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Why Inductances are positive?

  1. Aug 18, 2012 #1
    Exploiting quasistatic approximation, if one wishes to calculate self-inductance of any loop, he is led to the following double line integral:

    [itex]\oint\oint\frac{d\vec{l_{1}}\cdot d\vec{l_{2}}}{r}[/itex],

    where [itex]r[/itex] is the distance between the length elements [itex]\vec{dl_{1}}[/itex] and [itex]\vec{dl_{2}}[/itex].

    Is this integral always positive? If so, what would be the mathematical treatment associated to prove its positivity?
     
  2. jcsd
  3. Aug 18, 2012 #2

    marcusl

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    In a loop, dl1 and dl2 are nearly colinear (thus positive dot product) when r is small. When they are oppositely directed (negative dot product), r is large. The positive contributions to the integral will always outweigh the negative, giving a positive result.
     
  4. Aug 18, 2012 #3
    Thanks ! The argument is quite illuminating. But it still bears some risks for it might run into troubles for topologically complicated loops which orient in a bizarre manner. Am looking for a rigorous approach.
     
    Last edited: Aug 18, 2012
  5. Aug 18, 2012 #4

    Philip Wood

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    But someone ought to say 'thank you' for an argument which, naive or not, is clear, simple and manifestly correct for simple loops.
     
  6. Aug 18, 2012 #5
    Amended.
    Thanks to you too...
    :)
     
  7. Aug 19, 2012 #6

    Jano L.

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    I am afraid the above integral is positive infinity (1/r integrated down to r=0), so it is not really applicable. Such definition with mathematical curve instead of wire works only for mutual inductance of two nonintersecting circuits.

    In your case, when the loop is closed, the self-inductance L can be defined for quasistationary processes as the number L that gives magnetic energy when electric current flows through the loop:


    [tex]
    \int \frac{1}{2} \mathbf H\cdot \mathbf B ~dV = \frac{1}{2} LI^2.
    [/tex]

    Since the energy and I^2 are positive, L has to be positive.
     
  8. Aug 20, 2012 #7
    Hi Jano,
    I will be glad if you could elaborate on why the integral can't be a negative infinity.

    Also, going by the definition you provided for L, doesn't the integral term on the left only indicate the total magnetic energy of all space only if it contains linear media so that $$\mathbf H || \mathbf B$$? Or should we take the definition to a more general case correcting the integral to

    $$\frac{1}{2\mu_0}\int B^2 ~dV $$

    ignoring the energy involved in magnetization? But if done so, I think we shall still have some troubles...

    I think the proper definition should be:

    $$ LI ~dI = \int \mathbf {\partial H} \cdot \mathbf B ~dV$$

    Correct me if I am mistaken.
     
    Last edited: Aug 20, 2012
  9. Aug 20, 2012 #8

    mfb

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    The divergence comes from the regions around l1 = l2, where the dot product is always positive for smooth wires.
    However, I would expect an additional minimal separation of the wires in the integration, comparable to the size of the wire itself.
     
  10. Aug 20, 2012 #9

    marcusl

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    Topology cannot matter since parts of a physical loop (which by definition has continuous tangents everywhere) that are close must be parallel and will contribute more positive than distantly related negative portions.

    There is an apparent infinity in your integrand for self-inductance, BTW, but it is integrable. The infinity is typically handled as follows: We can substitute for current [tex]\vec{i}=\frac{1}{\mu}\vec{\nabla}\times\vec{B}, [/tex] and for the i'/v term
    [tex]\vec{A}=\frac{\mu}{4\pi}\int{\frac{\vec{i'}}{r}dV},[/tex] both into the expression for energy. After suitable manipulations, this becomes
    [tex]W=\frac{1}{2\mu}\int{B^2 dV}.[/tex]
    Since B is proportional to I, self inductance is defined as
    [tex]W_{11}=\frac{1}{2}Li^2.[/tex]
    This is, by the way, another way to see why self inductance is always positive. Energy is always a positive quantity, and the current squared is also always positive.
     
  11. Aug 20, 2012 #10

    Jano L.

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    If the medium around the wire is to be described by non-linear relation between H and B, things get little complicated.

    Let us at least assume that H is a monotonic function of B, possibly different in different places (wire, air,...)

    The increase of the magnetic energy will be

    $$
    \delta E_{mag} = \int_V \mathbf H\cdot \delta \mathbf B ~dV
    $$

    One can define magnetic energy by

    $$
    E_{mag}(I) = \int_V \int_0^I \mathbf H\cdot \frac{\partial \mathbf B}{\partial I} ~dI~dV ,
    $$

    but if H(B) is not linear, it won't be quadratic in I, so we cannot define inductance.

    The derivation is a little bit tricky. Try to read the first chapters of the book by Panofsky & Phillips, they discuss energy in these quasistationary phenomena quite well. If you'll get stuck, I'll help you.
     
    Last edited: Aug 20, 2012
  12. Aug 21, 2012 #11

    marcusl

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    Inductance still can be defined in this case, but it becomes current dependent. The effect was used by a company called Vari-L in the 1950's and 60's for their line of variable RF inductors. Passing a DC current through the winding shifted the B-H curve to a point where the incremental permeability was low (compared to zero bias). The inductance seen by a small RF signal thus depended on the point set by the DC bias, which could be varied.
     
  13. Aug 21, 2012 #12

    Jano L.

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    That is interesting Marcus, thank you.

    For emroz, it is perhaps good to say then that such an inductance has to be defined by the usual way

    $$
    drop~V = L(I)dI/dt,
    $$

    where drop is across the wire in the direction of current. Because L depends in I, the energy is then not a simple quadratic function of current.
     
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