SELF | 30 - 31 - 32 |
S.B. Karavashkin |
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30 | |
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(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 l = const by the condition of forming the surface of the region V . Introduce the time phase shift between the function (, 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 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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31
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 |
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(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 | |
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(24) |
The first integral of the right-hand sum not containing the phase shift 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 t determines the smallness, not variation of the value). 32After that the under-integral expression (, t) can be written as |
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(25) |
where | |
Passing to the limit at t 0 in (25), we come to the integral | |
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(26) |
Substituting (26) to (24), we yield the required: | |
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(27) |
Seeing Fig. 2, where on the boundary S2 | |
we yield | |
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(28) |
Contents: / 15-16-17 / 18-19-20 / 21-22-23 / 24-25-26 / 27-28-29 / 30-31-32 / 33-34-35 / 36-37-38 / 39-40-41 / 42-43-44 / 45-46-47 /