ELECTRODYNAMIC QUALITIES OF METALS. 
77 
Table III.—Cast Cobalt Bar. Lower end at first a true North Pole. 
Operations. 
Magnetometer 
readings. 
Differences. 
Total 
magnetization. 
Cobalt bar placed in position 
Zero 460 
952 
+ 492 
-492 
146 lbs. on ... 
895 
- 57 
-435 
„ off ... . 
885 
- 10 
-425 
10 “ ons ” and “ offs ” 
867 
- 18 
-407 
146 lbs. on ... 
882 
+ 15 
-422 
,, off ... . 
867 
- 15 
-407 
Bar struck a few blows with 
a mallet. Magnetism re¬ 
versed . 
370 
-497 
+ 90 
146 lbs. on ... 
395 
+ 25 
+ 65 
„ off ... . 
370 
- 25 
+ 90 
„ on ... . 
395 
+ 25 
+ 65 
„ off ... . 
174 lbs. on ... 
373 
- 22 
+ 87 
403 
+ 30 
+ 57 
,, off ... 
375 
- 28 
+ 85 
10 “ ons ” and “ offs ” 
375 
0 
+ 85 
249 lbs. on ... 
Bar here broke, 
but being held 
in its place 
gave a reading 
of 360 
232. In Table I. it is stated that when the iron bar was placed in position the 
image on the scale of the magnetometer was driven off the scale in the negative 
direction. A controlling magnet, 15 centims. long, placed at right angles to the 
magnetic meridian in a horizontal line passing through the centre of the needle, was 
used to bring the image to the division 709 on the scale. When the nearer end of 
this magnet was at a distance of 23 centims. from the needle, the image rested at the 
division 323 on the scale, and when the magnet was brought G centims. nearer to the 
needle, at the division 773 on the scale. Hence, taking the tangent of the angle of 
deflection as equal to the angle itself, we may reckon the deflection caused by bringing 
the iron bar of Table I. into position as 1369 scale divisions, which may be taken 
as measuring the magnetism of the bar." This increased during twelve successive 
* The method of calculation is as follows:—Let A denote the deflection produced by placing the bar in 
position; D and D' the readings with the controlling magnet introduced, with its centre at the distances 
r and r' from the magnet; a the half length of the magnet; and B a constant depending on the con¬ 
trolling magnet; then we have, 
D = A + 
B 
B 
and similarly 
(r-a ) 3 (r + «) 
4 Bar' 
o—A + 
4B ar 
A 3 —a 3 ) 3 
D' = A + - 
A' 3 —a 2 ) 3 
Eliminating B between these two equations, and solving for A we get 
D + IT D-D' r (r' 3 - a 3 ) 3 + r' (r 3 - a 3 ) 3 
A ~ 2 2 r A' 3 — a?Y — r'{r~ — a~)~ 
By taking for a the virtual half length of the magnet or the distance of either pole from the centre of 
