Semicircular ring falling through a Magnetic Field

[collinsmark]

Wait, hold on.

I agree that the positive charges would be pushed toward the right. .

But if you took the wire out of the magnetic field, attached it to a battery such that the battery's positive terminal was on Q and the negative terminal on M, the current (thus positive charges) would flow through the wire to the left.

So you should re-think that last part.

[Edit: If you'd like to think about it another way, the positive side of the induced emf is going to push the positive charges in the wire away from it. Which side of the wire are the positive charges being pushed away from?]

The ring doesnot form a closed loop. So you cannot hypothetically take the wire out of the magnetic field and attach it to a battery as that is applicable only in the case of closed loops .

You can do anything you want to hypothetically, hypothetically speaking.

What I was attempting to do was illustrate an important point. There is an emf induced on the semicircle, even if it is isn't a closed loop. Just like a battery will have a positive and negative terminals on it, even if it doesn't happen to be connected to anything at a particular time.

It's not valid to state, "a battery, standing by itself, does not have positive or negative terminals because its not conducting current." The truth is that it has an emf across the terminals, and it also has a polarity, whether it is conducing current or not.

And one method to determine the battery's polarity is to determine which direction the current would flow when connected in a closed loop, even if that process is done hypothetically.

Here, the wire is not analogous to the battery itself. But the wire will have an emf induced across it as if a battery was connected to it, even if there is no current. If you had a voltmeter connected to the moving wire, you could measure this induced emf.

I thought about one more possibility while thinking about the problem.
What will be the e.m.f induced int the semi-ring if the velocity is parallel to the diameter?
Will it be same as above - 2BRv ?

You and Dadface have discussed this already to some extent, but I think there is some confusion that I want to clear up.

There is an induced emf across the wire, in this second situation (where the wire moves to the side instead of down). But it's not induced between points M and Q. The potential between points M and Q is zero. But there is an non-zero induced emf between points N and M, which is the same emf induced between points N and Q. And as you have already calculated, the value of the emf is Brv. I just wanted to make sure we're all on the same page there. I'll leave it up to you to determine the polarity of the emf.

 

[zorro]

What I was attempting to do was illustrate an important point. There is an emf induced on the semicircle, even if it is isn't a closed loop. Just like a battery will have a positive and negative terminals on it, even if it doesn't happen to be connected to anything at a particular time.

It's not valid to state, "a battery, standing by itself, does not have positive or negative terminals because its not conducting current." The truth is that it has an emf across the terminals, and it also has a polarity, whether it is conducing current or not.

And one method to determine the battery's polarity is to determine which direction the current would flow when connected in a closed loop, even if that process is done hypothetically.

I agree with you.
What confused me was your statement -

But if you took the wire out of the magnetic field, attached it to a battery such that the battery's positive terminal was on Q and the negative terminal on M, the current (thus positive charges) would flow through the wire to the left.

So you should re-think that last part.

If you don't connect the battery, the positive charges move towards right (the force acting is towards right) and get concentrated there creating a higher potential.

I agree that if a battery is attached with positive terminal on Q and negative terminal on M, the positive charges would flow towards left ( the point Q will be still at higher potential). This method is fine if you just want to find out the polarity. You cannot conclude that the force acting on the positive particles is towards left.

Its towards right.

 

[collinsmark]

If you don't connect the battery, the positive charges move towards right (the force acting is towards right) and get concentrated there creating a higher potential.

I agree that if a battery is attached with positive terminal on Q and negative terminal on M, the positive charges would flow towards left ( the point Q will be still at higher potential). This method is fine if you just want to find out the polarity. You cannot conclude that the force acting on the positive particles is towards left.

Its towards right.

Yes! that's right! We're both in agreement (and always were) that the positive charges are forced to the right. There is no disagreement on that point.

What I was requesting that you rethink is this statement:

The positive charges will be pushed towards right. So Q is at a higher potential.

That's what I disagree with.

The positive charges feel a force toward the right because Q has the lower potential, not higher. The force on the charges is an effect of the emf, not the cause.

