Faraday Law of Inductance

Thread Starter

ben sorenson

Joined Feb 28, 2022
90
Hello, I was wondering on the Faraday Law Calculator where DT is the rate of change. The value entered would be how fast the electricity is being turned on/off correct? So if you had a signal of 60hz would the rate be seconds with a value of 60?
 

BobTPH

Joined Jun 5, 2013
5,803
It is a differential. I assume dI/dt, which is the change in current per unit time. For example,
3 Amps per second.
 

WBahn

Joined Mar 31, 2012
27,478
Hello, I was wondering on the Faraday Law Calculator where DT is the rate of change. The value entered would be how fast the electricity is being turned on/off correct? So if you had a signal of 60hz would the rate be seconds with a value of 60?
When you say, " the Faraday Law Calculator," are you referring to some website? If so, it would be helpful to have a link to it so that we don't have to guess what they are referring to.

DT (or, more usually, dt) is usually interpreted as a small amount of time. It is normally seen as part of a ratio, such as di/dt, which would be a small change in current divided by a small change in time, which would then be described as the rate of change of current.

It is an instantaneous value describing what is happening at a particular moment in time. This is often approximated by the average change over a very small amount of time.

This is entirely different from the number of cycles of something per second, which is what hertz measures.
 

WBahn

Joined Mar 31, 2012
27,478
You, of course, meant
V = -L di/dt​
The minus sign is very important as it shows the generated voltage opposes the change in current.
While Faraday's Law has the minus sign, reflecting Lenz's Law, the constitutive equation for an inductor does not.

By the passive sign convention, the symbolic voltage across an inductor is defined to be positive at the terminal where the symbolic current enters.

If you apply a positive voltage across an inductor having no current in it, then current flow begins in the direction of positive current and increases at a rate proportional to the applied voltage, hence there can't be a minus sign.

If you were to charge up a 10 H superconductor inductor to say +20 A of current (using whatever method you wanted to) and then reduced the voltage across it to zero so that it remains charged at 20 A and then put a resistor 100 Ω resistor across it, no current would flow in the resistor (because it has zero voltage across it). If you now open circuited the voltage source, the inductor will create whatever voltage is needed to keep 20 A flowing in it. This means creating a voltage across the resistor of 2000 V. But what is the polarity of this voltage? It has to be positive at the end of the inductor that current is flowing out of relative to the other end in order to get 20 A flowing in the correct direction through the resistor, so the voltage across the inductor has to be negative. But what is di/dt? Since energy is being drawn out of the inductor and dissipated in the resistor, the magnitude of the current must be decreasing which, since it is +20 A, means that di/dt must be negative. That is where the negative sign for the voltage comes from, not from a minus sign in the equation. For completeness, di/dt = -200 A/s
 

MrAl

Joined Jun 17, 2014
9,555
Hello, I was wondering on the Faraday Law Calculator where DT is the rate of change. The value entered would be how fast the electricity is being turned on/off correct? So if you had a signal of 60hz would the rate be seconds with a value of 60?
I think we need to see what formula you are using but more generally Faraday's Law of Induction is:
E=-N*dPhi/dt
where
E is the emf produced over time,
N is the number of turns (or just loops of wire),
Phi is the flux linking the turns N,
dt is the elemental time which can sometimes be approximated with a very short time period,
dPhi/dt is the time rate of change of the flux.

So what this says basically is that as the flux changes, a voltage is produced, and the amplitude of that voltage is proportional to how fast the flux changes. That means even with the same level of flux you can get a higher voltage if it changes faster than if it changes slower.

There are related laws that you may be looking for that relate the frequency to the voltage and induction as used in what is sometimes called the Transformer Equation. Interestingly, frequency is in the denominator so the higher the frequency the less peak induction.
There is a relationship here because higher frequencies change faster than lower frequencies and thus the flux will follow. This means that higher frequencies will cause more voltage. You really have to get into this for it to be any use to you though.
 

