Ok. I’m feeling pretty stupid. This should be simple. Is this ambiguous, or is it just me?
A long,straight wire carries a current of 740 mA. A thin metal rod 45 cm long is oriented perpendicular to the wire and moves with a speed of 1.8 m/s in a direction parallel to the wire. What are the size and direction of the emf induced in the rod if the nearest point of the rod is 3.5 cm away from the wire, and if the rod moves in a direction parallel to the current?
I keep reading and reading. I drew two geometries both of which seem to match what the problem says. I show them below. (I chose to make the “thin” rod as circular. It’s probably flat, but that’s not really my question.) Can any of you point to anything in the problem statement that makes it clear they mean one, the other, or something entirely different?
| Interpretation 1. | Interpretation 2. |
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| Metal rod perpendicular to wire? Yes. Moving parallel to wire? Yes. Nearest point 3.5 cm away? Yes. Middle of rod is is above the wire. Moving parallel to current? Yes. |
Metal rod perpendicular to wire? Yes. Moving parallel to wire? Yes. Nearest point 3.5 cm away.Yes. Left edge of rod is to right of wire. Moving parallel to current? Yes. |
Both geometries seem like decent AP Physics C level problems. They just aren’t the same problem.
Update for Mark
Lucia, is there any particular reason the rod is centered in Interpretation 1? Couldn’t the rod be in the same orientation but shifted either left or right and still fit the criteria? Would this be Interpretation 3 or Interpretation 1.5?
Lucia,
Both appear valid to me. I would have come up with interpretation A, I’m not sure B would have occurred to me until later. If at all. I’m not sure why I would have come up with A, but I’m pretty confident that’s what I would have thought.
Imagine if you had to work the solution, enter it into a box and your feed back was just red “x” or green “check”.
So far about 1 in 5 of this kids problem seem to have this level of ambiguity. The Amazon reviews of the book by people who say they are students are very negative. One of them complained about ambiguous questions. Well… uhmmm… yeah.
BTW: Consider this. By ’emf generated’ do they mean between the two ends? The center and one end? (One could ask between front and back faces, but the thickness is not given.)
Hmm. The more you discuss it the more unsure I am. Maybe you’re right, the thing for me to do is to actually try to solve the problem.
🙁 which means I have to go look up how to calculate this. Because I’ve forgotten how to do this.
Hmm, so I compute the magnetic field around the wire, and then I figure out induced emf from moving a conductor through the magnetic field, is that the idea? [Edit: sort of rhetorical. I’m pretty sure that’s the idea, but correct me if somehow that’s doing it the hard way or something]
Yes. Yep. Assuming they mean the student to provide the voltage difference from end to end, integration may or may not be required. Understanding symmetry may or may not be required.
Mark,
As far as I can see, solving the problem(s) doesn’t resolve the ambiguity in the problem statement.
I do think “emf” means voltage difference from end to end. But for case A, from center to end is interesting.
Mark: For the case on the left, think “symmetry”. Then you can bag the trig and scriblings.
Well, I get that symmetry helps in that the magnetic flux [edit: magnetic flux density] at the two endpoints are going to be the same, since the distance from each of the two endpoints to the wire is the same.
Really the only thing that’s tripping me up is figuring out the direction of everything. I think I got the magnetic field, that’s easy enough, but. I try to figure out how to describe the conductor moving nearby and keep losing myself in the middle of it.
[Edit: not how to describe the conductor moving. What the conductor moving does – what direction the EMF is I think.]
Meh I think I get it but I’m going to sanity check for awhile. 🙂
For geometry A (rod above), if they want voltage difference from end to end, you are done as soon as you figure out they voltage is the same at both ends. You enter zero: Done.
If they wanted the voltage difference between the center and an end, there is some trig to do. You only care about the force from center to end. So you only care about the “up/down” component of the magnetic field at the rod. So: you use |B| = \mu i /2 \pi r. But use trig to find “r” the distance from the wire to the point on the rod and use trig to find the “up/down” component of the magnetic field. The… integration.
I even though I can interpret the question as finding the biggest \delta V, I really, really find that strained. So I don’t think anyone needs to do it.
On the other hand, with arrangement B, there is no trig just easy integration of dr/r from end to end. The answer is not zero.
So: the answers are different. Either geometry is a perfectly valid problem for a student. You want them to both recognize when symmetry means the don’t need to do any more work and to be able to do the integration. So I don’t think doing the problem helps me guess which geometry is the one they mean.
That said: even figuring out the answer revealed which geometry they meant to describe, it seems to me that problem should be written so you can figure it out from the problem statement.
Aww crud. I was doing that wrong. :/
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But actually, knowing that I get zero end to end in case A would make me question my interpretation. I’d wonder at that point about the professor; is he/she the sort to give me a problem that cancels out like that.
I’m guessing they want interpretation 2 so that you can unambiguously determine a nearest point (the answer is proportional to ln(L)?)
The geometry is ambiguous. They should state what point of the rod is 3.5 cm from the wire. Replace nearest with center and you get 1. Replace nearest with end and you get 2. But any point on the rod could be 3.5 cm from the wire, so there are an infinite number of solutions. There’s really no excuse for not realizing that the problem as described has multiple answers.
It’s horribly ambiguous. My initial guess would have been that (A) was intended, but there’s nothing to actually indicate that. But there are, of course, lots of other interpretations: e.g. the rod could be above the wire as in (A) but not arranged symmetrically above it. And that makes me think that (B) was actually what was intended.
RB,
I don’t see how that resolves any ambiguity. If it’s (A) all points on a line dividing the left and right side of the bar center of unambiguously the nearest points. If its version (b) all points on the plane of the left edge are unambiguously nearest points. In both cases it’s more than one point. But the answer is the same for all of those points, so I don’t see how that helps.
The ambiguity lies in whether it’s A or B.
Jonathan,
Yes. There are an ∞ number of interpretations.
My guess is they want B too.
Sadly, this is a “webassign”. I don’t know if you are familiar with that. But it’s pretty frustrating if you can’t even figure out what the question is.
Jonathan,
(A) was what first popped in my mind. But I favor (B) as “what they mean. Yes… it could be an infinite other things in between. But I figure at least they mean one of two extremes.
Lucia,
Both are thin or I understood both to be essentially one-dimensional lines. So, the (unambiguously) left end (point) of the 45cm rod is said to be 3.5cm from the thin long wire in interpretation B?
Steve
There are an infinite number of positions that could fit the criteria. I just showed two extremes. I figure some readers would notice these are not the only two. (You did and Jonathan did.)
On tests I do tend to lean toward at least guessing the limiting cases. So I figure they mean one or the others of these.
Note I also interpreted
as parallel and pointing in the same direction. “i” and “v” pointing in opposite directions are still “parallel”.
RB
I see what you mean.
If they said the rod was “in the plane of the wire, then the left (or right) edge of the rod is closest. So that’s two candidates. (The choice makes no difference to the answer though.)
If they said “above” you would need to assume something. I picked “center”. So in that sense you are correct. There are more choices because you could slide the rod right or left and still have the closest point.
Still… that’s quite a bit of reasoning to decide one might be more likely!
Lucia,
Yes, it is quite a bit of reasoning. A picture’s worth a thousand words etc.
I agree with Steve. Imagine the surface of all points 3.5 cm from the rod. It’s a cylinder 3.5 radius, with ends that are hemispheres also 3.5 cm radius. It seems to me that, with axes perpendicular, the wire could touch any point on the cross-section of that surface that is in a plane perpendicular to both. 1 and 2 are special cases.
Lucia,
What the heck keeps them from drawing a diagram?
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It is a stupid question…. and doubly stupid because it does not specify the dimensions of the wires, nor specify if you are allowed to assume infinitely narrow wires. As written, there are an infinite number of possible interpretations, and a corresponding infinite number of calculations ranging in difficulty from ‘zero by inspection’, to complicated multi-dimensional integration. The more the student understands about the subject, the more frustrated he/she will be by the question’s ambiguity. Of course, if it is a multiple choice question, then the test author’s ‘assumed geometry’ may be pretty obvious from the potential answers. Good test takers are usually pragmatic.
Nick,
We all agree with Steve. I did picked special cases to show. I wanted to see if anyone could point to something in the text that would pin down the one and only case it must be. But clearly, the text is consistent with an ∞ number of cases.
I think your verbal description catches all of the possible cases. I’m just reasonably confident they must mean either (a) or (b).
I think RB actually sort of untangled the reason why it’s probably (b). The way I would explain his argument (which he didn’t say quite this way) is “If they did mean B, they might bungle a description that way. Because in B, it’s not immediately obvious that there are all those other cases. But if they meant A it’s pretty obvious that all you need to do is shift A to the right or left and the closest point is still 3.5 cm. So if they did mean A, it would have occurred to them to elaborate.”
I’m pretty sure if they meant one of those other intermediate cases they would have also said more.
FWIW: Both A and B make good AP Physics C physics problems. The others not so much. The reason is A & B are just enough to ensure the student knows and understands the physics, knows how to set up the math and can do it. The test is timed, the goal is not to divide the kids who are terrific at ‘fidgeting with digits’ from those who “know physics”. So I’m pretty sure it’s A or B.
So “A” or “B” are about what you expect on the test these kids hope to pass. All the others ones…. they are cases a physics prof who lacks insight might concoct.
SteveF,
The AP Phys C test itself, which they are prepping for has a multiple choice section and a free response. They publish the free response questions every year and they are good problems. Always clear.
The multiple choice they don’t publish. But my guess is those are clear.
High School’s give kids “AP Phys C” classes with different teachers having different approaches to what to do. These homeworks are rarely multiple choice, though some are. This one was not multiple choice. So you can’t reverse engineer from that.
