SELF

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S.B. Karavashkin

30

(22)

between the picked out section of the field line and the time interval is true. And for all filaments of current of the picked out region  deltabig.gif (843 bytes)l = const   by the condition of forming the surface of the region  V .

Introduce the time phase shift between the function  vectorF.gif (853 bytes)(vectorr.gif (839 bytes), t)   entering the picked out region and leaving it by the time variable. This introduction does not mean the non-simultaneity of measuring the flux of function through the region surface, it determines the degree of process delay at the exit of region relatively to the process taking place at its entrance, so far as  deltabig.gif (843 bytes)t    is determined by the linear measurement of the region and propagation velocity of the field process. Thus it cannot serve as a variable, because it would be the determining factor in case of non-simultaneous measurement of flux of the function passing through the surfaces of the selected region  V .

According to the said, consider the flux of function passing through the surfaces of  V . As the statement of problem, the surface  S  consists of three components:

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where  S1  and  S2   are the butt surfaces of the picked out region, and Sl  is its lateral surface. Consequently, the complete flux through the surface S is

(23)

In (23) the time shift of vector function and the absence of flux through the lateral surface are taken into account. Transforming this expression in accordance with (22), we yield

(24)

The first integral of the right-hand sum not containing the phase shift  deltabig.gif (843 bytes)t  vanishes in the source-free field, because, in this case, the condition of zero vector function divergence in stationary processes is true. The second right-hand integral generally is not zero and can be easily substituted for the integral over the space. To do so, divide the picked out region to   n  sub-regions having the height (along the line of current) equal to

where  n  can be any integer number larger than 1. (Here and further  delta.gif (843 bytes)t  determines the smallness, not variation of the value).

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After that the under-integral expression  deltabig.gif (843 bytes)vectorF.gif (853 bytes)(vectorr.gif (839 bytes), t)   can be written as

(25)
where
Passing to the limit at  delta.gif (843 bytes)t arrow.gif (839 bytes)0   in (25), we come to the integral

(26)

Substituting (26) to (24), we yield the required:

(27)

Seeing Fig. 2, where on the boundary  S2

we yield

 

(28)

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