# Can Lenz's rule really be used to determine the direction of induced current?

Faraday's Law states that the induced emf in a closed circuit is the negative of the rate of change of magnetic flux through the surface enclosed by the loop. To obtain the direction of the induced current, we may use Lenz Law, which states that emf will be induced so as to oppose the change that caused it, namely the change in magnetic flux. But I have a doubt regarding this.

The flux that appears in Faraday's law is not just the flux due to the external magnetic field, but rather the net flux due to both the external field and that produced by the current itself. So, the "change" that causes the induced current is due to the induced current as well. That said, how do we know which direction the emf will be induced in?

What if the flux due to the current has a greater magnitude that the flux due to the external field? Does that mean that the emf would now be induced to counteract this flux? I am confused.

• The induced current is born to oppose its sourc, why will it add to the flux of the external magnetic field? – Lelouch Jul 10 '16 at 7:20
• On the contrary it will decrease the flux due to the external field – Lelouch Jul 10 '16 at 7:21
• Maxwells equation refers to the total B field, which includes that of the induced current as well, but for practical purposes we neglect it due to its small contribution. – Lelouch Jul 10 '16 at 7:57
• Yes, but how can we be sure that the flux change ($\Delta \Phi_i,$ say), due to the induced current is less than the 'original' flux change ($\Delta \Phi_o,$ say)? That's the kernel of the question, surely. – Philip Wood Nov 23 '18 at 11:20

Suppose that, at time t = 0, a steadily rising, externally-generated magnetic flux, given by$$\Phi_{ext}=\lambda t$$begins to pass through the coil, which we assume to be in a closed circuit of total resistance $$R$$.

The induced emf, $$\mathscr{E},$$ will be given by$$\mathscr{E}=-\frac{d}{dt}(\Phi_{ext}+\Phi_i),$$in which $$\Phi_i$$ is the flux due to the current, $$\mathscr{E}/R$$. In fact, we can write$$\Phi_i=\frac{L}{R}\mathscr{E}$$in which L is the coil's self inductance (to be assumed constant).

If we substitute from the first and last of these equations into the middle one, separate variables and integrate between $$t=0$$ (when $$\Phi_{i}=0)$$ and $$t=t$$ we find $$\Phi_{i}=-\lambda \frac{L}{R} (1-e^{-\frac{R}{L}t}).$$ Differentiating, $$\frac{d\Phi_{i}}{dt}=-\lambda\ e^{-\frac{R}{L}t}$$that is $$\frac{d\Phi_{i}}{dt}=-\frac{d\Phi_{ext}}{dt}\ e^{-\frac{R}{L}t}$$

We've shown that the rate of change of the induced flux is opposite in direction to that of the external flux, but that (because the exponential factor drops from 1 to 0 as time goes on) its magnitude is never greater than that of the external flux. Therefore we've saved Lenz's law (even if we imagine that the law is referring to the opposing of change in external flux, rather than change in total flux).

Notes

1. At time $$t=0$$ it would seem that the rates of change are equal and opposite.

2. This approach seems to me somewhat clumsy. I'd guess that there's a much slicker method.

• This looks like the right approach to me, but I'm not sure how you got to your fourth equation. Can you fill in a step or two there? – Michael Seifert Nov 23 '18 at 14:48
• Also, I agree there should be a "slicker method", probably based on the underlying differential equation$$\frac{d\Phi_i}{dt} + \frac{R}{L} \Phi_i = - \frac{d \Phi_{ext}}{dt}$$and an argument about the signs of $\dot{\Phi}_i$ in the solutions of said equations. – Michael Seifert Nov 23 '18 at 14:50
• @Michael Seifert (a) I first solved the DE to get $\Phi_i$ as a function of $t$. Then I differentiated it. If I have a moment or two I'll put in some more steps. (b) My thoughts entirely, though I wondered if we had to assume that the external flux changed linearly with time. I can't believe that this assumption is necessary, but having played for a while I couldn't come up with anything quite as convincing as what I posted. – Philip Wood Nov 23 '18 at 15:05

Yes it can, in all cases with a 100% certainty. As to why, the best way to understand would be to imagine, have a physical model, conception in the mind about how electrons behave and what makes them to do so.

The short answer (without going into much detail about electrostatic force):

• When the Aether theory was made and accepted, thinkers said, that for the observed effects to work (like thunder and any kind of EM wave propagation in general), there have to be a medium with determined properties, that are even smaller than electron and proton, lets say sub-atomic particles (not exactly correct, but lets stick to it)
• It fills the known universe, and it is the basic building material in many form for other complex, bigger, slower particles (like electron and proton which form atom that is 90%+ empty for its occupied operational space - that is how you make a bigger, less stable particle out of smaller, more active ones, like electron-proton->atom->molecule->cell->body..etc), and responsible for their interaction, and behavior (like electrostatic forces between so called positive and negative particles)

• putting the proton aside for a moment, the analogy for the interaction of electrostatic forces, magnetic fields (which is none other, than the directed movement of the same particles that responsible for the electrostatic forces) upon the electron, would be like how air can act upon a wind generator. They are made up of the same substance, yet with a very different structure, gives a very different physical properties ( the building material for the air molecules, electrons and protons could hardly operate a wind generator with its specific engineered way, it was designed for moving air molecules just as electron and proton was made for a specific way to operate with a specific medium).

• when you move or change the magnetic field around an electron( or many), that means the electron is not sitting in a stable, balanced evironment as before, but get a push/force from one side. That makes the electron moving. We already know that when electron move in this particle mass, because of the electron specific curved structure they spin counterclock wise, meaning to the left if you see it from behind. However this rotational movement generate a circular motion of the aether particles around the electron while freely rotating, like a swung pinwheel in a child's hand, and that rotating particle mass in defined as the generated magnetic field of that electron. This magnetic field will always weaken the one made the electrone move.

• Meaning: out of the 1000 or so particle you used to force the electron into movement some of them will be caught up in the generated rotating particles by the spining and forward moving electron, and not reaching, thus not contributing to the electron'S additional movement/acceleration reaching a balance, where the changing magnetic field at any moment will be equal to the one generated by the moving electron.

Answering the last question, flux to the current can never be greater in magnitude than flux to the external field, and even if it is for a moment, it will weaken the force acting upon the electron, slowing it down, and generating less 'counter'flux and there goes the achived balance. A flux of an external field always force the electron to the best it can to move. The electron is designed (and not just the electron) to make an effort to counter this change, (might as well call it inertia), otherwise it would be an escalating reaction, causing some serius problem to the universe stability. One of the fundamental law of the universe.

Tip: In a coil, determine the magnetic field poles, that want to induce the current (increasing, decreasing, south-north polarity direction...etc). Lenz's law states that the induced current movement will be so, that the magnetic field generated by it would oppose the flux, what made the electrons move. We know that electrons spin left (looking from behind or should i say into the paper) when moved. if you add all of them in a coil turns circulating, you get a net magnetic field inside and outside of the coil (dipole orientation depends on the direction of the current) . The induced current direction is, when the external and the induced flux oppose each other. Not that difficult, just picture it in your mind.