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Magnetic field outside the solenoid

  1. Dec 26, 2015 #1
    1. The problem statement, all variables and given/known data
    I found solution to problem 257 from "300 Creative Physics problems" hard to understand. In that problem we have very long solenoid with coil wounded in one layer. Data such as: density of turns, magnetic field at distance of 5 cm form the axis of solenoid and diameter (1 cm) was given. We hevo to calculate field inside the solenoid. Solution states, that "Because of the symmetry of arrangement, the magnetic field produced by the coil outside the solenoid is similar to that of a straight, current-carrying wire, as it is shown in the figure on page 474 [picture attached]."

    2. Relevant equations


    3. The attempt at a solution
    Then, author used that fact and applied Maxwell second law to circle of radius 5cm. IMO, quoted text is meaningless. Magnetic field lines, which are eascaping interior of solenoid must "return" and re-enter that (Gauss law for magnetism). Also, they will never be perpendicular to the lines inside the solenoid, so they cannot be treated as circular field (produced by straight conducting wire). Please help!
     

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  3. Dec 26, 2015 #2

    Hesch

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    I think you are absolutely right. With the given distances ( 5cm from x-axis, diameter = 1cm ), you can never substitute the coil by a single straight wire along the center-axis.

    Maybe you could substitute the coil by a lot of straight wires, placed along the center-axis, but in a distance = 0.5cm. But then there is no symmetry any longer!
    Using Maxwell ( I think you mean Ampères law? ) it must be remembered that the mean value of the H- / B-field is found along the circulation path, not the field at some distinct location.

    You have not mentioned how dense the windings are placed, but maybe you would get a more accurate result by substituting the coil by a tube, wherein there is a circular current around the center-axis, then cut the tube into small pieces with the width dx, use Biot-Savart, and integrate along the tube to calculate the accumulated field at a distance = 5cm from the solenoid. You can "guess" a circulation current density in the tube = 1A/m, to be corrected having found an incorrect result at distance = 5cm. Then again calculate the field at the center ( distance = 0 ) by means of Biot-Savart, with the corrected current density.
     
    Last edited: Dec 26, 2015
  4. Dec 26, 2015 #3
    Thanks for reply, it helped.
     
  5. Dec 26, 2015 #4

    TSny

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    At a distance of 5 cm from the axis of the solenoid, I believe the B field will be circular about the axis of the solenoid to a good approximation. That is, the field will be approximately the same as the field produced by a long straight wire located at the axis of the solenoid and carrying the same current as the solenoid. So, I think the solution given in the book is probably OK.

    However, I don't understand the figure where the straight line that carries the current ##I_{wire}## is drawn at the location shown in the figure.
     
  6. Dec 27, 2015 #5

    Hesch

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    The practical B-field will not be circular around the axis. It will follow a circulation path ( not circular ) around the windings.
    So I don't agree to, that it's a good approximation:

    upload_2015-12-27_10-58-9.jpeg

    versus

    upload_2015-12-27_10-59-55.png
     
  7. Dec 27, 2015 #6

    TSny

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    What do you get if you integrate ##\vec{B}\cdot d\vec{s}## around the circular Amperian path shown below?
    (The figure is not drawn to scale: we are assuming a long, tightly wound solenoid and the radius of the Amperian path is ten times the radius of the solenoid.
     

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  8. Dec 27, 2015 #7

    jtbell

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    The problem statement says "very long solenoid" which usually means we assume it is "infinitely long". In that limit, there's no way for the field lines to "escape" and "re-enter" the solenoid. If the solenoid is merely "very long", the returning field lines spread out tremendously so the field associated with those returning field lines is negligible.

    I think the key here is that the current does not travel perfectly azimuthally around the axis of the solenoid. (If it did travel perfectly azimuthally, how would it get from one end of the solenoid to the other?) The current has a longitudinal component which produces an azimuthal component of the field.
     
    Last edited: Dec 27, 2015
  9. Dec 27, 2015 #8

    Hesch

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    You get the mean value of the B-field along the Amperian path, not the B-field at a specific location.

    Ampères law may only be used to calculate H-field at a specific location under absolute symmetrical conditions, such as within a toroid or around a straight wire.

    All small pieces of windings in the solenoid are placed skew to the center axis. Thus a solenoid is not symmetrical to a circular Amperian path around the center axis of the solenoid.
     
    Last edited: Dec 27, 2015
  10. Dec 27, 2015 #9

    TSny

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    Yes. But at a distance of ten times the radius of the solenoid, the variation the field as you go around the path will be small compared to the magnitude of B itself. So, to a good approximation you can consider the component of B that is tangent to the path as constant when integrating around the path. The component of B that is parallel to the axis of the solenoid and the component of B that is radially away from the solenoid will be negligible compared to the component that is tangent to the path of integration.
     
  11. Dec 27, 2015 #10

    Hesch

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    Well, that's the problem: The component that is parallel to the axis is the dominating component, and should not be neglected. The component that is tangent to the circular path around the center axis is the one to be neglected.

    Do you disagree with this figure, showing how the real field will be? ( Well something like that ).

    upload_2015-12-27_10-58-9-jpeg.93712.jpg

    If you sketch a circular path within the solenoid, perpendicular to the center axis, the resulting B-field will be zero, because no current will pass through the path.

    If you use Biot-Savart at a point on the center axis, the result will be a B-field (almost) parallel to the center axis, that is not zero, unless you stretch the solenoid so that it will almost become a straight wire.
     
