# Inductance of a coil (negative inductance?)

1. Nov 20, 2012

### eehsun

Hi everyone,
I encountered a problem as a part of the solution of which I needed to get the i-V relation across a solenoid to gain some intution in the other parts of the problem, which is the well known expression
VL=LdiL/dt, where V and i are referenced with respect to the passive sign convention.
Nothing tricky here, just basic stuff - however I wanted to quickly verify this before I moved on in the problem. The following is a simple MSpaint sketch that I have just created for illustrative purposes

The rest of the details are in the above picture. The question I had in mind is why I am not able to derive the simple relation VL=LdiL/dt for an inductor (solenoid in this case), even though I referenced everything in line with the passive sign convention?

Thanks!

Last edited: Nov 20, 2012
2. Nov 20, 2012

### eehsun

3. Nov 21, 2012

### eehsun

Any ideas? Something I may be overlooking?

4. Nov 25, 2012

### Astronuc

Staff Emeritus
This may help.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html#c2

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indcur.html#c1

http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Inductance/Solenoid.html [Broken]

Last edited by a moderator: May 6, 2017
5. Nov 26, 2012

### eehsun

I will give this some more thought and follow it up thereafter.
Many thanks for the response!

Note: I know how an inductor should behave under a time-varying excitation current and in this context I know why it makes sense for an inductor to have the particular i-V relation that it does have, V = Ldi/dt. I just have a problem showing this through Maxwell's eqns, but I think I'll hopefully manage to see the error in my approach if I give it some more thought.

6. Nov 27, 2012

### Gordianus

The voltage drop across an inductor is -L dI/dt and not L dI/dt.
Your first drawing is O.K. if dI/dt>0 and you consider the initial point is where the current enters the inductor and the output where the current exits the inductor. From that drawing you can easily see the voltage drop is negative (look at the plus and minus signs). That expalins everything.