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Inductance of a straight wire

  1. Jul 12, 2012 #1

    htg

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    It is zero, contrary to published formulas. I assume that we are talking about a single wire, not about a twisted bunch of wires. Why is the misconception so popular?
     
  2. jcsd
  3. Jul 12, 2012 #2
    Please elaborate. Why is it zero? Why would the published info be wrong if it is so obviously zero? Why do you get it & not the scientific community? Are you that much smarter than them, or do you have exclusive access to info we are unaware of? I'm just curious how a whole body of investigators, researchers, & practitioners can be so clueless, while the plain & simple truth is obvious to you but not the experts.

    You're making quite a claim, can you back it up? If there is merit to your claim, it may be merely semantics. How do you define inductance? What geometry do you use for the computations? What is wrong with published data? Please explain.

    Claude
     
  4. Jul 12, 2012 #3
    Say what?!!!
     
  5. Jul 12, 2012 #4

    vk6kro

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    Straight wires are used as inductors all the time at microwave frequencies. Usually as tracks on a printed circuit board.

    Even at 50 MHz a wire of 2 inch length can cause oscillation due to its inductance.

    Here is a calaculator for various wires:
    http://www.consultrsr.com/resources/eis/induct5.htm
     
  6. Jul 12, 2012 #5
    "How do you define inductance?"
    This is a good starting point..... perhaps there is some confusion about the definition of inductance that explains everything about this post.
    Can someone give a definitive definition of inductance and we can take it from there.
     
  7. Jul 12, 2012 #6

    htg

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    In usual circuits there is a return path, so you practically always have a loop, which actually has inductance.
    To see why a straight wire has zero inductance, consider two charged balls, a switch - e.g. a MOSFET, and the straight wire. From the Biot-Savart-Laplace law you can calculate the magnetic field generated by the current in one part of the wire and see that it does not induce any voltage in another part of the wire.
     
  8. Jul 12, 2012 #7

    jim hardy

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    then why will a straight wire antenna resonate?
     
  9. Jul 12, 2012 #8

    htg

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    A dipole has capacitance, which, when connected to proper inductance gives you an LC circuit, which resonates at certain frequency.
     
  10. Jul 12, 2012 #9
    So you are saying a wire that is open does not form a circuit ( loop), thereby there is no inductance.

    So by the same analogy, you take a 10uH inductor and hang in the air, there is no connection and no circuit. Then there's no inductance? As it only has inductance if you have it in a circuit that form a loop where current flow.

    Same thing as your gold coin at home don't worth anything as it only worth a lot if you sell it. If you don't sell it, there is no value!!!! OK, I'll give you my address, please sent them to me, there is no value, it's only take up room and is dead weight!!!!:rofl:
     
  11. Jul 12, 2012 #10

    jim hardy

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    and i always figured that capacitance was the return path.

    The directors & reflectors in a yagi will resonate unconnected to anything.

    Inductance is flux linkages per ampere
    and the amperes definitely flow
    and i don't see why flux wouldn't link a straight conductor.

    it's just counterintuitive, that's all.
     
  12. Jul 12, 2012 #11

    jim hardy

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    http://www.ee.scu.edu/eefac/healy/indwire.html


    L = [itex]\Phi[/itex]/I

    Being an old guy, that above is based on a 1908 work appeals to me.
     
  13. Jul 12, 2012 #12
    But the definition of inductance, as correctly stated in a post above, is flux linkage per current. An inductance of 1 henry equals 1 weber-turn per amp. Induced voltage is computed per Faraday as v(t) = -N*d(phi)/dt. Induced voltage does not define inductance. Since N*phi = L*i, then v(t) = -N*d(phi)/dt = -d(L*I)/dt = -L*di/dt - i*dL/dt.

    But dL/dt = 0 for an inductor of fixed value, leaving only the 1st term in the equation above. Nonetheless, L = N*phi/i, is how L is defined.

    You don't need a closed loop to have a current and/or a flux. It's been correctly stated above that capacitance can "close a loop" that is otherwise "open". Displacement current does not require a closed path, actually it is really an open path type of current. Inductance is well defined for an open circuit.

    Also, with wire tables, one can define the geometry as a pair of parallel wires, with a given spacing, and then tabulate an inductance per unit length. It is understood as conduction current in as closed path with a specific value of spacing. As the loop is lengthened the inductance increases linearly.

    Anyway, like I said, I would advise you & others against taking the viewpoint that for decades/centuries countless pros examined a topic but all missed something that is obvious to you. Even a very smart person is not in league of their own. If we held a science trivia contest with the world's best engrs, physicists, chemists, etc., I doubt that 1 person is heads & shoulder superior to all others. It isn't that way.

    I'll elaborate where necessary. BR.

    Claude
     
  14. Jul 12, 2012 #13
    Any circuit in which a change of current is accompanied by a change of flux, and therefore by an induced emf, is said to be 'inductive' or to posess 'self inductance'.
    It is impossible to have a perfectly 'non-inductive' circuit.
    In cases where the inductance has to be reduced to the smallest possible value....for instance, in resistance boxes, (I have one by my side !!!) the wire is bent back on itself so that the magnetising effect of one conductor is neutralised by that of the adjacent conductor.
     