On these high school classes: One kid I tutor for AP Phys C has no book at all which is a mistake on many levels. But his teacher picks great problems that are clear and dovetail with the content of the AP Phys C. Those problems seem to be from the books I will call “Halliday and Resnick” (which is now Walker) or “Sears and Zemanski” (which is now Young.)
This kid has a book. I don’t know if the problems come from that book. Most their assigned problems are on the curriculum for the AP C, but some are not. Not infrequently the problems are ambiguous. By “not infrequently” I mean there seems to be at least one problem with this level of ambiguity in each problem set of 8 or so. (Note however, I have only had this student a short while. So this may just be for the current sections.)
I would have thought that perpendicular means wire and rod are in same plane which can only happen as in interpretation 2. But then they screw it up with “nearest” point on rod rather than “near end” of rod.
choosing interpretation 1 requires you to add information to the question ie assumption that rod is centered on wire at offset 3.5cm.
so it seems to me that if there is a discrete interpretation, namely number 2, anything else requires additional assumptions.
SteveF:
FWIW: if the problem wants Δ V from end to end (or even from center to end) the thickness of the rod doesn’t matter.
So I take that as a very strong clue the question writer meant Δ V along the length. But yes, it would be better to say “from end to end” rather than have someone infer they can’t mean “from top to bottom” or “from front to back” because they didn’t give dimensions which means there is no answer to that question no matter how you interpret the geometry.
For now, I’m mostly glad that I got confirmation that the wording of the question is very ambiguous.
Lucia,
” if the problem wants Δ V from end to end (or even from center to end) the thickness of the rod doesn’t matter.”
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I don’t think that is in general correct. When the rod is positioned as in your figure A, a very thick rod would not have a defined single voltage “from center to end”. If you measured at different points on the flat, thick ends you would (I think) measure a very slight range of voltages, depending on the conductivity of the metal and the thickness of the rod. In addition, as the thickness of the rod increases (in your figure A) the ‘average’ distance of the rod from the current carrying wire would increase with rod thickness (3.5 cm + the radius of the rod), and so change even the average voltage you would measure on the ends. The infinitely thin wire approximation simplifies the calculation a lot.
Lucia –
I’m with jferguson in thinking that the use of “perpendicular” implies co-planarity, hence interpretation #2. In configuration #1, the relationship between the two lines is better described as “skew” because they do not lie in the same plane.
But I’m a big fan of a diagram here. Failing that, the problem could have been phrased without ambiguity — or at least less! — by saying that the near end of the rod is x distant from the wire.
SteveF
Yes. On the face of the end the voltage would vary. By “end to end” I mean carefully picking points so they are along a line parallel to the lengthwise axis.
HaroldW,
As a matter of “literal interpretation” “perpendicular” does not imply co-planarity. But I think the question writer had coplanarity in mind. Saying so would help. Adding the word “coplanar” into the question statement wouldn’t require very many words.
I suspect the problem is the authors didn’t have independent people work the problems. The book is on edition 3 so you’d think feedback from students might make them realize there is a problem. It’s hard to believe I’m the first to read it and not know which geometry they meant. But who knows?
In instance 1, is the result affected by whether or not the rod is centered on the wire? for example if the 3.5 dimension assumed that the rod is offset by it’s length, it seems as if you would have less mass near the wire. I would have thought when you started accumulating other variables with this problem you might have thought that if there is a manifestation with the fewest possible variants, that might be what was intended.
Had they written that the rod was 3.5cm distant from a wire and perpendicular to the plane occupied by the wire you would get instance 2. But you still wouldn’t know if the rod was centered lengthwise on the perpendicular plane occupied by the wire.
This is tangential to this discussion, but I started running into geometry problems with my colleagues in Chicago in the early ’80s and found that U of I had stopped requiring Descriptive Geometry at Engine School. I think Washington U dropped it about the same time. Maybe this is incorrect.
I screwed up above
if you do this you’d specify instance ONE. Sorry.
jferguson
Yes. If the rod is centered over the wire, the answer is zero. If it’s shifted over it’s not zero. The rod centered over the wire is a very good physics problem. So is the other case I drew an image for.
IIT dropped engineering graphics at some point. People started using CAD/CAM.
Hi Lucia,
the answer to Instance I suspect is more revealing of whether the test taker understood the thing. Seeing that it makes it much more difficult to decide which of the two possibilities was intended.
There was a brief outbreak of testing for draft priority in the ’60s. Tests must heave been prepared on the East coast. One question involved the commute in from Wheaton on the Burlington. The only way you could have arrived at the answer was to know the sequence of stations. It wasn’t given. Fortunately it was the only question missing necessary information so I finished everything else and then spent the last 15 minutes trying to arrive at an answer without knowing where La Grange was.
Couldn’t be done.
(Off topic, but interesting): The cold fusion fiasco, much discussed here a few months back, has apparently collapsed (surprise!) without a bit of fusion. The rich (and very unwise) US investor who founded a company called Industrial Heat, to develop the scam fusion claims of Mr Rossi, has now refused to make a scheduled payment of US$89 million on the scam technology (apparently finally coming to his senses!). This led to a lawsuit by Rossi and his company, Leonardo Corporation. Industrial Heat says:
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“Industrial Heat has worked for over three years to substantiate the results claimed by Mr. Rossi from the E-Cat technology – all without success. Leonardo Corporation and Mr. Rossi also have repeatedly breached their agreements.”
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Scams usually end with the mark impoverished, the scammers enriched, and lawyers taking a percentage of the action (sorry JD Ohio). What needs to happen to end this nonsense is a US based district attorney needs to arrest and prosecute the scammers for fraud. Sadly, that will probably not happen.
Heh. I looked at that in March. I remember a commenter saying “Feb or March”, so I was waiting for April 1. That said, watching for that was not a priority.
I am not surprised to that “all without success” features in this discussion.
To be fair to lawyers, a world without lawyers looks like “Game of Thrones”. Seriously. Yes, it’s true that adjudications of disputes can’t actually create anything. But having no way to adjudicate them other than killing people or promoting revolution isn’t helpful. I suspect JD Ohio and JosephW know this.
I suspect the scam was well thought out to avoid the possibility of any danger arising from US based district attorneys. Likely they thought of how to make themselves untouchable by EU officials. So… yeah…… They probably milked the investors for what they could get. Caveat emptor as they say.
Lucia,
” I suspect JD Ohio and JosephW know this.”
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As does my oldest son, an intellectual property lawyer. I really have nothing against lawyers, though I do wish there were less need for adjudication.
I haven’t read or seen Game of Thrones, though I probably would enjoy it from what I’ve heard. I do enjoy the Icelandic sagas, which often center around lawsuits…in a country that had laws, but no government. (Hrafnkel the Priest of Frey features Samr Bjarnisson…a sort of proto-lawyer. I like him because his first advice to his “client” is “don’t get involved in a stupid lawsuit, when there’s a good settlement on the table” — but he fights it to win anyway when the guy insists.) Their procedures are on the robust side, and you do get the occasional pretrial brawl…
I’m all for SteveF’s last statement and think the problem is more often to be found in the laws than in the lawyers, though there are clients who could use some reining in, too.
If I read ‘long’ as ‘infinite’, and (irrespective of size and orientation of the rod) the linear speed of the ‘rod’ does not change (relative to the wire current), surely the answer is always ZERO.
Reasoning: The ‘isobars’ of the ‘magnetic field’ are cylindrical, i.e. axially and radially symmetric about the current path. Parallel motion of the rod will not change the ‘field-line cutting-pattern’, ever.
It requires a change in the ‘cutting pattern’ to induce an emf.
Joseph W,
One political problem in “The Game of Thrones” is there seems to be no rule of law and no way to resolve any disputes other than force. There are no lawyers because there are no courts.
As much as people like to complain about lawyers, if we are to have systematic ways to deal with disputes there are going to be people who become skilled at presenting those in the framework of the method for dealing with that. In our system, that’s largely lawyers to present cases and judges to adjudicate.
It beats having your champion hacking down the other sides champion or sending your assassin, or going to war.
But yeah, in our system sometimes lawyers pretty much sucking everything out of both sides. (And some class action suits…. what can I say. But part of that is the laws for class action are perverse. The should require that the legal teams can’tmake more than some fraction of what real honest to goodness members of the class get in real honest to goodness $$– not coupons.)
OzWizard,
No. It’s sufficient to cut them. You don’t need a change in the cutting.
lucia:
There are times I wonder.
Interesting because I’ve read that Game of Thrones is inspired by the Wars of the Roses…when there certainly were courts and lawyers in existence. (In fact, in his younger days, I’m told that Richard III was an obsessive litigator over land claims.) But certain parties who could back their claims with force preferred an alternate route (common in the Icelandic sagas also)…or took it up when they didn’t have other routes.
Thus, claiming a superior title to the throne itself, by the most lawful argument you please, amounted to treason; so trying to sue for the crown would be most unwise. Bollingbroke in Shakespeare’s Richard II (in the incident that leads to the wars) has had his lands seized by the king after his temporary banishment…he shows up with an army and justifies himself that way:
I am denied to sue my livery here,
And yet my letters-patents give me leave:
My father’s goods are all distrain’d and sold,
And these and all are all amiss employ’d.
What would you have me do? I am a subject,
And I challenge law: attorneys are denied me;
And therefore, personally I lay my claim
To my inheritance of free descent.
But I think you’ve got the sense of it — we’re more an effect than a cause.
