42 Force on a Magnetic Particle in a Magnetic Field. 
Thus p=HH, where H is the maximum rate of increase of 
magnetic intensity. 
Draw a line perpendicular to both the current density and 
magnetic intensity,-its direction being that of translation of 
a right-handed screw rotated from current to magnetic 
intensity, and let its direction cosines be W’, wu”, v’’. Then 
if C be the current density, and ¢@ the angle between OC 
and H, 
pr’ = pr+4rCH sin dX”, : 
pe =pet+4rC8H sin oh pw”, 
pv =pv+4rCH sin gv"; 
“. p? =p’? + 8rCHp sin GAN” 4+ py” + vv’) + 167°C? H? sin’ h; 
or if w denote the angle between (A, m, v) and (X’, w’’, v’), 
p= H?[H? + 8aU sin $ cos WH + 167°C? sin? 4]. 
The force F, as has been shown above, lies in the plane 
through (A, w, v) and (A”, w”, v’”). Let it make an angle wy’ 
with (X”, w”, v’’); then it follows at once that 
p’ cos Ww’ —p cos w= 47rCH sin ¢. 
Thus 
fo pec aN ee cos y+ darV sin | ‘ 
| H? + 87C sin d cos WH + 167(” sin? ¢ |” 
When C=0, or 6=0, y’=y, the direction of F in both 
these cases being that in which the magnetic intensity 
increases most rapidly. 
If v=volume of the particle and & the susceptibility supposed 
small, it is easily shown that 
F=/vH[H?+ 86 sin cos H+ 167°C? sin? 6]}, 
which reduces to the well-known expression 
kvHH, when ¢ or C vanishes. 
The expression for cos yy’ leads to the very simple result 
that F is the resultant of two forces, one, equal to kvHH, 
along the direction in which the magnetic intensity increases 
most rapidly, and the other, equal to 4akvHC sin @, along the 
line perpendicular to the current and magnetic intensity. 