MrAl

Joined Jun 17, 2014
9,555
As the TS specifically mentioned Faraday, I stand by my assertion the minus sign is relevant. :)
The point being made by Mr. WBaln was that you do not use that sign in regular circuit analysis, you use it in pure physics.
Regular circuit analysis is different in that we use conventional current flow and assume that many things are positive that really are not. It simplifies the understanding of some things. It probably started way back when it was thought that current really did flow from the most positive to the most negative when really electron current is the opposite.
So the form in network circuit analysis would be:
V=L*di/dt
which has no minus sign while in physics it would be:
V=-N*dPhi/dt
which includes the minus sign.
So it's not like you are wrong it is just not necessary in regular circuit analysis unless you need it for some physical reason (such as maybe electrolysis which often deals with electron current flow). If you tried to make it negative in regular circuit analysis though you might end up with some unreal values because the other conventions would conflict with that definition and thus you might end up with a negative voltage when in real life it is positive :) or worse yet a positive voltage when it is really negative and thus blow out your million dollar solar panel :)
 

WBahn

Joined Mar 31, 2012
27,478
It has nothing to due with electron flow or conventional current. Physicists use conventional current flow because they define current as the time rate change of charge. It's the electron-flow crowd that pretend electrons are positively charged and then throw in magical mystery minus signs when they need to patch things up.

Faraday's Law has the minus sign because it is relating voltage to a change in flux, while the constitutive equation for an inductor is relating voltage to current in the inductor.
 

MrAl

Joined Jun 17, 2014
9,555
Some laws are written in terms of electron flow not conventional current flow. When you go to measure something you have to know that.
I'll see if i can find some write up on this on the web or in one of my physics books of old.
 

WBahn

Joined Mar 31, 2012
27,478
Look in 99.9% of physics textbooks and you will see that current is consistently defined as the flow of charge, where charge is a signed quantity that can be either positive or negative. All of the laws are developed in terms of charge, not of electrons. There is no sense current ever being defined as the flow of electrons and somehow being positive in the direction that electrons are moving.

It all starts from Coulomb's Law, which states that the force on charge Q1 exerted by charge Q2 is given by

\(
\textbf{F}_{12} \; = \; \frac{kQ_1Q_2}{r^2} \hat{\textbf{r}}_{12}
\)

where the r^ is the unit vector from Q1 to Q2.

Since F_12's (symbolic) direction is in the same direction as r_hat, it is always a factor pointing away from Q1. But because Q1 and Q2 are both signed signed quantities, the magnitude of F_12 may be either positive or negative.

The electric field at a point is then defined as a vector quantity such that the force on a point charge, q, located at that point is

\(
\textbf{F} \; = \; q \textbf{E}
\)

Note that, again, q is a signed quantity. If q has positive charge, the force on it is in the direction of the electric field at that point, while if it is negative, the force is in the opposite direction. The electric field itself is completely independent of the test charge and depends on neither it's sign nor magnitude.

Next comes the definition of electrostatic potential, which is defined such that the difference in electric potential of Point B relative to Point A is defined by the external work that must be done, per unit charge, on a point charge, q, that experience only a force due to an electric field and the external force, to move it from Point A to Point B with no change in the kinetic energy of the charge. A very slight bit of algebra and calculus results in

\(
V_B \; - \; V_A \; = \; \int_B^A {\textbf{E} \; \cdot \; \textbf{ds}}
\)

Note, yet again, the there is no mention of electrons. Voltage is defined in terms of electric fields, which are defined in terms of generic charge distributions.

Then comes the definition of current, which is simply

\(
i \; = \; \frac{dq}{dt}
\)

Yet once more, no mention of electrons. It is the time-rate-change of charge. Period.

Up next is the definition of magnetic field. This is defined such that the magnetic force, F_m, felt by a charge q that is moving with a velocity v relative to a magnetic field B feels a force given by

\(
\textbf{F}_m \; = \; q \textbf{v} \; \times \; \textbf{B}
\)

Again, not distinction regarding electrons. The definitions are all in terms of charge, be it positive, negative, or zero.

This defines B in terms of how it effects moving charges, but next is the Biot-Savart Law, which relates moving charges to the magnetic field they create at some point P.

\(
d\textbf{B} \; = \; = \frac{\mu_0}{4\pi} \; \frac{i \; d\textbf{l} \; \times \hat{\textbf{r}}}{r^2}
\)

where dl is a differential vector in the direction of the current (which, again, is merely defined as the rate of change of charge) and r is a unit vector pointing from the current element to the point where B is being evaluated.

