A metallic rod of 1m length is rotated with a frequency of 50 rev/s, with one end hinged at the centre and the other end at the circumference of a circular me tallic ring of radius 1m, about an axis passing through the centre and perpendicular to the plane of the ring (see figure). A constant and uniform magnetic field of 1T parallel to the axis present everywhere. What is the emf between the centre and the metallic ring?

Explanation

Method I
As stated in this question 6, during rotation, electrons are shifted to the end Q, creating an emf. Under steady state, emf acrosss a length dr is `d varepsilon=Bvdr`. Using equation `varepsilon=Blv`, the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by
`therefore varepsilon=intdvarepsilon=underset(0)overset(R)intBvdr=underset(0)overset(R)intBomegardr=(BomegaR^(2))/(2)`
But `v=omegar`. This gives `epsilon=(1)/(2)xx1.0xx2pixx50xx(1^(2))=157V`
Method II
Considering the positions OP and OQ, the emf produced depends on rate of change of flux which itself is proportional to rate of sweeping of area of the sector. If `theta` is the angle between the rod and the radius of the circle at P at time t, the area of the sector OPQ is given by `piR^(2)xx(theta)/(2pi)=(1)/(2)R^(2)theta`, where R is the radius of the circle. Hence, the induced emf is `epsilon=Bxx(d)/(dt)[(1)/(2)R^(2)theta]=(1)/(2)BR^(2)(d theta)/(dt)=(BomegaR^(2))/(2)`
[Note: `(d theta)/(2)=omega=2piv`]
The same expression we got earlier.