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Formula of Solenoid Magnetic B-field at Ending, Middle and Outside

  1. Aug 19, 2015 #1
    Can anyone way in on the correct Formulas of a Solenoid Magnetic B-field at Ending, Middle and Outside?

    I have the following so far but some of them seem to contradict one another .


    Photo #1:
    Screen_Shot_2015_08_19_at_1_12_32_PM.png

    Photo #2:

    Screen_Shot_2015_08_19_at_1_13_24_PM.png

    Photo #3:
    Screen_Shot_2015_08_19_at_1_14_25_PM.png

    Photo #4:

    Screen_Shot_2015_08_19_at_1_16_22_PM.png

    Photo #5:

    Screen_Shot_2015_08_19_at_1_16_59_PM.png

    Photo #6:

    sol_10.jpg
     
  2. jcsd
  3. Aug 20, 2015 #2

    Hesch

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    There is some confusion in the above:

    "n" is not a number of turns (photo #1). n = dN/dL (windings per length). You can see this substitution in photo #2.

    The idea in photo #2 is to choose a very narrow and very high circulation path exactly in the middle of the solenoid, where the magnetic fields must be symmetric.

    Here Ampere states that circulationH⋅ds = N * I , and due to the narrow/high circulation path and due to the symmetry, all of the MMF must be induced in the part of the path that follows the x-axis, thus Hx,middle = N * I / L = n * I → Bx,middle = μ * n * I. If you calculate the MMF in a path at the end of the solenoid, some of the accumulated MMF will be induced in the vertical parts of the path, thus the horizontal part of the H-field following the x-axis will be weakened.

    In general you cannot use Amperes law to calculate the H- or B-field at some other location than where symmetry rules, because Amperes law speaks of a mean value of the H-field, integrating along a circulation path. Choosing this path asymmetric as to the magnetic field, your calculations will be wrong.

    ( We are back to Biot-Savart ). I have seen your question in your earlier thread, but I have been busy with other things: ( Crashed my car and my PC-screen broke down ). :sorry:
     
    Last edited: Aug 20, 2015
  4. Aug 20, 2015 #3
    Hi Hesch.

    You got me all lost .

    First hope you are OK with respect to crashing your car and I'm sorry about the PC screen.

    So this is the scenario I am looking to solve. I have a Electric Steel Core, a Coil in the center of it and all other relevant info .

    Amperes(i) = 5
    Coil Cross Section Radius = 30mm
    Coil Length = 100mm
    Electric Steel Core Cross Section Radius = 10mm
    Electric Steel Core Length = 200mm
    Number Of Layers = 4
    Number Of Turns Per Layer = 21
    Total Number Of Turns = 84
    AWG = 5
    Diameter = 4.621
    Diameter with Insulation = 4.7219

    I am looking for the Magnetic Strength in Tesla
    1) 5mm away(air permeability) from core(Electric Steel) ending on both sides
    2) at Core Ending no both sides .

    Photo #3,#5: Provide it ?
     
  5. Aug 20, 2015 #4

    Hesch

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    Well, mayby it's a "sunday-chat" but so many things had to be repaired anyway: Airconditon did not work properly due to a malfunctioning cooler. It was changed to a new one because it was crashed in the accident. Then I had some dents in a fender which was also changed. The car was insuranced, so as for a nearly renewed car I will have to pay 800$. That's ok. ( Hope not that the insurance company reads this ). :nb)

    From my profile you can see that I'm from Denmark and therefore have difficulties in reading abbreviations, etc. So:

    What is "AWG"?
    What do you mean by "electric steel core"? Do you mean a core with a high relative permeability, μr , say 1000?
    What do you mean by "a Coil in the center of it" ( the center of the steel core ). Is the coil surrounded by the steel core? Please make a sketch.

    Comment: The magnetic (field) strength ( H-field ) is measured in unit [A/m]. The B-field (magnetic induction) is measured in Tesla (flux density, Wb/m2). I'm not sure: It's your language.

    It's some hard questions you have made, but I have some suggestions as to the answers, combining Amperes law with Biot-Savart. I'll have to think about it until tomorrow.
    Local time: 2:20 AM.
     
  6. Aug 20, 2015 #5

    Well i just replaced my car alternator today . it was $45. not so bad. I'm happy things for you are ok now .

    American wire gauge = AWG (but i think it doesn't matter in our case, i provided the info just incase)
    Electric steel core = yes it is, high relative permeability, Please see table. (https://en.wikipedia.org/wiki/Permeability_(electromagnetism))
    Please see image out here (http://forum.lawebdefisica.com/atta...7be38080238be3&attachmentid=6274&d=1357591573)
     
  7. Aug 21, 2015 #6

    Hesch

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    First of all you draw a circulation path like the one in photo #4, just wider ( larger x ) so that the path crosses the point P.

    Using Amperes law you can easily calculate the mean value of H: Hmean1 = ( N * I ) / s , s is the length of the path.

    Biot-Savart says, as for vacuum ( boldface means vector ):

    dB = ( μ0 * ids x r ) / ( 4π*r3 ) →

    dH = ( ids x r ) / ( 4π*r3 )

    Of course you should be able to find Hmean1 again by integration of ( H(s)⋅ds ) / s along the exactly same path in vacuum.

    Now you insert electric steel ( μr = 4000 ) as core in the solenoid, and the H-field will be disorganized. Anyway just integrate through the path, but when you pass through the steel, you divide H(s) by μr. Call the result of the integration Hmean2.

    Of course you will find that Hmean2 ≠ Hmean1. Say that Hmean2 = k * Hmean1 , ( k will be less than 1 ).

