Potential difference with an inductor

As far as I know, the potential difference between two points is defined as the negative line integral of electric field between those 2 points: $$\Delta V=-\int d \ell\cdot\mathbf E$$

I also know that when magnetic field changes, the curl of electric field is not 0 and potential difference makes no sense.

But when we have an inductor in a RLC circuit, then people always say that there is a potential drop across the ends of inductor. But since the magnetic field is changing potential difference, this should make no sense.

• The responses to physics.stackexchange.com/q/15402 will be helpful to you. – David H Jun 25 '13 at 10:36
• I lost you when you started the second sentence. Could you please re-frame your question to make it more understandable? Thanks :D – mikhailcazi Jun 25 '13 at 12:12
• can you please tell me what is wrong in second sentence? – Sahil Chadha Jun 25 '13 at 12:18
• @mikhailcazi, pretty sure Sahil is referring to $\partial\mathbf{B}/\partial t = \nabla\times\mathbf{E}$. – Kyle Kanos Jun 25 '13 at 12:46
• well kyle kanos is right and there is a minus sign in the above equation. – Sahil Chadha Jun 25 '13 at 12:50

While i think Lubos answer leaves nothing conceptualy to add, it is on too high level, if you have problem to understand nonconservative fields. Your question is actually freshman topic so i think at first you should understand this in freshman fashion and then move to Lubos answer.

Therefore you might use a very nice resource from OCW MIT by Walter Lewin

It basically tells you to always use Faraday's law. In this framework, as you said, when we have not-changing magnetic field, we get that electric potential drop on any loop is 0 (which textbooks call Kirchhoff II law). When we have changing magnetic field, than going in the direction of the current (so that you won't have to think about sign in inductance part), you write that the right-hand-side of Faraday's law (that is negative derivative of magnetic flux with respect to time) is simply =$-L\,dI/dt$ Than you do left-hand-side, remembering that ideal conductor has no resistance - therefore there is no electric field in it.

If you still don't get it, here is another resource, or the full lecture

It is not true that "potential difference makes no sense when the curl of $\vec E$ is nonzero". Instead, the scalar and vector potentials $\Phi,\vec A$ may always be defined and the electric field may be expressed in terms of these variables as $$E = -\nabla \phi - \frac{\partial A}{\partial t}$$ In fact, these potentials aren't unique – any other choice of the potentials related to the original one by a gauge transformation is equally good to calculate the same $\vec E$ and $\vec B = \text{curl }\vec A$.

In an rlc circuit, one may always confine the nonzero $\vec A$ to a small vicinity of the inductor which also determines the right gauge transformation. Then the potential difference across any element of the circuit is well-defined and the sum of these differences over a closed loop is zero.

• when we talk about RLC circuit we never use vector potential – Sahil Chadha Jun 25 '13 at 12:42
• That's what I wrote, too. – Luboš Motl Jun 26 '13 at 5:19
• inside a pure inductor electric field is zero if i travel inside an inductor and calculate the line integral of electric field it should be zero so why people are always saying that there is a potential drop of -Ldi/dt – Sahil Chadha Jun 26 '13 at 6:02
• Dear @Sahil, I feel that I have answered the same question about 3 times already. There is a potential drop across the ends of the inductor and there's no electric field inside the inductor and these two facts don't contradict each other because in the inductor, the electric field is not the gradient of the potential. Instead, it is given by the formula I wrote down. Have you at least tried to read it or have you predecided that there is a contradiction in physics and nothing will change your mind? – Luboš Motl Jun 27 '13 at 16:16