Talk:Inductance/Archive 1

‹See TfM›

Has anyone considered writing these articles for people who don't already understand what this is?

After all, what use is it to anyone? Someone who already understands this does not need to turn to this article. And those who have no idea but want a basic introductory explanation of what it is are left out. Great, so it is magnetic flux. Ok, now I go to that page and am inundated with another page full of abstruse verbage. Around I go, grasping at circular references, looking for some kind of conceptual foothold to begin with, but left with nothing but abstract mathematical symbols and explanations that may as well be in Greek. I mean, electical engineers aren't the only people interested in what this stuff is.

I couldn't agree more. This is supposed to be an encyclopedia entry, not a college level course. I had exactly the same problem you did. All I wanted was a basic explanation that was a little more complete than my vague grasp of the principles, but instead was bombarded with advanced material that belongs in a technical reference, with every link leading to more of the same. Most of the people who contributed to this ought to go find somewhere else to circle-jerk. 208.42.94.130 (talk) 08:25, 14 November 2008 (UTC)

Could someone please explain inductance in layman's terms? —Preceding unsigned comment added by 68.121.33.140 (talk) 03:02, 20 December 2008 (UTC)

Let me amplify this a bit. I have no problem with being presented with the math, eventually, but if I am trying to understand the subject, I need a handhold to grasp to get my feet up off the ground and start climbing. It takes a few more words and metaphors to get this across to someone who doesn't already know the subject intimately, and it only takes a quick reference card to get the subject across to those who already get it. So, if someone who was able to claw their way through the math and jargon to a real conceptual understanding of this material would be willing to spend a little time to help bring the rest of us along, it would be great. The power of Wiki is that the space is not really limited by anything, so this does not need to be an either-or proposition, please keep the technical depth, but provide a layman's framework in which to learn and understand the material as well. Visierl (talk) 14:42, 26 December 2008 (UTC)

Approximations in dependence on length of solenoid

The equation on the inductance of a solenoid states that it is inversely proportional to "l" (the length of the solenoid). This implies that inductance would be infinite for a zero-length solenoid. Certainly a single-winding solenoid could be made with a length arbitrarily close to 0. I think this equation is an approximation with assumptions (eg. $l\gg {\sqrt {A}}$). Could these assumptions please be stated? Jurgen Hissen 21:23, 18 May 2006 (UTC)

Need pictures

This page needs a picture of an inductor, and a drawing representation of it (showing turns), and the usual circuit representation for it. Maybe I'll draw these sometime. Fresheneesz 07:47, 22 November 2005 (UTC)

Mutual Inductance needs explaining

Mutual inductance needs a picture, needs to explain the dot convention, and give a picture for this. I might make these sometime as well. Also, mutual inductance need not occur between circuits - but can occur within circuit. Two nearby inductors can even have mutual inductance while remaining also in series. Fresheneesz 07:47, 22 November 2005 (UTC)

Error?

I can't even read the first sentence of this article. Shouldn't it read something like, inductance is the property of a device to produce a voltage proportional to a varying electrical current?

The expression given at the beginning, $v=-L{\frac {di}{dt}}$ is only correct for a circuit with a fixed geometry. If the circuit is contorted while current is flowing through it, then the change in geometry creates a back-emf, also, as shown below.

${\mathcal {E}}=-{\frac {d\Phi _{i}}{dt}}=-{\frac {d(L_{i}I_{i})}{dt}}=-({\frac {dL_{i}}{dt}}I_{i}+{\frac {dI_{i}}{dt}}L_{i})$

Question

(same poster) is the above assertion true? I have found no occurance in any reference texts.

Answer

I suppose one could look at it that way.

But the traditional way of analyzing motor/generators is to assume a fixed inductance for each part (the inductance of the rotating coil, and the inductance of the stationary coil), and calculating a separate "back emf" as a function of the speed of the motor.

I've always thought of a motor as "two fixed-geometry inductors that move relative to each other".

I've honestly never thought about considering a motor as "one big variable-geometry inductor". It's a perfectly self-consistent way to look at it.

But if I had some spinning motor, and someone asked me for "the inductance" of the motor, I'd probably just measure the voltage and current and plug in

$L={\frac {-v}{\frac {di}{dt}}}$

, completely ignoring the "variable inductance" of the motor taken as a whole.

--DavidCary 08:01, 29 Oct 2004 (UTC)

What is R in the mutual inductance equation? It also would help to clarify that ds is the increment along the curve.

-- Colin

Minor point

Is there a reason lower-case 'i' is used to denote the current here?

Yes, lower case i denotes current. : ) . Its that simple - its the standard way to denote current. Fresheneesz 07:41, 22 November 2005 (UTC)

Actually, upper case I denotes current. But it is a convention to use a lower-case i whenever the subject matter is regarding an ever-changing current. This generalizes to I for DC stuff and i for AC stuff.

Actually, that's not quite complete. Typically, I with capital subscript denotes quiescent current values while i with lower-case subscript denotes AC or small-signal current values. Then, i with capital subscript denotes total (quiescent plus small-signal) current values. Finally, I with lower case subscript denotes phasor current values. Alfred Centauri 23:55, 30 August 2006 (UTC)

What is the plural of a Henry?

I just pluralized "henry" and noted that I typed it "henries". But another appearance of the plural later on in the article was spelled "henrys". Is there an official plural for this unit? (Or is this like the ever-simmering British English versus American English debate?) If you know, could you please edit the article to make the two plurals consistent?

Atlant 11:42, 15 Jun 2005 (UTC)

NIST says it's henries. The IEEE says it's henrys. The IEE in 1947 (quoted in OED2) said it was henrys. The OED2 says it can be either. The Oxford Dictionary for Writers and Editors says henries. Personally, as a practising engineer, I am used to seeing henries and have never seen henrys used by an experienced writer, whether American or British. --Heron 19:55, 15 Jun 2005 (UTC)

I agree with your experience; thanks for confirming this and for cleaning up the article.Atlant 11:42, 16 Jun 2005 (UTC)

I don't think there is a plural. The unit Henry is similar to the unit Ohm or Hertz. All three don't have plurals. There is no Ohms or Hertzs, and there is no Henries or Henrys. It's Henry, names after Joseph Henry (see: Henry (unit)). THeeren (talk) 18:20, 16 January 2008 (UTC)

The article on the International System of Units specifically says it is henries (not Henries), at least in English, citing [NIST Special Publication 811: Guide for the Use of the International System of Units (SI)]. This is consistent with the general rule; see for example the article on the English plural, English Irregular Plural Nouns, or UNIT S4: YS OR IES?. There it states that for common nouns ending in a consonant followed by 'y', the 'y' changes to 'i' and the plural adds 'es', but that for proper names, the plural always ends in 'ys'. The plural of Henry should be Henrys, as in There are two Henrys in the class and the article on the War of the Three Henries (977–978) should be renamed, but the general rule says the plural of an object called a henry should always be henries. —Archimerged (talk) 02:17, 21 April 2008 (UTC)

Is mutual inductance good?

I'm still missing a feeling for what actually mutual inductance is. Is it something we want to have in a transformer? What would happen if it was very small (i.e. MI is zero), what, if it was a very big number (mutual inductance going to infinity)? Thank you, --Abdull 08:00, 21 July 2005 (UTC)

It depends. :-) If you're designing a transformer, mutual inductance is essential as its the method by which power is coupled from winding to winding. If you're designing a digital bus that transmits many bits in parallel, mutual inductance is bad as you probably want each of the bit lines to operate completely independently of the other bit lines and so they shouldn't couple energy from one line to its neighbours. Similarly, if you're using an assortment of filter inductors on a pc board, the odds are high that you don't want them inter-coupling with each other.

Atlant 14:33, 21 July 2005 (UTC)

Requested merge with Electromagnetic induction

I'm not sure why the merge request was made, but I think it's clear from the articles (or any physics textbook) that they're two separate concepts. -- SCZenz 14:21, 31 August 2005 (UTC)

Agree Light current 16:38, 31 August 2005 (UTC)

Me too. I call the request a silly thing. --Heron 17:13, 31 August 2005 (UTC)

Me three. However, I would hesitate to say that these subjects are separate concepts since the inductance of a conductor is a result of EM induction. Still, these subjects certainly deserve separate articles as EM induction is even more general than the induction article describes in its opening sentence. Alfred Centauri 22:23, 31 August 2005 (UTC)

Yes, and also the toasting of toast is due to a toaster. That doesn't mean they're not separate concepts, it just means they're related concepts--so they have separate articles that link to each other. -- SCZenz 22:32, 31 August 2005 (UTC)

Proposed change of definition of inductance

I propose that the defn should read;

''Inductance is that property of a conductor (including an inductor) that admits a current through it that is proportional to the time integral of the voltage applied across it.

I know this seems more complicated but I believe it is more relevant to isolated inductances (not mutual inductance or induced voltages) and avoids the confusion of induced 'back emf' . Also the usual causal relationship is that you apply a voltage to the inducatnce in the first place but only then does the current ramp up. A current cannot magically already exist in an isolated indutance with no external changing magnetic field. Any comments on this idea?? Light current 15:49, 1 September 2005 (UTC)

Disagree. More complicated, and less standard, is probably not a good idea--insofar as this is a general encyclopedia. -- SCZenz 15:57, 1 September 2005 (UTC)

I retract my earlier suggestion and wish to replace the definition by one that is more general, accurate but simpler.

