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

• faheemahmed6000
In summary, the conversation is about a question regarding Maxwell's "A Treatise on Electricity and Magnetism" and specifically about "Ampere's Force Law". The question is divided into two parts, with the first part asking about the outcome of a special case and the second part asking for a derivation of equation 21 from equation 15. The conversation also mentions the use of LaTeX on the site and provides a link for an introduction to it.

#### faheemahmed6000

Hi everyone here. I have my question in the following attached pdf file

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• confusion ampere.pdf
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Sorry I didn't posted my question directly. It was because of unavailability of Maths symbols

This site supports ##\LaTeX## ...
enclose your expression (say) \xi\nu with opening double-# and closing double-# to produce ##\xi\nu##

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:

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.