  15. Jul 12, 2012 #14
    My text books define self inductance in terms of induced emf/rate of change of current.
    If the magnetic circuit has constant reluctance then this definition is equivalent to
    L = flux linkage/current

    ???which post defined inductance????
     
    Last edited: Jul 12, 2012
  16. Jul 12, 2012 #15
    The above post by Jim Hardy gave the definition for inductance. By the way, flux linkage per unit current is the basis, & emf = rate of change of current times inductance is derived. L = N*∅/i is the basic definition of inductance. In units, 1 henry = 1 weber-turn/amp. Straight conductors can easily carry current & have an associated flux linkage.

    For a straight conductor, inductance is easily definable. No credible reference says otherwise. Again, if it were so simple that intuition alone can crack it, it would have been cracked in the latter 19th century.

    Claude
     
  17. Jul 12, 2012 #16
    You must be a theorist.

    Short wires called bond wires are used in integrated circuits to connect the bond pads on the surface of the integrated circuit to the pins on the package. These wires are not coils. They present an inductance of roughly 1 nH/mm. You ignore this at your peril.

    If you worked for me, and you tried to explain that there was no way your amplifier was oscillating because the wires have zero inductance, I'd fire you. Empiricism trumps half-cocked theory.
     
  18. Jul 12, 2012 #17
    So it is confirmed that straight wires have an inductance of 1nH/mm.
    How many employees do you have carlgrace?? How many have been fired because of their half-cocked theoretical ideas ?
    Do you employ any 'theorists'; ??
     
  19. Jul 12, 2012 #18
    Well it is confirmed that wires of a specific thickness have a specific inductance per unit length. I think it varies depending on material and geometry but it is always there.

    I was reacting to the OP's aggressiveness and certitude even while holding a notion that is widely known to be false. I say the OP is a theorist (I shouldn't have, it was rude) because this is a trait I have mostly seen in theorists (although of course not always). For example, there were strong theoretical arguments for why silicon transistors were not possible using the contemporary processing technology of the 1950s. That is, until Texas Instruments built them anyway.

    I work in academia now so I don't have any employees. At one time I lead a design team. I haven't seen anyone fired because of half-cocked theoretical ideas, but I have seen people eased out of their jobs due to inability to learn from empirical data and therefore change their views. Being inflexible is an express-lane to failure in the semiconductor business.

    There are many, many theorists where I work. I also worked with some system architects when I was in industry. Theory is extremely valuable, getting high off your own supply and "proving" things that are obviously false is actually destructive in practice. Look at the OP's
    post invoking Biot-Savart to "prove" a MOSFET doesn't have any inductance in the channel. For pete's sake!

    My post was too flippant, but my point was that while simple equations are useful, but they don't necessarily "mean" anything, especially when there is contradictory experimental evidence available.
     
  20. Jul 12, 2012 #19
    You do not employ anyone but you feel that you are in a position to determine who should be employed !!!
    You are rude
    Dont quote irrelevant facts about silicon transistors !!!
    "theory is extremely valuable" .... many here will value this statement
    ....."they dont necessarily "mean" anything, especially when there is contradictory experimental evidence available".
    Can you back this up with hard evidence or is it a subjective, personal opinion?
    Have you read the guidelines to these forums.... agreemernt with text books ????
    If you do not like what you hear there are complaint procedures.
     
  21. Jul 12, 2012 #20
    I did employ people until a few years ago. Do you think that as soon as I took a position in academia I had to stop drawing conclusions based on my own previous experience?
    Yeah I apologized about that. I shouldn't have said that about theory.
    I don't believe it is irrelevant. My point for saying that was showing there are a lot of pitfalls in trusting in "theory" too much. It has direct, specific consequences. TI became the world leader in transistor manufacturing because they chose not to buy into the contemporary state of semiconductor theory.
    Yes, I can back it up. For example,
    • the OP used correct, but simplified reasoning to show there was no inductance in a straight wire. The OP is wrong. That is hard evidence. Do you think the OP is correct?
    • Standard pn-junction theory (at the level the OP was using to "prove" the nonexistance of straight wire inductance) doesn't predict avalanche breakdown. So, based on the level of theory the OP was using, the OP would categorically state "Avalanche Breakdown is a misconception". The OP would be wrong. This is not a strawman, since you asked for a concrete example.
    Here's another concrete example:
    • In the mid-1990s a lot of researchers believed that it would be impossible to build high-precision analog circuits in deep submicron CMOS technology because of excess thermal noise due to hot electron effects. This effect is called Drain Induced Barrier Lowering. Rather than give up, people built precision circuits anyway. We had a situation in the early 2000s where theory "proved" it was impossible, yet we had devices in the lab operating. Today, the theory has caught up and the equations modified to show that there is less excess thermal noise in very deep submicron circuits because of something now called the Reverse Short Channel Effect. This is all in the literature.
    My point here is that learning a little theory and then making strident statements about "misconceptions" is not only stupid, it's dangerous.
     
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