    Last edited: Dec 27, 2015
  12. Dec 27, 2015 #11

    TSny

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    No, the field that is parallel to the axis is very weak due to the vast spreading out of the field lines as they leave one end of the solenoid and return to the other end. For a long solenoid, the circumferential component (tangent to the circular path) will be the dominant field.

    Yes, I disagree with it if it is meant to depict the total field of a long solenoid made from a winding of wire. It leaves out the circumferential field. Pictures like this treat the current in the solenoid as purely circumferential and neglect the component of current parallel to the axis of the solenoid. A solenoid made from a winding of wire will always have a component of current parallel to the axis. No matter how tight you wind the wire, any cross-section of the solenoid will be pierced by the current in the wire.
     
  13. Dec 27, 2015 #12

    Hesch

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    I said: (almost) parallel to the center axis.

    OK, give me some time and I will do a numerical calculation ( tomorrow ).

    Say:

    B = 1 mT at 5 cm from the center of axis.
    Diameter of windings: 1 cm.
    Density of winding: 1 mm per winding.
    Length of solenoid: 1 m. ( 1000 turns, very long ).
    Integration path: Exactly through windings using Biot-Savart.
    Integration steps: 360 per winding ( 1 degree ).
    Calculate the B-field at the center of the solenoid ( with direction ).

    Is this example OK?
     
  14. Dec 27, 2015 #13

    TSny

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    Yes, looks good. Not sure about the number of integration steps per winding, you might have to play around with that.

    Are you going to use the program to calculate the field at 5 cm from the axis to show that the field there is strongest in the direction that would be tangent to the Amperian circular path of radius 5 cm?
     
  15. Dec 27, 2015 #14

    Hesch

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    I'm going to calculate the field at 5 cm from the center of axis by means of Biot-Savart with direction, and with a current = 1A.
    In the exercise, the B-field is known at this location. In my example, B = 1mT. Now, say I calculate B = 0.23 mT at the location, I will correct the current to be:

    I = 1A * 1 mT / 0.23mT = 4.3478 A.

    Then I will calculate the B-field at the center of the solenoid with that current. ( That's the purpose of the exercise ).

    I expect that the directions of the B-fields at the two locations will be parallel to the center-axis within 0.1°. That's what I mean by "almost parallel"

    I will attach the source-code ( Pascal ) for inspection.

    Then you may try to calculate the B-fields at the two locations by means of Amperes law.
     
  16. Dec 28, 2015 #15

    TSny

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    OK

    We'll see. I believe you'll find that the direction of the B field at the exterior point will be tangent to the Amperian path to within a few degrees. It would be even closer to the tangent direction for a longer solenoid.
     
  17. Dec 28, 2015 #16

    TSny

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    Within the solenoid, the resulting component of the B-field that is tangent to the circular path will be zero (or negligible), while the component parallel to the axis of the solenoid will of course be strong inside the solenoid.

    Yes, agreed.
     
  18. Dec 28, 2015 #17

    Hesch

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    I have made my program, and it seems to work.

    I'm really surprised, because outside the solenoid, the B-field whirls around the solenoid at an angle = 89.11° to the center axis.

    All the figures of B-fields close to a solenoid must be redrawn !

    What I don't understand is that a magnetic field line inside the solenoid passes through perfectly straight. When it comes out, it will turn around and will start whirling around the solenoid 123 times. Having passed the solenoid outside, it will turn around again, will reenter the solenoid and will somehow in there, find its own tail again.

    How can it do that? Magnetic curves form a closed loop - exactly. After all this whirling around, they must find their own tail again.

    Well, anyway, my program can calculate the solenoid, described in #12 within 8 sec. A shorter solenoid ( like the one in #16 ) could be calculated much faster ( 100 times ). So maybe some day ( over night ) it could calculate a figure ( new release ) that shows the correct magnetic field.

    I have tested my program with a current = 250A, where the theoretical B-field at the center should be 0.1*π [T] ≈ 0.314159 [T]. My program calculated 0.31414 [T], ( maybe because the solenoid is not infinite long, after all ), so it should be OK.

    My result of the exercise is that the B-field at the center of the solenoid is ≈314 times stronger than outside ( at a distance = 5cm ).
     
    Last edited: Dec 28, 2015
  19. Dec 28, 2015 #18

    TSny

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    Good. I decided to try numerical integration of the Biot-Savart law with Mathematica. I also got 89.11o for the external point of the solenoid with your parameters. So, the component of the field that is parallel to the solenoid axis is only about 1.6% as strong as the component that points in the tangential direction to the circular path around the solenoid.

    Yes, they always draw the figures corresponding to current that flows only azimuthally around the axis of the solenoid without any current flowing along the axial direction.

    I was not able to find any computer drawn magnetic field lines that are accurate for a wire wound solenoid. In the process of searching, I did run across interesting comments in a few papers published in The American Journal of Physics. See some excerpts attached below which I tried to keep brief enough to avoid copyright issues.

    Interestingly, magnetic field lines due to steady currents do not always form closed loops, despite what many textbooks claim. The lines can be quite complex.

    Your program appears to be working great!

    Yes, ideally the field would be stronger at the center by a factor of 100##\pi##.
     

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    Last edited: Dec 28, 2015
  20. Dec 29, 2015 #19
    Hesch, can you share source code of your program?
     
  21. Dec 29, 2015 #20

    Hesch

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    Yes, later today. I will have to comment it. :smile:

    It's written in Borland Turbo Pascal, version 7, ( love it ), so you must have a "DosBox", or you may rewrite it
    in "C" or whatever, ( or in assembly to "tune" the speed ).
     
    Last edited: Dec 29, 2015
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