The thorny way of truth - Free Energy Community

MarinovThe second integral has been calculated on a computer and its value is''f" = (lO'^/^rfT '^'"^^"^^ ,,, = 0.2493 pN/A^ (9)'^(1 +cos(a+6)}l/2However the first integral(lO-V^)tT ""^"^"^fXi = '^^i;^{1 - cos(a-6)}l/2(10)is improper as for a = 6 it has a peculiarity and thus it can be not calculated on acomputer. Moreover, I shall now show that the first integral is converging to infinity.Indeed, let us choose some specific value for a (a = Const) and let us calculate theintegral depending on B, denoting it by *(a). For simplicity let us assume a = tt/2. Insuch a case the integral depending on g will beit/2/P*(tt/2) = / (1 - 1sinB)"^^ sin6d6 =.'^Ini*^ + (l+sin6)^/^} - {»^/2)ln(l - sing) - 2{l+sin6)^/2 ir/2The last integration is complicated but the reader can easily persuade himself thatthe integration is right, as if he will differentiate the expression on the right of (11),he will obtain the integrand in the integral on the left.All terms in the expression on the right are limited when the limits and 7r/2 willbe substituted except for the term ln(l-sin6); by substituting here B = tt/2, we obtainln(l - 1) = InO = - oo, so that the integral (tt/2) is tending to + «.After considering the case from a critical physical point of view, I found the followingway for calculating the pushing force f'l,:Let us put two concentric half-circles from both sides of the initial half-circledrawn in Fig. 2, respectively with radii R + xR and R - xR, where x is a quantity smallwith respect to 1. Let currents 1/2 flow along any of these half-circles in parallel tothe current I flowing in the initial half-circle. One can easily realize that the forcepushing the U-form Ampere bridge will be the sum of the forces with which the currents1/2 in any of the secondary half-circles act on the current I in the initial half-circlewhen x tends to zero.