Page:Elementary Text-book of Physics (Anthony, 1897).djvu/365

§ 296] exerted by a magnet pole on an element of the circuit is therefore $$\frac{mil}{r^2} \sin \alpha,$$ and this force urges the element to move at right angles to itself and to the line joining it with the magnet pole. Since the action between the pole and the circuit is mutual and the work done dependent only on their relative displacements, the force which each element of the circuit exerts on the pole is also equal to $$\frac{mil}{r^2} \sin \alpha,$$ and tends to urge the pole to move at right angles to the plane containing it and the element of the circuit. This action on the magnet pole is the same as that deduced by Biot from his study of the force between a pole and a long straight current.

We will apply this theorem to determine the force due to a circular current on a magnet pole placed at a point on the line drawn normal to the plane of the circuit through its centre. The force on the circuit, and therefore the force on the pole, has been shown to be equal to $$i \textstyle \sum \displaystyle \sin \theta \cos \phi.$$ In the case before us $$H = \frac{m}{R^2},$$ where $$m$$ is the strength of the magnet pole and $$R$$ the distance from the pole to the circuit. Since the elements of the circuit are symmetrical with respect to the pole, the force on the pole is along the line joining it to the centre of the circuit. The angle $$\theta$$ therefore equals $$\frac{\pi}{2}$$ and $$\sin \theta = 1;$$ the angle $$\phi$$ is the angle between the radius of the circle and $$R;$$ and $$\phi = \frac{r}{R},$$ where $$r$$ is the radius of the circle. The sum of all the elements of the circuit equals the circumference of the circle, or $$2\pi r.$$ The force on the pole is therefore equal to $$\frac{2\pi mir^2}{R^3}\cdot$$

If the magnet pole be placed at the centre of the circle, so that $$R = r,$$ the force on it becomes $$\frac{2\pi mi}{r}\cdot$$ Let the radius of the circle be the unit length, or one centimetre; the force acting on the magnet pole is then $$2\pi mi,$$ and if the magnet pole be the unit pole,