ME. G. F. C. SEARLE ON PROBLEMS IN ELECTRIC CONVECTION. 
693 
current at points near the axis, but a negative ^-component at points near the 
equatorial plane. The volume-integral is therefore less than it would be in a more 
suitably chosen space. If we take a very small circular cylinder, whose axis is in 
the axis of motion and whose length is very great compared with its radius, we 
shall clearly get rid of the “ back ” current. The volume-integral of the ^-component 
of the current in such a cylinder can best be calculated by means of the theorem 
that the line-integral of the magnetic force round any circuit is 4 77 times the current 
flowing through any surface bounded by the circuit. 
Let the charge q be in motion at speed u along the axis of x. Then by (45) the 
resultant magnetic force at the point x, p is 
H" 
qug 
u. 
j* 
2 
where p 2 = y~ + z 3 . 
The total x-current flowing across the section of the cylinder of radius p made by 
the plane x = x, is therefore 
X" 
1 gup 
The volume-integral, when 2 l is the length of the cylinder, is therefore 
When p is infinitely small compared with l we have simply 
j C" dco = qu, 
as we should have expected. 
The volume integrals of the other components of C” are clearly zero, so that 
fc «{'dco = 0, * jc s'da = 0. 
If, now, F denote the force per unit charge, we see from (64) that its value is"* 
and its components are 
F = E + VuB 
(65), 
* The accents, being no longer needed, have been omitted. The quantities E and B are the values 
which would obtain at any point if the unit charge, which has been supposed to be placed there, were 
removed. 
