# Bar and rail in magnetic field

1. Oct 12, 2011

### dimpledur

1. The problem statement, all variables and given/known data
Couple questions regarding the following image:

What is meant when the problem states that "steady state can be achieved"? Will current be induced in this set up due to the acceleration to the right? For example, would I go Fg1-Fg2-Ffriction= Fmagnetic, and from there calculate the velocity of the bar at steady state? Will the bar even be moving to the right in steady state?
I cant seem to find any notes regarding this. How does one determine the direction? I found this type of question easier when there was an actual current through the system..

Any help is appreciated.

2. Oct 12, 2011

### dimpledur

Okay, I've come to realize steady state refers to the velocity, meaning dv/dt. Could someone please check my work for the following question solving for the velocity of the bar.

Please note that the question states the bar is 1m long, but the rails are only 0.75m apart. l would equal 0.75 for this question then right?

m1g+μmg+IlB=m2g

note: I= lemfl/R=Blv/R

thus, v=gR(m2-μm-m1)/(l^2*B^2)

v=[(9.81)(1)(10-0.5-5)]/(0.75^2*0.5^2)
v=313.9 m/s directed along the rail.

Did I do that right? Thanks.

Last edited: Oct 12, 2011
3. Oct 15, 2011

### rdmusic16

I got the same.

4. Oct 19, 2011

### sandy.bridge

I wouldn't merely state "on the rail" as a sense of direction for the velocity vector. You need to distinguish between right and left. As for determining the direction, determine which direction the bar will accelerate due to the external forces, and once steady state is achieved, the velocity vector will be in the same direction. Steady state in this sense merely means $$\frac{dv}{dt}=0$$.