    But you can make Hmean1 = Hmean2 by dividing all the H(s) ( found by Biot-Savart ) by k. Of special interest is of course H(P).

    I think that will work. What do you think ? :smile:

    PS: You should use numeric integration. Say you divide the the coil into 84000 ds's ( 1000 ds's per turn ) and the magnetic path into 10000 ds's, the computer will have to calculate Biot-Savart 840E6 times. Say that each calcultion is done in 10 μs, the computer will do the job in about 2½ hour. So set Amperes (i) = 1, and so on. You can always multiply the result by 5 afterwards. Don't use excel, it will never end.
     
    Last edited: Aug 21, 2015
  8. Aug 21, 2015 #7

    Okay , this is way too complicate for my level of expertise at the moment . can we break it down a little?

    Magnetic field strength is one of two ways that the intensity of a magnetic field can be expressed. Technically, a distinction is made between magnetic field strength H, measured in amperes per meter(A/m), and magnetic flux density B, measured in Newton-meters per ampere (Nm/A), also called teslas (T).

    can i assume that the B is inside of a material like a magnet or a solenoid for example , and H is outside in a none uniform permeability like air?

    if we break down what I am trying to achieve into two parts . can i calculate the solenoid B field at its ending using the formula in Photo #3? and then take the result and find the H or B field at a distance X away from solenoid ending ? maybe by viewing it as a magnet like in this case (http://www.ibsmagnet.com/knowledge/flussdichte.php)

    I am trying to use the result later for generator calculation . where EMF(V) = L(Length of wire) x MPS(Meter Per Second) x T(Tesla)

    but if i find the H then i won't be able to use it in the EMF function . it brings me back to the need for B field at distance from solenoid ending .

    any idea ?

    can we use this version ? (http://spiff.rit.edu/classes/phys313/lectures/sol/sol_f01_long.html)
     
    Last edited: Aug 21, 2015
  9. Aug 21, 2015 #8

    Hesch

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    No, an analogy is that the H-fields corresponds to electric voltage and that the B-field corresponds to electric current density. Flux corresponds to electric current.
    An airgap is a magnetic resistance and an iron-core is a magnetic conductor.
    No, because there is only magnetic symmetri at the middle of the coil, which is a precondition of the calculations. In photo #3 you can only calculate Bmiddle. The calculations in #6, using Amperes law, are just used to "adjust" the Biot-Savart calculations.

    It's like assuming that a resistance is 1.5Ω. Then we calculate that if we put 2A through this resistor, we should have 3V across it. But there is 7V across it, so the resistance must be 7V / 3V * 1.5Ω = 3.5Ω.

    The relation:

    B = μ * H

    is always valid, so having calculated H and knowing μr , you can always find B.
     
    Last edited: Aug 21, 2015
  10. Aug 21, 2015 #9
    But hold on. Why are you saying photo #3 gives us only B-middle ?

    It clearly states :
    B is the magnetic field, in teslas, at any point on the axis of the solenoid. The direction of the field is parallel to the solenoid axis.

    x1 and x2 are the distances, on axis, from the ends of the solenoid to the magnetic field measurement point, in meters.
     
  11. Aug 21, 2015 #10

    Hesch

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    Sorry, I mixed up photos #2 and #3.

    But as for #3, the formula is only valid with a constant μ0 ( or constant μ ). When you insert a core with μr = 4000, the μ = μr * μ0 is not constant along the x-axis, and not constant inside/outside the solenoid.
     
  12. Aug 21, 2015 #11
    Why is it not constant in the middle or at any point in the solenoid? After all it is a formula of a solenoid (of some kind of core )


    If you look at photo #1: (and if you try to use the same data in both photo #1 and #3 you will get the same results for both ending and middle or any point inside the solenoid )

    Note : There is no change in material (permeability)
    Here μ=μrμ0=kμ0 and n is no of turns per unit length.
    Please see link (http://physics.stackexchange.com/questions/95725/magnetic-field-of-a-solenoid-at-the-poles)
     
  13. Aug 22, 2015 #12

    Hesch

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    No, it's a formula of a solenoid with no core ( or inside a material with constant μ ). As for your solenoid, μ will change when a path crosses the end of the core or the surface of the cylinder, with a factor 4000. This will change the all over magnetic field (much).
    Yes, but the data will change, regarding middle/end of the solenoid, thus the magnetic fields will change. Again: Of course the magnetic field will change, inserting some core in the solenoid. Otherwise I could ask: Why is it inserted?

    When you insert a core, the H-field is weakened inside the solenoid, but we know that Hmean must be N*I / s, thus the H-field outside the core must be stronger ( and changes shape ). The all over B-field will get stronger.
     
  14. Aug 22, 2015 #13
    Looking at photo #3 again as It clearly states :
    B is the magnetic field, in teslas, at any point on the axis of the solenoid. The direction of the field is parallel to the solenoid axis.
    x1 and x2 are the distances, on axis, from the ends of the solenoid to the magnetic field measurement point, in meters.

    then if i have a core μ. the measurement of points x1 and x2 along the axis and with in the core which is longer then the coil winding length then the results would be correct as core is constant. ?
     
  15. Aug 22, 2015 #14

    Hesch

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    There is no core in any of the photos. Here is a picture of a solenoid with core:
    simple-electromagnet2.gif
    and here without a core:
    300px-Solenoid,_air_core,_insulated,_20_turns,_%28shaded%29.svg.png
    So you have to calculate with other formulas than given in the photos.
     
    Last edited: Aug 22, 2015
  16. Aug 23, 2015 #15

    So this is what i have decided to do. i will go with no core and change the set up . therefor i can use the formula as is with x1,x2

    what a road we walked till now .
     
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