Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/329

 

the dimensions of the transverse section, we first determine the geometric mean of the distances of every pair of points of the section by the method already described, and then we calculate the coefficient of mutual induction between two linear conductors of the given form, placed at this distance apart.

This will be the coefficient of self-induction when the total current in the coil is unity, and the current is uniform at all points of the section.

But if there are $n$ windings in the coil we must multiply the coefficient already obtained by $n^{2}$, and thus we shall obtain the coefficient of self-induction on the supposition that the windings of the conducting wire fill the whole section of the coil.

But the wire is cylindric, and is covered with insulating material, so that the current, instead of being uniformly distributed over the section, is concentrated in certain parts of it, and this increases the coefficient of self-induction. Besides this, the currents in the neighbouring wires have not the same action on the current in a given wire as a uniformly distributed current.

The corrections arising from these considerations may be determined by the method of the geometric mean distance. They are proportional to the length of the whole wire of the coil, and may be expressed as numerical quantities, by which we must multiply the length of the wire in order to obtain the correction of the coefficient of self-induction.

Fig. 45. Let the diameter of the wire be $d$. It is covered with insulating material, and wound into a coil. We shall suppose that the sections of the wires are in square order, as in Fig. 45, and that the distance between the axis of each wire and that of the next is $D$, whether in the direction of the breadth or the depth of the coil. $D$ is evidently greater than $d$.

We have first to determine the excess of self-induction of unit of length of a cylindric wire of diameter $d$ over that of unit of length of a square wire of side $D$, or

$2\log {\frac {R\;\;\mathrm {for\;the\;square} }{R\;\;\mathrm {for\;the\;circle} }}$

${}=2\left(\log {\frac {D}{d}}+{\frac {4}{3}}\log 2+{\frac {\pi }{3}}-{\frac {11}{6}}\right)$

${}=2\left(\log {\frac {D}{d}}+0.1380606\right)$.