It’s hard to say what you could do to make class actions for money much better for the class members (injunctions are a different story). I haven’t really studied that debate since law school, but Brian Fitzpatrick’s Empiricial Study of Class Actions Settlements and Their Fee Awards — easy to find online — studies awards in 2006-07 and says lawyers fees overall amounted to 15% of the total. (Judges have the power to scrutinize the fees already and sometimes do.)
If even the top awards for a large class are on the order of $200-300, squeezing that 15% just isn’t going to generate that much…especially if the goal is to punish and deter the defendants who did the wrong, rather than to bankrupt them (in which case lawyers fees for the bankruptcy would chew out even more…). But if someone knows a better way I’m all for it.
Lucia (#145592): As a matter of “literal interpretation†“perpendicular†does not imply co-planarity.
Actually, I think the *literal* interpretation does so imply: e.g. this definition: “A line is perpendicular to another if it meets or crosses it at right angles (90°).” And there can be no such intersection without co-planarity. In interpretation #1, the two lines do not cross; they are skew to one another.
It is an *extended* definition of perpendicular which includes configuration #1 — the planes normal to the two lines are perpendicular to each other. I think all of us would describe the rod & wire as perpendicular. Only a geometry student would take issue.
JosephW,
Does that include cases that settle out of court? Generally when we hear the ridiculous ones its that lawyers settled out of court. The attorney’s got fees. Members of the class get coupons. Stuff like this
https://en.wikipedia.org/wiki/Coupon_settlement
Lucia, class actions are very weird creatures. From what I can see consumers get virtually nothing, lawyers get a lot and the corporations dislike them. The justification for class actions is that the injury is so small but in large numbers that a class action has to be instituted to enforce the right that is being violated. One time I tried to use airline coupons I received in a class action settlement and was told that I had to show up at the airport in person to use them. What a joke!
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The only indication that they aren’t a complete scam in my mind is that it does appear that defendants really don’t like to defend them. See for instance, the recent Alaburda case http://www.businessinsider.com/anna-alaburda-is-in-court-to-sue-thomas-jefferson-school-of-law-2016-3
JD
Yes, in fact, it’s all about settlements. (Most civil litigation settles, just as most criminal cases end with plea agreements.)
Joseph W,
I have not thought about it for a while, but it always seemed to me there were two perverse things about class actions lawsuits:
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1) The financial costs for each member of the class is often pretty small, as little as a few dollars or less, and few (or none) would ever be motivated to act to recover their own damages. These damages seem to me to fall in the ‘de minimus’ range… like a bottle of wine which actually contains 10 ml less wine that the stated volume.
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2) The motivation for the action seems (too often) a lawyer or lawyers who see an opportunity to ‘represent’ thousands (or millions) class members, and pocket huge fees.
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It seems to me once the motivation of huge fees is reduced, the number of class actions would fall. For example, legislation could set fees for class action suits at “customary and reasonable” hourly rates ($250 – $450 per hour?) rather than a percentage of the total class payment. I doubt this would ever actually happen, of course, but I think it would definitely reduce the number of class action suits.
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There is good reason to keep companies honest (eg. not deliberately meter less wine into the bottle than the stated volume), but that might be better handled via a district attorney’s office rather than a class action suit. And the DA can prosecute criminally if there is willful intent to defraud or endanger.
SteveF
The law needs to stipulate in cash payments to class members and it needs to stipulate the attorneys can only get their money after class members do.
I don’t mean after all class members do. Say for example 50% of class members have gotten their whatever is due them. At that point the lawyers can get 50% of what is due them, and so on. It shouldn’t matter why class members didn’t get what is due them. If, for example the settlement dictated they had to fill out long lengthy forms and none filled them out. They don’t get their money. Then the lawyers don’t either.
And the law limiting legal fees to a percentage of the amount of actual cash money distributed to members of the class needs to apply to out of court settlements.
I’m sure many class action suits are fair. But we all know of some that are ridiculous. To the extent that class action suits are already fair, the limits would have no effect. But it would get rid of some of these ridiculous ones.
Hmm… I just thought of something that might matter. Cases like “UBER” where the issue is employment status where part of the desired remedy is not to get the class repaid for the past but to get the status changed going forward.
Still, it seems to me that if Uber’s business model is incorrect, lots of real honest to goodness individuals are owned money. If not, they aren’t. It wouldn’t seem quite right if an individual who is owned money for past work, but who subsequently found a job elsewhere doesn’t get what (s)he is owned with the only “remedy” being that in the future the status is corrected.
It is connected with the Lorenz force and homopolar generator.
The force on a moving charge in a magnetic field.
Lucia: “I’m sure many class action suits are fair.”
I’m not. Virtually all of the class actions that I come into contact with are good for the attorneys but meaningless to the so-called clients. I don’t claim to be any kind of expert on these kinds of matters though.
JD
Regarding class actions, if the puzzle is how to remedy injustice spread thinly over a large population the answer is the same as how to provide protection the population in any fashion against private or public entities; the use of media exposure.
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A company’s brand or a politician’s reputation is certainly as valuable to them than their shareholder’s or taxpayer’s money. Most of the efficacy of class actions or the FDA (of EPA) could be performed by organizing public databases. It’s what the ACA online marketplace should be about if we had politicians that understood the power of anonymous feedback as well as the founders of Ebay and Uber did. An Angies’ list or Yelp done with official coordination of the state and proper checks and authentications would make the basis for half the law suits, as well as half the government agencies, go away, while swiftly and fairly remedying all market inefficiencies.
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For the poor souls that were victimized many times the sweet justice of having their injustice memorialized in the public record is plenty enough compensation, and much more than most get.
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As I think JD Ohio and Joseph W. would agree, a large portion of settlements are based on compensating the legal teams for keeping matters out of the public domain.
HaroldW,
That must be it. It must be that perpendicular does imply coplanar. I didn’t think it did so.
I don’t think it is that bad people in class actino lawsuits get little return. They didn’t get harmed much, and they don’t do much, so it makes sense to me they wouldn’t get much. I recently filled out a form to participate in a class action settlement. Assuming the settlement goes through, the most I could expect to get is $50. Odds are I’d get less.
It’s a pretty trivial amount, and I’m sure the lawyers are making a decent bit off the settlement, but I have no problem with that. They’re the ones doing pretty much all the work. If there were any victims who did a lot more work to drive the lawsuit/settlement forward, I’d hope they’d get more out of it than I would.
Brandon,
Getting $50 in actual money is fine. And if the class is large, and the lawyers can get fees equal to 10-20% of N*$50 with N being the number of members in the class who got $50, the lawyers could make a lot of money. But at least members of the class got something.
The problem is in some cases, negotiations have settled for coupons that can only be redeemed if you buy from the company who ‘injured’ you. So to get your “cents off” you have to buy something from the company. Generally speaking if a company rips me off or injures me in some way, I don’t consider getting a coupon for a 10% discount of future products they sell to be a benefit to me. I usually don’t ever want to buy their products again. Heck, I’m not even sure I would consider getting a “free” something from that company much of a benefit. (Might depend on what it is. But I can use money to buy things I pick. )
In these cases, the lawyers may make a lot. And basically, the people in the class got nothing of value to them.
To the engineers and architects (I hope this is OK with Lucia — if not remove the post) reading the board.
I am selling my house, which I bought 15 years ago. When I bought it, the basement had diagonal hairline cracks, which my inspector told me not to worry about. I haven’t noticed any changes (obviously, there could have been very small changes that I am not aware of) and had no problems with the basement leaking. (the house is on top of a small knoll) The cracks are about 1/16″ wide and are diagonal. The inspector said that 2 basement foundation walls were pushed in by about 1/2″ (The house was built in 1962)
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If there are real problems, I am happy to fix them. If the inspector, is being super cautious to cya, I don’t feel I should have to do anything. My question is how should I check out this inspector’s opinion and what should I be looking for. If anyone, can help, I would appreciate it.
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Again, Lucia, if you think this is improper, please take it down. However, I suspect there are some very knowledgeable people here and hope I can benefit from their knowledge.
JD
JD,
It’s fine, I just don’t know anything about construction! Someone else might though.
Lucia, Faraday’s Law: “The emf induced is equal to the rate of change of magnetic flux linkage or the rate of flux cutting.”
Look at the magnetic flux pattern as concentric cylinders (rather than ‘lines’) and you will see my point more clearly. They are magnetic potential ‘surfaces’, rather than ‘lines’, and there is no ‘change’ in the cutting pattern. i.e. rate of change of magnetic flux linkage = 0.
JD, I was an architect until i came to my senses and retired. It’s hard to know what to suggest without seeing them, but if they aren’t leaking you may be ok. In any case, I wouldn’t touch them but that is because I think fresh repairs tend to unnerve buyers.
I sold my parents’ 1964 house in Chicago in 2013. It had diagonal cracks running from the window-wells down to the slab in the basement. They had appeared the first year parents were in the house and in my opinion were due to incorrect placement of the steel reinforcement at the corners of the windows and concrete with too much water – induces shrinkage. These cracks had leaked, but were sealed from the outside by an outfit that did that sort of work and the leaking was cured.
The buyers’ inspector told them not to worry about it.
You might find out who does repairs on this sort of thing out your way, have one of their people look at them and give you some idea what could be done, should be done, and what it might cost.
If you have to deal with this problem to make a sale, the best thing is to agree on what the work is worth, give the money to the buyer and let him/her have it done – keeps you out of the middle if there is a workmanship problem.
Replacing a water heater or furnace or ac unit discovered to be problematic during a sales inspection is probably ok – you can probably agree on who should do it and you can have it done yourself but on structural or roofing repairs it’s much better to let the buyer do it.
If they can’t – don’t know how to get something like that done, maybe their real estate ‘professional’ can help them out.
