OF ENERGY IN THE ELECTROMAGNETIC FIELD. 
349 
angle between its direction and that of <§, the magnetic intensity, the direction cosines 
L, M, N of the line perpendicular to the plane containing (§,' and are given by 
t _ h'/d — Q'y . i\ ,r _ PV—R'g _ _ Q 'a — V'/3 
(S >£> sin 0 ’ _ (&'-§ sin 6 ’ x (£ «£) sin 6 
so that the surface integral becomes 
sin 0(LZ+Mm + Nn)tfS. 
If at a given point clS be drawn to coincide with the plane containing (S' and it 
then contributes the greatest amount of energy to the space ; or in other words the 
energy flows perpendicularly to the plane containing (Sf and <§, tire amount crossing 
unit area per second being (£'<§ sin 6/±tt. To determine in which way it crosses the 
plane take (S' along 0 z, <§ along O y. Then 
P' = 0 Q'=0 S=1 
a=0 = 1 y—0 
and if sin 0=1 
L= 1 M = 0 N = 0 
If now the axis Ox be the normal to the surface outwards, 1= 1, m = 0, n= 0, so that 
this element of the integral contributes a positive term to the energy within the sur¬ 
face on the negative side of the yz plane ; that is, the energy moves along xO, or in the 
direction in which a screw would move if its head were turned round from the positive 
direction of the electromotive to the positive direction of the magnetic intensity. If 
the surface be taken where the matter has no velocity, (S' becomes equal to (S, and the 
amount of energy crossing unit area perpendicular to the flow per second is 
electromotive intensity x magnetic intensity x sine included angle 
47T 
Since the surface may be drawn anywhere we please, then wherever there is both 
magnetic and electromotive intensity there is flow of energy. 
Since the energy flows perpendicularly to the plane containing the two intensities, 
it must flow along the electric and magnetic level surfaces, when these exist, so that 
the lines of flow are the intersections of the two surfaces. 
We shall now consider the applications of this law in several cases. 
