a uniform magnetic field of induction `B` fills a cylindrical volume of radius `R`. A rod `AB` of length `2l` is placed as shown in . If `B` changing at the rate `dB//dt`, the emf that is produced by the changing magnetic field and that acts between the ends of the rod is

Options

A. (a) `(dB)/(dt)lsqrt(R^(2) - l^(2))`
B. (b) `(dB)/(dt)lsqrt(R^(2) + l^(2))`
C. ( c) `(1)/(2)(dB)/(dt)lsqrt(R^(2) - l^(2))`
D. (d) `(1)/(2)(dB)/(dt)lsqrt(R^(2) + l^(2))`

Answer

A

Explanation

(a) Consider a point on the circumference of a circle of radius `r (r lt R)`. Let `E` be the elctric dield along the tangents to the circle. Then `E` is given by `E = (r )/(2)(dB)/(dt)`

Now consider a point `P` in the rod and a small distance `dl = PQ` along `AB`.
`epsilon = int vec(E). vec(dl) = int(r )/(2)(dB)/(dt)dlcos theta`
`= ((OM))/(2)(dB)/(dt)int dl = sqrt(R^(2) - l^(2))/(2)(dB)/(dt)2l`
`rarr` `epsilon = sqrt(R^(2) - l^(2))((dB)/(dt))l` [`:.` `int dl = 2l`]