A thin semicircular conducting ring of radius `R` is falling with its plane verticle in a horizontal magnetic inducting `B`. At the position `MNQ`, the speed of the ring is `V` and the potential difference developed across the ring is

Options

A. zero
B. `(BvpiR^(2))/2` and `M` is at higher potential
C. `piRBV` and `Q` is at higher potential
D. `2 RBV` and `Q` is higher potential

Answer

D

Explanation

Induced motional `emf` in `MNQ` is equivalent to the motional `emf` in an imaginary wire `MQ` i.e., `e_(MNQ)=e_(MQ)=Bvl=Bv(2R)`
`[l=MQ=2R]`
Therefore potential difference developed across the ring is `2RB_(V)` with `Q` at higher potential.
`e_(MNQ)=e_(MQ)=Bvl=Bv(2R)`
`[l=MQ=2R]`