Задачи по векторному анализу. Михайлов В.К - 65 стр.

           íåäîñòàòî÷íî äëÿ åãî ïîòåíöèàëüíîñòè; â ýòîì ñëó-
           ÷àå íåîáõîäèìî âûïîëíåíèå êðèòåðèÿ 1.
     Ïðèìåð 1. Ìàãíèòíîå ïîëå ïðÿìîãî òîêà i, òåêóùåãî âäîëü
îñè z, èìååò âèä:
               ρ µi y           
               B = 0 − 2 , x2 ,0 , ãäå ρ 2 = x 2 + y 2 .
                  2π  ρ ρ 
Óñòàíîâèòü, ÿâëÿåòñÿ ëè îíî ïîòåíöèàëüíûì.
     Ðåøåíèå. Ñíà÷àëà ïðîâåðèì âûïîëíåíèå êðèòåðèÿ 2. Íå-
                                           ρ
ñëîæíûå âû÷èñëåíèÿ ðîòîðà ïîëÿ B äàþò:
                        ρ            ρ           ρ
                  (rot B ) x = (rot B ) y = (rot B ) z = 0 .
                          ρ
Îäíàêî íà îñè z ïîëå B íå ñóùåñòâóåò, ñëåäîâàòåëüíî, îáëàñòü
                     ρ
îïðåäåëåíèÿ ïîëÿ B ÿâëÿåòñÿ íåîäíîñâÿçíîé, è ïîýòîìó çäåñü
íåîáõîäèìà ïðîâåðêà âûïîëíåíèÿ êðèòåðèÿ 1. Äëÿ ýòîãî âûáå-
ðåì êîíòóð Ñ â âèäå îêðóæíîñòè, îõâàòûâàþùåé îñü z : x = Rcosα,
y = Rsinα . Òîãäà
                        ρ ρ
                      ∫ B ⋅ dl = ∫ ( Bx dx + By dy ) =
                         C          C



             µ 0i 2π  R sin αR sin α R cos αR cos α 
                                                     dα = µ0i ≠ 0 .
             2π ∫0 
         =                           +
                             R2             R2       
                                           ρ
Ñëåäîâàòåëüíî, âî âñåì ïðîñòðàíñòâå ïîëå B íå ÿâëÿåòñÿ ïîòåí-
öèàëüíûì, ò. å. äàííîé òî÷êå ïðîñòðàíñòâà íåëüçÿ îäíîçíà÷íî
ñîïîñòàâèòü îïðåäåëåííûé ïîòåíöèàë ϕ , òàê êàê ïðè îáõîäå âîê-
ðóã îñè z ýòî ÷èñëî óâåëè÷èâàåòñÿ íà µ0i :
                             ϕ M = (ϕ M ) o + Nµ 0 i ,
ãäå N — ÷èñëî îáõîäîâ âîêðóã îñè z âäîëü âåêòîðíîé ëèíèè
      ρ
ïîëÿ B .
     Ïðèìåð 2. Ýëåêòðè÷åñêîå ïîëå äëèííîé ðàâíîìåðíî çàðÿ-
æåííîé íèòè, íàòÿíóòîé âäîëü îñè z, èìååò âèä:
                               ρ           y 
                               E = k  x2 , 2 ,0 ,
                                      ρ   ρ    



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