ELECTRIC CIRCUITS 89 



66. Example, (a) Find the inductance of an endless solenoid 

 in the form of a ring, Fig. 49. 



n = number of turns in solenoid = 1000. 



r = mean radius of the ring = 10 cm. 



I = 2irr = mean length of the flux path through the 



solenoid. 



A = sectional area of the solenoid = 5 sq. cm. 

 i = current in solenoid in c.g.s. units. 



The flux through the solenoid is 



m.m.f 4:irni,. 



</> = = -- = -77-7- lines, 

 reluctance I/A 



the flux per unit current is 



the interlinkages of flux and turns per unit current is 



I 



and is equal to the inductance of the circuit in c.g.s. units; thus, 

 the inductance of a circuit is proportional to the square of the 

 number of turns. 



Substituting the values given above the inductance of the 

 solenoid is 



4 X 3.14 X (1000) 2 X 5 



2X3.14X1^" UtS; 



and the inductance in practical units is 



L = 10 6 X 10- 9 = 0.001 henry. 



(b) Find the energy stored in the magnetic field of the solenoid 

 when the current has reached a value / = 10 c.g.s units. 



The energy stored when any current i is flowing is equal to the 

 work done in building up the current i against the counter e.m.f. 



of inductance e = -7- ; it is therefore 



W = feidt = f X j t idt= X C l idi 



i 2 

 = X ^ergs. 



