Chap. 10] 



addition to the vertical, and 



/ 



ELECTRICAL METHODS 



777 



Zo = j^ (sin a!2 ± sin ai) sin ipi 

 lOr 



Yo = r—- (sin o!2 ± sin ai) cos <pi 

 lOr 



^ (10-49c) 



Xo = 0, 



where Yo is the horizontal component at right angles to the line Ei-E^ , 

 and Xo is the component parallel with it. 



If the line E1-E2 is inclined at an angle ^2 from the horizontal, all 

 components are effective and 



Zo = -r- (sin Q!2 dz sin ai) sin cpi cos <p2 

 lOr 



Yo = TTT (sin «2 ± sin ai) cos ipi 

 lOr 



Xo = — - (sin 0:2 ± sin «]) sin ^1 sin ^ 

 lOr 



(10-49d) 



The field of a horizontal loop (see Fig. 10-99c) may be calculated by 

 adding the fields of its straight portions, so that 



Zo = 



1 



+ 



1 



+ 



+ 



ri sin 2ai r2 sin 2a2 ra sin 2ai n sin 



-^1 



5m 2a4j 



(10-49e) 



At the center of a square loop with sides a, ri = r2 = rs = r4 = r = a/\/2. 

 Then the sum in the bracket is 4\/2/a, and the intensity 



Zo = 



47 V2 



5a 



(10-49/) 



In the central portion the field is very nearly uniform. Hence, formula 

 (10-49/) may be applied in a fairly large area. In hilly country these 

 formulas remain the same, provided the loop is laid out in a plane on a 

 slope and the vertical component is measured at right angles to the ground 

 surface. The field in the center of a rectangle with the sides a and b is, 

 by application of eq. (10-49e), 



Zo = ^Va^ + 6^(i + iy 



(10-49s) 



