Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/171

The current $i$ enters along $AB$, and divides in the circular trough into two parts, one of which, $x$, flows along the arc $BQP$, while the other, $y$, flows along $BRP$. These currents, uniting at $P$, flow along the moveable conductor $PO$ and the electrode $OZ$ to the zinc end of the battery. The strength of the current along $OP$ and $OZ$ is $x+y$ or $i$.

Here we have two circuits, $ABQPOZ$, the strength of the current in which is $x$, flowing in the positive direction, and $ABRPOZ$, the strength of the current in which is $y$, flowing in the negative direction.

Let ${\mathfrak {B}}$ be the magnetic induction, and let it be in an upward direction, normal to the plane of the circle.

While $OP$ moves through an angle $\theta$ in the direction opposite to that of the hands of a watch, the area of the first circuit increases by ${\frac {1}{2}}OP^{2}.\theta$, and that of the second diminishes by the same quantity. Since the strength of the current in the first circuit is $x$, the work done by it is ${\frac {1}{2}}x.OP^{2}.\theta .{\mathfrak {B}}$, and since the strength of the second is $-y$, the work done by it is ${\frac {1}{2}}y.OP^{2}.\theta {\mathfrak {B}}$. The whole work done is therefore

depending only on the strength of the current in $PO$. Hence, if $i$ is maintained constant, the arm $OP$ will be carried round and round the circle with a uniform force whose moment is ${\frac {1}{2}}i.OP^{2}{\mathfrak {B}}$. If, as in northern latitudes, ${\mathfrak {B}}$ acts downwards, and if the current is inwards, the rotation will be in the negative direction, that is, in the direction $PQBR$.

492.] We are now able to pass from the mutual action of magnets and currents to the action of one current on another. For we know that the magnetic properties of an electric circuit $C_{1}$, with respect to any magnetic system $M_{2}$, are identical with those of a magnetic shell $S_{1}$, whose edge coincides with the circuit, and whose strength is numerically equal to that of the electric current. Let the magnetic system $M_{2}$ be a magnetic shell $S_{2}$, then the mutual action between $S_{1}$ and $S_{2}$ is identical with that between $S_{1}$ and a circuit $C_{2}$, coinciding with the edge of $S_{2}$ and equal in numerical strength, and this latter action is identical with that between $C_{1}$ and $C_{2}$.

Hence the mutual action between two circuits, $C_{1}$ and $C_{2}$, is identical with that between the corresponding magnetic shells $S_{1}$ and $S_{2}$.

We have already investigated, in Art. 423, the mutual action