Inductance is defined as the magnetic flux produced due to unit current through any surface bounded by a closed circuit. Principles of Applied Electricity, by AM Howatson ,Chapman& Hall 1969 (no ISBN found)

This means simply L=phi/i (orL= d(phi)/di). After all, capacitance is defined as how much charge per volt C=Q/V. This would say how much flux per amp--- perfectly consistent.

Any comments? Light current 16:18, 1 September 2005 (UTC)

When I rewrote the first sentence in the Capacitance article, I wrote "Capacitance is a measure of the amount of electric charge stored for a given electric potential." So, my suggestion would be for you to write "Inductance is a measure of the amount of magnetic flux produced for a given electric current". I would follow that with a few sentences to explain what that means. This is what I did for the Capacitance opening paragraph but I see that another Wikipedian has deleted that explanation. Anyhow, I like the track that you are on here. Go for it... Alfred Centauri 15:48, 2 September 2005 (UTC)

Not to throw a monkey wrench into this general agreement, but consider the case of a device that has reactive properties like a traditional inductor or capacitor, but which does not use magnetic or electrical field mechanisms to do so. Would people say that "That's not an inductance" just because it doesn't use a magnetic field, or with that in mind would the above suggestion have merit? Would a better rephrase of the corresponding capacitance description not be "Inductance is a measure of the amount of electric potential stored for a given electrical current?" (Though in both cases I'd say it should be "stored/required" not just "stored.") A quick followup statement could establish that magnetic fields are the common method of storage, and present the flux-based definition. (Oh, and yes, to Light Current, a current can indeed pre-exist in an isolated inductor with a static magnetic field, for example an SMES.) (71.233.165.69 00:41, 26 June 2006 (UTC))

Please give an example of a device that has reactive properities not based on magnetic or electric field mechanisms. Also, please give an example of an inductor storing electric potential. Alfred Centauri 03:48, 26 June 2006 (UTC)

The device is well known: it a gyrator. But there are electric fields in any electrical or electronic circuits. LPFR 13:29, 8 October 2006 (UTC)

Does L = N dphi/di?

According to fundamental principles, the inductance L is defined as:

$L={\frac {\lambda }{i}}=N{\frac {\phi _{M}}{i}}\,$

Does it follow that:

$L=N{\frac {d\phi _{M}}{di}}\,$ ?

Well, let's see...

$N\phi _{M}=Li\,$

$N{\frac {d\phi _{M}}{dt}}=L{\frac {di}{dt}}+i{\frac {dL}{dt}}$

$L=(N{\frac {d\phi _{M}}{dt}}-i{\frac {dL}{dt}}){\frac {dt}{di}}$

$L=N{\frac {d\phi _{M}}{di}}-i{\frac {dL}{di}}$

Moral of the story - use the the formula from first principles or if not, state your assumptions, e.g.:

$L=N{\frac {d\phi _{M}}{di}}$ for constant L

Besides, the opening sentence refers to magnetic flux and current, not their differentials. I recommend beginning with the (always correct) non-differential form and then derive the simplified differential form by explicitly stating the caveat 'for constant L'. Alfred Centauri 22:58, 3 September 2005 (UTC)

Good point. I have modified intro. Light current 23:21, 3 September 2005 (UTC)

Just looked at your equations again. They show that L has to be constant with varying i, so this means a linear inducatnce, but does not exclude a time varying inductance! Interesting! Light current 00:56, 4 September 2005 (UTC)

Consider this. If L varies with i and i varies with time, L is an implicit function of time so dL/dt is non-zero. Conversely, if L is a function of time, dL/di is only zero if i is constant. Put another way, if L is constant and i varies, we get v= Ldi/dt. If i is constant and L varies, we get v = idL/dt. If both L and i are functions of time, we get the sum of the terms. Bottom line, L only equals dphi/di if L is constant! Alfred Centauri 01:46, 4 September 2005 (UTC)

Yes, that is all perfectly clear to me. I agree Light current 01:49, 4 September 2005 (UTC)

Physical significance of term 'lambda'

Could AC state the physical quantity to which lambda refers. Its new to me. Light current 01:09, 4 September 2005 (UTC)

My pleasure. $\lambda$ is called the flux linkage. If a flux $\Phi$ links N turns of a coil, the flux linkage is $N\Phi$ and denoted by $\lambda$: [1]Alfred Centauri 02:06, 4 September 2005 (UTC)

OK Thanks! Never heard it called 'lambda' before Light current 02:12, 4 September 2005 (UTC)

The "Definition" section is mathematically correct, but it may be misleading to the casual reader. It implies that the inductance of a coil is simply proportional to the number of turns where, in fact, it is proportional to the turns squared. This is because the flux term $N\Phi$ has a "hidden" ampere-turns term which causes the current to cancel so that the inductance of an air coil becomes a function of turns squared, permeability, and coil geometry. Although this gets cleared up in the formula for an N-turn solenoid in the "Permeability" section, I think this example should be brought up to the "Definitions" section to make it clearer (similar to the way "Capacitance" is defined in the Wiki article in terms of permittivity, area and separation distance). I made a number of changes that, hopefully address this - added an "Inductace of Solenoid" section and removed the "Permeability" section (since there's already a Wiki article on permeability). Feel free to mercilessly change as appropriate... Bert 13:25, 10 April 2006 (UTC)

Counter electromotive force

I would argue that any counter emf is identical to the applied emf in all cases where there is no externally generated magnetic field impinging on the coil. From this stand point I maintain that this counter emf is in fact ficticious. There is only one emf acting across a coil --- that is the applied emf - end of story. The applied emf is the cause, and the current and resulting magnetic field are the effects. The applied emf is countered by the reactance of the coil (giving the effect of inductance) not by some magical 'back emf'. The back emf story would be just like saying that the emf of a battery connected to a resistor is counteracted by a back p.d generated by the resistor somehow. Comments invited. Light current 04:19, 4 September 2005 (UTC)

Consider an ideal wire (0 resistivity) formed into a circle (or a square or whatever) where the ends are connected to an ideal voltage source. According to arguments you and 'O' have made on the Current source talk page, this is an inconsistent circuit (5 = 0). Yet, from our discussions here, we know that there is a voltage between the ends of this wire and that the current through the wire is finite and a linear function of time. How can this be?

This can be true because the wire has some inductance and the voltage is applied 'suddenly'. The case 'O' and I were dicussing had zero inductance Light current 18:56, 4 September 2005 (UTC)

In other words, circuit analysis ignores EM effects. As an aside, consider this. If you connect the terminals of a voltage source together with a wire or short circuit or whatever you wish to call it, you have necessarily created a loop (a circuit) which must have inductance. You can reduce the inductance by reducing the area of the loop but to get the inductance to zero, the area must go to zero but then everthing is just one point so potential difference loses any meaning. 208.61.125.43 20:09, 4 September 2005 (UTC)

I know. But in our thought experiment we were (at least I was) assuming an infinitely short, short circuit?? and a zero loop area. Do you buy that one? ;-) Light current 20:12, 4 September 2005 (UTC)

Let's try it from a different angle. Replace the voltage source with a current source. This may raise some eyebrows, but an ideal current source does not produce any voltage. The voltage across a current source is a determined by and produced by the external circuit. If this current source produces a constant current, the voltage across the source is zero.

Agreed Light current 18:56, 4 September 2005 (UTC)

However, if the current source produces a current that is changing with time, there is a voltage across the source. Where does this voltage come from? The answer is the voltage comes from the non-conservative electric field produced by the changing magnetic flux through the surface bounded by the wire that is produced by the current through the wire.

Agreed Light current 18:56, 4 September 2005 (UTC)

Imagine that the current direction is clockwise and that the current is increasing. The electric field induced is anti-clockwise. That is, the induced electric field is against the current - the charge is flowing 'uphill'. That is, work is required to change the current.

Agreed - this is the voltage that the current source generates to drive its current thro the inductance of the loop which we have both agreed exists. Light current 18:56, 4 September 2005 (UTC)

Nope - the (ideal) current source doesn't generate voltage - it generates current.

208.61.125.43 20:09, 4 September 2005 (UTC)

Maybe if you look at it the other way around. Does an ideal voltage source generate current? If not, where does the current come from. The way I look at it, the ideal voltage source has to generate the appropriate current to keep the voltage constant. The ideal current source generates the appropriate voltage to keep the current constant. Both have to be able to supply power to the load so I think generate is a fine description. Gah4 (talk) 00:40, 4 December 2008 (UTC)

I know it generates current and the voltage existing on the resistor is a by product of that current- but the prime mover in this case is still the current source. Why invent an independent source of voltage rthat would not exist if it were not for the current source. If feel this line of argument could take us up a blind alley with no way out! BTW 208.61.125.43 are you a new user or just forgotten to log in??Light current 20:22, 4 September 2005 (UTC)

Regarding your statement about a back emf from a resistor - that is exactly what KVL states! A voltage source produces a voltage in the absence of an external circuit. If you connect an external resistive circuit, KVL says the sum of the voltages around the circuit must be zero. The voltage across the resistor is a result of the current through the resistor.

