Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/169

490.] a sufficient number of portions connected by flexible joints. Hence we conclude that if by the displacement of any portion of the circuit in a given direction the number of lines of induction which pass through the circuit can be increased, this displacement will be aided by the electromagnetic force acting on the circuit.

Every portion of the circuit therefore is acted on by a force urging it across the lines of magnetic induction so as to include a greater number of these lines within the embrace of the circuit, and the work done by the force during this displacement is numerically equal to the number of the additional lines of induction multiplied by the strength of the current.

Let the element $$ds$$ of a circuit, in which a current of strength $$i$$ is flowing, be moved parallel to itself through a space $$\delta x$$, it will sweep out an area in the form of a parallelogram whose sides are parallel and equal to $$ds$$ and $$\delta x$$ respectively.

If the magnetic induction is denoted by $$\mathfrak{B}$$, and if its direction makes an angle $$\epsilon$$ with the normal to the parallelogram, the value of the increment of $$N$$ corresponding to the displacement is found by multiplying the area of the parallelogram by $$\mathfrak{B} \cos \epsilon$$. The result of this operation is represented geometrically by the volume of a parallelepiped whose edges represent in magnitude and direction $$\delta x$$, $$ds$$, and $$\mathfrak{B}$$, and it is to be reckoned positive if when we point in these three directions in the order here given the pointer moves round the diagonal of the parallelepiped in the direction of the hands of a watch. The volume of this parallelepiped is equal to $$X \, \delta x$$.

If $$\theta$$ is the angle between $$ds$$ and $$\mathfrak{B}$$, the area of the parallelogram is $$ds \cdot \mathfrak{B} \sin \theta$$, and if $$\eta$$ is the angle which the displacement $$dx$$ makes with the normal to this parallelogram, the volume of the parallelepiped is

$$ds \cdot \mathfrak{B} \sin \theta \cdot \delta x \cos \eta = \delta N$$.

Now $$X \, \delta x = i \, \delta N = i \, ds \cdot \mathfrak{B} \sin \theta \, \delta x \cos \eta$$, and $$X = i \, ds \cdot \mathfrak{B} \sin \theta \cos \eta$$ is the force which urges $$ds$$, resolved in the direction $$\delta x$$.

The direction of this force is therefore perpendicular to the parallelogram, and is equal to $$i \cdot ds \cdot \mathfrak{B} \sin \theta$$.

This is the area of a parallelogram whose sides represent in magnitude and direction $$i \, ds$$ and $$\mathfrak{B}$$. The force acting on $$ds$$ is therefore represented in magnitude by the area of this parallelogram, and in direction by a normal to its plane drawn in the direction of the longitudinal motion of a right-handed screw, the handle of which