CONDUCTIVITIES OF METALS aND ALLOYS AT LOW TEMPERATURES. 391 
Corrections due to the Leads to the Platinum Resistance Coils. 
It has been stated (p. 383) that only 29 + centims. of the total length, 30'6 centims., 
of platinum wire were actually wound on the sleeve; the remainder, l - 5 centims., 
served to make connection to the No. 22|- double silk-covered copper wires leading 
to the resistance bridge. 
Owing to its comparatively large cross section the whole of the copper wire may be 
assumed to have the temperature of the enclosure, but the temperature in each 
0’8 centim. of platinum wire will vary from that of the enclosure to that of the coil 
of which it is the continuation. This will make the mean temperature of the wire as 
determined by its resistance slightly less than the temperature of that part of it on 
the sleeve, and will also lead to the conduction of a small amount of heat from the 
rod to the leads. We have therefore to determine the magnitude of these effects. 
If we measure x along the platinum wire from its junction with the copper wire, we 
have for the temperature excess v at x the equation 
,, sinh ax 
V = V 0 — 7 T-7 , 
sinh at 
where v" 0 is, as before, the temperature of the wire on the .sleeve, and a =■ 
where p is the perimeter, q the cross section, of the wire, k the conductivity of its 
material, and h the external conductivity. 
The mean value v of v is given by 
/ phV 12 
\qkJ ’ 
V = v" 0 
cosh al — 1 
al sinh al 
- v" 
— V 0 
sinh \al 
al cosh \al 
v" 0 
sinh T^al 
\al 
cosh \al. 
For the platinum wire used, p measured over the silk = 0 - 046 centim., h — 0'00022 
(mean), q = 0'00012 sq. centim., k = 0T66 ; therefore a = (-—) = 0'7l, and 
\ I i/ i/ / 
l — 0'8 centim. Hence 
v" 0 1'013 
n 
k-al = 0'28 and v = — x = 0'487v 0 
2 2 1'039 
The resistance of the Wire will therefore be the same as if, at each end of it, the 
length on the sleeve were increased by V = 0"487Z = 0"365 centim. and the projecting 
ends removed. The effective length of the wire as used is, therefore, not the total 
length, 30 - 6 centims., but a shorter length, 29‘83 centims. Hence 
temperature excess of wire on sleeve = 30‘6/29"83 = 1"026 times mean 
temperature excess of whole wire. 
Hence, in the original equation for k ((4), p. 386), we must take 
{v A , Vb) = (1 + 29 h)(v" A , v\) 
(A, A) = 1'026 (v A , v B ), 
and 
