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8th April 2019, 17:18 #1
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Faraday disc voltage calculation formula
Hi folks, so my problme is I've been given the formula actually more than 1 version of the formula but I can't seem to solve it because my maths are very bad.
Can you please provide me a formula for calculating the voltage from a Faraday disc? I would also be happy if someone could show the calculation process using the parameters that I will give here so that I see how they insert into the various parts of the formula?
B field strength 1 Tesla, rotation = 1000 rpm/min , disc radius 15cm.
I seem to have problems because the voltage in each next segment of the disc is different due to larger radius having higher angular velocity and I can't incorporate this into the equation.

8th April 2019, 17:49 #2
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Re: Faraday disc voltage calculation formula
I googled "voltage produced by faraday disk"
item below the videos: pdf title current flow patterns in a faraday disc semantic scholar
european journal of physics Eur. J. Phys 25 (2004) 171183 (13 pages)
section 2 starting on page 173 answers your question
it appears the integral you showed is missing an angular velocity factor

8th April 2019, 19:33 #3
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Re: Faraday disc voltage calculation formula
Yes I found the formula there but that doesn't help me fully because I still don't know how to calculate this.
Compared to ordinary induction in a loop of wire this formula seems complicated, can someone please make the calculations using my provided parameters in a step by step process so that I could see how it's done?
thanks.

8th April 2019, 21:12 #4
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Re: Faraday disc voltage calculation formula
the calculation is essentially done
w (greek lower case omega) is the angular velocity of the rotating faraday disc, in radians per second
frequency in hertz, f, is related by 2 pi f = w
the grrek upper case phi with the B subscript is the magnetic flux = B A cos(theta)
where B is the magnetic field strength, A is the area of the faraday disc, and theta is the angle between the
magnetic field and the normal to the surface of the faraday disc.
this calculation gives you the voltage produced by rotating the faraday disc.

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9th April 2019, 06:29 #5
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Re: Faraday disc voltage calculation formula
well then which do I use do I turn my rpm/min into radians per second and use that or convert radians to rpm/min and use that?
Also how do I solve the integral and what do these small letters mean "b,a" on the upper and lower side of the integral?
r would be radius ok but what is d? also what is E?
it is frustrating I have talked to many mathematicians and nobody has simply showed me the calculation or explained each symbol I guess they simply think that some things should be self evident but their not if the person looking at them is not mathematically minded.

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10th April 2019, 03:10 #6
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Re: Faraday disc voltage calculation formula
always do the calculations in SI units, which means radians/second
Figure 2 of the pdf shows everything you want regarding the physics involved.
The faraday disc is rotting at angular velocity
The magnetic field is perpendicular to the faraday disc and pointing toward up, out of the page.
Charge Q feels 2 forces, the electric force, eE and the magnetic force, evB. E, v and B are vectors.
The between the v and the B indicates the vector product of the two vectors, v and B.the arrows indicate the direction of the electric force and the magnetic force.
The arrow pointing at the v indicated the direction of the velocity at the point on the disk, at that instant.
Moving a conductor in a magnetic field forces some of the electrons that are free to move about in the conductor to move. In this case with this geometry, the free electrons move toward the inside edge. This sets up a potential difference, called an ElectroMotiveForce, or EMF, between the inside and outside of the disc. It also produces an electric field in the disc that opposes the movement of the electrons toward the outer edge. Hence the electric force point toward the outside edge, while the magnetic force points toward the inner edge.
The disc has an inner radius of a, (the radius of the axel) and an outer radius of b.
At some point, the electric force and the magnetic force are equal, and the electrons in the disc are at equilibrium.
Equation 2 page 173 shows the relationship between the radial component of the electric field Er , and the vertical component of the magnetic field, Bz.
The next line down shows the potential between the rim and the axel:
V is the potential (or voltage)
=
Er is the electric field on the radial direction (as shown by the arrow in figure 2)
The dr indicates that one integrates this over the radius, r.
The small a and b are the limits of integration. Recall they are the inner and outer radius of the disc
Because of equation 2, the next = replaces Er with the equivalent expression in terms of the magnetic field.
note the 2 pi in the numerator and the denominator – they’re amount to multiplying by 1
the B integral:
since Bz is constant, take it out of the integral
now integrate 2 pi r dr from the axel to the rim (a to b) – this is the surface area of the disk, less the area of the axel.
Thus this B integral is the surface are of the disc times the magnetic field through the disc, or the magnetic flux, Greek upper case phi with a B subscript.
So the voltage the faraday disc produces becomes equation 3, where the notation was changed from V, electric potential, to curly E, the EMF noted above.
As the paper continues for several more pages, it discusses the details.
as for your frustration, i suggest ( in the most emphatic but gentlest terms possible) that you learn some math and physics

