# Faraday's law - does the induced current's magnetic field affect the change in flux?

I've had this conceptual problem with Faraday's law and inductance for a while now.

Take the example of a simple current loop with increasing area in a constant field (as in this answer). So Faraday's law states that the increasing flux (due to the increasing area) causes an EMF and hence a current. Due to the minus sign in Faraday's law, or by Lenz's law the direction of the current is such that the magnetic field it creates opposes the external field.

Now why do we never consider the magnetic field created by the induced current, when calculating the change in flux? All workings I have seen always calculate the flux from $\underline{\mathbf B} \cdot \underline{\mathbf A}(t)$. Why is $\underline{\mathbf B}$ not adjusted by the induced magnetic field? Is it just that small that we can neglect it unconditionally?

I have the same problem with self-inductance in AC circuits (although, maybe if I understood the above problem, this would become apparent to me as well). Say we start from current $I=0 \text{A}$. Then the EMF in the circuit increases (but is still very low), which increases $I$, which in turn creates an increasing $\underline{\mathbf B}$ inside the coil. Wouldn't the induced counter-EMF be much greater than the external EMF applied to the circuit? And if so, how come there is current moving in the first place, if the slightest increase in EMF causes a counter-EMF which acts to stop the current?

Is it just that I am looking at idealised situations or that the magnitudes of the external and induced effects differ greatly? Or do I have a conceptual misunderstanding about how (self-)inductance works?

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I see two questions here. The first is why self-inductance is not considered when solving Faraday's law problems, and the second is why an EMF can ever produce a current in a circuit with non-zero self-inductance. I will answer both of these in turn.

# 1. Why self-inductance is not considered when solving Faraday's law problems

Self inductance should be considered, but is left out for simplicity. So for example, if you have a planar circuit with inductance $L$, resistance $R$, area $A$, and there is a magnetic field of strength $B$ normal to the plane of the circuit, then the EMF is given by $\mathcal{E}=-L \dot{I} - A \dot{B}$.

This means, for example, that if $\dot{B}$ is constant, then, setting $IR=\mathcal{E}$, we find $\dot{I} = -\frac{R}{L} I - \frac{A}{L} \dot{B}$. If the current is $0$ at $t=0$, then for $t>0$ the current is given by $I(t)=-\frac{A}{R} \dot{B} \left(1-\exp(\frac{-t}{L/R}) \right)$. At very late times $t \gg \frac{L}{R}$, the current is $-\frac{A \dot{B}}{R}$, as you would find by ignoring the inductance. However, at early times, the inductance prevents a sudden jump of the current to this value, so there is a factor of $1-\exp(\frac{-t}{L/R})$, which causes a smooth increase in the current.

# 2. Why an EMF can ever produce a current in a circuit with non-zero self-inductance.

You are worried that EMF caused by the circuit's inductance will prevent any current from flowing. Consider the planar circuit as in part one, and suppose there is a external emf $V$ applied to the circuit (and no longer any external magnetic field). The easiest way to see that current will flow is by making an analogy with classical mechanics: the current $I$ is analogous to a velocty $v$; the resistance is analogous to a drag term, since it represents dissipation; the inductance is like mass, since the inductance opposes a change in the current the same way a mass opposes a change in velocity; and the EMF $V$ is analogous to a force. Now you have no problem believing that if you push on an object in a viscous fluid it will start moving, so you should have no problem believing that a current will start to flow.

To analyze the math, all we have to do is replace $-A \dot{B}$ by $V$ in our previous equations, we find the current is $I(t) = \frac{V}{R} \left(1-\exp(\frac{-t}{L/R}) \right)$, so as before the current increases smoothly from $0$ to its value $\frac{V}{R}$ at $t=\infty$.

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So in short, the self-inductance accounts for transient regimes, which is often omitted in lecture discussing quasi-DC effects. Without $L$ (but then also without $R$), the current establishes instantaneously. Since $L/R$ is the natural time scale by dimension argument, one can wonder why we need to introduce both $L$ and $R$ in the non-instantaneous argument. This is because an inductance can store energy, which must be dissipated. An other way to see this is saying that $L$ introduces delay in the circuit, which must be compensated by advance (capacitance) or dissipation. –  FraSchelle Apr 30 '13 at 7:20
Thanks a lot guys, that's really helpful! I'll play around with Mathematica a bit; I guess some plots will help my understanding as well. (Also, wow, did you actually make an account just answer that? Welcome to physics.stackexchange.com!) –  Martin Büttner Apr 30 '13 at 7:54
Actually I made my account to ask a question on the Mathematica stack exchange. But I figured I would try to answer a few other questions. It's kind of fun in a way. –  NowIGetToLearnWhatAHeadIs Apr 30 '13 at 11:47

Yes you're correct . We willingly don't consider it , it is perfectly correct . However , like in a DC circuit , we think current is established instantaneously . It isn't , first there is a very Induced EMF , and no current flows in circuit . and then there's falling of the EMF with time and it is only at$\infty$ time , that we get the so called current . In circuit theory we assume this phenomena didn't happen or that we have seen the circuit after $\infty$ time .

And how big that $\vec{B}$ is depends on the battery in your circuit , which is generally ignored as convention.

Dr. Walter Lewin discusses this too in one of his lectures .

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I intuitively feel that there is some relation between self-inductance and inertia. Actually, self-inductance is basically an electromagnetic inertia. The only difference is that moving mass is closed, concealed inside the massive body, electric inertia has its guts outside -- we can see its lines of force. If flux shows nothing more than the current, then increase of current (related with EMF) must correspond with related increase in flux. Basically what you see, is increase in current whether you measure by ampermetre or magnetic field. Applied EMF, current does not jump to infinity (assume no resistance, for simplicity). Self-inductance serves as inertia to prevent this.

Think it as mass. You apply force, the mass resists. You may think that it reacts as much to prevent any speed increase. Yes, mass creates the opposite force (EMF) to prevent acceleration. It is the same question, IMO.

Here is the feedback: you apply force. Mass reacts by creating counterforce (3rd law of Newton, "To every action there is always an equal and opposite reaction"). Now, you think, wouldn't the induced counter-EMF be much greater than the external EMF applied to the circuit? And if so, how come there is current moving in the first place, if the slightest increase in EMF causes a counter-EMF which acts to stop the current? Do you recall your thoughts? Why don't you ask the same question regarding Newton mechanics?

This question is discussed here Why body starts moving when force is applied?

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