Potential difference with an inductor

Question

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 http://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
@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
Related: http://physics.stackexchange.com/questions/136431/application-of-kirchhoffs-laws-in-circuits-with-inductors — Kyle Kanos, Dec 15 '14 at 16:55

score 4 · Answer 1 · answered Jun 25 '13 at 12:22

4

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.

answered Jun 25 '13 at 12:22

Luboš Motl

179,018

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 05:19
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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 06:02
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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
@Lubos, you said 'there's no electric field inside the inductor'. However if the current in an inductor changes we know from Faraday law that an emf is induced in it, which implies that there does exist a field in it. – Kashmiri Oct 18 '20 at 17:11
Right, Yasir, it's hard to restore the precise context in each comment written in 2013. I was probably assuming some stationary assumption in that sentence. Is that true that even with the changing magnetic fields, "j" current in the inductor is proportional to "E"? No current means no "E". Of course both are nonzero in some coils with changing mangetic fields. – Luboš Motl Oct 19 '20 at 16:38

score 3 · Accepted Answer · edited Aug 15 '14 at 20:00

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

Potential difference with an inductor

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