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Mutual Inductance of two current elements

  1. Jun 19, 2010 #1
    I've been working my way through an old scientific paper on inductance calculation, and there's a fundamental principle I'm having a problem with.

    Suppose there are two filamentary conductors of any arbitrary shape. Now consider an infinitesimal element dl1 somewhere on the first conductor, and another element dl2 somewhere on the second conductor. My understanding is that the contribution to the mutual inductance of the two conductors due to dl1 and dl2 is equal to the dot product of dl1 and dl2 divided by the distance r which separates them. Then Integrating twice with respect to dl1 and then dl2 should give the mutual inductance of the two conductors (Neumann integral). Is this correct so far?

    If so, then my understanding of the dot product is not very good. Logically, it seems to me that the magnitude of dl1 should be multiplied by the magnitude of dl2 and the result then multiplied by a reduction factor which accounts for the projection of dl1 onto dl2, and and vice versa. My understanding of dot product is that the x, y and z components of the first element are multiplied by the x, y and z components respectively of the second element. While this would account for the the fact that the elements dl1 and dl2 may not be parallel, it doesn't fully account for how the elements may be located in space and the projection of one upon the other, which I think must play a part in the calculation. Can someone shed some light on this?
     
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  3. Jun 19, 2010 #2

    gabbagabbahey

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    So far so good.

    What makes you think that the dot product doesn't account for the projection of dl1 onto dl2? The projection of one vector onto another is defined by the dot product!

    As for accounting for where the elements are located in space, that is taken care of since dl1 and dl2 are both measured from the origin and hence describe the location (and orientation) of the current elements relative to that origin.
     
  4. Jun 19, 2010 #3
    Hi Gabbagabbahey. Thank you for the explanation.

    The detail that has me confused can probably be best illustrated in the following diagram:
    MutIndDiagram.png
    In both cases, the elements dl1 and dl2 are parallel, and hence have the same direction cosines and therefore the same dot product. They are the same distance r apart in both cases. Does that mean that the mutual inductance is the same in both cases?

    My original thought was that in Case 1, the projection of dl1 and dl2 on each other would be zero, and therefore the mutual inductance in Case 1 would be zero.

    Perhaps this is the point that I don't understand. Isn't the dot product dependent only on the direction cosines of the two vectors, and not their displacement from the origin?
     
    Last edited: Jun 19, 2010
  5. Jun 20, 2010 #4
    The mutual induction between two non-closed current elements (thin filament) are NOT well-defined
    Induction requires a closed circuit or something with finite volume
    The derivation of the formula I suppose you are dealing with is as such:
    First,calculate the magnetic flux through circuit 2 caused by current in circuit 1:

    [tex]\Phi = \oint_{{l_2}} {{{\vec A}_1}} \cdot d{{\vec l}_2} = \frac{{{\mu _0}}}{{4\pi }}\oint_{{l_2}} {d{{\vec l}_2}} \cdot \left( {\oint_{{l_1}} {\frac{{{I_1}d{{\vec l}_1}}}{r}} } \right) = {I_1}\frac{{{\mu _0}}}{{4\pi }}\oint_{{l_1}} {\oint_{{l_2}} {\frac{{d{{\vec l}_1} \cdot d{{\vec l}_2}}}{r}} } [/tex]

    Then,by definition

    [tex]M = \frac{\Phi }{{{I_1}}} = \frac{{{\mu _0}}}{{4\pi }}\oint_{{l_1}} {\oint_{{l_2}} {\frac{{d{{\vec l}_1} \cdot d{{\vec l}_2}}}{r}} } [/tex]

    M equals the integral does not mean that dM equals the [tex]\frac{{{\mu _0}}}{{4\pi }}\frac{{d{{\vec l}_1} \cdot d{{\vec l}_2}}}{r}[/tex]
     
    Last edited: Jun 20, 2010
  6. Jun 20, 2010 #5
    Okay fair enough. So then as long as the individual x, y and z components of the current vector of dl1 are multiplied with the respective x, y and z components of the current vector of dl2, and then the magnitudes of these three products are added together this will give the correct dot product?

    Then the result of integrating will give the correct mutual inductance?

    Hence, the dot products for both Case 1 and Case 2 in my example will be the same value (assuming that current is flowing in the same direction in each case)?
     
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