Induced EMF and current for a bar sliding on a circular rail

[FranzDiCoccio]

Homework Statement

The red (conducting) bar in the drawing rotates ccw about a vertical axis through its extreme O, while its other extreme is in contact with the blue horizontal circular (conducting) rail. The point O and is connected by a conducting wire with resistance R to a point C on the rail.
A uniform magnetic field B is present in the region. The B field points in the same direction as the rotation axis of the bar, upwards.
Use Faraday's law to find the initial emf in the circuit . Find the current initially flowing through the circuit.

Relevant Equations

Faraday's law (but I am not sure how to go about it)
Motional emf (but it is not the strategy suggested by the problem)

On the left: my copy of the illustration in the problem.
On the right: top view, with the angle.
The problem gives the magnitude of the magnetic field, the radius of the rail, the resistance of the resistor, the initial rotational frequency of the bar.
I am able to obtain the given solutions, but 1) I cannot figure out how to do that with Faraday's law 2) I'm not sure this problem even makes sense.

So, my reasoning is the following: if only the bar was present, I could find its "motional emf", or better, the potential difference between its extremes. Starting from the "translation" case
$$V_r = v_r B r$$
I'd integrate along the radius
$$V =\int_0^r v_r B r dr= \int_0^r \omega B r^2 dr = \frac{\omega B}{2} r^2$$
Then, closing my eyes, I'd apply Ohm's law to find the current:
$$I =\frac{V}{R}$$
These formulas do give the proposed solution.

Is this correct, though? According to my solution the current should flow from O to A. But then, where does it go?
The problem says nothing about the resistance of the rail. In fact, the solution for the current seems to suggest it has no resistance at all.
Assuming that it has a tiny resistance, I'd say it splits at A, and part of it travels ccw and the other part cw.
How much goes one way depends on the angle theta, assuming a uniform resistivity of the rail.

Using Faraday's law is a bit of a nightmare.
If I consider the area spanned by the circuit and the "shorter" arc, I get the same result as above.
$$\Phi =\frac{\theta r^2}{2} B$$
and hence (dropping Lenz's minus)
$$V =\frac{\omega r^2}{2} B$$
The current goes from A to C clockwise.
But what of the area spanned by the longer arc and the circuit? Shouldn't I consider that as well?
It seems to me that I obtain the same current, but flowing through the longer arc, counterclockwise?

Is there a reason for considering one arc only? If so, what happens when the bar completes one cicle? Is the area still increasing?

If not, wouldn't be the current twice as the one given in the solution? What, the two currents flow in opposite directions and meet at C?
Or else, perhaps the current is the one given in the solution, and it is divided equally into the two arcs?

Whatever it is, I think that the situation in this problem was not well pondered, especially considering that it is supposed to be high school level.
Thanks a lot for your help.

 

[Cutter Ketch]

Yes, you have to consider both paths. Yes clockwise through the shorter arc and counter clockwise through the longer arc. Note that that gives the same direction through the red arm and the load. They add.

 

[rude man]

Homework Statement::

Is this correct, though?

Yes.

According to my solution the current should flow from O to A. But then, where does it go?

Current goes O-A-C-O.

The problem says nothing about the resistance of the rail. In fact, the solution for the current seems to suggest it has no resistance at all.

That's what I would assume too.

Assuming that it has a tiny resistance, I'd say it splits at A, and part of it travels ccw and the other part cw.
How much goes one way depends on the angle theta, assuming a uniform resistivity of the rail.

There's no "split". Current goes per the loop above.

Using Faraday's law is a bit of a nightmare.
If I consider the area spanned by the circuit and the "shorter" arc, I get the same result as above.
$\Phi =\frac{\theta r^2}{2} B$

and hence (dropping Lenz's minus)

$V = \omega r^2 B/2$

The current goes from A to C clockwise.