One other thing. I NEVER let anyone I was selling a place to find out I had any connection with construction. Possibly wrongly, I worried that if a problem developed the buyer might suppose I should have know about it.
It might also be a good idea to look over the documentation required for a sale if you haven’t been involved in one in fifteen years. It makes quite a pile. I can’t remember the specifics but i felt as a seller I was held responsible for a number of things related to the condition of the house that i shouldn’t have been. A lot of this is due to federal regulation. We sold our own place in 2003 and the pile of paper was about a third of what it became by 2013.
I can expand on this if it would help. Of course it could be more unhelpful.
Oz, look at the bottom here for how you can get the answer for a fixed magnetic field for example either using a Lorentz force on a charge or using the rate of change of flux enclosed.
Thanks a lot Jferguson for your very useful comment & to Lucia for letting me go off-topic.
…
I kept my original post simple. To expand a bit.
….
The inspector stated: “A hairline cracks are present in the rear and left side foundation walls (basement) at the rear left corner. At the lower sections of this crack, staining and efflorescing was noted (in the corner). This is indicative of prior moisture penetration through this crack.”
…
After the inspection, I took a picture of the corner, and there is what I suppose is white efflorescing near the diagonal crack. The efflorescing (if that is what it is) is mostly not in contact with crack, but is just small blotches near the corner. One strip splotch starts above a diagonal crack and vertically intersects with it going about 3 inches below it, but it does not follow the path of the diagonal crack. (In fact none of the splotches follow the path of the diagonal crack.)
…
If there are splotches there, I assume there must have been water at some time. Also, there is a little dark spot in the bottom corner. All I can say is that in the 15 years I lived in the house there were 0 problems with water leaking. My thoughts are that possibly before I moved in there may have been minor leakage problems. Or possibly there was a small amount of water/moisture in the concrete in this area, but it never turned into a leak.
…
One thought I have is that the corner is right under a major downspout and that when I moved in to the home there was an extender from the downspout. It is possible that originally the downspout didn’t take the water far enough from the house and that after the 3 foot extender was put in (which was far enough to take the water to a downslope away from the house), the problem stopped. My discussion here raises the issue in my mind of whether there is any way to test the splotches for their age or to test any dark areas for their age or whether there was recent moisture.
…
Additionally, is the 1/2 inch inward slope, a real problem or not? If so, can it be fixed by anchor bolts to the mud sill as was suggested by the inspector. I would add that I agree with you and don’t want to do fresh repairs if I can avoid it.
Thanks very much again for your help. JD
lucia:
That would be pretty ridiculous. I wouldn’t sign onto a settlement like that. Heck, I’d probably opt out of the settlement on principle. It’s easy enough to do, and even if I would have no intention of suing over the same matter, it’d at least make my views on the matter clear.
In case my last comment wasn’t clear, when there is a class action lawsuit on a matter, you cannot file a lawsuit separately without opting out of the other suit. If you simply fail to sign onto a suit/settlement you were contacted about, it is assumed you are forfeiting your right to sue over the matter.
So if you want to file your own lawsuit separate from a class action lawsuit, when you’re contacted about the class action suit, you need to go on record that you are opting out. Usually you can do it just by sending a letter to the lawyers who contacted you.
Brandon
Yes. But given the junk mail overload, I think the “you’re roped in unless you opt out” is a not-quite right feature of class actions too. This is particularly so since you have very little say about settlements, negotiations, strategy and so on. Also: people have lives. And many people either don’t understand all these consequences for the rights merely because someone else decided to initiate a suit and then send them a piece of mail that could easily be overlooked or pitched.
Another way to handle case 1: Transform to the rest frame of the rod. In that frame there is still a magnetic field, but the rod doesn’t care about it, because v is now 0. What is interesting is the electric field, generated by the line-charge which is generated by the difference in linear number density of the electrons and the copper nuclei of the wire. This is a 1/r electrostatic field (you don’t need to know the magnitude), and this is what will have to create the EMF between the ends of the rod.
It is now obvious that the two ends will have no EMF generated if they are at the same radial distance from the wire; and the answer would be just the same if it were not a rod, but a twisted contorted length of wire. So, no symmetry, nor integration needed; nor any visualization (which I find harder) neither.
This transformation will also work for case 2; but it doesn’t really save any effort in that case. You still end up doing the same integral.
Of course, relativistic transformations of electromagnetic fields may be out-of-scope for AP Physics. But it is covered in Purcell’s freshman-level text for electromagnetism.
JD, the real question is whether your walls are any different from the others in the neighborhood. Unless concrete is intended to be moisture resistant – proper admixtures and technology not generally employed in the ’60s – the moisture will come through the wall if the outside ‘waterproofing’ (if there is any) has failed. The efflorescence is salts which the moisture flow brings with it and leaves when it evaporates. Likely the cracks go through the wall and the outside black stuff has been breached (term of art).
My guess is that your neighbors all have the same conditions. to go back to my earlier thought, it would be good to discover who the better guys are who deal with these problems in your area and scope out your choices. That way you can’t be blindsided by a potential buyer who wants you to fix it.
I had an offer on a house which would require some serious cosmetic work at great expense. I agreed to have the work done but raised the price by double the cost – sort of an aggravation fee. The buyer apparently thought I was going to include it and both his and my real estate ‘professional’ accused me of bad faith negotiation and hinted that I could get into legal trouble for this.
Although i never sold to anyone who caused trouble after the sale, i always worried about it and tended not to agree to petty requests for repairs or raised the price. I did not want to sell to someone who would be trouble. Even though there is usually a real-estate person between you and the potential buyer you can usually get a feeling for trouble in the offer – too much fussy stuff.
After the episode above, I always did FISBOs but always with an attorney who did real estate work.
One last thing, we’re in contraction mode at 73. We rent a delightful townhouse in a park like setting in south florida for about what the maintenance fee would be if it were a condominium. We will never again own.
Our earlier homes, except for the one in Philadelphia, were good investments. I’m not sure that buying today except in some exceptional places will do as well. on the other hand, you have kids, and you may have to own in order to get a good school system.
I apologize to you and the other readers if this is too much.
Neal,
Relativity is covered in his book too. But the AP Physics C test does not. So it’s best for teachers to avoid discussing that. They need every week they have to cover the material that is on the test. The tests are administered the first week of May, the do cover RL and RC circuits and unless it went by me these kids haven’t hit RL circuits yet. (Not that they are hard… but they do need to cover them.)
Calendar.
http://apcentral.collegeboard.com/apc/public/exam/calendar/index.html
Neal,
As you might be interested in comparing to Australia, here is info.
This is a content list of what’s on the APC EM test:
This is from:
http://media.collegeboard.com/digitalServices/pdf/ap/ap-physics-c-course-description.pdf
The test is 50% multiple choice and 50% free response. In my view, the best way to gauge the “flavor” of the type of questions is to see the published AP tests.
They are here:
https://apstudent.collegeboard.org/apcourse/ap-physics-c-electricity-and-magnetism/exam-practice
To some extent (and likely a large one) relativity is generally deferred until the 3rd semester of physics in the “Physics for Science and Engineering”. This is done to make it possible for the engineers be prepared to start their engineering courses (circuits, statics/dynamics, thermo, heat transfer and so on.) If any significant time is devoted to relativity during the second semester (EM), that generally means some other EM topic is given short shrift (LC, LR, LRC circuits? eddy currents? Dielectrics? Take your pick.) the physics base for some engineering courses is deferred to the point where electrical engineers can’t start on their engineering course until they are juniors. (Or the alternative is the Electrical Engineering department tells physics they’ll teach their own EM classes.)
Meanwhile doing relativity 3rd semester doesn’t set the physics curriculum for actual physics majors back. So it’s usually done 3rd semester.
Other things they don’t look at in 1st 2nd semester: Transform current flowing in a motionless wire into a moving reference frame. So while the method of transforming into a reference frame moving with the bar might seem easier to physics students once they are introduced to that, most students taking AP Physics c, will never have seen the transform. Explaining that solution would require a day or two of practice problems where discuss how to transform, do a few, and then do it in an application.
The AP shifts the freshman classes to high school– but still, it follows what is actually in (or supposed to be in) a typical 1st and 2nd semester Sci&Eng college physics class. In the end, the problems on the AP C test can all be done applying classic methods. The computed integrals are never more difficult than sin/cos, polynomials or 1/r. Otherwise, they just need to recognize “B is constant, no matter what the shape of the loop, $latex \int dA =A$ and so on. If they don’t immediately see that something like symmetry makes two ends of an integral cancel, usually better to guide them to just do the 1/r integral rather than transforming current flowing in a wire to as has a very high chance of appearing in a problem on the AP, and while it can be avoided in some problems, it can’t be in others.
Practicing recognizing symmetry is very useful for problems on the test because this happens quite a bit on the test. So encouraging them to visualize and use symmetry is very useful. Of course, if a student can’t do that, you try to find another way. But I find many can at least see symmetry in simple cases. Plus, really, they are going to see this again in calculus and so on. So it’s a good thing to get used to.
Suggesting solution methods that involve topics they haven’t covered merely to avoid integrating 1/r is generally doesn’t help them much.
Neal,
More specific:
Learning objectives for AP Physics C are in here (B too, but that’s been replaced by Physics 1.)
http://apcentral.collegeboard.com/apc/public/repository/ap05_phys_objectives_45859-1.pdf
( I havent’ found a nice checklist for AP Phys 1. Just a long document from which one could make a checklist.)
I agree that doing a relativistic transformation is a bit out-of-the-way for preparing someone for the AP Physics test. But it’s fun for people that like relativity, and requires no computation at all (if you’re relatively familiar with relativity).
.
