I have some serious troubles with understanding the EMF in inductor. Assume we change the current from value $I_0$ to zero by rotating the tumbler of the source. Then why should we consider the EMF induced by inductor would act like on the right picture and not like on the left one? There are opposites cases of electric potentials distribution and they seems result in the similar current flow. But the first one is not realized in life enter image description here

Or why do not we consider the case when the current is induced like in the 3rd bottom figure?

enter image description here


1 Answer 1


Do not use plus/minus signs to show direction of electromotive force or induced current. They will mislead you. Also, electromotive force is an independent force in the circuit that isn't captured by the concept of potential; EMF in an inductor isn't necessarily directed from terminal with higher potential (+) towards the terminal with lower potential (-).

The correct direction of EMF can be denoted by an arrow parallel to the component (inductor) and can be derived from the Lenz law: its direction is such that current contributed by this EMF is in direction that counteracts(but not completely) the change of current that is actually happening.

So if current is going from top to bottom, and decreasing in time, EMF has to point upwards.

  • $\begingroup$ Thus, the 2nd picture on the right is correct in that case, isn't it? And is the last one false because the current can't change its direction like this due to Lenz law? $\endgroup$ Commented Jan 30, 2021 at 15:21
  • $\begingroup$ Both 1st and 2nd picture show EMF direction correctly (using the arrow). If plus/minus signs are meant to show electric potential, then the 1st picture has them correctly. 2nd does not. $\endgroup$ Commented Jan 30, 2021 at 15:44
  • $\begingroup$ As for the 1st picture on the left, I can't figure out the reason why the positive potential, that should be very high at the top side of the inductor after turning the EMF supply off, doesn't produce the current flow towards the positive pole of the source? For example if we had the 1000V voltage spark after the turning the source off the current would be flow to the 10V source potential. Where is the problem in my reasoning? $\endgroup$ Commented Jan 30, 2021 at 18:18
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    $\begingroup$ Before the voltage source is disconnected, electric potential of the top terminal of the inductor is the same as potential of the bottom terminal (same as potential of the + terminal of the source). Difference of potential is only on the resistor. Then right after the voltage source is disconnected at its + terminal, the current through the inductor starts decreasing but continues to flow. Because no current flows through the top terminal (disconnected), it is getting more and more negatively charged and the bottom terminal is getting positively charged. $\endgroup$ Commented Jan 30, 2021 at 19:13
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    $\begingroup$ So potential of the top terminal decreases (compared to + terminal of the source) and when it is low enough, the electrostatic field in space between the source terminal and the disconnected inductor terminal gets so strong some negative charge from the inductor can jump over to the source and a spark occurs. This stops the current decrease and even may start to increase the current again. $\endgroup$ Commented Jan 30, 2021 at 19:14

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