A Moveable Cross wire and a Function of Time

[RedPotato]

Homework Statement

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?

Homework Equations

EMF= −N dΦBdt
I = ε/R
F = I ∫ dL X B

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.

 

[anathema]

Thank you for your post. I believe your calculations for parts (a) and (b) are correct. For part (c), we can use the equation F = ma to determine the acceleration of the wire. We know the force acting on the wire (0.02 N) and we can calculate the mass of the wire using its length and the density of the material it is made of. Once we have the acceleration, we can use the equations of motion to determine the velocity of the wire as a function of time. We can also use the equation F = ILB to determine the current as a function of time, since the magnetic field and length of the wire are constant. I hope this helps. Let me know if you have any further questions. Good luck with your calculations!