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Magnetic field of a solenoid

  1. Oct 30, 2009 #1
    What would be the magnetic field of a solenoid if the wire of the solenoid itself is made up of another solenoid? I would be glad to hear any explanation about it.
  2. jcsd
  3. Oct 30, 2009 #2


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    One would have an azimuthal field (from the smaller diameter windings) superimosed on an axial field (of the larger windings).
  4. Oct 30, 2009 #3


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    Thread moved to Classical Physics.

    Hmmm, it seems to me there would be only the azimuthal field within the windings. No appreciable axial field, since that is outside the solenoid that makes up the windings.
  5. Oct 30, 2009 #4

    Vanadium 50

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    On the inside of the inner solenoid. In the region between the inner and outer coils, you have the return flux of the inner solenoid superimposed on the axial field of the outer solenoid. I think the technical term for that field configuration is a "big fat hairy mess".

    If you're really masochistic, you can calculate that when the two solenoids aren't coaxial.

    I think if the original question were clarified, the answers would be clearer as well.
  6. Oct 30, 2009 #5


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    My understanding of the solenoid's configuration is something like this:


    Vanadium seems to have a different picture in mind, so yes some clarification by the OP would be helpful.
  7. Oct 31, 2009 #6
    That seems to be exactly what the OP is describing.

    I think that both fields will exist, but the axial magnetic field will be weakened a result of reduced axial alignment between magnetic dipole moments. If one passes changing direct current going one direction in the windings, I think an axial POLAR electric field will be produced by the changing azimuthal magnetic flux, and that this electric field could be in excess of the axial polar magnetic field that is produced by the changing azimuthal electric flux. If the power density is high enough, this could have some interesting properties, such as the ability to accelerate negative ions in one direction.
    Last edited: Oct 31, 2009
  8. Oct 31, 2009 #7
    Smythe "Static and Dynamic Electricity" 3rd edition pages 296-297 solves the field in a helical solenoid.
    Bob S
  9. Oct 31, 2009 #8
    I think two magnetic fields will be in this system: one in the primary coil and another in the secondary. They will not be superimposed but separated - confined inside the corresponding coils.
  10. Oct 31, 2009 #9
    The magnetic field B is continuous, it exists both inside and outside a solenoid. The return flux for the field inside a solenoid is outside.
    Bob S
  11. Oct 31, 2009 #10
    The magnetic field outside a regular solenoid is much weaker than inside so we can speak of a field jump within the wire "wall". I do not speak of the magnetic field inside a wire.
  12. Oct 31, 2009 #11


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    Bob, can you describe the main features of the field? Or at least confirm/dispell my suspicions about the field, that it:

    1. Is oriented azimuthally, within the coils of the smaller outer solenoid, and
    2. A weaker field is within the large solenoid, along the axis. (It would be weaker because of the smaller turns-per-length of the large solenoid, I suspect.)
    3. A much weaker field, taken to be zero in the infinite-length limit, is outside both solenoids.
  13. Oct 31, 2009 #12


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    Definitely a field within the smaller outer solenoid. I keep going back and forth on whether there is an appreciable field inside the larger solenoid. At first I thought not, since it lies outside the outer solenoid. But now I am thinking that the usual argument with taking

    loopB·dr = μo Ienclosed

    for a loop that is partially inside, partially outside the solenoid, tells us there is an axial field inside the large solenoid.
  14. Oct 31, 2009 #13
    I think there will be two spiral fields of different tensions, the spiral step being determined with d/D. (The wire thickness is taken to zero). If d<<D then both will be azimuthal, with no axial components. By the way, outside the big coil the field will be also spiral, that's my feeling.
    Last edited: Oct 31, 2009
  15. Oct 31, 2009 #14


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    Yeah, I agree about an azimuthal field outside both coils. But I think it's axial inside the large main coil.


    In fact, a normal solenoid would have an azimuthal field outside as well.

    I was confused at first from always reading that the field is essentially zero outside of a standard solenoid. But this isn't true, it is just the axial component that is zero outside the solenoid. By Ampere's Law there would be an azimuthal field of μoI/2πr, the same as for a normal wire.

    At any rate, the smaller outer coil will have this "azimuthal" field (azimuthal w.r.t. it's "axis"), which would result in an axial field inside the large main solenoid.

    So, to summarize, I think we would have:

    1. An axial field within the large main coil, of strength
    B = μo n1 I​
    where n1 is the turns-per-length of the main coil.

    2. An azimuthal field within the smaller outer coil, of strength
    B = μo n2 I​
    where n2 is the turns-per-length of the small coil

    Note that n1 < n2, so that field #1 is weaker than field #2.

    3. Finally, an azimuthal field outside both coils, of strength

    B = μo I / 2πr ​
    where r is the distance to the central axis. This field would be the weakest of the three, since typically

    r >> 1/n1 > 1/n2

    where 1/n is the spacing between adjacent turns in a coil.

    If both coils are of the same handedness, then fields 2 & 3 are in the same direction. If they are of opposite handedness, the fields point in opposite directions.
    Last edited: Oct 31, 2009
  16. Nov 1, 2009 #15
    I vote for three spiral fields. The actual arrangement of wires is extremely important.
  17. Nov 7, 2009 #16
    now how if the same solenoid (which is made up of another solenoid) i.e the 1st solenoid is hollow in the inside and we have a liquid metal flowing within it? (considering fluidity, waves, magnetohydrodynamics)
  18. Nov 7, 2009 #17
    There is no effect on the flowing liquid (conducting) metal in a constant B field along the axis of a solenoid, in part because the direction of flow is parallel to the field and dB/dt = 0. There are propulsion systems using crossed E and H fields perpendicular to the desired direction of flow, using the Lorentz force F = l(I x B) . See paragraph 3.1.1 in
    There are also generators (homopolar) based on the Lorentz equation. See

    [added] There is the possibility that a constant B field aligned with the direction of flow of a conducting liquid might reduce turbulence for high Reynold's numbers (just a guess).
    Bob S
    Last edited: Nov 7, 2009
  19. Nov 12, 2009 #18


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    If the axial field was existing, I strongly doubt it would correspond to the equation you mentionned. The proposed arrangement is not just a solenoid, it is a solenoid inside a solenoid. As we know, a moving electron generates a magnetic field perpendicular to the direction of movement. For the axial field to correspond to this equation, electrons would have to jump transversally from one position in a loop to the same position in the next loop and so on ... then, an axial field would be perpendicular to the electron movement. In our case, the field is perpendicular to the electron movement, which means that there will be an azimutal field inside and around the small coil (your point 2), spiralling around the axis, but as there are no electrons moving directly around (spiralling) the main axis, there should be no corresponding field.

    You can also see it this way: If this arrangement was really creating an axial field, then using toroidal transformers would be somehow useless because the goal is to avoid spreadiong the magnetic field. A toroidal transformer is a small solenoid that is wind in a single loop of a big solenoid. Then an axial field should be detected exiting from both sides of the transformer. I tested it a few years ago (at school) ... although our measurement apparatus was far from a professionnal level, no axial field could be detected. When I asked for the mathematical demonstgration, my teacher told me that the standard equation (the one you used) cannot be used (for the reason I mentionned), and that the mathematical proof was too complicated for me (I guess he didn't want to get into it :smile: )

    Of course, the small solenoid field would exit on both side of it and link to the other side. If an iron rod is inserted in the main axis (and none in the smal coil axis), then the small solenoid field would go through the rod, but this is not an axial field caused by the big "solenoid" (which is not really a solenoid)

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