Here is where we start to disagree (I think). The voltage across the resistor is identically equal to the voltage of the source (KVL). BUT -- the causal relationship is that the voltage source must first have a voltage at its terminals in order to drive current thro the resistor to create the pd (which of course is the original emf of the source). This is the circular argument to which I refer.!! THe voltage source is Boss, the resistor is completely passive. Light current 18:56, 4 September 2005 (UTC)

You are correct - we disagree. Consider the following scenario: A loop of ideal conductor with a resistor connected to two points on the loop. Induce a current in the loop in some way, e.g., push a magnet towards the loop and then stop. There is a constant current in the loop. Now, cut the loop in between the two points where the resistor is connected. A step of voltage appears across the resistor that gradually decays to zero. Where did this voltage come from? 208.61.125.43 20:09, 4 September 2005 (UTC)

No applied voltage/ current from the source. Therefore, voltage spike comes from energy stored in magnetic field. No problem! Light current 20:28, 4 September 2005 (UTC)

Once again, look to the ideal current source (a pump for charge). The voltage across the resistor is generated by the charge pumped through the resistor not by any voltage produced by the current source.

Yes, this is the normal action of a resistor. In this case the prime mover is a current source and the resistor has no choice but to develop a pd across itself. equal to IR Light current

The current source pumps charge (electrons) through the resistor but the electrons collide with the structure of the resistor. This slows the electrons which causes a increase in the electron density behind these slowed electrons. But, this increased density creates an electric field that accelerates these electrons. Thus, the potential difference across the terminals of a resistor is due to the electric field produced when electrons slow down inside the resistor. 208.61.125.43 20:09, 4 September 2005 (UTC)

Which came first; chicken or egg.??Light current 00:53, 5 September 2005 (UTC)

Thus, I cannot agree with your statement that the induced voltage is fictitious (you must be thinking of something analogous to the fictitious 'centrifugal force').

Again, perhaps I have not made myself clear in the article and my previous 'rantings'!. I did not say that induced voltage was ficticious (ie voltage induced from an external changing magnetic field) - I said that any concept of 'back emf' in an isolated inductor generating its own changing magnetic field due to an externally applied voltage is a myth. How can you say that the voltage you apply to a lone inductor is induced? It is not - it is applied! After all there can be only one emf/voltage appearing across the ends of an indutor of inductance at any one time. 88.110.246.253 18:56, 4 September 2005 (UTC)

When you get excited, your spelling goes to hell! But, to answer your question above consider this. An electric field cannot exist inside a perfect conductor. (Why? because the charge inside the conductor rearranges itself to cancel any internal field) But, if you connect a voltage source to the ends of a perfect conductor there is no way for the charge to re-arrange itself to cancel the field. The only way to cancel the field inside the conductor is to superimpose an opposite field. This opposing field is none other than the induced electric field from a changing current and, if you do the vector products correctly, you'll find that the direction of this field is opposite the field of the voltage source. Thus, to cancel the field inside the conductor, there must be a changing current that creates the opposing induced field. 208.61.125.43 20:09, 4 September 2005 (UTC)

One last thing, this article is entitled 'Inductance'. As in the case of the capacitance article, it will be easy to confuse the physical phenomenon of inductance (capacitance) with the properties of the ideal and/or physical 'inductor' (capacitor). Let's try to keep it focused on inductance. Alfred Centauri 14:56, 4 September 2005 (UTC)

I Agree that we should follow Uncle Alberts maxim (everthing should be as simple as possible -- but no simpler). Here I think the idea of 'back emf' (which I believe has not been taught for over 30 years now at university level), is confusing to everyone (including me). Its a lot simpler to say that for an Isolated inductor connected to a voltage source, V, that V=Ldi/dt. The sign is covered by the sign of di/dt. ie increasing current requires a positive voltage to be applied and to cause the current to decrease requires a negative voltage to be applied. Is this not correct? 18:56, 4 September 2005 (UTC)Light current 18:59, 4 September 2005 (UTC)

I was taught back-emf when I took college E&M just 3 semesters ago. We used Wangsness, which though not a standard I think, is comparable in mathematical rigor to Jackson as far as I've seen. I'll report back on that next semester when I'm taking graduate E&M. But anyway, the concept does simplify understanding AFAI am concerned. Figuring out the difference between the applied field and the field affecting inductance is quite simple this way, and is its main use. ub3rm4th 16:51, 28 September 2005 (UTC)

It's simpler but unfortunately leaves a lot of questions unanswered. For example, consider the following simple circuit - an inductor and a resistor connected together. Assume that at t=0, there is a non-zero current in the circuit. What creates the emf that drives this current? Remember, this inductor is isolated from any external fields. 208.61.125.43 20:09, 4 September 2005 (UTC)

The emf is induced in the inductor by its collapsing magnetic field. I have no problem with that. But remember that there is only one source of energy here: in the mag field and cct follows the induction law as if the field were created by another coil somewhere. BTW my spelling is always bad cos I try to type too fast! Light current 20:37, 4 September 2005 (UTC)

In para 3 of Properties of inductance we ahve a completely circular argument. This para seems to be therefore patent nonsense Light current 04:25, 4 September 2005 (UTC)

OK. I've read the latest version of the article and I think I understand what is bugging you - it's the minus sign for the emf in the equations. However, simply because this bothers you is not sufficient reason to rewrite an article using some questionable arguments. I think these changes will be reverted.

If, after I have had a chance to explain my thoughts, you still wish to revert the article, I would ask that you copy the controversial material here so that any other interested parties can still see it and comment on it. Thanks Light current 19:07, 4 September 2005 (UTC)

I suggest reading the article on emf. I rewrote a portion of it that I think directly applies to this negative sign business. --

OK I'm going to read that now Light current 19:07, 4 September 2005 (UTC)

I have read that article now, and I think I agree with all of it. But the so called negative sign business is , I assume, the fact that a battery has an opposing electric field across its terminals when open circuited. This is true. However', the opposing electric field would not be there if the battery was not there (causality) . This also reminds me of Newtons Law of motion (cant remember which one) that action and reaction are equal and opposite Light current 19:19, 4 September 2005 (UTC)

Consider a staight wire segment that is not connected at the ends. If this wire is moving perpendicular to a uniform magnetic field, charge will flow in the wire. The direction of this flow is the direction of the emf that causes the flow. For example let's say the magnetic field points 'out' of the page and the wire is moving to the left. The resulting emf points down. This means there will be a conventional current 'down' (or an electron current 'up'). However, this current cannot continue forever since the ends of the wire are unterminated. As the electrons accumulate at the top end of the wire, an electric field pointing 'up' forms. That is, this electric field is in the opposite direction of the induced emf. At some point when enough electrons have accumulated at the top end of the wire, the electric field from this charge accumulation exactly cancels the induced emf. What is the voltage across this wire? The voltage is due to the electric field so the polarity is - at top, + at bottom. But, this is exactly opposite the induced emf. So, if you equation is for the induced emf, there should be negative sign. If you equation is for the voltage due to the opposing electric field, there should not be negative sign. Alfred Centauri 16:26, 4 September 2005 (UTC)

Yes I think I agree with this, although I don't really think its relevant to my argument about inductance. Light current 19:24, 4 September 2005 (UTC)

confusion between emf and induced voltage

Here is something I wrote for the Electromotive force article:

Regardless of how it is generated, emf causes an electric current through a circuit connected to the terminals of the source. For example, the chemical reaction that separates electric charge onto the two terminals of a battery proceeds as long as there is an external circuit through which electrons can flow from the '-' terminal to the '+' terminal and thereby recombine with the positive ions.

However, if an external circuit is not connected, an electric current cannot exist. Thus, between the terminals of the source, there must exist an electric field that exactly cancels the generated emf. The source of this field is the electric charge separated by the mechanism generating the emf. For example, the chemical reaction in the battery proceeds only to the point that the electric field between the separated charges is strong enough to stop the reaction. This electric field between the terminals of the battery creates an electric potential difference that can be measured with a voltmeter. The value of the emf for the battery (or other source) is the value of this 'open circuit' voltage.

   As there is no external circuit , Kirchoffs voltage does not apply, and hence the sum of the voltage around the circuit (which does not exist) does not need to equal zero.  The "emf" is simply an "emf" which has the potential to motive electrons around circuit.       

Nonsense. Check your facts before making such absurd claims. Consider the fundamental principles underlying KVL.

The use of the term emf is in decline but it is still found in introductory and technical level texts on electricity. Within Electrical Engineering, the term emf is occasionally used for a voltage produced by electromagnetic induction. However, the term induced voltage is preferred.

  I would hope that the idea of dropping the term "emf" never eventuates. There is no other suitable word(s) to replace it, for the concept that it conveys. IE a "force that motivates electrons!"    The term "emf" is not directly interchangeable with "voltage produced by electromagnetic induction" or back emf. There is a big difference.

The volt of course is the unit of emf , potential difference or voltage drop. It is in fact the term "voltage" and likewise the word "amperage" which are badly abused. Extending the logic of doing away with emf is akin to doing away with "mmf"(magneto motive force" What a horrible thought!

Bottom line - emf causes charge to move - the strength of the emf is measured by the voltage required to stop the charge flow. Alfred Centauri 17:42, 4 September 2005 (UTC)

If I place a resistor accross the emf, are you als implying that the "voltage" accross the resistor will stop the voltage flow.

