Physics Q&A Game: Exploring Faraday's Law with Marcus' Polar Route Question

[chroot]

Just like the Astronomy Q&A game in the Astronomy forum, let's play a physics Q&A game here. The astronomy game has certainly taught me some very neat facts -- maybe this one will do the same.

I'm going to post one question posed by marcus:

"A favorite version of the question chroot just answered is the one about the airplane flying the polar route. It is going 200 meters per second and its wingspan is 30 meters----what is the voltage difference between the two wingtips? I am not asking this question because it is a cousin of the one asked by Ivan, just recalling it. the plane is in a region where the Earth's magn. field is roughly vertical and of such and such a strength etc etc."

It seems that there should be no voltage. Faraday's law dictates that

E = -N d[phi]/dt

Since [phi] is defined as the flux: [phi] = B * A, where B is the magnitude of the magnetic field and A is the area enclosed by the loop. In this problem, neither A nor B is changing -- so the induced voltage should be zero.

Am I right? Was it really just a trick question?

- Warren

 

[Ivan Seeking]

-[30X200][50X10^-6][+-30]= +-9 volts

 

[Tyger]

My answer

When the plane takes off a voltage will be induced across the wings, and it will be sustained in flight. But you can't measure it with a voltmeter across the wingtips because the very same induction will occur with the meter leads. However, if the plane can land with the charge on the wingtips remaining, it can be measured on the ground with an electrometer, which is a very sensitive voltmeter.

 

[Ivan Seeking]

Originally posted by Tyger
When the plane takes off a voltage will be induced across the wings, and it will be sustained in flight. But you can't measure it with a voltmeter across the wingtips because the very same induction will occur with the meter leads. However, if the plane can land with the charge on the wingtips remaining, it can be measured on the ground with an electrometer, which is a very sensitive voltmeter.

I bought perfect test leads for my perfect voltmeter. They have them down at the gedunken supply house.

Re: "There are more things in heaven and earth, Horatio, Than are dreamt of in your philosophy."
I like you already!

 

[marcus]

Originally posted by Ivan Seeking
-[30X200][50X10^-6][+-30]= +-9 volts

30 meter wingspan x 200 meter/sec speed
x 50 microTesla (the vertical component of the geomagn. field)

let's see how that multiplies out

it comes to 0.3 volts

but you have an extra factor of 30 you multiplied in
which brings it to 9.0

You are permitted to edit your answer if it is a typo!

If it is not a typo then explain please the factor of "[+-30]"
You may certainly be right to have it. I am often forgetting
details and there may be some obvious thing I am missing.
I got this problem from an old edition of Halliday Resnick
but don't remember the exact numbers

 

[Ivan Seeking]

Originally posted by marcus
30 meter wingspan x 200 meter/sec speed
x 50 microTesla (the vertical component of the geomagn. field)

let's see how that multiplies out

it comes to 0.3 volts

but you have an extra factor of 30 you multiplied in
which brings it to 9.0

You are permitted to edit your answer if it is a typo!

If it is not a typo then explain please the factor of "[+-30]"
You may certainly be right to have it. I am often forgetting
details and there may be some obvious thing I am missing.
I got this problem from an old edition of Halliday Resnick
but don't remember the exact numbers

Shoot no. I was integrating the emf over the length of the wings. twas thinking E and not ε

 

[Ivan Seeking]

canni still go next? canni canni please please

 

[marcus]

Originally posted by Ivan Seeking
canni still go next? canni canni please please

Your turn!

sorry I got preoccupied with something else and just got
back

please proceed

(this game was partly your idea----the teaser problem---which chroot generalized, you and chroot make the rules I think)

 

[damgo]

It's the Hall effect, Warren. :) Actually a real-life voltmeter should work fine, I think -- just stretch the leads out, doesn't matter if you even touch the wing or not.

 

[chroot]

I'm still not sure how this answer was arrived at.

- Warren

 

[chroot]

Ohhh.. the Hall effect. Hmph.



- Warren

 

[Ivan Seeking]

Originally posted by marcus
Your turn!

sorry I got preoccupied with something else and just got
back

please proceed

(this game was partly your idea----the teaser problem---which chroot generalized, you and chroot make the rules I think)

yippie!

The dA/dt was just the width of the wings multiplied by the speed of the plane.

Ok. I don't know if this will be really easy, or really hard, or somewhere in between. A True Story.

The answer to a very complex physics or mathematics problem [I don't remember the exact nature of the problems] was to be calculated on one of the world's top supercomputers of the time; circa 1975. Two approaches were considered. The first and preferred approach was to calculate an extremely precise answer, and next was to approximate the answer. The best answer would require about a year of CPU time. The approximation could be calculated in relatively short order: in a matter of days.

