- #1
sebqas
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The EMF for the Faraday disc is calculated as
[tex] \epsilon = \int_0^R \textbf{v} \times \textbf{B} \cdot d\textbf{l} [/tex]
the integration is between the disc axis (0) and its edge (R is the disc radius).
As for uniformly rotating disc [tex]\boldmath v = \omega \times r [/tex] and B-field is orthogonal to the disc plane, one gets famous result:
[tex] \epsilon = 1/2 \omega B R^2 [/tex]
However such calculation works only if the integrand (v x B) is curl-free, then the integration does not depends
on the contour and the EMF is well defined. But anybody can imagine that instead of solid disc we have a
conducting fluid flowing out from the center and rotating. In this case v has radial component, curl(v x B) is
non-zero and the integral for EMF depends on the shape of the contour.
So then, what is the actual EMF ?
[tex] \epsilon = \int_0^R \textbf{v} \times \textbf{B} \cdot d\textbf{l} [/tex]
the integration is between the disc axis (0) and its edge (R is the disc radius).
As for uniformly rotating disc [tex]\boldmath v = \omega \times r [/tex] and B-field is orthogonal to the disc plane, one gets famous result:
[tex] \epsilon = 1/2 \omega B R^2 [/tex]
However such calculation works only if the integrand (v x B) is curl-free, then the integration does not depends
on the contour and the EMF is well defined. But anybody can imagine that instead of solid disc we have a
conducting fluid flowing out from the center and rotating. In this case v has radial component, curl(v x B) is
non-zero and the integral for EMF depends on the shape of the contour.
So then, what is the actual EMF ?
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