Does it matter which arc? Does $\omega = \frac {d\theta} {dt}$ depend on which arc? You had it right when you got to V !

BTW in general whenever it comes to moving media like your rod you should stick with your first method V = vBr. You can get in a trap wit Faraday in those cases. So I don't approve of your instructor asking you to go with faraday even though in this case it works.

 

[Cutter Ketch]

There's no "split". Current goes per the loop above.

There are two paths from A to C and you claim the current only flows through one of them? On what do you base this remarkable assertion?

 

[Cutter Ketch]

Does it matter which arc? Does ω=dθdtω=dθdt \omega = \frac {d\theta} {dt} depend on which arc? You had it right when you got to V !

Yes, it matters very much when the answer is “both”. By the way, this means he did not get it right when he calculated V

Retraction: he did get it right, and the two loops don’t add. The emf calculates the same with either loop and the emf is available to both loops, but it is the same emf

 

[Cutter Ketch]

Faraday’s law is not a nightmare. It is very simple, and has the advantage here over what you did of not requiring integration. If you think that is unimportant, look carefully at the integration you posted. I think you will see some remarkably bad math and a leap to an answer you wanted that isn’t what the integral produces. (Answer correct, integral set up incorrectly). With Faraday’s law you can essentially write the answer down without any manipulation.

 

[rude man]

There are two paths from A to C and you claim the current only flows through one of them? On what do you base this remarkable assertion?

The path I gave O-A-C-O comprises two zero-ohm paths. The circular rail is a short (zero ohms) so the long way around AC is in parallel with the short way around AC. Zero ohms || zero ohms = 0 ohms. The current path is thru the resistor and the bar and equals emf/R. Nothing very remarkable.

The OP probably made a typo in his/her integral because the result is correct.

 

[rude man]

Yes, it matters very much when the answer is “both”.

No it doesn't. Emf = $d\phi/dt$ independent of $\theta. \omega$ is independent of $\theta$.

 

[FranzDiCoccio]

Faraday’s law is not a nightmare. It is very simple, and has the advantage here over what you did of not requiring integration. If you think that is unimportant, look carefully at the integration you posted. I think you will see some remarkably bad math and a leap to an answer you wanted that isn’t what the integral produces.

Sorry about that, I had the time to post the question only very late at night, and I got ahead of myself with the maths. In an infinitesimal element along the bar
$$dV = v B dr = \omega B r dr$$
and
$$V = \int_0^r dV =\omega B \int_0^r r dr = \frac{\omega B r^2}{2}$$

As to the rest of the problem I guess I was confused by the "bifurcation" at A, and by the rail being a short.
I was wondering why the current should go through one arc only and, in that case, which one?
Also, the area of one circular sector was growing, whereas that of the other was shrinking. I agree that the rate of change is the same.
A further thing that confused me in the FNL approach was the question "what happens when the bar has completed one turn?".

I tried wrapping my head around this problem by considering a slightly different one.

Suppose you have a rectangular loop (yellow), in a uniform magnetic field orthogonal to it (green). Place a conducting bar orthogonal to the longer sides of the rectangle, so that it connects them (red). Now slide the bar along the long sides, so that it stays connected and parallel to the short sides.
We can also assume that maybe the short sides have some resistance.

The bar would have a motional emf $vB\ell$. If the bar could be considered as a emf generator $V= vB\ell$, a current $I$ would flow through it. This would split into two currents, flowing through the two resistances $R_1$ and $R_2$. I'd say
$$I_1 = \frac{V}{R_1}, \quad I_2 = \frac{V}{R_2}$$

Assuming this is correct, and I'm not missing something, how would I go about solving the same problem via FNL?

I can see that in the rectangular loop, choosing either "half rectangle" would give the same induced emf (equal to the motional emf).
But when it comes to calculating the current, I cannot just ignore one of the loops, can I?
I'd probably need to do something similar to what I did above, i.e. considering a "fictional" emf generator located "on the bar".

