248 



THE MAGNETIC CIRCUIT 



[ART. 70 



Prob. 7. Show that the required flux density in the air-gap of a 

 lifting electromagnet (Fig. 58) can be calculated from the expression 

 B = 15.7\ / oF/A, in kl/sq.cm., where F is the rated supporting force, 

 in metric tons, A is the area of contact in sq. dm., and a is the factor 

 of safety. 



Prob. 8. Show that in an armored tractive magnet (Fig. 60) the 

 tractive effort F varies with the air-gap s according to the law ^s 2 = 3.08 

 kg-cm.. when the excitation is 2000 amp.-turns and the cross-section 

 of the plunger and of the stop is 12 sq.cm. Assume the leakage and the 

 reluctance of the steel parts to be negligible. 



Prob. 9. Referring to the preceding problem, what is the true 

 average pull between the values s = l and s=4 cm., and what is the 

 arithmetical mean pull? Ans. 0.77 and 1.63 kg. 



Prob. 10. Indicate roughly the principal paths of magnetic leakage 

 in Fig. 60, and explain the influence of the leakage upon the tractive 

 effort, with a small and a large air-gap. 



Prob. 11. The flux between two thin and high bus-bars, placed at 

 a short distance from each other, has the general character shown in 

 Fig. 61. Calculate the force per meter length that pushes the bus-bars 

 apart when, during a short-circuit, the estimated current is 50 kilo- 

 amperes. Ans. About 800 kg. per meter. 



Prob. 12. Deduce an expression for the magnetic pull due to the 

 eccentric position of the armature in an electric machine (Fig. 62). 

 A certain allowance is usually made for this pull in addition to the 

 weight of the revolving part, in determining the safe size of the shaft. 



Solution : Since the pull is proportional 

 to the square of the flux density, we 

 replace the actual variable air-gap 

 density by a constant radial density 

 acting upon the whole periphery of the 

 armature and equal to the quadratic 

 average (the effective value) of the 

 actual flux density distribution. Let 

 this value be B e ff kl.per sq. cm. when 

 the armature is properly centered. 

 Let the original uniform air-gap be a, 

 and the eccentricity be e. Since a and 

 e are small as compared to the diameter 

 of the armature, the actual air-gap 

 at an angle a from the vertical is 

 approximately equal to a scos a. Neg- 

 lecting the reluctance of the iron parts 



of the machine, the flux density is inversely as the length of the air- 

 gap, so that we have 



B a = B e ffa/(a-e cos a) = B e ff/[l - (s/a) cos a]. 

 Let A be the total air-gap area to which B refers; then, according 



ae 



FIG. 62. An eccentric armature' 



