# How does Faraday's law include self inductance?

• Conductivity
In summary, the current in a 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.

#### 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?

Conductivity said:
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)

Conductivity said:
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.

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?

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.

NascentOxygen said:
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?

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.

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
NascentOxygen said:
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?

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
Conductivity said:
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
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?

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

## 1. What is Faraday's Law and how does it relate to self-inductance?

Faraday's Law states that a changing magnetic field will induce an electromotive force (EMF) in a conducting loop. This EMF can lead to a current flow, which is known as self-inductance.

## 2. How does self-inductance affect the behavior of circuits?

Self-inductance can cause a delay in the response of a circuit to a changing input, known as inductive reactance. It also can create back EMF, which can limit the current flow in a circuit.

## 3. What factors influence the strength of self-inductance?

The strength of self-inductance is influenced by the number of turns in the conducting loop, the size and shape of the loop, and the magnetic properties of the material used in the loop.

## 4. How can self-inductance be calculated or measured?

Self-inductance can be calculated using the formula L = NΦ/I, where L is the inductance, N is the number of turns in the loop, Φ is the magnetic flux, and I is the current. It can also be measured using an inductance meter.

## 5. What are some real-world applications of Faraday's Law and self-inductance?

Faraday's Law and self-inductance have many practical applications, including in transformers, motors, generators, and electronic circuits. They are also used in wireless charging technology and in medical devices, such as MRI machines.