Physics Q&A Game

chroot

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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

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

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Re: My answer

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

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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 dont remember the exact numbers
 

Ivan Seeking

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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 dont 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

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canni still go next? canni canni please please
 

marcus

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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

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

- Warren
 

chroot

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Ohhh.. the Hall effect. Hmph.

:wink:

- Warren
 

Ivan Seeking

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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

Re: My answer

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.
 
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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

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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

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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!
 
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Re: Re: My answer

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

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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

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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::smile:

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?
 
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My answer

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

chroot

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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

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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

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Re: My answer

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.
 

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