# How exactly does an inductor resist change in current?

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?

• 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. Commented Mar 14, 2022 at 13:27