A NEW CURRENT WEIGHER, ETC. 543 



as far as terms involving R* or R*p*. Hence, to the same order, 



If F be the force parallel to the axis, experienced by the helical filament defined 

 by h = 0, y = 0, and if F 7 be the force on the helical wire when carrying the same 

 current, we have 



where the force on the helix h, y is expanded by TAYLOR'S theorem. On integration 

 the first term yields F u exactly, since the total current is i. For the other terms we 

 may use (6), and may replace (A + A)" 1 by A" 1 hA.~ t + lt 2 A.~ 3 and p*(A. + h)~* by 

 />'A~ S . When we substitute for h and y from (7) and integrate, we obtain 



F _ F gKB'f.l/d'F PF\_ 1 JF 1 *F/R'_jV. 1 ^F/K'_^\1 

 A \B\dx* dy'/ 4A</.r A'f/o-'Us 8/^P^\16 8/J' 



Using (4), and inserting the value of K, we find 



F ,_ F _RV R'-2p'\HF/ R'-2?A RPrPFl 

 8A\ 4 A* /IcArA 2 A 8 / 8A2j' 



or, as far as the terms involving R 4 or R*p*, 



F'-y R * Ji 3(R'-2j) a )]rfF R 4 d'F 

 "8AI 4A ]dx. 24A a (/x 8 ' 



This expression includes all the terms up to R 4 or R^r arising from the differential 

 coefficients of not greater than the second order in the Taylor expansion in (9). 



In the case of the ampere balance it is unnecessary to go beyond terms involving 

 R 1 . To this order we have 



F' = F -^^ .......... (10). 



8 A 



It is easy to give a physical interpretation to this result. For, if we take a helical 

 filament of radius A 2, with its ends in the same planes as the centres of the terminal 

 sections of the helical wire, the force on it is 



The first two terms of this series will be the same as the terms shown in (10) if 

 2 = R*/8A. Hence, as far as correcting terms involving R 3 , F' is the force on a helix 

 of radius A R a /8A. Thus, the force experienced by a wire helix of mean radius A 

 is the same as that experienced by a filamentary helix of radius A R J /8A. It is 