I know this can be confusing. But perhaps this link might clear up some of the confusion. Pay careful attention to the polarity of the induced emf, and the corresponding induced current in the figures.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c2"

 

[zorro]

The positive charges feel a force toward the right because Q has the lower potential, not higher. The force on the charges is an effect of the emf, not the cause.

I know this can be confusing. But perhaps this link might clear up some of the confusion. Pay careful attention to the polarity of the induced emf, and the corresponding induced current in the figures.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c2"

I completely agree with the matter given in that site.
But the basic difference here is that the wire does not form a closed loop. What you said will be true if the wire forms a closed loop ( I think so ).
Do you have any material that provides an explanation of induced e.m.f. in an open loop?

 

[collinsmark]

I completely agree with the matter given in that site.
But the basic difference here is that the wire does not form a closed loop. What you said will be true if the wire forms a closed loop ( I think so ).
Do you have any material that provides an explanation of induced e.m.f. in an open loop?

Yes, the site I gave you provides about the best explanation available. The first diagram in the linked page shows the polarity of a rod in an open loop.

The piece of knowledge to take home with you at the end of the day is that the induced emf is the same whether the loop is open or closed. It doesn't matter if there is an open or closed loop. The induced emf is independent of that.

 

[Dadface]

Nice link collinsmark.Do you see it now Abdul.Closed loop or open loop,semi circular shape,rectangular shape or any other shape,the theory needed to calculate the emf is the same.There is an initial vertical flow of charge resulting in a charge difference and therefore an emf between the top and bottom of the circuit.

 

[zorro]

The positive charges feel a force toward the right because Q has the lower potential, not higher. The force on the charges is an effect of the emf, not the cause.

What confused me so much was about the difference between induced electric field and electric field due to the movement of charges.
We don't define a potential for an induced electric field. Moreover, an induced electric field is independent of any charge.
The potential we were talking about was due to the electric field created by the movement of charges. Refer the figure.

Both the electric fields are along the same direction. Is Q still not at a higher potential?

 

[hikaru1221]

Hi all,

@Abdul Quadeer: You're self-contradicting.

Moreover, an induced electric field is independent of any charge.

If it's independent of the charge or the flow of charge, then why does it depend on whether the loop is open or closed?

I would like to use the term "developed E-field" instead of "induced E-field", as it is not a real E-field and also not formed by magnetic induction, as observed in the reference frame of the ground: \vec{E}^*=\vec{v}\times\vec{B} .
The developed emf: e = \int_M^Q \vec{E}^*d\vec{l} = \int_M^Q (\vec{v}\times\vec{B} )d\vec{l} , independent of the charge and also the flow of charge.
This is a true emf. We don't care about what kind of force developed inside the emf "source". As long as the electron is propelled and given some amount of energy when moving through this "source", it has emf, by definition.
By the way, electric potential is defined for the external circuit in this case. I'm not sure whether it's valid for the semi-ring or not.

@collinsmask: I think this reasoning of yours is a fallacy:

But if you took the wire out of the magnetic field, attached it to a battery such that the battery's positive terminal was on Q and the negative terminal on M, the current (thus positive charges) would flow through the wire to the left.

The wire (and plus the B-field) is a mechanism which has the emf. It is NOT an external circuit. I think the correct equivalent circuit where we connect a resistor to QM to form a closed loop should be like in my picture. In this case, I think Abdul Quadeer was right.

 

[zorro]

@Abdul Quadeer: You're self-contradicting.

If it's independent of the charge or the flow of charge, then why does it depend on whether the loop is open or closed?

Yes I was wrong.

@collinsmask: I think this reasoning of yours is a fallacy:

The wire (and plus the B-field) is a mechanism which has the emf. It is NOT an external circuit. I think the correct equivalent circuit where we connect a resistor to QM to form a closed loop should be like in my picture. In this case, I think Abdul Quadeer was right.

That was what I meant when I said it doesnot form a closed loop to Collinsmark.

The ring doesnot form a closed loop. So you cannot hypothetically take the wire out of the magnetic field and attach it to a battery as that is applicable only in the case of closed loops .

