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Sila na nabijenu česticu koja se giba magnetskim poljem

Sila na nabijenu česticu koja se giba magnetskim poljem. Lorentzova sila. F A = BIl sin . I = enSv. F A = nSlBev sin . nSl = N. F A = NBev sin . Općenito je Lorentova sila:. F L = Bev sin . F L = BQv sin . Sila je najjača kada se nabijena čestica giba okomito na silnice:.

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Sila na nabijenu česticu koja se giba magnetskim poljem

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  1. Sila na nabijenu česticu koja se giba magnetskim poljem Lorentzova sila FA = BIlsin I = enSv FA = nSlBevsin nSl = N FA = NBevsin Općenito je Lorentova sila: FL = Bevsin FL = BQvsin

  2. Sila je najjača kada se nabijena čestica giba okomito na silnice: sin= 1 FL = BQv Sile nema kada se nabijena čestica giba paralelno silnicama: sin = 0 FL = 0

  3. Pravilo desne ruke: za pozitivno nabijenu česticu za negativno nabijenu česticu

  4. Primjer: Proton kinetičke energije 0,5MeV uleti okomito na silnice homogenog magnetskog polja indukcije 0,1T. Kolika je Lorentzova sila na proton? Masa protona iznosi 1,6710-27 kg, a naboj1,610-19C. Rješenje: Ek = 0,5 MeV  = 0,5106 eV = 0,51061,610-19 J = 810-14 J B = 0,1 T m = 1,6710-27 kg e = 1,610-19 C F = ? F = Bev F = 1,610-13 N

  5. Magnetohidrodinamički (MHD) generator pozitivnipol S ioni mlaz ioniziranog plina N elektroni negativnipol

  6. Zadatak: Elektron ubrzan naponom 150 V uleti u homogeno magnetsko polje indukcije 0,2 T okomito na silnice. Kolikom silom magnetsko polje djeluje na elektron? Masa elektrona je 9,110-31 kg, a naboj 1,610-19 C. Rješenje: U = 150 V B = 0,2 T m = 9,110-31 kg e = 1,610-19 C F = ? F = Bev Ek = W F = 2,310-13 N

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