5 1 6.] ACTION OF A CLOSED CIRCUIT ON AX ELEMENT. 155
The coordinates of points on either current are functions of s or of/.
If F is any function of the position of a point, then fre shall use the subscript (s o) to denote the excess of its value at P over that at A, thus F M = F P -F A .
Such functions necessarily disappear when the circuit is closed.
Let the components of the total force with which A P acts on A A be ii X, ii Y^ and ii Z. Then the component parallel to X of
��the force with which dx acts on ds will be ii - =-7 ds dtf.
��Hence - Jt+tf+fJ . (13)
ds ds r
Substituting the values of R, S, and S from (12), remembering
that -,, ,. , , ,. dr /1/n
ss r-, (14)
��and arranging the terms with respect to , m, n, we find
��-JT7 - ~v -J-, -}-,
dsds l r 2 ds * ds
��Since A, B, and C are functions of r, we may write
��the integration being taken between r and oo because A, B, C vanish when r = oo.
Hence A + S =-*, and C = -. (17)
��516.] Now we know, by Ampere s third case of equilibrium, that when / is a closed circuit, the force acting on ds is perpendicular to the direction of ds } or, in other words, the component of the force in the direction of ds itself is zero. Let us therefore assume the direction of the axis of x so as to be parallel to ds by making I = 1 , m = 0, n 0. Equation (15) then becomes
��i/sd/~ ds ds
To find -- , the force on ds referred to unit of length, we must