160 



J. K. Roberts: 



In order to compute the values of p and X of Equation (1) for this 

 ♦case two points on the graph are used. 



When N=:10, i/^=10.4xl0^ and when N = 20, i/^=12.7x 10'' 



In the same way as before we get 



p = 4.63xl0* E.M.U.\ ,ox 



A = 382cm. / ^^^ 



Using these values of X and p the couples which should correspond to 

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



N = 15, ,/.= 12-33x10" and N = 25, i/.^ 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 above 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) 242 (4) 



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



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

 ance of the copper from end to end, which we may call cr, is that if 

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

 replaced, 



R^2/«a- 



