Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/326

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691.] In calculating the electromagnetic action of a current flowing in a straight conductor of any given section on the current in a parallel conductor whose section is also given, we have to find the integralRhwhere $$dx\, dy$$ is an element of the area of the first section, $$dx^\prime \, dy^\prime$$ an element of the second section, and $$r$$ the distance between these elements, the integration being extended first over every element of the first section, and then over every element of the second.

If we now determine a line $$R$$, such that this integral is equal to Rhwhere $$A_1$$ and $$A_2$$ are the areas of the two sections, the length of $$R$$ will be the same whatever unit of length we adopt, and whatever system of logarithms we use. If we suppose the sections divided into elements of equal size, then the logarithm of $$R$$, multiplied by the number of pairs of elements, will be equal to the sum of the logarithms of the distances of all the pairs of elements. Here $$R$$ may be considered as the geometrical mean of all the distances between pairs of elements. It is evident that the value of $$R$$ must be intermediate between the greatest and the least values of $$r$$.

If $$R_A$$ and $$R_B$$ are the geometric mean distances of two figures, $$A$$ and $$B$$, from a third, $$C$$, and if $$R_{A+B}$$ is that of the sum of the two figures from $$C$$, then Rh

By means of this relation we can determine $$R$$ for a compound figure when we know $$R$$ for the parts of the figure.

692.]

(1) Let $$R$$ be the mean distance from the point $$O$$ to the line $$AB$$. Let $$OP$$ be perpendicular to $$AB$$, then Rh