No, but thanks for playing. Last time I checked, voltage doesn't flow.

The emf , will still be there and there will be a voltage drop accross the resistor. We have a circuit and KVL must be (and will be) satisfied.

KVL will be satisfied whether there is a resistor there or not.

Agree --but cant see relevance to my argument about an isolated inductor :-) Light current 19:27, 4 September 2005 (UTC)

Present State of Page

Just having looked at the page for the first time since you altered it, I am quite relieved that, (expecting the worst after reading your posts), you have left so much of my material there. Thanks!:-). As you know I'm still confused by the minus sign and the page explanation does nothing to help me. Why must we refer to two voltages equal but of opposite sign when one would do (as the sign is already included in the di/dt or the dØ/dt)? The second one is identically equal to the first in the case of an isolated inductor fed by a switched voltage source -- is it not? The building of the magnetic field does indeed present reactance to the voltage source and this is what we expect from an inductor. However, to say that the 'reactance' is caused by the changing magnetic field the coil is in the process of generating - and this then creates another voltage equal in magnitude but opposite in sense to the applied one, is IMHO an unneccessarily complicated and circular argument for presentation to engineering (or even physics) students Light current 19:53, 4 September 2005 (UTC)

I'd like to compose an answer to this and your other comments but every time I try to save, I get an edit conflict so let me know when your done for awhile and then I'll respond! User:Alfred Centauri 20:12, 4 September 2005 (UTC)

"Why must we refer to two voltages equal but of opposite sign when one would do". But will one do? Look, the integral of the electric field along a closed path is the negative time rate of change of the magnetic flux through the surface bounded by that path. No matter how much you fuss, this is the convention.

I'm going to agree with that for now. You mean the general integral form of Maxwells equs (from Faradays Law) Presumably you mean magnetic flux density- but lets not argue about that. I have to agree with your above statement.

Er, no. I mean magnetic flux. The integral form relates the closed contour integral of (E dot dS) to the negative time rate of change of the area integral of (B dot dn) but this is that integral is just the total magnetic flux through the surface bounded by the contour. It's the differential form that relates -dB/dt to the curl of E. Alfred Centauri 13:00, 5 September 2005 (UTC)

Im not trying to fuss or complain, I'm just saying that the difference between 'applied emf' and 'induced emf' is extremely confused and needs clarifying. They are (necessarily?) opposite in sign. If we can come up with unambiguous, simple definitions for capacitance, inductance and compare and contrast their similarities/ differences we will have done what most textbooks on electrical engineering have failed to do so far. 88.110.246.253 23:40, 4 September 2005 (UTC)

If you connect a source of emf (a battery for example) to a conductor (a short circuit), the circuit is, by your own words, inconsistent. One of the reasons for this is that an electric field cannot exist inside a perfect conductor so there cannot be a voltage drop across such a conductor, right? How do you reconcile this?

Butting in - the battery has an internal resistance, so the currrent is limited. -Baruch Atta

The electric field due to the voltage across the battery must be exactly cancelled by an electric field in the opposite direction along the path of the conductor. In this way, the field inside the conductor is zero as required. Where does this opposing field come from? It is the induced E field from a changing magnetic flux (inside the loop formed by the conductor) that is itself created by a changing current in the conductor. That is, in order to have zero field inside the conductor, there must be a changing current through the conductor. This is not a circular argument nor is it patent nonsense. If there were no induced emf, the only solution for the circuit described above is for infinite current to exist in the conductor.

Now, if you integrate just the induced E field along the conductor, the result (in volts) is equal and opposite to the applied voltage (or emf). This is not an indication that the emf is fictitious. On the contrary, this is the only value the integral could have if there is to be no field insided the conductor and therefor a finite current

I really think the problem here and in textbooks is that the student is not adequately prepared for the 'weirdness' of a non-conservative electric field.

<!del--- Forgive me butting in here. I agree that the non conservative field is what confuses people. It confuses me. Even the name is intimidating. Is there a simpler name such as changing, not steady or something more friendly that could be used?? Light current 15:02, 5 September 2005 (UTC)--->

Ive just looked at my college notes on Field Theory and guess what? - the topic of non conservative fields is not covered nor even mentioned!. No wonder I've been having trouble all these years understanding electromagnetic induction! Light current 16:49, 7 September 2005 (UTC)

As I said below, the concept of potential difference is useless for non-conservative fields because the 'potential difference' between two points depends on the path you take between the points. That is, the potential difference is not unique. Alfred Centauri 13:00, 5 September 2005 (UTC)

Isnt this just because the field is changing with time?Light current 15:04, 5 September 2005 (UTC)

If I integrate the electric field along some (not necessarily close) path, I get a positive voltage if I integrate in the direction of the field. If there is a changing magnetic flux into the page and I integrate in an anti-clockwise direction along a closed path, I get a positive voltage. Thus, the field is anti-clockwise. This means that if free (positive) charge is in the region, the charge will move in an anti-clockwise direction. If instead, there is a loop of wire that is open on say, the left side, charge will flow anti-clockwise in the conductor until it is stopped by the upper end and accumulates there until what - until the electric field of this accumulated charge stops the current. If you place a voltmeter across the ends of the conductor with the red lead on top, you will measure a positive voltage but this voltage is due to the charge that accumulated from the current driven by the field in the opposite direction. This is a fundamental physical fact. The emf (the mover of charge) is in the opposite direction that the field due to the accumulated charge is in. But, the emf came first! It was the emf that drove the charge in the first place. The field from the accumulated charge eventually stopped the current. Alfred Centauri 21:43, 4 September 2005 (UTC)

Before I try to digest the above answer, could you say whether you agree that the sign of any induced voltage is (in some part) determined by the sign of di/dt or dØ/dt? Thanks Light current 22:06, 4 September 2005 (UTC)

Yes. Alfred Centauri 22:21, 4 September 2005 (UTC)

I assume your reply means that you agree (not that you could say whether you do or not)Sorry to be pedantic! :-) Light current 23:07, 4 September 2005 (UTC)

emf vs induced voltage

Ive just been looking at the page again. The statement I have trouble with is:

${\frac {d\lambda }{dt}}=-{\mathcal {E}}=v$

where ${\mathcal {E}}$ is the Electromotive force (emf) and $v$ is the induced voltage. Note that the emf is opposite to the induced voltage.

You see, the way I see it, neither one is defined and the difference between these 2 voltages is not stated. It probably makes complete, unambiguous sense to you, but I'm afraid it doesnt to me (Im not saying its wrong -- just that I dont undersatnd it as it stands). If we were to define what we mean by:

a) the emf

b) the induced voltage

maybe our problems would go away!! What think you? Light current 00:10, 5 September 2005 (UTC)

The emf always exists when there is a changing magnetic field. However, the induced voltage may not exist! If there is no charge in the vicinity of a changing magnetic field, the will not be current nor will there be an electric field due to the presence of or accumulation of charge. However, the closed contour integral of the electric field will nonetheless be non-zero. Observe that the integration of the electric field (V/m) over distance (m) has units of Volts. The $100,000 question is: how do you measure this? Integrating along a closed contour means that your beginning point and your end point are the same so a potential difference between two points that are the same is difficult to define. In fact, it is impossible.

Is this pd not obviously zero!? Light current 13:30, 5 September 2005 (UTC)

It is obvious to me (and should be to you) that it not zero, it is instead 'undefined'. Potential difference only has meaning if the the work done in moving a charge from one point to another is independent of the path taken. In a non-conservative field the work done depends on the path which means that the work along a closed path can be non-zero. Recall that volts is Joules / Coulomb. When you integrate the electric field along a path, the result is the work done in move unit charge along that path. In a conservative field, the work done along any closed path is identically zero. This is not the case for the induced electric field. Alfred Centauri 13:47, 5 September 2005 (UTC)

Sorry ,I missed the point about the field changing. You are absolutely correct. Apologies! Light current 14:23, 5 September 2005 (UTC)

The induced emf is non-conservative so the concept of potential is no longer meaningful (recall that the gradient of a potential field must be conservative!). One way to measure the strength of the emf is by measuring the amount of charge the emf can separate. As charge moves due to the non-conservative induced field, a conservative field is developed between the separated charge. When this conservative field stops the movement of charge, the potential between the separated charge is exactly equal (and opposite) to the emf in Volts. This measurable voltage is the induced voltage and is the potential difference created by the charge the emf has separated. But, it must be kept in mind that this conservative field is in the opposite direction to the induced non-conservative field! If this explaination doesn't help, I've got a couple of other angles but I don't want to introduce too many ideas at once. Alfred Centauri 03:45, 5 September 2005 (UTC)

OK Im beginning to see that:

a) ${\frac {d\lambda }{dt}}=-{\mathcal {E}}$ (we must accept this from previous discussion and Mawells laws etc)

b) the induced emf can exist even with no moveable charges present (ie in an insulator)--

agreed. Alfred Centauri 00:33, 7 September 2005 (UTC)

c) If, however, a source of moveable charges exists, these charges will be acted upon by the emf and they will move(necessarily doing work- but lets forget about work at the moment)) to counteract the emf.

agreed though let me clarify that the work is done on the charge by the emf. The charge gains kinetic energy as a result. Alfred Centauri 00:33, 7 September 2005 (UTC)

d) the induced emf acts like a battery connected to a conductor with the conductor representing the body containing the moveable charges.

