Page:Elementary Text-book of Physics (Anthony, 1897).djvu/364

350 the circuit, or $$N - N'$$, is equal to the flux of force through the cylindrical surface. Now let $$\theta$$ represent the angle between $$s$$ and $$l;$$ the area traversed by $$l$$ during its displacement is then $$sl \, \sin \theta.$$ Let $$\phi$$ represent the angle between the normal to this area and the direction of the magnetic force $$H$$ acting through it. Then $$H \cos \phi$$ is the component of the magnetic force normal to this area, and $$Hsl \sin \theta \cos \phi$$ is the flux of force through this area. The flux of force through the cylindrical surface is therefore given by $$\textstyle \sum \displaystyle Hsl \sin \theta \cos \phi = N - N'.$$ The energy lost by the displacement, or $$i(N - N'),$$ is equal to $$i \textstyle \sum \displaystyle Hsl \sin \theta \cos \phi$$ and since all parts of the circuit are displaced through the same distance $$s,$$ this loss of energy is equal to the work which would be done on the circuit by a force acting in the direction of $$s$$ and equal to $$i \textstyle \sum \displaystyle Hsl \sin \theta \cos \phi,$$ or by a force acting on each element of the circuit equal to $$i Hsl \sin \theta \cos \phi.$$ We may therefore consider the action of the magnetic field on the circuit as the resultant of an action of the magnetic field on each element of the circuit.

The magnitude and direction of the resultant force which acts on each element may be found as follows: The force $$i Hsl \sin \theta \cos \phi$$ is equal to zero when $$\sin \theta = 0,$$ or when $$s$$ and $$l$$ coincide with each other; it is also equal to zero when $$\cos \phi = 0,$$ or when the direction of $$H$$ lies in the surface described by $$l.$$ The resultant or maximum force which acts on the element is therefore at right angles to $$l$$ and to $$H;$$ the element $$l$$ is urged to move at right angles to itself and to the magnetic force. The magnitude of the force acting on an element is obtained by supposing the element displaced in this direction, that is, along the normal to $$l$$ and $$H.$$ In this case we have $$\sin \theta = 0,$$ and $$\cos \phi = \sin \alpha,$$ where $$\alpha$$ is the angle between the element $$l$$ and the direction of the magnetic force $$H.$$ Substituting these values, the resultant force on the element is found to be equal to $$iH \sin \alpha.$$

In the special case in which the magnetic field is due to a single magnet pole of strength $$m,$$ we have $$H = \frac{m}{r^2},$$ where $$r$$ is the distance from the pole to the element of the circuit. The force