The cost of CPU time caused this issue to be scrutinized very closely. It was decided that in spite of the money, the approximate answer was the best choice. What is the physical motivation [the physics] for their choice?

 

[Alexander]

Originally posted by Tyger
... However, if the plane can land with the charge on the wingtips remaining, it can be measured on the ground with an electrometer, which is a very sensitive voltmeter.

Charge can't remain (separated when plane is on the ground) because plane is made of metal.

 

[Alexander]

Originally posted by damgo
It's the Hall effect, Warren. :) Actually a real-life voltmeter should work fine, I think -- just stretch the leads out, doesn't matter if you even touch the wing or not.

No, voltmeter won't show anything (because emf [vB]l is evenly distributed along a wire l).

 

[marcus]

Originally posted by Alexander
No, voltmeter won't show anything (because emf [vB]l is evenly distributed along a wire l).

I often disagree with you Alexander or else am put off by your argumentatitive tone of voice (I do not remember which it is) but here I must commend you and voice agreement.

This is not normally considered an example of the Hall effect.

It is a standard textbook example that comes in right after the Faraday law. Halliday Resnick has it as an antenna of a car driving where the field is horizontal and the vertical antenna is cutting across field lines.
Giancoli has it with airplane wings cutting across vertical field lines.

The basic law (as in so many things) is the Lorentz force.
(Which also explains the classical Hall effect! but the problem is elementary and appears long before the Hall effect is discussed.)

A voltmeter could not be used, I reckon, to measure the voltage difference between the wingtips in flight (for the reason Alexander gave----no current would flow thru the meter altho the voltage difference would be there)
A conductor is not necessarily a constant potential thing when it is moving thru a B field as someone said, maybe Alexander.

Lorentz force is so basic, almost the F=ma of E/M
Lorentz force law is really the definition of the E and B fields, it is part of the groundwork.

dropping out the electrostatic part of the law it just says
F = qv X B

The B arrows near the north geomagnetic pole, which is a "south" pole---paradoxically---point DOWN.

So conduction electrons in the wing feel a force to the right

The voltage is higher on the left wingtip, where positive charge carriers would concentrate if there were some.

In Giancoli the problem comes right after the example of the bar sliding along two parallel rails, in a uniform vertical field

The rails merely permit the voltage difference between the ends of the bar to be measured (it is a Faraday law loop including the rails)

The airplane wing is the bar with the two rails removed.

You could put rails in the sky and have the airplane swoop down and fly along them with its wingtips touching them so that the voltage could be measured but the FAA and the Airline's Pilots Association would not allow this so one must trust in Faraday and believe on grounds of faith.

 

[damgo]

^^^ Well, you would need 'leads' with a great deal of capacitance; then charges will build up at the ends and cause a voltage difference across the voltmeter inputs. emfs in 'series' along each lead act just like one big emf, remember.

 

[marcus]

Originally posted by damgo
^^^ Well, you would need 'leads' with a great deal of capacitance; then charges will build up at the ends and cause a voltage difference across the voltmeter inputs. emfs in 'series' along each lead act just like one big emf, remember.

not a bad idea. As I picture your idea it involves remote switches that can be used to isolate parts of a conductor running the length of the wing
one might even "freeze" the voltage differerence in carefully insulated elements inside the airplane wing and keep
the difference "on hold" so to speak until the plane could
land and have it measured.

the humble textbook writers did not bother to discuss how the voltage could be measured!

 

[Tyger]

Originally posted by Alexander
Charge can't remain (separated when plane is on the ground) because plane is made of metal.

So let's make the plane out of fiberglass with metal wingtips and have a wire and a switch between them. We can close the switch when the plane takes off and open it before it lands, keeping the charge on the wingtips. And I see marcus thought of this too.

 

[Alexander]

Originally posted by marcus
I often disagree with you Alexander or else am put off by your argumentatitive tone of voice (I do not remember which it is) but here I must commend you and voice agreement... (blah blah blah)...

Markus, why this long post? Any relevance to a voltmeter reading? Looking across this long post I could not find what you state about voltmeter. Can you be concrete in you posts, without mentioning all what textbooks have cover-to-cover?

Voltmeter with its leads span as airplane wings will show zero reading despite that there may be quite high potential between tips of leads. (And I can prove that if needs be.)

From you post I understood that you claim opposite. Well, then prove it.

