In a multiple choice question a simple circuit of only a battery and an inductor the length of the inductor is doubled. The question is how does the velocity of the electrons change. Big spoiler alert: the velocity decreases. My reasoning was that in the equation $V=-\dot{I}L$ the inductance doubles so the change in current must halve in order for the voltage to remain constant. Thereby the current and the velocity of electrons decreases. Is this correct so far? Because the solutions say: inductance increases so the resistance increases and (by Ohm's law I guess) the current decreases. However, I do not understand why the resistance (there is no resistor) should change if the inductance increases?


2 Answers 2


In real life an inductor consists of a coil of wire (with or without a laminated iron core). So a real inductor has both resistance and inductance. If you double the inductance by increasing the length of wire on the coil, then the resistance will increase (roughly 1.4 times). All the same, your question should have made it clear that you were dealing with a real inductor, not an ideal or 'pure' inductor. And it should also have told you that the inductance is doubled by adding more turns to the coil.

Your question should also have made it clear whether it wanted a comment about electron drift velocity just after the inductor is connected to the battery, or after a steady current is achieved. Your answer is suited to the former (though it looks as if you should include a resistive term in your equation); their answer will also apply in the latter case, though, as I've said, I think they should have warned you that it was not a pure inductor.

You'll have gathered that I don't think much of the question!


In simple terms the effect of a real inductor can be thought of as being due the resistance of the inductor (the wires have resistance) and the inductance of the inductor.
If the current is steady (dc) then the inductance of the inductor plays no part in controlling the current in the circuit.
So doubling the inductance by winding more turns will increase the inductance and the resistance.
The resistance has increased because there is a longer length of wire which makes up the inductor.
If the current is steady then only the increase in resistance will affect the current which will be decreased if the emf of the battery stays constant.
A reduced current would mean that the drift speed of the electrons would be smaller.

If you are going to consider the time it takes for the current to reach a steady value after completing the circuit then the important parameter is the time constant of the circuit which is $\frac{\rm inductance}{\rm resistance}$.
If the length of wire is doubled and this doubles the number of turns then the inductance increases by a factor $2^2=4$ and the resistance increases by a factor $2$ and the time constant increases by a factor $2$..

In the equation which you referred to $V = L \frac {dI}{dt}$ the $V$ is the voltage induced across the inductor due to the current in the circuit changing.
If the current is not changing then $V=0$.

  • $\begingroup$ Question for OP: Does the original question say the inductance is increased by adding more turns? (For example, it could also be increased by inserting a core into an air-core coil) $\endgroup$
    – The Photon
    Commented Jul 20, 2018 at 15:59
  • $\begingroup$ No, the inductance is increased by length. $\endgroup$ Commented Jul 27, 2018 at 16:18
  • $\begingroup$ @Eman I am not sure what you meant by your comment but it did alert me to the fact that the self inductance is proportional to $\rm{turns}^2$ and not $\rm turns$ as I originally implied. $\endgroup$
    – Farcher
    Commented Jul 27, 2018 at 21:40
  • $\begingroup$ What I mean is: the question says that the inductance is doubled by doubling the length of the inductor. It does not say anything about turns. I do not know if they were implying that the number of turns is proportional to the length or not. They just say that the length is doubled. $\endgroup$ Commented Jul 28, 2018 at 10:00
  • $\begingroup$ @Eman Thank you for the clarification. Overall it goes down as yet another question which needs improvement? $\endgroup$
    – Farcher
    Commented Jul 28, 2018 at 10:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.