60 ADVANCED ELECTRICITY AND MAGNETISM. 



of I amperes in the coil of Fig. 47 is exactly the same as to have nl 

 amperes in the coil of Fig. 46. Now with a current of nl in 

 Fig. 46 the total kinetic energy is %L'(nI) z , and with a current 

 of I amperes in Fig. 47 the total kinetic energy is %L"P; and 

 these two amounts of energy are equal according to the above 

 statement. Therefore we have: 



or 



w 2 !/ = L" (i) 



where L' is the inductance of the coil in Fig. 46 and L" is the 

 inductance of the coil in Fig. 47. Therefore according to equa- 

 tion (i) the inductance of the coil in Fig. 47 is n z times as great 

 as the inductance of the coil in Fig. 46. 



39. The inductance of a transmission line. The inductance 

 of a two-wire transmission line in henrys per mile is: 



/>-\ 

 L = 0.001483 logio ( 1 



(i) 



where D is the distance of the wires apart center to center, 

 and R is the radius of each wire; D and R must both be 

 expressed in the same unit. This equation is only approximately 

 true, as will appear in the following discussion. 



Figure 48 is a sectional view of the two wires W and W" of 

 a transmission line with an outflowing current of J abamperes 

 in one wire and an equal inflowing current in the other wire. 

 In order to determine the inductance in abhenrys of one mile of 

 the line it is sufficient to calculate the magnetic flux $ which 

 passes between* the wires for the given current /. Then <I> = LI 



$ 

 or L = j according to Art. 31. 



Now the magnetic flux between the wires is the flux which 

 crosses the plane which is represented by the dotted line PP 



* The flux which crosses through the material of the wires is neglected in this 

 discussion, and therefore equation (i) is only approximately true. 



