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

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1. May 22, 2016

### faheemahmed6000

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

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2. May 22, 2016

### faheemahmed6000

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

3. May 22, 2016

### robphy

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

4. May 22, 2016

### Staff: Mentor

5. May 22, 2016

### faheemahmed6000

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.

6. May 23, 2016