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PROFESSOR J. H. POYNTING ON ELECTRIC CURRENT AND THE 
then from (6) 
or 
or from (l) 
(ii) 
whence the velocity of propagation of magnetic induction is also equal to I/a/K/t. 
It would seem that in some cases, such as that of the field surrounding a straight 
wire with a steady current, the electric intensity may be regarded as entirely due to 
the motion of magnetic induction, and its components will therefore he C ~, 
But in other cases it would seem that the electric induction cannot be wholly due 
to the motion of magnetic induction, and we must therefore introduce the terms 
involving \b. If, for instance, the electric and magnetic intensities were inclined at an 
angle 9, we should have to suppose the electric intensity E to be produced by the motion 
of the component of magnetic induction I perpendicular to E, viz., /xl sin 6, the other 
component /xl cos 6 being at rest. To produce intensity E, E tubes must cut unit 
length in the direction of E per second; and since the value of the magnetic induction 
is gl sin 6, this requires a velocity v, given by v./xl sin 9— E or r=E//xIsin 9. Now 
we can easily imagine a case where E and I coincide, as, for instance, a condenser 
with its planes parallel to the axis of a wire carrying a current, and its terminals 
connected with two points in the wire. Here I sin 9=0, and v is infinite. Or we 
have to suppose the electric intensity to be produced by the movement of tubes of 
induction of no intensity with infinite velocity, a statement without physical meaning. 
But it is, perhaps, worth noting that if we suppose that the electric intensity is 
produced by the motion of magnetic induction, and that the magnetic intensity is 
produced by the motion of the electric induction, each carrying their energy with 
them, the right quantity of energy crosses the unit area. 
For E magnetic tubes, with I sin 9 unit cells per unit length, will carry across unit 
El sin 6 
Sir 
, or half the energy which actually 
area in the plane of E and I a quantity 
crosses the plane. If I sin 9 is due to the motion of electric tubes, then I sin ^/4 tt 
tubes must cut unit length in the direction of I sin 9 per second. The number of 
unit cells per unit length is E, and therefore the motion of the tubes will carry a 
quantity of energy 
El sin 6 
87T 
, or the other half actually crossing 
The equations which have been obtained in the foregoing manner by the aid of the 
