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Consider the following experiment which is performed in vacuum away from all G fields and EM fields.

A wire is connected to a motor that helps to swing (rotate it around one end only) it. Let’s consider the motor to be ideal. The wire is made of a conducting metal( it does not conduct heat). And it is also very rigid. Now since we know that the motor is going to rotate the wire, the end of the wire will experience the maximum centrifugal force( frame of reference is the wire itself) which is opposite to the end attached to the motor. Since the atoms in the metal are rigid and will not move, the electrons can freely move. Therefore the electrons will start to move towards the end of the wire and positive charges will move towards the other end. This will create a potential difference across the wire.

So my question is simple, is it possible to rotate a wire and create a potential difference across it which we can measure?

physics2000
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1 Answers1

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Let's imagine that you create a wire that swings fast enough to create an acceleration of 100m/s^2.

The force required to push an electron against that is $$ F = ma = 9.11 \times 10^{-29}\text{N}$$

The electric field that cancels the acceleration is $$ E = \frac Fq = 5.6\times 10^{-10} \text{V/m}$$

You would have to measure fractions of a nanovolt over a meter, and do it without allowing your measurement device to be subject to the same acceleration.

Not doing experiments like this, I suspect it would be easier to design something that measured acceleration and charge on isolated electrons that it would be trying to measure voltages on a macroscopic, accelerating wire.

EDIT:

Nonetheless, thanks to Ben's answer in another question, I now know a similar sort of experiment has been done and reported on, more than 100 years ago. The Stewart-Tolman_effect was measured not by the pure centrifugal force of a wire, but by rotating several loops of wire and rapidly stopping it. When the wire is stopped, the electrons inside the wire continue moving, causing a temporary potential difference that can be measured.

BowlOfRed
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  • Thanks. But I suggest that we would have more than 10^10 electrons in this wire. Since this is in the classical sense, do you have any idea what would it be in the quantum mechanical sense? – physics2000 Mar 22 '18 at 18:15
  • The electric field that is generated is essentially independent of the number of electrons. You get the same voltage on a thick wire as a thin one. No idea what corrections you need for QM. I would suggest you not accept an answer for a period of time (at least a day). Acceptance of one answer can discourage other (potentially better) answers from being written. You can change or withdraw acceptance. – BowlOfRed Mar 22 '18 at 18:23
  • thanks so much. I will upvote it though – physics2000 Mar 22 '18 at 18:24
  • one small question though, why would the electric field be independent of the number electrons? Higher the charge density higher will be the electric field, isn’t it? – physics2000 Mar 22 '18 at 18:31
  • The wire is neutral. So a higher electron density implies a higher proton density as well. The electrons will move a shorter distance in the gravitational field to create the same electric field, but will stop accelerating at the same field strength. – BowlOfRed Mar 22 '18 at 18:35