Energy flows in the electromagnetic field (Поток энергии в электромагнитном поле)

The energy flows take place in the time dependent electromagnetic fields. We just have seen that Poynting’s theorem claims. If the electromagnetic field depends on time it may cross the surface and so the total energy that was concentrated inside the volume may decrease or may increase as well. But also, we shall see that in static field nevertheless energy flow exists. The energy transmitted along the wires of the line is propagated in the dielectric, as well as the energy emitted by the antenna. Wires and cables only serve as guide for this energy.

Now, let’s consider two wire transmission lines which is loaded and the load is a resistor.

We should consider an electromagnetic field which is induced by all these wires. ∂W over ∂t (∂W/∂t). By the way in this case, in principle does not matter the time dependence of this current, it may be static current which does not depend on time but may be current that changes in time. According to the Poynting’s theorem ∂W/∂t may be expressed in this form.

What is V and S? We shall surround the space with the resistor in the load by the surface S and everything which is inside will be volume V. So, what we shall have inside the volume? The first term corresponds to the power losses inside the conducting material and that will be the resistor. Also, we should take in account that energy of the electromagnetic field may cross the surface which is described by this area S and this first term with V. We can find or we can express this integral from the Poynting vector as the sum of these two components:

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So, we can conclude that increase in the electromagnetic field energy stored inside the volume and release of the thermal energy in the line wires and the load can occur only due to the influx of electromagnetic energy which enters this volume, which crosses this area S. If the current is static then the field distribution inside that volume will be constant. So ∂W/∂t will be equal to zero as well. In this case all energy which is dissipated (рассеянный) in this resistor will cross this surface and may be calculated as a vector product of E and H. The dissipated energy of this element can be very small, let’s neglect the energy, which is dissipated inside the wires, but the energy comes not through the wires, but through the space which surrounds these wires, lines here is only "subguides" (вспомогательные направляющие), guides along which the energy of the electromagnetic fields propagates (распространяется) and change their position.

Transmission of energy in a dc line (Передача энергии в линиях постоянного тока)

All the energy entering the volume V through the surface s is released as heat in the receiver. The energy of the electromagnetic field released in the conductor in the form of heat penetrates them through the surface of these conductor. In this case we shall not neglect the resistivity of the wires themselves like we have done in the previous case. In such a case when we shall consider only some part of the transmission line, only some sample. And we can say that the energy which is dissipated inside the wire should be exactly equal to the energy which crosses the area of the cylinder which is a part of the wire. The right part its energy that dissipated inside. It might be expressed as squared current and resistance of this part of the wire. Here we assume that the current is distributed uniformly of the cross section of this wire and this power losses should be identical to this:

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Magnetic field, induced by the wire, circulate around this wire, the field intensity, if there is some voltage applied to this wire, may be split into two parts. One part is the field which is normal to the surface of this wire, and the component of electric field which is parallel to the wire. This component must exist because the Ohm’s Law here is E_τ is equal to J/γ and so we shall consider both vectors: H, just circulate around the surface, that’s why in this point it is directed like it shown here, and E_τ, and we can see that these two vectors initiate a flow of the energy just inside the wire. So, in such a case energy crosses the cylindrical border of this wire and enters and then disperse inside the wire, it is transformed to the serval (хоть убей, я сам не понял, что он говорит) energy. In principle, it is easy to find exact expression of these vectors if you transform these reflections into the formulas. S - the Poynting’s vector, is E cross H (S=E x H) they are normal to each other that’s why we can simply multiply H and E_τ. H is the field which is induced by very long wire its current I/(2πr) (r₀ is a radius of this wire). I·R/l – is the resistance of the wire. I²·R is Power, 2π·r₀·l is an area of this surface. So, finally we get relations which is – energy which crosses the surface of the wire is the energy which is dissipated inside the same wire.