Whereas it always takes me some time to get oriented around 3-dimensional visualization: pointing my thumb, right-handed twirling for cross-products. I haven’t gotten better at it over the years.
Neal,
The right hand rule is awkward. I think that’s features that makes it difficult for students.
I usually do things the classic way. I never did much relativity and certainly didn’t continue to do it.
For simple geometries (like above even with the ambiguity) can get around the awkwardness of the right hand rule by sketching a coordinate system, finding “i,j,k” direction vectors and just doing $latex \vec(v) \times \vec{B} $ mathematically. So there are other options for students at the “AP C level”.
There are no options for the kids in “traditional physics” or “Conceptual Physics” in Highschool. Their teacher will not have taught them “cross product” and if you see them once a week to help them, you would spend the entire time discussing the “math of cross products” which they still won’t get and all the class presentations will use “the hand”.
Right now, I’ve found the students who are most confused are the ones whose teacher decided to teach the right hand rule several ways. One set of kids have a teacher who– by all accounts– alternates between the “v-thumb, fingers-B, direction of palm F”, the “curl from v toward B, see where thumb is” and the really awkward “hold your fingers in three directions” method. Evidently, he also throws some sort of “left hand rule” info something. I haven’t actually been to his classes — these are just what it looks like from questions students have and notes and so on. So the teachers seemed to be alternating between 4 rules.
In principle, “here are a lot of methods, chose the one you find easiest to understand” seems ok. But in practice….. no.
I agree – using unit vectors to find cross products used to be my preferred method.
lucia,
But electricity and magnetism is all about special relativity. The relationship between charges and magnetic fields are what they are so that the Universe works the same way for observers moving relative to each other. You can even derive Maxwell’s equations using only special relativity and Coulomb’s Law.
Dewitt,
I’m certainly not going to dispute that. And I’ll admit one could say E&M is “all about special relativity if their view is “I want to fully and truly understand the workings universe. God Particle. Quarks. Big Bang and all that. Don’t distract me with applications” There is nothing really wrong with that goal and the fact that it some people’s goal is good because that basic understanding does ultimately help drive more and niftier applications.
But some people have other goals or priorities — they want to move on to applications — and there is nothing wrong with those goals.
And historically somehow people like volta, tesla, edison, Franklin and so on and so on managed to do an awful lot before Einstein came up with special relativity. So one can take the point of view that electricity and magnetism isn’t real “all about” special relativity in the sense that you don’t need to know special relativity to do an awful lot of very nifty things in EM without knowing anything about special relativity.
For what it’s worth, we can also derive the ideal gas law and things like heat and mass diffusion constants from statistical mechanics principles. I would hardly advocate having high school kids wait to learn the ideal gas law until they’ve progressed to that level of physics. And I really woudn’t say ideal gas law is “all about” statistical mechanics just because I can derive it using statistical mechanics.
And the fact is: being able to derive Maxwell’s equations from special relativity and Coulomb’s law, while nifty, is not something students need to be able to do to start learning an quite a bit of material of practical importance to Electrical Engineering. So I think it makes sense that most schools defer special relativity even if some thing EM is “all about” special relativity.
lucia,
You don’t have to actually derive Maxwell’s equations using special relativity or introduce Lorentz transforms to use coordinate transform as Neal did. All you have to do is mention the basic principle of special relativity, that physics must work the same for observers moving relative to each other, to say: look at the problem from the point of view of the rod rather than the wire.
Dewitt,
Anyway, if someone wants to do the problem by shifting into another reference frame, that’s fine. But I don’t think merely saying “that physics must work the same for observers moving relative to each other, to say: look at the problem from the point of view of the rod rather than the wire.” would permit students to work this problem if they otherwise couldn’t.
The problem is easy to do in the reference frame given.
Dewitt,
For gauging what students can do:
1) They know Maxwell’s equations in integral form.
2) They don’t know what a divergence or curl are.
3) The do know how to perform surface and line integrals
4) They do not know how to transform the integral form of anything into the differential form of anything.
Yup, wouldn’t have helped me.
Mark,
I just wouldn’t ever be motivated to do it that way. I don’t really see how it makes anything easier. In the current frame, you think about the force balance on a charge carrier (i.e. electron) in the rod. You know it’s moving with the rod. It must experience a magnetic force. But to move with the rod, it must experience another force as well. That must be electric.
(1) $latex F_M = q_e \vec{v} \times \vec{B} $
(2) $latex F_E = q_e \cdot \vec{E} $
You do the right hand rule to sort out directions and find the magnitude of E is
(3) $latex E = v B $.
The all are expected to know the magnetic field around a current carrying rod:
So
(4) $latex E(r) = v B \mu_o/(2 \pi r) $ where r is radial distance from the rod.
They know the definition of potential between the ends.
They use right hand rule for direction. For case B, the geometry is easy– no trig, squares or square roots. Then they integrate. Done.
Even if students knew partial derivatives (which is calculus III) and know the divergence and curl (calculus III) I don’t see how transforming to find the electric field in the frame moving with the rod to get (4) is easier than doing steps 1-3 and remembering the relation for magnetic field around a current carrying wire. (Mind you, they could derive that too, They are expected to be able to do so. Calculus I is generally a co-requisite for mechanics, Calc II for EM. Calc III for the third required course in Physics.)
And bear in mind: I did 1-3 ploddingly because I like to highlight every conceptual step. Some students would get to (3) right away without thinking of the magnetic and electric forces separately.
I suspect people forget when they learned what. Tutoring actually forces you to remember. Some methods of solving problems may become your “favorite”, but that doesn’t mean a Phys II student will be able to do it.
I’m looking at Classical Electrodynamics by J.D. Jackson – used in graduate level physics/engineering courses and the special theory is chapter 11. For many EE/applied physics programs, the special theory is not very important to know. The principle of simple Galilean relativity though helps for some problems such as the time=relative distance/relative speed type problems.
I heartily concur that it’s best not to make someone initially learning electricity and magnetism slog through a relativistic treatment.
Still, there can be embarrassing consequences, even for us non-scientists. Having been spared the relativistic treatment back when I took the compulsory science course, I later in life ended up asking an expert witness, “So how would that charged particle know if it were moving with respect to a constant uniform magnetic field?”
(Since we were paying him he pretended not to think I was stupid for asking the question.)
Joe Born,
No matter how you present physics, some people aren’t going to know something.
If your expert witness was not stupid, he would know the reason you were paying him as an expert witness is that he knew that stuff. I really don’t think scientists and engineers are astonished that other people don’t know how to deal with transforming reference frames.
I mean… above, I mentioned the Uber class action suit. I’ve read a few news articles, and some of the arguments. But…well do I know employment law? Nope. Do I know contract law? Nope. Do I have a good idea about what can and can’t constitute a class in a class action? Not really. I’m interested in seeing how that case pans out, but I have very little idea about the law.
(And when I have looked up the IRS checklist for determining contractor vs employee and so on, my conclusion about whether Uber drivers are employees or contractors is “dunno”.)
I’m not recommending that either. But I think the brighter students might be motivated to look deeper if they knew that E&M was related to special relativity. And that’s all you really need to say.
I’m used to seeing statements that special relativity is only important at velocities approaching the speed of light. While you don’t need special relativity to solve E&M problems, I don’t see why it is not a good idea to at least mention that it applies. And GPS wouldn’t work without General Relativity to calculate corrections to the satellite clocks for the difference in the gravitational field at orbital altitude. You don’t need to explain the details of how to do those calculations.
Probably right. And even if they don’t dig deeper right away, they can file the fact away so that they can look it up if it later becomes important to them.
Say Lucia, thanks for posting this, I just noticed it. If I can find my scrap notes maybe this will help me see what I was doing wrong.
Well, actually I’m pretty sure I know what I was doing wrong, it was more conceptual than mechanical. Still, I appreciate it, gives me a chance to go back through it.
[Edit: I agree with the commenters who think that mentioning special relativity would motivate the brighter students to look into this. But I also still think the original point holds; it doesn’t help a student who can’t figure out the simple standard version of the problem]
DeWitt
I doubt mentioning special relativity will motivate brighter students any more than some other topic in EM. Of those enrolled in physics, I don’t think being interested in special relativity is correlated with being more or less bright. Some bright students taking physics have no particular interest special relativity. At. All. They only care about it to the extent that it affects a problem. Others think it’s soooper cool. But which care about what doesn’t seem to have much to do with being “bright”.
That said, there is nothing wrong with someone mentioning that if you use special relativity and Coulomb’s law you can get all the other laws. It takes no more than 2 minutes to mention that when wrapping up Maxwells equations and some students might find that interesting.
No one is saying never discuss special relativity. The issue is when to discuss it.
Interestingly, your example has nothing much to do with Electricity and Magnetism. Gravity is not an electricity or magnetism. I’m pretty sure gravitation is considered mechanics. I think time dilation due to the satellite moving is also not an EM issue; it’s really a problem in kinematics… right? (Real q.)
Also, I don’t see how people not knowing this about GPS is a huge problem. It’s not really clear that many people not knowing the nuts and bolts of how to design a functional GPS system is a huge problem. Certainly engineers and physicists not specifically knowing until Physics III isn’t a huge problem. Physics III is requires. They learn it then. Not seeing a problem here.
Meanwhile quite a bit of harm could arise if one suggested students actually use the ideas from special relativity when solving AP Phys C problems. Time spent encouraging its use would almost certainly require a teacher to drop a topic that is in the curriculum and on the test to introduce some material that is not. (They’d probably drop inductance and LR and LC circuits.)
