Question

A conducting frame abcd is kept in a vertical plane. A conducting rod of of mass m and length `l` can slide smoothly on it remaining always horizontal. The resistance of the loop is negligible and inductance is constant having value `L`. The rod is left from rest and allowed to fall under gravity and inductor has no initial current. A magnetic field of constant magnitude Bis present throughout the loop pointing inwards. Determine

(a) position of the rod as a function of time assuming initial position of the rod to be `x=0` and vertically downward as the positive x-axis.
(b) the maximum current in the circuit.
(c) maximum velocity of the rod

Answer 1

Correct Answer - A::B::C
`(Ldi)/(dt)=Bvl`..(i)
`mg-Bil=m(dv)/(dt)`
differentiating,`-Bl(di)/(dt)=m(d^(2)v)/dt^(2)`...(ii)
From (i) & (ii) `(d^(2)v)/(dt^(2))=(B^(2)l^(2))/(mL)v=-omega^(2)v`
Where `omega=(Bl)/sqrt(mL) thereforev=A sin omegat`
`(dv)/(dt)=Aomega cos omega t , t=0 rArr (dv)/(dt)=g`
`therefore g=A omega therefore A=g/omega`
`therefore v_(max)=A=g/omega`
`therefore v=q/omega sin omega t therefore (dx)/(dt)=v=q/omegat`
`therefore x=g/omega^(2)(1-cos omega t)`
From (ii) `Bil=mg -m(dv)/(dt)=mg-m Aomega cos omegat`
`i_(max)=(mg+mAomega)/(Bl)=(mg+mg)/(Bl)=2mg//(Bl)`