1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Self-inductance of solenoid

  1. Apr 1, 2014 #1

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Consider a solenoid of length l, no. of turns N, n = N/l, area A, and negligible resistance. In its middle is placed a single-turn loop of same area (well, slightly smaller) in the solenoid's middle. The loop has, for the moment, a very high resistance so there is no appreciable loop current. The solenoid is sinusoidally excited with voltage v1. ø11 is the flux in the solenoid due to the current i1 in the solenoid, and ø21 is the flux in the loop generated by the current i1 in the solenoid. Obviously, ø21 = ø11.

    So this would seem a tightly coupled ideal transformer since all the flux generated by the solenoid also cuts the area of the loop, so there is no leakage flux nor turns resistances (the loop resistance forms the load for the transformer & so is external to the transformer).

    Easily shown:
    L1 = self-inductance of the solenoid = Nø11/i1 = μ0n2Al,

    and mutual inductance M = ø21/i1 = μ0nA.

    Also, M = k√(L1 L2) where L2 is the self-inductance of the loop. Since the system is tightly coupled, k ~ 1.

    This then says L2 = M2/L1 = μ0A/l
    but this cannot be since it says L2 is a function of a property of the solenoid (namely l). In fact, computing the self-inductance of the loop is known to be exceedingly difficult, requiring elliptical integrals and what-not.

    Now, an objection could be raised: since there is negligible current in the loop, the loop's self-inductance is meaningless. So let's now impose finite resistance R to the loop:

    This is then a transformer with a secondary load R and secondary (loop) current i2. The equations for this transformer, assuming zero solenoid resistance, are

    v1 = L1 di1/dt + M di2/dt
    v2 = -i2*R = L2 di2/dt + M di1/dt.

    Invoking phasors and using the derived expression for M gives

    V2/V1 = μ0nAR/{μ0 n2lAR - jω(μ0 n2 lA L2 - μ02 n2 A2)}
    = 1/{N -(jω/R)(NL2 - μ0nA)}

    and invoking the knowledge as before that v2/v1 must = V2/V1 = 1/N (ideal transformer again) we see that

    NL2 = μ0nA
    which again says L2 = μ0A/l as before when i2 was = 0.

    So that's my dilemma: how can L2 simply depend on the length l of the solenoid and its area when we all know L2 is a very difficult compute.
     
  2. jcsd
  3. Apr 2, 2014 #2

    UltrafastPED

    User Avatar
    Science Advisor
    Gold Member

    Why does the difficulty of a computation matter to a result? "Nature" comes up with results which otherwise require difficult computations all the time ...
     
  4. Apr 2, 2014 #3

    rude man

    User Avatar
    Homework Helper
    Gold Member

    This completely misses my point, which is that the self-inductance of the loop computes two entirely disparate ways, giving two entirely disparate answers. In the case of the loop within the solenoid it turns out to be a simple function of its (the loop's) area and the length of the solenoid, which makes no sense.

    Whereas the stand-alone computation of the loop inductance from basics is extremely complex, is not a function of its presence in an inductively coupled system, and in fact is a function not only of its area but even the radius of the loop wire.

    This should give you an idea of the complexity of computing the loop self-inductance:

    http://www.technick.net/public/code/cp_dpage.php?aiocp_dp=util_inductance_circle

    Note that the formula is an approximation. An accurate result is close to prohibitively difficult. Compare that with L2 = μ0A/l!
     
  5. Apr 2, 2014 #4

    UltrafastPED

    User Avatar
    Science Advisor
    Gold Member

  6. Apr 2, 2014 #5

    rude man

    User Avatar
    Homework Helper
    Gold Member

  7. Apr 3, 2014 #6

    rude man

    User Avatar
    Homework Helper
    Gold Member

  8. Apr 3, 2014 #7

    UltrafastPED

    User Avatar
    Science Advisor
    Gold Member

  9. Apr 3, 2014 #8

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Thanks Ultra. I sent a couple of e-mails to MIT in an attempt to contact one of the three profs in charge, and asked to forward to someone else if they're not available.

    In case you're interested, here's what I posted in one e-mail relating to that chapter 11:

    Specifically, I question the derivation of L2 between equations 11.2.15 and 11.2.16. Is the coil the same length as the solenoid? If not, the derivation does not make sense to me.


    Will let you know if I get a response.

    rude man
     
  10. Apr 6, 2014 #9

    rude man

    User Avatar
    Homework Helper
    Gold Member

    ***********************************************************************

    Since originally posting I have drawn some tentative conclusions:

    1. The formula for mutual inductance M = k√(L1 L2) holds if and only if, when calculating the separate inductances as if the other coil were absent, the magnetic path is the same for both coils.

    This obviously obtains for example in a toroid transformer but not in the case I describe above.

    2. Similarly, if a sinusoidal primary voltage, perfect coupling, and a real secondary load are assumed, there is no imaginary component in the secondary voltage if and only if the condition in (1) obtains. Alternatively said, a step input of voltage gives a step output of voltage if and only if the above-said magnetic paths are identical.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Self-inductance of solenoid
  1. Self Inductance (Replies: 9)

  2. Self inductance (Replies: 2)

Loading...