Revolving disc in a magnetic field

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SUMMARY

The discussion centers on calculating the magnetic field strength of a rotating disc, specifically confirming a rotational speed of 9337 rpm. The equation derived for the electromotive force (emf) is expressed as ε = (Bωr²)/2 = Bπr²f, where B represents magnetic field strength, ω is angular velocity, r is the radius, and f is the frequency. The method involves integrating the incremental emfs along a radial slice of the disc, providing a clear approach to understanding the relationship between these variables.

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Bolter
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Homework Statement
Calculate the number of revolution being made per minute (rpm) by the fly-wheel
Relevant Equations
See image attached below
Screenshot 2020-02-19 at 17.04.37.png

I know that there is a known equation to calculate the magnetic field strength of a rotating disc which I have made use of in here

IMG_3900.JPG


Do you agree that revs per minute of the disc turns out to be 9337 rpm?

Thanks for any help! Much appreciated
 
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I'm sure someone else can give a more rigorous derivation, though to derive the initial equation I considered a radial slice of the disk and integrated up all of the incremental emfs along that slice.

That is, ##\int d\varepsilon = \int Bv dr = \int B\omega r dr##

which gives ##\varepsilon = \frac{B\omega r^{2}}{2} = B\pi r^{2}f##

But to answer your question, yes, I agree.
 
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etotheipi said:
I'm sure someone else can give a more rigorous derivation, though to derive the initial equation I considered a radial slice of the disk and integrated up all of the incremental emfs along that slice.

That is, ##\int d\varepsilon = \int Bv dr = \int B\omega r dr##

which gives ##\varepsilon = \frac{B\omega r^{2}}{2} = B\pi r^{2}f##

But to answer your question, yes, I agree.

This is a really nice and simply way of looking at it which I hadn't thought of :)
 

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