Line Integral,convenient for certain geometries. It's no coincidence the integrand we have is equivalent to the Electric Field. A non-zero line integral around a loop directly implies a non-conservative Vector Field is in play.


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