APPENDIX I. 



field, i. e., a field the ordinates of which are positive, in Fig. 105, to 

 the left of a, and an outward field to the right of a. It is easy to 

 show that the collective action of all the currents represented by the 

 curve / will be to produce a field as shown by the sinusoidal line A. 

 This curve must obviously pass through the point b, because the mag- 

 netizing effects on both sides of this point are equal and opposite. 

 For the same reason the curve must pass through a. That the curve 

 must be sinusoidal is easily proved, as follows: Let i be the current 

 per centimetre of circumference in b, and let r be the radius of the 

 armature ; then the current through a conductor distant from b by the 

 angle a, will be i cos a per centimetre of circumference. If we take 

 an infinitesimal part of the conductor comprised within the angle d o, 

 the current will therefore be 



di = i r cos a d a, 



and the magnetizing effect in ampere-turns of all the currents com- 

 prised between the conductor at b, and the conductor at the point 

 given by the angle a will be 



di = i r sin a, 



i 



and since the conductors on the other side of b act in the same sense, 

 the field in the point under consideration will be produced by 2 r 

 sin a ampere-turns, i being the current per centimetre of circum- 

 ference at b. 



Since for low inductions, which alone need here be considered, the 

 permeability of the iron may be taken as constant, it follows that the 

 field strength is proportional to ampere-turns, and that consequently 

 A must be a true sine curve. 



When starting this investigation, we had assumed that the field 

 represented by the curve B is the only field which had a physical 

 existence in the motor ; but now we find that the armature currents 

 induced by B would, if acting alone, produce a second field, repre- 

 sented by the curve A. Such a field, if it had a physical existence, 

 would, however, be a contradiction of the premise with which we 



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