# Conducting bar on a rail question

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Disclaimer: this question is really just for my own edification in preparation for a test.

In the figure above, we have a conducting bar that is placed onto two sloped conducting rails. The box labeled V is a voltage source. Lastly there is a magnetic field starting from the "base", rising up to the top of the plane of the computer screen. So basically the B field is perpendicular to the base, which is labeled in the figure.

My question pertained to Faraday and Lenz's Law. Say the bar is set at the very bottom of the bar at f-a.

My first question is: If there is a voltage applied between points a and f, why does bar begin to climb the slope? Eventually coming to rest at an elevated point?

My logic is as follows: the current created by the voltage opposes the external magnetic field, reducing its strength somehow. According to Lenz's law, this decrease in magnetic flux must be opposed by an induced magnetic field and induced current... so the conducting bar climbs the slope to create both the induced magnetic field and induced current. And it does so by rising up the slope, increasing the area. And using the right hand rule, the direction of the current induced by the rising conducting bar would be from a-->x-->y--> f? Is this logic correct?

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Disclaimer #2: The next question follows on the assumption that all of my logic above the dotted line is correct...

My second question is: which direction is the current (initiated by the voltage) going if the conducting bar is rising? My logic is that the current (initiated by the voltage) MUST be reducing the field strength, as I have explained above the dotted line. By using right-hand rule, this means that the current initiated by the voltage MUST be going from f --> y --> x --> a --> f.

Is this the correct direction and logic?

Thank you so much PF! I have been learning from yall for years! :) you guys never fail me :) this site is so helpful. I wish every subject had a site like this. Someday, I'm sure.

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My logic is as follows: the current created by the voltage opposes the external magnetic field, reducing its strength somehow. According to Lenz's law, this decrease in magnetic flux must be opposed by an induced magnetic field and induced current... so the conducting bar climbs the slope to create both the induced magnetic field and induced current.

Letz's law needs changing magnetic flux. Your magnetic field is constant. Initially, the bar is stationary, so the loop formed by the bar and the rest of the circuit is also constant. So the flux is constant.

Even if it were not constant, the bar would not magically climb the slope. A force must act on so that it could go up. What force could that be?

.Scott
Homework Helper
Letz's law needs changing magnetic flux. Your magnetic field is constant. Initially, the bar is stationary, so the loop formed by the bar and the rest of the circuit is also constant. So the flux is constant.

Even if it were not constant, the bar would not magically climb the slope. A force must act on so that it could go up. What force could that be?
What is changing is the current. It will climb until the it generates enough force to be repelled by the magnetic field. But I am not clear on why the current would not continue to build.

Right, the current is changing as .Scott said.

Since there is current due to voltage, this would mean that there is a second magnetic field created right? And say this current created a magnetic field that opposed the external magnetic field. Would this be the reason the bar rises? I mean, if the current were to create a magnetic field that opposed and decreased the strength of the external magnetic field, this would be considered magnetic flux? Is that right?

And if this IS magnetic flux, an induced magnetic field would need to exist according to Lenz's laws. My conclusion is that the rising bar creates this needed induced magnetic field. By rising up the slope, it increases the area of the circuit.

Is this right? :/ Motional emf is a subject I am very weak on. I've looked through several textbooks at the library...and still the subject is pretty scary to me!

Also to .Scott: wouldn't the current stop when the bar reaches the end of the rails?

I say that because my textbook says that this is the only reason we use rotating coils in real life, because having infinitely long rails to produce current is impractical.

BruceW
Homework Helper
My question pertained to Faraday and Lenz's Law. Say the bar is set at the very bottom of the bar at f-a.

My first question is: If there is a voltage applied between points a and f, why does bar begin to climb the slope? Eventually coming to rest at an elevated point?
yes, I think the bar does climb the slope (as long as the current is in the correct direction). But I can't think of how to explain it with Faraday's law. I think it is simpler to explain with the idea of the Lorentz force.

The passage says it definitely does climb it, but gives no explanation as to why. I think they assume I should know :/ lol

BruceW
Homework Helper
personally, I can only understand it by thinking about the Lorentz force law on a current-carrying wire. have you learned about this yet?

What is changing is the current. It will climb until the it generates enough force to be repelled by the magnetic field. But I am not clear on why the current would not continue to build.

The current will be naturally limited by the resistance of the bar and the power supply. Anyway, Lenz's law seems irrelevant here, it says nothing about the force acting on the bar.

I also believe that the interaction with the magnetic field produced by the circuit itself should be neglected, in most lab setups the self-force on conductors is negligible. Not in a rail gun, though.

dlgoff
Gold Member
I also believe that the interaction with the magnetic field produced by the circuit itself should be neglected, in most lab setups the self-force on conductors is negligible.
Yep. Maybe the problem has an external magnetic field that the OP has missed?

Yep. Maybe the problem has an external magnetic field that the OP has missed?

It is there :)

Lastly there is a magnetic field starting from the "base", rising up to the top of the plane of the computer screen. So basically the B field is perpendicular to the base, which is labeled in the figure.

This problem is way over my head. :/ *white flag* I surrender. Hopefully it doesn't come up on the MCAT!

dlgoff
Gold Member
It is there :)
oops

This problem is way over my head. :/ *white flag* I surrender. Hopefully it doesn't come up on the MCAT!

Have you studied the Lorentz force?

I took physics a while back and am only just refreshing before the MCAT. I remember the equation F = Eq + qvB.

Very well. Note the second term on the right hand side has charge and velocity. Can you express that as current?

Very well. Note the second term on the right hand side has charge and velocity. Can you express that as current?

qvB = IxB? Where x is distance?

BruceW
Homework Helper
yep. often you'll see it written as BIL (where L is x). You can work out the angle at which the bar will stop moving, if you also consider that there is a gravitational force downwards, and a normal force perpendicular to the rails. That is the most likely question they will ask you, I think ;)

.Scott
Homework Helper
Also to .Scott: wouldn't the current stop when the bar reaches the end of the rails?
According to the problem, the bar stops "at an elevated point" - I presume that to be short of the top. Should the current ever be interrupted by the bar reaching the end of the rails, the result would be explosive. The magnetic field would collapse and all the accumulated energy would be directed in heating the air in the conduction gap.

We're presuming that there is no resistance. That's the way these problems work.

What is happening is that the external magnetic field is actually pushing the rod back and the rail is directing it upward. There is nothing to stop the current from climbing, but the amount of force require to climb becomes greater and greater as the slope increases - eventually matching the rate at which energy can be pumped into the circuit given the inductance.

I didn't quite catch on to the whole system last night. I must be a little more awake tonight.