I'm not so sure about this one. The emf acts to accelerate charge around a closed loop. If there is not a closed loop, charge will accumulate somewhere creating a potential difference. This potential difference is a measure of the strength of the emf. The chemical reaction inside the battery is the emf. The voltage across the battery terminals is a result of the charge driven by the emf accumulating on the terminals of the battery. Alfred Centauri 00:33, 7 September 2005 (UTC)

e) Therefore, the voltage induced in the conductor must be in opposite sense to the emf in order to obey KVL

Kind of. See my answer to d). Alfred Centauri 00:33, 7 September 2005 (UTC)

Is this a fair summary of your argument so far? Light current 13:52, 5 September 2005 (UTC)

very close - I think we getting to the same page! Alfred Centauri 00:33, 7 September 2005 (UTC)

Emf exam

Consider a ring of resistive material with uniform resistivity. Assume that there is a magnetic flux changing at a constant rate through the surface area enclosed by this ring. This changing magnetic flux induces a constant electric field that in turn drives a constant current through the ring of resistive material.

Recall that the line integral of the induced electric field around the path traced by the ring gives the emf (in volts) induced by the changing magnetic flux.

Answer the following questions:

(1) Is it possible to measure the total emf directly with a voltmeter? If so, explain how to do this. If not, propose how to measure the total emf.

If the ring had a break in it, and the ring totally enclosed the flux, then it would be easy to measure the induced emf, by connecting the meter across the break but ensuring that the voltmeter leads are outside the ring.

However I dont think this is the question you are asking is it?

Anyway, even with a continuous resistive torus, it should be possible.

first try: connect leads of an ideal voltmeter to 2 points diametrically opposed on the ring and double your reading?.(making sure that the volt meter leads were parallel to the mag flux (so no induction in them).

second try: use a CT to measure the current in the ring and multiply by the rings circumferencial resistance?(but I guess you'll disallow this 2nd answer as the exam is on emf!Light current 14:03, 5 September 2005 (UTC)

It is my opinion that a voltmeter will read zero volts between any two points on the ring as long as the ring is unbroken. This result is the only result consistent with KVL and question 2. Thus, I believe that your 'second try' answer is correct. Alfred Centauri 23:54, 6 September 2005 (UTC)

(2) KVL says that the sum of the voltage rises and voltage drops around a circuit must equal zero. Does KVL apply here? Explain your answer.

The induced emf is distributed evenly around the ring and does not appear at a specific point in space. Ie it is given by -int E.dl around a closed path. Now KVL is normally quoted for circuits and voltages (not emfs). Since however, we have a source of movable charges, these charges will respond to the emf and move giving rise to a current. THe charge will move at such a rate that, at each and every point in the ring, the induced voltage delta(v) = I delta(R), (where delta means an infinitely small amount).KVL does appear to apply here. At each and every point in the ring, the pd dropped (developed?) across an infinitesimal part of the resistor(ring) must equal the induced emf over that part of the ring.?Light current 14:15, 5 September 2005 (UTC)

It is true that emf is not a point function precisely because it is defined as the integral of the electric field along a closed path. However, KVL is directly aimed at emf because it is emf that drives charge around a closed circuit. A conservative electric field cannot do this! Nonetheless, your answer is correct otherwise and consistent with the idea that a voltmeter reads zero volts between any two points on the ring for the very reason you gave that the emf is exactly balanced by Idr.Alfred Centauri 00:16, 7 September 2005 (UTC)

(3) Assume the current in the ring is anti-clockwise. If a voltage source were instantly inserted in series with the ring, what is the value and polarity of the voltage required to stop the current in the ring?

Presumably one would have to insert the source equal in voltage to the original emf to counteract the original emf because this is the prime mover in this cct.Light current 14:15, 5 September 2005 (UTC)

Correct. The voltage source should generate a current equal and opposite to the current driven by the emf. Now, doesn't this bother you? After all, you can now measure a voltage across the resistive ring but there is zero current!!!. What happened to Ohm's law? Alfred Centauri 00:19, 7 September 2005 (UTC)

Not too bothersome. We can measure the emf by the above CT method, OR (Ive just thought of this one) adjust the voltage source until the CT o/p reads zero. Then the applied voltage equals the emf. So what we're measureing is the induced emf not the voltage. Anyway, the emf has been backed off so no current flows in the ring - Ohms law is NOT violated -- no problem! Am I correct?Light current 01:10, 7 September 2005 (UTC)

(4) With this voltage source in circuit and with zero current in the ring, magically adjust the resistivity of the material to one-half its former value. Does the current change? Explain. If the current does change, what is the new value of the current?

A No, the current does not change, because the inserted voltage source still balances the induced emf.Light current 14:15, 5 September 2005 (UTC)

Correct. Alfred Centauri 00:22, 7 September 2005 (UTC)

(5) Answer (4) again with the resistivity magically adjusted to zero.

A Again the answer is that the current remains zero because the inserted voltage has been adjusted to completely neutralise the original emf.

Again, correct. But once again, doesn't this bother you? You have a voltage across a short circuit and zero current! How can you explain this? Alfred Centauri 00:22, 7 September 2005 (UTC)

Doesn't bother me as much as it would have afew days ago! There is an emf generated in this short cct by the changing mag flux. If this emf were not cancelled by my inserted voltage source, then I presume an infinite current would flow. Is this the same thing as a s/c turn in a transformer? The only weird thing to me now is that you can have an emf exist in s/c! Light current 00:49, 7 September 2005 (UTC)

I hadn't considered a shorted turn. Hmmmm... Sure! The short reduces the path integral for the emf by the circumference of one turn so now there is less emf and the balance between the applied voltage and the emf is lost. The result - LOTS of current. Good point! Alfred Centauri 01:05, 7 September 2005 (UTC)

Looks like I have been almost completely rehabilitated - doesnt it?? ;-) Light current 00:33, 7 September 2005 (UTC)

I hope you learned as much as I did. You've asked some great questions that made me think about some concepts much more precisely than I have in the past. Thanks! Alfred Centauri 01:05, 7 September 2005 (UTC)

These are excellent questions but I'm not sure if I've got all the answers completely correct. ;-) Light current 14:15, 5 September 2005 (UTC)

Alfred Centauri 13:49, 5 September 2005 (UTC)

Is the idea of Induction universal??

Relativity

Now that you have become comfortable with the balance of forces, let me thow you a curve ball. [BTW I though all balls were curved!! ;-)Light current 17:26, 7 September 2005 (UTC)]

According to Einstein's theory of gravity (General Relativity), the only force on the apple is the push from the deformation. This force causes the apple to be accelerated which is why the apple has weight. So, according to GR, the forces aren't in balance. Enjoy! Alfred Centauri 16:06, 7 September 2005 (UTC)

Im trying to keep off relativity at the moment as I'm sure I could be easily shot down. However, for those who are interested in relativity, I found the translation of Einsteins 1916 book (originally in German)('Relativity', pub Routledge Classics 2002, ISBN 0-415-25538-4/5) to be much easier to read than all those Teach yourself relativity, and relativity made simple books. Give it a try! Light current 17:14, 7 September 2005 (UTC)

Not reading below (since it seems unconnected to this question), what are you trying to say exactly? I can only assume that you are referring to Newton's third law (because, as a current freshman physics tutor, that is what is being taught currently it's what jumps to mind), but nothing there is being violated. ub3rm4th 16:51, 28 September 2005 (UTC)

The postulate - Version 1

All the previous arguments (on induction) seem to be leading us down a path of saying that:

Whatever force can exist in the universe, whether mechanical, electromotive (emf), magnetic(changing or not), electric (changing or not), gravitational etc, that force, by its very presence, INDUCES an equal and opposite force on any object to which it is applied. If it is not applied to an object, then the force just exists on its own like an emf in an insulator. But, when applied to an object, that force MUST be resisted by the object (whatever it is) and it must be resisted in a way that leads to the continued existence of the object (ie its not crushed by gravity, it creates an induced voltage to stop current rising to infinity etc,etc)

Would anyone agree with this extreme generalistation of mine?? Light current 15:46, 5 September 2005 (UTC)

As far as I can tell, what you're saying is that things that exist tend to be in equilibrium states, because if they don't they change until something happens so that they are. Thus if an object couldn't withstand compression from gravity, it would collapse until some force appeared that could. (There's even an exception: a black hole.) Likewise, some circuits come to equilibrium because of induction. But I wouldn't really call the general phenomenon you're describing a generalized induction. -- SCZenz 00:31, 7 September 2005 (UTC)

In your order. Yes!, Yes, (I knew someone would mention black holes!!) and Yes. OK, SCZenz THanks for your input :-) Light current 01:22, 7 September 2005 (UTC)

Well, I'm not sure about your wording here but I think you are close to an insight that is quite sophisticated. It turns out that if you start with just the electrostatic force law and then make this force law compatible with special relativity, you get Maxwell's equations. That is, magnetic effects and induction effects are consequences of observing electric effects from a relatively moving reference frame. Not suprisingly, there are analogs to magnetism and induction for gravity. When it comes to the other two natural forces - QCD and electro-weak - I'm much less certain as there isn't a classical (macroscopic) description of these forces that I'm aware of. Alfred Centauri 16:00, 6 September 2005 (UTC)