 

[marcus]

Originally posted by Alexander


Voltmeter with its leads span as airplane wings will show zero reading despite that there may be quite high potential between tips of leads. (And I can prove that if needs be.)
.

Me too, this is what I was saying. You did not understand.

Originally posted by Alexander

From you post I understood that you claim opposite. Well, then prove it.

Then you REALLY misunderstood me---180 degrees. Well that happens. Better luck next time.

My post was opposed to using the Hall effect (which damgo referred to) as an explanation. Sorry it confused you!

 

[Ivan Seeking]

Well, I see we've all been very busy while I was away.

If you are all ready to get back to the next question? I'm asking not pushing:

The answer to a very complex physics or mathematics problem [I don't remember the exact nature of the problems] was to be calculated on one of the world's top supercomputers of the time; circa 1975. Two approaches were considered. The first and preferred approach was to calculate an extremely precise answer, and next was to approximate the answer. The best answer would require about a year of CPU time. The approximation could be calculated in relatively short order: in a matter of days.

The cost of CPU time caused this issue to be scrutinized very closely. It was decided that in spite of the money, the approximate answer was the best choice. What is the physical motivation [the physics] for their choice?

Too vague or have you just been too busy?

 

[Tyger]

My answer

Machine reliability. The machine would have a high chance of breaking down and losing all the data in a year.

 

[chroot]

Originally posted by Ivan Seeking
It was decided that in spite of the money, the approximate answer was the best choice.

I'm not sure I understand. The approximate answer is cheap, right? Why then would they decide to do it "in spite of the money," if it's the cheap solution?

- Warren

 

[Ivan Seeking]

Originally posted by chroot
I'm not sure I understand. The approximate answer is cheap, right? Why then would they decide to do it "in spite of the money," if it's the cheap solution?

- Warren

Another motivation was realized.

 

[Ivan Seeking]

Originally posted by Tyger
Machine reliability. The machine would have a high chance of breaking down and losing all the data in a year.

No one was concerned about the machine breaking down. But I did respond to your post.

 

[Alexander]

Originally posted by marcus
...


Lorentz force law is really the definition of the E and B fields, it is part of the groundwork.

No, Lorents force is not definition of electric nor nagnetic field. Electric field is defined as a force on a static charge (which, is derived in QED to actually be a result of creation- of virtual photons around one electric charge and disappearance of virtual photon around another charge), and magnetic field is defined as the specific term arizing from Lorents transformation of electric field from one reference system to another (namely - B is just the vector product of electric field with velocity of reference system [vxE]/c2[/).

Lorentz force is so basic, almost the F=ma of E/M

Really?

Here is a simple problem on understanding of Lorents force. Two protons move in perpendicular direction. At some moment of time vector of velocity of one of them is directed toward another one. Imagine L letter, first proton is in upper point and its velocity is directed down, second is in lower point right under it, and its velocity is directed to the right). What is the Lorents force F = q[vxB] of the lower proton on the upper one? Well, lower proton creates magnetic field B directed (in the vicinity of the upper electron) "out of the screen", which when vectorly multiplied by the velocity vector of upper proton gives the Lorents force on upper proton directed to the left.

Now, upper proton also creates a magnetic field. The magnitude of this magnetic field is non-zero everywhere except on the axis of motion of the upper proton (= on the line of its velocity vector). Therefore, lower proton happens to be in ZERO megnetic field (because this proton is located directly on the path of the upper proton) and therefore the magnitude of Lorents force on it is ZERO.

But we have a contradiction here: One objects (lower proton) exert force on another one, but gets no reaction force back.

How so? Momentum can't conserve if one objects exerts force on another and is not subjected by the same but oppositely directed force.

Can you resolve this apparent paradox (taking into consideration that Lorents force is "so basic" for you)?

 

[chroot]

Originally posted by Ivan Seeking
No one was concerned about the machine breaking down. But I did respond to your post.

Uh maybe because they wanted the answer fast so they could publish their papers before the other guys and get their PhD's...

- Warren

 

[Ivan Seeking]

Originally posted by chroot
Uh maybe because they wanted the answer fast so they could publish their papers before the other guys and get their PhD's...

- Warren

Hee hee. No. There was a physical reason for their choice.

 

[chroot]

No offense, Ivan, but there's really not enough information here to make a guess. And this sort of "riddle" really isn't the sort of question I was hoping to see here.

- Warren

 

[Ivan Seeking]

Originally posted by chroot
No offense, Ivan, but there's really not enough information here to make a guess. And this sort of "riddle" really isn't the sort of question I was hoping to see here.

