If we write the vector product S = (E x H), so this defines the orthogonal components ofthe two vectors E and H. This is indicated by the x between the two vectors. We call thisvector product
S = (E x H )
the Poynting vector of energy flow. In fact, however, is its unity, as shown in Eq. (9)follows VA/m2, ie surface density of the performance. With a distinction between aPoynting vector and Poynting vector Si within Sat outside the current-carrying wires.
Direction assignment: The vectors are E, H and S in that order, a legal system. That is,E points in the direction of one vector ex, H in the direction of one vector ey, ez S thenshows in the direction of one vector.
, And in a cylindrical wire is used sensibly and the previously drawn cylindrical coordinates with the same legal nimble in the following order each one associatedvectors: er, ea and ez, so he x ea = ez.
We now consider first the fields inside the wire. As a result of the wire resistance of the power generated in the axial direction already mentioned above voltage drop. Thisreferred to the unit length, the component of the electric field Ez is in the conductor. Themagnetic field strength is quite necessary for the associated conduction current density,the magnetic form that is both within and outside a single round wire circles. We havetherefore wire the component Ha. Because of the legal system of x ea = ez he is from mismatching vectors ea = ez x - it. This means, however, that the Poynting vector hasthe following components Ez and Ha in the wire inside only the component-Sr. Thismeans, however: the power of the electromagnetic field inside the wire flows along the wire axis. There will be zero, as is evident from the radial component of Sr Si in the wire: Sr = Ez Ha(r) = Ez Jz p r2/ 2pr = Ez r Jz / 2
Jz is the conduction current density Jz = I / p r2, which we assume here to be constantover the wire cross section. Sr is therefore a pure loss of power density. If it is integrated over the entire surface of the wires, we get the power loss in the wire Pv = I2. R. Otherservices, particularly that which is transported to the consumer (the boxes), comes in thewire inside does not! We are investigating now the dielectric around the wires.
Vectors are tangential to the drawn lines of the field distribution of the parallel wire line.
At the wire surface, the vector E two components: the relatively small component Ez inthe longitudinal direction of the wire. You have we been treated; He also a componentperpendicular to the axis of the wires leading from bridge cables to a spatial fielddistribution towards the mating conductor with the smaller Potenial. The magnetic fieldvectors in the dielectric are perpendicular to the vectors with the components he andperpendicular to the axis of the wires.
This is the vector product Sa = Er er x Ha ea = Er Ha ez
This was to be shown. All the energy is not transported to the consumer in the dielectricand transport in the wire inside. And on closer inspection by the inclusion of Ez is clearthat the energy transported vector from Eq. (11) at a slight angle to the wire axis isbecause the power dissipation in the wires from the dielectric current and the dielectricenergy transported is taken.