Sign of voltage across inductor

[ElectroWhat]

I need help understanding the sign of voltages across an inductor.

Homework Statement

There is a current source providing current I to a resitor of resistance R in series with an inductor of inductance L, of which the other end is grounded. The voltage at the current source is V1 and the voltage at the inductor/resistor point is V2.

Explain the sign of V2 if the current is ramped up.

Homework Equations

I know from Lenz's law that
Emf = - d Bflux / dt = - L dI/dt.

The Attempt at a Solution

From Lenz's law, if the current is increasing, an Emf will be created in the opposite direction of the current increase. So the Emf will be pointing from ground to the V2 point. So the voltage will be increasing from V2 to ground ... so the voltage is NEGATIVE?

But that doesn't make any sense. Here's what I'm having trouble with:

This would mean the current is not only flowing AGAINST the Emf, but actually increasing against it.

What am I doing wrong?

 

[ehild]

"So the Emf will be pointing from ground to the V2 point. "
Right.

"So the voltage will be increasing from V2 to ground ... so the voltage is NEGATIVE?"

No, why do you think so? Assuming that the current flows through the resistor towards the inductor, (see pic) the negative emf means a voltage source which opposes this current, that is, positive pole at the resistor/inductor point with respect to the ground: V2 is positive.

ehild

 

[ElectroWhat]

I can't see your attachment. It says "pending approval" on my end.

"So the Emf will be pointing from ground to the V2 point. "
Right.

If the force is pointing from ground to the V2 point, then the current is flowing against the force. No?

One of Maxwell's equations is:
\nabla \times \vec{E} = - \frac{\partial}{\partial t} \vec{B}
Using stoke's theorem this gives:
\oint_{\partial S} \vec{E} \cdot d\vec{l} = - \frac {\partial}{\partial t} \int_S \vec{B}\cdot d\vec{a}

So the change in current will create a change in magnetic field, which will create an electric field against the change in current.

It really does appear that the electric field is pointing from ground to the V2 point, and the current is flowing against the electric field. How is this possible?

And if the electric field is pointing from ground to the V2 point, which means going from ground to V2 is going down in potential, then V2 must have a lower potential than ground. V2 is therefore negative.

Something is wrong here, but I can't find the problem with the logic.

 

[JustinLevy]

The voltage across an inductor is:
V = L \frac{dI}{dt}

The question only gives you enough details to get the sign of the voltage, but that is all you wanted anyway.

Or are you asking how to derive that from Lenz's law?

Hmm... I'm not good at tracking down sign errors. I don't really see what you've done wrong there. Sorry. All I can verify is that it is indeed wrong as you suspect.

Hopefully someone else can see what is wrong and lead you through it.