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 = NB
r2...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