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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.

direction of Vl and I

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.

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    \$\begingroup\$ I also need answer to this question...i have very vague idea about this. \$\endgroup\$
    – user19579
    Oct 28, 2013 at 9:27
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    \$\begingroup\$ @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!" \$\endgroup\$ Oct 28, 2013 at 9:37

4 Answers 4

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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.

schematic

simulate this circuit – Schematic created using CircuitLab

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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. enter image description here
Image source

What about electric field lines?
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.

enter image description here

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

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

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    \$\begingroup\$ Faraday law is the statement about the electromotive force (EMF), not the potential difference. \$\endgroup\$
    – dmitryvm
    Jul 31, 2016 at 17:14
  • \$\begingroup\$ @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? \$\endgroup\$
    – Junius
    Jul 31, 2016 at 19:33
  • \$\begingroup\$ @dmitryvm Correct. Updated the text. 10x. \$\endgroup\$ Jul 31, 2016 at 20:43
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The formula with a minus (-) is for the electromotive force (EMF). Without a minus, it's for the voltage drop (or potential difference).

The voltage drop uses the convention used in electronics that the current goes from + to - (using the flow of positive charges --> holes), so it's measured from the positive terminal (the one with more potential) to the negative one (the one with less potential).

In my understanding, the electromotive force does not use this convention (else, why would it be the contrary of the voltage drop which uses the positive charges convention?) --> it uses the electrons flow (negative charges), so it's measured from negative to positive terminal.

By the way, EMF is present only when the component is a generator. It's not just the opposite of the voltage drop --> it's only mentioned (so exists only) when the component is actually creating itself a difference in potential of its both terminals (hence its name, electromotive force). If the component is like a resistor, there's actually no EMF, since it's not producing no difference in potential on its terminals by itself (it's being subject to a difference of potential created by some generator, which has the EMF, or it wouldn't generate any difference in potential).


Therefore, for your example

Formula without a minus:

\${v_L} = L\frac{{di}}{{dt}}\$

The derivative of the current is positive, since current is rising (derivative and current in the same direction), so \$v_L\$ will be positive. \$v_L\$ is voltage drop (or potential difference), so it's measured from + to -. If it's positive and the current is going from up to down, then the + will be up and the - will be down.

Formula with a minus:

\$\epsilon =-L\frac{di}{dt}\$

Imagine for a second that the minus is not there. Then,

\$\epsilon =L\frac{di}{dt}\$

Again, the derivative is positive. The electromitive force is measured from negative to positive and the current is going from up to down. So, if the result is positive, that means the EMF is measured on the same direction as the current. Therefore, up would be a - and down would be a + (because it's measured with the flow of electrons, not holes).

But getting back to the actual formula with a minus, then what I just said is inverted --> + up and - down. And the idea still remains: the efm is measured from - to + (so down to up) and the voltage drop from + to - (so up to down, as in your drawing).

2 notes:

  • In the first case, the inductor would be working as a load (which it partially is) - being energized. So the current is trying to go from the + to the - from inside the inductor, like with all passive components, like a resistor.
  • In the second case, the inductor would be working as a generator (which it partially is) --> so the current is trying to go from the + to the - by outside of the inductor, like with all generators. This doesn't happen in a resistor. There's no EMF generated on it (it doesn't create by itself a difference in potential on its terminals).

Related to these notes (even though I believe off-topic to the question): https://physics.stackexchange.com/questions/184541/if-induced-voltage-back-emf-is-equal-and-opposite-to-applied-voltage-what-dri.


There are already answers here and one was accepted, but I thought I'd write it, since I've been trying to understand how inductors and capacitors store energy (I'm a student), and I thought maybe this approach (hopefully correct) might help understanding as it did with me. If anything here is wrong please someone tell me.

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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.

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