I do realize that the first circuit is probably not equivalent to this one, with the rail being a short.
But I still do not understand entirely why the current would flow along one of the arcs only.
Wouldn't there be an induced magnetic field? Can I really ignore one of the loops, in this respect?

Of course I was exaggerating when I said "a nightmare". I think I can handle the FNL law in most of the cases. Somehow I was (and perhaps I am still) struggling a bit with the one in the OP.

 

[etotheipi]

I do realize that the first circuit is probably not equivalent to this one, with the rail being a short.
But I still do not understand entirely why the current would flow along one of the arcs only.
Wouldn't there be an induced magnetic field? Can I really ignore one of the loops, in this respect?

I believe what @rude man is saying is that the two possible paths from A to C, each of resistance zero ohms, are in parallel with each other. As in circuit theory, we can simplify the maths by replacing these two loops with an imaginary equivalent loop also of resistance zero, from A to C, since this has no effect on the current through the resistor or the emf produced across the rotating rod.

If you were to set up the experiment and put an ammeter on both arcs, you'd measure a current in both.

 

[rude man]

I tried wrapping my head around this problem by considering a slightly different one.
View attachment 258598
Suppose you have a rectangular loop (yellow), in a uniform magnetic field orthogonal to it (green). Place a conducting bar orthogonal to the longer sides of the rectangle, so that it connects them (red). Now slide the bar along the long sides, so that it stays connected and parallel to the short sides.
We can also assume that maybe the short sides have some resistance.

The bar would have a motional emf $vB\ell$. If the bar could be considered as a emf generator $V= vB\ell$, a current $I$ would flow through it. This would split into two currents, flowing through the two resistances $R_1$ and $R_2$. I'd say
$$I_1 = \frac{V}{R_1}, \quad I_2 = \frac{V}{R_2}$$

That is correct.

You did not specify that the long sides are zero-resistance rails. Obviously you meant to.

Just replace your sliding bar with a battery of voltage $Bv\ell$ , then it will be obvious how the current travels.. You will get what you got.

This again shows the danger and confusion likely if you use Faraday (FNL) with moving media such as your sliding bar. Avoid it. Tell your instructor what I said and if he/she wants an example of where FNL absolutely does not work in moving media I can post here.

 

[FranzDiCoccio]

That is correct.

You did not specify that the long sides are zero-resistance rails. Obviously you meant to.

Yes, that's right. I wanted to avoid that a whole loop with zero resistance, because that was one of the things that misled me in the original problem.

Just replace your sliding bar with a battery of voltage $Bv\ell$ , then it will be obvious how the current travels.. You will get what you got.

This again shows the danger and confusion likely if you use Faraday (FNL) with moving media such as your sliding bar. Avoid it.

So you're saying that whenever there is a moving part, the best course of action is to fall back onto motional emf.

Tell your instructor what I said and if he/she wants an example of where FNL absolutely does not work in moving media I can post here.

There is no instructor. I found this problem in a book. So, in asking to face the task using FNL , the book is actually misleading me? It figures... a lot of problems in this textbook are wrong, or misleading... But that's the book we use.

I would like to see the example you're mentioning, if possible. I'm curious about that.

Also, could I ask one more question? There something that sometimes puzzles me.
Let me refer to the problem with the rectangular loop, for the sake of simplicity.
In that case, due to the current flowing through it, the red sliding bar would experience a Lorentz force opposite to the direction of $v$, that is the same as any non-moving wire carrying the same current in the same direction, placed in the same $\vec B$ field.

So, aren't we somehow ignoring the fact that the bar is moving? The conventional charges moving through the wire are not really moving "downward" in the drawing, but diagonally, right and downward.
That would give rise to a Lorentz force that is orthogonal to such direction and to the field, so down and left.

