28 



THE MAGNETIC CIRCUIT 



[ART. 13 



kl/sq. cm. One-half of the average length of the path in the steel is 

 in., and in the cast iron 20.5 cm. Hence, with reference to the 

 magnetixation curves, we find for one-half of the magnetic circuit (the 

 other half being identical, it is sufficient to calculate for one-half) : 



amp.-turns for steel core 65 X 37.3 = 2425 



amp.-turns for one air-gap 0.2 X 0.8 X 17200 = 2752 



amp.-turns for the cast-iron yoke 180 X 20.5 = 3690 



Total 



8867. 



Ans. The exciting current is 8867/450 = 19.7 amperes. 

 Prob. 9. In the solution of the preceding problem the effect of leakage 

 is disregarded. It is found by experiments on similar electromagnets 



FIG. 5. An electromagnet (dimensions in centimeters). 



that the leakage factor is equal to about 1 .2, that is to say, the flux in the 

 upper yoke is 20 per cent higher than that in the lower one. This means 

 that out of every 1200 lines of force in the upper yoke 1000 pass through 

 the lower yoke as a part of the useful flux, and 200 find their path as a 

 leakage through the air between the limbs, as shown by the dotted lines. 

 Calculate the exciting current required in the preceding problem, assuming 

 (a) that the total leakage flux is concentrated between the two air-gaps 

 along the line aa\ (6) that it is concentrated along the line 66, at one- 

 third of the distance from the bottom of the exciting coil, that is 6.33 

 cm. from the air-gaps. 



Ans. (a) 44.2 amperes ; (6) 40 amperes. 



Prob. 10. Show that 'it is more correct in the preceding problem to 

 assume the leakage flux concentrated at one-third of the distance from 

 the bottom of the exciting coils, than at the center of the coils. 