- Warren

Yes. Perhaps I failed at formulate a good question out of this. I really didn't mean to make a riddle as much as to provide hints. Oh well, I will fess up and punt to you chroot. The answer is:

It was found that over the period of one year, the chance of an error due to bombardment by cosmic rays was nearly 1. As a result, this error would have made the exact answer less reliable than the approximation. Actually Tyger was getting close. Also, I would think that the chances for other errors could also be significant.

Here's to you chroot.

 

[chroot]

Well, hell, Ivan, machine reliability clearly includes this.

In fact, cosmic ray strikes are the most common cause of single-bit errors in modern computer memories -- hence the common SECDED technology -- "single error correction, double error detection." I was thinking "cosmic rays" the entire time, but thought that "machine reliability" included it!

The advance of satellite technology, I thought, had made the cosmic ray argument well-known.

When you said a "physical reason," I thought you meant something about the problem being solved -- not something about the physical machine. :)

In any event, it is certainly an interesting thing!

Why don't you ask another question?

- Warren

 

[Ivan Seeking]

Originally posted by chroot
Well, hell, Ivan, machine reliability clearly includes this.

In fact, cosmic ray strikes are the most common cause of single-bit errors in modern computer memories -- hence the common SECDED technology -- "single error correction, double error detection." I was thinking "cosmic rays" the entire time, but thought that "machine reliability" included it!

The advance of satellite technology, I thought, had made the cosmic ray argument well-known.

When you said a "physical reason," I thought you meant something about the problem being solved -- not something about the physical machine. :)

In any event, it is certainly an interesting thing!

Why don't you ask another question?

- Warren

Since Tyger included the specific reference to breakdowns, I tried to point to his answer but felt that he was otherwise off course.

Darn. That may be the only other interesting thing I have ever learned. O.K. How about this, I should say up front that I have an answer, but my confidence in this answer is not quite 100% [comes from a so-so professor]. So I will play this one by ear. At some appropriate point I will interject my two cents worth.

The total force due to friction between two materials is calculated as the normal force times the coefficient of friction between the materials. Why do racing cars use wide tires?

 

[Ivan Seeking]

Originally posted by chroot
The advance of satellite technology, I thought, had made the cosmic ray argument well-known.

This is what I was not sure about. Since I have heard little discussion about this topic outside of NASA, I feared this could be an extremely obscure argument, or that this proof may have been a cosmic error! I was waiting to get run up the flag pole after giving my answer!

 

[On Radioactive Waves]

Why do racing cars use wide tires?

that easy!? greater surface area ensures that it is harder to reach the static coefficiant of friction.

am i right?

 

[Ivan Seeking]

Originally posted by On Radioactive Waves
that easy!? greater surface area ensures that it is harder to reach the static coefficiant of friction.

am i right?

I don't think I understand your answer. Can you splain.

 

[On Radioactive Waves]

pressure of the tire to the surface is force exerted per unit area. the normal force is not a point source force but one distributed evenly over the area of interaction. greater area means more friction // less slipping.

 

[Ivan Seeking]

Originally posted by On Radioactive Waves
pressure of the tire to the surface is force exerted per unit area. the normal force is not a point source force but one distributed evenly over the area of interaction. greater area means more friction // less slipping.

The frictional force f is calculated as μN. Where N is the normal force [the weight of the car], and μ is the coefficient of friction. Sorry.

 

[On Radioactive Waves]

are you talking about the static or kinetic coefficient of friction?

ffk = [mu]kN

ffs is less than or equal to [mu]sN

sorry i don't know how to do subscripts

 

[Ivan Seeking]

Originally posted by On Radioactive Waves
are you talking about the static or kinetic coefficient of friction?

ffk = [mu]kN

ffs is less than or equal to [mu]sN

sorry i don't know how to do subscripts

Both apply but it doesn't matter. In either case here, we only need to total weight of the car, and the applicable coefficient of friction in order to calculate the frictional forces.

 

[On Radioactive Waves]

so the answer is not greater traction? sorry i didnt explain my answer well. peace

 

[Ivan Seeking]

Originally posted by On Radioactive Waves
so the answer is not greater traction? sorry i didnt explain my answer well. peace

How does the performance of the vehicle benefit from wide tires? Clearly we do get some traction advantage but why; by what physical mechanism or process? A simple application of the fritional force theory seems to yield no advantage. Peace on you too.

 

[On Radioactive Waves]

hmm is it something to do with torque?

