322 St. John — Wave lengths of Electricity on Iron Wires. 



2 v" 



ism, § 295) that approximately y 2 == T -- 1 , where £- 2 is the square 



of the frequency, and 1/ is the self-induction for very rapid 

 oscillations, and C the capacity of the system. It is easy from 

 this to calculate an approximate value for the ratio between 

 the self-induction per unit length of the iron and copper cir- 

 cuits. 



Let L = the self-induction of the copper per unit length. 

 Let L' = the " " " iron " " 



Let C =the capacity of either per unit length. 



Using as a basis of calculation the data from the third maxi- 

 mum of the curves in fig. 4 of the plate, the total length of the 

 copper circuit (diameter 0'1201 cm ) is : 



The sides,.. 562-5x2 = 1125 cra 



The closed end, = 30 



The equivalent of the end capacities, 62x2= 124 



1279 

 For the iron (diameter - 1186 cm ) the length is : 



cm 



The sides, 553x2 = 1106 



The closed end, = 30 



The equivalent of the end capacities, 62x2 = 122 



1258 



Since the two circuits have the same frequency the products 

 of the self-induction and capacity are equal. 



1258 2 L' C= 1279 2 LC 



L' 



^=1-034 

 Li 



In the same manner for 



( Copper (diameter 0-08840 cm ) L' , 



( Iron (diameter 0-08847 ) L 

 ( Copper (diameter 0-07836 ) L'_ . 

 J Iron (diameter 0-07850 ) L 



By the use of Lord Rayleigh's formula for inductance under 

 very rapid oscillations, it is easy now to calculate a value for 

 the permeability of the iron. 



Lord Hayleigh's formula is 



MW® 



