Force on a metal bar across two wires in a magnetic field

[aolivias]

Homework Statement

two parallel wires. The distance between the wires is d from each other and they are connected by a capacitor with the capacitance C. The constant magnetic field B is applied and is perpendicular to the plane of the wires. We put a metallic bar across the parallel wires and start pulling it with force F parallel to the wires. What is the acceleration of the bar?

Relevant Equations

I also attached my work to a file so it can be readable. I also did it here below:

I so far did C=Q/V,
I=dQ/dt,
I=c*dV/dt
F=ma, F=-IlB. I set these equations equal to each other:
ma=-IlB, In the next step I substituted c*dV/dt for I and got:
m*dv/dt=-(cdVlb)/dt
then, I substituted a into dv/dt and solved for a and then for final answer:
a=-(cvlb)/(mt)

I am not sure if this is correct or if I am on the right path.

I need help with the problem above

 

[TSny]

F=ma, F=-IlB. I set these equations equal to each other

F = ma should be written as ∑F = ma. The left side is the vector sum of the forces acting on the bar.
The force F_mag = IlB is the magnitude of the magnetic force acting on the bar. But there is also the applied force, F, acting on the bar.

The current is produced by the induced emf, ε, due to the changing magnetic flux through the circuit. So, you will need to bring in Faraday's law of induction. How is ε related to the potential difference, V, across the capacitor?

The problem statement is not clear. I guess you are meant to neglect any electrical resistance in the wires and in the bar.

 

[rude man]

I need help with the problem above

I agree with post 2 except when it comes to moving media I prefer the Blv law (l=distance, v=velocity) to faraday's law for finding the emf. You can get into trouble with faraday's law in moving media. The Blv law is based on the Lorentz force which has a different basis than faraday's.

Otherwise, just play with the various relations as post 2 suggests.