Yes I agree this wording needs a lot of work. Unfortunately I am no expert in nuclear physics (or physics in general for that matter - I'm an engineer). I just thought that after my discussions with [User:Alfred Centauri], this business of actions being equal and opposite seems to make sense to me within my limited sphere of knowledge. If it turned out to be a (fairly) general principle, I'm sure it would aid science understanding tremendously. One example I can quote is that of Newtons 3rd? law -to every action there is an equal and opposite reaction. It took me years to see how this could be true in the case (lets say) of an apple resting on the ground. THe apple has the force of gravity acting upon it,and is therfore attracted to the centre of mass of the earth. Since the apple doesnt move, there must be an equal and opposite force acting upwards on the apple. So far so good. THe big question is: how does the earth know what force to exert in response to the apple. The answer came to me one day. When the apple rests on the earth, the earth is slightly deformed by the attractive forces. The more the earth (and apple I suppose)is deformed, the greater its 'upthrust'. Hence the earth is deformed just enough so that the reaction force exerted by the earth is exactly equal to the 'weight' of the apple and a state of mechanical equilibrum then exists. The 'bigger question' is: does this sort of thing happen between other forces and objects or between two different forces like emf and induced voltage. I'm sure people could think of many more examples. Food for thought!!! Light current 02:58, 7 September 2005 (UTC)

You're basically right about the apple on the earth, although I think most of the "deformation" is actually at the contact point between the electron clouds on the bottom of the apple and those on the top layer of ground. You only have to scrunch those a little before they produce a lot of force per area. -- SCZenz 05:04, 7 September 2005 (UTC)

Are you implying here that the reaction force is generated by electrostatic repulsion? Light current 05:09, 7 September 2005 (UTC)

Uh, yes, I believe I am. That is to say, the normal force--the force perpendicular to a surface which keeps objects from falling through--is indeed due to electrostatic repulsion. All contact forces are--when you touch something solid, your hand doesn't go any further because of repelling electron clouds. -- SCZenz 08:11, 7 September 2005 (UTC)

OK then, but in an insulating material, is it not true that electrons are tightly bound to their atoms. If so how can there be an electron cloud? Light current 13:40, 7 September 2005 (UTC)

In quantum mechanics, the uncertainty principle forces objects with large momentum, such as the electron in an atom, to have very imprecise locations. Even tightly-bound electrons can be found in any location in their orbital--"tightly bound" just means that they're associated with only one nucleus and it's very hard to remove them. (You're thinking of the sea of free electrons in a metal, which has much larger extent, of course.) -- SCZenz 14:41, 7 September 2005 (UTC)

Ok. So regardless of how it happens, in the macroscopic case of the apple on the ground, one could say that the reaction force is induced by the gravitational force. Yes?. The upthrust of a fluid is induced by the dispalcement of that fluid (Archimedes principle). Voltage is induced to counteract emf in a changing magnetic field. Charges are induced on the plate of an electroscope to balance the electric field of a charged object brought close by. Heres a biggie -- electric flux is induced by the presence of charge???. Do you agree with all the above?. If not - say why not Light current 15:24, 7 September 2005 (UTC) Just noticed in my college notes that electric flux used to be called the electric induction.Light current 17:02, 7 September 2005 (UTC)

I think that you might be using the word 'induced' too loosely here. Electromagnetic induction is very clearly defined as the creation of a force field by the flow of something. Gravity doesn't induce a reaction force on the apple any more than the engine in a car induces a wall in front of you. In the Newtonian view, gravity accelerates each particle in the apple in the same way. If nothing is in the way, the apple accelerates. If something is in the way, the apple doesn't accelerate as a whole. Why? As SCLenz says, the electrostatic repulsion between the outermost electrons of the atoms of the apple and the ground is more than enough to stop the acceleration.

Possibly Im using the word 'induction' too loosely here , but it will do until someone comes up with a better word!. Light current 00:54, 8 September 2005 (UTC)

I disagree with this statement. You're using "induces" to mean "causes." So your "generalized principle of induction" is just that forces cause other forces. It's extremely general, and as we've been discussing, not always true. -- SCZenz 02:32, 8 September 2005 (UTC)

If its not always or generally true, then the postulate is erroneous. However, I dont think we can say that yet--Light current 17:09, 8 September 2005 (UTC)

What's your current statement of the postulate, and in what domain does it apply? (Yes GR, or no GR?) With a firm statement, I can think of a counterexample if there is one. -- SCZenz 17:16, 8 September 2005 (UTC)

Apply a force to anything. There is always a reaction (Newton was quite right). Scrub 'induces' if you like and substitute 'causes'. It makes no difference to my postulate.Light current 03:14, 8 September 2005 (UTC)

That's not what Newton's "equal and opposite reaction" is. It means that if body A exerts a force on body B, then body B exerts the same force on A. It does not mean that if body A has a force in it, something will show up to exert and opposite force on body A. That, as I've been trying to give examples of, isn't always true. Consider, for example, a rocket flying off into space. Nothing opposes the force from its motor, so it simply leaves earth's gravity and goes elsewhere. -- SCZenz 04:57, 8 September 2005 (UTC)

The rocket example is especially interesting. Newtons 'equal and opposite' can be restated like this: the momentum of the center of mass of a system is unchanged by forces internal to the system. For the rocket, the system is the rocket, fuel, AND the exhaust. Before the engine is ignited, the center of mass of the rocket has zero momentum in the rocket's reference frame. When the fuel burns, the product(s) of combustion exits the engine bell at high speed but this exhaust has mass and thus momentum. For the momentum of the center of mass to remain unchanged, the rocket must have an equal but opposite momentum. Alfred Centauri 12:45, 8 September 2005 (UTC)

Think of a spring. If you push on the spring there is a force proportional to how far you've compressed the spring. At some point, the restoring force of the spring balances your push and now you have equal and opposite forces. Did your pushing induce this force? I suppose you could say that. I'm just not sure what that perspective buys us. Give me an example of how to use this idea of yours. Alfred Centauri 19:35, 7 September 2005 (UTC)

What it buys us is the notion that all forces in the universe have an equal and opposite reaction. I have yet to think of an application where a new insight might be revealed - but give me time!!. I was hoping that others may latch onto this idea and suggest applications of this postulate. Light current 01:04, 8 September 2005 (UTC)

I'll more or less second that. You're using the word "induction" in it's general-purpose sense, in which "X induced Y" more or less does mean "X caused Y." But that's not what "induction" means in physics. Anyway, it's not universally true that all forces on an object cause an opposite, balancing force to appear. Planets in orbit have no balancing force.

The Postulate Version 2

All forces existing in the universe, cause an (induced)reaction to be generated by any object or force field or medium to which the original force is applied. This reaction force is equal in magnitude and opposite in direction to the original force.

Counterexample. To get us away from gravity, and whether it's a force or not, I'll use the good old electric force. Suppose we have a uniform electric field E, as would be produced (for example) between two large charged parallel plates. A charged particle (with charge q and mass m) placed in this field has a force F = qE. In accordance with F = ma, the particle has acceleration a = qE/m. There is no other force on the particle, as we can see from the fact that such a particle will in fact accelerate. -- SCZenz 18:51, 8 September 2005 (UTC)

Note on my counterexample above. Of course, it is true that the particle exerts an equal and opposite force on the plates producing the electric field, but that's just an example of Newton's Third Law. (Since your statement doesn't say what the force applies to, it might be read as synonymous with that law--which would mean it was correct, but not new.) -- SCZenz 18:51, 8 September 2005 (UTC)

Ok But were you aware that Newtons 3rd law can be applied to every physical force in the universe?. I was not. I was used to only thinkng of it in the terms of mechanics (as was Newton of course). Light current 22:23, 8 September 2005 (UTC)

I actually hadn't thought about it before. In terms of the classical electric and gravitational forces, it follows because F = GmM/r² and F = kqQ/r² include the same properties of both particles, and are defined to apply to both. In terms of all forces in the universe, this one made me (and two of my labmates) think for a few minutes. The answer is: yes, it applies to every force in the universe. The logic is as follows:

1. All forces, at a deep level, are conveyed by the exchange of particles; one particle emits a photon and the other absorbs it (see Feynman diagram for illustrations of this).
2. Since momentum is conserved in all particle exchanges, the momentum lost by one particle has to be gained by the other(s).
3. Since force is the rate of change in momentum per time, and the time for the exchange is the same for both particles, this means the force is equal and opposite.
4. Since all interactions between objects can be built up from interactions between particles, the 3rd law holds for everything.

