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I Confusion in Maxwell's derivation of Ampere's Force Law

  1. May 22, 2016 #1
    Hi everyone here. I have my question in the following attached pdf file
     

    Attached Files:

  2. jcsd
  3. May 22, 2016 #2
    Sorry I didn't posted my question directly. It was because of unavailability of Maths symbols
     
  4. May 22, 2016 #3

    robphy

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    I doubt anyone else will read your file.
    This site supports ##\LaTeX## ...
    enclose your expression (say) \xi\nu with opening double-# and closing double-# to produce ##\xi\nu##
     
  5. May 22, 2016 #4

    jtbell

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  6. May 22, 2016 #5
    I am reading Maxwell's "a treatise on electricity and magnetism, Volume 2, page 156" about "Ampere's Force Law". I have some confusion in the following pages:

    Ampere2.PNG Ampere3.PNG

    My question is of two parts:
    1.
    Equation 20, i.e. ##P=\dfrac{B+C}{2r}## is the outcome of special case (i.e. l=1, m=0, n=0)

    But in Page 156, Article 517, Maxwell says: "We can now eliminate P, and find the general value of ##\dfrac{dX}{ds}## and uses this formula (i.e. ##P=\dfrac{B+C}{2r}##) in the general case.

    However in the general case, where 0 < l, m, n < 1, and hence
    ##\dfrac{d^{2}X}{dsds'}=l\left( \frac{dP}{ds'}\xi^{2}-\dfrac{dQ}{ds'}+(B+C)\dfrac{l'\xi}{r}\right) +m(...)+n(...)\neq0##
    (since direction of X is not in the direction of ds)

    therefore,
    ##\dfrac{dX}{ds}=l\left[ (P\xi^{2}-Q)_{(s',0)}-\int\limits_0^s' (2Pr-B-C)\dfrac{l'\xi}{r}ds'\right] +m\int\limits_0^s'(...)ds'+n\int\limits_0^s'(...)ds'##
    Now in this general case, how can we get ##P=\dfrac{B+C}{2r}##.

    If ##P\neq\dfrac{B+C}{2r}## in general case, what does Maxwell mean by "We can now eliminate P, and find the general value of ##\dfrac{dX}{ds}##"
    2. How can one get equation 21 from equation 15. Please give a lengthy derivation.
     
  7. May 23, 2016 #6
    Please somebody answer my question.
     
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