20 PRINCIPLES OF ELECTRICAL DESIGN 



where the index " is added to the symbol B to indicate that inch 

 units are used and that the density is not expressed in gausses. 



Knowing B for a given sample of iron, the value of the per- 

 meability /i can be found, and R i calculated by putting the 

 numerical values in the above equation. The necessary m.m.f. 

 for this portion of the magnetic circuit (i.e., path (1) only) is 

 $1 X Ri gilberts. 



The actual procedure would be simplified by using the curves 

 of Fig. 3 thus: 



Referring to the upper curve (for steel), the ampere-turns per 

 inch required to produce a flux density of 100,000 lines per square 

 inch is seen to be 80, and since the iron portion of path (l)is 

 25 in. long, the ampere-turns required to overcome the reluctance 

 of iron only are 80 X 25 = 2,000. 



For the air portion of path (1), we have, 



m.m.f. = $1 X reluctance of air gap 

 or 



0.4?nS/ = 1,000,000 X 4 Q ^ 6 45 



whence 



SI = 1,560 



The total SI for path (1) are therefore 2,000 + 1,560 = 3,560 

 or, 



m.m.f. = 0.4T X 3,560 = 4,470 gilberts. 



Observe now that this m.m.f. is the total force which sets up 

 the magnetic flux in path (2), or, in other words, it is the differ- 

 ence of magnetic potential which produces the flux of induction 

 in the two parallel paths (1) and (2). To calculate the total flux 

 in path (2) we have, 



$ 2 = m.m.f. X PZ 



2ft V 6 45 

 = 4,470 X i * 2 ' 54 = 227,000 maxwells 



The total flux in limb (3) under the exciting coil is, therefore, 

 $ 3 = ^ -f- <|> 2 = 1,227,000. The density in this core is, 



" 3 = -I* = L^ 7 229 = 61)35 o lines per square inch. 



^T-3 Z\J 



The necessary ampere-turns per inch (from Fig. 3) are 8, and 

 the SI for path (3) are 8 X 50 = 400. 





