Question

In the figure, a conducting rod of length `l=1` meter and mass `m = 1 kg` moves with initial velocity `u = 5 m//s`. On a fixed horizontal frame containing inductor `L = 2 H` and resistance `R = 1 W`. `PQ` and `MN` are smooth, conducting wires. There is a uniform magnetic field of strength `B = 1T`. Initial there is no current in the inductor. Find the total charge in coulomb, flown through the inductor by the time velocity of rod becomes `v_(f) = 1 m//s` and the rod has travelled a distance `x = 3` meter.

Answer 1

Correct Answer - A::B::C
Let `i_(1)` and `i_(2)` be the current through `L` and `R` at any time `t`
`therefore i=i_(1)+i_(2) rArr (Blv)/R=i_(2)` and `Blv=L(di_(1))/(dt)`
Force on conducting rod `=m(dv)/(dt)=-ilB=-(i_(1)+(Blv)/R)lB`
`rArr mdv=-lBi_(1)dt-(B^(2)l^(2))/Rv dt`
`rArr mint dv=-lBinti_(1)dt-(B^(2)l^(2))/R int v dt`
`rArr m(v_(f)-u)=-lBQ-(B^(2)l^(2))/R x`
(`v_(f)`=velocity when it has moved a distance `x`)
`rArr Q=((B^(2)l^(2))/Rx-m(v_(f)-u))/(Bl)=1C`.