160 



J. K. Roberts: 



111 order to compute the values of p and \ of Equation (1) for this 

 case two points on the graph are used. 



When N=10, ^}/=lOA x 10«, and when N = 20, ./.= 12.7 x 10" 



In the same way as before we get 



p = 4.63xl0^ E.M.U.\ ,<.> 



X = 382cm. / ^' 



Using these values of A and p the couples which sliould correspond to 

 •different values of N can be calculated from formula (1) to be : — 

 N = 15, i/.= 12-33 X 10''' and N = 25, ./.^ 12-29 x 10« 

 These are plotted in Fig. 2 in the calculated curve. 



In order to apply these results to the design of a new apparatus, 

 it is necessary to compare the values of p as calculated al)ove with 

 the values of the resistances from end to end of the two cylinders 

 which were used. If we do this we get the following : — 



P/Kesistance of copper cylinder =(a) 17-6 and (b) 24-2 (4) 



Tiie values of X obtained were A = (a) 92-5 and (b) 382 (5) 



llie reason for the fact that p is about twenty times the resist- 

 ance of the copper from end to end, which we may call a, is thai if 

 r is thf resistance of each of the 2.n strips by which the copper is 

 replaced, 



R^2«o- . 



