# Understanding motional EMF

## Main Question or Discussion Point I understand how motional EMF is produced, the magnetic force separates the charges in the rod to create an EMF. I don't understand though, why the current would flow from a lower potential to a higher potential.

I'm sorry I've made like 3 threads in the past 2 days, trying to contribute where I can too.

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The v in your diagram is the velocity of the motion, not the voltage.
Ya that's right. I just took this off of google images. If the bar is moved to the right it produces a magnetic force upwards causing a potential difference. In this case the potential will be higher at the top than the bottom but the current will move counter clockwise. On the bar the current will be moving upwards towards a higher potential. How is this possible?

Drakkith
Staff Emeritus
If higher potential means a more positive voltage then current flows towards it because current is composed of negative charges which are attracted towards that side.

If higher potential means a more positive voltage then current flows towards it because current is composed of negative charges which are attracted towards that side.
Ya current in circuits also really goes from negative to positive but by convention we always say that the current is positive. this breaks convention no?

BruceW
Homework Helper
Ya that's right. I just took this off of google images. If the bar is moved to the right it produces a magnetic force upwards causing a potential difference. In this case the potential will be higher at the top than the bottom but the current will move counter clockwise. On the bar the current will be moving upwards towards a higher potential. How is this possible?
if the bar is moved to the right, and the magnetic field is into the page (as the drawing suggests), then there will be an anticlockwise current, and a force on the bar to the left, due to Lenz's law. So I agree with the drawing. But there is no potential difference here. There is magnetic field but no electric field, so the potential difference is zero.

edit: and yes, there is a magnetic force on the charges, which is what causes them to move anticlockwise. This is what the 'emf' means. So the 'emf' arrow in the picture just means the force due to the magnetic field is pushing the charges in the bar upwards. The emf due to magnetic field is not related to potential, instead it is just the force integrated over the length of wire.

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if the bar is moved to the right, and the magnetic field is into the page (as the drawing suggests), then there will be an anticlockwise current, and a force on the bar to the left, due to Lenz's law. So I agree with the drawing. But there is no potential difference here. There is magnetic field but no electric field, so the potential difference is zero.

edit: and yes, there is a magnetic force on the charges, which is what causes them to move anticlockwise. This is what the 'emf' means. So the 'emf' arrow in the picture just means the force due to the magnetic field is pushing the charges in the bar upwards. The emf due to magnetic field is not related to potential, instead it is just the force integrated over the length of wire.
There is an electric field. That's why there's current running through the conductor. The magnetic force causes the positive charge in the bar to move up and the negative ones in the bar to move down. This causes a potential difference in the bar where Va>Vb in my last picture. This is the emf that gives rise to the current that flows. It flows towards increasing potential, however because the current is counterclockwise.

BruceW
Homework Helper
hold on, hold on. In the bar itself, there is no electric field. There is just magnetic field, and there is a current due to the velocity of the bar through the magnetic field (i.e. a force due to the magnetic field, with no electric field). Now, the rest of the circuit is stationary, and there isn't any magnetic force on the charge carriers in the rest of the circuit (except due to Lenz's law, but we can ignore that to first order). So, I agree that in the rest of the circuit there will be an electric field, due to a build-up of charge carriers at $a$ and a lack of charge carriers at $b$ (if we have positive charge carriers). And this electric field will cause the charge carriers to flow around the rest of the circuit, so that current is the same at all points in the circuit. So in the rest of the circuit, the electric field will be pointing anticlockwise to maintain the current. And then there will be a potential across the bar, which means that a steady current will flow.

So, ok there is an electric field caused by build-up of charge. But this electric field will be anticlockwise, if we have positive charge carriers. So the potential will be large at $a$ and small at $b$. Also, the emf is a force, integrated over the length of wire. So for the bar, the emf will indeed be pointing upwards, if we have positive charge carriers (which is the convention) since the magnetic force on these charge carriers will be upwards.

edit: so, once there is electrostatic equilibrium (which occurs extremely quickly), there will be a steady current through the bar, meaning that the emf on the charge carriers in the bar is zero overall. i.e. there is a magnetic emf upwards, but there is now an electric emf downwards due to the resistance of the circuit, which prevents an infinite current from flowing.

second edit: and in the picture, it looks like they just show the magnetic emf. This is pretty standard, I think it is usual to just talk about the emf due to external fields. i.e. the electric emf is due to an internal field between the charge carriers themselves, but the external emf is due to the external magnetic field.

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hold on, hold on. In the bar itself, there is no electric field. There is just magnetic field, and there is a current due to the velocity of the bar through the magnetic field (i.e. a force due to the magnetic field, with no electric field). Now, the rest of the circuit is stationary, and there isn't any magnetic force on the charge carriers in the rest of the circuit (except due to Lenz's law, but we can ignore that to first order). So, I agree that in the rest of the circuit there will be an electric field, due to a build-up of charge carriers at $a$ and a lack of charge carriers at $b$ (if we have positive charge carriers). And this electric field will cause the charge carriers to flow around the rest of the circuit, so that current is the same at all points in the circuit. So in the rest of the circuit, the electric field will be pointing anticlockwise to maintain the current. And then there will be a potential across the bar, which means that a steady current will flow.

