And P=Fvm, I = blvmr+R.
Solution: VM = p (r+r) BL = 4× (2+2) 2× 0.5m/s = 4m/s. 。
(2) When the speed is 2m/s, the induced electromotive force E=BLv=2×0.5×2V=2V,
Current I = er+r = 22+2a = 0.5a,
Ampere F = BIL = 2× 0.5× 0.5N = 0.5N 。
The tensile force on the metal bar is f = PV = 42n = 2n.
According to Newton's second law: F-F =ma.
Get a=F? F an m = 2？ 0.50.2m/s2 = 7.5m/s2.
(3) In this process, we can know from the kinetic energy theorem that:
Pt+W A = 12 MVm2？ 12mv02
And w =-(QR+QR) =-2qr =-6.4j. 。
Solve t=mvm2? mv02？ 2wA 2p = 0.2× 42？ 0.2× 12+2×6.42×4s = 1.975s
(4) There are124+15×12 =131.5 squares between the graph line and the horizontal axis.
The corresponding area is131.5× 0.2× 0.1n.s = 2.63 n.s, that is? ? F amp △ t = 2.63 N s
So q =? ? I.△t=？ ? F A △ TBL = 2.632× 0.5c = 2.63c .
Answer: (1) The maximum speed of a metal rod is 4m/s. 。
(2) When the speed of the metal bar is 2m/s, the acceleration is 7.5m/s2. ..
(3) The corresponding time t of this process is 1.975 seconds.
(4) The amount of electricity passing through the resistor R within 4)0 ~ 3s is 2.63C 。