Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/341

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704.] We might deduce the value of $$M$$ in this case from the expansion of the elliptic integral already given when its modulus is nearly unity. The following method, however, is a more direct application of electrical principles.

Let $$A$$ and $$a$$ be the radii of the circles, and b the distance between their planes, then the shortest distance between the arcs is

We have to find $$M_1$$, the magnetic induction through the circle $$A$$, due to a unit current in $$a$$ on the supposition that $$r$$ is small compared with $$A$$ or $$a$$.

We shall begin by calculating the magnetic induction through a circle in the plane of $$a$$ whose radius is $$a - c$$, $$c$$ being a quantity small compared with $$a$$ (Fig. 49).

Consider a small element $$ds$$ of the circle $$a$$. At a point in the plane of the circle, distant $$\rho$$ from the middle of $$ds$$, measured in a direction making an angle $$\theta$$ with the direction of <$$ds$$, the magnetic force due to $$ds$$ is perpendicular to the plane, and equal to

If we now calculate the surface-integral of this force over the space which lies within the circle $$a$$, but outside of a circle whose centre is $$ds$$ and whose radius is $$c$$, we find it

If $$c$$ is small, the surface-integral for the part of the annular space outside the small circle $$c$$ may be neglected.

We then find for the induction through the circle whose radius is $$a - c$$, by integrating with respect to $$ds$$, provided $$c$$ is very small compared with $$a$$.

Since the magnetic force at any point, the distance of which from a curved wire is small compared with the radius of curvature,