# The sign in the formula for the voltage across the inductor

This question is about voltage induced in a inductance.

I know the rate of change of current induces EMF in the inductor. The direction of the induced EMF will be opposite to the change that is causing it. The EMF induced is given by the following formula.

$$V_l = -L \frac{dI}{dt}$$

The negative sign indicates direction of induced emf across the coil is opposite the current change. In the picture I have drawn current direction and $V_l$ direction.

If $\dfrac{dI}{dt}$ is postive, $V_l$ will be negative according to formula right? Or does the direction of $V_l$ has to be reversed?

In many Internet references I found $V_l = L \dfrac{dI}{dt}$. Which one is correct?

I hope they reversed $V_l$ reference points and removed the negative sign in the formula. Please clarify.

• @user19579: Please don't forget that comments are for suggesting improvements to the question so that the author can edit it. The upvote button is there to help you express "woo, me too!" – RedGrittyBrick Oct 28 '13 at 9:37

The way you drew it, I and $\frac{dI}{dt}$ are both positive in the direction of the arrow and the correct expression is

$V_L = L\frac{dI}{dt}$

The negative sign depends on the direction you define your voltage and current. simulate this circuit – Schematic created using CircuitLab

TLDR
The formula is not derived from Maxwell–Faraday equation, because the induced electric filed is circular and electric potential is not defined for such a field. Instead, the formula comes from change in energy of magnetic field inside the inductor.

What causes confusion?
Consider a solenoid. Assume there is current flowing through it.
Apply Maxwell–Faraday equation in an integral form: $$\oint {\vec Ed\vec l = - {\partial \over {\partial t}}\int\!\!\!\int {\vec Bd\vec s} }$$ $$\int\!\!\!\int {\vec Bd\vec s} = \Phi = L{{dI} \over {dt}}$$ $$\oint {\vec Ed\vec l = - L{{dI} \over {dt}}}$$

Now, since $$\oint {\vec Ed\vec l} = {V_{beginning}} - {V_{end}} \equiv \Delta V\:\:(1),$$ the potential (voltage) difference between the beginning (where the current flows in) and the end of inductor (where the current flows out) is $$\Delta V = - L{{dI} \over {dt}}.$$

Not true. Why?
Look at the electric and magnetic fields inside solenoid. Image source

They are circular. Electric field produced by changing magnetic field inside solenoid is an example of a so-called non-conservative field (link, equation 372), because line integral around a closed loop is non-zero ($\oint {\vec Ed\vec l = - L{{dI} \over {dt}}}\ne 0$). Electric potential cannot be defined for a non-conservative electric field as it is done in the formula (1). It means the formula $\Delta V = - L{{dI} \over {dt}}$ is wrong.

Correct derivation and correct formula
Energy of magnetic field stored in the inductor $$W = {1 \over 2}L{I^2}$$ Change in the magnetic field energy per unit of time $${{dW} \over {dt}} = LI{{dI} \over {dt}} \:\:(2)$$ Assume the current started increasing. It means the energy of the inductor goes up (${{dW} \over {dt}} = LI{{dI} \over {dt}}>0$), that is, the inductor consumes energy from the electric system (increasing its magnetic field energy).
Assume, for lumped element model, one wants to replace an inductor with a power source. Amount of energy consumed by such a power surce per unit of time is given by $${{dW} \over {dt}} = I{V_L}\:\:(3)$$ $${V_L} \equiv {V_ + } - {V_ - } = {V_A} - {V_B}\:\:(4)$$

Note the direction of current at the picture above and polarity of the source. The current is flowing into positive terminal, what means that the power source consumes energy (not supplies).

Now combining formulas (2), (3) and (4) you get the right formula

$${V_L} = L{{dI} \over {dt}}$$

• Faraday law is the statement about the electromotive force (EMF), not the potential difference. – dmitryvm Jul 31 '16 at 17:14
• @dmitryvm: I read in "introduction to electromagnetics", that for EM cases, electromotive force is equal to voltage. And is'nt voltage by definition a difference of potentials? – Junius Jul 31 '16 at 19:33
• @dmitryvm Correct. Updated the text. 10x. – Sergei Gorbikov Jul 31 '16 at 20:43

The simple answer is that the voltage induced across the inductor as a result of the dI/dt change of current MUST BE opposite to the voltage applied by the current source. If this were not the case, a runaway voltage condition would be created. We know from simple observation that this is not the case. If it were the case, we could simply place an inductor in series with a voltage source and get an increase in supply voltage. This is clearly not what happens.

You can think of it in terms of voltage "drop". In simple circuits using only passive components all of the components in a single circuit loop will always drop voltage while being fed from a power source (i.e. current or voltage). The voltage drops around the loop will equal the supply voltage at every instant in time. So that, all of the components in the loop will have a voltage drop ( usually expressed as a negative voltage) across them.

When the supply voltage is removed, that's a different situation. The inductor voltage will reverse because the dI/dt factor is moving in the opposite direction, or polarity.