I was reading an article about partial and total loop inductances from this site and in a section where for calculation of total inductance of a rectangular loop is given as:

 Figure 1

In each portion of the loop we assign a partial inductance value as well as partial mutual inductance between all parts of the loop. (In this case, we only show the partial mutual inductance of the parallel sections, since perfectly perpendicular conductors will not have significant mutual inductance.)

I don't understand:

  1. Why mutual inductance is taken into consideration for this model (for total inductance calculation) or why mutual inductance is even important when circuit is not isolated electrically?
  2. Why perfectly perpendicular conductors will not have significant mutual inductance as mentioned by the author?

Assuming the inductance is a wire ~0.2 to 1uH/m depending on l/r or length to radius ratio of the wire and not a discrete small coil L then the radiating coupling fields of M will cancel on each opposing side.

While orthogonal fields don’t couple (or affect each other, then each side will have an M value thus 2M, and depending on gap or length if square then simplifies to 4M which can be a small ratio of L. Inductance.

The M is small as you can imagine compared to a transformer where the coil gaps are almost touching ( except for magnet wire insulation winding method and core mu ) so in XFMR’s M approaches 0.99 .

But for a wire square loop where the H field flux density are attenuated greatly on the othe side thus M coupling inductance is also a small value yet measurably different than 4x L for length, l in a straight wire.

This L value changes as the length approaches a 1/4 wave or harmonic of this and is ignored in this formula.

But if the link doesn’t make sense to you, neither will this answer.

Saturn PCB tool has a tab for this inductance loop.

  • \$\begingroup\$ +1,hi @Tony Stewart EE75,"then the radiating coupling fields of M will cancel on each opposing side." But isn't this cancellation occur when both parllel wire carry current in same direction ? But here Current in parllel wire will be in opposite direction and flux gets added instead of cancellation ? \$\endgroup\$
    – user215805
    Apr 10 at 7:02
  • \$\begingroup\$ "While orthogonal fields don’t couple" can you please give an example or hint , how orthogonal fields don't couple ? \$\endgroup\$
    – user215805
    Apr 10 at 7:05
  • \$\begingroup\$ The vectors don’t align up so no crosstalk unless not exactly orthogonal. Or some other fringe effect from nearby parts. have you ever tried to look at loop or dipole antenna nulls from the end view? \$\endgroup\$ Apr 10 at 16:56
  • \$\begingroup\$ No I don't have much idea about dipole antenna , that is why i was expecting an answer which is easy to understand, although i understand your some points but still not enough to clear my doubt \$\endgroup\$
    – user215805
    Apr 11 at 12:06
  • \$\begingroup\$ There may be some fringe effects from near field, but H flux that is orthogonal to wire does not add to inductance it must be parallel and inverse gap square-root reduced in coupling \$\endgroup\$ Apr 11 at 14:10

Let's start with the definition of inductance.

''An electric current flowing through a conductor generates a magnetic field surrounding it. The magnetic flux linkage \$\Phi_{B}\$ generated by a given current \$I\$ depends on the geometric shape of the circuit. Their ratio defines the inductance \$L\$''


Now you want to find the amount of flux linkage through the own coil.

1.why mutual Inductance is taken into consideration for this model (for total Inductance calculation) or why mutual Inductance is even important when the circuit is not isolated electrically?

Now flux generated by one line couple with another line which defines the mutual inductance. When you want to calculator total flux linkage by the whole loop, you must consider the total interaction including the coupling between different line segments. Considering line 1 (i.e. \$L_{p1}\$),

\$ \Phi_1 = \Phi_{11}-\Phi_{21}-\Phi_{31}-\Phi_{41}\$ where \$\Phi_{ij}\$ is flux linkage on \$i^{th}\$ line segment due to the flux generated by \$j^{th}\$ segment. Note that the negative sign comes from Lenz's law.

Now, let's consider Ampere’s Law: A current caring conductor generates a magnetic field around it as below:

enter image description here

2.why perfectly perpendicular conductors will not have significant mutual inductance as mentioned by author ?

Now consider having perfectly perpendicular conductor (in red below): enter image description here

Note that the field lines crated by first current element does not couple with the second one as the field lines and current element \$I_2\$ are parellel to each other. (refer Faraday's law.) This means in the above equation \$\Phi_{12}=\Phi_{14}=0\$, i.e.

\$ \Phi_1 = \Phi_{11}-\Phi_{31}=\Phi_{11}-\Phi_{13}\$


\$ \Phi_2 = \Phi_{22}-\Phi_{42}=\Phi_{22}-\Phi_{24}\$

\$ \Phi_3 = \Phi_{33}-\Phi_{13}\$

\$ \Phi_4 = \Phi_{44}-\Phi_{24}\$

Now, the total flux linkage will be \$\Phi_{B}=\Phi_{1}+\Phi_{2}+\Phi_{3}+\Phi_{4}\$.

\$ \Phi_B = \Phi_{11}+\Phi_{22}+\Phi_{33}+\Phi_{44}-2\Phi_{13}-2\Phi_{24}\$

\$ \frac{\Phi_B}{I} =\frac{\Phi_{11}}{I}+\frac{\Phi_{22}}{I}+\frac{\Phi_{33}}{I}+\frac{\Phi_{44}}{I}-2\frac{\Phi_{13}}{I}-2\frac{\Phi_{24}}{I}\$

This is what you have!

enter image description here


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