Calculate the rate at which the flux linked with the generated area changes with time when a rod length l is (a) translated (b) rotated in a uniform field of induction B as shown in fig.

Explanation

(a) Component of velocity perpendicular to the rod `= upsilon sin theta`. Therefore, in time t, area traversed `= l xx upsilon sin theta xx t`.
`:.phi = B (l x upsilon sin theta xx t) cos 0^(@) = B l upsilon t sin theta`
`(d phi)/(dt) = B l upsilon sin theta`
(b) In time t, if `theta` is the angle traced by the free end, then
`:.` Area swept, `A = pi l^(2) xx ((theta)/(2 pi)) = (1)/(2) l^(2) theta`
`phi = B ((1)/(2) l^(2) theta) cos 0^(@) = (1)/(2) B ,l^(2) theta`
`(d phi)/(dt) = (1)/(2) B l^(2) (d theta)/(dt) = (1)/(2) B l^(2) omega`