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Magnetic Fields In Semi-Infinite Solenoid's

  1. Sep 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose your physics lab class lasts for a very long time - long enough for you to wind a semi-infinite solenoid. (That spool of copper wire is the gift that keeps on giving.) What is the on-axis magnetic field at the end of the solenoid closest to you (ie., not at infinity)?

    2. Relevant equations
    1. i think: http://www.netdenizen.com/emagnet/solenoids/Image34.gif
    2. mu(0)*n*i
    3. http://img.sparknotes.com/figures/2/288a4611d51a3a7ce874c4a906855ac9/latex_img33.gif

    3. The attempt at a solution
    - well i thought that i could use formula 1, but there is a L term in that, and i thought that it wouldnt work because this is a semi-infinite solenoid.
    - the formula from spark notes (4pi*i*n)/(c) , i had never ever seen before so i was scared to go for it.

    i dont really know how to apply any of these formulas to the question.



    heres a problem from spark notes thats practically the same except it asks to find the other end:
    A semi-infinite solenoid is a solenoid which starts at a point, and is infinite in length in one direction. What is the strength of the magnetic field on the axis of the solenoid at the end of a semi-infinite solenoid?

    Solution for Problem
    To solve this problem, we use the superposition principle. If we put two semi- infinite solenoids end to end, we have an infinite solenoid, and the field strength at any point in the infinite solenoid is (4pi*i*n)/(c) . By symmetry, the contribution of each semi-infinite solenoid is equal, so the contribution of one semi-infinite solenoid must be exactly one half of the magnetic field in an infinite solenoid, or

    B = (2pi*i*n)/(c)

    This problem displays the power of the superposition principle, which simplifies what would be a complex calculation.

    Last edited: Sep 30, 2011
  2. jcsd
  3. Sep 30, 2011 #2
    So you have an answer and everything is OK? The magnetic field is one half that of an infinite solenoid.
  4. Oct 1, 2011 #3
    so really the answer would be 0.5*mu(0)*i*n

    i think i get it. superposition of two semi-infinite solenoids counts as a infinite solenoid. so i guess if you find the magnetic field at the non infinite end of a semi-infinite solenoid then it would be just half of the formula.

    sweet. i get it now :D
    Last edited: Oct 1, 2011
  5. Oct 1, 2011 #4
    Another way to think of it which is similar. Set up an integral from plus to minus infinity for the sum magnetic field of the individual coils of the solenoid at some point on the axis. By symmetry of the integral the integral from zero to infinity will be half the integral from plus to minus infinity.
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