A metal rod `1/(sqrt(pi))` m long rotates about one of its ends in a plane perpendicular to a magnetic field of induction `4 xx 10^(-3)T`. If the emf induced between the ends of the rod is 16 mv, then the number of revolutions made by the rod per second is

Options

A. 3 rps
B. 4 rps
C. 5 rps
D. 6 rps

Answer

B

Explanation

The induced emf `e = (d phi)/(dt) = d/(dt)(AB) = B(dA)/(dt)`
In one rotation the rod traces out a circle of radius l or an area `A = pi l^(2)`.
`:.` In n rotations it will trace out an area of `n pi l^(2)`
`:. e = B(dA)/(dt) = B cdot (n pi l^(2))/(dt)`
But `n/(dt) = ("No. of rev. performed")/("time") = "frequency" f`
`:. e = B f pi l^(2) = Bf cdot A`
`:. F = (e)/(B pi l^(2)) = (16 xx 10^(-3))/(4 xx 10^(-3) xx [pi((1)/(sqrt(pi)))^(2)])`
`= 4/(pi xx 1/(pi)) = 4 r.p.s.`