I tried to understand the problem b) and c).[/B]
Faraday's law: ∇xE = - ∂B/∂t
emf ε = Bdv
Force : F =ma, Lorenz's force F=q(vxB) ==> ma = IdB
Power : power of battery = εI, mechanical power of the wire = Fv
The Attempt at a Solution
I think I solved a), where the magnetic force on the wire is F = IdB = ma and the speed v = at = IdBt/m.
I'm now tried to understand b) and c). First, I assume the initial speed is 0 at t=0 when the generator is replaced by a battery. The battery's power is εI, and it must be equal to the power consumed by the wire which is Fv = IdBv. Then the speed of the wire is v = ε/dB.
Also using Faraday's law I get to the same conclusion : ∇xE = - ∂B/∂t >>> ε = -∂(flux)/∂t = Bdv
It doesn't feel natural. The battery is connected at t=0, and in no time there the wire goes at speed v=ε/dB? So I think it is nonsense. But I don't know where I choose a wrong way. Help me plz.
I guess the answer to c) is that the current has to be 0, because when the terminal speed has been reached, there is no acceleration, then no force F=IdB=0. Am I right at this?