You can attach a battery only in the case of a closed loop.

 

[collinsmark]

@collinsmask: I think this reasoning of yours is a fallacy:
[...]
The wire (and plus the B-field) is a mechanism which has the emf. It is NOT an external circuit. I think the correct equivalent circuit where we connect a resistor to QM to form a closed loop should be like in my picture. In this case, I think Abdul Quadeer was right.

Hello hikaru1221,

You've got your resistor in the wrong place. We're calculating the induced emf across the moving wire itself. Essentially, the moving wire is the resistor.

For a more concrete example, for illustrative purposes only, let's replace the moving wire/resistor with a polar device, such as an LED. If you want the moving LED to light up it needs to be placed in the correct polarity, and that means M is positive and Q is negative.

That was what I meant when I said it doesnot form a closed loop to Collinsmark.

The ring doesnot form a closed loop. So you cannot hypothetically take the wire out of the magnetic field and attach it to a battery as that is applicable only in the case of closed loops .

You can attach a battery only in the case of a closed loop.

Hello Abdul,

Again all of this is hypothetical to illustrate the point of why M is positive and Q is negative. You can claim that Q is the positive terminal if you wish and you are completely free to do so, but you will dinged on your homework and tests if you are ever asked to specify the polarity. I know this subject can be a little confusing, which is why hypothetical situations and thought experiments are so useful.

So hypothetically, if we removed the magnetic field, and replaced it with a battery, the equivalent circuit is shown below, for the situation where a second wire (outside the magnetic field) is connected between M and Q. (The LED is left in the circuit for illustrative purposes only). The voltage drop goes positive to negative from M to Q. Thus M is the positive terminal, and Q is negative, when considering the moving semicircle of wire itself.

[Edit: Now, if you wish to model the circuit without the second wire (open-loop), such that there is no current, then the above circuit is still the equivalent circuit! It's just that the LED/resistor has infinite resistance. But there is still an efm induced across it the wire. But there is no current because of the infinite resistance (an open loop has infinite resistance, btw), and that's why M is positive and Q is negative when discussing the emf induced across the semicircle of wire itself.]

[Edit: Another reason why some of this can be so confusing is because the correct answer of what the polarity is depends heavily on question of the the polarity of "what".

• If asked "what is the polarity of induced emf across the wire moving in the magnetic field, you should give the answer as I described above. In this case, M is positive and Q is negative. That's because the question is all about the moving wire itself. And the answer must have focus given to the moving wire.
• If instead, the whole system is to be taken as a "black box" that is to function as an electrical generator which has terminals M and Q, and you are asked, "what is the polarity of terminals M and Q on this black box generator?" the answer in this case is that Q is positive and M is negative. That's because the question doesn't involve the moving wire at all per se, and the attention is given to components external to the system.]

 

[hikaru1221]

Hello hikaru1221,

You've got your resistor in the wrong place. We're calculating the induced emf across the moving wire itself. Essentially, the moving wire is the resistor.

The emf is developed on the wire, so the wire is a source itself, regardless of whether its resistance is zero or non-zero. If zero, it's an ideal source; if not, it's simply equivalent to an ideal voltage source in series with a resistor. The resistor I put in the picture was to form AN EXTERNAL CIRCUIT so that THE LOOP IS NOW CLOSED; this resistor is not the internal resistor of the wire. In the lumped model (this is the term that electrical engineers use), the emf is always attributed to a source.

Now let's forget the wire for a while. Say, the wire is some two-ported mechanism where we know for sure a current will go out from terminal Q; we don't care whether it's a source or anything. If we connect an external resistor to Q and M, since the current goes from Q, through this resistor, to M, the potential of Q must be higher than the potential of M, by Ohm's law.

For a more concrete example, for illustrative purposes only, let's replace the moving wire/resistor with a polar device, such as an LED. If you want the moving LED to light up it needs to be placed in the correct polarity, and that means M is positive and Q is negative.

