Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/187

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.

ch ds

��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

(16)

��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

7 Y

To find -- , the force on ds referred to unit of length, we must

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