That’s not such a big problem at a University where each school gets to decide which topics are important and what order they wish to present them. But if a high school teacher does that, a higher fraction of their students will fail the AP test because they won’t earn the points on questions involving the topic on the test. If they fail, the students will then be forced to take Phys I or Phys II over even though they covered the bulk of the material in those course And that will happen because missing a topic like “LR and LC circuits” resulted in a failing grade.
Failing the AP Phys C tests will not encourage the students to continue in physics and could cause some of them to decide to apply their talents in areas where they did pass the AP. Perhaps French. Or Economics.
lucia,
Continuing to flagellate a deceased equine: Time dilation is explained by the same math that shows that magnetic fields must exist if electrostatic fields exist and vice versa, invariance under a Lorentz transform. But that’s special relativity. GPS requires not only a time dilation correction because the satellites are moving, but a general relativistic correction because they are in a different accelerated frame than clocks on the surface of the planet.
But if there’s a test, you have to teach to it. The Feynman Lectures were an attempt to restructure the teaching of basic college physics. It didn’t work out all that well.
DeWitt,
Yes. But it doesnt’ show either must exist at all. The transform is perfectly consistent with neither existing. So it seems to me EM is a separate ‘thing’.
I suspect it didn’t work out because the Feynman structure don’t dovetail well with engineering programs. If Physics departments followed that program, engineering departments would start running their own introductory physics classes to ensure their students are introduced to to topics in EM and ME at a depth and breath and level people in Engineering departments hope students will see before they start statics, dynamics circuits and so on.
If the students don’t get the appropriate depths and breadth in correct topics the engineering departments have to teach them anyway so they’ll see no point in having the take the classes taught in Physics.
jferguson: There is a god. We had a good rain today from about 9:00 a.m. until 10:30 p.m. I filmed my basement at 11 p.m. and, as I knew, there was no water. This will help a lot. The implication from the inspector’s report was that I had a leaky basement. My video disproves that. Thanks for your pointers.
JD
Hi JD, the cure may have been getting the downspout discharge away from the foundation wall. …little things.
good luck with the sale.
john
jferguson, JD Ohio,
Yup, I have three times solved a “wet basement” problem by simply installing piping (underground) to carry downspout discharge well away from the foundation.
DeWitt,
“deceased equine’.
.
Indeed.
SteveF, JD,
Garrison keillor observed that property is the enemy of leisure. Not than either of you appear ever to have any… leisure that is.
SteveF:
I had an issue with a house on a slope, where inevitably it would get water pooling against one wall during heavy rain storms. It really wasn’t possible to regrade the slope to prevent this from happening.
I addressed the issue by digging down along the foundation on the side, tarring the exposed wall to reduce the amount of leaking, then adding a french drain system to carry the water off instead of letting it pool against the foundation.
[I also redirected the water from the downspouts into the drainage system.]
I hired a couple of college students do to most of the digging, and paid them by the hour. An excavator wasn’t an option here, because you really don’t want excavators operating by your foundation. [But excavator costs can be surprisingly pretty low—generally overall costs would be less than even cheap student labor—-if you know somebody who has the equipment, and you aren’t hiring them through a contractor.]
Total cost, including labor, was under $1000. The “main ingredient”, flexible drain pipe, goes for less than $1/foot. So the material costs are quite modest. Start to finish, you could do a project of this scale, once you’ve recruited the labor, in under two weeks.
Carrick, what you and SteveF did is what should have been done when the places were built.
jferguson & others — as I tried to make clear before, I never had problems in 15 years that I lived in the house. I was assuming that the inspector knew what he was talking about. He may not have. Of course, it is also possible, that there were problems before I bought the house, and they were taken care of by the extension.
JD
jferguson:
Yeah I agree. But this is an older house—my “starter home”—built back in the day where nobody seemed to care whether basements leaked or not (I am guessing it was originally used to store fruits and tubers).
One of the odder architectural features of this home was the floor decking was 3/4″ x 2 1/2″ red oak tongue & grove (no subfloor below this), which went from exterior wall to exterior wall, no breaks.
The walls were actually built on top of the oak decking. The walls are 2-1/2 studs prebuilt assemblies (think “Sears home”), with 1/4″ plywood cladding on both sides.
They put carpet over the oak flooring (of course) and the wallpaper… OMG I’ve never seen such ugly stuff.
We ended up installing drywall over the existing walls (because the plywood was too irregular to paint over), painting and putting in modern trimming. (My Dad and I did about 90% of this work.) Once we ripped up the carpet and refinished the flooring, the aged hardwood was gorgeous.
Anyway, by the time we were done, it was a very cozy home. But the basement was the least of the problems (though it added a lot of great storage area to a tiny house).
Carrick, that sounds a very neat house. you don’t suppose it actually was a kit house, do you?
My Dad thought it might have bee a kit house. Kinda hard to explain the construction method otherwise.
Looks to Eli that without a ground the EMF difference between the ends of the rod is zero, thus the geometry is irrelevant.
Eli (#145799) –
You’re going to have to explain why you think EMF is zero without a ground.
Lucia,
I admit my previous error and present one final attempt.
There is an infinite number of possible rod configurations and they range through all possible orientations between the two diagrams you presented initially.
Start from your “Interpretation 2” (or I2, for short), looking from ‘above’ that diagram, along the direction of the current (call it the ‘plan view’ of I2, which has the rod pointing radially, away from the wire). Now rotate the outer end of the rod either CW or CCW from that position (keeping the inner end at the correct distance from the current wire) until you approach one of the two possible “Interpretation 1” (i.e. I1, or symmetrical) positions. The inner rod end will have to move slightly outwards as you approach the extreme positions to achieve each of the two possible I1 orientations. So far, so good.
In either of the available I1 orientations, I deduce that there will be two voltage gradients, however, they will be of equal magnitude but opposite ‘sense’, and occur along the opposite halves of the rod, meeting at a common (+ve?) voltage level in the centre. In other words, there will be no emf (or potential difference) between the ends of the rod.
In the I2 position, the inner end will be +ve and the outer end -ve.
These deductions are based solely on the knowledge that “parallel currents attract each other”. [I leave the emf magnitude calculations to others. I have assumed the current arrow is conventional current (i.e. +ve particle flow direction)].
Harold, because the antenna is a conductor and all points are at the same EMF relative to each other. If the antenna was a semi-conductor or there was a connection to a resistor that was grounded on the other side, then the problem becomes more difficult.
However, IEHO the question is whether the questioner knew that and purposely set up a trick question (yes Eli had to think about it for a while) or just was incompetent remains open till we see the answer sheet.
Harold, (something hung up in processing was sent) but the point is that the antenna is a wire and there is no potential difference across a conductor.
Eli: conductor only means high conductivity, not infinite conductivity. Ground is just a reference point on that wire.
For the purposes of a problem like this conductors ARE perfect. Otherwise you have to specify the resistivity/conductivity which is not done, and in any case would not make a significant difference. The question appears to ask for the voltage difference btw the ends of the wire, not the difference to the ground plane, which is what Eli answered. Zilch
Potential difference is just integral E.dl and E = v x B
You seem to be asking what is the value of the induced current.
Eli,
Perhaps that’s why the problem is posed as a thin metal rod developing an emf, not an ideal wire.
Then he has to specify many properties of the rod/wire.
Eli –
I think it’s been too long since you took a test. It’s a simple problem if you don’t try to make it complicated. That is, once you decide that Interpretation #2 is the geometry you should use.
Eli,
You might want to check out Freedman and Young’s University Physics. Check out the section called “Motional Electromotive Force”.
Hi!
Think I would solve this problem like this:
1) Long straight wire: according to standard phrasing in textbooks like this, this implies infinitely long, making it easy to find the magnetic field as function of radial distance from the wire.
2) Perpendicularity: the property of being perpendicular clearly implies that the lines are in the same plane, otherwise they will never meet at right angles:
https://en.wikipedia.org/wiki/Perpendicular
Hence interpretation 2.
3) Lorentz force: the electrons in the rod experiences a force from moving through the magnetic field. This force is:
F = q(E+(vxB))
Please add vector bars / boldface in your minds where applicable.
https://en.wikipedia.org/wiki/Lorentz_force
4) Stationary charges: Since no current is flowing through the bar, the electric field inside the bar has to generate a force on the charges of the same magnitude, but opposite sign, as the vxB term.
5) The Back-EMF is then the integral of the electric field over the rod, from end to end.
WIth the best regards,
Johan
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Tell Eli how to develop a potential difference across a conductor pls.
Problem asks for the emf IN the rod. not of the rod wrt the ground plane.
Eli,
Correct. They ask in the rod not with respect to the ground plane. Would you like me to send you pdfs of the pages in Freedman and Young that discuss how to do it? I’m sure you can find it in your library Freedman and Young “college physics” in the library. These are the authors who replaced the old Sears and Zemansky.
I have the 13th edition. You’ll find Motional EMF on page 970. That will explain how. Otherwise, you can examine the two problems here:
http://labman.phys.utk.edu/phys222core/modules/m5/motional_emf.htm
Or here:
http://www.sparknotes.com/testprep/books/sat2/physics/chapter16section1.rhtml
This is a very common problem in Freshman Engineering Physic and variations appear on the AP. The only tricky bit here is being certain what geometry they meant.
lucia, ah, but what if they weren’t asking after motional EMF when they said:
But were really asking about the size and direction of the emf induced in the rod moving with a speed of 1.8 m/s… by something other than that motion?
My answer: They should be fired as test writers.
Eli: “Tell Eli how to develop a potential difference across a conductor pls.”
Answer: induce regions of positive and negative charge.
Eli, you seem to have in mind a rule of electrostatics, but this problem involves dynamics. Consider this lecture, section 14.2.
It seems like Eli has something like a parallel plate capacitor involving electric fields only in mind. Until he explains his objections, it is hard to understand where the confusion lies.
