at Low Pressures. 261 



as perfect absorbers of the ions and they must consequently take 

 up a negative charge in the same way as the glass wails of the 

 tube. This is no doubt the explanation of the disturbance which 

 the electrodes produce in the discharge. Near the walls of the 

 tube the luminosity appears to be less intense than at the centre 

 of the tube, but this appearance is probably partly due to the 

 greater thickness of glass through which the discharge near the 

 walls is seen. 



The fact that the Hall effect was found to be proportional to 

 the magnetic field and independent of the current shows that the 

 charges which the walls of the tube and the electrodes no doubt 

 take up are too small to influence the measurements made. For 

 since the mean value of n is proportional to the current and (as is 

 shown below) the ratio of n at E to n at E' is independent of the 

 current, hence the difference between n at E and n at E' must be 

 proportional to the current, so that increasing the current ought 

 to have increased the difference between the charges on the two 

 electrodes necessary to make the numbers of positive and negative 

 ions striking then equal, if these charges had been appreciable. 

 According to these considerations the smaller the currents used 

 the more reliable the results obtained should be, and this is 

 exactly what was found experimentally to be the case. 



When the magnetic field is being applied it is easy to see that 

 the luminosity increases in intensity as one passes horizontally 

 from one side of the tube to the other. 



The above theory of the Hall effect shows that this should be 

 the case, for we have 



£.*-»>+ WHX 



%.f z = n(-Z+WX). 



Putting -r^ = -r- 2 = 3 and adding these equations give 

 2/3^ = nHX{k 1 + k 2 ). 



TTV 



Hence log n = — -=- (fc, + k 2 )z + A. 



For a tube of square cross section with its sides respectively 

 parallel and perpendicular to the magnetic field and sides of 

 length a, 



f a 

 I = a \ idz, 



Jo 

 where / is the total current through the tube. 



