(1) when K is connected to 1, the metal bar ab in the magnetic field just to remain stationary, the bar's gravity and the amperometric force is balanced, there are mg = BIL
and I =
< td style="border-bottom:1px solid black">E
| R+r |
The joint solution, R =
?r
(2) When K is connected to 2, the bar reaches the steady state with uniform motion, then we have
mgv=
,
and the induced electromotive force E′= padding-bottom:1px;font-size:90%">mgR0
| B2L2 |
①
According to the theorem of momentum we have : mgt-B
Lt=mv
and q=
t
And the induced charge q=
t,
=
,
=
, △Φ=BLs, and the association q=
The joint solution is t=
+
.
(3) The switch K is suddenly connected to 3, the capacitor is charged, the circuit is charged with current, ab bar is subjected to an upward amperometric force, let the instantaneous acceleration be a, according to Newton's second law to get
mg-BiL = ma ②
And i=
=
, and U= E=BLv, which gives i=
=CBLa③
From ②③, we get: mg-BL?CBLa=ma
Solve that a=
? ④
See that the rod has constant acceleration and moves in a uniformly accelerated straight line.
When ab descends a distance s, let the speed of the rod be v′, then v′2-v2=2as ⑤
Let the power stored in the capacitor be △E, then according to the conservation of energy we get
mgs+
mv2=
mv′2 + △E⑥
Interacting ①④⑤ font-size:90%">B2L2mgCs
| m+B2L2C |
.
Answer: (1) The resistance R of the sliding varistor connected to the circuit is
-r.
(2) The magnitude of the stabilization velocity is
, the time required for the process of dropping s is
+< table cellpadding="-1" cellspacing="-1" style="margin-right:1px">
| B2L2s |
| mgR0 |
.
(3) The rod makes an acceleration of a =
the uniformly accelerated straight-line motion, ab and then fall the distance of s, the capacitor stores the power is
.