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(B) When the metallic disc rotates,the free electrons within the disc experience a centrifugal force directed radially outward,given by $F_c = m\omega

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a point on the rim of the disc is at a lower potential than its centre

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(B) When the metallic disc rotates,the free electrons within the disc experience a centrifugal force directed radially outward,given by $F_c = m\omega^2 r$,where $m$ is the mass of the electron,$\omega$ is the angular velocity,and $r$ is the distance from the axis of rotation.Due to this centrifugal force,the free electrons migrate from the centre towards the rim of the disc.As a result,the rim of the disc accumulates a net negative charge,while the centre becomes positively charged.Since the potential $V$ is related to the electric field $E$ by $E = -dV/dr$,and the electric field is directed from the positive centre to the negative rim,the potential decreases as we move from the centre to the rim.Therefore,a point on the rim of the disc is at a lower potential than its centre.

$A$ thin metallic disc is rotating with constant angular velocity about a vertical axis that is perpendicular to its plane and passes through its centre. The rotation causes the free electrons in the disc to redistribute. Assume that there is no external electric or magnetic field. Then,

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