Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/251

to represent the quantity AB . i₂, the parallelepiped contained by this line, by AM, the magnetic induction, and by AA', the displace ment, will represent the work done during- this displacement.

For a given distance of displacement this will be greatest when the displacement is perpendicular to the parallelogram whose sides are AB and AM. The electromagnetic force is therefore represented by the area of the parallelogram on AB and AM multiplied by i₂, and is in the direction of the normal to this parallelogram, drawn so that AB, AM, and the normal are in right-handed cyclical order.

On Four Definitions of a Line of Magnetic Induction.

597.] If the direction AA', in which the motion of the sliding piece takes place, coincides with AM, the direction of the magnetic induction, the motion of the sliding piece will not call electromotive force into action, whatever be the direction of AB, and if AB carries an electric current there will be no tendency to slide along AA'.

Again, if AB, the sliding piece, coincides in direction with AM, the direction of magnetic induction, there will be no electromotive force called into action by any motion of AB, and a current through AB will not cause AB to be acted on by mechanical force.

We may therefore define a line of magnetic induction in four different ways. It is a line such that

(1) If a conductor be moved along it parallel to itself it will experience no electromotive force.

(2) If a conductor carrying a current be free to move along a line of magnetic induction it will experience no tendency to do so.

(3) If a linear conductor coincide in direction with a line of magnetic induction, and be moved parallel to itself in any direction, it will experience no electromotive force in the direction of its length.

(4) If a linear conductor carrying an electric current coincide in direction with a line of magnetic induction it will not experience any mechanical force.

On General Equations of Electromotive Force.

598.] We have seen that E, the electromotive force due to induction acting on the secondary circuit, is equal to $-{\frac {dp}{dt}}$, where

$p=\int {\left(F{\frac {dx}{ds}}+G{\frac {dy}{ds}}+H{\frac {dz}{ds}}+\right)ds}.$

(1)