Ui = . (×)
ter:
= (Bx, 0, Bz), = (0, l, 0) in = (v, 0, 0)
sedaj izračunamo vektorski in skalarni produkt in dobimo:
Ui = vlBz =...V
F = Fm = IilB =...N
in pospešek:
ma = IilB => a = =...
Ui = . (×)
torej poznam tudi tole velikost:
Ui = v cos()lB sin()
Uit = = NBS = NBr2...Vs
N1 = 100, l1 = 0, 8m, r1 = 5cm, I1 = 2A, ter N2 = 1000 in r2 = 3cm in = 30o
a) A = = =...J
b) = L1I1 =...Vs
c) Uit = N2B1S2cos = =...Vs
l1 = 0, 7m, N1 = 100, = 3, N2 = 200 in r2 = 2cm
Ui = =...V
Nt = 1000, rt = 10cm, lt = 50cm, lp = 10cm in It = 10A
Največja napetost ki se inducira je:
Ui = . (×) = - lp2Bt
vem pa tudi, da zaradi nihanja velja enačba:
M = - Jp
=> Fgr'sin - Fmr' = - Jp
če velja sin = , ap = - x in x = rt
mpgr'x - IilpBtr'2 = - Jpap
izpeljava je predolga, a ne prezapletena, rešitev:
= -
=> Ui2 = lp4Bt2
dobimo kvadratno enačbo in njeni rešitvi:
Ui1 = - +
in
Ui1 = - -
Fg = Fm
torej:
IiaB = 4Sag
B = 4Sg
potem:
Bav =
=> v =
me 2007-11-05