Cool! -- SCZenz 22:45, 8 September 2005 (UTC)

LC: your postulate version 2 is, as worded, nonsense. It seems to be claiming that there are no net forces in the universe. So, why don't you try to clarify what it is you mean by a 'reaction force'. Give me an example of a an induced reaction force using Coulombs electric force law or Newton's gravitational force law as these are the only known long range forces in the universe. Try to be precise with your example please. Alfred Centauri 20:02, 8 September 2005 (UTC)

I think that comment is a little harsh AC,. After all, it is merely a generalisation of Newtons 3rd law as SCZenz has pointed out and is therfore not nonsense. I'm sure you accept Newtons laws as valid. --Light current 23:41, 8 September 2005 (UTC)

A) Apple on the ground. Ground reacts to gravity force by induced force pushing up. B) Fluids react to displacement by induced upthrust. C) Electric Flux is induced by a charge. --Light current 22:38, 8 September 2005 (UTC)

Take a rocket in space. Force on rocket F=ma. this is also equal to the force on the exhausted material. Momentum is conserved in the system. since Force = rate of change of momentum and momentum is not changing, then yes there is no net force in the system but the rocket still accelerates --Light current 22:51, 8 September 2005 (UTC)

The rocket has a force on it, and accelerates. The rocket+fuel system has no net force, and does not accelerate. -- SCZenz 22:53, 8 September 2005 (UTC)

Correct. Would you care to expand on that?--Light current 22:57, 8 September 2005 (UTC)

SCZenz: I have a hunch that what LC is trying to say is that ultimately, all forces are internal. What force is there 'external' to the universe? Alfred Centauri 20:02, 8 September 2005 (UTC)

No I think I was trying to say that all forces are matched. SCZenz puts it very well in his last post. --Light current 22:55, 8 September 2005 (UTC)

The Postulate Version 3

Newtons third law is applicable to every conceivable force in the universe. --Light current 23:12, 8 September 2005 (UTC)

OR THe sum of forces in a closed system is zero --Light current 23:29, 8 September 2005 (UTC)

Or, as I said before, all forces are ultimately internal - there are no forces external to the universe. You do see this don't you? Newton's law of inertia says momentum doesn't change unless there is a net force. I can't conceive of a momentum associated with the universe (can you?) so there can not be a net force in the universe - all forces in the universe must cancel. Well done - you're a deep thinker now! Alfred Centauri 00:04, 9 September 2005 (UTC)

Thanks for the compliment. Yes I do see what you mean. Correct: sum of all forces in universe is zero. I think I'll quit while I'm ahead! --Light current 00:09, 9 September 2005 (UTC)

Well, I don't think it was obvious that the third law applies to every force in the universe. But it does end up being equivalent to conservation of momentum, which as far as we know holds for everything. So I'd say we all learned something, even if we didn't really get a new physics principle out of it. -- SCZenz 00:19, 9 September 2005 (UTC)

LC is very good at making statements and follow-up statements that challenge us to think very clearly about how to answer. I for one have, as a result of trying to answer LC, come to understand the material much better than before. Thanks, LC! Alfred Centauri 00:57, 9 September 2005 (UTC)

If I'd seen this [2] earlier then we could have made a few short cuts. C'est la vie!

NOTHING beats working through the problem on you own. It doesn't matter if your new insight is old news to someone else, the thrill of having the 'light bulb go off' due to your own reasoning is what 'it' is all about. Alfred Centauri 01:03, 9 September 2005 (UTC)

Yes,you are correct. If it aint hurtin', it aint workin'. And no pain - no gain! etc --Light current 01:22, 9 September 2005 (UTC)

The postulate in Celestial Mechanics

Im sorry to disagree with you but planets in orbit do have a balancing force. Its called centrifugal force. This is what stops then falling into the sun!. Light current 01:08, 8 September 2005 (UTC)

Centrifugal force is a fictitious force--it's useful calculationally, but there's no physical mechanism creating it. Instead, it's a byproduct of considering a rotating (and thus non-inertial) reference frame (i.e. the one in which the earth is stationary). If you consider a more-or-less ineratial frame, in which the earth is orbiting and the sun is stationary, we don't fall into the sun because the earth has a large velocity perpendicular to the force of gravity. (It's always falling "past" the sun.) -- SCZenz 01:49, 8 September 2005 (UTC)

The postulate in General Relativity

You don't even have to use a rotating frame to see a fictitious force. When you step on the accelerator in your car, the acceleration is forward (hopefully) yet you feel as if a force pushes you backward into your seat. That is acceleration forward 'feels' like a force is pushing backwards. This force you 'feel' doesn't exist - it is fictitious. Now, to make the connection to GR, imagine that you are sitting in a car seat that is facing up. You feel as if you are being pushed back into your seat with a force equal to your weight. But, by the above reasoning, this implies that you are being accelerated up. Yet, you know that you are stationary on the surface of the Earth. According to GR, (this will raise some eyebrows) spacetime itself is accelerating downwards toward the center of the Earth. To be 'stationary' in spacetime would mean that you would also (free) fall towards the center of the Earth. Since the surface of the Earth prevents you from doing this, the surface of the Earth is, in effect, accelerating you. That is, your worldline is not on a geodesic of the spacetime. Wait -what's that I smell? Is it LC's brain short circuiting??? ;<) Alfred Centauri 02:52, 8 September 2005 (UTC)

Yes it is!! My brain is completely fried. I shall need some sleep to repair it, but, as Arnold once said...'I'll be back!!!' (Space-time continuum -- what the hell does that mean??) Light current 03:03, 8 September 2005 (UTC)

The earth may be accelerating you upwards, but gravity is accelerating you down wards. The accelerations (and forces of course) balance. This is what I have been saying. So, as an addition to the postulate, can we say that all accelerations balance?? BTW if you are freely falling in a gravitational field, you feel no forces. But to an outside observer, a force is obviously acting on you and you react to that force by accelerating (F=ma)[ or changing your momentum]along the nearest geodesic. Light current 15:05, 8 September 2005 (UTC)

Alfred's point was that gravity doesn't count as a force in GR--you're just moving on a "straight line" through curved space when you're falling, so if you're standing on earth you're accelerated upwards. An outside observer using general relativity does not, in fact, think that someone falling has a force on them. ;) -- SCZenz 15:28, 8 September 2005 (UTC)

If you're claiming all accelerations balance, then you're saying all objects are in inertial reference frames, and that they go in straight lines. This is certainly not true, whether you take the Newtonian or General Relativistic perspective on gravity being a force. (We've given examples already.) -- SCZenz 15:28, 8 September 2005 (UTC)

Hmmm! (or Grrrrrr!). I dont know enough about celestial mechanics to argue with you. I will leave you to work out the validity (or not) of my postulate regarding planets etc! All I would say is that, is in curved space, orbiting planets do indeed stay in orbit. Light current 02:08, 8 September 2005 (UTC)

Ahh, yes. But in General relativity (i.e. curved space), an apple sitting on the surface of the earth has an unbalanced force upwards. ;) -- SCZenz 02:18, 8 September 2005 (UTC)

If this is true, what is the effect of the 'unbalanced' force?. If it cannot act on anything (due to the way you look at things in GR), then it seems similar to an 'emf' in an insultor. Light current 15:44, 8 September 2005 (UTC)

The point is the apple should naturally move through the earth on a geodesic, which is the general relativistic analog to a "straight line" in flat space. The effect of the force is to keep the apple on the surface, which in GR is an accelerated reference frame. I freely admit it's not at all obvious why this is a useful way of looking at things, but most good popular books on GR should help, if you get the chance. -- SCZenz 16:02, 8 September 2005 (UTC)

You mean the force is stopping the apple falling thro the surface? (which is its natural tendency in curved space). If this is what you mean then, it is balancing the effect of the curved space. Now youre going to say that the effect of curved space is not a force. But it depends how you look at it. Space is only curved by the presence of mass. In GR, the force of gravity has been replaced by curved space. So then you can say that there is an unbalnced force!. If you are to replace the gravitational force, you must also replace the reaction force by something equivalent!( inverse curvature ath the point of contact?) --Light current 16:57, 8 September 2005 (UTC)

This appears to be similar to the question that [User:Alfred Centauri] was asking me on my talk page. I have declined to answer him until I have read Einstein's theory again. I therefore respectfully decline to answer your version of the question at the moment lest I appear foolish! Dont forget, I'm not a brilliant physicist (or even an ordinary one). Anyway, I thank you and AC for taking interest in my deficient state of knowledge and helping me to see the errors of my ways! I'm just trying to learn more about the nature of the universe (it keeps me out of the bars anyway!) Light current 02:38, 8 September 2005 (UTC)

The postulate in Quantum Mechanics

I would like to avoid the quantum mechanical aspects at the moment if you are agreeable. The reason being that I know very little about quantum mechanics and I feel that the general law of induction should be established or debunked at the macroscopic level first. Light current 01:55, 8 September 2005 (UTC)

Uh, ok. -- SCZenz 02:18, 8 September 2005 (UTC):

There's no force that keeps an electron in the lowest orbital of a hydrogen atom from going lower--it's just stuck there because no lower energy is allowed by quantum mechanics. I will, of course, certainly agree that physical systems displaced from equilibrium are often brought back into equilibrium by forces resulting from the displacement. -- SCZenz 19:56, 7 September 2005 (UTC)

Derive: W=1/2 * L *delta{i*i}

http://fr.wikipedia.org/wiki/Inductance#Puissance_emmagasin.C3.A9e

pf：

with $P=i^{2}*R$ , we take $i^{2}$

thus ($i^{2}$)’ = 2ｉ．ｉ’

＝2ｉ’．ｉ

that refers to

${\frac {d(i^{2})}{dt}}=i\cdot {\frac {d(i)}{dt}}+{\frac {d(i)}{dt}}\cdot i=2{\frac {d(i)}{dt}}\cdot i\,$

by

$P=u\cdot i=L{\frac {di}{dt}}\cdot i\,={\frac {1}{2}}\cdot L{\frac {d(i^{2})}{dt}}\,(=L*{\frac {di^{2}}{2dt}})$

then by

$P={\frac {dW}{dt}}$ .........(basic definition)

therefore,

${\mathbf {d} W=P*dt}$

$dW={\frac {1}{2}}L*d{i^{2}}$

Hence, ${\int }$：$W={\frac {1}{2}}\cdot L(i_{tf}^{2}-i_{ti}^{2})\,$ ........getting the proof.

