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anonymous

  • one year ago

On the "Inductance of a Coil" page: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indcur.html#c2 the formula that is depicted includes "A" (the "area of the coil"), without delineating WHAT exact "area" is involved, and, in what 'units'. Is it the full surface area on the *interior*, or on the *exterior*, or both, or is it the cross-sectional area covered by the core (the air, if air-core), or...?

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  1. IrishBoy123
    • one year ago
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    |dw:1434819007297:dw| note \(\Phi = BA\), \(\Phi \) is flux

  2. anonymous
    • one year ago
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    Thanks for that reply... still unsure whether your sketch is depicting (specifically) cross-sectional area of the 'core' itself, or "core + windings". To *ADD* to the over-all problem, I remain uncertain as to whether the formula shown (link in original post) deals strictly with a single-layer coil ONLY...(?) I have *scoured* literally scores of sites (and several engineering books), and tried numerous "online calculators" trying to ascertain HOW to estimate (i.e., calculate) the inductance of a large/multi-turn coil (e.g., several hundred turns of, let's just say for example, #10 AWG magnet wire (core diameter, 10-inches)) ...something similar to the attached image ((from: http://info.ee.surrey.ac.uk/Workshop/advice/coils/air/area.xhtml)) ...though larger diameter air-core. Inductance (L) is needed, in order to determine its Inductive Reactance, in order to begin to *estimate*, ultimately, its current draw. \[X_{L} = 2pifL\] Then (current): \[I = \frac{ V }{ \sqrt{X _{L}^{2}+R ^{2}} }\] (1st time with equation-inserter, here; pi doesn't seem to co-operate)

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  3. IrishBoy123
    • one year ago
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    N in the original linked equation is number of turns. is that what you are looking for? \(\Phi = N \times A \times B \ \) by A, i mean the **cross-sectional area of the 'core' itself** but i see your point. if you had - hypothetical example - very very thick wire in coil, would you need to use a bigger diameter than just the core cross section? i don't think so because it is the B field from (A) the the surface of / tangential to the coil to (B) the centre of the core that is "passed down" the core. having said all that, look for "Nagaoka coefficient" in this link: https://en.wikipedia.org/?title=Inductor this "might" help.... for \(\pi\) in latex type \*( \pi\) without the * or \(X_L = j \ \omega \ L\) is \*(X_L = j \ \omega \ L\) without the *

  4. anonymous
    • one year ago
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    Thanks for the (formula-input) pointers. Still, at this point, I am juggling efforts using several different formulas [and, with the one from THIS site, linked in my original post above, I *remain* unsure as to the UNITS required for "A" (area)] ... and, I am not achieving any "realistic" answers from ANY of said formulas. If you look at the image (link below) given by Harold Wheeler in his 1928 paper "Simple Inductance Formulas For Radio Coils" THAT image slightly resembles my coils, wherein "a" is about 6 inches, "b" is 2-5/8 inches, and "c" is (so far) 2-1/2 inches. [see]: http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=1669896 Obviously, this coil is NOT for a radio <grin>, and, I am needing to 'nail' a correct inductance value (at anywhere between 200 and 480 or-so turns), in order to be able to finish the (current-requirement) calculation, as stated in previous post above. "PS": I am uncertain as to whether the 'Nagaoka coefficient' plays any part here, based on the Wiki comment: "K is approximately unity for a coil which is much longer than its diameter and is tightly wound using small gauge wire (so that it approximates a current sheet)" Any electrical engineers on-board yet...?

  5. radar
    • one year ago
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    Guessing if you the same units for area as you do the length you will be O.K. The formulae in my book used meters.

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