Graduation of the Sonometer. 61 



the number of turns in the two ends at least sufficiently nearly 

 to apply to some of the results. 



Let M x be the coefficient of induction of the larger coil on 

 the movable one, M 2 that of the smaller, the former having m 

 turns, the latter n. When the movable coil was 200 millims. 

 from the large and 47 millims. from the small coil, since there 

 was no induced current, 



mM 1 =nK 2 . 

 Applying formula (1), we have c for the larger coil 



= v/200 2 + 25 2 = 201-5, 

 and for the smaller coil 



c = v/47 2 + 25 2 = 53-2, 

 b being the same for both. Then 



m f 1 _ 3 / 25 \ 2 15 / 25 V, t 

 (201-5)H2 4V201-5/ + 8 V201-5/ J 



n (1 3/25 \ 2 15/25 \ 4 35/25 \ 6 

 ~ (53-2) 3 t 2 4V53-2/ + 8 ^53-2/ 8 V53-2/ 



, 2835 / 25 Y s 



Multiplying each side by 2 and finding the successive terms, 



122 



10 9 



6645 



mx ^ \l- -02308 + -00088- &c.} 



or 



^=43-6. 



{1 - -33123 + -18286--09422 + -02633, 



I have applied the formula to the results for various metals 

 given by Prof. Hughes in a table in his paper. In the table 

 below, in the first column are Prof. Hughes's numbers, i. e. dis- 

 tances from the point of no induction. In the second are 

 numbers proportional to mMi-nMj ; where M 1? M 2 are the 

 coefficients of induction of two simple coils calculated on the 

 above hypothesis, m and n the number of turns in the two 

 respectively. In the third column are the resistances for bars 

 of the metal 100 millims. long and 1 millim. in diameter 



