# Faraday's law and "infinite induction"

I may have confused after thinking too much about Faraday's law. If an emf is induced in a circuit due to some changing magnetic field, the induced current will be in a direction such that the "induced magnetic field" opposes the original magnetic field (Lenz's law).

So wouldn't the induced magnetic field also generate an induced current that opposes it, and that current would also generate a magnetic field and so on... so it seems like there would be infinitely many induced magnetic fields and currents. This is particularly true if the original magnetic field is a sinusoid so it's infinitely differentiable.

Am I misinterpreting Faraday's law? Is Faraday's law just a description of how the electric and magnetic fields can coexist, so we don't have to worry about "infinite induction"?

• "opposes the original magnetic field (Lenz's law)." is inaccurate. "opposes the change in the original magnetic field" is better. If the original magnetic field decreases, the induced magnetic field will be in the same direction as the original field to oppose the change. Apr 20, 2014 at 1:51
• You are this close to discovering electromagnetic radiation... Apr 20, 2014 at 1:51
• "... so it seems like there would be infinitely many induced magnetic fields and currents." the infinity here is generated by the verbal description of the interaction between the current-carrying parts of the circuit. Similar infinity happens often in physics, for example description of reflection of light between two reflective surfaces also involves such infinity. Alternatively, one can assume there is just one current density and one magnetic field that must obey some differential equations. These two descriptions are related and both are useful. Apr 20, 2014 at 2:00
• @JánLalinský I understand your point but I am somewhat confused as to what I should plug in to Faraday's equation. For example if I know there is a sinusoidally varying magnetic field, $\mathbf{B}(t) = A cos(\omega t + \phi)$, which is inducing an emf in this circuit, then I can plug $d\mathbf{B}/dt$ into Faraday's equation but I'm not satisfied because it seems there is also the effect of the induced magnetic field(s) that I should also taking into account. So would my calculation be an approximation for the induced emf? Is it a good approximation in practice? Apr 20, 2014 at 3:51
• If you plug-in just external $\mathbf B$ in the Faraday Maxwell equation, you are neglecting self-inductance effects. Whether that is accurate enough depends on whether the magnetic flux due to self-inductance effect is lower than the magnetic flux due to external field. If the former is low enough, it can be neglected with good accuracy. In case of DC electromotor or a generator, I think self-inductance is quite important since there is dense wiring inside. Apr 20, 2014 at 10:14

Here is one way to think about it:

When a charged particle travels in a magnetic field, it experiences a force. If the particle is stationary but the field is moving, then in the frame of reference of the field the particle should see the same force.

Now let's take a conductor wound into a coil. In order to increase the magnetic field inside, I could take a dipole magnet and move it close to the coil. As I do so, magnetic field lines cross the conductor, and generate a force on the charge carriers.

It is a convenient trick for figuring out "what goes where" to know that the induced current will flow so as to oppose the magnetic field change that generated it. In the perfect case of a superconductor, this "opposing" is perfect - this is the basis of magnetic levitation. For resistive conductors, the induced current is not quite sufficient to oppose the magnetic field, so some magnetic field is left.

The point is that the flowing of the current is instantaneous - it happens as the magnetic field tries to establish in the coil. So it's not "Apply field in coil. Coil notices, and generates an opposing field. " - instead, it is "Start to apply field in coil. Coil notices and prevents field getting to expected strength".

Not sure if this makes things any clearer...

If you have a thin circuit with a total resistance $R$, and place it in an external (changing) $\vec{B}$ field, then there is flux through the ring.

First, there is flux $\Phi_1$ from the external, changing $\vec{B}$ field. Since that $\vec{B}$ field is changing, there is an emf due to that.

Second, the current from the ring itself produces it's own $\vec{B}$ field, so it's own flux, $\Phi_2$. In a quasistatic approximation, you might say that this flux is proportional to the instantaneous current, $I$, through the circuit and denote the proportionality by $\Phi_2=LI$. If the current is changing, then that flux is also changing, so there is an emf due to that.

If the circuit were moving there might be a third contribution to the change in flux, let's ignore motional emf for now.

So together we have a total emf: $\mathscr E = -d(\Phi_1+\Phi_2)/dt$. Based on the resistance, we have $$RI=\mathscr E=-d(\Phi_1+\Phi_2)/dt=-d\Phi_1/dt-LdI/dt.$$

This is a differential equation, and the solution depends on how the external $\vec{B}$ field is changing (to get $-d\Phi_1/dt$). Even if that original external magnetic field is sinusoidal, we still have the external $\vec{B}$ and the internal (self inducuced) $\vec{B}$ field to deal with. The differential equation for the solution just depends on both.