I'd like to make sure of this: the component of such force going "down" does not affect the motion of the bar, but rather is what is needed to maintain the current through it. This is because the charges can move freely "downward", but they cannot do so going left, because of the "boundary" of the bar.
So there is no force pulling the bar "as a whole" downward, is there?

After asking this question explicitly I'm almost sure that the answer is most likely "no, there isn't", and I'm tempted to delete it.
I'll leave it, though, just to be on the safe side.

Thanks for your help!

 

[rude man]

Yes, that's right. I wanted to avoid that a whole loop with zero resistance, because that was one of the things that misled me in the original problem.

You can't have a loop with total resistance = 0.This would be a singularity in which the loop's induced B field would be equal to the externally applied B field and the bar would not be movable no matter how great the applied force. In such problems not only do you have to assume finite R but large enough R so that the induced B field is << externally applied B field. Otherwise you'd have to solve the differential equation iR = d$\phi$/dt - L di/dt with $\phi$ the externally applied field and L the inductance of the loop. Finding that L for a single loop involves math even beyond advanced undergraduate physics.
So L di/dt = (d/dt) (Induced B flux) must be << external flux rate.
(And even that is a simplification, ignoring the effect of the induced field coupling back to whatever produces the external field!, i.e. mutual inductance which you may encounter later)

So you're saying that whenever there is a moving part, the best course of action is to fall back onto motional emf.

Absolutely.

There is no instructor. I found this problem in a book. So, in asking to face the task using FNL , the book is actually misleading me? It figures... a lot of problems in this textbook are wrong, or misleading... But that's the book we use.

I'm saying it's generally a bad idea to use FNL for moving media. If you must use FNL in this case you can use d$\phi$/dt = (area swept by the bar in unit time) x B. I'm sure that was the book's intent. Still, avoid FNL for moving media when allowed to. Use $Bv\ell$
[quote}
I would like to see the example you're mentioning, if possible. I'm curious about that.
[/quote} See attached pdf file.

Also, could I ask one more question? There something that sometimes puzzles me.
Let me refer to the problem with the rectangular loop, for the sake of simplicity.
In that case, due to the current flowing through it, the red sliding bar would experience a Lorentz force opposite to the direction of $v$, that is the same as any non-moving wire carrying the same current in the same direction, placed in the same $\vec B$ field.

So, aren't we somehow ignoring the fact that the bar is moving?

Where would the current come from if the bar didn't move? The bar doesn't know that it's moving, just that it carries current i. And so F = Bi$\ell$. where i = B$\ell$v/R.

The conventional charges moving through the wire are not really moving "downward" in the drawing, but diagonally, right and downward.
That would give rise to a Lorentz force that is orthogonal to such direction and to the field, so down and left.

I'd like to make sure of this: the component of such force going "down" does not affect the motion of the bar, but rather is what is needed to maintain the current through it. This is because the charges can move freely "downward", but they cannot do so going left, because of the "boundary" of the bar.
So there is no force pulling the bar "as a whole" downward, is there?

No there isn't but it's an EXCELLENT question. Without going into the gory details you'll have to accept the fact that there is no net downward force to the bar. Charge is moved along the bar by the induced emf B$\ell$v only..
If you can get holds of Resnick & Halliday I can point you to that detail. For now just accept $\vec F = i \vec l ~x~\vec B$ gives magnitude and direction of F.

OK I attachd the pdf file.

 

[Cutter Ketch]

Assuming this is correct, and I'm not missing something, how would I go about solving the same problem via FNL?

Hey, I love this. This is the exact right analogy, and this exactly what I was thinking of when I retracted my earlier boner. The answer is simple. Faraday’s law must be true for both loops. Calculating either loop shows an emf along the moving wire. And, of course, the emfs calculated with either loop are equal because it’s the same emf.

So, to recap, you are welcome to calculate with either loop. Faraday’s law has to be true in both loops simultaneously and they are both experiencing the same emf from the same moving part so you get the same answer either way. And, yes, the current takes both paths.

