A conducting light string is wound on the rim of a metal ring of radius `r` and mass `m`. The free end of the string is fixed to the ceiling. A vertical infinite smooth conducting plane is always tangent to the ring as shown in the figure. A uniform magnetic field Bis applied perpendicular to the plane of the ring. The ring is always inside the magnetic field. The plane and the strip are connected by a resistance `R`. When the ring is released, find

a. the curent in the resistance `R` as as function of time.
b. the terminal velocity of the ring.

Answer

B, D

Explanation

Let at time `t` velocity of ring be `v` (downwards)
`e=Bv(2r)=2Bvr`
(Two batteries of emf `2Bvr` are connect in parallel)

`:. i=e/R=(2Bvr)/R`
Now, `a=(mg-F_m-T)/m`
Here `F_m=2[(i/2)(2r)B]=2irB-(4B^2r^2v)/R`
`:. a=g-(4B^2r^2v)/(mR)-T/m`.............(i)
`alpha=(Tr)/(mr^2)=T/(mr)` .............(ii)
`a=ralpha=T/m`...............(iii)
From Eqn (i) ,( ii) and (iii) we get
`a=g/2-(2B^2l^2v)/(mR)`
or `int_0^v(dv)/(g/2-(2B^2r^2v)/(mr))=int_0^1dt`
or `v=(mgR)/(4B^2r^2)(1-e^(-(2B^2r^2)/(mR)))`
`i=(2Bvr)/R=(mg)/(2Br) (1-e^(-(2B^2r^2)/(mR)t))`
and `v_r=(mgR)/(4B^2r^2)`