A coil `ABD` of `N` turns `&` radius `R` carries a current of `1 Amp` `&` is placed on a horizontal tabel.`K` is a very small horizontal conducting ring of radius `r` placed at a distance `Y_(0)` from the centre of the coil vertically above the coil `ACD`.Find an expression for the `EMF` established when the ring `K` is allowed to fall freely.Express the `EMF` in terms of instanteous speed `v &` height `Y`.

Answer

B, C

Explanation

`B=(mu_(0)IR^(2)N)/(2[R^(2)+y^(2)]^(3//2))`
`epsilon=-(dphi)/(dt)=-A(dB)/(dt)=-pir^(2)(dB)/(dt)`...(1)
here `(dB)/(dt)=-(mu_(0)IR^(2)N)/2xx3/2 1/([R^(2)+y^(2)]^(5//2))xx2yxx(dy)/(dt)`
`=-(3mu_(0)IR^(2)N)/2 (yv)/([R^(2)+y^(2)]^(5//2))` put this value in eq. (i) and get Ans.