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B Solenoid Magnetic Field

  1. Apr 30, 2016 #1
    I am confused about how the magnetic field of a solenoid is calculated in this image. sol.gif
    Why is only the current going into the page included in calculating the magnetic field? Shouldn't the field be twice the amount calculated because there is a magnetic field generated by the part of the turns the bottom with current going out of the page? And I don't understand why the vertical components of the path are taken to be zero. Although it is said that the field is perpendicular, the diagram shows that there is a portion of the field parallel to the vertical path.
     
  2. jcsd
  3. Apr 30, 2016 #2

    Hesch

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  4. Apr 30, 2016 #3
    Yes. But it doesn't answer my question.
     
  5. May 1, 2016 #4

    Hesch

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    Ampere's law is used.
    It says that only the current passing through the area of the rectangle is to be included in the calculation:

    circ H ⋅ ds = N * I

    They are not taken to be zero, but they cancel each other due to symmetrical reasons. The formula is only valid as for the center of the solenoid.

    The upper side of the rectangle is assumed to be so far away, that it has no influence, so this horizontal component is taken to be zero.
    You are the one to sketch the rectangle.
     
  6. May 1, 2016 #5
    Yes I understand that. But if you use ampere's law that way, the circular winding of the wire seems to have no bearing on the field inside the wire. I thought that the field inside the solenoid is increased because of the winding if the wire.
     
  7. May 1, 2016 #6

    Hesch

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    I'm not using Ampere's law that way, just using Ampere's law.

    If you want a "bearing", you must use Biot-Savart's law. It doesn't need symmetry, infinite length of the solenoid, and so on.

    Biot-Savart is accurate in case of finite length of the solenoid. Ampere is an approximation.
     
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