To my understanding, the electric field inside an inductor is zero if the inductor is made of an ideal wire. According to this post, this happens because the induced field is canceled out by the electrostatic field due to charges that develop on the surface of the wire.

Now suppose that the current through the inductor begins to decline. Then according to Faraday’s law, an emf will be induced that tries to resist this decline in current. But my question is, how can it resist the change in the current if the induced field is just canceled out by the electrostatic field that develops due to the charges on the surface? What's the exact mechanism that resists the change in current in terms of electric fields?


1 Answer 1


The emf opposing the change in current is not generated by an electrostatic field. When the current is flowing, there is a magnetic field. When you start to turn down the current, the change in current changes the strength of the magnetic field. Because of Maxwell's equations (specifically, Faraday's Law), the changing magnetic field creates an electric field that creates an emf that opposes the change in current.

You may ask what mechanism underlies Faraday's law. The answer is that there is no mechanism underneath, as far as we know this equation is (and, Maxwell's equations in general are) one of the basic elements of the world. There's an excellent video featuring Feynman about this I recommend if you have time: https://www.youtube.com/watch?v=36GT2zI8lVA

  • $\begingroup$ Thanks for the answer. According to the post I linked, the electric field that is induced by the changing magnetic field is canceled out by an electrostatic field that develops in response to the induced field. If the two fields cancel each other out, how does the inductor resist the change in current? $\endgroup$ Commented Mar 14, 2022 at 13:09
  • 1
    $\begingroup$ Ah, I see. I'll have to come back to this, currents with an ideal conductor can be tricky (because if $R=0$ then $V=IR$ implies $I=\infty$ if $V\neq 0$). As that answer says, for a real conductor, there is not perfect cancellation between the emf and static field. So, the perfect conductor case is kind of a pathological mathematical problem as opposed to a physics problem. I think probably the answer is that the current flows on the boundary of the conductor and there isn't a cancellation of fields on the boundary, but I am not 100% sure off the top of my head. $\endgroup$
    – Andrew
    Commented Mar 14, 2022 at 13:27
  • $\begingroup$ Great, thanks for the comment. If you come up with any further insights, I'd appreciate it if you'd share them! $\endgroup$ Commented Mar 14, 2022 at 16:34

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