Again, nothing has been mentioned about electrons, let alone treating them in any way different because we long ago declared them to have a negative charge.

Next is magnetic flux through a surface, which is defined as the surface integral of the component of the magnetic field normal to the surface.

\(
\phi_m \; = \; \int_S \textbf{B} \; \cdot d\textbf{S}
\)

Now we finally get to Faraday's and Lenz's Law, which says that any change in the total magnetic flux for a surface bounded by a closed path induced an emf along that path.

\(
\mathcal{E}_i \; = \; - \; \frac{d\phi_m}{dt}
\)

That minus sign is the combination of two factors. The first is Lenz's Law and the second is the definition of voltage along a path as defined previously.

Note, again, nothing changes if we decide to have our charge carries be electrons because everything is defined in terms of the forces on an arbitrary charge that is a signed quantity.

This brings us to inductance.

If you crank through the consequences of the above definitions, you will see that the magnetic flux through an isolated loop is proportional to the the current, i, in the loop. We call the proportionality constant the inductance of the loop and traditionally use the symbol L

\(
\phi_m \; = \; Li
\)

Notice there is no minus sign here. The minus sign due to Lenz's Law has NO applicability to this equality. This is not talking about changing anything or electric potentials of any kind. It is merely stating that the magnetic flux through a surface is proportional to the current that defines that surface.

When you crank this through and note the difference in the definitions of induced emf and the constitutive equation for an inductor per the passive sign convention, you get

\(
V \; = \; L \; \frac{di}{dt}
\)
 

MrAl

Joined Jun 17, 2014
9,555
Hello,

I have to say that's a very nice post and very clear and it's nice that you took the time to write that and make it so clear and i mean that.
Also, it doesnt matter what i was talking about it's still very informative, even though there are other things not being talked about yet.
<joking mode ON>
So if that's what makes up the universe, then i guess i can easily see that some kind of particles are flying off of the back of my CRT oscilloscope screen right into the electron gun. Oh wait, i mean the charge vacuum catchers mitt (ha ha). I guess that also means that in a bubble chamber all the particle paths curve in the same direction (ha ha).
<joking mode OFF>
I still like your post it's nice to see all that in one place at the same time.
The Biot-Savart Law is one of my favorites, i was able to use that fairly fundamental idea to develop a "magic" inductance expression that in theory can solve every inductance problem such as self inductance and mutual inductance. I was bothered by a book i purchased that had a huge amount of information in it about inductance but did not go into the theory behind all the charts and tables too well, although it did a little bit. I figured it was time to develop a universal expression, although the integrations are not always easy to perform.
 
Last edited:

WBahn

Joined Mar 31, 2012
27,478
Hello,

I have to say that's a very nice post and very clear and it's nice that you took the time to write that and make it so clear and i mean that.
Also, it doesnt matter what i was talking about it's still very informative, even though there are other things not being talked about yet.
<joking mode ON>
So if that's what makes up the universe, then i guess i can easily see that some kind of particles are flying off of the back of my CRT oscilloscope screen right into the electron gun. Oh wait, i mean the charge vacuum catchers mitt (ha ha). I guess that also means that in a bubble chamber all the particle paths curve in the same direction (ha ha).
<joking mode OFF>
I still like your post it's nice to see all that in one place at the same time.
The Biot-Savart Law is one of my favorites, i was able to use that fairly fundamental idea to develop a "magic" inductance expression that in theory can solve every inductance problem such as self inductance and mutual inductance. I was bothered by a book i purchased that had a huge amount of information in it about inductance but did not go into the theory behind all the charts and tables too well, although it did a little bit. I figured it was time to develop a universal expression, although the integrations are not always easy to perform.
Despite your marking it as "joking mode", it still underscores the inability to distinguish between charge and charge carriers. When an electron beam goes from the gun to the CRT screen, the charge carries are going from the gun to the screen. But the charge that is going from the gun to the screen is negative. So go ahead and define the current as being in the direction of the charge carriers (i.e., from the gun to the screen) (i.e., "electron flow"). The measure of current is CHARGE per unit TIME, not CHARGE-CARRIERS per unit time. So if you have 6.24x10^18 electrons flowing toward the screen per second, how much CHARGE is flowing toward the screen every second? Simple, that many electrons have a charge of -1.00 coulombs, which means that you have a current of -1.00 coulombs per second, which by definition is -1.00 amperes. That negative sign does NOT mean that the charge carriers are moving away from the screen! It is completely consistent with with negatively charged charge carriers moving toward the screen.

