Voltage between centre and edge of conducting disk

Part (b) is an example of a situation where an E field is maintained in a wire by currents induced by a changing B field. The changing B field is not "doing work" on the charges in the wire in the sense of converting energy from the B field to the electrons. The energy is coming from the work done by the power supply that is driving the changing B field.
  • #1
cromata

Homework Statement


A metal disc of radius R rotates with a constant velocity ω about its axis. FInd the potential difference between the centre and the rim of the disc if:
a)the external magnetic field is absent
b)the external magnetic field of induction B is directed perpendicular to the disc
c)Why is there a difference between results in those two cases if rotE=0 for consant magnetic field (3rd Maxwell equation).

Homework Equations


Δφ=∫Edr
Fmag=qv×B

The Attempt at a Solution


a)Free electrons of metal disc are rotating, so there has to be some force that is responsible for centripetal acceleration of electrons. That force exists because free electrons go to the rim of the disc and create electric field, so we have: Fcp=e*E, E=m*ω^2*r/e. So potential diference is Δφ=∫Edr (from 0 to R), Δφ=mω^2*r^2/2e.
b)I think this question is undefined because it's not same if B and ω are parallel or antiparallel. Let's suppose that they are parallel, then magnetic force on electron is F=Beωr (directed to the centre of the disc), so we have Fcp=B+E, and from here we can calculate E and then use Δφ=∫Edr... but i think i got something wrong in this b)part and i don't know the answer on last question
 
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  • #2
cromata said:
a)Free electrons of metal disc are rotating, so there has to be some force that is responsible for centripetal acceleration of electrons. That force exists because free electrons go to the rim of the disc and create electric field, so we have: Fcp=e*E, E=m*ω^2*r/e. So potential diference is Δφ=∫Edr (from 0 to R), Δφ=mω^2*r^2/2e.
There is no magnetic field. How can electrons go to the rim of the disc?
cromata said:
b)I think this question is undefined because it's not same if B and ω are parallel or antiparallel. Let's suppose that they are parallel, then magnetic force on electron is F=Beωr (directed to the centre of the disc), so we have Fcp=B+E, and from here we can calculate E and then use Δφ=∫Edr... but i think i got something wrong in this b)part and i don't know the answer on last question
Have a read.
https://www.google.co.in/url?sa=t&s...QIHDAA&usg=AFQjCNH9N69CnzBKuK7WS0vocM10Eop1Uw
 
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  • #3
Thx, but what is causing potential difference/electric field in a) if electrons don't go to the edge of the rim?
 
  • #4
cromata said:
Thx, but what is causing potential difference/electric field in a) if electrons don't go to the edge of the rim?
There is no potential difference.
 
  • #5
This problem is from I.E.Irodov: Problems in general physics, and solution in a) says the same thing that i said
Screenshot_2017-07-01-13-22-17.png
 
  • #6
I understood Irodov's solution but I am not sure I can comment on it.

I request @TSny to take a look.
Edit: I see TSny has been offline for a long time.
@gneill, could you please comment on this?

I have some strange feeling about this questiono0)..
 
Last edited:
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  • #7
cromata said:
b)I think this question is undefined because it's not same if B and ω are parallel or antiparallel. Let's suppose that they are parallel, then magnetic force on electron is F=Beωr (directed to the centre of the disc), so we have Fcp=B+E, and from here we can calculate E and then use Δφ=∫Edr... but i think i got something wrong in this b)part
I'm not quite sure what your specific question is here. Perhaps you can rephrase it?
In any realistic experimental setup, the centripetal force on an electron is negligible compared to the magnetic force when you have a moderate B field present. So, you may neglect the centripetal force in part (b).

and i don't know the answer on last question
I'm sot sure what question (c) is getting at. It is true that rotE is zero for parts (a) and (b). But I don't see why that would make anyone think that the emf's for (a) and (b) should be the same.
 
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  • #8
-I think that c) part was there just to confuse, (because when you have non-static field you can obtain different potential differences between same two points if you choose 2 diferent paths to go from one point to another. ,you can see in this video from 35th minute what I mean) but a) and b) are two separated situations so c) question actually makes no sense.
-and thanks for part b). So I can just use Ee=Bev and calculate E and Δφ (because Fcp is negliglbe)?
 
  • #9
cromata said:
-I think that c) part was there just to confuse, (because when you have non-static field you can obtain different potential differences between same two points if you choose 2 diferent paths to go from one point to another. ,you can see in this video from 35th minute what I mean)
Yes, the video is very interesting. Lewin's demo shows how nonconservative E fields can produce nonintuitive results for volt meter readings in circuits.
but a) and b) are two separated situations so c) question actually makes no sense.
Right. The E field is conservative in (a) and (b). But they are, nevertheless, different situations
-and thanks for part b). So I can just use Ee=Bev and calculate E and Δφ (because Fcp is negliglbe)?
Yes, I believe so.
 

1. What is the definition of voltage between centre and edge of conducting disk?

Voltage between centre and edge of conducting disk refers to the difference in electric potential between the centre and the edge of a circular disk made of a conductive material. It is a measure of the amount of electrical energy that is needed to move a unit of charge from the centre to the edge of the disk.

2. How is the voltage between centre and edge of conducting disk calculated?

The voltage between centre and edge of conducting disk can be calculated using the formula V = (Q / εA) ln(R/r), where V is the voltage, Q is the charge on the disk, ε is the permittivity of the material, A is the area of the disk, R is the radius of the disk, and r is the distance from the centre of the disk to the point where the voltage is being measured.

3. What factors affect the voltage between centre and edge of conducting disk?

The voltage between centre and edge of conducting disk is affected by the charge on the disk, the permittivity of the material, the area and radius of the disk, and the distance from the centre of the disk to the point where the voltage is being measured. Changes in any of these factors will result in a change in the voltage between centre and edge of conducting disk.

4. Why is the voltage between centre and edge of conducting disk important?

The voltage between centre and edge of conducting disk is important because it helps us understand the behavior of electric fields and how they interact with conductive materials. It is also useful in designing electrical circuits and devices, as well as in various applications such as electrostatic painting and electroplating.

5. How does the shape of the conducting disk affect the voltage between centre and edge?

The shape of the conducting disk can affect the voltage between centre and edge in two ways. Firstly, the area and radius of the disk will directly affect the voltage, as seen in the formula V = (Q / εA) ln(R/r). Secondly, the shape of the disk can also affect the distribution of charge, which can in turn affect the voltage. For example, a disk with a pointed edge will have a higher voltage at the edge compared to a disk with a smooth edge.

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