 

[FranzDiCoccio]

OK I attachd the pdf file.

Thanks a lot. Your example is interesting indeed. The vector form of the equations is perhaps too advanced for High School, but motional emf should suffice to analyze the problem, right?

So the magnetic field would produce effects only in the bottom side of the loop, and in the portions of the vertical sides that "dip" in the field.
In the bottom side a motional emf would appear, as if a generator was present there.
The Lorentz force in the vertical portions would not amount to an emf, though, because it would push the charges in a direction that is perpendicular to the wire.
The current induced by the emf on the bottom side would not interact with the field in the vertical sides, because they would be in the same direction.

So I'd say that the overall effect would be an emf in the bottom side, a "viscous" force opposite to the velocity and a slight "pull" of the loop in a direction that is perpendicular to both the initial velocity and the field.
Is this correct?

So, to recap, you are welcome to calculate with either loop. Faraday’s law has to be true in both loops simultaneously and they are both experiencing the same emf from the same moving part so you get the same answer either way. And, yes, the current takes both paths.

Ok but when I have to calculate the current I should place a fictitious generator somewhere, and that could go only on the bar, not anywhere else on the loop (as it would be the case if there was only one loop).
So I agree that the FNL approach is a bit "slippery", and that the motional emf is more reliable.

 

[FranzDiCoccio]

Hi... me again
I was thinking that the general advice "use motional emf if there are moving parts" should be made more precise, something like "use motional emf if there are moving parts changing the area of more than one loop".
Indeed I recalled situations where the moving parts are hinged, or move in a non trivial way, like in the sketch below

In the first case the red bar keeps in contact with the yellow L-shaped conductor, while it slides along it so that the red angle in the triangle becomes small. Let's say that the resistance is all in the red bar.

In the second case there loop is a parallelogram made by four hinged bars with some resistance.

I'd say that FNL is more convenient in these cases, given that there is only one loop and the motion of the parts is not simple.

Would you agree?

 

[rude man]

Consider a rectangular loop of finite resistance oriented as shown. Also assume a uniform B field surrounding the bottom side of length L of the loop but not the top. B is perpendicular to the bottom side of the loop.
Now, impart a velocity v to the coil along the normal to the loop and the B field. There will be an emf generated along the bottom side of the loop but none in the other three.
What is the emf of the bottom side? Note that ∂B/∂t = ∂φ/∂t = 0 everywhere inside the loop.
So Faraday’s law is inapplicable and Maxwell's equations do not cover this situation unless del x E = -∂B/∂t is modified.
And the answer is that a term ## \nabla ~x~ (\mathbf{ (v x B)} $must be added to the right-hand side, yielding$ or emf = BLv when E is integrated over L. This second term is based on the Lorentz magnetic force qv x B and is not related to Maxwell’s equations for stationary media including Faraday’s law.
Note that the BLv law works fine for this situation while Faraday’s law does not.
Cf. following two illustrations:

Hi... me again
I was thinking that the general advice "use motional emf if there are moving parts" should be made more precise, something like "use motional emf if there are moving parts changing the area of more than one loop".
Indeed I recalled situations where the moving parts are hinged, or move in a non trivial way, like in the sketch below

View attachment 258626
In the first case the red bar keeps in contact with the yellow L-shaped conductor, while it slides along it so that the red angle in the triangle becomes small. Let's say that the resistance is all in the red bar.

In the second case there loop is a parallelogram made by four hinged bars with some resistance.

I'd say that FNL is more convenient in these cases, given that there is only one loop and the motion of the parts is not simple.

Would you agree?

As I said, no, I would not go with FNL in any moving situation.

Hey, I love this. This is the exact right analogy, and this exactly what I was thinking of when I retracted my earlier boner. The answer is simple. Faraday’s law must be true for both loops. Calculating either loop shows an emf along the moving wire. And, of course, the emfs calculated with either loop are equal because it’s the same emf.

