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Magnetic field expresion question

  1. Aug 31, 2009 #1
    http://i29.tinypic.com/2rdiuye.jpg

    i cant understand how they got the first expression

    in a normal coil the distance is r
    but here they have a root in the denominator
    like they used Pythagorean theorem and multiplying by sinus

    and i cant understand where they do it here in order to get the
    field expression
    ??
     
  2. jcsd
  3. Sep 2, 2009 #2
    The first expression is the magnetic field due to a current loop of current NI, radius R1 at a point on the axis of symmetry a distance x from the centre. (That is, because the solenoid is so far away we treat it just as a current loop). You can calculate this directly from the Biot-Savart law.

    I can't quite follow what you're saying, but you are right in that the square root comes from a sine; by (cylindrical) symmetry the only component of the field that isn't cancelled acts along the axis of symmetry. So we just take this component to get the sine.

    (Have a look in your textbook for the magnetic field due to a current loop for the details).
     
  4. Sep 2, 2009 #3
    i have this formula for a magnetic inside a coil

    [tex]
    B=\frac{\mu _0NI}{l}
    [/tex]

    this is a cross section of the system.
    to have a mutual induction
    i would need to have some current in one of the wires

    i dont have it here

    ??
     
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