8 
dmdm3; 
p” 
and the portion of this resolved in the horizontal plane, is 
dm a 
x OP. 
The moment of this force to turn the suspended magnet is 
dpa x OP sinOPC = CP x ae cd 
OE x OC sin OCP. 
Hence, putting OC = a, CP=7, and OCP =90° - uw, the whole 
moment of the force of the iron bar is 
| | rdmdm 
a@ COS U a 
p? 
Now, PP?= P’O?+ OP?. Or, putting OC’=h, CC’=e, CP’=7’, 
p?=(h-r)? +a? +7? - ar sin u= e+ r°+r?— 2 (hr’'+ ar sin uv). 
: wheel ; ' 
Accordingly, expanding e according to the ascending powers 
a, es : ; 
of oe integrating and making 
fr-dm=M,, {r"dm'= M’,, 
and observing that, on account of the symmetrical distribution 
of free magnetism in the magnet and bar, J/,, and M’, vanish 
when z is an even number,—we have, for the moment of the 
force of the iron bar, 
3ah M3 hA\ M; a3 
— ps MM'cos u whi-g5[Gr (1 _ a) ap (1-7 Sain u) | 
M's h? M’, M3 ae h? 
ao al!-6 a) + 2 ag (- 9“ sintu-35) 
M; a. 
dae ~ oe 2 r 
+ ue (1 18 2 sin w) | &e. 
“‘ This formula is unfortunately not convergent, and is, con- 
sequently, of no use in the present investigation. In fact, 
aS) 
M’; M' 
ar is of the same order of magnitude as e”, a as e4, and so 
