456 Prof. W. A. Jenkins on 



Details of the Experiment.— The tube, mounted as before 

 on a board capable of rotation, was adjusted until it was 

 exactly parallel to the Earth's field. The method used for 

 this — that of adjusting the position of the tube until, when 

 a large current is sent round the solenoid, no deflexion 

 of the magnet occurs — was described in the previous- 

 paper. 



The solenoid tube was then rotated through {in angle of 

 90°. This was done by means of a telescope and scale and 

 two mirrors at right angles to each other, mounted in a 

 suitable position on the board carrying the solenoid. The 

 actual angle between the mirrors was 89° 25' 30", but so 

 long as the angle is definitely known the fact that it is not 

 90° makes no difference to the experiment. 



The distance of the telescope and scale from the mirrors 

 was 210 cm., and as a rotation of 1° gave a motion of 75 

 divisions of the scale, an accuracy in the determination of 

 the angle of rotation of 12" was obtainable, for a motion 

 of j of a division could easily be followed. 



The source of light, plane reflecting glass, scale and 

 telescope were then arranged as shown in diagram 1 and 

 an image of the scale reflected from face (J obtained. 

 A current was now sent through the tube and adjusted until 

 the deflexion of the magnet was such that the mirror-face D 

 reflected the same mark of the scale on to the cross-wires of 

 the telescope as did the mirror C. The current strength 

 was then measured by means of a Kelvin Balance which was- 

 in the circuit, and the current was found to be '01945 ampere. 



The balance will measure the current to '00001 ampere, 

 and is a very convenient instrument for carrying out the 

 experiment to a fair degree of accuracy. When high 

 accuracy is desired, an electrical method similar to the one 

 described in the previous paper can be used. 



The angle between the »mirrors D and C had previously 

 been found by mounting the system on the table of an 

 accurately calibrated spectroscope. It was found to be 

 44° 29' 15". Thus we get H = F cot 0, or allowing for 

 the fact that the angle of rotation was not 90°, 



TT _ F cos44°29 / 15" 

 sin 44° 56' 15" 



= F -99209. 

 where F = 4:irnC cos a, * being half the angle subtended at 



