Energetic budget and oneness

1) Theorem of Poynting:

Multiplying the for H and la per - E, evidences to 1° the member the term that is worth therefore integrating all the equation on one volume V contornato from one S surface pu² to apply the theorem of the divergence to this last term and to write therefore isolating sources to 1° the member obtains where

indicates the power yielded from sources to the field, is the power dissipated for Joule effect from the conduction currents, is the power that goes to vary the energy stored in the electromagnetic field and is the power that flows through the S surface that encloses volume V.

 

2) Theorem of Poynting for not dissipative means in the harmonic source case:

Considering the single current electrical worker impressa, , is had, being the not dissipative means has and therefore replacing the previous expressions of and and H in the theorem of general Poynting it is had

which is simplified being and taking advantage of the trigonometrical formulas therefore obtains

whose valor medium it is and it is worth 0 in the case of , , in how much are periodic functions having period After all obtains that extension as the valor medium of the power distributed from sources is equal to the valor medium of the power that flows through the S surface, such power distributed from sources can be active or passive to second of the phase-difference y_and.

 

3) Theorem of Poynting for a metallic covering:

Remembering that for a conductor it is had and orthogonal to the surface and tangent H to the same one, tangent to the surface is had and therefore its product to climb with the normal school is 0, from which is had moreover considering the medium amounts can be thought null also the which had contribution to replacing therefore in the term of source and in the term of dissipation due to the currents of conduction ed it is had: .

 

4) Theorem of Poynting for the coaxial cable:

A system constituted from a generator is considered that feeds a cargo by means of a coaxial cable, the analysis is carried out considering 3 various sections of the system:

to)       the theorem of Poynting is written on the volume to shape of rubber ring within to the cable and containing the generator inner to it, has where 1° the term is various from zero solo in the containing volume the generator, supposing_g //and_g and and the constant has that, applying the trigonometrical formulas and calculating the average on T is reduced to while for 2° the member it is had that the 1° integral mediated it is cancelled while the 2° gives back the outgoing medium power from the S surface_, considering the maximum distributed power is had but the flow of the carrier of Poynting is not null only to the inside of the coaxial cable (…for the property of the metallic conductors) and therefore it is had

b)       income draft Comprises all the coaxial cable ad.eccezione.della and of escape, being absent for hypothesis dissipations and sources are had that the flow that enters from the income section is equal to the flow that exits from the escape section, .

c)       writes the carrier of Poynting on the volume to shape of rubber ring within to the cable and containing the cargo R_c externally to it, this time is absent the currents impresse and the flow can be considered like source that propaga along the coaxial one

after all is had.

 

5) Theorem of oneness:

It characterizes of the conditions under which the solution of the equations of Maxwell is only in one volume V of linear means for times t > t₀ . They are:

to)       the tangential member of and or H for t > t ₀ must be assigned on the surface that contorna V.

b)       E e H are assign to you in V for t = t₀ .

It is demonstrated for absurdity supposing that two solutions and ₁exist , H₁ e and₂ , ₂H , considering the fields difference e whose tangential members are null on the S surface and therefore are null also the correspondent carrier of Poynting as also the sources J _i and J_m are null . Calculating the theorem of Poynting is obtained but to time t₀ for the hypothesis 2) according to member is worth 0 thereforefor t >t ₀ would have to become negative, contraddicendo its nature of dissipated power therefore positive and therefore we are joints to an absurdity and therefore the two solutions are coinciding.