 

[marcus]

Originally posted by Ivan Seeking
How does the performance of the vehicle benefit from wide tires? Clearly we do get some traction advantage but why; by what physical mechanism or process? A simple application of the fritional force theory seems to yield no advantage. Peace on you too.

wider tires allow one to inflate the tires to a lower pressure
and still support the car


the weight of the car is equal to the footprint area of the tires
multiplied by the air overpressure

with wider tires, it is more practical to have a large footprint area and thus a lower pressure (for the same weight of car)

intuitively lower pressure will contribute to better adherence to the road (small irregularities won't cause vibration that bounces the tire out of contact)

 

[Ivan Seeking]

Originally posted by marcus
intuitively lower pressure will contribute to better adherence to the road (small irregularities won't cause vibration that bounces the tire out of contact)

Can you elaborate on this direction of thought?

 

[Ivan Seeking]

Originally posted by marcus
(small irregularities won't cause vibration that bounces the tire out of contact)

What does this matter if the frictional force calculated is independent of surface area?

 

[marcus]

Originally posted by Ivan Seeking
Can you elaborate on this direction of thought?

hi Ivan, I don't really believe the normal force is constant

I think irregularities in the road cause the wheel to bounce up and down and at times nearly leave the road

If I were designing a racing vehicle I would want low pressure in the tires so I could have a tight suspension (with little or no play). And so you could pick up the tire a ways before it would actually lose contact with the road.

i would probably want a large footprint to even out little irregularities in the road and to retain contact even with a lot of up/down motion

this is guessing, you asked me to elaborate.

 

[Ivan Seeking]

Originally posted by marcus
lower pressure will contribute to better adherence to the road (small irregularities won't cause vibration that bounces the tire out of contact)

Well, I think I'll call this good enough. I was really looking for any of the following the references:

1).The coefficient of friction is higher for a soft tire:
Buried in this argument are any number of ideas that ultimately lead to the final value of the coefficient of friction. For example, the tire deforms and can then push against vertical surfaces. But this is all that happens at the microscopic level. So really we end up with a number of different levels of "contact" characteristics between the surfaces. After rationalizing this a bit we realize that all arguments boils down to the measured coefficient of friction for a particular tire configuration; e.g. pressure, temperature, stickiness, surface relaxation times, etc. etc. etc. The more one thinks about this, the more variables seem to crop up.


2).There is more rubber to remove:
To insure maximum friction, we typically [historically] want to remove as much rubber from the tire as possible. Obviously, as "On Radioactive Waves" almost started to touch on, if we add rubber by making a thicker tire, we create greater angular inertia than with a wider tire. This then affects the acceleration performance of the car, in addition perhaps to making for an impractical tire design.

I am sure other variables exist that I have never even considered. In any event, good enough, here's to you marcus.

 

[marcus]

Originally posted by Ivan Seeking
good enough, here's to you marcus.

In or around 1990 the BIPM, under the international authority of the General Conference on Weights and Measures (CGPM), established new Electrical Standards known as the 1990 volt and the 1990 ampere.

These are defined independently of the metal kilogram prototype in Paris and are in fact based solely on the atomic clock together with two quantum effects.

What are the two (low temperature) quantum mechanical effects used to define V₉₀ and A₉₀ ?

Or, equivalently if you prefer to look at it that way, the effects used to define V₉₀ and Ω₉₀ .

The definitions are based on ADOPTING exact values for two fundamental physical constants. What are they?

PS: it is sort of analogous to defining the meter by adopting an exact value for the speed of light---the official way since 1983.
You can't measure the speed of light in vacuo any more because it has a decreed exact value---in the 1990 electrical context there are two more constants which you can't measure.
'These are the electrical standards used in practice, though not yet finally made official.

The adopted values are among those listed at the NIST fundamental constants website in the "adopted values" section.

To sum up----the 1990 electrical standards have no logical connection to the metal kilogram and the Newton force based on it---or on the old force between infinite parallel wires definition of the ampere. How are they defined?

 

[nbo10]

The quantum magentic flux unit is used, and I believe that is defined with h-bar and e.


JMD

 

[marcus]

Originally posted by nbo10
The quantum magentic flux unit is used, and I believe that is defined with h-bar and e.


JMD

Yes! but that is only one fundamental constant and only enough to define one unit-----the 1990 ampere.

be more complete and specific.

How, for example, is the 1990 volt defined? Using the atomic clock plus what quantum device?

good, but not yet complete



BTW if you do dimensional analysis of hbar/e²
you will find that (if you carefully do it in SI units only and
follow SI conventions) you will get a ratio of voltage to current.
But that sort of ratio is called a resistance.
check out the adopted constants at the NIST website
nist.gov, if I remember right---the world's premier "fundamental
constants" website