In this example of yours, you already forgot to include a source. If there is no source, there will be no current. The LED alone won't make it equivalent to the emf.
The difference between LED (or any similar device) and the source (the emf) is that while LED is an energy absorption device, the source is generally an energy provision device (except more complicated cases where two or more sources or emf's are involved).

[Edit: Another reason why some of this can be so confusing is because the correct answer of what the polarity is depends heavily on question of the the polarity of "what".

• If asked "what is the polarity of induced emf across the wire moving in the magnetic field, you should give the answer as I described above. In this case, M is positive and Q is negative. That's because the question is all about the moving wire itself. And the answer must have focus given to the moving wire.
• If instead, the whole system is to be taken as a "black box" that is to function as an electrical generator which has terminals M and Q, and you are asked, "what is the polarity of terminals M and Q on this black box generator?" the answer in this case is that Q is positive and M is negative. That's because the question doesn't involve the moving wire at all per se, and the attention is given to components external to the system.]

Abdul Quadeer is asking about the potential - the nature of the field and the phenomenon, not some convention on the polarity (The polarity is just the way people name it after all!). I also have never heard of polarity of emf. Anyway I am not sure if we can even define a potential for such "field" (\vec{E}^*=\vec{v}\times\vec{B} - this "field" only exists inside the wire and is not an independent entity, so I would rather call it "force per unit charge". This term makes even more sense when it comes to emf, as for any voltage source, it is the "force per unit charge" inside the source which develops the emf, regardless of the nature of the force). So I would rather pay attention to the external circuit.

 

[collinsmark]

Abdul Quadeer is asking about the potential - the nature of the field and the phenomenon, not some convention on the polarity (The polarity is just the way people name it after all!).

Looking back to the original inquiry in post #9, I see that you are right. And for that I apologize if I caused a lot of undue confusion.

I agree that in the general sense, point Q has the higher potential. By that I mean that a given positive charge q will have higher potential energy at point Q than it would at point M. If I contradicted that in any of my previous posts, I am sorry.

For some reason or another, I had it on my mind that Abdul was asking about the polarity (direction, if you will) of the emf induced on the wire. Maybe it was because I working on two many things at once. I don't remember.

That being said, the induced emf is a voltage and it does have a polarity. If Abdul is ever asked to place this induced emf on a figure showing a moving wire, and label the + and - signs, I want him to get the right answer. And in this case, the '+' sign of the induced emf, on the moving wore itself, corresponds to M and the '-' sign to Q.

This is because, hypothetically, if there is any current due to the induction, it will flow through the moving wire from M to Q. '+' to '-'.

Take a look at the link that I referred to a few posts ago. (I'll link it again here),
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c2"

Look at the first figure on the page, where it describes "induced voltage." Notice the polarity of the induced voltage.

Also look at the second and third figures on the page, and take note of the '+' and '-' signs.

This "hyperphysics" page is from the University of Georgia State University. But I don't think they are alone in using this convention in labeling the induced emf in this way. I've seen this sort of thing in other textbooks too.

I think the crux of this seeming contradiction stems from the different questions, "if you are a moving wire, how does the rest of the world seem to you?" and "if you are the rest of the world how does a moving wire seem to you?" It seems this convention of putting the '+' and '-' signs such that the current flows from '+' to '-' though the moving wire is a result of the former question.

 

[hikaru1221]

I agree on the polarity I believe Abdul Quadeer would understand more than he asked for after your post

I think the crux of this seeming contradiction stems from the different questions, "if you are a moving wire, how does the rest of the world seem to you?" and "if you are the rest of the world how does a moving wire seem to you?" It seems this convention of putting the '+' and '-' signs such that the current flows from '+' to '-' though the moving wire is a result of the former question.

I'm no expert at this issue, but since you mentioned this, I would like to point out for Abdul Quadeer that all the reasonings above are in the reference frame of the ground. It would be a different story if we change to the frame of the moving wire.

 

[zorro]

Thank you collinsmark and hikaru1221 for enlightening me with your thoughts.
I got it all now. As hikaru1221 said - I understood more than what I asked.