ON STANDARDS OF ELECTRICAL RESISTANCE. 169 
If the angular motion of the coil be 30, the work spent in keeping up the 
rotation against the electromagnetic force is 
; HyG cos ad8+MyK cos (@—¢) dé. 
Since this work is exactly consumed in keeping up or increasing the current, 
we must have 
HyG cos 0d6+MyK cos (6 —9)d0=Ry’ dt+Lydy. 
Since 6=wt and a the solution of this equation is 
y= Pile => = {GH (R cos 6+ Lw sin 6)+ KM (R cos (0@—¢)+ Lw sin (6—¢)) } 
4 Ce-te > 
the last term becoming insensible soon after the beginning of the experiment. 
We can now find the equation of motion of the magnet. 
_ Let A be its moment of inertia, MH; the torsion of the fibre per unit of 
angular rotation, then 
es $ —<MKy cos (—6)—MH (sin $ +79). 
Substituting the value of y and separating terms in 0, we find 
@y 1 MKw 
A Wes SR 4 La? { GH cos (+ Lw sin 6) +kur —MH (sin ¢— 1) 
1 MKw 
2R?4 1a? 1 
4KM(R cos 2(0—9)-+ Iw sin 2 (0—»)) | 
Tn order that ¢ may continue as it does nearly constant, the part indepen- 
dent of @ must vanish, or 
oR { GH(R cos ¢ +Lw sin ¢) +kuR| _MH (sin ¢ + Tg) =0. 
This gives the following quadratic equation for R, 
1 
+ { GH(Rcos(20—$)+ Lwsin(20—9)) 
GK KM\ 1 GKI.? 
oa» aie a we 
: reer Ce a oo oe 7 
sin 
The solution of this equation may be expressed to a sufficient degree of accu- 
racy as follows :— 
GK 
- —1)t 
+See fae nt Ee ) ar “yt. 
To determine the quantities occurring in this equation, we must measure 
the dimensions of the coil, the strength of the magnet, and the force of ten- 
sion of the fibre. 
~ 1st. Dimensions of the coil. 
Let a=mean radius of the coil.... 6.1.0.6... 2.0. eee = 0:1566 metre. 
m=number of windings of wire ............6: =307 
1 =effective length of wire=2rna.............. =302-063 metres. 
b=breadth of section of coil perpendicular to the 
plaumiak the: coil: ical micte wailed oi os oi eat = °0185 metre. 