10th April 2019, 11:01 #7
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Re: Faraday disc voltage calculation formula
thanks but no thanks , I appreciate your time and input and it made something bit clearer but still I can't solve this god damn equation
When I said I really lack maths I meant it.
I guess I will try to find my school private maths teachers and ask her to show me the full calculation because I need someone to show me how to calculate this otherwise I don't understand shit.

10th April 2019, 13:10 #8
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Re: Faraday disc voltage calculation formula
ok pardon my outbursts, I'm trying one more time.
so assuming I will learn the integral , there are still blindspots in this equation that I don't know how to get to.
Er, how do I get the E field value and then multiply it by r?
ok rad/s divided by 2pi would be ok for me that I understand.
So what do I do with the dr? it comes directly after the Er and also after Bz, what do I do with these parts ?
also what do I do when the integral is between parts of equation like in the second part of the equation where it is in the middle?
Oh and also how do I get the phi aka flux through area value? Do I simply take the supposed strength of my magnet in Teslas and convert that value to phi but I guess there is something else involved like surface area.
thanks

10th April 2019, 14:24 #9
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Re: Faraday disc voltage calculation formula
Please read the attachment
1 members found this post helpful.

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10th April 2019, 18:13 #10
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Re: Faraday disc voltage calculation formula
I went through your comments and althought not fully at least now I have a feeling which parts of the formula match which physical phenomena on the disc which is important, please see my calculations and tell me whether i'm on the right track, also can you do the first part and the overall formula for me? Just calculations I want to see how they match up in real life then it would be much easier for me to learn.
My parameters are as follows.
B=1 T
rpm/min =3000 which is 31.41 rad/sec
disc radius (to the rim) 12cm , disc inner radius (axle) 2.5 cm (PS. I mistakenly used diameter as radius but for the calculations part it doesn't matter)
PS. I find ti interesting that the second and third part of the equation match exactly, I wonder why is that in the case if my calculations are correct.
But anyway if you came this far for me I would be happy if you would show the first part of the formula with a real calculation using the numbers I gave as my parameters,
thanks.

10th April 2019, 19:05 #11
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Re: Faraday disc voltage calculation formula
this statement
is a tautology.
each of the four items is the same as the other three.
once you calculated one, you have calculated all.
like this:
5 = 3 + 2 = 10/2 = 9  4
all of these four things is 5. all four are equal. once you calculated one, you calculated them all.
you did the math to find the third one.
then you did the math, exactly the same math, to find the fourth
if they were different, i would say you made a mistake.
since they agree, you did the same work twice.
i make the assumption you can push the buttons on a calculator
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10th April 2019, 20:31 #12
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Re: Faraday disc voltage calculation formula
Hmm, but whats the point of showing a formula that is actually the same formula with almost the same variables just shuffled around a bit in three different ways?
The only reason I could think if in some cases one of those variables is not know so that one can use the other one to get the result?
and what do I do with the result? it's a number 4325 and those can;t be volts so what else is missing here?

