A metal disc of radius `R = 25 cm` rotates with a constant angular velocity `omega` = `130` rad `s^(-1)` about its axis. Find the potential difference between the center and rim of the disc if
(a) the external magnetic field is absent,
(b) the external uniform megnetic field `B = 5.0`mT directed perpendicular to the disc.

Explanation

(a) As the metal disc rotates, any force electron also rotetes with it with same angular velocity `omega`, and that's why an electron must have an acceleration `omega^(2) r` directed towards the dis'c centre, where `r` is sepration of the electron from the centre of the disc. We know from Newton's secound law that if a particle has some acceleration. We also know that a chagred particle can be indluenced by two fields electric and magnetic. In our problem magnetic field is absent hence we reach at the conslusion that there is an electric field near any electrons and is directed opposite to the acceleartion of the electron.
If `E` be the electric field strength at a distance `r` from the centre of the disc, we have from Newton's secound law.
`F_(n) = m w_(n)`
`eE = m r omega^(2)`, or, `E = (m omega^(2) r)/(e)`
and the potential difference,
`varphi_(cen) = varphi_(rim) = int_(0)^(a) vec(E).d.vec(r) = int_(0)^(a) (m omega^(2) r)/(e) dr`, as `vec(E) uarr darr d vec(r)`
Thus `varphi_(cen) = varphi_(rim) = Delta varphi = (m omega^(2))/(e) (a^(2))/(2) = 3.0 nV`
(b) When field `vec(B)` is present by defintion, of motional e.m.f. :
`varphi_(1) - varphi_(2) = int_(1)^(2) -(vec(v) xx vec(B)).dr vec(r)`
Hence the sought potential difference,
`varphi_(cen) - varphi_(rim) = int_(0)^(a) -v B dr = int_(0)^(a) -omega r B dr`, (as `v = omega r`)
Thus `varphi_(rim) - varphi_(cen) = varphi = (1)/(2) omega B a^(2) = 20 m V`
(In general `omega lt (eB)/(m)` so we can neglect the effect discussed in (1) here).