So, to recap, you are welcome to calculate with either loop. Faraday’s law has to be true in both loops simultaneously and they are both experiencing the same emf from the same moving part so you get the same answer either way. And, yes, the current takes both paths.

Aren't you concerned that the area of one loop is increasing while the other area is decreasing, and emf = -area x B?

 

[FranzDiCoccio]

I read your example from the file you attached, and I think I understood it.
I do not understand why you copied it in a post. Anyway, latex went wrong at some point, and the images are missing.

As I said, no, I would not go with FNL in any moving situation.

Uhm... but motional emf does not seem very convenient in the two cases I proposed.
There is more than one piece moving, and they are translating and rotating.
FNL seems quicker and I do not see any real danger, in those cases, since there is only one loop, and its area is changing.

[edit] Would you really examine the motional emf for every moving part? How would you go about the case with many moving parts?

Aren't you concerned that the area of one loop is increasing while the other area is decreasing, and emf = -area x B?

Although in the case of the rectangular loop I do agree with you in being wary of the FNL law, I do not see any inconsistency with that.

I expect that the induced currents would give rise to induced field contrasting the change in the magnetic field flux, in accordance with Lenz's law. So the current in the loop with increasing area (left) will flow clockwise , while the current in the loop with decreasing area (right) will flow counterclockwise.
This is precisely what you find if you use motional emf.

Personally, I find that the minus in FNL law could be a bit misleading, unless you introduce an explicit reference.

 

[rude man]

So I'd say that the overall effect would be an emf in the bottom side, a "viscous" force opposite to the velocity and a slight "pull" of the loop in a direction that is perpendicular to both the initial velocity and the field.
Is this correct? [/quote/]
Force is perpendicular to current direction and B, not velocity and B. So there is no "pull" in the direction you say.

Without vector notation you can write, for each side, emf = B$\ell v ~sin(\theta) ~cos(\phi)$ where B, v and $\ell$ are the magnitudes of each of the vectors $\mathbf {B, \ell, v}$. $\theta$ is the angle between $\mathbf v$ and $\mathbf \ell$ and $\phi$ is the angle orthgonal to the plane defined by $\mathbf v$ and $\mathbf \ell .$. From this you can determine emf for all 4 sides. This quantity is non-zero for the bottom horizontal side only. Once you learn vectors you will see what a relief it is not to have to deal with expressions like the above. Further, you can easily compute in different coordinate systems, not just cartesian. This is a huge advantage when particular problems lend themselves to a particular coordinate system.

That's for emf. For non-vector force magnitude it's $F = B \ell i ~cos(\beta)$ where $\beta$ is the angle between $\mathbf \ell$ and $\mathbf B$. You use the right-hand rule to determine the direction of force; with the fingers curling from $\ell$ to $\mathbf B$, the thumb points in the direction of $\mathbf F$. The only side seeing force is the horizontal one immersed in the B fild because it's the only one where B and v are perpendicular. As you would expect, the force on that side opposes the direction of motion.

 

[FranzDiCoccio]

Force is perpendicular to current direction and B, not velocity and B. So there is no "pull" in the direction you say.

Right... perhaps I see my mistake. I was focusing on the free charges inside the "vertical" sides. I assumed that, since they have a velocity $\vec v \perp \vec B$ they would be pulled in the direction of $\vec v \times \vec B$, and they would sort of "drag" the wire in the same direction.

I think I overlooked the fact that, as these charges move, they "effectively" cause the same amount of opposite charge of the to move in the opposite direction, assuming that the wire was not charged to begin with.

I think I was sort of misled by a textbook "derivation" or "justification" which explains the Lorentz force on a current carrying wire as the result of the Lorentz force on the charges moving through it.

So the movement of the "vertical" sides only causes a tiny motional emf that is orthogonal to the field (and wire) and to the velocity.
This displacement of charges in the (dipped portion) of the vertical sides has no net effect on the entire loop, because for each possible force there is an opposite force canceling it.

