Vol. 6, 1920 
PHYSICS: C. BARUS 
61 
corresponding to the current will appear. The absolute value of the set 
thus depends upon the past history of the iron, however varied this may 
have been, with the understanding that the set produced by the maximum 
current in either direction is characteristic of the strength of that current. 
Differential values, in other words, remain the same. 
3. Data. — Finally the numerical equivalents of the observations made 
may be stated. The elastic torsional coefficient for a tube of the dimensions 
given may be computed with the usual equation, using the differential 
form dT/dr = 2Trnr^d/l (mean radius r and length /) and taking the rigidity 
as w = 8.2 X 10^^. If T is the torque corresponding to the twist of d radians, 
the relation was found to be 
T = 10^ X 4.7^. 
On the interferometer, if the breadth of ray parallelogram is 6 = 10 cm. 
and is the rotation of the mirror mm' around a vertical axis, correspond- 
ing to the displacement of mirror AN (the mirror being at i = 45 ° to the 
rays), the relation will be, since = AA/" cos t/b, 
Ad = 0.071 AAT. 
Hence, if we assume that the resistance to magnetic set is the same as the 
elastic resistance for the same twist A^, the above values of AN will 
correspond to the following data : 
Current AA^ X 10* T X 10"^ E E/vol. 
10 Am. 31 1.03 1.13 .10 
' 20 Am. 65 2.16 4.97 .43 
The energy, E, potentialized by the magnetic set is computed as T Ad/2. 
From the volume of iron in the tube, 11.5 cm^., the data of the last column 
follow. 
My conception of this phenomenon is that of two concentric circular 
fields in opposite directions, one within the other on the outside of the 
tube, introducing a rather intense vortex sheet in the thin walls of the tube, 
in which the circular fields terminate. 
4. Longitudinal Field. — ^The longitudinal strain produced by a longi- 
tudinal field is well known. The question may be asked whether in this 
case, in a free iron bar, there is any corresponding torsion. This was 
easily answered by slipping a helix over the steel tube. Feeding it with 
I amperes, the micrometer displacement AN was successively 
/= 1.5 4.5 9.4 1.5 Am. 
105 X AiV = 15 10 10 5 Cm. 
This means, no doubt, that the residual torsion left by the last experiments 
is being eliminated. 
1 Advance note from a report to the Carnegie Institution of Washington, D. C. 