--HydrogenSu 10:52, 25 December 2005 (UTC)

Sources

I just wanted to put up a couple of my sources for putting up some transformer equations in the mutual inductance section. V_secondary = V_primary N_secondary/N_primary. And convsersely the curreent I_secondary = I_primary * N_primary/N_secondary

[3] [4]

Fresheneesz 09:49, 14 April 2006 (UTC)

The symbol for current

The symbol for current is (I), so why is (i) used here instead? GoldenBoar 18:40, 8 May 2006 (UTC)

If we are referring to a DC circuit, then I is the symbol for current. If we are referring to a general time varying current, then i(t) or just i is used. There are a number of other variations that involve subscripts. Alfred Centauri 00:53, 9 May 2006 (UTC)

L - History of the symbol for inductance

I'm no expert, but the old book I have (1937) states that L stands for "Linked Flux", it is not homage to Lenz. I don't want to just go and arbitrarily edit the page, but here is my reference.

p.291 Electrical Engineering Vol 1 Chester L. Dawes, S.B., A.M. McGraw-Hill —The preceding unsigned comment was added by Ninthbit (talk • contribs) 22:52, 9 February 2007 (UTC).

Simplified Solenoid Inductance Formula

I have been playing with the formula for self-inductance L of a solenoid, and find that it can usefully be rearranged in terms of the lengths of the wire and the solenoid. Note that the formulae ignore end-effects, so are approximate.

$L={\mu _{0}\mu _{r}l_{w}^{2} \over 4\pi \ l_{s}}$ H

which simplifies to

$L=0.1{\mu _{r}l_{w}^{2} \over l_{s}}$ μH

For an air-cored coil this simplifies further to:

$L=0.1{l_{w}^{2} \over l_{s}}$ μH

where

μ₀ is the permeability of free space (4π × 10^-7 henrys per metre)

μ_r is the relative permeability of the core (unitless)

l_w is the length of the wire in metres.

l_s is the length of the solenoid in metres.

Nice, eh? I don't suppose it's original but I haven't seen it elsewhere. I don't think I have made a mistake. GilesW 22:05, 31 May 2007 (UTC)

---

REAL WORLD PROBLEM! —Preceding unsigned comment added by 167.102.133.216 (talk) 19:57, 18 January 2008 (UTC)

Yes, it's very nice, and exactly what I was looking for! Now, maybe you can help me with a real-world problem. I am erecting shortwave antennas. I find that my dipole antenna is resonant at too high a frequency. I know that I could add length to my dipole to lower the frequency. But that would mean getting out the soldering iron and doing repairs outside. I rather just add a lumped inductance somewhere in the middle of each arm of the dipole by taking the antenna wire and rolling it around a piece of plastic pipe. My question is will doing this lower the resonant frequency? Because as I roll the inductor, I am reducing the length of the antenna by the same amount. Specifically, the antenna is now resonant at about 5.0 MHz, and I would like it to be resonant at 3.8 MHz. If anyone takes up this challenge, please email me at jcotton@excite.com and let me know that this has been attempted. Thanks for looking —Preceding unsigned comment added by 167.102.133.216 (talk) 19:55, 18 January 2008 (UTC)

Misleading Equation

Section: Inductance of a solenoid

The equation for inductance displays as L = (μ0 / μr)(N^2 * A) / l, which looks incorrect. When it confused my derivation of relative permeability (μr), I realized there might be a mistake. L = (μ0 * μr * N^2 * A) / l yields the correct results, which I confirmed by deriving μ directly from B/H. I consulted both electronic machinery (transformer) text and electromagnetic text. I am new to this, so I thought someone else should look at it and confirm this before any editing. Thank you for your help. Amanda.Tighe 18:14, 4 June 2007 (UTC)

You're right. I fixed it. Thank you for pointing it out. --Heron 19:48, 4 June 2007 (UTC)

$\lambda$ vs. $\Phi$

The page is inconsistent about the use of $\lambda$ vs. $\Phi$. I'd suggest explaining that distinction in the intro and getting mutual inductance out of the intro, to be explained later.Ccrrccrr 11:43, 3 August 2007 (UTC)

Table with inductances for simple circuits

To be really useful, there should be a table with formulas for wire pair, wire plus wall, coaxial cable (with and without skin effect), rectangle, ring, solenoid ...

BTW, the style of some sections should be checked and corrected by someone with native english, it sounds queer to me. —Preceding unsigned comment added by 84.56.8.130 (talk) 16:42, 4 November 2007 (UTC)

Section 'Inductance of any shaped loop' should be removed

This is redundant to the section 'Vector field theory derivation', it is an intermediate step to derive the von Neumann formula, and the expression gets singular in the self inductance case. —Preceding unsigned comment added by Rdengler (talk • contribs) 16:50, 24 November 2007 (UTC)

SECTION Inductance of simple electrical circuits

CONCERNING the formula for Single layer solenoid: I have no idea what the term with the O is? Is it a zero? If so, why include it at all? Is it another term? If so, what? —Preceding unsigned comment added by 167.102.133.216 (talk) 20:27, 18 January 2008 (UTC)

This is a standard notation in mathematics, see Big O Notation. The complete expression is

$L={\frac {\mu _{0}r^{2}N^{2}\pi }{l}}\left\{1-{\frac {8w}{3\pi }}+\sum _{m=1}^{\infty }\left({\frac {1\cdot 3...\left(2m-3\right)}{2\cdot 4...2m}}\right)^{2}{\frac {2m-1}{2m+2}}2^{2m+1}\left(-1\right)^{m+1}w^{2m}\right\}$

This is the inductance of a cylinder with a constant current around its surface.

—Preceding unsigned comment added by Rdengler (talk • contribs) 17:02, 21 January 2008 (UTC) 

• That seems to give a different number for the coefficients than provided in the article. --207.70.169.36 (talk) 23:02, 1 April 2009 (UTC)

The formula for inductance of pair of parallel wires applies only for d>>2a, i.e. no proximity effect. The formula for inductance of parallel wires, high frequency, incorporates the proximity effect. 192.91.147.35 (talk) 21:16, 30 January 2008 (UTC)

The inductance of a pair of parallel wires exactly is $(l\mu _{0}/\pi )\left(\ln {\left(d/a\right)}+1/4\right)$ in the low frequency case for all d>2a - there is no proximity or skin effect.

Remarks about the definition of magnetic flux

I have removed 'per coil turn' from the definition of inductance.
What is the magnetig flux through a circuit consisting of a thin wire? It is proportional to the number of flux lines that have crossed the wire and now are "captured". This implies that the flux through a circuit with n turns is n times the flux through one turn. Defining the flux geometrically (area times field strenght) leads to inconsistencies and complications. Consider a circular loop with two turns in a constant magnetic field and deform the circuit to a shape looking like an 8. The flux doesn't change.

The magnetic flux through a circuit with a continuous current distribution may be defined by replacing the current distribution with a bunch of thin wires, each carrying a small current. —Preceding unsigned comment added by Rdengler (talk • contribs) 07:41, 19 August 2008 (UTC)

Addition of simple formula

Ths page does a great job explaining the complex concept of inductance, yet I beleive the most important thing to mention is that inductance is the ratio of magnetic flux created by an inductor to the change in current that induces it.

The formula

L = Φ/i

where L is the inductance in Henries, Φ is magnetic flux in Webers, and i is the current in Amperes, is the simplest formula I know to give a rudimentary understanding of inductance. Is this correct? If it is, I will add it to the page to allow beginners to understand better. --Skyfinity (talk) 23:07, 29 November 2008 (UTC)

Hear hear -- I came here to make exactly this point. I'm not a physicist but a mathematician, and I could not understand from the page the DEFINITION of inductance, only about manifestations of it. The capacitance page gets it right. akay (talk) 15:31, 14 January 2009 (UTC)

Inductor symbol error

In the diagram, the arcs that form the symbols of the inductors are the wrong way round. The convex sides of the arcs should face each other.

Where a single inductor with a core is depicted, the arcs face the core depicted by parallel lines.

I was brought up on BS3939, superseded many years ago by EN60617-4 and the BS version of it.

see: EN60617-4:1996 Graphical symbols for diagrams. Basic passive components (ISBN 058026745 8). There are numerous engineering publications that use those symbols, mostly correctly(!). GilesW (talk) 21:35, 15 January 2009 (UTC)

'Mutually inducting inductors'

I think 'Coupled inductors' would be more technically correct than 'Mutually inducting inductors' in the text of the diagram (and anywhere else, if applicable).GilesW (talk) 21:41, 15 January 2009 (UTC)