INDUCTION INSTRUMENTS 



445 



soidal currents are employed, h A = k A I A sin ut. The field due 

 to the second coil, B, will be h B = k B I B sin (co 0) where & 

 is the time-phase difference of the two currents. If the coils are 

 of equal magnetic strengths (k A I A = k B I B ) and are placed with 

 their planes perpendicular, and if the two currents are in time 

 quadrature (ft = 90), the fields at any instant are as shown in 

 Fig. 257. 



ut 



FIG. 257. Showing components of rotating field. 

 The resultant field is 



= k A I A , a constant. 



H = vk 2 A P A sin 2 ut + k 2 A P A cos 

 That is, the field at is of constant intensity. 

 at any instant is given by the angle y; 



Its direction 



tan y = 



sn 



cos cat 



= tan ut 



y = at. 



The resultant field, therefore, rotates with a constant angular 

 velocity, co, making one complete rotation in the time required 

 for one complete cycle of the current. 



If a copper cylinder be suspended by a thread so that its axis 



