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

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704.]
TWO PARALLEL CIRCLES.
309
Coefficient of Induction of Two Parallel Circles when the Distance between the Arcs is Small compared with the Radius of either Circle.

704.] We might deduce the value of 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.

First Approximation.

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

= .


Fig. 49.
We have to find , the magnetic induction through the circle , due to a unit current in on the supposition that is small compared with or .

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

Consider a small element of the circle . At a point in the plane of the circle, distant from the middle of , measured in a direction making an angle with the direction of <, the magnetic force due to 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 , but outside of a circle whose centre is and whose radius is , we find it

.

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

We then find for the induction through the circle whose radius is , by integrating with respect to ,

,

provided is very small compared with .

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