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A question in my book was given as

A conducting disc of radius R rotates about its axis with an angular velocity $\omega$. Then the potential difference between centre of mass of disc and its edge is (no magnetic field is present)

The answer stated that:

  1. There will be a potential difference present
  2. The value of which was obtained by using $eE = m\omega^2r$ (not sure why this holds true) for an electron $e$ in the disc and then obtaining the electric field and finding the potential difference by integration -$E.dr$

My question: While deriving the motional emf in a rotating disc in magnetic field, we consider only the flux change (or say the $(v\times B).dl$) responsible for the induced emf given as $\frac{1}{2}B\omega r^2$. Why do we not consider the emf obtained due to the above phenomenon aswell since it happens even in the absence of magnetic field?

Extra but related question: Why does $eE = m\omega^2r$ hold true for an electron in a conducting disc? What stops it from moving radially?

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Both these questions have been asked (and answered) within the last 3 or 4 weeks on this site. The gist of the answer to your main question is that the 'centrifugal emf' is very much smaller than the 'magnetic emf'. You should show this for yourself by putting figures into your formulae. [Note that a magnetic field of 1 T would be a very strong field.]

To answer your extra question: the free electrons do move outwards at first, as there is no force to provide them with a centripetal acceleration. But as they move outward they create a charge separation: negative to the outside of the disc; positive to the inside. These charges create an electric field, $E$, that does provide the required centripetal acceleration. Hence the equation that you've quoted: $๐‘’๐ธ=๐‘š๐œ”^2๐‘Ÿ$.

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