412 Prof. E. Edlund on the Theory 



and the point c in the direction cn — vr. Let us now suppose 

 the magnet divided into elementary magnets, of which we 

 have to consider only those situated at the circumference of 

 the magnet, since all the others may be treated in a similar 

 manner. If we call M the magnetic intensity per unit of 

 length of the circumference, the intensity which corresponds 

 to the differential of the angle u will be equal to Mrdu. We 

 will now calculate by means of the formula (6) the magnitude 

 of the induction which is produced in the element of the 

 circuit As, when this element and the ring of the magnet are 

 in rotation in the same direction with the angular velocity v. 

 Let us give to the element Mrdu of the magnet a velocity 

 vr in the opposite direction to the rotation, and to the element 

 of circuit As a velocity ak of the same magnitude and in the 

 same direction. The relative velocity between the two ele- 

 ments will not be thereby modified. The element of the 

 magnet Mrdu enters thus into rest, and As moves within the 

 resultant of ag and ak. The resultant ah consequently repre- 

 sents the magnitude and direction of the velocity which the 

 element of the circuit As receives in this manner. Since en 

 and ak are parallel, the angle oak = cmo = 90°— ?/. But the 

 angle oag being 90°, the angle kag ought consequently to be 

 = u. The horizontal line ac is equal to VR 2 + f 2 + 2Rrcos u, 

 and the cosine of the angle cao or t f will consequently be equal 



R + rcost* 

 to — . , , _ -. femce ag — vti and ak or ah = vr, 



VB 2 + ^ + 2RrcosM v u 



the resultant ah will be equal to v^W + r 2 + 2Hr cos w, and 

 the cosine of the angle hag will be equal to 

 v(R + r cos u) 

 v \/ R 2 + r 2 + 2Rr cos u 



But since these two angles are both acute they ought to have 

 the same magnitude. The line ag being at right angles to 

 the vertical plane which passes through the line oa, the re- 

 sultant ah must also, for the reason that the angles named are 

 of equal magnitude, be perpendicular to the vertical plane 

 which passes through ac, and consequently also perpendicular 

 to the line uniting c to the element of circuit As. The dis- 

 tance between As and c is evidently = \/H 2 + R 2 -f- r 2 -f 2Rr cosu. 

 The element of the magnet Mrdu acts normally to the plane 

 which passes through this element, and the direction of motion 

 of the element of circuit ah. To obtain from this the effect 

 along the vertical element As, we must multiply by the cosine 

 of the angle which the normal mentioned makes with the ele- 

 ment As: now this angle is equal to that which the latter 



