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) Centripetal force required for circular motion of electron is generated by a radial electric field caused by the redis- tribution of the electrons in the disc.
`F = eE = mromega^(2) rarr E = (mromega^(2))/(e)`
From `dV = -(m omega^(2))/(e)r dr`
rarr `int_(v_(1))^(v_(2))dV = -(momega^(2))/(e)int_(0)^(R)r dr`
`V-(1) - V_(2) = (momega^(2)R^(2))/(2e) = ((9.1 xx 10^(-31))(130)^(2)(0.25)^(2))/((2)(1.6 xx 10^(-19))`
`=3.0 xx 10^(-9)V = 3.0n V`
`V_(1) gt V_(2)`, i.e., potential at center is more than the potential at edge.
(b) A disc may be assumed to be made up of a large number of radial, conducting, differential elements rotating with angualr velocity `omega` about the center of disc `O`. This ,
`V_(center) - V_(edge) = (1)/(2)BR^(2)omega`
Now, `V_("center") gt V_("edge")` for anticlockwise rotation and `V_(egde) gt V_("center")` for clockwise rotation.
(a)
Substituting the values, we have
`V_("center") - V_(edge) = (1)/(2) xx 5.0 xx 10^(-3) xx 0.25 xx 0.25 xx 130`
`= 0.02 V = 20mV`