INDUCTANCE AND CAPACITY. 



361 



Case L Long solenoid. Consider a coil of wire wound in a 

 thin layer on a long cylindrical rod of non-magnetic material, 

 wood, for example. Let r be the mean radius of the windings, 

 / the length of the coil, and Z the number of turns of wire. 

 The field intensity in the coil is &C = ^.irZijl* and the area of the 

 opening of the coil is Trr 2 , so that the flux through the opening 

 is 4?rV 2 Z/y/, and for the flux turns, we have the value 47rV 2 Z 2 z//, 

 which, divided by i, gives the value of L as above explained. 

 Therefore 



4?rV 2 Z 2 

 L = -. (6) 



in which the inductance L is expressed in centimeters. To re- 

 duce to Henrys, the right-hand member of equation (6) must be 

 divided by IO 9 . 



Case II. Coil wound on an iron core. A coil of Z turns 

 of wire is wound on an iron ring / 

 centimeters in circumference (mean), 

 and s square centimeters in sec- 

 tional area, as shown in Fig. 2. 

 The coil produces through the ring 

 a magnetic flux which is equal to 

 the magneto-motive force of the 

 winding divided by the magnetic re- 

 luctance of the ring.t The mag- 

 netomotive force of the winding is 

 47rZ/', and the magnetic reluctance 

 of the ring is Ij^s, i being a given value of current in the coil, 

 and IJL being the permeability of the iron. Therefore 



or 



Fig. 2. 



fJtS 



* A proof of this equation may be found in Nichols and Franklin's Elements of 

 Physics, Vol. 2, p. 119. 



f See Appendix A of the first volume of this text. 



