Since the force is always at right angles to the direction of motion of the ion, the speed of the ion will not be altered by the action of this force; and if the ion is projected with a velocity v in a direction at right angles to the magnetic force, and if the magnetic force is constant in magnitude and direction, the ion will describe a curve in a plane at right angles to the magnetic force. If [rho] is the radius of curvature of this curve, m the mass of the ion, mv²/[rho] must equal the normal force acting on the ion, i.e. it must be equal to Hev, or [rho] = mv/He. Thus the radius of curvature is constant; the path is therefore a circle, and if we can measure the radius of this circle we know the value of mv/He. In the case of the rapidly moving negative ions projected from the cathode in a highly exhausted tube, which are known as _cathode rays_, the path of the ions can be readily determined since they make many substances luminous when they impinge against them. Thus by putting a screen of such a substance in the path of the rays the shape of the path will be determined. Let us now suppose that the ion is acted upon by a vertical electric force X and is free from magnetic force, if it be projected with a horizontal velocity v, the vertical deflection y after a time t is ½ × et²/m, or if l is the horizontal distance travelled over by the ion in this time we have since l = vt, Xe l² y = ½ -- --. m v² Thus if we measure y and l we can deduce e/mv². From the effect of the magnetic force we know e/mv. Combining these results we can find both e/m and v. [Illustration: FIG. 13.] The method by which this determination is carried out in practice is illustrated in fig. 13. The cathode rays start from the electrode C in a highly exhausted tube, pass through two small holes in the plugs A and B, the holes being in the same horizontal line. Thus a pencil of rays emerging from B is horizontal and produces a bright spot at the far end of the tube. In the course of their journey to the end of the tube they pass between the horizontal plates E and D, by connecting these plates with an electric battery a vertical electric field is produced between E and D and the phosphorescent spot is deflected. By measuring this deflection we determine e/mv². The tube is now placed in a uniform magnetic field, the lines of magnetic force being horizontal and at right angles to the plane of the paper. The magnetic force makes the rays describe a circle in the plane of the paper, and by measuring the vertical deflection of the phosphorescent patch at the end of the tube we can determine the radius of this circle, and hence the value of e/mv. From the two observations the value of e/m and v can be calculated.