Bar and rail in magnetic field

[dimpledur]

Homework Statement

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 can't 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.

 

[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.

 

[rdmusic16]

I got the same.

 

[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 $$\frac{dv}{dt}=0$$.

 

[DeeNos]

"Steady state" refers to a state where all variables in the system have reached a constant level and there is no further change over time, such as in this case where the bar has reached a constant velocity. In this setup, the bar will experience a force in the direction of the magnetic field and will accelerate to the right until it reaches steady state. This acceleration will induce a current in the bar, and the direction of this current can be determined using the right-hand rule. As for the direction of the bar's motion, it will indeed be moving to the right in steady state as there is a net force acting on it in that direction. I would recommend reviewing the principles of electromagnetism and the right-hand rule to better understand this problem.