Dr. R. A. Houstoun on Magnetostriction. 



81 



BC and DA adiabatics. On AB a quantity of heat q is taken 

 in ; on CD a quantity q-\-dq is given out. The area of the 

 figure is equal to the work done on the wire in the cycle and 

 this is equal to dq. Hence 



^=AC=AK=AGxBJ. 



Also q/T = dq/dT. Therefore, eliminating dq> 



AGxBJ=^T/T. 

 But q, the heat received in the isothermal AB, is equal to 

 bdF, i. e. hBJ. Substituting, 



' b_ _ AG 

 T ~ dT ' 



But AG is the alteration of ^, F constant in going from the 

 one isothermal to the other ; hence 



b _ c^ 

 T " BT' 



The second relation can be proved in an analogous manner 

 by using the B, H diagram. 



If* hysteresis be not disregarded, the cycle becomes 

 irreversible, and in place of q/T = dqldTwe must use 



q _ q + dq 

 T T + rfT 



0. 



Also the figure is no longer a parallelogram as the shape of 

 the isothermals and adiabatics is different according to the 

 direction in which the wire is being put through the change. 

 In order to prove the third relation, assume that the wire 

 is put through the cycle represented in the two following- 

 diagrams . 



Fig. 2 







F^F 







B 



The two diagrams 

 ABCD 



The temperature is kept constant, 

 represent the same cycle, the points ABCD in the one 

 figure corresponding to the points A'B'C'D' in the other 



Phil. Mag. S. 6. Vol. 21. No. 121. Jan. 1911. G 



