THE DIRECT-CURRENT MOTOR CHAPTER I THE INDUCTION FACTOR WHEN a straight conductor carrying a current is placed in a uniform magnetic field in a plane at right angles to the magnetic lines, it experiences a force urging it to move in that plane at right angles to its own length. If H is the intensity of the field in lines per square centimetre, the length of the conductor in centimetres, and i the current in the conductor measured in amperes, the force on the conductor is given by The sign of ƒ can be changed by reversing the sign of either H or i; hence the direction of the force can be reversed by altering the sign of the field or of the current. If both are altered at the same time, the direction of the force is unchanged. B Fig. 1 is a section of the armature and pole pieces of a dynamo at right angles to the shaft. A and B are two surface conductors, one on each side of the plane through the centre containing the brushes. The currents in these conductors are in opposite directions, but the lines of force due to the magnets are in the same direction. Hence the forces acting on the conductors constitute a couple. If any number of conductors are arranged evenly round the armature, the action of the field on the current in the conductors on one side of the brush plane, combined with that on the other side, results in a couple tending to turn the armature about its axis. Let there be A surface conductors, and let their axes lie on a cylindrical surface of radius r and length l centimetres, as shown in Fig. 2. The distance between any drical area enclosed by them is square centimetres. If 2πι Adn dn lines enter the cylindrical surface between them, 2πrl will be the intensity of magnetisation at the surface of the 360 FIG. 2 cylinder, that is, the number of lines per square centimetre crossing the tangent plane to the cylinder at that point. In Fig. 3, d represents the axis of one of the conductors lying in the cylindrical surface, and we assume that the lines of force enter the cylinder in the direction indicated by the sheaf of arrows, so that the plane at right angles to the lines of force makes an angle of 0 degrees with the tangent plane at d, the line edc representing a portion of the cylindrical surface in a plane at right angles to the shaft. The force f on the conductor is, as we have seen, Hli10-1, where I is the number of lines per square centimetre in the plane making degrees with the tangent at d. Substitute for H its value in terms of the intensity in the Ө force acts in a direction making 0 degrees with the tangent at d. Hence the resolved part of the force in the direction The sum of the tangential forces on all the conductors round one side of the armature is thus AiN 2TTY 10-1, where N |