Revolving disc in a magnetic field

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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
Printed problem text: a fly-wheel 1.5 m in diameter revolves on a horizontal axle in the magnetic meridian, induced emf between axle and rim 5.5 mV, find rpm given horizontal Earth field B = 2.0 × 10^-5 T.

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

Handwritten derivation on lined paper: EMF = B pi r^2 f rearranged to f = EMF / (B pi r^2), with values 5.5 x 10^-3 V, 2 x 10^-5 T and r = 0.75 m, giving f = 155.618..., then omega = 2 pi f = 977.777... rad/s and conversion to about 9337 rpm.


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 :)