Did we determine what the voltage is in Interpretation 1 which seems the simplest? I usually envision the theoretical solution by first supposing an experimental first look. I can make a long wire carrying 740 ma. I can also make a 45 cm rod (a coat hanger will do) and with a wooden spacer sweep it along at a good approximation to the 1.8 m/s.
But now I am stuck measuring the voltage, which I take to be end-to end. The wire leads from the ends back to the voltmeter will also pass through the magnetic field, cancelling the induced voltage if the voltmeter is near the center of the rod. Curiously, this would be zero volts – in practice.
Thinking more of a dipole with the meter at the center, I also thought of the two-pieces with one half-piece having the flux lines arching out and the other have piece arching in, cancelling. I think this is pretty much what OzWizard said on April 14, 2016 at 12:20. Symmetry argument. The answer is zero volts.
Bernard Hutchins —
Agreed, the configuration of interpretation #1 results in a net of zero volts end-to-end. See #145557 (lucia) and #145655 (Neal J. King) above.
It’s an interesting point about trying to measure the voltage. I’d have to think a bit about how a standard DC voltmeter would operate when in motion in a magnetic field. (Now I’m thinking about interpretation#2, where we expect to see a non-zero emf.) Its wires would be subject to the same motional emf as the rod. Plus a circuit would be made…hmmm.
Reply to HaroldW at 7:31 AM
Thanks HaroldW for sorting this out for me. I tried the experiment on an 8 foot long wire with a full amp. The “rod†was just a 2 foot clip lead with high impedance digital multi-meter clipped to the ends, held and moved along the wire as best I could. I ran the meter leads behind my neck so as to keep them away from the field of the wire. Nothing! You put a clip lead across a voltmeter and you get zero.
A general rule of “exams-man-ship†is that if you are given too many complicated details, the answer is 0, infinity, or 1. Called a “trick question†– if you missed it!
Over 20 years ago a former EE student of mine came to my door. He was and is the only EE I ever heard of taking a professional license exam. He had some sort of prep book and questions, one of which I remember and which I have drawn here:
http://electronotes.netfirms.com/ENWN35FigJ.jpg
The question was to give Vout. He knew they probably wanted –Vin[ (1/sC2) / (1/sC1)] = -Vin(C1/C2). But this was too clever by half! Since there is no DC path to the (-) input, the tiny but finite input bias current will charge the caps, so Vout will be at + or – supply (clipped). Which did they want? I suspect that these writers of an exam for practicing engineers didn’t know what REALLY happens. I never found out.
Bernie
HaroldW,
I imagine if the wires connecting to the voltmeter are parallel to the magnetic field, there should be no interaction? It might be useful then to imagine a stationary bar and stationary voltmeter and a big sliding magnet. An ideal voltmeter has infinite impedance. Incidentally, if the rod was not a conductor there should be no free electrons that experience the force.
to HaroldW at 11:49 am
Do you mean the voltmeter wires are parallel to the wire itself. The wires do eventually have to close back at the meter. The field about the wire is circular I believe. I don’t understand being parallel to a circle! I don’t understand your big sliding magnet. Do you mean a wire carrying a current with the wire itself moving along its length? Or is it a thin band with poles parallel to the thin dimension? Hard experiment!
ok, ignore that ..
Sorry for mixing up Harold and RB.
Bernard Hutchins —
No problem with the mixup…you gave the time of post which made it abundantly clear what you were responding to.
.
I’ve thought a little more about the voltmeter. I think it will read zero even when a motional EMF is present. Let’s first consider what happens when you put the voltmeter right on top of the rod, connecting each input to a different end of the rod. Think of the voltmeter as a large resistance with some method of measuring the current through that resistor. The wires of the voltmeter will experience the same emf-of-motion as the rod, but no current will flow, just as no current flows in the rod. Hence the voltmeter will read zero volts.
.
What about using long leads so that the voltmeter is far from the current, in a region of low magnetic field? The induced emf goes as $latex \vec{v} \times \vec{B} \cdot \vec{dl} $ for an increment $latex \vec{dl} $ of the lead. That would be zero if one could orient the leads parallel to B…but as has been pointed out, that leads in a circle around the wire, and one can’t connect the two ends that way. Or parallel to v…but as v is parallel to the wire, that doesn’t get towards a lower magnetic field (and also doesn’t allow for the ends to connect.) I think that however one connects the leads, the net emf ( $latex \int \vec{v} \times \vec{B} \cdot \vec{dl} $ ) will come out the same.
Neal J. King’s method (#145655) provides a different, and more rigorous, method of coming to the same conclusion. Net emf = f(r1) – f(r2), where r1 & r2 are the distances of the respective ends from the wire. [We can compute that f(r) is proportional to log(r) and current and velocity, but that’s not important.]
To HaroldW at April 19th, 2016 at 6:35 pm
I guess one problem is that we are talking about measuring a voltage across the ends of the rod where we are at the same time arguing that that voltage difference is zero (back to back batteries). However, by all rights, we should assume the difference is non-zero and demonstrate that it is zero for the specified geometry.
Now, if this voltage difference between the ends WERE non-zero, then would there not be an ELECTRIC field from one end to the other, EXTERNAL to the wire, with field lines looking like those of a bar magnet.
Attempts to measure the voltage difference with conductors will distort (destroy?) the electric field? Is it the case (as you perhaps suggested) that no matter how we arrange the attaching leads we destroy the field?
Is it (perhaps) the case, that if the geometry is shifted such that the center is not closest to the wire, that the “two batteries” are still equal in voltage, the shorter length being in the stronger regions of the magnetic field, and vice versa?
I can’t quite put this together. In fact, I’m not even close!
Bernie
Bernard Hutchins –
In my #146174 above, I was thinking about the case of non-zero emf (that is, Interpretation #2) although that was perhaps not made clear. Despite the presence of a non-zero emf, a voltmeter being carried along with the rod will read zero. The reasoning is that its conductive leads will be subject to the same motion effects as the rod.
.
However, a *stationary* voltmeter should be able to indicate the emf. My original thought experiment had the voltmeter connected across the ends of the rod, and as the rod moves down the page, wires play out behind it, with the moving bits oriented parallel to the rod’s velocity vector, hence subject to no motional emf. A more sensible version is that shown on page 3 of this lecture which I previously recommended to Eli. [Where’d he go, by the way?] The rod is slid across a pair of conductive rails; where the diagram shows a resistor, imagine a voltmeter instead. In this arrangement, the voltmeter will indicate the emf correctly.
To HaroldW at April 20, 2016 5:11 am
I am thinking (at the moment) that the rod is a capacitor! The electric field about it is analogous to that of, for example, a neat parallel plate capacitor – it just flays about into space.
If you needed a metal rod, but only had an ordinary capacitor, you could solder the leads together, dissolve off the plastic and remove the insulators, separate the “plates” and roll them out to a rod. But the rod is a capacitor that has its leads shorted together in the middle. Impossible to charge, or to measure the voltage on it, unless you break it in two. Connected, it can only have a voltage (and electric field) when moving in a magnetic field.
This doesn’t help a lot, but gives me a better idea of the voltage not being end to end through the conductor (impossible) but rather looping around outside, and collapsing instantaneously when the rod stops moving.
Your attempt to have the conducting test leads as parallel rails was similar in function to my looping the wires behind my neck to keep them out of the magnetic field as best I could.
I need to look at the lecture you suggested (thanks) and will get back to this later.
Bernie
It’s a trick question. It makes no difference how the bar is oriented.
The rod experiences a magnetic field due to the current in the wire, however, although the rod is moving, it’s position relative to the wire is constant. The field that the rod experiences is constant and there is therefore no induced current or voltage.
You could as well set the rod next to a bar magnet. Would you measure a voltage, as long as the magnet remains stationary in relation to the rod? If you put them in a box and took them on an airplane would you measure a voltage? during takeoff and landing? I’m skeptical if you do.
The moment you placed the rod next to the magnet (or wire), you induced a current, which caused a voltage, which could have been measured (or stored in a capacitor) at that time. This was due to the work involved in placing the conducting rod into the magnetic field, but since the rod is a conductor the current flowed back again after the rod stopped moving in relation to the magnetic field.
If you could measure a voltage between the ends of the rod in a stationary state, then you could use that voltage to power something. If this were true, then generators wouldn’t need to rotate in order to generate electricity. It would be a very different world if this were possible – but not one where we could survive.
This problem continues to vex me. I think what I wrote at: Bernard Hutchins (Comment #146232) April 20th, 2016 at 11:19 am, about the rod being a capacitor, is correct. Further, several (at least three) videos which I have subsequently found, such as:
https://www.youtube.com/watch?v=HrJEUTRM3JM
may well have the right answer (voltage) but a faulty explanation. I COULD WELL BE WRONG.
First of all, I believe the answer to Lucia’s posed “Interpretation 1†is that the voltage is zero, for the reason that we have a B flux that is equal parts up and down. Batteries back to back. Otherwise, a difficult calculus problem. A trick question. I (and the videos) am interested in the moving rod in a UNIFORM B field. (Walk before you run.)
In this case, the charges (electrons or “holesâ€) are forced by the B field (qvB where q is the electron charge and v is velocity) toward the ends. The videos say (vaguely) that the charges are at the ends and that any charges inside the rod (not moved to ends) are balanced by an E field (qE) so E = vB. They then say the E field corresponds to a voltage difference V=EL so V= -BvL. Possibly (likely?) the right answer.
