# How does Faraday's law include self inductance?

Conductivity
A varying magnetic flux causes an induced voltage (according to Faraday-Lenz law) that generates a current that generates a magnetic field that generates another magnetic flux in the same circuit. Does the Faraday-Lenz law include this self induction? In other words why can't we take that induced magnetic field and use it again inside lenz law?

Another question, When you connect a battery to a circuit, and we graph Current versus time. The current slowly rises to the maximum current possible is this due to magnetic effects? or just transient state until the charges settle into steady state?

Jony130
A varying magnetic flux causes an induced voltage (according to Faraday-Lenz law) that generates a current that generates a magnetic field that generates another magnetic flux in the same circuit. Does the Faraday-Lenz law include this self induction? In other words why can't we take that induced magnetic field and use it again inside lenz law?
I don't understand, could you please explain what you mean?

But maybe this will help
Why is there any voltage even present across the inductor? We always accept a voltage across a resistor without argument because we know Ohm’s law (V = I × R) all too well. But an inductor has (almost) no resistance it is basically just a length of solid conducting copper wire (wound on a certain core). So how does it manage to “hold-off” any voltage across it?
In fact, we are comfortable about the fact that a capacitor can hold voltage across it. But for the inductor, we are not very clear!
A mysterious electric field somewhere inside the inductor! Where did that come from?
It turns out, that according to Lenz and/or Faraday, the current takes time to build up in an inductor only because of ‘induced voltage.’ This voltage, by definition, opposes any external effort to change the existing flux (or current) in an inductor. So if the current is fixed, yes, there is no voltage present across the inductor, it then behaves just as a piece of conducting wire. But the moment we try to change the current, we get an induced voltage across it. By definition, the voltage measured across an inductor at any moment (whether the switch is open or closed) is the ‘induced voltage.’

http://booksite.elsevier.com/samplechapters/9780750679701/9780750679701.PDF (page 22 Understanding the Inductor)

When you connect a battery to a circuit, and we graph Current versus time. The current slowly rises to the maximum current possible is this due to magnetic effects?
Yes, this rise is due to magnetic flux build up. This rise "stops" if we saturate the core (and jump to Vin/R_coil) or current reach value equal to Vin/R_coil or some other element (if we have an ideal inductor ) limits the current.

Conductivity
Like isn't there an induced magnetic field going through the loop? Why not take it into account while calculating the change in flux? as if it was a 2nd magnetic field?

About current, assume we turn off all the magnetic effects and there are no inductors. If we graph I and T will we get the same shape roughly like rising to maximum?

Staff Emeritus
Current in the coil modifies the magnetic field the coil finds itself in, and the field due to the current is directed so as to oppose that which gave rise to the current. In any case, the voltage induced is always determined by the rate of change of the net flux, the vector sum of all.

In a circuit where there is no inductance, current is free to instantly jump from one value to another. Current doesn't have inertia, though you could say that current flowing in an inductor does seem to have inertia, although this is a property of the inductance rather than the current.

Conductivity
Current in the coil modifies the magnetic field the coil finds itself in, and the field due to the current is directed so as to oppose that which gave rise to the current. In any case, the voltage induced is always determined by the rate of change of the net flux, the vector sum of all.

In a circuit where there is no inductance, current is free to instantly jump from one value to another. Current doesn't have inertia, though you could say that current flowing in an inductor does seem to have inertia, although this is a property of the inductance rather than the current.
"Rate of change of net flux". So why do we only consider the change of flux (Not from the loop, outside source of magnetic field) only?
For example, dB/dt of a magnet affect a loop without considering the rate of change of the magnetic field induced by the loop?

Staff Emeritus
If there is not going to be much current in the coil (due to a high series resistance) then where the applied field is strong you may overlook B due to the coil's weak current. Otherwise, it will need to be taken into account.

Homework Helper
Gold Member
I wonder if it might help to put this question into a context?
It may be that in some cases people do consider the effect of secondary currents, but in a question where one is only asked to calculate the induced emf, no secondary current is required to flow?

With a transformer the flux produced by the primary depends on the primary voltage and the primary inductance. A secondary emf is caused. If then a secondary current is allowed to flow, this would indeed counter the primary flux, but then the primary back emf would fall, allowing additional primary current flow to maintain the flux at the original level.

• cabraham
Conductivity
If there is not going to be much current in the coil (due to a high series resistance) then where the applied field is strong you may overlook B due to the coil's weak current. Otherwise, it will need to be taken into account.
But if let's say it is strong compared to to the applied B, Just for curiosity. In school, we always overlook the effect of that induced MF

How would they calculate the rate of change of the induced MF as it depends on the applied MF? You would have to use the law twice or what?

Staff Emeritus
You'd combine Ohm's Law and Faraday's Law into a differential equation involving ##\mathsf{ \vec B}## and ##\mathsf{ \dfrac {d\vec {B}}{dt}}## and solve it.

• Conductivity
Homework Helper
Gold Member
But if let's say it is strong compared to to the applied B, Just for curiosity. In school, we always overlook the effect of that induced MF
You'll need to use differential equations then, as NascentOxygen said earlier.

In practice, this situation is observed in current transformers. In a CT, the net flux in the core is equal to primary flux-secondary flux. When the CT secondary is shorted or connected to a very low resistance burden, the secondary flux is almost equal to the primary flux. This leaves only a little flux in the core and hence the voltage induced in the secondary is also very small. If the secondary resistance is increased, secondary current decreases and hence, total flux in the core increases. This means the secondary voltage also increases. Further if you open circuited the secondary of a CT, there would be no secondary current and net core flux would be the primary flux only, which is very high. This will induce very high voltage across the CT secondary and could destroy the coils.
Hence, an important precaution in CT operation is "never open circuit the secondary of a CT while in operation".

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• Conductivity
Conductivity
Could you give an example? Just to confirm?
But don't you have to use faraday's law neglecting the induced magnetic field to get a differential equation of the net field?

cabraham
The self inductance is included in Faraday's law, but it is subtle. The flux density "B" is the NET flux density, which is the superposition of external & internal due to loop current. For a superconductor ring immersed in an external B, the induced loop current generates an internal B equal & opposite to Bext. The net is zero. Since voltage around loop = -N*d (phi)/dt, there is zero E field inside the superconductor ring, consistent with Ohm's law, J = sigma*E.

Claude