110 Mr. G. K. Winter on the Ratio of maximum 



and this is only rendered practicable by joining the end of one 

 layer to the beginning of the next, which is necessarily at the 

 other end of the coil. This may be done either by making the 

 connexion outside the coil (which is only possible when the wire 

 is thick and the layers consequently not very numerous), or the 

 wire may be carried across over the layer just wound; but the 

 economy of this plan is doubtful, and it is rarely adopted ; it in- 

 troduces extra resistance into the circuit, and space is lost by 

 the wire in the next layer having to bridge over the connecting 

 wire. 



Let r= the radius of the wire, 



5= the thickness of the silk or other insulator, 

 m= the sectional area of the space to be filled with wire, 

 /= the average length of one convolution, 

 c= the specific conductivity of the wire to be used, that 

 of pure copper at 0° Centigrade being unity, and 

 p and X= respectively the radius and length of a wire of pure 

 copper equal in resistance to the given external re- 

 sistance. 



It is at once evident that the sectional area taken up by each 

 convolution, if the wire be wound according to the first method 

 (fig. 1), willbe4(r + s) 2 . 



It is also equally evident that the sectional area taken up by 

 each convolution, if the second method of winding the wire be 

 adopted, will be 



2{r + s) V'4[> + s)*-(r + *)* = 2 </3(r + s)*, 



in which the coefficient 4 is replaced by 2 \/Z = 3*4641. This 

 coefficient, which depends entirely upon the method of coiling 

 the wire, we will call a, either value being substituted for it in 

 the numerical solution of the problem, according to the circum- 

 stances of the case. 



With regard to the third method of winding the wire (fig. 3), 

 namely in layers insulated from each other by a layer of paper or 

 other insulator, it will be sufficiently correct for practical pur- 

 poses if we consider s increased by one quarter of the thickness 

 of the insulating layer. 



The sectional area taken up by each convolution being a(r + s) 2 , 



TYl 



the number of convolutions will be — r~« 



a(r + s) 2 



The area of the copper wire will be r 2 7r ; hence the resistance 



of the galvanometer = 9 , — - — ry 

 D rVfl(r-M) 2 



c\ will be the length of a wire of the same conductivity as that 



