MR. G. F. C. SEARLE ON PROBLEMS IN ELECTRIC CONVECTION. 
705 
At the fixed point through which 0 is passing, the forces are increasing at the 
rates 
E 
dt 
u 
cos OE n 
a u 
dJL 
dt 
u 
— cos 
ci T u 
The direction of d'E/dt is along OE, and that of dH/dt is along* II, i.e., per¬ 
pendicular to the plane of the paper and towards the reader. Thus E p acts along 
OE, and has the value pK 0 rfa ; pYuB acts along OQ in the plane of the paper, and 
K c/E 
has the value pupH 0 r/a ; — V B acts along OL at right angles to OE, and has the 
value — Kpu cos i|/E 0 H 0 r/47r« 3 . The direction of G- EE pdJL/dt is outwards from the 
paper, so that ATJ-D acts along OM at right angles to OE. The value of YGD is 
— K pit cos t//E 0 H 0 r/47r« 3 . 
Integrating these forces with regard to r, and remembering that pa = <x, we find 
for the normal pull 
N = \ E 0 cr cos 0 — ^ oymH 0 sin xfj — K/mE () H 0 sin 6 cos \fj/4rr. 
Making use of cr = KE 0 cos d/47r and employing the relations given in § 17 between 
E, H, d, (p, and \p, and noting that i }j -\- 6 = <f> the expression easily reduces to 
N = ^ (cos 3 0 - sin 3 d) - ^ • 
07 T 07T 
This is identical with the result obtained from the Maxwell stress. 
In finding the tangential stress, we know that E + YuB is normal to the surface, 
so that the first two terms may be disregarded. For the tangential stress we thus 
obtain 
T = K pu cos i//E (J H 0 cos 0/4tt = Kh 3 E 0 3 cos xjj cos 6 sin </>/47ri> 3 
KE 0 2 . . 
= —— cos d sin d. 
47T 
This acts on the E side of F, and is therefore identical in direction and magnitude 
with the force derived from the Maxwell stress. 
Normal Pull .—If we express H in terms of E and i/>, and d in terms of ifj, and 
write /3 for vd/v' 2 , we shall find that 
N = normal pull = 
KE 0 3 2 /3 3 COS 4 — COS 3 yjr (0 + (B~) + 1 — /3 
" .Stt 1 - (2/3 - /3d COS 3 yfr 
or in terms of cr 
2ira~ 2/3 3 COS 4 -y/r — COS 3 yjr (/3 + /3 3 ) +1 — /3 
~K (1-/3 cos 3 ^) 3 
4 x 
MDCCCXCVI.—A. 