 

[rude man]

Hi... me again
I was thinking that the general advice "use motional emf if there are moving parts" should be made more precise, something like "use motional emf if there are moving parts changing the area of more than one loop".
Indeed I recalled situations where the moving parts are hinged, or move in a non trivial way, like in the sketch below

View attachment 258626
In the first case the red bar keeps in contact with the yellow L-shaped conductor, while it slides along it so that the red angle in the triangle becomes small. Let's say that the resistance is all in the red bar.

In the second case there loop is a parallelogram made by four hinged bars with some resistance.

I'd say that FNL is more convenient in these cases, given that there is only one loop and the motion of the parts is not simple.

Would you agree?

Yes I agree.
There is an alternative interpretation to the $B\ell$v law which is "cutting flux in unit time". It too is not based on the FNL law (i.e. Maxwell) but derives from the Lorentz force. In fact I said for the rotating bar you can do that problem in this way: as it rotates, the bar is cutting flux at a rate equal to the area swept by the bar in unit time, times B. If you can make this association between cutting flux and the FNL law then the FNL law is OK and often simpler, as your sliding bar shows. In the sliding bar the area change within the loop in unit time does equal the flux cut in unit time times B. Good question!

BTW my rectangle example can also be interpreted as "cutting flux" in addition to the B$\ell$v law since they are fully equivalent.

*************************************************************************************************************************************

There are so many posts in this thread so far that I'm a bit confused as to what was directed to whom.

I must have inadvertently posted my text for my rectangle example in addition to the file, didn't mean to. Just go with the file.

I also made a comment about your rotating bar which was directed at Cutter, not you ("you" being Franz). I definitely don't like FNL there. Unless you apply an alternative for the Blv law which is "cutting flux in unit time" which is just as valid as Blv and which I mentioned. In this case "cutting flux in unit time" is the same as pretending the FNL law applies but it's not generally legit, as my rectangle example shows.

 

[rude man]

Right... perhaps I see my mistake. I was focusing on the free charges inside the "vertical" sides. I assumed that, since they have a velocity $\vec v \perp \vec B$ they would be pulled in the direction of $\vec v \times \vec B$, and they would sort of "drag" the wire in the same direction.

Why? If $\mathbf v$ is $\perp \mathbf B$ then by Lorentz there is no force in the $\mathbf v$ direction. There is an inward force on all four sides as I previously described, or should have.

I think I was sort of misled by a textbook "derivation" or "justification" which explains the Lorentz force on a current carrying wire as the result of the Lorentz force on the charges moving through it.

This is correct!
Lorentz F = qvB but i = q/t or q = it so qvB = i(vt)B = i$\ell$B. If you'll forgive the lack of vector notation. Assume $v \perp B \perp \ell$.

So the movement of the "vertical" sides only causes a tiny motional emf that is orthogonal to the field (and wire) and to the velocity.
This displacement of charges in the (dipped portion) of the vertical sides has no net effect on the entire loop, because for each possible force there is an opposite force canceling it.

No "tiny motional emf" along the vertical sides. Can't we just leave it that the vertical sides are not cutting any flux, so no emf? But you're right, any forces not obeying F = Bil are nulled, and there are several.

The interaction of charges moving both in the v direction and along the wire direction is pretty complicated, involving various force components along v and i all as you say canceling each other (thank goodness), drift velocity of the charges due to collisions, ...; I refer you again to Resnick & Halliday,sect. 35-7 (I have the 1966 rev.)

 

[Cutter Ketch]

Aren't you concerned that the area of one loop is increasing while the other area is decreasing, and emf = -area x B?

Increasing area vs decreasing area means clockwise emf vs anti clockwise emf which, as it happens, is the same direction through the moving wire! All happy as it must be because Faraday’s law has to be true in each loop.