at Low Pressures. 259 



The theory of the Hall effect in salt solutions has been worked 

 out by Donnan {Phil. Mag., Nov. 1898), and Larmor (Aether and 

 Matter, p. 301) gives the theory for a completely dissociated salt 

 solution. 



In air at low pressures the amount of dissociation is relatively 

 very small and the ionisation does not depend merely on the 

 concentration as in salt solutions. 



Nevertheless it is easy to show that, provided the influence of 

 the walls of the discharge tube is neglected, Larmor's theory for 

 completely dissociated salt solutions applies without modification 

 to the uniform positive column. 



Let v x (= k-^X) be the velocity along the tube of the positive 

 ions, and v 2 (= k 2 X) that of the negative ions. 



Then if i is the current density at any point 



i = Xe (A^i! + Jctfiy), 



where n x and n 2 are the numbers of positive and negative ions 



in unit volume respectively, and e is the charge carried by 



an ion. 



Let the axis of the discharge tube AB be denoted by x, and 



the perpendicular horizontal direction of the Hall effect by z. 



dX 

 Then -=— = 0, and since Z is small and probably uniform, we 



may put -=— = 0, whence %= n 2 , and i = Xne {k x + k 2 ). 



In the uniform positive column the ionisation is everywhere 

 equal to the recombination or q = an 2 where q is the rate of 

 ionisation and a the constant of recombination. Now the most 

 important peculiarity of the positive column is that X is inde- 

 pendent or nearly so of i, so that 



q = an 2 = a 



X 2 e* (k, + k 2 y 



is simply proportional to the square of the current density. Hence 

 if owing to the Hall effect the transverse distribution of the 

 current is changed the ionisation will adjust itself automatically 

 to the new conditions and still be everywhere equal to the recom- 

 bination. 



The ionisation and recombination may therefore be neglected 

 and the calculation becomes identical with that for a completely 

 dissociated salt solution. The result for this case as Larmor 

 shows (loc. cit.) is 



Z = \HX{h-h). 



If z is reckoned positive in the direction in which a current going 



