694 
MR. G. F. C. SEARLE ON PROBLEMS IN ELECTRIC CONVECTION. 
F, = E, 
Fo Er, — rB 3 
F 3 = E s + mB 2 
since u is parallel to the axis of x. 
Writing the equations in terms of II instead of B we have 
(66), 
F = E + /xVuII. 
F 2 — E i} Fo = E 3 — /x'allg, F 3 = Eg + /lull-, . 
(®')> 
( 68 ). 
Inserting the values of E and H in terms of 'P and B from equations (26) to (31), 
we find at once 
T , 09 T1 d9 ,, d9 
F 3=- 
or, 
chj 
F — — V9 
dz 
■ (69), 
• (70). 
Thus it appears that though there is no proper potential from which the electric 
force can be derived, yet there is a potential for the mechanical force experienced by 
a moving charge. The electric force is really the mechanical force experienced by a 
unit charge at rest , while the force — V'P is the mechanical force experienced by a 
unit charge moving at the same speed as the system which gives rise to E and H. 
Mechanical Force on a Moving Pole. 
11 . In exactly the same manner we should find that if the mechanical force 
experienced by a unit magnetic pole moving with the system be denoted by R, then, 
R = H - IttYuD = E - KVuE.(71), 
so that its components are 
Hi = Hj, R, 2 = H 2 + KwE 3 , Rg = Hg — KmE 2 . . . (72). 
Inserting the values of E and H in terms of 'P and B from equations (26) to (31), 
we find 
Ri = - 
dx 
or 
_ dB 
r 2 = — — 
dy 
R = — VB 
Ro = 
<m 
dz 
. . . . (73), 
. . . . (74). 