So, ok there is an electric field caused by build-up of charge. But this electric field will be anticlockwise, if we have positive charge carriers. So the potential will be large at $a$ and small at $b$. Also, the emf is a force, integrated over the length of wire. So for the bar, the emf will indeed be pointing upwards, if we have positive charge carriers (which is the convention) since the magnetic force on these charge carriers will be upwards.

edit: so, once there is electrostatic equilibrium (which occurs extremely quickly), there will be a steady current through the bar, meaning that the emf on the charge carriers in the bar is zero overall. i.e. there is a magnetic emf upwards, but there is now an electric emf downwards due to the resistance of the circuit, which prevents an infinite current from flowing.

second edit: and in the picture, it looks like they just show the magnetic emf. This is pretty standard, I think it is usual to just talk about the emf due to external fields. i.e. the electric emf is due to an internal field between the charge carriers themselves, but the external emf is due to the external magnetic field.

Wowow my brain. I wish my my teacher explained it like this though. He just talked about how the bar wants to maintain equilibrium (not even electrostatic, just talking about lenz' law and bar resisting change) because that's how nature is rather than explaining anything about the magnetic force in the bar. It wasn't even in the original coursebook, that picture's from an older book that I got from a friend because young & freedman always explains things a lot better. I wrote the exam yesterday and would have went on without understanding what emf fundamentally is. I just thought it was synonymous with net voltage. In fact he told us that the reason they called voltage emf was because they misunderstood it as a force. God damn...

Nah now that I think about it I'm probably remembering it wrong, he seemed like a knowledgeable guy.

So to make sure I'm understanding this correctly, the voltage created by the magnetic field carries current from a to be in the U part of the bar here and then the magnetic force carries the positive charge up to $a$. The resistance of the wire creates an emf going clockwise which comes to equilibrium with the emf from the magnetic force. When this happens, the current reaches a steady state (assuming velocity is kept constant). There's still no electric field in the bar at this point, charge continues to be carried by the magnetic emf?

BruceW
Homework Helper
Wowow my brain. I wish my my teacher explained it like this though. He just talked about how the bar wants to maintain equilibrium (not even electrostatic, just talking about lenz' law and bar resisting change) because that's how nature is rather than explaining anything about the magnetic force in the bar.
yeah, Lenz' law is another separate thing which will happen. The electrostatic forces (which cause the current to be the same at all points in the circuit) are much stronger than the forces due to Lenz' law. (unless the acceleration of the bar is extremely quick, but that situation would be very hard to analyze).

Regtic said:
I wrote the exam yesterday and would have went on without understanding what emf fundamentally is. I just thought it was synonymous with net voltage. In fact he told us that the reason they called voltage emf was because they misunderstood it as a force. God damn...

Nah now that I think about it I'm probably remembering it wrong, he seemed like a knowledgeable guy.
I don't know the historical use of the word emf... But it is now usually meant to mean the integral of the electromagnetic force over the length of wire. So it is not the same as potential difference, because it also includes magnetic force. I think sometimes emf is used to mean just potential difference. But in my experience, it is more often used to mean the integral of the electromagnetic force over the length of wire.

Regtic said:
So to make sure I'm understanding this correctly, the voltage created by the magnetic field carries current from a to be in the U part of the bar here and then the magnetic force carries the positive charge up to $a$. The resistance of the wire creates an emf going clockwise which comes to equilibrium with the emf from the magnetic force. When this happens, the current reaches a steady state (assuming velocity is kept constant).
yeah. that sounds about right. (ignoring any effects of Lenz' law, to keep things simple). The electrostatic force between charge carriers will cause the charge carriers to move through the U part of the circuit. And the total emf will be zero, due to the internal electrostatic force created due to the resistance of the circuit.

Regtic said:
There's still no electric field in the bar at this point, charge continues to be carried by the magnetic emf?
well, as you said in the last bit, the resistance of the wire creates an emf going clockwise, and the current reaches a steady state. So this electrostatic force also exists in the bar, since current must be the same at all points in the circuit. In other words, the electrostatic force due to the resistor will cancel out the magnetic force so that there is a steady current in the bar also.

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well, as you said in the last bit, the resistance of the wire creates an emf going clockwise, and the current reaches a steady state. So this electrostatic force also exists in the bar, since current must be the same at all points in the circuit. In other words, the electrostatic force due to the resistor will cancel out the magnetic force so that there is a steady current in the bar also.
Ahhh ok thank you makes sense. Very cool.

BruceW
Homework Helper
I hope it helped. It can be confusing, because often people just talk about the emf's due to external fields. But there are also the electrostatic forces between charge carriers in the wire, which means the current has the same value all the way around the loop, in this example. So you might be tempted to think that there is an almost infinite current in the bar, and a low current through the resistor. But this will not happen, due to the internal forces between charge carriers. In other words, if there starts to be a build-up of charge, the forces between charges will cause the current to be the same all the way around the loop.

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