The problem is that the videos show (explicitly in the one linked above) the E field as being INSIDE the conducting rod along its length, with the charges at the ends. I believe there is no E field inside a conductor, and that the charges are only on the surface. The actual E field flays about OUTSIDE the bar and in fact, does balance the B force. I don’t know exactly what the field looks like but I suspect it resembles the magnetic field of a bar magnet.
It is perhaps a case where a complicated (integration) calculation yields a simple result. [ We are used to these – like the gravitational field inside the Earth looking like a point mass.]
In my April 20 post I “morphed†a capacitor into a rod. The electric field of the capacitor was of course confined (fairly neatly in this case) in the free space, external to the conducting “platesâ€, fringing a bit at the edges. The charges were on the surfaces of the plates. In the morph, the E field become more distorted but remains outside the conductor.
I think that’s only true for high frequency fields. If the field were only on the outside, then it would seem that conductivity would not be a function of area. But it is. A hollow rod is less conductive than a solid rod with the same diameter.
DeWitt at 1:47
I think the conductor is assumed perfect in the problem. One of the videos says exactly that.
Bernie
To: KenW (Comment #146405) April 23rd, 2016 at 11:31 am
The conducting rod is moving relative to a constant field. That’s HOW a generator WORKS! Here the magnet happens to be an electromagnet. The voltage is net zero because of different portions of the conductor seeing opposite fields, one up, the other down. That’s the “trick”.
Bernard, the rod is not moving relative to the magnetic field.
If you were sitting in a little room inside the rod, with no window, measuring the magnetic field – then you would measure the same field all the time – whether you were moving or not.
This is true as long as the rod does not accelerate, and the current in the wire remains constant – both of which require an external force being applied to the system. Only then would you notice anything.
KenW: “If you were sitting in a little room inside the rod, with no window, measuring the magnetic field – then you would measure the same field all the time” and “The field that the rod experiences is constant and there is therefore no induced current or voltage”
True, in the rod’s frame of reference, the magnetic field is constant. However, there is an electric field emanating from the wire, of magnitude
$latex \vec{E}’ = \vec{v} \times \vec{B} $
which will generate the same force as the magnetic field. One can’t run away from force by moving (at a constant speed, anyway).
A description of how this arises can be found here, in section 12.4. 12.4.3 compares the magnetic force (in the frame where the rod is moving) to the electric force (in the rod’s rest frame).
This approach to looking at the problem was mentioned earlier in #145655.
Re: Bernard Hutchins (Comment #146407)
Let’s assume the conductor is a cylinder. I agree with you that the field inside the conductor is zero for a perfect conductor, for electrostatic equilibrium reasons (Gauss’ law etc). For the same reason, the induced charge cannot be distributed over the entire surface of either end (D.da = charge within a surface) but would be at the periphery. If so, it should be permissible for the electric field lines to run along the outer surface of the cylinder. Anywhere along the length of the cylinder, except at either end, D.da=0. so there is no charge except along the periphery of either end.
to: HaroldW (Comment #146443) April 24th, 2016 at 4:33 am
You said to KenW:
True, in the rod’s frame of reference, the magnetic field is constant. However, there is an electric field emanating from the wire, of magnitude
\vec{E}’ = \vec{v} \times \vec{B}
which will generate the same force as the magnetic field.
(1) What is “the rod” and what is “the wire”?
(2) How does the E field emanate? Starting at a (+) charge at the surface, leaving normal to the surface, looping through space, and terminating on a (-) charge back on the surface, again normally?
(3) Or what are you saying? The E field you denote as a vector (cross product) but call a magnitude.
(4) Do you suppose, perhaps, that changing the reference frame just makes things harder?
Bernie
to: RB (Comment #146506) April 25th, 2016 at 8:38 am
I think the E field lines loop out into space on (+) charges, perpendicular to (not along) the surface and return to negative charges, also normally. The field lines form a pattern not totally unlike a dipole, but of course distorted by the conductor.
Suppose you moved the rod in the B field, “charging” it, and then cut the rod in the middle before stopping. The two halves would be oppositely charged, and could be pounded flat into parallel plates, a capacitor with flaying (fringing) fields, all external to the conductors, MOST of it between the plates.
If we reconnect the conductors (plates) we discharge the capacitor, except as it is maintained by the B field being crossed by the connecting wire, when it again moves.
Bernie
Bernie Hutchins:
See Example 1 here . There will be a fringing E field also in addition to the tangential one, but there is an E field along the surface that cancels out the vxB everywhere inside the plate at equilibrium so that you are left with charge only at the top and bottom.
to: RB (Comment #146593) April 26th, 2016 at 7:55 am
Thanks – good lead – that paper (1998!) by Lorrain et al. I will read the whole thing later. It references Corson & Lorrain, and Dale Corson was one of my Physics professors (later President of Cornell). So you pulled the right string (or wire !).
Bernie
Bernard Hutchins (#146565)
[I’m skipping over your (1) & (2) because the Lorrain et al. paper will undoubtedly answer those questions better than I could.]
“(3) Or what are you saying? The E field you denote as a vector (cross product) but call a magnitude.”
Yup, my mistake. In a previous version of the comment I had written something like “magnitude of v cross B” before I decided to wade in and get all Latex-y. E is, of course, a vector, and you should just ignore the “of magnitude”.
“(4) Do you suppose, perhaps, that changing the reference frame just makes things harder?”
I would agree; I think it’s easier to stay in the lab frame and include the v cross B term. But KenW’s comment (#146441) talked of what happens in the rod’s frame of reference, where v is necessarily zero so there is no v cross B term. Which is fine as far as it goes, but if you go there, you must also transform the E,B fields. In the end, one will get the same answer (scalar emf will be the same in both reference frames). The point I was trying to make is that you can’t just set v to zero; there is another effect of looking at the problem from the rod’s frame of reference, perhaps unrecognized by KenW.
Okay, at the risk being of really, really dense – and perhaps flat out wrong – the whole problem with The Question seems to be that the frame of reference is ambiguous!
from my view:
consider an endless wire in infinite space:
Through the wire flows a current.
Somewhere near the wire is a thin conducting rod.
Imagine the rod is stationary. There is no motional emf in the rod.
The orientation of the rod is immaterial.
The amount of current in the wire is immaterial.
okay so far?
Now – imagine that the rod remains stationary and that the wire is moving along its length. (in empty space, who can tell the difference?) This changes the current flowing relative to the rod – but from the perspective of the rod, there is no change.
Now the tricky part: An observer moving relative to the rod WILL measure a field in the rod, and what (s)he measures will depend on the vector of the observer’s motion! This is the case if the observer is, say, attached to the earth. (is that right?)
Reading the question, I don’t think that it’s specific where we are. We have emf or no emf – depending on your frame of reference (and maybe your state of mind).
Perhaps this was deliberate, and explains the lack of precise information about the orientation of the rod. This would be a sign of a trick question (but that’s speculation about the motives of the author).
As for myself, I’ve about reached the point of holding contrary opinions simultaneously, so I’m open to any interpretation at this point.
thanks
KenW got it right. The rod moves along the flux lines so no lines are cut and no emf is induced.
RobWansbeck, I’m glad you think I’m right, but I’m not as cocksure as I was a couple of days ago! In my mind I’ve rationalized away the “vâ€, for the moment anyway.
KenW,
HaroldW has already provided some links. The idea is that there is an electromagnetic field, and depending on the frame of reference, you can get a different mixture of E and B fields so that the physics is the same. You can see this in the example of a particle that always moves closer to a current-carrying wire as explained in the Feynman lectures .
RB, thanks for the Feynman link. I will try to digest that, and review HaroldW’s links again.
RB, okay, got it. Feynman explains it so that even I (think i) understand it!
The positive and negative charge densities in a moving current carrying wire vary because they have different velocities – and therefore occupy different volumes. This creates an electric field which has exactly the same effect on a stationary nearby charged particle as the as the B field would have on a moving one. The net force on the particle is always the same for any net velocity.
This is comforting, but GEEEEEZ, anybody could not know that!
My problem was being too fixated on the B field and the right hand rule. This is fine for everyday use, but only half of what you need to know to understand the situation in question. Now most of the rest of this thread makes sense too.
Thank you HaroldW, Bernard, and especially RB for the Feynman link! That will keep me busy for a while now.
KenW,
You are welcome. Of course, Maxwell wasn’t aware of the embedded ideas of special relativity when he came up with his equations and constructed elaborate mechanical models using the ether concept. This is what Feynman says about Maxwell’s equations in Chapter 18 of the lectures.
This is behind a paywall, but the Lorentz force can be derived from Maxwell’s equations without having to explain the underlying mechanism of how the wire appears to be charged.
Actually, this might be a better reference – you get the same force without digging into the details of the machinery at the charge level.
As a mathematician, I use/teach the definition that lines must intersect to be called perpendicular, even in 3-space. The reason for this is that an angle must be formed using parts of the line, not a projection of the line.
Hence, the Interp 1. would fail the defn.
Interp 2 requires that the additional interpretation that the line the rod lies on intersects the wire. Then it would satisfy the problem.
As a footnote, I believe the induced emf is the electric potential, delta V from one end to the other. Since Lorentz say F=qv x B the work needs to summed over the length of the rod. Since B is proportional to I/r or B=cI/r the integral becomes cv times the integral for 1/r from 3.5 to 48.5 so the answer would appear to be the appropriate constant*I*1.8*(ln(48.5)-ln(3.5)) provided I got the units right!
Sorry if this is a duplicate of a previous observation. I only read the comments up to about dawn on 4/9. I think option 2 was intended,
but not for any reason from physics or mathematics. I wondered how such a huge ambiguity could creep in. My guess, just a guess, is the problem had a picture much like option 2, when originally considered. Then the picture was eliminated and the text expanded to try to describe the picture.