But the "electron flow" crown universally would say that the current is 1.00 A toward the screen. They will insist that one coulomb per second is hitting the screen. But now ask them what the charge on the screen screen will be after one second if there was no return path (assuming the screen was initially uncharged and also assuming that the current could be maintained for one second in the face of the growing charge) and they will say -1.00 coulombs. But how can that be? How you you deposit one coulomb per second of charge on something that was initially uncharged and, after one second, have a negative charge? The only way is through the application of a magical mystery minus sign to cover up the improper math.
 

MrAl

Joined Jun 17, 2014
9,555
Despite your marking it as "joking mode", it still underscores the inability to distinguish between charge and charge carriers. When an electron beam goes from the gun to the CRT screen, the charge carries are going from the gun to the screen. But the charge that is going from the gun to the screen is negative. So go ahead and define the current as being in the direction of the charge carriers (i.e., from the gun to the screen) (i.e., "electron flow"). The measure of current is CHARGE per unit TIME, not CHARGE-CARRIERS per unit time. So if you have 6.24x10^18 electrons flowing toward the screen per second, how much CHARGE is flowing toward the screen every second? Simple, that many electrons have a charge of -1.00 coulombs, which means that you have a current of -1.00 coulombs per second, which by definition is -1.00 amperes. That negative sign does NOT mean that the charge carriers are moving away from the screen! It is completely consistent with with negatively charged charge carriers moving toward the screen.

But the "electron flow" crown universally would say that the current is 1.00 A toward the screen. They will insist that one coulomb per second is hitting the screen. But now ask them what the charge on the screen screen will be after one second if there was no return path (assuming the screen was initially uncharged and also assuming that the current could be maintained for one second in the face of the growing charge) and they will say -1.00 coulombs. But how can that be? How you you deposit one coulomb per second of charge on something that was initially uncharged and, after one second, have a negative charge? The only way is through the application of a magical mystery minus sign to cover up the improper math.

I think what you are saying is that if you reverse one thing you have to reverse everything, and then everything works again.
But i fear this means that we have to ignore the fact that electrons actually do flow only one way.
 

WBahn

Joined Mar 31, 2012
27,478
I think what you are saying is that if you reverse one thing you have to reverse everything, and then everything works again.
But i fear this means that we have to ignore the fact that electrons actually do flow only one way.
What we have to NOT ignore is that circuit equations describe the flow of charge, not of charge carriers. They say nothing about what direction the charge carriers are physically moving. That's true for vacuum tubes. That's true for electron beams and ion beams. That's true for semiconductors. When we need to focus on the movement of the charge carriers, then we can, and do, focus on the movement of the charge carriers. But when we do, we are working with the movement of charge carriers, not the movement of charge. The one instance I am aware of where the circuit equations are sensitive to the nature of the charge carriers is the Hall Effect.
 

MrAl

Joined Jun 17, 2014
9,555
What we have to NOT ignore is that circuit equations describe the flow of charge, not of charge carriers. They say nothing about what direction the charge carriers are physically moving. That's true for vacuum tubes. That's true for electron beams and ion beams. That's true for semiconductors. When we need to focus on the movement of the charge carriers, then we can, and do, focus on the movement of the charge carriers. But when we do, we are working with the movement of charge carriers, not the movement of charge. The one instance I am aware of where the circuit equations are sensitive to the nature of the charge carriers is the Hall Effect.
Ok great that is what i was trying to get at. I was going to say that 'charge' does not magically flow from one place to another. It moves from a place of higher potential to lower potential, and you cant get higher or lower without adding or removing electrons. If we shot 10 negative charges at the screen and 10 positive charges the net would be 0, but we could not do that without knowing that electrons have a negative charge.
Maybe i am not explaining my point right though. I tend to go toward the most physically 'real'.
Granted we dont need this for V=L*di/dt no problem there.
 
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