# A Moveable Cross wire and a Function of Time

1. Apr 18, 2012

### RedPotato

1. The problem statement, all variables and given/known data

A circuit with a moveable cross wire is indicated in attached figure. The resistance is 2.5 Ω, the battery voltage is 5 V, the moveable wire length is 12 cm, and there is a uniform, constant magnetic field into the plane of the circuit (the page) with magnitude 0.3 T. (a) What is the current if the moveable wire is locked into position? (b) What is the force on the moveable wire at the moment it is released from the locked position? (c) If there is an opposing force of 0.02 N on the wire, how fast will it move as a function of the time after release?

2. Relevant equations

EMF= −N dΦBdt
I = ε/R
F = I ∫ dL X B

3. The attempt at a solution

I believe I got parts a and b (which could be wrong). Since the wire isn't moving initially there is not magnetic flux, so I just calculated the current to be the given Voltage divided by the Resistance (2 A). Then I calculated the force with F=ILB and go .072 N. What I'm confused about is how to get the function of time with regard to the changing are once the wire starts moving. Please help and also verify I did the first 2 parts correctly.