10th April 2019, 21:36 #13
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Re: Faraday disc voltage calculation formula
V is the unknown to solve for.
the E integral is derived from the relationship between V and E
the B integral is derived from the relationship between the electric force and the magnetic force acting on the electron free to move in the faraday disk
the magnetic flux statement is the answer one should use because it is the derived formula for the voltage produced by a faraday disk
it consists of a bunch of things easy to determine:
B, which we set by the choice of magnet,
A, the area of the disk, set by the choice of a, the axle radius and b the disk radius
and omega, the rate at which we rotate the disk
A 1 Tesla magnitude field is a big magnetic field and may require a superconducting magnet
the 3000 rpm or 314.15 rad/sec is a fairly big rotation rate.  it looks like you lost a decimal place somewhere
(side question  what is rpm/min?)
did you convert a and b from cm to m? that would make a big difference
everything has to be in SI units.
conservation of energy always applies.
for all of the energy you get out of the faraday disk, more has to be put into it.
some of the energy goes to just turning the disk.
some goes to the heating of the bearings and the heating the disk due to the motion of the electrons, etc
the rest is useful energy
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11th April 2019, 07:38 #14
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Re: Faraday disc voltage calculation formula
sure the same laws of resistance heating etc apply to the disc as well as any other generator that works with induction.
rpm/min is just a short for RPM in one minute.
3000 rpm is not that high , a speed easily achievable by a universal motor.
So i turned everything into meters and yes indeed it was more like 314 rad/sec than 31, so now my result is 4.468 volts with a field of 1 T.
One thing I do wonder, I don't see the formula taking into account the airgap length between the magnet and the disc which would influence the flux strength which would directly influence the output voltage?
One more thing, what if my excitation field instead of being DC aka a permanent magnet or DC electromagnet is an AC electromagnet with a certain AC frequency, what changes in the voltage output calculation then and how do I calculate it?
Thanks, you've been very kind to me and helpful so far.Last edited by Salvador12; 11th April 2019 at 07:55.

12th April 2019, 18:22 #15
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Re: Faraday disc voltage calculation formula
rpm/min is just a short for RPM in one minute.
so rpm/min is revolutions per minute per minute?
the airgap question can be "eliminated" by measuring the B field at the disk.
If you used AC to create the magnet, first thought says you get AC output instead of DC output.
whether you have V RMS or V Peak will depend on whether you used B RMS or B peak.
have not done the math

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12th April 2019, 19:45 #16
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Re: Faraday disc voltage calculation formula
yes sure if AC field then AC output, I was just wondering whether something changes with the AC field or does the peak AC field strength equals the same output voltage as if the same strength DC field would be used.
rpm/min was meant simply as that revolutions per minute.

2nd May 2019, 16:19 #17
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Re: Faraday disc voltage calculation formula
My friend calculates the formula with the addition of introducing the "magnetic induction in the airgap" and he gets a much lower result than I do, I got given my input parameters 4.8 volts which seemed a bit much for a faraday disc he got something on the order of milivolts.
How do I know how to calculate the voltage correctly?
Can you take a look at the formula he uses , well two of them and please tell me what to make of it ?
I have attached two formulas , the upper one is supposedly for disc voltage calculation and the lower one well I'm not sure can you please explain?
The numbers are already put in the lower one which he gave
thanks

2nd May 2019, 17:36 #18
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Re: Faraday disc voltage calculation formula
the top equation looks like the magnetic flux
no idea what the bottom one is
if the B in the numerator and the B in the denominator are both magnetic field, then the number at the end is
some sort of area
recommendation:
all the symbols we use have implied units, such as magnetic field, B, has units of Teslas
when one calculates, the letters get replaced with the appropriate numbers
when you write down a number, write down its units.
what you do to the numbers you must also do to the units.
after processing the numbers AND he units, one gets the result
if the units are wrong, there's a mistake

2nd May 2019, 20:58 #19
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Re: Faraday disc voltage calculation formula
The idea here is this as much as I could understand, that the formula you gave me and also the one he has in the upper part calculates the voltage of the disc assuming that the field strength (in my case 1 T ) is the field strength seen by the disc, assuming one uses a coil with a high permeability core and very little flux is lost.
the lower formula as I understand is taking into account the loss if one uses and air core solenoid between the field strength at the solenoid and the field strength that goes through the disc also taking into account the airgap which is not considered in the upper formula also I guess not in the formula presented in this thread.
I guess this makes sense now. What do you think could it be correct?
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