When a rod of length l is rotated with angular velocity of `omega` in a perpendicular field of induction B , about one end , the emf across its ends is

Options

A. `Bl^(2)omega`
B. `0.5 Bl^(2)omega`
C. `Bl omega`
D. `0.5 Blomega`

Answer

B

Explanation

Induced e.m.f. `(e) = (d phi)/(dt) = B cdot (dA)/(dt)`
But `dA = pil^(2) and dt=T and T = (2 pi)/(omega)`
`:. e = (B cdot pi l^(2))/((2pi//omega))=1/2 Bl^(2)omega`.