A copper disc of diameter 20 cm is rotating uniformly about its horizontal axis passing through the centre with angular frequency 600 rpm. A uniform magnetic field of strength `10^(-2)T` acts perpendicular to the plane of the disc. Calculate the induced emf between its centre and a point on the rim of the disc.

Explanation

Diameter of the disc = 20 cm
`therefore` Radius, r = 10 cm = 0.1 m
Angular frequency, n = 600 rpm
`therefore` Angular speed,
`omega=(600xx2pi" rad")/(60 s)=20pi" rad."s^(-1)`
Let us take a small segment dx on the disc at a distance x from its centre. Length of dx is so small that speed of all the points on this segment is considered to be the same, which is, v' = `omegax`.
Therefore, motional emf across dx,
`de=v'Bdx` [where B is the magnetic field]

`therefore` Total induced emf between the centre and a point on the rim of the disc,
`e=int_(0)^(r)v'Bdx=int_(0)^(r)omegaxBdx=(1)/(2)omegaBr^(2)`
`=(1)/(2)xx20pixx10^(-2)xx(0.1)^(2)`
`=0.00314V=3.14mV`